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APPLICATIONS OF GROUP THEORY TO ATOMS,MOLECULES, AND SOLIDS

The majority of all knowledge concerning atoms, molecules, and solids hasbeen derived from applications of group theory. Taking a unique, applications-oriented approach, this book gives readers the tools needed to analyze any atomic,molecular, or crystalline solid system.

Using a clearly defined, eight-step program, this book helps readers to under-stand the power of group theory, what information can be obtained from it, and howto obtain it. The book takes in modern topics, such as graphene, carbon nanotubes,and isotopic frequencies of molecules, as well as more traditional subjects: thevibrational and electronic states of molecules and solids, crystal-field and ligand-field theory, transition-metal complexes, space groups, time-reversal symmetry,and magnetic groups.

With over a hundred end-of-chapter exercises, this book is invaluable for grad-uate students and researchers in physics, chemistry, electrical engineering, andmaterials science.

T H O M A S W O L F R A M is a former Chairman and Professor of the Departmentof Physics and Astronomy, University of Missouri-Columbia. He has founded ascience and technology laboratory for a major company and started a companythat manufactured diode-pumped, fiber-optic transmitters and amplifiers.

S I N A S I E L L I A LT I O G L U is a former Chairman and Professor of Physics at theMiddle East Technical University in Ankara, Turkey. He has been a recipient ofHumboldt and Fulbright Fellowships. Currently he is a Professor of Physics andDirector of Basic Sciences at TED University in Ankara.

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APPLICATIONS OF GROUP THEORYTO ATOMS, MOLECULES,

AND SOLIDS

THOMAS WOLFRAMFormerly of the University of Missouri-Columbia

S INASI ELLIALTIOGLUTED University, Ankara

Formerly of Middle East Technical University, Ankara

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University Printing House, Cambridge CB2 8BS, United Kingdom

Published in the United States of America by Cambridge University Press, New York

Cambridge University Press is part of the University of Cambridge.

It furthers the University’s mission by disseminating knowledge in the pursuit ofeducation, learning, and research at the highest international levels of excellence.

www.cambridge.orgInformation on this title: www.cambridge.org/9781107028524

c© T. Wolfram and S. Ellialtıoglu 2014

This publication is in copyright. Subject to statutory exceptionand to the provisions of relevant collective licensing agreements,no reproduction of any part may take place without the written

permission of Cambridge University Press.

First published 2014

Printed in the United Kingdom by CPI Group Ltd, Croydon CR0 4YY

A catalogue record for this publication is available from the British Library

Library of Congress Cataloguing in Publication dataWolfram, Thomas, 1936–

Applications of group theory to atoms, molecules, and solids / Thomas Wolfram, Sinasi Ellialtıoglu.pages cm

ISBN 978-1-107-02852-4 (hardback)1. Solids – Mathematical models. 2. Molecular structure. 3. Atomic structure. 4. Group theory.

I. Title.QC176.W65 2013530.4′1015122–dc23

2013008434

ISBN 978-1-107-02852-4 Hardback

Cambridge University Press has no responsibility for the persistence oraccuracy of URLs for external or third-party internet websites referred to

in this publication, and does not guarantee that any content on suchwebsites is, or will remain, accurate or appropriate.

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Contents

Preface page xi

1 Introductory example: Squarene 11.1 In-plane molecular vibrations of squarene 11.2 Reducible and irreducible representations of a group 121.3 Eigenvalues and eigenvectors 271.4 Construction of the force-constant matrix from the

eigenvalues 301.5 Optical properties 31References 34Exercises 35

2 Molecular vibrations of isotopically substituted AB2 molecules 392.1 Step 1: Identify the point group and its symmetry operations 392.2 Step 2: Specify the coordinate system and the basis functions 392.3 Step 3: Determine the effects of the symmetry operations on

the basis functions 412.4 Step 4: Construct the matrix representations for each element

of the group using the basis functions 412.5 Step 5: Determine the number and types of irreducible

representations 422.6 Step 6: Analyze the information contained in the

decompositions 422.7 Step 7: Generate the symmetry functions 432.8 Step 8: Diagonalize the matrix eigenvalue equation 502.9 Constructing the force-constant matrix 502.10 Green’s function theory of isotopic molecular vibrations 522.11 Results for isotopically substituted forms of H2O 60

v

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vi Contents

References 62Exercises 62

3 Spherical symmetry and the full rotation group 663.1 Hydrogen-like orbitals 663.2 Representations of the full rotation group 683.3 The character of a rotation 723.4 Decomposition of D(l) in a non-spherical environment 753.5 Direct-product groups and representations 763.6 General properties of direct-product groups and representations 793.7 Selection rules for matrix elements 833.8 General representations of the full rotation group 85References 88Exercises 88

4 Crystal-field theory 904.1 Splitting of d-orbital degeneracy by a crystal field 904.2 Multi-electron systems 954.3 Jahn–Teller effects 116References 119Exercises 119

5 Electron spin and angular momentum 1235.1 Pauli spin matrices 1235.2 Measurement of spin 1265.3 Irreducible representations of half-integer angular

momentum 1275.4 Multi-electron spin–orbital states 1295.5 The L–S-coupling scheme 1305.6 Generating angular-momentum eigenstates 1325.7 Spin–orbit interaction 1385.8 Crystal double groups 1505.9 The Zeeman effect (weak-magnetic-field case) 153References 155Exercises 156

6 Molecular electronic structure: The LCAO model 1586.1 N-electron systems 1586.2 Empirical LCAO models 1626.3 Parameterized LCAO models 1636.4 An example: The electronic structure of squarene 1686.5 The electronic structure of H2O 182

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Contents vii

References 188Exercises 189

7 Electronic states of diatomic molecules 1937.1 Bonding and antibonding states: Symmetry functions 1937.2 The “building-up” of molecular orbitals for diatomic

molecules 1987.3 Heteronuclear diatomic molecules 206Exercises 209

8 Transition-metal complexes 2118.1 An octahedral complex 2118.2 A tetrahedral complex 227References 237Exercises 237

9 Space groups and crystalline solids 2399.1 Definitions 2399.2 Space groups 2449.3 The reciprocal lattice 2469.4 Brillouin zones 2479.5 Bloch waves and symmorphic groups 2499.6 Point-group symmetry of Bloch waves 2529.7 The space group of the k-vector, gs

k 2589.8 Irreducible representations of gs

k 2599.9 Compatibility of the irreducible representations of gk 2609.10 Energy bands in the plane-wave approximation 265References 276Exercises 276

10 Application of space-group theory: Energy bands for the perovskitestructure 28010.1 The structure of the ABO3 perovskites 28010.2 Tight-binding wavefunctions 28210.3 The group of the wavevector, gk 28310.4 Irreducible representations for the perovskite energy bands 28410.5 LCAO energies for arbitrary k 29810.6 Characteristics of the perovskite bands 300References 301Exercises 302

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viii Contents

11 Applications of space-group theory: Lattice vibrations 30411.1 Eigenvalue equations for lattice vibrations 30511.2 Acoustic-phonon branches 30911.3 Optical branches: Two atoms per unit cell 31411.4 Lattice vibrations for the perovskite structure 32011.5 Localized vibrations 327References 334Exercises 334

12 Time reversal and magnetic groups 33712.1 Time reversal in quantum mechanics 33712.2 The effect of T on an electron wavefunction 34012.3 Time reversal with an external field 34112.4 Time-reversal degeneracy and energy bands 34212.5 Magnetic crystal groups 34612.6 Co-representations for groups with time-reversal operators 35012.7 Degeneracies due to time-reversal symmetry 357References 361Exercises 361

13 Graphene 36313.1 Graphene structure and energy bands 36313.2 The analogy with the Dirac relativistic theory for massless

particles 36813.3 Graphene lattice vibrations 369References 381Exercises 381

14 Carbon nanotubes 38314.1 A description of carbon nanotubes 38414.2 Group theory of nanotubes 38614.3 One-dimensional nanotube energy bands 39314.4 Metallic and semiconducting nanotubes 40114.5 The nanotube density of states 40314.6 Curvature and energy gaps 406References 407Exercises 407

Appendix A Vectors and matrices 410

Appendix B Basics of point-group theory 415

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Contents ix

Appendix C Character tables for point groups 430

Appendix D Tensors, vectors, and equivalent electrons 442

Appendix E The octahedral group, O and Oh 449

Appendix F The tetrahedral group, Td 455

Appendix G Identifying point groups 462

Index 465

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Preface

The majority of all knowledge accumulated in physics and chemistry concerningatoms, molecules, and solids has been derived from applications of group theory toquantum systems.

My (T.W.) first encounter with group theory was as an undergraduate in physics,struggling to understand Wigner’s Group Theory and Its Application to the Quan-tum Mechanics of Atomic Spectra (1959). I felt there was something magical aboutthe subject. It was amazing to me that it was possible to analyze a physical systemknowing only the symmetry and obtain results that were absolute, independent ofany particular model. To me it was a miracle that it was possible to find some exacteigenvectors of a Hamiltonian by simply knowing the geometry of the system orthe symmetry of the potential.

Many books devote the initial chapters to deriving abstract theorems before dis-cussing any of the applications of group theory. We have taken a different approach.The first chapter of this book is devoted to finding the molecular vibration eigen-values, eigenvectors, and force constants of a molecule. The theorems required toaccomplish this task are introduced as needed and discussed, but the proofs of thetheorems are given in the appendices. (In later chapters the theorems needed for theanalysis are derived within the discussions.) By means of this applications-orientedapproach we are able to immediately give a general picture of how group theory isapplied to physical systems. The emphasis is on the process of applying group the-ory. The various steps needed to analyze a physical system are clearly delineated.By the end of the first chapter the reader should have an appreciation for the powerof group theory, what information can be obtained, and how to obtain it. That is,the “magic” of group theory should already be apparent.

In addition to the essential, traditional topics, there are new topics, includingthe electronic and vibrational properties of graphene and nanotubes, the vibra-tions of isotopically substituted molecules, localized vibrations, and discussions ofthe axially symmetric lattice dynamics model. The energy bands and vibrational

xi

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xii Preface

normal modes of crystals with the perovskite structure are also discussed indetail.

The material in this book was developed in part from group-theory courses andfrom a series of lectures presented in courses on special topics at the University ofMissouri-Columbia. It is appropriate for science and engineering graduate studentsand advanced undergraduate seniors. The ideal reader will have had a course inquantum mechanics and be familiar with eigenvalue problems and matrix algebra.However, no prerequisite knowledge of group theory is necessary.

This book may be employed as a primary text for a first course in group the-ory or as an auxiliary book for courses in quantum mechanics, solid-state physics,physical chemistry, materials science, or electrical engineering. It is intended as aself-teaching tool and therefore the analyses in the early chapters are given in somedetail. Each chapter includes a set of exercises designed to reinforce and extend thematerial discussed in the chapter.

Thomas Wolframand

Sinasi Ellialtıoglu

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1

Introductory example: Squarene

In this chapter we illustrate the solution of a simple physical problem in order tofamiliarize the reader with the procedures used in the group-theoretical analysis.Since this is the initial chapter, we shall give nearly all of the details involved inthe analysis. Some readers familiar with group theory may find that the discussionincludes too much detail, but we would rather be clear than brief. In later chaptersless detail will be required since the reader will by then be familiar with the analysismethod.

Theorems from group theory are stated and discussed when employed in theanalysis, but the proofs of the theorems are not presented in this chapter. Read-ers interested in the proofs can find them in Appendix B or refer to a number ofexcellent standard group-theory texts [1.1].

The procedures employed in this chapter are simple but somewhat tedious andnot the most efficient way to analyze the simple example discussed. However, theseprocedures will prove extremely valuable when we are faced with more complexproblems. Therefore the reader is encouraged to work through the details of thechapter and the exercises at the end of the chapter.

1.1 In-plane molecular vibrations of squarene

As an introductory example we consider a fictitious square molecule we shall call“squarene”. The squarene molecule, shown in Fig. 1.1, lies in the plane of the paperwith identical atoms at each corner of a square. Our aim in this chapter is to usegroup theory to assist in determining the vibrational frequencies and motions ofthe normal modes (eigenvalues and eigenvectors) of the molecule.

In general the molecular vibration problem may be cast into the form of a matrixeigenvalue problem,

F ξ = ω2 M ξ,

1

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2 Introductory example: Squarene

1 2

34

Figure 1.1 The squarene molecule.

where F is a matrix of “force constants” and ξ is a column vector whose compo-nents are the vibrational displacements of the atoms. The matrix M is a diagonalmass matrix, Mi j = mi δi j where mi is the mass of the atom with coordinatecomponent ξi and ω is the angular frequency of vibration. This equation may betransformed to a standard eigenvalue form,

F′ η = ω2η,

with F′ =M−1/2 F M−1/2 and η =M1/2 ξ .In the case of squarene all the masses are the same, so M can be replaced by a

scalar, m, the common mass of the four atoms. Therefore the eigenvalue equationwe are concerned with is the 8× 8 matrix equation

F ξ = mω2ξ. (1.1)

Solutions of (1.1) consist of a set of eight eigenvalues, m(ωk)2, and a set of eightassociated eigenvectors, ξ k (k = 1, 2, . . ., 8), that specify the relative displacementsof the atoms. The solutions of (1.1) will include three “body modes”, namely trans-lation in the x-direction, translation in the y-direction, and rotation about an axisperpendicular to the plane of the molecule at the center of the square. The angularfrequencies of these three body modes may be taken as zero because there are no“restoring forces” when a molecule is translated or rotated as a unit. Therefore, forour two-dimensional squarene molecule there will be five vibrational modes andthree zero-frequency, body modes.

One approach to the molecular vibrations of squarene would be to postulate aforce-constant model, F, and solve the resulting eigenvalue problem. In its mostgeneral form, F will have 64 non-zero matrix elements. If no simplifications aremade, the diagonalization of an 8× 8 matrix could be daunting without the use ofa computer. If approximations are employed to simplify F, it will not be evidentwhether the results are constrained by symmetry or by the assumptions made in thepostulated force-constant model. The use of group theory not only greatly simpli-fies the analysis, but also yields the most general form possible for F and providesphysical insight into the nature of the normal modes of vibration.

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1.1 In-plane molecular vibrations of squarene 3

Often, the normal-mode frequencies can be measured by spectroscopic means;infrared or Raman spectroscopy, for example. However, the force constants (thematrix elements of F) can not be measured directly. They are usually inferredby using a specific model and choosing the constants so that the eigenfrequenciesmatch the observed frequencies. This method does not always produce a unique setof force constants, and other spectroscopic data must be employed; for example,the frequencies of isotopically substituted molecules and/or elastic constants.

In Chapter 2 we shall investigate how group theory can be used to predictisotopic frequency shifts. In fact, in some cases the frequencies of an isotopic mole-cule can be expressed in terms of frequencies of the non-isotopic molecule [1.2]without ever having to specify the force-constant matrix.1

There is a general order to the way in which group theory is employed to ana-lyze an eigenvalue problem. We shall label the various steps as “Step 1” through“Step 8”.

1.1.1 Step 1: Identify the point group and its symmetry operations

At rest, the squarene molecule has the symmetry of a square. For simplicity weconfine our discussion to the two dimensions of the plane of the molecule. (Thereis no loss of generality in this assumption because the in-plane vibrational modesdo not mix with the motions perpendicular to the plane of the molecule (see Exer-cise 1.16).) The operations that rotate or reflect an object into a configurationindistinguishable from its original configuration are called the covering opera-tions and constitute a group. For squarene these operations are rotations about anaxis perpendicular to the plane of the molecule and reflections through lines in theplane of the molecule. For example we can rotate the square by 90 degrees or anymultiple of 90 degrees about an axis perpendicular to the plane passing throughthe center of the square. So there are four rotations: 90-degree, 180-degree, 270-degree, and 360-degree rotations. Rotation by 360 degrees is taken to be the sameas “doing nothing”. The “doing-nothing” operation is usually called the “identity”operation and denoted by the symbol E . There are seven other symmetry opera-tions. If we draw a diagonal line across the square and reflect the corners throughthis line, the square remains unchanged. There are two different diagonal lines thatpass through the center and two corners of the square. Another pair of reflectionlines passes through the middle of the square and bisects two opposite sides. Thesesymmetry operations are shown schematically in Fig. 1.2. The point group cor-responding to these operations is “C4v”. (In three dimensions the group is D4h .

1 Assuming harmonic motions and neglecting very small changes in the force constants due to isotopesubstitution.

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4 Introductory example: Squarene

(a)

1 2

34

E

(b)

C4, C2, C34

(c)

σv1

(d)

σv2

(e) σd1 (f)

σd2

Figure 1.2 The squarene molecule and its symmetry elements.

Methods for identifying the group of a molecule are discussed in Appendix G.) ForC4v there are in total eight symmetry elements.2

(1) E = a 0-degree or 360-degree rotation.(2) C4 = a 90-degree rotation.(3) C2 = a 180-degree rotation.(4) C3

4 = a 270-degree rotation.(5) σv1 = reflection in a diagonal line.(6) σv2 = reflection in a diagonal line.(7) σd1 = reflection in a line cutting the sides of the square.(8) σd2 = reflection in a line cutting the sides of the square.

For the remainder of this chapter we shall regard the symbols E , C4, σv1 etc. assymmetry operators that act on the molecule, not the coordinate system.3 Thus, forexample C4 rotates the molecule and its nuclear displacement by 90 degrees in aclockwise manner.

The formal definition of a group is given in Theorem 1.1.

2 Inversion through the center of the square (i.e., r → −r) is a symmetry operation. However, the effect ofinversion is the same as the effect of a C2 rotation in this case and so may be omitted.

3 There is a distinction between the symmetry elements and the symmetry operators, but the group of theoperators is isomorphic to the group of the elements, which means that the two groups are equivalent grouptheoretically.

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1.1 In-plane molecular vibrations of squarene 5

Theorem 1.1 (Definition of a group)1. A group is a set of elements that are closed under group multiplication. That is,

the product of any two elements in the group produces another element of thegroup. If A and B are in the group then so is AB.

2. The associative law holds: A(BC) = (AB)C.3. There is a unit or identity element, E, such that E A = AE = A.4. Every element in the group has an inverse element that is also in the group.

If A is an element then A−1 is also an element in the group, and AA−1 =A−1 A = E.

5. The word “operator” may be substituted for the word “element” in items 1through 4 to define a group based on the operators associated with the elements.

A bit later we shall verify that the eight symmetry elements identified abovesatisfy all of the conditions of a group as defined in Theorem 1.1.

1.1.2 Step 2: Specify the coordinate system and basis functions

The coordinate system to be employed in the analysis is often a set of Cartesiancoordinates,4 but there are many other choices; Euler angles, internal coordinates,cylindrical coordinates, and spherical coordinates, to name a few. In general thechoice of basis functions depends on the physical phenomena to be analyzed. Inthe case of molecular vibrations we are concerned with the motions of each of theatoms comprising the molecule. Therefore an appropriate choice of basis functionsis the atomic displacement vectors. To measure the displacements of each atomfrom its equilibrium position we label the corner positions of the square (not theatoms of the molecule) and construct X–Y Cartesian coordinate systems anchoredto these corner positions as shown in Fig. 1.3(a).

1.1.3 Step 3: Determine the effects of the symmetry operationson the basis functions

We consider how a vector representing the displacement of one of the atoms istransformed when the molecule is subjected to one of the symmetry operations.Note that the X–Y coordinate systems as well as the position numbers of the cor-ners are fixed. Imagine the molecule placed on top of the square in Fig. 1.3(a)and rotated or reflected under the action of a symmetry operation. If we rotate themolecule clockwise by 90 degrees (the C4 operation) the atoms of the moleculemove to different corner positions as indicated by Table 1.1.

4 In chemistry texts internal coordinates are often used. See reference [1.3].

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6 Introductory example: Squarene

(a)

1 2

34

X1 X2

X3X4

Y1 Y2

Y3Y4

(b)

X1

Y1

rx(1)

ry(1) r

C4=⇒

X4

Y4rx(4)

ry(4) r−

Figure 1.3 (a) The coordinate system. Here 1, 2, 3, and 4 label the fixed cornersof the square, and Xi and Yi label the fixed coordinate axes. (b) The effect of C4rotation on the displacement vector, r, on an atom initially at corner 1.

Table 1.1 The effect of C4 on squarene

Atom’s initial corner Atom’s final corner

1 42 13 24 3

Consider a vector displacement r of an atom initially located at corner 1. The com-ponents of this vector in the X1–Y1 coordinate system are rx and ry , respectively,as shown in Fig. 1.3(b). After rotation the displacement vector r (anchored to theatom) is now centered on the X4–Y4 coordinate system associated with corner 4.In the X4–Y4 coordinate system, the X4 component is ry and the Y4 component is−rx . In symbolic notation,

C4 rx(1)→−ry(4),

C4 ry(1)→ rx(4),

meaning that the C4 operation transforms the component of r along the X1 axisinto a component along the −Y4 axis and transforms the component of r along theY1 axis into a component along the X4 axis. The displacement vector of any of theatoms of squarene will transform in the same way, so, in general,

C4 rx(i)→−ry( f ),

C4 ry(i)→ rx( f ),

where “i” and “ f ” refer to the initial and final corners given in Table 1.1.

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1.1 In-plane molecular vibrations of squarene 7

Table 1.2 The effect of σv1 on squarene

Atom’s initial corner Atom’s final corner

1 12 43 34 2

Table 1.3 The action table for squarene

E C4 C2 C34 σv1 σv2 σd1 σd2

rx rx −ry −rx ry ry −ry −rx rxry ry rx −ry −rx rx −rx ry −ry

1 1 4 3 2 1 3 2 42 2 1 4 3 4 2 1 33 3 2 1 4 3 1 4 24 4 3 2 1 2 4 3 1

Next consider the action of the reflection operation, σv1. For this symmetryoperator we have the results shown in Table 1.2. For σv1 the displacements aretransformed so that rx(i)→ ry( f ) and ry(i)→ rx( f ). Symbolically,

σv1 rx(i)→ ry( f ),

σv1 ry(i)→ rx( f ),

where the initial and final corner numbers are given in Table 1.2.A convenient device for determining the effects of symmetry operations is the

action table shown in Table 1.3. The columns are labeled by the group operators.The rows are labeled by rx and ry and the corner numbers. The intersection of acolumn and a row is the action of the column operator on the displacements (firsttwo rows) and the corner labels (last four rows). The table makes it easy to generatethe effect of any operator. For example, consider σd1 acting on the displacementvector initially at corner 2. From the table (column 7) we see that under action ofσd1 corner 2→ corner 1, rx →−rx , and ry → ry . Therefore, rx(2)→−rx(1) andry(2)→ ry(1). Using the action table the effect of any operator of the group on thebasis functions can be determined.

Another useful device is the group multiplication table shown in Table 1.4. Inthis table the rows and columns are labeled by the operators of the group, Oi

(i = 1, 2, . . . , 8). The intersection of the i th row with the j th column is the oper-ator obtained by the sequential application of the operators in the order Oi O j .The order is important because Oi O j is often different from O j Oi . The group

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8 Introductory example: Squarene

Table 1.4 The group multiplication table for C4v

C4 E C4 C2 C34 σv1 σv2 σd1 σd2

E E C4 C2 C34 σv1 σv2 σd1 σd2

C4 C4 C2 C34 E σd2 σd1 σv1 σv2

C2 C2 C34 E C4 σv2 σv1 σd2 σd1

C34 C3

4 E C4 C2 σd1 σd2 σv2 σv1

σv1 σv1 σd1 σv2 σd2 E C2 C4 C34

σv2 σv2 σd2 σv1 σd1 C2 E C34 C4

σd1 σd1 σv2 σd2 σv1 C34 C4 E C2

σd2 σd2 σv1 σd1 σv2 C4 C34 C2 E

multiplication table is easily generated from the action table. For example, C4

transforms rx , ry, 1, 2, 3, 4 into −ry, rx , 4, 1, 2, 3, respectively. If we now applyσd1 to this result then −ry, rx , 4, 1, 2, 3 is transformed into −ry,−rx , 3, 2, 1, 4.Therefore, σd1C4 transforms rx , ry, 1, 2, 3, 4 into −ry,−rx , 3, 2, 1, 4. On lookingat the action table, we see that this is the same action as σv2. Thus σd1C4 = σv2.This result may be seen as the operation at the intersection of the row labeled σd1

and the column labeled C4 in the group multiplication table. Similarly, one maysee from the same table that C4σd1 = σv1, which shows that with a change of orderin multiple operations one does not necessarily obtain the same result.

The action table can also be applied to products such as rx(1) ry(2), [rx(1)]2,or products of more than two displacements. For example, C4 rx(1) ry(2) →−ry(4) rx(1) and C4 [rx(1)]2 → [ry(4)]2.

1.1.4 Step 4: Construct matrix representations for each elementof the group using the basis functions

It will prove useful to employ an eight-component column vector so that we canshow the effect of a symmetry operation on all the atomic displacement vectors atthe same time. To do this, we use, as the components of the 8-vector, the rx and ry

displacement components of all four atoms. Using the action table we have, in oursymbolic notation,

C4

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

rx(1)ry(1)rx(2)ry(2)rx(3)ry(3)rx(4)ry(4)

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦⇒

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

ry(2)−rx(2)

ry(3)−rx(3)

ry(4)−rx(4)

ry(1)−rx(1)

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦. (1.2)

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1.1 In-plane molecular vibrations of squarene 9

The rows of the column vectors in (1.2) are labeled from top to bottom by ex(1),ey(1), ex(2), ey(2), ex(3), ey(3), ex(4), and ey(4), unit vectors along the fixedCartesian coordinates in Fig. 1.3(a). It is important to understand the meaning of(1.2). The rows of the column vectors are labeled by the unit vectors of the fixedcoordinate system. Therefore a column vector such as⎛

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

000c000

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

is a displacement of length c for the y-component on corner 2. The column vector⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

d000000

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

is a displacement of length d for the x-component on corner 1. What equa-tion (1.2) means is that a clockwise rotation, C4, rotates the molecule so that thex-component of the molecular displacement on corner 1 after rotation appears asthe −y-component on corner 4. In other words, we assign to the new vector (right-hand side) the value −rx(1) for its y-component on corner 4. This is indicated bythe arrow in (1.2). Similarly, we assign to the new vector the value ry(4) for itsx-component on corner 3. In more compact notation,

�(C4) Vi = V f , (1.3)

where �(C4) is an 8 × 8 matrix.5 The eight-dimensional column vector, Vi , hasas its components the entries of the column on the left-hand side of (1.2). Thecomponents of the column vector, V f , are the entries of the right-hand side of (1.2).

�(C4) can easily be constructed. For example, (1.2) shows that rx(1) is trans-formed to −ry(4), therefore the first column of �(C4) has only one entry, namely

5 An equally valid method of describing the group and its representations is to fix the molecule and rotatethe coordinate framework. Clearly, rotating the coordinate axes counter-clockwise with the molecule fixed isequivalent to fixing the coordinate axes and rotating the molecule clockwise. However, the usual convention inmost texts is rotation of the molecule or function rather than the coordinate axes.

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10 Introductory example: Squarene

a “−1” in the eighth row. Similarly, ry(3) is transformed to rx(2), so the sixth col-umn of �(C4) has a single entry, a “+1” in the third row. Proceeding in this wayproduces the 8× 8 matrix

�(C4) =

1−1

1−1

1−1

1−1

Matrix multiplication of Vi by �(C4) produces the vector V f . Using the actiontable we can readily construct matrices for each of the eight operations of the C4v

group. These matrices are shown in Table 1.5.The eight matrices in Table 1.5 form a matrix representation of the C4v group.

In the parlance of group theory we say that the displacements rx( j) and ry( j)( j = 1, 2, 3, and 4) are basis functions for an eight-dimensional representation ofthe group C4v. The symbols rx and ry without regard to the corner numbers forma two-dimensional representation of C4v. Finally, we note that the corner numbers(the last four rows in Table 1.3) form the basis for a four-dimensional representationof C4v.

From the group multiplication table we can find the inverse of any representationmatrix by finding the intersection that produces the identity element. For example,the inverse of C4 is C3

4 and the reflection operators are their own inverses.Inspection of the representation matrices in Table 1.5 reveals a number of prop-

erties. (1) All of the matrices are unitary, meaning that the inverse, �(Oi )−1, is the

transpose, complex conjugate of �(Oi ), where Oi denotes the i th operator of thegroup. Since, in this case, all the matrix elements are real, the inverse is simplythe transpose. (2) The rows (or columns) of �(Oi ) are orthogonal to each other;that is,

h∑s=1

�(Oi )rs �(Oi )ts =h∑

s=1

�(Oi )sr �(Oi )st = δr t ,

where h, the order of the group, is the number of elements in the group and �(Oi )rs

is the rs matrix element of �(Oi ). (3) The determinant of each matrix is +1. Theresults (2) and (3) are consequences of (1), namely of the matrices being unitary.

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1.1 In-plane molecular vibrations of squarene 11

Table 1.5 Matrix representations of the group C4v: (a) through (h),representation matrices based on the displacement coordinates of squarene

(a) �(E) =1

11

11

11

1

(e) �(σv2) =−1

−1−1

−1−1

−1−1

−1

(b) �(C4) =1

−11

−11

−11

−1

( f ) �(σv1) =1

11

11

11

1

(c) �(C2) =−1−1−1−1

−1−1−1−1

(g) �(σd1) =−1

1−1

1−1

1−1

1

(d) �(C34) =

−11

−11

−11

−11

(h) �(σd2) =1−1

1−1

1−1

1−1

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12 Introductory example: Squarene

To verify that the matrices in Table 1.5 form a proper representation of the group,one can simply multiply them together to show that they obey the properties of agroup representation given in Theorem 1.2.

Theorem 1.2 (Properties of a matrix representation)1. For any operators A, B, and C in the group there are associated matrices �(A),

�(B), and �(C) such that if A B = C then �(A) �(B) = �(A B) = �(C).(This requires that �(A)−1 = �(A−1).)

2. For every point group there exists a representation consisting of unitary matri-ces. Unitary matrices are square matrices for which �(A)−1 is the complexconjugate, transpose of �(A).

It follows from Theorem 1.2 that matrices given in Table 1.5 obey the samegroup multiplication table (Table 1.4) as do the operators with �(Oi ) instead of Oi

labeling the rows and columns.

1.2 Reducible and irreducible representations of a group

A group can have many matrix representations. If we perform the same similaritytransformation, T−1 �(Oi )T, on all of the matrices of Table 1.5 we will obtain adifferent set of 8× 8 matrices, �(a)(Oi ), that also represents the group. (This couldbe written as T−1 � T = �(a), where it is understood that the similarity transforma-tion, T, is applied to all of the representation matrices.) Two representations relatedby a similarity transformation are said to be equivalent. If we use different basisfunctions we may obtain a representation whose dimensionality is different fromthat of �. For example, we could base a representation on how the atoms of themolecule move without regard to atomic displacements. That would generate �(b),a 4 × 4 matrix representation of C4v. Finally we could combine all these repre-sentations into a 20× 20 representation by placing �(Oi ), �(a)(Oi ), and �(b)(Oi )

as blocks along the diagonal to form the matrix �(c)(Oi ). This process is shownschematically in Fig. 1.4(a). �,�(a), �(b), and �(c) are all valid representationsof the group. �(c) is composed of three uncoupled representations (unconnectedblocks of matrices). If an arbitrary similarity transformation were performed on�(c), the block-diagonal form would usually be lost even though the transformedrepresentation would be equivalent to �(c).

We see from the previous paragraph how larger representation matrices can becomposed from smaller ones. The reverse process can also be accomplished. Givenan N×N matrix representation, �N , it may be possible to transform all its matricesinto block-diagonalized forms by a similarity transformation � = S−1 �N S. That

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1.2 Reducible and irreducible representations of a group 13

Γ(c) =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

Γ

(8×8)

(8×8)

(4×4)

0

0

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

ΓN (Oi) =

Γ(1)(Oi)

Γ(2)(Oi)

Γ(m)(Oi)

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

a12

a22a21

a11 . . .

. . .

. . .

a1N

a2N

aNNaN2aN1

......

...

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

0

. . .

0

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

Γ(a)

Γ(b)

S−1 ΓN (Oi) S =

(a)

(b)

(c)

Figure 1.4 (a) The (20×20) representation �(c) = �+�(a)+�(b), (b) the N×Nrepresentation �N , and (c) decomposition of �N into smaller representations bya similarity transformation.

is, by a similarity transformation we can decompose the �N representation intosmaller representations. Symbolically, we have

�N ≡ �(1) ⊕ �(2) ⊕ · · · ⊕ �(m). (1.4)

Equation (1.4) is an equivalence statement meaning that there exists a similaritytransformation that will decompose �N into the series of uncoupled blocks shownon the right-hand side. In more detail,

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14 Introductory example: Squarene

�N (Oi ) = �(1)(Oi )⊕ �(2)(Oi )⊕ · · · ⊕ �(m)(Oi ) (i = 1, 2, . . . , h).

The symbol⊕ means that the matrix representation which follows is a block alongthe diagonal of �N after a suitable similarity transformation. The order of repre-sentations on the right-hand side of (1.4) is irrelevant. Each of the representationmatrices �N (Oi ) consists of m uncoupled blocks or submatrices. Such a trans-formation is shown schematically in Fig. 1.4(c). For simplicity the decompositionformula of (1.4) is often written as

�N = �(1) + �(2) + · · · + �(m), (1.5)

with the understanding that simple matrix addition is not implied. We will use thisnotation when there is no chance for confusion.

Since each �(i) appearing in (1.5) is itself a representation, it may be possi-ble to decompose it into smaller blocks. Eventually, minimum-sized blocks mustbe achieved. These minimal blocks are said to be irreducible representations.(We shall use the abbreviation “IR” for “irreducible representation” throughoutthis book.)

Theorem 1.3 (Definitions of reducible and irreducible representations)• A representation that can be reduced or decomposed by a similarity trans-

formation into smaller, uncoupled, submatrix representations is a reduciblerepresentation.• Conversely, a representation that can not be reduced by any similarity trans-

formation into smaller, uncoupled, submatrix representations is an irreduciblerepresentation.

For most science problems the basis functions normally employed to describethe system are bases for a reducible representation, �. The role of group theoryis to provide a method of finding the matrix S such that the similarity trans-formation S−1 � S = �S produces a representation that has only IRs along thediagonal of each �S(Oi ). The notation S−1 � S = �S is symbolic. It means thatthe same similarity transformation is applied to every representation matrix of �,that is S−1 �(Oi )S = �S(Oi ) (i = 1, 2, . . . , h), and every �S(Oi ) has the sameblock-diagonal form consisting of submatrices that are irreducible representationmatrices.

The columns of S or the rows of S−1 are the symmetry functions for the system.These symmetry functions are linear combinations of the original basis functions.When a physical problem is expressed in terms of its symmetry functions theeigenvalue equation is block-diagonal and a great simplification of the mathemat-ics usually results. A method for deriving the symmetry functions is discussed in“Step 7” below in Subsection 1.2.6.

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1.2 Reducible and irreducible representations of a group 15

Symmetry functions are usually not eigenfunctions. If the decomposition of therepresentation based on the physical coordinates contains the same IR r times,then there will be r sets of symmetry functions for that IR and the eigenfunctionswill be linear combinations of these symmetry functions. On the other hand, if thedecomposition contains a particular IR only once, then the basis functions for thatIR are also eigenfunctions. This is a rather amazing result. It means that some ofthe eigenfunctions of a physical system may be entirely determined by symmetry,independently of the details of the system’s Hamiltonian or force-constant matrix.6

We shall see examples of this as we proceed with the analysis of the vibrations ofsquarene.

1.2.1 Step 5. Determine the number and types of irreducible representations

The matrix representation, �, we have constructed for squarene is reducible intounconnected IR blocks by the similarity transformation S−1 � S, where S is thematrix of symmetry functions. The reduction of � to diagonal blocks need notresult in distinct IRs, and in many cases the same (or equivalent) IR will occurmore than once among the diagonal blocks. The number and types of IRs thatoccur among the diagonal blocks depend on the basis functions used to generate �.

What is the connection between the IRs of the C4v group and the physicalproblem of the molecular vibrations? The connection is this: The similarity trans-formation S which block-diagonalizes the reducible representation � will alsoblock-diagonalize the matrix F of (1.1).

To determine the number and type of IRs contained in � we need to explore afew other properties of a group.

1.2.2 Classes of a group

An important feature of a group is its classes. The definition of a class is given inTheorem 1.4.

Theorem 1.4 (Definition of a class)• Two operators, A and B, are conjugate if A = C BC−1, where C is any member

of the group. A class is a set of all of the distinct operators of a group that areconjugate to each other.• From the definition of a representation, Theorem 1.2, it follows that two rep-

resentation matrices �(A) and �(B) belong to the same class if �(A) =�(C) �(B) �(C)−1, where �(C) is a matrix representation of any operator ofthe group.

6 A valid Hamiltonian is not arbitrary. It must commute with the operations of the group. Similarly, F mustcommute with the operations of the group.

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16 Introductory example: Squarene

To find the classes of C4v we can make use of the group multiplication table(Table 1.4). The inverse of any operation can be found from the table by observingwhich product produces the identity, E :

E−1 = E,

C−14 = C3

4 ,

C−12 = C2,

(C34)−1 = C4.

Also it is evident that any reflection operator is its own inverse; that is, σ−1i = σi .

Clearly, E is in a class by itself since AE A−1 = E for all A in the group. To find theother classes we start with C4 and form the products Oi C4 O−1

i (i = 1, 2, . . . , h).This procedure yields

E C4 E = C4, σv1 C4 σv1 = σv1 σd2 = C34 ,

C4 C4 C−14 = C4, σv2 C4 σv2 = σv2 σd1 = C3

4 ,

C2 C4 C−12 = C2 C4 C2 = C4, σd1 C4 σd1 = σd1 σv2 = C3

4 ,

C34 C4 (C3

4)−1 = C3

4 C4 C4 = C4, σd2 C4 σd2 = σd2 σv1 = C34 .

Therefore C4 and C34 form a class. Continuing this procedure with the other

elements of the group shows that there are five classes. They are as follows:{E}; {C4,C3

4}; {C2}; {σd1, σd2}; and {σv1, σv2}. Similarly, for the representationmatrices the classes are {�(E)}; {�(C4), �(C3

4)}; {�(C2)}; {�(σd1), �(σd2)}; and{�(σv1), �(σv2)}.

1.2.3 The character of a representation matrix

Another important property of a representation matrix is its trace. The trace of amatrix is the sum of the diagonal elements of the matrix. The trace of a represen-tation matrix is called its character. The usual symbol for a character is χ and thecharacter of �α(Oi ) is denoted by χα(Oi ) or χα

i .If A and B are in the same class then A = C BC−1 for some element C in

the group. This requires that �(A) = �(C) �(B) �(C)−1 and therefore �(A) and�(B) are related by a similarity transformation. The trace of a matrix is unchangedby a similarity transformation and therefore the character of �(A) must be equal tothe character of �(B). From this result it follows that all representation matricesbelonging to the same class have the same character.

The character table for the C4v group is given in Table 1.6. The first columnof the table lists the names of the IRs of the C4v group. The next five columnsgive the characters (the χs) of the classes of C4v for the five different IRs. Thelast row lists the results for �, our 8× 8 representation for squarene obtained from

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1.2 Reducible and irreducible representations of a group 17

Table 1.6 The character table for C4v

Classes

� (IR name)E

χ(E)2C4χ(C4)

C2χ(C2)

2σvχ(σv)

2σdχ(σd) Basis functions

A1 1 1 1 1 1 z, x2 + y2, z2

A2 1 1 1 −1 −1 Rz

B1 1 −1 1 1 −1 x2 − y2

B2 1 −1 1 −1 1 xyE 2 0 −2 0 0 (x, y), (xy, yz), (Rx , Ry)�(8× 8) 8 0 0 0 0 (rx (i), ry(i)), i = 1 to 4.

the traces of the representations in Table 1.5. A shorthand notation is used in thecharacter table. Instead of listing all of the elements of a class, only one elementis given and the number of elements in the class is specified. For example, theclass {C4, C3

4} is denoted by 2C4, meaning that there are two elements in the classto which C4 belongs. Since all the members of a class have the same character,it is not necessary to specify each element. The last column of Table 1.6 lists afew functions that can serve as basis functions for the corresponding IR. Charactertables for the different point groups are given in Appendix C.

The character table for C4v gives all the information we need to determine thenumber and types of IRs of our 8×8 representation, �. However, to find the decom-position of � and to construct the S matrix that block-diagonalizes the eigenvalueequation we need a few more results from group theory.

1.2.4 Orthogonality of the characters of an irreducible representation

C4v has five classes. If the characters of the classes of a given IR of C4v are con-sidered as components of a five-dimensional unit vector, five normalized vectorscan be constructed, one for each of the IRs of C4v. We define the kth component ofsuch vectors by

e αk =

√Nk

hχα

k , (1.6)

where k = 1, 2, . . . , 5 denotes the class, Nk is the number of elements for the kthclass, χα

k is the character of the kth class of the αth IR, and h is the number ofelements in the group. The vectors are

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18 Introductory example: Squarene

e A1 = 1√8

⎛⎜⎜⎜⎜⎜⎝

1√2

1√2√2

⎞⎟⎟⎟⎟⎟⎠ , e A2 = 1√

8

⎛⎜⎜⎜⎜⎜⎝

1√2

1−√2−√2

⎞⎟⎟⎟⎟⎟⎠ ,

(1.7)

e B1 = 1√8

⎛⎜⎜⎜⎜⎜⎝

1−√2

1√2

−√2

⎞⎟⎟⎟⎟⎟⎠ , e B2 = 1√

8

⎛⎜⎜⎜⎜⎜⎝

1−√2

1−√2√

2

⎞⎟⎟⎟⎟⎟⎠ , eE = 1√

8

⎛⎜⎜⎜⎜⎝

20−2

00

⎞⎟⎟⎟⎟⎠ ,

where the order of the components (top to bottom) is the same as the order of theclasses in the character table.

It is easily verified that these vectors are orthogonal to each other, meaning thatthe scalar product (dot product) of any two different vectors is zero, (eα, eβ) =(1/h)

∑Nck=1 Nk (χ

αk )∗ χβ

k = δαβ . The orthogonality of the characters of the IRs is ageneral feature of a group. A statement of this property is given in Theorem 1.5.

Theorem 1.5 (Orthogonality of the characters of irreducible representations) Thecharacters of the representation matrices of the IRs of a group are orthogonal toone another:

1

h

h∑i=1

(χαi )∗ χβ

i = δαβ, (GT1.5a)

where χαi is the character of �α(Oi ) and h is the number of elements in the group.

Since operations belonging to the same class have the same character,

1

h

Nc∑k=1

Nk (χαk )∗ χβ

k = δαβ, (GT1.5b)

where Nk is the number of elements in the kth class and Nc is the number of classesin the group.

Theorem 1.5 is the general statement of the result we found for the charactersin (1.7). To proceed with our analysis we also need the results from group theorygiven in Theorem 1.6.

Theorem 1.6 (Properties of irreducible representations)1. The number of different irreducible representations of a group is equal to Nc,

the number of classes.

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1.2 Reducible and irreducible representations of a group 19

2. The sum of the squares of the dimensions of all the distinct IRs of a group isequal to the number of elements in the group:

Nc∑i=1

l2i = h,

where li is the dimension of the i th IR and h is the number of elements in thegroup.

We see from the character table for C4v that there are five IRs. Since the characterE is equal to the dimensionality of the IR, it is clear the first four IRs are one-dimensional and the last is two-dimensional. The first result quoted in Theorem 1.6tells us that there are five IRs since there are five classes for C4v. The second resultof Theorem 1.6 tells us that the sum of the squares of the dimensions of the five IRsmust be 8. There can’t be a three-dimensional IR (since 32 = 9 is already greaterthan 8) and the smallest dimension possible has li = 1. So we need five integers,all of which are either 1 or 2 and whose squares add up to 8. The only solution is12 + 12 + 12 + 12 + 22 = 8.

Our next step is to determine the number and types of IRs contained in our8×8 representation, �. In general, if � is any representation of the group we knowthat there exists a similarity transformation, S−1�(Oi )S = �′(Oi ), where �′(Oi )

is block-diagonal with various IR matrices along the diagonal. The character of�′(Oi ) is equal to the sum of the traces of the IR blocks along the diagonal. Sincethe trace of �(Oi ) is unchanged by the similarity transformation, its trace is equalto the trace of �′(Oi ). Therefore, it follows that

χ�i =

Nc∑α=1

nαχαi (i = 1, 2, . . . , h), (1.8)

where χ�i is the character of the representation matrix �(Oi ) and χα

i is the characterof the αth IR representing the operation Oi . The coefficients nα are the number oftimes the αth IR occurs among the diagonal blocks of �′(Oi ). Notice that the nα

coefficients are the same for all of the different operations Oi . This is what is meantwhen it is stated that S−1� S block-diagonalizes all of the representation matricesin the same way.

If we multiply both sides of (1.8) by the factor (1/h)(χβ

i )∗ and sum over all of

the operations of the group, we obtain

1

h

h∑i=1

(χβ

i )∗ χ�

i =Nc∑α=1

nα 1

h

h∑i=1

(χβ

i )∗χα

i =Nc∑α=1

nα δαβ = nβ. (1.9)

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20 Introductory example: Squarene

To obtain the final result in (1.9) we used the equation (GT1.5a) from Theorem 1.5.Equation (1.9) can be rewritten in terms of the characters of the classes:

nβ = 1

h

Nc∑k=1

Nk (χβ

k )∗ χ�

k , (1.10)

where the sum of k is over the classes, Nk is the number of elements in the kthclass, Nc is the number of classes, χ�

k is the character of the kth class of �, and χβ

k

is the character of the kth class of the βth IR.To see how (1.10) works, we apply it to our representation for squarene. With

the help of the character table (Table 1.6), we find that

nα = 1

h[χ∗E χα

E + 2(χ∗C4χα

C4)+ χ∗C2

χαC2+ 2(χ∗σv χ

ασv)+ 2(χ∗σd

χασd)],

n A1 = 1

8[ 8× 1+ 2(0× 1)+ 0× 1+ 2(0× 1)+ 2(0× 1) ] = 1,

n A2 = 1

8[8× 1+ 2(0× 1)+ 0× 1+ 2(0×−1)+ 2(0×−1)] = 1,

nB1 = 1

8[8× 1+ 2(0×−1)+ 0× 1+ 2(0× 1)+ 2(0×−1)] = 1,

nB2 = 1

8[8× 1+ 2(0×−1)+ 0× 1+ 2(0×−1)+ 2(0× 1)] = 1,

nE = 1

8[8× 2+ 2(0× 0)+ 0×−2+ 2(0× 0)+ 2(0× 0)] = 2.

(1.11)

The decomposition of � into IRs is then, symbolically,

� = �A1 + �A2 + �B1 + �B2 + 2� E . (1.12)

Since the E IR is two-dimensional, it requires two basis functions. That is, S−1

will have two rows of symmetry functions for each E IR. To distinguish thetwo functions they are called the “row-1” and “row-2” basis functions of E . (AnN -dimensional IR will have N rows of symmetry functions.) For squarene E occurstwice in the decomposition, so there will be two different row-1 and two differentrow-2 symmetry functions in S−1.

1.2.5 Step 6: Analyze the information contained in the decomposition

One of the most important results of group theory is the result stated below inTheorem 1.7.

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1.2 Reducible and irreducible representations of a group 21

Theorem 1.7 (Matrix elements of a commuting operator) If H is a Hermitianoperator that commutes with the operations of the group, the matrix elements ofH vanish between basis functions for different IRs or between basis functions fordifferent rows of S−1 (or columns of S) of the same IR.

Let Vαj be a basis vector for the j th row of the αth IR and Wβ

k be a basis vectorfor the kth row of the βth IR, then

〈Vαj |H|Wβ

k 〉 =MV W δ jk δαβ,

where MV W is a constant that depends on the two functions.Note that V and W can both be basis functions for the same row of the same IR

even if they are different functions.

For our vibration problem Theorem 1.7 tells us that there are no non-zero matrixelements of F between symmetry functions that belong to different IRs or todifferent rows of S−1 (or columns of S) of the same IR.

A great deal of information can be gleaned from the decomposition of �

(Eq. (1.12)). First, we note that �A1 , �A2 , �B1 , and �B2 are one-dimensionalIRs and � E is a two-dimensional IR. The number of degrees of freedom is(4 × 1 + 2 × 2) = 8. We know that the similarity transformation S which block-diagonalizes � must also block-diagonalize the force-constant matrix, F, of (1.1).Thus S−1F S will have four 1 × 1 blocks. Since, according to Theorem 1.7, therecan be no matrix elements between basis functions for different IRs, the columns ofS corresponding to these one-dimensional blocks must be eigenvectors. For thesefour normal modes the eigenvectors are the same as the symmetry functions andare completely determined, independently of the values of the force constants. Theeigenvalues (mass times frequencies squared), of course, depend on the force con-stants, since they are diagonal matrix elements of S−1F S corresponding to the A1,A2, B1, and B2 vibrations.

We also know that S−1F S has two (2 × 2) E-representations. Since E appearstwice in the decomposition of �, there will be two different symmetry functionsthat are the bases for row 1 and two that are the bases for row 2 of E . There can bematrix elements between two functions that are bases for the same row of the sameIR. Therefore S−1F S could have what appears to be a 4× 4 block.

1.2.6 Step 7: Generate the symmetry functions

In order to determine the S matrix we need to find the symmetry functions. Group-theoretical results can be used to generate the symmetry functions starting from anarbitrary function of the displacement coordinates. The theorem that allows us todo this is given below in Theorem 1.8.

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22 Introductory example: Squarene

Theorem 1.8 (The symmetry-function-generating machine: The projection opera-tor) Let V be an arbitrary column vector and generate the vector Vα

m according tothe rule

Vαm =

h∑k=1

�αmm(Ok) [�(Ok)V]. (GT1.8)

Then Vαm is a basis function for row m of the αth IR or a null vector.

In Eq. (GT1.8), �αmm(Ok) is the mth diagonal matrix element of the αth IR for

the kth operator of the group and �(Ok)V is the function obtained by multiplyingthe function V by the matrix �(Ok). If we choose V to be any linear function of thedisplacement coordinates for squarene, Vα

m (when normalized) will be a symmetryfunction that is a basis for row m of the αth IR. Therefore Theorem 1.8 providesa direct method for obtaining the symmetry functions that are needed in order toblock-diagonalize the eigenvalue matrix equation of (1.1).

To simplify the notation we shall use Vxk (Vyk) to indicate a column vectorhaving a single entry of “1” in the row corresponding to x (y) displacement on thekth corner.

As an example we generate the symmetry coordinate that transforms accordingto the B1 irreducible representation. �B1 is one-dimensional, so the matrix elementsare the same as the characters listed in Table 1.6. We select as our arbitrary columnvector, Vx3, where

Vx3 =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

00001000

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦.

Using the action table (Table 1.3) and character table (Table 1.6), we have

VB1 ∝[(1)�(E)+ (−1)�(C4)+ (1)�(C2)+ (−1)�(C3

4)

+ (−1)�(σd1)+ (−1)�(σd2)+ (1)�(σv1)+ (1)�(σv2)

]Vx3.

(1.13)

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1.2 Reducible and irreducible representations of a group 23

(a) A1 (b) A2

rotation

(c) B1 (d) B2

(e) E row 1

V E1a

translation

(f ) E row 1

V E1b

(g) E row 2

V E2a

(h) E row 2

V E2b

translation

Figure 1.5 The squarene molecule and its symmetry functions.

Expressing VB1 as a normalized column vector, we have

V B1 = 1√8

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

−1−1−1

1111−1

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦. (1.14)

We could have chosen different displacements for V and generated the same result.Since �B1 appears only once in the decomposition, V B1 must be an eigenvectorof the molecular vibration problem of (1.1). Figure 1.5(c) shows schematically therelative motions of the atoms for this normal mode.

Now let us find the symmetry functions that transform according to E . The rep-resentation is two-dimensional and, according to (1.12), it is contained twice inthe decomposition of �. This means that there must be four symmetry functions(2 IRs × 2 rows). To use the symmetry-function-generating machine we need thediagonal matrix elements of the IR for E . According to the character table (the lastcolumn of Table 1.6) X and Y (or the displacements rx and ry) can be used as basisfunctions for a two-dimensional E representation.

We have already done the work for this representation. The first two rows ofTable 1.3 show how rx and ry transform under the group operations. We have, forexample, that

C4 rx →−ry,

C4 ry → rx ,

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24 Introductory example: Squarene

Table 1.7 IR matrices for the E representation

� E (E) =[

1 00 1

]� E (C4) =

[0 1−1 0

]

� E (C2) =[ −1 0

0 −1

]� E (C3

4) =[

0 −11 0

]

� E (σd1) =[ −1 0

0 1

]� E (σd2) =

[1 00 −1

]

� E (σv1) =[

0 11 0

]� E (σv2) =

[0 −1−1 0

]

from which it follows that C4 has the 2× 2 IR,

�E(C4) =[

0 1−1 0

].

All of the 2× 2 matrices for the E representation are shown in Table 1.7.To generate a row-1 symmetry function we proceed as we did in finding V B1 ,

except now we use the diagonal elements, � E(Oi )11, instead of the characters togenerate the function. The matrix elements are given in Table 1.7. Since there areseveral coordinates, we shall label these vectors according to the IR and the rowof the matrix element used to generate the function. Using Vx1 as our arbitraryfunction and the first-row diagonal matrix elements, we obtain

VE1a ∝

[(1)�(E)+ (−1)�(C2)+ (−1)�(σd1)+ (1)�(σd2)

]Vx1

=[Vx1 + Vx2 + Vx3 + Vx4

]and the normalized vector is

VE1a =

1

2

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

10101010

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦.

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1.2 Reducible and irreducible representations of a group 25

The physical motion represented by this vector is translation in the x-direction asshown in Fig. 1.5(e). If we use Vy1 to generate a symmetry function, we find

VE1b =

1

2

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

010−1

010−1

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦.

VE1a and VE

1b are the two row-1 symmetry functions. The vectors are orthogonalto each other. The remaining two E-functions transform according to the row-2symmetry functions of the E irreducible representation. For the row-2 vectors weuse the (2,2) diagonal elements with Vx1 and Vy1 as the generating functions. Weobtain

VE2a =

1

2

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

10−1

010−1

0

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦, VE

2b =1

2

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

01010101

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦. (1.15)

The vector VE2b represents a translation in the y-direction.

Using Vx1 as the arbitrary function, we find the A2 symmetry function to be

VA2 = 1√8

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

1−1

11−1

1−1−1

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦.

The motion associated with VA2 , shown in Fig. 1.5(b), is a rotation about a z-axisthrough the center of the square.7 This result could have been anticipated from the

7 The VA2 mode represents a rotation in the limit of small displacements.

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26 Introductory example: Squarene

Table 1.8 The S matrix. The top row labels the symmetry functions according tothe IR to which they belong. The left-most column lists the displacementcoordinates and the bottom row lists the normalization factors. The center 8× 8is the S matrix.

VA1 VA2 VB1 VB2 VE1a VE

1b VE2a VE

2b

rx1 1 1 −1 1 1 0 1 0ry1 1 −1 −1 −1 0 1 0 1rx2 −1 1 −1 −1 1 0 −1 0ry2 1 −1 1 −1 0 −1 0 1rx3 −1 −1 1 −1 1 0 1 0ry3 −1 1 1 1 0 1 0 1rx4 1 −1 1 1 1 0 −1 0ry4 −1 −1 −1 1 0 −1 0 1Normalization factor 1/

√8 1/

√8 1/

√8 1/

√8 1

212

12

12

character table (Table 1.6), which lists Rz (rotation about the z-axis) as the basisfunction (last column) for the A2. All of the symmetry functions for squarene areshown schematically in Fig. 1.5.

The symmetry functions can be used to construct the S and S−1 matrices neededfor the similarity transformation that block-diagonalizes the eigenvalue matrixequation (1.1). The 8×8 matrix S is given in Table 1.8. The inverse, S−1, is simplythe transpose of S.

1.2.7 Step 8. Block-diagonalize the eigenvalue matrix equation

A vibrational normal mode must leave the center of mass stationary. This can beexpressed by the conditions ∑

i

mi ri = 0, (1.16)

where ri is the vector displacement of the i th atom with mass mi . It can be seenfrom Fig. 1.5 that for squarene all of the symmetry functions except the translationsleave the center of mass unchanged. Often, however (see Exercise 1.8), Eq. (1.16)can be applied as a constraint to assist in the analysis.

Center-of-mass considerations

The similarity transformation carries the F matrix of (1.1) into the block-diagonalform. We shall denote this block-diagonal matrix by the symbol, F, where

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1.3 Eigenvalues and eigenvectors 27

F = S−1 F S. Since the A1, A2, B1, and B2 irreducible representations are one-dimensional, the diagonal matrix elements must be the eigenvalues. Thus, the firstfour diagonal entries in F are m(ωA1)2,m(ωA2)2,m(ωB1)2, and m(ωB2)2. We haveidentified the A2 mode as rotation and therefore ωA2 = 0. For the two 2× 2 blockswe use the fact that the eigenvectors must be linear combinations of the symme-try functions that transform according to the same row of the same IR. However,in this case, one of the row-1 symmetry functions, VE

1a , corresponds to translationof the molecule in the x-direction. The vibrational frequencies are unaffected bytranslation of the molecule as a unit and therefore we may assume that there is nomixing of the two symmetry functions. Similarly, VE

2b corresponds to translation inthe y-direction and hence the mixing of it with VE

2a may be taken to be zero also.From Fig. 1.5 it is clear that the vibrational motions represented by VE

1b and VE2a

are equivalent physically. In fact, application of the operator C34 to VE

1b producesVE

2a and therefore, according to Theorem 1.9, these two normal modes must bedegenerate; that is, ω E(1b) = ω E(2a) ≡ ω E . The last four diagonal elements of F

are 0, m(ω E)2, m(ω E)2, and 0. Therefore, F is completely diagonal.

Theorem 1.9 Let H be a Hermitian matrix that commutes with the elements ofthe group. If Vλ is an eigenvector of H with eigenvalue λ, then Oi Vλ is also aneigenvector with eigenvalue λ, where Oi is any element of the group.

1.3 Eigenvalues and eigenvectors

In the symmetry analysis of an eigenvalue equation of the type Hψ = λψ , it iscommon for the same IR to occur more than once in the decomposition of �. Con-sider the case for which �α is an m×m IR that occurs p times in the decompositionof �. There will be p × m symmetry functions: p symmetry functions for each ofthe m rows. The p (m × m) submatrices on the diagonal of �′ = S−1 � S willbe uncoupled from one another. However, it does not follow that H′ = S−1 H S

will automatically be block-diagonal since there can be non-zero matrix elementsbetween the same rows of the same IR and each row of �α occurs p times. Theproblem is easily corrected by rearranging the rows and columns of H′ so that all pfunctions belonging to a given row are grouped together. The rearrangement doesnot change the eigenvalues, but simply changes the order in which we list the sym-metry functions. After rearrangement H′ will have m (p × p) blocks, even though�′ = S−1 � S has p (m × m) blocks. This type of rearrangement is illustrated inFig. 1.6 for a case where p = 3 and m = 2.

Each of the m submatrices of H′ will produce p eigenvalues when diagonalized.There is no requirement of degeneracy among these p eigenvalues of one of thep × p matrices. However, each of the m submatrices will yield the same set of

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28 Introductory example: Squarene

(a) Three 2 × 2

1a 2a 1b 2b 1c 2c

Γ = S−1 Γ S =

1a × ×2a × ×1b × ×2b × ×1c × ×2c × ×

(b) Matrix elements between same rows

1a 2a 1b 2b 1c 2c

H = S−1

H S =

1a × × ×2a × × ×1b × × ×2b × × ×1c × × ×2c × × ×

(c) Two 3 × 3

1a 1b 1c 2a 2b 2c

HRearranged =

1a × × ×1b × × ×1c × × ×2a × × ×2b × × ×2c × × ×

Figure 1.6 Block-diagonal forms. (a) S−1 � S. The first-row functions are labeled1a, 1b, and 1c. The second-row functions are labeled 2a, 2b, and 2c. (b) H′ =S−1 H S before rearrangement. (c) S−1 HS after rearranging same-row functionsadjacent to one another.

eigenvalues. Consequently each of the p eigenvalues will be m-fold degenerate.For example, for squarene the two first-row eigenfunctions have eigenvalues of0 and m(ω E)2 and the second-row functions yield exactly the same eigenvalues.In principle, the two first-row (or the two second-row) symmetry functions forthe E IR could be mixed to form the eigenvectors. However, no mixing occursfor squarene because V1a is a translation of the molecule in the x-direction andtherefore does not mix with the V1b vibrational mode. Similarly V2b is a translationin the y-direction and therefore does not mix with the V2a vibrational mode.

If an IR occurs more than once in the decomposition �, the eigenvectors andeigenvalues of the final sub-blocks of H′ = S−1 H S usually can not be determined

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1.3 Eigenvalues and eigenvectors 29

by group theory alone (unlike in the case of squarene). However, the eigenvectorsmust be linear combinations of the symmetry functions that belong to the same rowof the same IR. This allows us to form eigenvectors in terms of mixing parameters.For example, if we have an uncoupled 2 × 2 block associated with the symmetryfunctions Vα

1a and Vα1b, the eigenvectors ψα

1r and ψα1s may be written as orthogonal

linear combinations,

ψα1r =

1√1+ |a|2

[Vα

1a + aVα1b

],

ψα1s =

1√1+ |a|2

[−a∗Vα1a + Vα

1b

],

(1.17)

where a is a “mixing” parameter that specifies the amount of mixing of thetwo symmetry functions which is needed in order to form the eigenvectors.Equation (1.17) defines a transformation, t, given by

t = 1√1+ |a|2

[1 a−a∗ 1

](1.18)

that diagonalizes the 2× 2 submatrix of H′. In general, a will not be zero and maybe complex. In the case of molecular vibrations a is sometimes determined by therequirement that the translations have zero frequencies or that the center of mass ofthe molecule be stationary. In other cases the mixing parameter can be determinedfrom experimentally observed frequencies.

Suppose H′ has the 2 × 2 submatrix R, with elements R11, R12, R21, and R22.(Because the submatrix must be Hermitian, R21 must be the complex conjugateof R12.) The similarity transformation t−1Rt leads to a diagonal submatrix for theproper choice of a,

t−1R t =[

λr 00 λs

], (1.19)

where λr and λs are the eigenvalues of ψα1r and ψα

1s , respectively. For molecularvibrational analysis these eigenvalues would be mass-weighted, squared frequen-cies, λr = m(ωr )2 and λs = m(ωs)2. The matrix, R, can then be expressed in termsof the eigenvalues and the mixing parameter,

R = t[

λr 00 λs

]t−1 = 1√

1+ |a|2(

λr + |a|2λs aλs − a∗λr

a∗λs − aλr λs + |a|2λr

)

=[

R11 R12

R∗12 R22

].

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30 Introductory example: Squarene

Table 1.9 The F matrix is diagonal with four distinct eigenvalues: m(ωA1)2,

m(ωB1)2,m(ωB2)2, and m(ω E)2. The zero-eigenvalues F22,F55, and F88

correspond to rotation (A2), translation in the x-direction (V E1a) and translation

in the y-direction (V E2b), respectively.

m(ωA1)2

0

m(ωB1)2

m(ωB2)2

0

m(ω E )2

m(ω E )2

0

The point of this discussion is that it is often useful to express the matrixelements of H in terms of the eigenvalues and symmetry-function mixingparameters.

1.4 Construction of the force-constant matrix from the eigenvalues

Our symmetry analysis of squarene shows that there are at most four distinct vibra-tional frequencies. Group theory can not tell us what these frequencies are, and tocarry the analysis further we have two choices: (1) construct a model for the force-constant matrix F and determine the frequencies in terms of the force constants, or(2) use experimentally measured frequencies to calculate the set of force constantsthat will yield these frequencies. Since the force constants are not observable, weprefer to use the second approach.

The diagonalized F matrix is given in Table 1.9. Now that we know the elementsof F and the S matrix we can construct the F matrix in terms of the vibrationalfrequencies since S F S−1 = F. Recalling that S−1 is simply the transpose of S

and carrying out the matrix multiplication yields the force-constant matrix given inTable 1.10.

There are 64 force-constant matrix elements, but only four independent param-eters, the frequencies, m(ωA1)2,m(ωB1)2,m(ωB2)2, and m(ω E)2. The form of F

given in Table 1.10 is the most general form that (1) is consistent with symmetryrequirements, (2) commutes with all of the operations of the group, and (3) haszero frequencies for the translation and rotation modes.

The alternate approach to analyzing the vibrational problem is to constructan empirical force-constant model and then diagonalize the matrix to find its

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1.5 Optical properties 31

Table 1.10 The force-constant matrix deduced from vibrational frequencies

r + v s −t − v u −r + v −s t − v −u

r + v −u t − v −s −r + v u t − v

r + v −s t − v u −r + v s

r + v −u −t − v s −r + v

r + v s −t − v u

Fji = Fi j r + v −u t − v

r + v −s

r + v

r = 18 m[(ωA1)2 + (ωB1)2 + (ωB2)2], s = 1

8 m[(ωA1)2 − (ωB1)2 + (ωB2)2],t = 1

8 m[(ωA1)2 + (ωB1)2 − (ωB2)2], u = 18 m[(ωA1)2 − (ωB1)2 − (ωB2)2],

v = 12 m(ω E )2.

eigenvalues. The force constants can then be adjusted until the eigenvalues achievea best fit to the observed frequencies. However, special constraints must beemployed in order to ensure that the translation and rotation modes have zerofrequencies. Chemists often use internal coordinates [1.3] rather than Cartesiancoordinates; for example, “bending” (changes in the angles between adjacent sides)and “stretching” (changes in distances between atoms) coordinates. This approachleads to a more complex form of the eigenvalue problem. Use of internal coordi-nates does, however, reduce the size of the eigenvalue equation. Some scientistsfind the description physically appealing because the “stretching” and “bending”force constants for a particular chemical bond are roughly the same for differentmolecules with similar bonds.

1.5 Optical properties

Infrared active normal modes

We can determine which of the vibrational modes are infrared active and/or Ramanactive. To do this we consider Theorem 1.7 and the result

〈V αj |H|W β

k 〉 =MV W δ jk δαβ. (1.20)

If H is a constant, say unity, then (1.20) gives

〈V αj |W β

k 〉 =MV W δ jk δαβ, (1.21)

which means that vectors belonging to different IRs or to different rows of the sameIR are orthogonal.

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32 Introductory example: Squarene

When a molecule in its ground state absorbs optical radiation a vibrational modemay be excited if the photon energy is approximately equal to the vibrationalenergy. Usually, the vibrational frequency is in the infrared spectral range. Whetheror not a given vibrational mode can be excited depends largely upon the symmetryof the eigenvector.

A normal mode of molecular vibration involves the displacements of the nucleifrom their equilibrium positions. When the nuclei move they may create an electricdipole due to the redistribution of electronic charge. The absorption of infraredradiation may occur when the electric dipole moment for the entire molecule due tonuclear displacements is non-zero. The electric dipole moment for the molecule is

μ =N∑i

qi ri ,

where N is the number of atoms of the molecule, qi is the effective charge, and ri

is the displacement of the i th atom. The electric dipole moment μ is a vector andits components μx , μy , and μz will transform under the operations of the groupaccording to the IRs whose basis functions are x , y, and z. It follows that thecomponents of μ belong to the same IRs as the translation modes of the molecule.For squarene (in two dimensions) the character table (Table 1.6) indicates that μx

and μy belong to the E IR of the C4v group.The quantum-mechanical probability of a transition from the molecular ground

state to a state with one quantum of vibrational energy is proportional to [1.3]∑α

∑k

|〈V0|μ|Vαk〉|2, (1.22)

where V0 is the molecular vibrational ground state, μ is the dipole moment oper-ator for the molecule, and Vα

k is the eigenvector for the state with one quantumof energy in the normal mode that belongs to the kth row of the αth IR. Thevibrational ground state, V0, transforms under the operations of the group as aconstant, so for the purpose of determining whether the integral vanishes we needonly know whether 〈μ|Vα

k〉 vanishes due to symmetry. Therefore (1.21) requiresthat Vα

k belongs to a translation IR, since otherwise the matrix element must van-ish. Since μ is antisymmetric with respect to the center of symmetry, we can alsosee the 〈μ|Vα

k〉 must vanish unless Vαk is also antisymmetric. That is, a dipole tran-

sition from the ground state (symmetric) must be to an antisymmetric vibration.In general a dipole transition can occur only if the initial and final vibration statesare opposite in parity. In addition, if the total electric dipole moment is zero thetransition is forbidden.

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1.5 Optical properties 33

Theorem 1.10 (Infrared-active vibration mode) For single quantum excitationsfrom the vibrational ground state to a final vibrational state, a vibrational modecan not be infrared active unless it belongs to the same IR as x or y or z. If thetotal electric dipole moment due to the mode is zero, the transition is forbidden.The final state must be antisymmetric with respect to the center of symmetry if oneexists.

This rule is called the selection rule for infrared absorption. It is almost trivial toapply. In the case of squarene the functions labeled VE

1b and VE2a are the only modes

that belong to the same IR as a translation mode. Therefore VE1b and VE

2a couldpossibly be infrared active. However, they are not because the dipole moment itselfvanishes in this case; that is, μ = ∑

qi ri = q∑

ri = 0 for all of the vibrationalmodes.

Raman-active normal modes

If the optical radiation incident on a molecule has energy greater than the energyof the normal modes of the molecule it is still possible for a vibrational transitionto occur. If the optical radiation has angular frequency ω0 and a normal mode hasangular frequency ω1, in some cases the scattered optical radiation is observed tohave angular frequencies ω0±ω1. In this Raman process the light excites a normalmode (Stokes) or absorbs (anti-Stokes) the energy of a thermally excited normalmode.

For the Raman (Stokes) process the quantum-mechanical probability of thescattered light having frequency ω0 + ω1 is proportional to [1.3]∑

α

∑k

|〈V0|αp|Vαk〉|2. (1.23)

The operator αp is the polarizability tensor. Symmetry-wise its elements behave asthe symmetric (under inversion) bilinear products of the coordinates, such as x2,y2, z2, xy, xz, and yz. In our two-dimensional example there is no z-coordinateso the relevant bilinear products are just x2, y2, and xy. Again, V0, the vibrationalground state, can be treated as a constant and the question is whether 〈αp|V α

k 〉vanishes due to symmetry. According to (1.21), a non-vanishing result requiresthat V α

k belong to the same IR as at least one of the products, x2, y2, and xy, ofαp. This information can be gleaned from the character table. For C4v the far-rightcolumn of the character table (Table 1.6) lists the basis functions for the IRs. We seethat the A1, B1, B2, and E vibrational modes have basis functions that contain suchbilinear products. However, the E modes do not have basis functions that transformas x2, y2, or xy, and therefore can not be Raman active. Another way to see this isthat the E modes are antisymmetric under inversion, while αp is symmetric.

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34 Introductory example: Squarene

Theorem 1.11 (Raman-active normal mode) A normal vibration mode can not beRaman active unless it belongs to an IR that has basis functions that are symmetricunder inversion and transform as bilinear functions of the spatial coordinates.

Selection rules that forbid transitions such as those stated in Theorem 1.10 andTheorem 1.11 are always true. However, the converse statements need not alwaysbe valid. That is because the rules use only the IR label, but the row label of theIR as well as other factors must also be considered in order to say that a matrixelement does not vanish. A more complete discussion of selection rules is given inChapter 3.

Center of symmetry

If a molecule is unchanged by inversion (r→ −r) through a center of symmetry,the normal modes can be classified as odd or even. Even modes are unchanged byinversion and odd modes go into the negative of themselves under inversion. (Theusual notation is g (gerade) for even modes and u (ungerade) for odd modes.)Clearly, the dipole-moment operator is odd under inversion because it is linear inthe displacement coordinates. Therefore any normal mode that is infrared activemust also be odd under inversion.

In the Raman case, the operator, αp, involves binary products of the displace-ments and is even under inversion. Therefore Raman-active normal modes mustalso be even under inversion. We can conclude that, for a molecule having aninversion center, no normal mode can be both infrared active and Raman active.

Our brief discussion of optical transitions included only transitions from theground state to some excited vibrational state with a single quantum of energy.Other sorts of transitions are possible. For example, transitions from the groundstate to more than one excited state, absorption of more than one quantum of vibra-tional energy, or even transitions from one excited state to another. These othertypes of transitions tend to be weaker in intensity, but can yield valuable spectro-scopic information. Whatever the nature of the transition, the selection rule canbe determined group-theoretically if the symmetry of the transition operator, theinitial state, and the final state are specified.

References

[1.1] M. Tinkham, Group Theory and Quantum Mechanics (New York: McGraw-HillBook Company, 1964);D. M. Bishop, Group Theory and Chemistry (New York: Dover Publications, Inc.,1993);E. P. Wigner, Group Theory and Its Application to the Quantum Mechanics of AtomicSpectra (New York: Academic Press, 1959).

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Exercises 35

[1.2] R. E. DeWames and T. Wolfram, “Vibrational analysis of substituted and perturbedmolecules I. The exact isotope rule for molecules”, J. Chem. Phys. 40, 853–860(1964).

[1.3] E. B. Wilson, J. C. Decius, and P. C. Cross, Molecular Vibrations: The Theory ofInfrared and Raman Spectra (New York: Dover Publications, 1980).

Exercises

1.1 Consider a square, planar molecule (e.g., XeF4) with four atoms of mass m1

on the corners of the square and an atom of mass m2 at the center of the squareas shown in Fig. 1.7.

m1 m1

m1m1

m2

X1 X2

X3X4

X5Y1 Y2

Y3Y4

Y5

Figure 1.7

The mass matrix of (1.1) can not be taken as a scalar since there are twodifferent masses. Show that the eigenvalue equation can be rewritten as

F′ η = ω2 η, (E1.1)

where F′ = M−1/2F M−1/2,M1/2 is the square root of the diagonal massmatrix, and η =M1/2 ξ .

1.2 Assuming F is a Hermitian matrix, show that F′ = M−1/2 F M−1/2 is also aHermitian matrix; that is, F′i j = (F′j i )

∗. (Since F′ is Hermitian we can useall of the results of Chapter 1 to solve the new matrix eigenvalue equation,(E1.1).)

1.3 Using the coordinate system shown in Fig. 1.7, construct the 10 × 10 repre-sentation matrices �(Oi ) (i = 1, 2, . . . , 8) for the operations of the group ofthe molecule. Use the fact that the central atom can only be transformed intoitself.

1.4 For the representation matrices in Exercise 1.3, show that the decompositioninto IRs is � = �A1 + �A2 + �B1 + �B2 + 3� E .

1.5 For the molecule of Exercise 1.1, use the symmetry-function-generatingmachine operating on the functions rx(1) + rx(5) and ry(1) + ry(5) toconstruct the A1, A2, B1, and B2 symmetry functions. Why are these

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36 Introductory example: Squarene

symmetry functions the same as those of squarene? Are these symmetryfunctions eigenfunctions?

1.6 For the molecule of Exercise 1.1, use the symmetry-function-generatingmachine operating on the functions rx(1), rx(5), ry(1), and ry(5) to constructthree row-1 and three row-2 E-symmetry functions.

1.7 For the molecule of Exercise 1.1, construct the x-translation and y-translationvectors ξTx and ξTy . Use these vectors to form ηTx and ηTy .

1.8 The decomposition of � (Exercise 1.4) contains the E IR three times.Therefore there are three symmetry functions for each row. One of the row-1 symmetry functions is the translation function ηTx . The two remainingrow-1 symmetry functions correspond to vibrations. (1) The displacements(ξ -functions) of these two row-1 functions must leave the center-of-mass sta-tionary, and (2) the corresponding η-functions must be orthogonal to all ofthe other η symmetry functions.(a) Show that the row-1 and row-2 η-functions for the E vibrations satisfying

conditions (1) and (2) are

row-1 functions row-2 functions

η1α = 1√N1α

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

10101010−μ

0

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦, η2α = 1√

N1α

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

010101010−μ

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

,

η1β = 1√5

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

010−1

010−1

00

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦, η2β = 1√

5

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

10−1

010−1

000

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦,

where N1α =√

4+ μ2 and μ = 4√

m1/m2.

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Exercises 37

(b) For the E IR, make a sketch of the components of each of the sixη-functions and give the relative amplitudes of the displacements.

(c) Make a sketch of the sum of η1α and η1β and the sum of η2α and η2β .1.9 Show that ηTx , ηTy , η1α, η2α, η1β , and η2β found in Exercise 1.8 form a set of

mutually orthogonal vectors. Which of these vectors could be mixed to formeigenvectors? What symmetry functions are already eigenvectors?

1.10 Eigenvectors ψk satisfy the equation

H ψk = λk ψk,

where H is Hermitian and commutes with the operations of the group. Showthat, if ψk is an eigenvector with eigenvalue λk , then φ = Oi ψk is also aneigenvector with eigenvalue λk , where Oi is any operation of the group.

1.11 Make a sketch of the block-diagonalized portion of S−1 F′ S for the E-modeblock if S is based on the symmetry functions of Exercise 1.8. Explain whythe two 2× 2 blocks must have the same eigenvalues.

1.12 For an arbitrary group G, form a set of h matrices by application of the fol-lowing rules. (i) Rearrange the group multiplication table so that E appearsonly on the diagonal. (ii) The representation matrix for �(Ri ) is obtainedby replacing Ri in the rearranged multiplication table by unity and all otherentries by 0.

For example, for the group C4 the rearranged multiplication table is

E C2 C4 C34

E E C2 C4 C34

C2 C2 E C34 C4

C34 C3

4 C4 E C2

C4 C4 C34 C2 E

�(C4) =1

11

1

�(C2) =1

11

1

Prove, for any group, that the matrices constructed by application of rules (i)and (ii) form a representation of G. This representation is called the regular

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38 Introductory example: Squarene

representation. (Hint: The columns are labeled by Ri and the rows are labeledby R−1

j , so �(Ra)i j can be written as �(Ra)R−1i ,R j

.)1.13 For the operator C4v use the regular representation (Exercise 1.12) to show

explicitly that �(C4)−1 = �(C−1

4 ).1.14 Let G be the set of all the operators of the group G (the order of the oper-

ators is irrelevant). Show that RG = G, where R is any operator of thegroup G. This is called the rearrangement theorem because the only effectof multiplying G by a member of G is to rearrange the order of the operatorsin G.

1.15 Prove that a molecule with three atoms has no out-of-plane vibrational modes.(Hint: Consider the translation and rotation modes.)

1.16 For the planar molecule, squarene, it is also true that the out-of-plane motionsdo not mix with the in-plane vibrational modes. The displacements rz(i) canbe used to form a representation of C4v.(a) Find the characters for the operations and decompose the representation.

Identify which IRs corresponds to body modes and which to vibrations.(b) Considering the symmetry of the out-of-plane vibration, argue why there

are no matrix elements between this mode and the in-plane modes.

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2

Molecular vibrations of isotopically substitutedAB2 molecules

In this chapter we illustrate how group theory can be used to find the normal modefrequencies of isotopically substituted molecules. We consider triatomic moleculesof the AB2 type with isotopic substitutions for either the A atom or the B atom,or both. In particular, we analyze and derive numerical results for H2O when deu-terium or tritium is substituted for hydrogen and when 18O is substituted for 16O.We begin by analyzing the vibrational modes of the AB2 molecule, and then go onto discuss the vibrational modes of isotopically substituted AB2 molecules.

2.1 Step 1: Identify the point group and its symmetry operations

A schematic representation of an AB2 molecule is shown in Fig. 2.1(a). In twodimensions the only symmetry operations are the identity, E , and 180-degree rota-tion about an axis through the A site, bisecting the B–B line. Reflection in the lineof the C2 axis gives the same result as C2 in two dimensions. The group is C2. Inthree dimensions, the elements include C2, a reflection in the plane perpendicularto the plane of the paper and containing the C2 axis, σv, and reflection in the planeof the paper, σ ′v. In three dimensions the group is C2v. Since the AB2 moleculehas no out-of-plane vibrational modes (Exercise 1.15), we shall use the smaller C2

group for analysis.

2.2 Step 2: Specify the coordinate system and the basis functions

The coordinate system, shown in Fig. 2.1(b), consists of Y –Z Cartesian coordinateserected at each corner of the triangle. (We use y–z instead of x–y because it iscustomary for the C2 axis to be chosen to lie along the z-axis). The displacementvectors, ri (i = 1, 2, and 3), of the atoms are the basis functions for the molecularvibrational eigenvalue problem,

F ξ =Mω2 ξ, (2.1)

39

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40 Molecular vibrations of isotopically substituted AB2 molecules

(a)A

B B

(b)

Y1

Z1

Y2

Z2

Y3

Z3

3

1 2

(c)

cm

3

1 2

α

zcm

t

d

(d)

cm

3

1 2

(e)

zcm = mA dmA + 2mB

r⊥z(1)= − tzcm

r⊥y(1)

r⊥y(1)

r⊥z(1)r⊥(1)

rcm(1)

cm

1

β

β

zcm

t

Figure 2.1 (a) The AB2 molecule. (b) The coordinate system. (c) The geometryand center of mass. (d) Rotation about the center of mass. (e) Displacements areperpendicular to the lines from the atom to the center of mass.

where F is a (6× 6) force-constant matrix, and ξ is a 6-vector whose componentsare the ryi and rzi displacements of the atoms. The subscripts are as follows: i = 1is ry1, i = 2 is rz1, i = 3 is ry2, i = 4 is rz2, i = 5 is ry3, and i = 6 is rz3. Thematrix M is a 6 × 6 diagonal matrix whose elements are Mii = m B for i = 1, 2,3, 4, and m A for i = 5 or 6, where m A and m B are the masses of atoms A and Brespectively. As discussed in Chapter 1, the eigenvalue problem can be transformedto the standard form

F η = ω2 η, (2.2)

where

F =M−1/2 F M−1/2,

η =M1/2 ξ.

Here, η is a 6-vector whose components are mass-weighted displacements, ηi =(mi )

1/2 ξi , where mi is the mass of the atom whose displacement component is ξi

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2.4 Matrix representations using the basis functions 41

Table 2.1 The action table for the AB2 molecule

E C2

ry ry −ryrz rz rz

1 1 22 2 13 3 3

(m A or m B). The 6-vector η transforms under the symmetry operations in the sameway as ξ .

Since the eigenvalue equation in η-space is in normal form, the eigenvectorsform an orthogonal set of functions, but the ξ -space functions do not, unless M isa scalar.

2.3 Step 3: Determine the effects of the symmetry operationson the basis functions

The action table for the displacements of the AB2 molecule is given in Table 2.1.Using this table, we have

C2

⎡⎢⎢⎢⎢⎢⎢⎢⎣

ry1

rz1

ry2

rz2

ry3

rz3

⎤⎥⎥⎥⎥⎥⎥⎥⎦⇒

⎡⎢⎢⎢⎢⎢⎢⎢⎣

−ry2

rz2

−ry1

rz1

−ry3

rz3

⎤⎥⎥⎥⎥⎥⎥⎥⎦. (2.3)

2.4 Step 4: Construct the matrix representations for each elementof the group using the basis functions

The 6× 6 matrix representation for C2 is

C2 =

−11

−11

−11

.

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42 Molecular vibrations of isotopically substituted AB2 molecules

Table 2.2 The character table for C2

� E C2 Basis functions Basis functions

A 1 1 Rz, z x2, y2, z2, xyB 1 −1 x, y, Rx , Ry xz, yz�(6× 6) 6 0 ryi , rzi , (i = 1, 2, 3)

The identity is the unit matrix,

E =

11

11

11

.

We shall refer to these matrices as the � representation.

2.5 Step 5: Determine the number and types of irreducible representations

The character table for C2 is given in Table 2.2. To find the decomposition of �

into the irreducible representations we employ (1.10) with Nc = 2, Nk = 1, andh = 2,

nβ = 1

2

2∑i=1

(χβ

i )∗ χi = 1

2

2∑i=1

χ∗i χβ

i , (2.4)

where nβ is the number of times the βth IR occurs, χi is the character of the i thelement of �, and χ

β

i is the character of the i th element of the βth IR. The resultsof (2.4) are obvious from the character table, nA = 3, and nB = 3; that is, � =3�A + 3�B.

2.6 Step 6: Analyze the information contained in the decompositions

There are six modes. Three modes belong to the A representation and three to theB representation. Both A and B IRs are one-dimensional. We know that thereare three body modes: Ty , translation in the y-direction; Tz , translation in thez-direction; and Rx , rotation of the molecule about an x-axis through the centerof mass of the molecule. From the list of basis functions in the character table itis clear that Ty and Rz both belong to B, while Tz belongs to A. Therefore, theremust be one B and two A vibrational modes. According to the optical selectionrules discussed in Chapter 1, all three normal modes are infrared-active becauseeach belongs to an IR of a translation mode. All of the vibrational modes are alsoRaman-active since both the A and the B IRs have basis functions that are binaryin the displacements and there is no center of symmetry.

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2.7 Generating the symmetry functions 43

ηAa ηA

b ηB3 ηB

TyηA

TzηB

Rx

ηAa × ×

ηAb × ×

ηB3 ω2

3

ηBTy

0

ηATz

0

ηBRx

0

Figure 2.2 The block-diagonal form of F′ = S−1F S. The “×” signs indicate thepossible non-zero matrix elements. The “0” entries indicate the zero-frequencybody modes. The “η” terms are the symmetry functions. The B vibrationalfrequency is designated as ω3. The two vibrational A symmetry functions aredesignated as ηAa and ηAb .

The decomposition of � determines the block diagonal form F′ = S−1F S, whereS is the matrix of the symmetry functions. The body modes, Ty , Tz , and Rx , willhave zero frequencies and do not mix with any of the vibrational modes. Thereforethe 3 × 3 block corresponding to the B IRs is completely diagonal. Figure 2.2shows the form of the block diagonal F′ matrix. The last three rows and columnshave been rearranged for convenience in the order ηBTy

, ηATz, and ηBRx

.

In Fig. 2.2, the rows and columns are labeled by the symmetry functions. ηAaand ηAb are the symmetry functions for the two vibrational modes belonging toA. These two symmetry functions are mixed to form the eigenvectors of the two Avibrational modes. ηB3 is the symmetry function for the B vibrational mode, and wedesignate its frequency as ω3. ηBRx

is the symmetry function for rotation, while ηBTy

and ηATzare the symmetry functions for translations. Figure 2.2 shows that obtain-

ing the eigenvectors and eigenvalues will require the diagonalization of a 2 × 2matrix.

2.7 Step 7: Generate the symmetry functions

2.7.1 The body modes

The translation symmetry functions are easily obtained. They correspond to unitdisplacements of each of the atoms in the y-direction or in the z-direction,

ξTy =1√3

⎡⎢⎢⎢⎢⎢⎣

101010

⎤⎥⎥⎥⎥⎥⎦ , ξTz =

1√3

⎡⎢⎢⎢⎢⎢⎣

010101

⎤⎥⎥⎥⎥⎥⎦ .

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44 Molecular vibrations of isotopically substituted AB2 molecules

(a) Rotation ηBR

cm

A

B B

3

1 2

(b) Translation ηBT

A

B B

3

1 2

x y

(c) Vibration ηB3

A

BB

3

1 2

(d) Vibration ηAa

A

B B

3

1 2

(e) Vibration ηAbc

A

B B

3

1 2

(f) Translation ηATz

A

B B

3

1 2

Figure 2.3 Symmetry functions and normal modes for AB2 molecules. r(i),i = 1, 2, and 3 are displacement vectors in ξ -space for the atoms at the siteslabeled 1, 2, and 3. ηAa and ηAbc are mixed to form two vibrational eigenfunctions.Other functions are eigenvectors. (a) Rotation, ηBRx

, |r(3)y/r(1)y | = −2m B/m A,

|r(1)z/r(1)y | = tanα/(m A d21 ). (b) Translation in the y-direction, ηBTy

equal dis-

placement in ξ -space. (c) Vibration symmetry function, ηB3 , |r(3)y/r(1)y | =2m B/m A, |r(1)z/r(1)y | = cotα. (d) Vibration symmetry function, ηAa , equal andopposite displacements. (e) Vibration symmetry function, ηAbc, |r(3)y/r(1)y | =2m B/m A. (f) Translation ηATz

, equal displacements in ξ -space.

On transforming these to η-space and normalizing the vectors, we obtain

ηTy =1√

m A + 2m B

⎡⎢⎢⎢⎢⎢⎢⎢⎣

√m B

0√m B

0√m A

0

⎤⎥⎥⎥⎥⎥⎥⎥⎦, (2.5)

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2.7 Generating the symmetry functions 45

ηTz =1√

m A + 2m B

⎡⎢⎢⎢⎢⎢⎢⎢⎣

0√m B

0√m B

0√m A

⎤⎥⎥⎥⎥⎥⎥⎥⎦. (2.6)

Our next task is to find the symmetry function, ηBRx, for rotation. The axis of

rotation for Rx must pass through the center of mass of the molecule. For AB2

molecules the center of mass lies on the C2-axis a distance zcm above the linejoining the B atoms, as shown in Fig. 2.1(c). To determine zcm, we require thatm A(d − zcm) = 2m B zcm, or

zcm =(

m A

m A + 2m B

)d. (2.7)

If we draw a line from the center of mass to each of the corners of the moleculethe displacement vectors for Rx must be perpendicular to these lines, as shownin Fig. 2.1(d). The vector from atom B at corner 1 to the center of mass isrcm(1) = tey1+zcmez1, where ey1 and ez1 are unit vectors along the Y1 and Z1 coor-dinate directions, and t is half the B–B length. A vector perpendicular to rcm(1)is r⊥(1) = zcmey1 − tez1. Similarly, r⊥(2) = zcmey2 + tez2. We can express thedisplacements of r⊥(2) in terms of the y-component of the displacement r⊥(1)y:

r⊥(2)y = r⊥(1)y,

r⊥(1)z = −(

t

zcm

)r⊥(1)y,

r⊥(2)z =(

t

zcm

)r⊥(2)y =

(t

zcm

)r⊥(1)y.

In order for the molecule to rotate about the center of mass∑i

Mi r⊥(i) = 0. (2.8)

This requires that

m A r⊥(3)y = −2m B r⊥(1)y,

r⊥(2)z = −r⊥(1)z = −(

t

zcm

)r⊥(1)y.

(2.9)

From the geometry in Fig. 2.1(c) it follows that the length t = d tanα, where α isone-half of the apex angle. Therefore,

r⊥(2)z =(

tanα

m Ad21

)r⊥(1)y, (2.10)

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46 Molecular vibrations of isotopically substituted AB2 molecules

with

d1 = 1√2m B + m A

. (2.11)

Now we can obtain the rotation function ξRx ,

ξRx ∝

⎡⎢⎢⎢⎢⎢⎢⎢⎣

1−tan α/(m Ad2

1 )

1tan α/(m Ad2

1 )

−2m B/m A

0

⎤⎥⎥⎥⎥⎥⎥⎥⎦. (2.12)

The proportionality sign is used in (2.12) because the vector is not normal-ized to unity. Since η = M1/2ξ , we can find the symmetry function (also theeigenfunction), ηRx ,

ηRx ∝√

m B

⎡⎢⎢⎢⎢⎢⎢⎢⎣

1−tan α/(m Ad2

1 )

1tan α/(m Ad2

1 )

−2√

m B/m A

0

⎤⎥⎥⎥⎥⎥⎥⎥⎦. (2.13)

The vector can be normalized to unity by dividing by the square root of the sum ofthe squares of its components, with the result

ηRx =d1d2√

2

⎡⎢⎢⎢⎢⎢⎢⎢⎣

m A Cα

−Sα/d21

m A Cα

Sα/d21

−2√

m Am B Cα

0

⎤⎥⎥⎥⎥⎥⎥⎥⎦, (2.14)

where d1 = (2m B + m A)−1/2, d2 = (m A + 2m B S2

α)−1/2, Sα ≡ sinα, and Cα ≡

cosα.

2.7.2 Vibration symmetry functions

B symmetry functions

We can make use of the symmetry-function-generating machine described inChapter 1.

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2.7 Generating the symmetry functions 47

Let f be an arbitrary function and generate the vector, f αm , according to the rule

f αm =

h∑k=1

�αmm(Ok) [Ok f ], (GT1.8)

then f αm is an (un-normalized) basis function for the mth row of the αth irreducible

representation. In (GT 1.8), �αmm(Ok) is the mm diagonal matrix element of the αth

IR for the Ok operation.For the B symmetry functions we use ry1, rz1, and ry3, separately as the

“arbitrary” functions to obtain

E(ry1)− C2(ry1) = ry1 + ry2,

E(rz1)− C2(rz1) = rz1 − rz2, (2.15)

E(ry3)− C2(ry3) = ry3 + ry3.

The three (orthogonal and normalized) B symmetry functions are

ηBa =1√

2m B

⎡⎢⎢⎢⎢⎢⎢⎢⎣

√m B

0√m B

000

⎤⎥⎥⎥⎥⎥⎥⎥⎦= 1√

2

⎡⎢⎢⎢⎢⎢⎢⎢⎣

101000

⎤⎥⎥⎥⎥⎥⎥⎥⎦,

ηBb =1√

2m B

⎡⎢⎢⎢⎢⎢⎢⎢⎣

0√m B

0−√m B

00

⎤⎥⎥⎥⎥⎥⎥⎥⎦= 1√

2

⎡⎢⎢⎢⎢⎢⎢⎢⎣

010−100

⎤⎥⎥⎥⎥⎥⎥⎥⎦, (2.16)

ηBc =1√m A

⎡⎢⎢⎢⎢⎢⎢⎢⎣

0000√m A

0

⎤⎥⎥⎥⎥⎥⎥⎥⎦=

⎡⎢⎢⎢⎢⎢⎢⎢⎣

000010

⎤⎥⎥⎥⎥⎥⎥⎥⎦.

We already have two of the three B symmetry functions, namely ηTy (Eq. (2.5))and ηRx (Eq. (2.14)).

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48 Molecular vibrations of isotopically substituted AB2 molecules

The third symmetry function, the vibrational symmetry function, ηB3 , must beorthogonal to ηTy and ηRx , and this requirement is sufficient to determine theeigenvector ηB3 . We can express ηB3 as a linear combination of ηBa , η

Bb , and ηBc ,

ηB3 = μηBa + ν ηBb + ρ ηBc , (2.17)

where μ, ν, and ρ are coefficients to be determined by orthogonality andnormalization conditions,

μ2 + ν2 + ρ2 = 1 (normalization),

μ = −√

m A

2m Bρ (orthogonal to ηTy ), (2.18)

μ =(

tanα

m A d21

)ν +

√2m B

m Aρ (orthogonal to ηRx ).

After a bit of algebra we find that

μ = √m A d2 Sα,

ν = √m A d2 Cα, (2.19)

ρ = −√2m B d2 Sα,

and

ηB3 =√

m B d2 Sα ηBa +√

m A d2 Cα ηBb −

√2m B d2 Sα η

Bc , (2.20)

or, as a column vector,

ηB3 =√

m A

2d2

⎡⎢⎢⎢⎢⎢⎢⎢⎣

−Cα

−2√

m B/m A Sα

0

⎤⎥⎥⎥⎥⎥⎥⎥⎦. (2.21)

A sketch of this vibration mode is shown in Fig. 2.3(c). Note that r(1) and r(2) liealong the lines joining the A and B atoms. It is easily verified that this mode leavesthe center of mass stationary.

Since ηBRx, ηBTy

, and ηB3 are eigenfunctions of F, we have succeeded in completelydiagonalizing the 3 × 3 B-block eigenvalue problem. Next we look at the 3 × 3,A-block problem.

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2.7 Generating the symmetry functions 49

A symmetry functions

We start by finding a set of symmetry functions using each of ry1, rz1, and rz3 as“arbitrary” functions. Applying the symmetry-function-generating machine gives

E(ry1)+ C2(ry1) = ry1 − ry2,

E(rz1)+ C2(rz1) = rz1 + rz2,

E(rz3)+ C2(rz3) = rz3 + rz3.

The corresponding symmetry functions are

ηAa =1√

2m B

⎡⎢⎢⎢⎢⎢⎢⎢⎣

√m B

0−√m B

000

⎤⎥⎥⎥⎥⎥⎥⎥⎦= 1√

2

⎡⎢⎢⎢⎢⎢⎢⎢⎣

10−1

000

⎤⎥⎥⎥⎥⎥⎥⎥⎦,

ηAb =1√

2m B

⎡⎢⎢⎢⎢⎢⎢⎢⎣

0√m B

0√m B

00

⎤⎥⎥⎥⎥⎥⎥⎥⎦= 1√

2

⎡⎢⎢⎢⎢⎢⎢⎢⎣

010100

⎤⎥⎥⎥⎥⎥⎥⎥⎦, (2.22)

ηAc =1√m A

⎡⎢⎢⎢⎢⎢⎢⎢⎣

00000√m A

⎤⎥⎥⎥⎥⎥⎥⎥⎦=

⎡⎢⎢⎢⎢⎢⎢⎢⎣

000001

⎤⎥⎥⎥⎥⎥⎥⎥⎦.

The translation symmetry function can be expressed in terms of these threesymmetry functions,

ηATz= d1

[√2m B ηAb +

√m A ηAc

]. (2.23)

The remaining two A modes are vibration modes, so they must be formed fromsymmetry functions that are orthogonal to the translation function. The functionηAa is already orthogonal to ηATz

, and we can form the last function, ηAbc, as thelinear combination

ηAbc = d1

[√m A ηAb −

√2m B ηAc

]. (2.24)

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50 Molecular vibrations of isotopically substituted AB2 molecules

Table 2.3 The S matrix: a is the mixing parameter,C = (1+ |a|2)1/2, Cα = cosα, Sα = sinα, d1 = (m A + 2m B)

−1/2,d2 = (m A + 2m B S2

α)−1/2, r = 1/(

√2C), t = (m A/2)1/2d2,

u = m1/2A d1, s = m1/2

B d1, and p = (2m B)1/2d2

ηA1 ηA2 ηB3 ηBTyηATz

ηBRx

ηA1 r −ar −t Sα s 0 tuCα

ηA2 aur ur −tCα 0 s −tu−1Sα

ηB3 −r ar −t Sα s 0 tuCα

ηBTyaur ur tCα 0 s tu−1Sα

ηATz0 0 pSα u 0 −puCα

ηBRx−2asr −2sr 0 0 u 0

These two symmetry functions (ηAa and ηAbc) are mixed to form the two vibrationnormal modes (i.e., the eigenfunctions). According to (1.17) of Chapter 1, we canwrite the two vibrational eigenfunctions as

ηAvib1 = ηA1 =1

C

(ηAa + a ηAbc

),

ηAvib2 = ηA2 =1

C

(−a ηAa + ηAbc

),

(2.25)

where

C =√

1+ |a|2.

2.8 Step 8: Diagonalize the matrix eigenvalue equation

The similarity transformation S−1 F S produces a block-diagonal matrix, Fd, whereS is the matrix with the symmetry functions as columns. The S matrix is shownin Table 2.3. Since all of our symmetry functions are also eigenvectors of F, thetransformed matrix Fd is completely diagonal as shown in Table 2.4. The diagonalelements of Fd are the eigenvalues of F, that is, the squares of the normal-modeangular frequencies.

2.9 Constructing the force-constant matrix

The force-constant matrix can be obtained by reversing the similarity transfor-mation: S Fd S−1 = F, where Fd is the diagonal matrix shown in Table 2.4.

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2.9 Constructing the force-constant matrix 51

Table 2.4 The diagonal matrix Fd = S−1 F S: ω1 = ωA1 , ω2 = ωA

2 , andω3 = ωB

3 . The column labels are the same as in Table 2.3.

ω21

ω22

ω23

0

0

0

Table 2.5 The force-constant matrix, F: C = (1+ |a|2)1/2, Cα = cosα,Sα = sinα, d1 = (m A + 2m B)

−1/2, d2 = (m A + 2m B S2α)−1/2, r = 1/(

√2C),

t = (m A/2)1/2d2, u = m1/2A d1, s = m1/2

B d1, p = (2m B)1/2d2,

v = (m Am B)1/2d2

2 , K ≡ a2ω21 + ω2

2, L ≡ ω21 + a2ω2

2, and M ≡ ω21 − ω2

2

r2L aur2 M −r2L aur2 M −vS2αω

23 −2asr2 M

+ t2S2αω

23 + t2SαCαω

23 + t2S2

αω23 − t2SαCαω

23

u2r2 K −aur2 M u2r2 K −vSαCαω23 −2usr2 K

+ t2C2αω

23 + t2SαCαω

23 − t2C2

αω23

r2L −aur2 M −vS2αω

23 2asr2 M

+ t2S2αω

23 − t2SαCαω

23

u2r2 K vSαCαω23 −2usr2 K

+ t2C2αω

23

Fi j = F j i p2S2αω

23 0

4s2r2 K

The S matrix is given in Table 2.3 and S−1 is the transpose of S. The resultingforce-constant matrix is shown in Table 2.5.

It can be concluded that the most general form of the force-constant matrix foran AB2 molecule is a 6×6 matrix with 34 non-zero force constants. However, therecan be at most four independent parameters: ω1, ω2, ω3, and the mixing parameter,a. Since F = M1/2 F M1/2, the force-constant matrix in ξ -space is easily obtained.The matrix elements of F are

Frs = Frs(mr ms)1/2. (2.26)

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52 Molecular vibrations of isotopically substituted AB2 molecules

We have now achieved a complete analysis of the normal modes of vibration ofAB2 molecules. In the next section we discuss how the frequencies of isotopicallysubstituted forms of AB2 molecules can be obtained.

2.10 Green’s function theory of isotopic molecular vibrations

Consider an arbitrary, isotopically substituted, molecule. The matrix N × Neigenvalue equation that determines the vibrational frequencies is

F ξ = ω2 M(i) ξ,

where M(i) is the diagonal matrix of the masses of the molecule with isotopicallysubstituted masses. We can write

M(i) =M+�M(i), (2.27)

with M the matrix of the normal masses and �M(i) = M(i) −M. The eigenvalueequation may be rewritten in the form[

F− ω2 I− ω2 �M(i)]η = 0, (2.28)

with

F =M−1/2 F M−1/2, (2.29)

�M(i) =M−1/2 �M(i) M−1/2, (2.30)

η =M1/2 ξ. (2.31)

Since both M−1/2 and �M(i) are diagonal, it follows that �M(i) is also diagonal.Its matrix elements are given by

[�M(i)]rs = δrs (�mr/mr ), (2.32)

where �mr is the difference between the isotopic and normal mass of the atomwhose coordinate corresponds to ξr . Equation (2.28) can be rewritten in the form(

F− ω2 I) [

I− ω2 G(ω2)�M(i)]η = 0, (2.33)

where

G(ω2) = (F− ω2 I

)−1, (2.34)

with I an N × N unit matrix and G(ω2) the molecular Green’s function. Theeigenvalues are determined by the vanishing of the determinant,

det{(F− ω2 I) [

I− ω2 G(ω2)�M(i)]} = 0, (2.35)

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2.10 Green’s function theory of isotopic molecular vibrations 53

where the notation det A indicates the determinant of the matrix A. Since det AB =det A det B, (2.35) yields two conditions,

det{(

F− ω2 I)} = 0, (2.36)

det{[

I− ω2 G(ω2)�M(i)]} = 0. (2.37)

Equation (2.36) is satisfied for ω2 = (ωk)2 (k = 1, 2, . . . , N ), where ωk is the kth

frequency of the normal molecule. The roots of (2.37) are the isotopically shiftedfrequencies, ω2 = (ω

(i)k )2. To make use of (2.37), we need the Green’s function

matrix elements.

2.10.1 Green’s function for molecular vibrations

Let U be the N × N matrix constructed with the eigenvectors of F as the columnsof the matrix. Then U−1 F U is completely diagonal, with matrix elements

{U−1 F U}rs = δrs ω2r . (2.38)

The molecular Green’s function is also diagonal when transformed by U,

G = {U−1 G(ω2)U}rs = − δrs

ω2r − ω2

. (2.39)

It follows that

G(ω2) = UG U−1, (2.40)

{G(ω2)}rs =∑

k

∑p

Urk [ω(AB2)2k − ω2)]−1 δkp U−1

ps

=∑

k

Urk Uks

ω2k − ω2

. (2.41)

In the first part of this chapter we obtained the eigenvectors for the molecular vibra-tions of an AB2 molecule in terms of the normal vibrational frequencies, ω2

1, ω22,

and ω23, and the mixing parameter, a. Equation (2.40) or Eq. (2.41) can be used

to construct the Green’s function for the AB2 molecule. In this case, the matrixof eigenvectors, U, is the same as the S matrix given in Table 2.3. The diagonalGreen’s function, G, is shown in Table 2.6.

The Green’s function in η-space is G(ω2) = UG U−1 = SG S−1, where theS matrix is given in Table 2.3. Carrying out the matrix multiplication gives theGreen’s function shown in Table 2.7.

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54 Molecular vibrations of isotopically substituted AB2 molecules

Table 2.6 The diagonal form of the molecular Green’s function, G(ω2). Emptysquares indicate matrix elements whose values are zero.

(ω21 − ω2)−1

(ω22 − ω2)−1

(ω23 − ω2)−1

−(ω2)−1

−(ω2)−1

−(ω2)−1

2.10.2 Isotopic frequencies of A(i)B2, B(i)AB(i), and B(i)A(i)B(i) molecules

There are three types of symmetrically substituted molecules: (1) A(i)B2,(2) B(i) AB(i), and (3) B(i) A(i)B(i), where A(i) or B(i) indicates the isotopic massesm(i)

A and m(i)B , respectively. For example, the normal (most abundant) form of the

water molecule has mH = 1 and mO = 16 in atomic mass units. Isotopicallysubstituted forms include molecules with mH = 2 (deuterium) or 3 (tritium) andmO = 18. We shall indicate the normal water molecule by H2O. For the hydrogenisotopes we will use the symbols D (for deuterium) and T (for tritium). Employingthe observed frequencies of H2O and one of the frequencies of an isotopically sub-stituted molecule allows us to determine all of the parameters in the theory. Oncethat has been accomplished, the force-constant matrix (Table 2.5) and the Green’sfunction (Table 2.7) are determined. Using the Green’s function, the vibrationalfrequencies of all types of isotopic substitutions can be calculated from (2.37).

A(i)B2 molecules

For the isotopically substituted species A(i)B2 the matrix �M(i) has only two non-zero elements:

(�M(i))55 = (�M(i))66 = m(i)A − m A

m A= m(i)

A

m A− 1 ≡ ε. (2.42)

Equation (2.37) gives

det{I− ω2 G(ω2)�M(i)} = [1− εω2G(5, 5)][1− εω2G(6, 6)] − ε2ω2G(5, 6)2

= 0. (2.43)

Since G(5, 6) = 0 (Table 2.7), the equations that determine the isotopic frequen-cies are

1− εω2G(5, 5) = 0 (B vibrational modes) (2.44)

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2.10 Green’s function theory of isotopic molecular vibrations 55

Table 2.7 Green’s function, G(ω2), for AB2 molecules. The parameters are asfollows: λ1 = 1/(ω2

1 − ω2), λ2 = 1/(ω22 − ω2), λ3 = 1/(ω2

3 − ω2),�1 = (λ1 + a2λ2), �2 = (a2λ1 + λ2), X1 = (λ1 − λ2), X2 = (λ1 + λ2),X3 = (λ3 + 1/ω2), Cα = cosα, Sα = sinα, C = 1/(1+ |a|2)1/2, r = 1/(

√2C),

d21 = 1/(m A + 2m B), d2

2 = 1/(m A + 2m B S2α), t2 = (m A/2)d2

2 , u2 = m Ad21 ,

s2 = m Bd21 , p2 = (2m B)d2

2 , and v2 = (m Am B)d42 .

r2�1 aur2 X1 −r2�1 aur2 X1 −vS2αλ3 −2asr2 X1

+ t2S2αλ3 + t2SαCα X3 + t2S2

αλ3 − t2SαCα X3 − u(s−uvC2α)

ω2

− s2+t2u2C2α

ω2

r2u2�2 −aur2 X1 r2u2�2 −vSαCα X3 −2usr2�2

+ t2C2αλ3 + t2SαCα X3 − t2C2

αλ3 − su/ω2

− s2+(t/u)2 S2α

ω2 − s2−(t/u)2 S2α

ω2

r2�1 aur2 X1 −vS2αλ3 2asr2 X1

+ t2S2αλ3 − t2SαCα X3 − u(s−uvC2

α)

ω2

− (s2+t2u2C2α)

ω2

u2r2�2 vSαCα X3 −2asr2�2

+ t2C2αλ3 − su/ω2

− s2+(t/u)2 S2α

ω2

p2S2αλ3 0

G(i, j) = G( j, i) − u2(1+p2)C2α

ω2

4s2r2�2

− u2/ω2

and

1− εω2G(6, 6) = 0 (A vibrational modes). (2.45)

Equation (2.45) takes the form

2m B + m A − 2m Bεω2

C2

(a2

ω21 − ω2

+ 1

ω22 − ω2

)+ εm A = 0. (2.46)

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56 Molecular vibrations of isotopically substituted AB2 molecules

With a bit of algebra (2.46) can be rewritten as a quadratic equation in thevariable ω2:

(ω2)2 − A ω2 + B = 0. (2.47)

It follows that the sum of the roots of (2.47) is equal to A and the product of theroots is B,

(ω(i)1 )2 + (ω(i))2 = m A

m(i)A

{ω2

1 + ω22

+ ε

C2

2m B (ω21 + a2 ω2

2)+ m A C2(ω21 + ω2

2)

2m B + m A

},

(2.48)

(ω(i)1 )2 (ω(i))2 = m A

m(i)A

(ω21ω

22)

2m B + m(i)A

2m B + m A, (2.49)

ε = m(i)A

m A− 1, (2.50)

where ω(i)1 and ω

(i)2 are the A vibrational mode frequencies of the A(i)B2 molecule,

and ω1 and ω2 are those of the normal AB2 molecule. Equation (2.49) is the well-known Teller–Redlich product rule [2.1] for isotopic substitution. The remainingisotopic frequency (B vibrational mode) is obtained from (2.44). That equationgives

[1−εω2 G(5, 5)] = 1−ε

[2ω2m Bd2

2 S2α

ω23 − ω2

−m Ad21−2m Bd2

1 m Ad22 C2

α

]= 0. (2.51)

The solution of (2.51) gives (ω(i)3 )2,

(ω(i)3 )2 = ω2

3

εm Ad22 + 1

1+ ε= ω2

3

m A

m(i)A

m(i)A + 2m B S2

α

m A + 2m B S2α

, (2.52)

where ε = m(i)A /m A − 1.

B(i)AB(i) molecules

The isotopic frequencies of B(i) AB(i) molecules are easily obtained from those ofB A(i)B molecule discussed in the previous section. Consider a molecule B(i)X B(i)

with m X = (m(i)B /m B)m A. Such a molecule is the same as the normal AB2

molecule except that the masses are all scaled by the factor m(i)B /m B . Therefore

the frequencies for this molecule are related to those of the normal molecule by

(ω2k)B(i)X B(i) = m B

m(i)B

(ω2k)B AB, k = 1, 2, and 3.

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2.10 Green’s function theory of isotopic molecular vibrations 57

If we now substitute a normal A atom for the X atom, we arrive at the B(i) AB(i)

molecule and we may apply the results of the previous section with the followingrules:

(a) m A is replaced by (m(i)B /m B)m A

(b) m(i)A is replaced by m A

(c) (ω2k) is replaced (m B/m(i)

B )(ω2k)

(d) m B is replaced by m(i)B .

Making these replacements gives us the following results:

(ω(i)1 )2 + (ω

(i)2 )2 = ω2

1 + ω22 +

ε

C2

2m B(ω21 + a2ω2

2)+ m AC2(ω21 + ω2

2)

2m B + m A,

(2.53)

(ω(i)1 )2(ω

(i)2 )2 = m B

m(i)B

(ω21ω

22)

m A + 2m(i)B

m A + 2m B, (2.54)

(ω(i)3 )2 = ω2

3

m B

m(i)B

m A + 2m(i)B S2

α

m A + 2m B S2α

, (2.55)

ε = m B

m(i)B

− 1. (2.56)

B(i)A(i)B(i) molecules

The frequencies of the final symmetrically substituted form B(i) A(i)B(i) can beobtained by the same method as was applied to B(i) AB(i). For B(i) A(i)B(i) wemake the following replacements in (2.48)–(2.50):

(a) m A is replaced by (m(i)B /m B)m A

(b) (ω2k) is replaced (m B/m(i)

B )(ω2k)

(c) m B is replaced by m(i)B .

This gives

(ω(i)1 )2 + (ω

(i)2 )2 = m A

m(i)A

{ω2

1 + ω22

+ ε

C2

2m B(ω21 + a2ω2

2)+ m AC2(ω21 + ω2

2)

2m B + m A

},

(2.57)

(ω(i)1 )2(ω

(i)2 )2 = (ω2

1ω22)

2m(i)B + m(i)

A

2m B + m A, (2.58)

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58 Molecular vibrations of isotopically substituted AB2 molecules

(ω(i)3 )2 = ω2

3

m Am B

m(i)A m(i)

B

m(i)A + 2m(i)

B S2α

m A + 2m B S2α

, (2.59)

ε = m Bm(i)A

m(i)B m A

− 1. (2.60)

2.10.3 Asymmetric isotopic substitution for B(i)AB molecules

For the isotopic molecule B(i) AB we have two non-zero matrix elements for�M (i),

(�M (i))11 = (�M (i))22 = m(i)B − m B

m B= m(i)

B

m B− 1 = ε. (2.61)

Equation (2.36) yields

[1− εω2 G(1, 1)][1− εω2G(2, 2)] − [εω2G(1, 2)]2 = 0. (2.62)

The Green’s function matrix elements and parameters are given in Table 2.7:

G(1, 1) = r2�1 + t2S2αλ3 − s2 + t2u2C2

α

ω2,

G(2, 2) = r2u2�2 + t2C2αλ3 − s2 + (t/u)2S2

α

ω2,

G(1, 2) = r2au X1 + t2SαCα X3.

Equation (2.62) leads to a cubic equation in the variable ω2 from which the isotopicfrequencies can be calculated. The details are not presented here, but may be foundin reference [2.2].

The mixing parameter

The square of the mixing parameter, a2, can easily be obtained from the resultsfor the B(i) AB(i) molecule. Use is made of (2.53) to solve for (ω

(i)2 )2 in terms

of (ω(i)1 )2. (ω(i)

2 )2 can then be eliminated from (2.54) and the equation solved forC2 (= 1+ |a|2). The result is

C2 = 2εm B(ω21 − ω2

2)

(ω(i)1 )2 + ω2

1ω22/(ω

(i)1 )2 − (ω2

1 + ω22)(2m B + m A + εm A)− 2m Bεω

22

,

(2.63)where

ε =(

m B

m(i)B

)− 1.

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2.10 Green’s function theory of isotopic molecular vibrations 59

Knowledge of the normal molecular frequencies, ω1 and ω2, of H2O and one ofthe isotopic frequencies, ω(i)

1 or ω(i)2 , of D2O is sufficient to determine the square

of the mixing parameter via (2.63). The sign of the parameter a is not determinedfor symmetrical substitution. However, it is determined in the case of asymmetricalisotopic substitution since a appears linearly in G(1, 2)2. Therefore knowledge ofω1, ω2, ω3, and one of the frequencies, ω(i)

1 or ω(i)2 , of the B(i) AB molecule is

sufficient to determine all of the frequencies of all of the isotopically substitutedforms of H2O. There is no need to construct any type of force-constant matrix. Onthe other hand, if one wishes to construct the force-constant matrix, it is completelydetermined (Table 2.5) in terms of the frequencies of the normal molecule and themixing parameter a.

2.10.4 The sum and product rules for AB2 and its isotopicallysubstituted species

Equation (2.62) can be expressed as a cubic equation in x = ω2 of the form x3 +A2 x2+ A3 = 0, where A2 and A3 are constants. In this form the constant A3 givesthe product of the roots. That is, A3 = [ω(i)

1 ω(i)2 ω

(i)3 ]2. This result is equivalent

to the Teller–Redlich product rule for B(i) AB isotopic frequencies. A sum rulerelating the sums of the frequencies of B AB, B(i) AB, and B(i) AB(i) can also beobtained [2.3]. This rule states that∑

ω2k(B AB)+

∑ω2

k(B(i) AB(i)) = 2∑

ω2k(B(i) AB). (2.64)

In (2.64) the symbol ω2k(B AB) indicates the frequencies of the normal molecule.

The symbols ω2k(B(i) AB(i)) and ω2

k(B(i) AB) indicate frequencies of the isotopicmolecules B(i) AB(i) and B(i) AB, respectively. This sum rule may be obtainedusing the Green’s function method; however, the calculation is rather messy. Thereis another way to obtain the sum rule of (2.64), which we will now discuss. Theforce-constant matrix in diagonal form is given by (2.37),

{U−1 F U}rs = δrs(ωr )2.

Therefore, Tr{U−1 F U} = ∑k ω2

k , but, since the trace of a matrix is not changedby a similarity transformation or a unitary transformation, it follows that TrF =∑

(F)kk = ∑ω2

k . The matrix F is related to F by the relation F = M1/2 F M1/2.The matrix F is the force-constant matrix in coordinate space and its elementsdo not change with isotopic substitutions (this is not true for F). The diagonalelements of F are

(F)kk = (M1/2 F M1/2)kk = mk Fkk .

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60 Molecular vibrations of isotopically substituted AB2 molecules

Therefore we can write ∑mkFkk =

∑ω2

k . (2.65)

To make this clearer we rewrite (2.65) in the form∑mk(XY Z)Fkk =

∑ω2

k(XY Z), (2.66)

where, for our AB2 molecule and its isotopically substituted species, XY Z can beB AB, B A(i)B, B(i) AB, or B(i) AB(i). Note that the matrix elements Fkk are thesame for all of the molecules being considered. On applying (2.66) we obtain∑

mk(B AB)Fkk =∑

ω2k(B AB)

= m B(F11 + F22 + F33 + F44)+ m A(F55 + F66),∑mk(B(i) AB(i))Fkk =

∑ω2

k(B(i) AB(i))

= m(i)B (F11 + F22 + F33 + F44)+ m A(F55 + F66),∑

mk(B(i) AB)Fkk =∑

ω2k(B(i) AB)

= m(i)B (F11 + F22)+ m B(F33 + F44)

+ m A(F55 + F66).

Because of the symmetry of the molecules we are considering, F11 = F33 andF22 = F44 (see Table 2.5). Therefore,∑

ω2k(B AB) = 2m B(F11 + F22)+ m A(F55 + F66),∑

ω2k(B(i) AB(i)) = 2m(i)

B (F11 + F22)+ m A(F55 + F66),∑ω2

k(B(i) AB) = m(i)B (F11 + F22)+ m B(F11 + F22)+ m A(F55 + F66),

from which it follows that∑ω2

k(B AB)+∑

ω2k(B(i) AB(i)) = 2

∑ω2

k(B(i) AB).

2.11 Results for isotopically substituted forms of H2O

In this section we apply the theory developed above to obtain results for isotopi-cally substituted variants of H2O. The physical and spectroscopic data needed toapply the theory are

ν1(H2O) = 3825.32 cm−1, m A(H2O) = 16, m A(H (18)2 O) = 18,

ν2(H2O) = 1653.91 cm−1, m B(H2O) = 1, α(H2O) = 105◦,ν3(H2O) = 3935.59 cm−1, m B(D2O) = 2, a = −0.741 41,ν1(D2O) = 2758.06 cm−1, m B(T2O) = 3, ω = 2πν.

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2.11 Results for isotopically substituted forms of H2O 61

Table 2.8 Experimentally observed and calculated frequencies for isotopicallysubstituted variants of H2O. All theoretical values are derived using the observedH2O frequencies and ω1 from D2O (frequencies with asterisks are the input data).The calculated value for a(H2O) is − 0.741 41. The angle α(H2O) = 105◦, andthe atomic masses are m A = 16, m B(H) = 1, m B(D) = 2, and m B(T) = 3.Frequencies are in units of cm−1. To convert to frequency in Hz, multiply by thevelocity of light, 2.9978× 1010 cm/s.

Frequencies in cm−1

Molecule[Reference]

ν1Calculated

experimental

ν2Calculated

experimental

ν3Calculated

experimental

H2O 3825∗ 1654∗ 3936∗[2.6] 3825 1654 3936D2O 2758∗ 1209 2925[2.6] 2758 1210 2884[2.7] 2669 1178 2788HDO 2819 1451 3883[2.8] 2824 1440 3890[2.9] 2724 1403 3707T2O 2299 1017 2436[2.10] 1017 2438[2.7] 2234 995 2367HTO 2363 1373 3882[2.10] 1359 3895D2O(18) 2746 1202 2862[2.8] 2784 1206 2889

Using the above data, we find from (2.67) that a2 = 0.5497. Calculations of asym-metric substitution show that a is negative. The calculated frequencies for H2Owith various substitutions of deuterium (D) and tritium (T) are shown in Table 2.8and compared with experimental data. The mixing parameter, a2, and the fre-quencies of various isotopically substituted forms of H2O, D2O, H2S, and D2Sare available [2.2]. Isotopically substituted XY3 molecules [2.4] and ethylene [2.5]have also been analyzed using the Green’s function method.

2.11.1 The force-constant matrix for H2O

Using the value of C2 and the experimental frequencies listed in Table 2.8 forH2O, we can calculate the numerical values of the force constants. Evaluation ofthe elements in Table 2.5 gives the force-constant matrix in Table 2.9.

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62 Molecular vibrations of isotopically substituted AB2 molecules

Table 2.9 The force-constant matrix (F in η-space) for H2O. The force constantsare in units of 4π2×106cm−2.

11.678 −4.417 1.247 −0.948 −3.236 1.341

3.556 0.948 2.626 0.867 −1.546

11.678 4.417 −3.236 −1.341

3.556 −0.867 −1.546

Fi j = Fji 1.618 0.000

0.773

References

[2.1] O. Redlich, “Eine allgemeine Beziehung zwischen den Schwingungsfrequen-zen isotoper Moleküln (nebst Bemerkungen über die Berechnung harmonischerKraftkonstanten)”, Z. Phys. Chem. B 28, 371–382 (1935).

[2.2] R. E. DeWames and T. Wolfram, “Vibrational analysis of substituted and perturbedmolecules I. The exact isotope rule for molecules”, J. Chem. Phys. 40, 853–860(1964).

[2.3] J. C. Decius and E. B. Wilson Jr., “Sum rules for the vibration frequencies ofisotopic molecules”, J. Chem. Phys. 19, 1409–1412 (1951).

[2.4] C. D. Bass, L. Lynds, T. Wolfram, and R. E. DeWames, “Vibrational analysis ofsubstituted and perturbed molecules. II. Planar XY3 molecules; application toBCl3–HBCl2–DBCl2”, J. Chem. Phys. 40, 3611–3619 (1964).

[2.5] T. Wolfram and A. Asgharian, “Generalized isotope rules for molecular vibrationswith application to A2 B4 planar molecules”, J. Chem. Phys. 74, 1661–1676 (1981).

[2.6] G. Hertzberg, Infrared and Raman Spectra of Polyatomic Molecules (New York:D. Van Nostrand, Inc., 1945).

[2.7] R. Lemus, “Vibrational excitations in H2O in the framework of a local model”,J. Mol. Spectrosc. 225, 73–92 (2004). P. F. Bernath, “The spectroscopy of watervapour: Experiment, theory and applications”, Phys. Chem. Chem. Phys. 4, 1501–1509 (2002).

[2.8] W. S. Benedict, N. Gailar, and E. K. Plyler, “Rotation–vibration spectra of deuter-ated water vapor”, J. Chem. Phys. 24, 1139–1165 (1956).

[2.9] A. Janca, K. Tereszchuk, P. F. Bernath, N. F. Zobov, S. V. Shirin, O. L. Polyansky,and J. Tennyson, “Emission spectrum of hot HDO below 4000 cm−1”, J. Mol.Spectrosc. 210, 132–135 (2003).

[2.10] P. A. Staats, H. W. Morgan, and J. H. Goldstein, “Infrared spectra of T2O, THO,and TDO”, J. Chem. Phys. 24, 916–917 (1956).

Exercises

Consider the AB2 molecule in three dimensions (Fig. 2.4). For simplicity useC2v as the covering group. The operations and character table for the groupare shown in Fig. 2.4.

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Exercises 63

σv plane

σ plane

C2

A

BB A

BB

Coordinate system

z1

x1

y1z2

x2

y2

z3

x3

y3

Γ E C2 σv σ′

A1 1 1 1 1

A2 1 1 −1 −1

B1 1 −1 1 −1

B2 1 −1 −1 1

Character table for C2v

Figure 2.4

2.1 Construct the action table for the atomic displacements rx , ry , and rz , and theatomic positions 1, 2, and 3.

2.2 Use the action table to find the characters of the operations of C2v for arepresentation(a) based on rxi (i = 1, 2, and 3),(b) based on ryi and rzi (i = 1, 2, and 3), and(c) based on rxi , ryi , and rzi (i = 1, 2, and 3).

2.3 Decompose each of the representations (a), (b), and (c) in Exercise 2.2 intothe IRs of C2v.

2.4 Using the results of Exercise 2.3, explain why the in-plane symmetry func-tions do not mix with the out-of-plane (i.e., perpendicular to the plane of themolecule) symmetry functions.

2.5 Find the out-of-plane symmetry functions using the symmetry-function-generating machine.(a) Use rx1 and rx3 as the generating functions.(b) Identify the symmetry function for the rotation Rz (about the C2-axis).(c) Construct the translation Tx for translation in the x-direction.(d) Transform the three symmetry functions to η-space and normalize them.(e) Express Tx in η-space.(f) Find the second B1, out-of-plane, normal mode by requiring it to be

orthogonal to Rz and Tx .

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64 Molecular vibrations of isotopically substituted AB2 molecules

(g) Show that the mode in part (f) corresponds to a rotation about a y-axispassing through the center of mass.

(h) Sketch the three out-of-plane modes.2.6 Consider a molecule with three atoms. The three atomic positions define a

plane. By considering the body modes, prove that a triatomic molecule hasno out-of-plane vibration modes.

2.7 The vibration frequencies of H2S are ω1 = 2,714 cm−1, ω2 = 1,215 cm−1,and ω3 = 2,732 cm−1, and the square of the mixing parameter is c2 = 0.8978.Calculate the isotopic frequencies for the D2S molecule.

2.8 Consider the planar molecule AB3 for which the B atoms are at the verticesof an equilateral triangle and the A atom is at the center of the triangle asshown in Fig. 2.5. The covering group is D3h; however, for this problem youmay use the D3 group.

C2 C2

C2

C3

A

B1

B2B3

x

y

z

D3 E 2C3 3C ′2

A1 1 1 1

A2 1 1 −1 Rz, z

E 1 −1 0 Rx, Ry

Figure 2.5

(a) Decompose the representation �(Z) based on the four out-of-planedisplacements, rzi = Zi (i = 1, 2, 3, and 4) into the IRs of D3.

(b) Identify the IRs for the body modes and the vibrations.(c) Find the eigenvectors for the three body modes.(d) Determine the out-of-plane, vibration eigenvector using orthogonality

requirements.2.9 Show that the isotopic frequencies of A(i)B3 are given by

1− ω2G44(ω2)ε = 0,

with ε = M (i)A /MA − 1, and G44(ω

2) = [(F− ω2)−1]44.

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Exercises 65

The matrix indices designated as i = 1, 2, and 3 refer to the out-of-plane dis-placements of the Bi atoms, and i = 4 refers to the out-of-plane displacementof the A atom.

2.10 Find an expression for the out-of-plane frequency of the A(i)B3 molecule interms of the out-of-plane frequency of the AB3 molecule.

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3

Spherical symmetry and the full rotation group

3.1 Hydrogen-like orbitals

The full rotation group, O(3), consists of all the real orthogonal transformationsthat leave a sphere invariant. The group includes proper rotations and inversion.The number of elements in the full rotation group is infinite, but the operatorsform a group that satisfies the definition given in Chapter 1 by Theorem 1.1. Towork with the full rotation group, it is useful to have basis functions for the irre-ducible representations. One set of basis functions is the set of eigenfunctions forthe hydrogen-like atom.

Hydrogen-like functions are eigenstates of Schrödinger’s equation for a sin-gle electron bound to a nucleus whose charge is +Ze. Similar functions are alsoobtained as one-electron, approximate solutions for a many-electron atom. In thiscase the charge Ze is the (effective) average charge experienced by the electron,including the “shielding” of the nuclear charge by the other electrons.

Schrödinger’s equation in the eigenvalue form is

(H0 − Ek)�k(r) = 0,

with

H0 = − �2

2me∇2 − Ze2

r+ Veff(r), r = |r|,

where H0 is a one-electron Hamiltonian and �k(r) is an eigenstate whoseeigenvalue is Ek. The Hamiltonian consists of the electron’s kinetic energy, theeffective nuclear Coulomb attraction, and an effective (spherically symmetric)potential, Veff.

The hydrogen-like eigenstates, called orbitals, are characterized by three quan-tum numbers, n, l, and m. The principal quantum number, n, can take on the values1, 2, 3, . . . The quantity l is the angular-momentum quantum number, and its range

66

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3.1 Hydrogen-like orbitals 67

θ

φ

x

y

z

r

−e

+Ze

Figure 3.1 The coordinate system for the hydrogenic orbitals.

of values for a given n is l = 1, 2, . . . , n − 1. Finally, m, the magnetic quantumnumber, takes on integral values between −l and +l. The energy corresponding to�nlm depends on the value of n, but not on l and m. (If spin–orbit interactions areconsidered the energy will also depend on l, but not on m). For our purposes it issufficient that the energy does not depend on m.

The orbitals can be separated into a radial function, Rnl(r), and an angularfunction, Y m

l (θ, φ),

�nlm(r) = Rnl(r) Y ml (θ, φ),

where n = 1, 2, 3, . . . ; l = 1, 2, 3, . . . , n − 1; and m = 0,±1,±2, . . . ,±l.The coordinate system defining the variables r , θ , and φ is shown in Fig. 3.1.Rnl(r) determines the radial shape of the wavefunction. The number of radial

nodes is n. The principal quantum number, n, determines the energy (eigenvalue)of the orbital, and orbitals having the same value of n are said to belong to thesame “shell”. The angular function, Y m

l (θ, φ), is one of an infinite set of orthogonalfunctions known as the spherical harmonics. The quantum number l determines theangular shape of the orbital. The s-orbitals have l = 0, the p-orbitals have l = 1,and the d-orbitals have l = 2. The magnetic quantum number m determines theorientation of the orbital with respect to the z-axis. Since m ranges from −l to +lin integral steps, it takes on (2l+1) integral values. Each of the spherical harmonicfunctions can be separated into a product of two functions:

Y ml (θ, φ) = Pm

l (cos θ)eimφ,

where Pml (cos θ) is an associated Legendre polynomial. The functions for l = 0,

1, and 2 are given in Table 3.1.

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68 Spherical symmetry and the full rotation group

Table 3.1 Spherical harmonic functions for l = 0, 1, and 2

Y 00 (θ, φ) = 1

2

√1/π Y−2

2 (θ, φ) = 14

√15/(2π) sin2 θ e−2iφ

Y−12 (θ, φ) = 1

2

√15/(2π) sin θ cos θ e−iφ

Y−11 (θ, φ) = 1

2

√3/(2π) sin θ e−iφ Y 0

2 (θ, φ) = 14

√5/π (3 cos2 θ − 1)

Y 01 (θ, φ) = 1

2

√3/π cos θ Y 1

2 (θ, φ) = − 12

√15/(2π) sin θ cos θ eiφ

Y 11 (θ, φ) = − 1

2

√3/(2π) sin θ eiφ Y 2

2 (θ, φ) = 14

√15/(2π) sin2 θ e2iφ

3.2 Representations of the full rotation group

In the previous chapters we used the operators OR to subject the nuclear displace-ments to a rotation, R, with the convention that a rotation was clockwise. Werotated the function(s) and kept the coordinate axes fixed. We also noted that thesame matrix representation of the group could be obtained by fixing the functionand rotating the axes in a counter-clockwise manner.

We now need to define a symmetry operator more generally. The conventionalsymbol for an operator that operates on a function of the coordinates is PR . Itsaction on a function f (x) is defined by the equation

PR f (x) = f (R−1x), (3.1)

where x represents a system of orthogonal coordinates, e.g., (X , Y , Z ), or, as inthe case of squarene, (X1, Y1, X2, Y2, . . . , Y4). PR f (x) is clockwise rotation of thefunction f (x) with the coordinates fixed and f (R−1x) is the function expressed inthe counter-clockwise-rotated coordinate system.

If R, S, T , . . . are the elements of a group G, the operators PR , PS , PT , . . . forma group that is isomorphic to G. This means that, if RS = T , then PR PS f (x) =PT f (x) = f ((RS)−1x) = f (S−1 R−1x). Note the order of the operators afterthe last equals sign. It is easy to make the erroneous argument that PR PS f (x) =PR f (S−1x) = f (R−1S−1x) (which is incorrect).

The appearance of R−1 on the right-hand side of (3.1) may seem a bit strangeat first. However, this definition of the action of the operator is consistent with themethod used in previous chapters. A simple example may serve to clarify this point.Consider a vector r in the x–y plane as shown in Fig. 3.2(a). R is a rotation by π/2about the z-axis (out of the plane of the paper). In the case of Fig. 3.2(a) we rotatethe vector r clockwise with the coordinate axes fixed. This is the method used inChapters 1 and 2. For this operation the x-component of r before rotation becomesthe negative y-component of r after rotation. Similarly, the y-component before

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3.2 Representations of the full rotation group 69

(a)

x

y

r

C4

Functionx

y

r

(b)

x

y

r

C4

Axesy

x

r

= x

yr

(c)

x

y

r

C−14

Axesy

x

r

= x

y

r

Figure 3.2 Rotation of a function versus rotation of the axes. (a) Clockwise rota-tion of r by π/2 with the coordinate axes fixed. (b) Clockwise rotation by π/2 ofthe axes with r fixed. (c) Counter-clockwise rotation by π/2 of the axes with rfixed. The components rx and ry in (a) are the same as rx ′ and ry′ , respectively,in (c).

rotation becomes the x-component after rotation. The matrix that accomplishesthis is (

0 1−1 0

). (3.2)

If we rotate the axes clockwise (Fig. 3.2(b)), we are making a change ofcoordinates described by x ′ = Rx; that is,

x ′ = R11 x + R12 y,

y′ = R21 x + R22 y, (3.3)

in which the x ′-axis is −y and the y′-axis is x . Clearly, the transformation coef-ficients are R11 = R22 = 0, R12 = −1, and R21 = 1. The matrix for thisoperation is (

0 −11 0

). (3.4)

In Fig. 3.2(c), we rotate the axes counter-clockwise. This gives x ′ = y andy′ = −x , so the matrix representing this transformation is(

0 1−1 0

). (3.5)

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70 Spherical symmetry and the full rotation group

From these different cases we can conclude that a rotation of the functionclockwise with the axes fixed produces the same representation matrix as a counter-clockwise rotation of the axes with the function fixed. In the above example wesee that the matrix for clockwise rotation of the axes is the inverse of the matrixfor counter-clockwise rotation. This relationship is always true since if we rotateclockwise through an angle and then counter-clockwise through the same angle wehave changed nothing. That is, R(cw) R(ccw) = E , from which it follows thatR(ccw) = R(cw)−1.

Returning to our discussion of the operator, PR , we see from its definition in(3.1) that PR operating on f (x) has the same matrix representation as rotating thefunction clockwise, since R−1 is a clockwise rotation of the axes.

Now let us look at the effect PR has on the spherical harmonic functions. Thefunction Y m

l has the property that if the polar axis is rotated the new function canbe expressed as a linear combination of the old functions,

PR Y ml (θ, φ) =

+l∑m′=−l

D(l)m′m(R) Y m′

l (θ, φ). (3.6)

The matrices D(l)m′m(R) form a representation of the group of rotations, as we shall

show below. Let � be a linear combination of spherical harmonic functions,

� =∑

m

Cm Y ml ,

where Cm is a constant coefficient in the expansion. We can represent � as acolumn vector with (2l + 1) rows: ⎛

⎜⎜⎜⎜⎜⎝

Cl

Cl−1

Cl−2...

C−l

⎞⎟⎟⎟⎟⎟⎠ .

In this case the rows of the column vector are the coefficients of the functions,Y l

l , Y l−1l , Y l−2

l , . . . , Y−ll . So the Y m

l s act as the unit vectors of a coordinate spacewith (2l + 1) dimensions. If we operate on �, we have

PR � =∑

m

Cm PR Y ml =

∑m

∑m′

Cm D(l)m′m(R) Y m′

l

=∑m′

(∑m

D(l)m′m(R)Cm

)Y m′

l =∑m′

C ′m′ Ym′l , (3.7)

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3.2 Representations of the full rotation group 71

where ∑m

D(l)m′m(R)Cm = C ′m′ . (3.8)

Equation (3.8) written out in matrix form for l = 1 is⎛⎜⎝ D(1)

11 (R) D(1)10 (R) D(1)

1−1(R)

D(1)01 (R) D(1)

00 (R) D(1)0−1(R)

D(1)−11(R) D(1)

−10(R) D(1)−1−1(R)

⎞⎟⎠⎛⎝ C1

C0

C−1

⎞⎠ =

⎛⎝ C ′1

C ′0C ′−1

⎞⎠ . (3.9)

Equation (3.9) is the same form as we had for the representation matrices inChapter 1, namely,

�(R)Vi = V f , (3.10)

where Vi is the initial vector, �(R) is the representation matrix for R, and V f isthe final vector (after rotating the function clockwise). In Chapter 1 we rotated thefunction, not the axes. In (3.9) D(1)(R) rotates the axes counter-clockwise, whichproduces the same matrix representation as rotating the function clockwise.

A word about notation is needed. We shall reserve the symbol D(l)(R) for therepresentation of PR based on the (2l + 1) spherical harmonic functions. Otherrepresentations of PR will be denoted by D(R) (D without a superscript (l)), or by�(R) or �α(R).

In general, if f j is one of a set of functions ( j = 1, 2, . . . , N ) for which

PR f j =N∑

r=1

Dr j (R) fr , (3.11)

then D(R) is a matrix representation of PR (not necessarily an IR). The order ofthe indices is essential in (3.11) if D(R) is to form a representation of the groupunder ordinary matrix multiplication. For the set of basis functions,

PS PR f j =∑

k

D(R)k j PS fk =∑

k

∑m

D(R)k j D(S)mk fm

=∑

m

{∑k

D(S)mk D(R)k j

}fm =

∑m

D(S R)mj fm . (3.12)

In (3.12) k and m take on the values 1, 2, . . . , N , where N is the number of basisfunctions. Equation (3.12) confirms that the D(R) matrices of (3.11) satisfy therequirements of a matrix representation of the PR operators.

The spherical harmonic functions, Y ml , provide basis functions for an IR of

the operators of the full rotation group. There is an infinite number of such

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72 Spherical symmetry and the full rotation group

(a)

x

y

z

x

(b)

x

y

z

x

z

β

(c)

x

y

z

x

Figure 3.3 Euler-angle rotations: (a) rotation about the z-axis by α, (b) rotationabout the y′-axis by β, and (c) rotation about the z′-axis by γ . The sequence (a) to(c) corresponds to the operations in the order Rz′(γ ) Ry′(β) Rz(α). The axes arepositioned in each figure so that the z- or z′-axis is vertical.

representations, one for each value of l, and each is an IR. The dimensional-ity of the representations, (2l + 1), is always an odd integer. Even-dimensionalrepresentations are discussed in Section 3.8.

Theorem 3.1 The (2l + 1)-dimensional representations of the full rotation group,D(l)(R), based on the spherical harmonic functions Y m

l (θ, φ), m = ±l, ±(l − 1),±(l − 2), . . . , 0, are irreducible. There are no other inequivalent irreduciblerepresentations.

The connection between the quantum numbers of the hydrogen-like orbitals andthe representations is apparent. Each angular quantum number, l, is the name of anIR of the rotation group. Each magnetic quantum number is the name of a partnerfunction of the IR; that is, Y m

l is the mth-row function for the lth IR.

3.3 The character of a rotation

It is useful to use Euler angles α, β, and γ to describe a rotation. The Eulerian rota-tions, illustrated in Fig. 3.3, are defined as counter-clockwise rotations as follows:start with the x–y–z axes then perform (a) a rotation about the z-axis through anangle α carrying x → x ′, y → y′, and z → z, followed by (b) a rotation about they′-axis through an angle β, carrying x ′ → x ′′, y′ → y′ and z → z′, followed by(c) a rotation about the z′-axis through an angle γ , carrying x ′′ → x ′′′, y′ → y′′,and z′ → z′.

The IR in terms of the Euler angles is denoted by D(l)(α, β, γ ), and (3.6) takesthe form

P(α,β,γ ) Y ml =

+l∑m′=−l

D(l)(α, β, γ )m′m Y m′l . (3.13)

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3.3 The character of a rotation 73

The Eulerian system has the property that the same rotated vector can be achievedby applying the rotations in the reverse order about fixed axes: start with the x–y–zaxes, then perform (1) a rotation about the z-axis through an angle γ , followed by(2) a rotation about the y-axis through an angle β, followed by (3) a rotation aboutthe z-axis through an angle α. The final rotated vector is the same as produced bythe Euler rotations of Fig. 3.3. Stated mathematically,

Rz(α) Ry(β) Rz(γ ) = Rz′(γ ) Ry′(β) Rz(α), (3.14)

where the rotations on the left operate on a vector in the fixed x–y–z axes system.For γ = 0, we have

Rz(α) Ry(β) = Ry′(β) Rz(α). (3.15)

If (3.15) is substituted into (3.14), we obtain

Rz(γ ) = R−1 Rz′(γ ) R, (3.16)

where R = [Ry′(β) Rz(α)]. Recalling that elements A and B belong to the sameclass if they are related by A = R−1 B R, where R is any member of the group(Chapter 1, Theorem 1.3) we can conclude that Rz(γ ) and Rz′(γ ) are in the sameclass. But, since we can choose the angles α and β at will, the axis z′ can bearbitrarily chosen. Therefore all rotations by an angle γ belong to the same classregardless of the axis of rotation.

Now suppose we choose a rotation about the z-axis through an angle α withβ = γ = 0. The operator P(α,0,0) acts only on the φ-dependent part of Y m

l (θ, φ),

P(α,0,0) Y ml (θ, φ) = Plm(cos θ)[P(α,0,0) eimφ] = e−imα Y m

l (θ, φ). (3.17)

The negative sign in the exponential results because P(α,0,0) rotates the functionclockwise, that is, takes φ into φ − α. We see that P(α,0,0) acting on Y m

l produces amultiple of the same function. Therefore the representation matrix D(l)(α, 0, 0) ≡D(l)(α) is diagonal,

D(l)(α)m,m′ = e−imαδm,m′, (3.18)

and for the character of the matrix we have

χ(l)(α) =+l∑

m=−l

e−imα = sin[(l + 1/2)α]sin(α/2)

. (3.19)

Since all rotations by an angle α about any axis are in the same class, it followsthat the characters of all such rotations are the same. Therefore, in finding the

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74 Spherical symmetry and the full rotation group

character of the matrix representing a rotation it is not necessary to specify the axisof rotation; only the angle need be specified. We may then write

χ(l)(A) = sin[(l + 1/2)A]sin(A/2)

, (3.20)

where χ(l)(A) is the character of the representation matrix for a rotation of thefunction through an angle A in radians about any axis. We note that the characterof a rotation by A is the same as that for a rotation by −A. Therefore the charac-ter is the same irrespective of whether we rotate the function or the coordinates.However, the matrix D(l) depends on the sign of A, and incorrect results may beobtained if one does not consistently use a single convention. We shall consistentlyuse the convention that PR rotates the function clockwise as implied by (3.1).

Approximate solutions of Schrödinger’s equation for atoms with many electronsmay be built up from combinations of hydrogen-like, one-electron orbitals. Thewavefunction may be, for example, a Slater determinant (or a linear combinationof Slater determinants) whose elements are one-electron, hydrogenic orbitals. Asmentioned previously, the energies of the orbitals do not depend on the magneticquantum number, m. This means that orbitals for a particular l are (2l + 1)-folddegenerate in a spherically symmetric environment. Since all the partner functions(basis functions for the rows of D(l)) have the same energy, linear combinationsof these functions also have the same energy. Certain linear combinations of thespherical harmonics are more convenient to work with than others. In particular, itis useful to take combinations that are real functions. The usual choices for thesefunctions are given in Table 3.2.

The combinations of Y ml s in Table 3.2 are symmetry coordinates for IRs of the

octahedral group, O .

Table 3.2 The s, p, and d hydrogen-like orbitals

s = Y 00 =√

1/(4π) dz2 = Y 02 =√

5/(16π) (3z2 − r2)/r2

dx2 =√

12 (Y

22 + Y−2

2 )

= √15/(16π) (x2 − y2)/r2

px = −√

12 (Y

11 − Y−1

1 ) = √3/(4π) x/r dxy = −i√

12 (Y

22 − Y−2

2 ) = √15/(4π) xy/r2

py = i√

12 (Y

11 + Y−1

1 ) = √3/(4π) y/r dxz =√

12 (Y

12 + Y−1

2 ) = √15/(4π) xz/r2

pz = Y 01 =√

3/(4π) z/r dyz =√

12 (Y

12 − Y−1

2 ) = √15/(4π) yz/r2

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3.4 Decomposition of D(l) in a non-spherical environment 75

3.4 Decomposition of D(l) in a non-spherical environment

For a given value of l the hydrogen-like orbitals Y ml (θ, φ) are basis functions for a

(2l + 1)-dimensional matrix representation of the rotation group. All of the partnerfunctions (those functions corresponding to different values of m) are degenerateeigenfunctions of the hydrogen-like Hamiltonian in (3.1).

If the environment of the atom is changed so that it is no longer sphericallysymmetric, some of the degeneracies may be lifted. Examples of changes in envir-onment include application of an electric or magnetic field or imbedding the atomin a crystal or molecule. In a non-spherical environment the symmetry is reduced,and the group of the covering operations is no longer the full rotation group.

Suppose G, a group for a non-spherical environment, consists of the elementsR1, R2, R3, . . . , Rh . The matrices D(l)(R1)mm′ , D(l)(R2)mm′ , D(l)(R3)mm′ , . . . ,D(l)(Rh)mm′ of (3.6) form a valid representation of G, but the representation mightnot be an IR. To find out whether it is, we can decompose D(l) into the IRs of thegroup G in the usual way (Chapter 1, Eq. (1.10)),

nβ = 1

h

Nc∑k=1

Nk χ l(Rk)∗ χβ(Rk), (3.21)

where nβ is the number of times the βth IR of G appears in the decompositionof D(l), Nc is the number of classes of G, and Nk is the number of elements inthe kth class. As an example, consider an atom in a crystal at a site of octahedralsymmetry. The character table for the octahedral group O is given in Table 3.3.

The last row of Table 3.3 shows the characters for the D(l) representation derivedfrom (3.20) for A = 2π/n. For the d-orbitals, l = 2 and we have χ(2)(C2) =χ(2)(π) = S(2)

2 = 1, χ(2)(C3) = χ(2)(2π/3) = S(2)3 = −1, and χ(2)(C4) =

χ(2)(π/2) = S(2)4 = −1. The decomposition of D(2) is as follows:

n A1 = 1

24

[1× 5× 1+ 8×−1× 1+ 3× 1× 1+ 6× 1× 1+ 6×−1× 1

]= 0,

n A2 = 1

24

[1× 5× 1+ 8×−1× 1+ 3× 1× 1+ 6× 1×−1+ 6×−1×−1

]= 0,

nE = 1

24

[1× 5× 2+ 8×−1×−1+ 3× 1× 2+ 6× 1× 0+ 6×−1× 0

]= 1,

nT1 = 1

24

[1× 5× 3+ 8×−1× 0+ 3× 1×−1+ 6× 1×−1+ 6×−1× 1

]= 0,

nT2 = 1

24

[1× 5× 3+ 8×−1× 0+ 3× 1×−1+ 6× 1× 1+ 6×−1×−1

]= 1,

D(2) = E + T2. (3.22)

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76 Spherical symmetry and the full rotation group

Table 3.3 The character table for the octahedral group, O. The last row showsthe characters for a reducible representation of O based on the sphericalharmonic functions, Y m

l (θ, φ). S(l)n = sin[(l + 1/2)2π/n]/sin(π/n).

IR name E 8C3 3C2 6C ′2 6C4 Basis functions

A1 1 1 1 1 1 x2 + y2 + z2

A2 1 1 1 −1 −1 xyzE 2 −1 2 0 0 {(x2 − y2), (3z2 − r2)}T1 3 0 −1 −1 1 {x, y, z}; {Rx , Ry, Rz}T2 3 0 −1 1 −1 {xy, xz, yz}D(l) 2l+1 S(l)

3 S(l)2 S(l)

2 S(l)4

The decomposition shows that in an octahedral environment the five-fold degen-erate d-orbitals split into two groups: a doubly degenerate E group and a triplydegenerate T2 group. Since each IR in the decomposition is contained once only,the symmetry functions constructed from linear combinations of the spherical har-monics will be eigenfunctions of a Hamiltonian having spherical or octahedralsymmetry. (The energies of Eg and T2g functions are degenerate for sphericalsymmetry, but are split in octahedral symmetry.)

The d-orbitals listed in Table 3.2 are bases for IRs of the O group. The functionsdz2 and dx2 are the basis functions for the E IR, and dxy , dxz , and dyz are the basisfunctions for the T2 IR.

3.5 Direct-product groups and representations

Consider two groups, Ga and Gb, that have no common elements other than theidentity and whose elements commute with one another. We can form a larger,composite group denoted by Ga×Gb. If Ga has elements Raj , j = 1, 2, 3, . . . , ha ,and Gb has elements Rbk, k = 1, 2, 3, . . . , hb, then the elements of the direct-product group, Ga × Gb, are of the form Raj Rbk . The total number of elements inGa × Gb is hahb. Such product groups are useful when the operators of the twogroups operate on different kinds of coordinates, or to extend a group to includenew elements. The first case occurs, for example, when dealing with vibronicstates, �(r, X) = φ(r)ψ(X), where φ(r) is a function of the electronic coor-dinates and ψ(X) is a function of the nuclear displacements. In this case, theoperators corresponding to Ga operate on electronic coordinates and the opera-tors corresponding to Gb operate on the nuclear displacements. The groups Ga andGb can be identical, but the operators operate on different types of coordinates. An

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3.5 Direct-product groups and representations 77

+

x1

y1

z1

x2

y2

z2

A1

v1

v2 A2

symmetryCenter of

Figure 3.4 A molecule with a center of symmetry. The atoms at A1 and A2are of the same type and in symmetry-equivalent positions. Under inversion adisplacement v at A1 becomes a displacement −v at A2.

example using the direct product to enlarge a group is the addition of the inversionoperator, i , to a group that does not contain i or an operation equivalent to i .

3.5.1 Inversion

The operation of inversion, denoted by the symbol “i”, carries a vector r into −r.For a molecule with a center of symmetry, “i” carries equivalent atoms A1 and A2

into one another. The displacement vector, v1, of A1 is carried into v2 of A2, wherev1 = −v2 as shown in Fig. 3.4.

The inversion operation will be a member of the covering group when a “cen-ter of inversion” or “center of symmetry” exists for the system being analyzed.However, in some cases the inversion operation may produce the same effect asa rotation or reflection, and as a result some groups do not explicitly contain theinversion-operation symbol. For example, in Chapter 1 we considered the groupC4v of the covering operations of a square. In that case the inversion operation (intwo dimensions) was equivalent to a rotation by 180◦ about an axis perpendicularto the plane of the square and the inversion operation is not explicitly included inC4v, but C4v × i would contain redundant operations.

The octahedral group, O , has 24 elements. When the inversion operationis included the extended group, Oh = O × i , has 48 elements. The set ofcorresponding operators, PRk and Pi Rk , forms a group isomorphic to Oh .

If G is a group with elements Rk (k = 1, 2, . . . , h), and none of the elementsis equivalent to i , then a larger group can be formed, whose elements are Rk and

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78 Spherical symmetry and the full rotation group

Table 3.4 The character table forthe inversion group

IR E i

�g 1 1�u 1 −1

Table 3.5 The character table for Oh = O × i

R i R

IR E 8C3 3C2 6C ′2 6C4 i 8iC3 3iC2 6iC ′2 6iC4 Basis functions

A1g 1 1 1 1 1 1 1 1 1 1 x2 + y2 + z2

A2g 1 1 1 −1 −1 1 1 1 −1 −1�g Eg 2 −1 2 0 0 2 −1 2 0 0 (x2 − y2), (3z2 − r2)

T1g 3 0 −1 −1 1 3 0 −1 −1 1 (x, y, z); (Rx , Ry, Rz)T2g 3 0 −1 1 −1 3 0 −1 1 −1 (xy, xz, yz)

A1u 1 1 1 1 1 −1 −1 −1 −1 −1A2u 1 1 1 −1 −1 −1 −1 −1 1 1 xyz

�u Eu 2 −1 2 0 0 −2 1 −2 0 0T1u 3 0 −1 −1 1 −3 0 1 1 −1 (x, y, z)T2u 3 0 −1 1 −1 −3 0 1 −1 1

i Rk (k = 1, 2, . . . , h). Clearly the larger group has twice as many elements asthe original group. Inversion and the identity form a group by themselves that isdenoted by the symbol i or S2. The character table for the i group is given inTable 3.4.

It is common to write that Gh = G × i , which means that Gh is the directproduct of the groups G and i . A basis function of an IR of Gh is labeled “g” or“u” depending upon whether it is even or odd under the inversion operation. Forg-functions, Pi f (r) = f (r); for u-functions Pi f (r) = − f (r).

The character table for Oh = O × i is easily obtained from the character tablesof O and i . It is essentially the direct product of the character tables of G and i , ascan be seen in Table 3.5.

The inversion operation that carries r into −r can be accomplished by a rotationby π about an axis through the origin, perpendicular to any plane containing r.Therefore the character of i for the spherical harmonic representation is sin[(l +1/2)π ]/sin(π/2) = (−1)l . As a result, the spherical harmonic functions for agiven l are symmetric (g) under inversion for even values of l and antisymmetric(u) under inversion for odd values of l.

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3.6 General properties of direct-product groups and representations 79

A useful theorem from group theory is that the character of the direct-productrepresentation is the product of the characters, χ(i R) = χ(i) χ(R). (We shallprove a more general result a bit later in this chapter.) Therefore,

χ l(i R) = χ l(R) for l even, (3.23)

χ l(i R) = −χ l(R) for l odd. (3.24)

Equations (3.23) and (3.24) can be used to assign “g” or “u” labels to thedecomposition of D(l) into the IRs of Oh . A few results are

s-orbitals D(0) = A1g, (3.25)

p-orbitals D(1) = T1u, (3.26)

d-orbitals D(2) = T2g + Eg, (3.27)

f -orbitals D(3) = A2u + T1u + T2u. (3.28)

From (3.26) and (3.27) we see that at a site of octahedral symmetry the p-orbitalsremain three-fold degenerate, but the d-orbitals split into two-fold degenerate Eg

states and three-fold degenerate T2g states. This information is available in the char-acter table for the Oh group (Table 3.5), which shows that dxy , dxz , and dyz are thebasis functions for the different rows of the T2g IR and that dz2 and dx2 are basisfunctions for the different rows of the Eg IR.

No IR appears more than once in the decomposition of D(2), therefore the sym-metry functions for the T2g or those for the Eg IRs are eigenstates of the originalO(3)-Hamiltonian plus any potential having octahedral symmetry. We know theyare eigenstates because there can be no non-zero matrix elements of the Hamilto-nian (including octahedrally symmetric terms) between bases for different IRs orbetween bases belonging to different rows of the same IR.

The above conclusions are valid only if the octahedral environment is repre-sented by a potential. For a molecule or solid, the octahedral environment is dueto the surrounding ligands. If the orbitals of the ligands that create the octahedralenvironment are included in the analysis there usually will be combinations of theligand orbitals that belong to the T2g (and Eg) IRs. If so, these combinations willbe mixed with the d-orbitals to form the eigenstates. In either case the T2g and Eg

states are not mixed so long as the symmetry of the system is spherical, O or Oh .

3.6 General properties of direct-product groups and representations

In general we can have the direct product of two groups if the operations of onecommute with the operations of the other and they have no elements in commonother than the identity. If the operators corresponding to the two groups operate on

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80 Spherical symmetry and the full rotation group

⎛⎝ A11 A12

A21 A22

⎞⎠ ×

⎛⎝

B11 B12 B13

B21 B22 B23

B31 B32 B33

⎞⎠ =

⎛⎝ A11B A12B

A21B A22B

⎞⎠

=

⎛⎜⎜⎜⎜⎜⎜⎝

A11B11 A11B12 A11B13 A12B11 A12B12 A12B13

A11B21 A11B22 A11B23 A12B21 A12B22 A12B23

A11B31 A11B32 A11B33 A12B31 A12B32 A12B33

A21B11 A21B12 A21B13 A22B11 A22B12 A22B13

A21B21 A21B22 A21B23 A22B21 A22B22 A22B23

A21B31 A21B32 A21B33 A22B31 A22B32 A22B33

⎞⎟⎟⎟⎟⎟⎟⎠

Figure 3.5 The direct product of two matrices: [A × B]ss′,t t ′ = Ast Bs′t ′ .

different kinds of coordinates, they do not have common operators even if the twogroups are the same.

Consider the direct-product group Ga×Gb with elements Ai B j and correspond-ing operators PAi PB j = PAi B j . If �a is a matrix representation of the operators ofGa and �b is a matrix representation of the operators of Gb, then �a × �b is amatrix representation of the operators of the direct-product group Ga × Gb. Thisstatement is a brief way of saying that

�(a×b)(Ai B j ) = �a(Ai )× �b(B j ), (3.29)

for all of the operators of Ga × Gb. The matrix elements are defined as follows:

[�(a×b)(Ai B j )]ss′,t t ′ = [�a(Ai )]st [�b(B j )]s′t ′ . (3.30)

The direct product of two matrices is illustrated in Fig. 3.5.Direct-product matrix multiplication is also similar to regular matrix multiplica-

tion, but the sum on the index is now a sum on a pair of indices:

[�(a×b)(Ai B j )�(a×b)(Ai ′B j ′)]nn′,mm′

=∑

k

∑k′[�a(Ai )]nk[�b(B j )]n′k′ [�a(Ai ′)]km[�b(B j ′)]k′m′

={∑

k

[�a(Ai )]nk[�a(Ai ′)]km

}{∑k′[�b(B j )]n′k′ [�b(B j ′)]k′m′

}= [�a(Ai Ai ′)]nm[�b(B j B j ′)]n′m′ = [�(a×b)(Ai Ai ′)(B j B j ′)]nn′,mm′ . (3.31)

The diagonal elements of [�(Ai B j )]nn′,mm′ are obtained when m = n andm ′ = n′,

[�(a×b)(Ai B j )]nn′,nn′ = [�a(Ai )]nn[�b(B j )]n′n′ . (3.32)

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3.6 General properties of direct-product groups and representations 81

The character of �(a×b)(Ai B j ), a quantity denoted by χ(a×b)(Ai B j ), is the sum ofthe diagonal elements,

χ(a×b)(Ai B j ) =∑

n

∑n′[�(a×b)(Ai B j )]nn′,nn′

=∑

n

[�a(Ai )]nn

∑n′[�b(B j )]n′n′

= χ�a(Ai )χ

�b(B j ). (3.33)

Equation (3.33) shows that the character of the matrix representation of PAi Bi forthe direct-product group Ga × Gb is the product of the characters of the matrixrepresentation of PAi and the matrix representation of PB j . Thus we can find thecharacters of the direct-product representations from the characters of the individ-ual representations. Some important properties of direct products are summarizedin Theorem 3.2.

Theorem 3.2 (Properties of direct-product representations)Notation and definitions

G(a×b) = Ga × Gb.

● The elements of Ga are Ai and the corresponding operators are PAi .● The elements of Gb are B j and the corresponding operators are PB j .● The elements of Ga × Gb are Ai B j and the corresponding operators are

PAi PB j = PAi B j .● �a(Ai ) is a matrix representation of PAi and �b(B j ) is a matrix representation

of PB j .● �(a×b)(Ai B j ) is a matrix representation of the direct-product operator PAi B j .● χ(a×b)(Ai B j ) is the character of the matrix representing PAi B j .● χ�a

(Ai ) is the character of the matrix representing PAi and χ�b(B j ) is the

character of the matrix representing PB j .

Theorem 3.2a. �(a×b) = �a × �b. This means that �(a×b)(Aai Bb

j ) = �a(Aai ) ×

�b(Bbj ) for all Aa

i and Bbj . The direct product of two representations is a

representation of the direct-product group.Theorem 3.2b. The direct product of two IRs is an IR of the direct-product group.

That is, if �a is an IR of Ga and �b an IR of Gb, then �a ×�b is an IR of Ga ×Gb.Theorem 3.2c. The number of classes in Ga × Gb is the product of the number

of classes in Ga times the number of classes in Gb.Theorem 3.2d. The character of a direct-product matrix representation is

the product of the characters of the two representations: χ(a×b)(Ai B j ) =χ�a

(Ai )χ�b(B j ).

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82 Spherical symmetry and the full rotation group

Theorem 3.2e. If C1 is a class of G1 and C2 is a class of G2, then C1 × C2 is aclass of G1 × G2.

3.6.1 Direct products of groups whose operators operate on differenttypes of coordinates

Let ξ1, ξ2, . . . , ξN be basis functions for a representation �a of Ga whose elementsare Ai , and let ζ1, ζ2, . . . , ζM be basis functions for a representation �b of Gb

whose elements are B j . The operators PAi operate only on the ξ -functions andthe operators PB j operate only on the ζ -functions. The N M functions, ξsζs′ (s =1, 2, . . . , N ; s ′ = 1, 2, . . . , M), are basis functions for a representation of thedirect-product group, �(a×b). Consider

PAi B j ξt ζt ′ = (PAi ξt)(PB j ζt ′) =∑

s

∑s′

�(Ai )st �(B j )s′t ′ (ξs ζs′)

=∑

s

∑s′

�(a×b)(Ai B j )ss′,t t ′ (ξs ζs′). (3.34)

For the last equality in (3.34) we used the definition of the matrix element �(a×b)ss′, t t ′

in (3.30). Note the order of the indices in (3.34). The rows are s and s ′ and thecolumns are t and t ′.

Equations (3.30) and (3.34) show that �(a×b) is a matrix representation of Ga ×Gb with basis functions ξs ζs′ .

There is an aspect of the direct product that should be noted. Theorem 3.2b tellsus that if �a and �b are IRs of Ga and Gb, respectively, then �a × �b = �(a×b) isan IR of Ga × Gb. However, �(a×b) is also a representation of Ga (or Gb), but it isa reducible representation with respect to Ga (or Gb).

3.6.2 Direct products of representations of the same group

A common use of the direct product of representations of the same group is in theanalysis of multi-electron problems and selection rules for transitions between twostates of a system. For example, for a two-electron system the wavefunction maybe composed of products of one-electron functions, ϕλ(r1) ϕν(r2), where λ and ν

represent the quantum numbers. The group of the Hamiltonian for the system isG H and its operators P H

R operate on the coordinates of ϕλ(r1) ϕν(r2). Explicitly,P H

R ϕλ(r1) ϕν(r2) = ϕλ(R−1r1) ϕν(R−1r2). If ϕλ(r1) is a basis function for therepresentation �1 of G H and ϕν(r2) is a basis function for the representation �2

of G H , then the product, ϕλ(r1) ϕν(r2), is a basis function for the direct productof these representations, �1 × �2. The two matrix representations �1 and �2 are

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3.7 Selection rules for matrix elements 83

representations of the same group, G H . �1 × �2 is usually a reducible representa-tion even if �1 and �2 are both IRs of G H . For example, if G H is the octahedralgroup, O , �1 = T1 and �2 = T2, then T1 × T2 decomposes into the following IRsof O : A2 + E + T1 + T2. What does this decomposition mean with regard to thefunction ϕλ(r1) ϕν(r2)? It means we can be certain that no part of ϕλ(r1) ϕν(r2)

transforms according to any IR not contained in the decomposition. Applying thesymmetry-function-generating machine to ϕλ(r1) ϕν(r2) will generate linear com-binations of products that transform according to the IRs in the decomposition orproduce a null function.

It is important to distinguish between the direct-product representation �(α×β) =�α(G)×�β(G) of the same group, G, and a representation �(Ga×Gb) of two groups,Ga and Gb, whose operators are distinct (except for the identity) and commutewith each other. If ϕα(r1) is a basis function for the αth IR of Ga and ϕβ(r2) is abasis function for the βth IR of Gb, then ϕα(r1) ϕβ(r2) is a basis function for anIR of Ga × Gb. On the other hand, if both ϕα(r1) and ϕβ(r2) are basis functionsfor IRs of the same group G, then ϕα(r1) ϕβ(r2) is usually not a basis function foran IR of G. That is, �(α×β) is usually a reducible representation. (Note that theelements of G × G do not form a group.)

3.7 Selection rules for matrix elements

As we have discussed previously in connection with the optical transitions, thequestion of whether a transition is forbidden by symmetry depends on a matrixelement of the type

〈 f αk (ξ)| fop(ξ)| f β

j (ξ)〉 =∫

f αk (ξ)∗ fop(ξ) f β

j (ξ) dξ, (3.35)

where f αk (ξ) is the initial state and f β

j (ξ) is the final state. In (3.35) f αk (ξ) and

f β

j (ξ) are eigenstates of a Hamiltonian or Hermitian operator whose covering

group is G. Here f αk belongs to the kth row of the αth IR and f β

j belongs to thej th row of the βth IR of G. The operator fop(ξ) could be, for example, the electricdipole of a molecule or the Raman tensor. In the most general case, fop could beany function of ξ . If the matrix element vanishes due to symmetry considerations,the transition from the initial to the final state is said to be symmetry-forbidden orjust forbidden.

Let us begin with a simple case for which fop(ξ) = f γ

i is a basis function for thei th row of the γ th IR of G. For convenience let us arbitrarily group the functionstogether in the following way:

〈 f αk (ξ)|F(ξ)〉, (3.36)

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84 Spherical symmetry and the full rotation group

where

F(ξ) = [ f γ

i (ξ) f β

j (ξ)]. (3.37)

From (1.21) we know that the matrix element will vanish unless F(ξ) belongs tothe same row of the same IR as f α

k (ξ), that is, the kth row of the αth IR.The function F(ξ) = f γ

i (ξ) f β

j (ξ) is a basis function for the direct-product rep-resentation �γ × �β . We can determine whether the function has the possibilityof belonging to the αth IR of G by decomposing �γ × �β into the IRs of G. Ifthe decomposition does not contain �α then we can definitely say that the matrixelement vanishes due to symmetry considerations. If the decomposition does con-tain �α we can not conclude that the matrix element is symmetry-allowed. To saythat the matrix element is symmetry-allowed requires that F(ξ) not only contains�α but also contains a part that transforms as the kth row of �α. How can wedetermine whether F(ξ) belongs to the kth row of the αth IR? One method is toapply the symmetry-function-generating machine to F(ξ). In this case we need toconsider ∑

Ri

�α(Ri )∗kk{PRi F(ξ)}, (3.38)

where the sum over Ri is over all of the elements of the group G.If the result of (3.38) is not zero, then the matrix element is not required by sym-

metry to vanish. If the result is zero, the matrix element is required by symmetryto vanish.

Care must be taken when employing symmetry arguments to matrix elements.Selection rules are usually only in terms of the IRs with no consideration for therow requirement. On the other hand, the negative statement is always valid. That is,if �γ ×�β does not contain �α the matrix element certainly vanishes. Another wayto express the selection rule is that a non-vanishing matrix element requires that

�α × �β × �γ must contain the totally symmetric IR of G. (3.39)

In the case of the Oh group the totally symmetric IR is A1g.A simple example serves to illustrate the problem. Consider the case where

f αk = x, f β

j = z, and f γ

i = y. These functions are all basis functions for the

T1 IR of the O group. The product f γ

i f β

k is a basis function for the direct-productrepresentation, T1 × T1 = A1 + E + T1 + T2. Since T1 × T1 contains T1 and f α

k

belongs to T1 we would conclude that the matrix element does not vanish becauseof symmetry if we considered only the direct-product decomposition. By contrast,if the symmetry-function-generating machine is applied to f γ

i f β

k in an attempt togenerate a kth-row T1-function we would obtain zero. It is easy to see why. Theproduct f γ

i f β

k = yz is the basis function for one of the rows of T2 and therefore

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3.8 General representations of the full rotation group 85

it provides no basis for the T1 representation of G. Thus in this example the matrixelement must be zero due to symmetry considerations despite the fact that it passesthe IR selection-rule test.

3.8 General representations of the full rotation group

Representations of the full rotation group based on the spherical harmonic func-tions provide the IRs for the full rotation group for integral values of the angularmomentum, i.e., for l = 0, 1, 2, 3, . . . Quantum systems with spin have totalangular momentum j = l + s that can take on half-integral values. In general, jcan take on the values 1

2 , 1, 32 , 2, 5

2 , . . . Therefore it is essential to have the IRsof both integral and half-integral angular momenta. The matrix representations forboth integral and half-integral angular momenta are based on the work of HermannWeyl. He developed the theory of continuous groups based on matrix representa-tions during the period from 1923 to 1938. His method for finding the IRs of therotation group is based on establishing a homomorphism between 2 × 2 unitarymatrices and the 3 × 3 matrices representing three-dimensional rotations. We willnot delve into the details of the theory, but will simply state the results. More detailsmay be found elsewhere [3.1, 3.2].

Theorem 3.3 The IR matrices of the rotation group for a given integral or half-integral value of j are

D( j)(α, β, γ )m′,m = e−im′αe−imγ

×∑κ

(−1)κ√( j + m)!( j − m)!( j + m ′)!( j − m ′)!

κ!( j + m − κ)!( j − m ′ − κ)!(κ + m ′ − m)!

×(

cos

2

))2 j−2κ−m′+m (− sin

2

))2κ+m′−m

.

D( j)(α, β, γ )m′m is a matrix element of the m ′th row and mth column of the j thIR for a rotation characterized by the Euler angles α, β, and γ . In Theorem 3.3the sum on the index κ includes only the integral values of κ that lie between(a) the larger of zero or m − m ′ and (b) the smaller of j − m ′ and j + m. Theresulting representation matrices for j = 0, 1

2 , 1, 32 , and 2 are given in Table 3.6.

These matrices rotate the basis functions of the IRs in clockwise manner throughthe Euler angles −α,−β, and −γ . For integral values of j the basis functions arethe spherical harmonics discussed previously in this chapter. For the half-integralvalues of j the basis functions are linear combinations of spin or spinor functions.

In Section 3.3 we showed that all rotations through the same angle about anyaxis belong to the same class and therefore have the same character. We consider

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86 Spherical symmetry and the full rotation group

a rotation about the z-axis so that β = γ = 0. For diagonal elements (m = m ′)all of the terms of Theorem 3.3 vanish except the term for which κ = 0. For thatterm, (−sin(β/2))2κ+m′−m = (−sin(β/2))2κ = (−sin(β/2))0 = 1 and the factorialcoefficient is also 1. Therefore we have

D( j)(α, 0, 0)mm = e−imα, (3.40)

where m takes on values from j to − j in unit steps. The trace or character is then

χ( j)(α) =j∑

m=− j

D( j)(α, 0, 0)mm = sin[( j + 1/2)α]sin(α/2)

. (3.41)

Equation (3.41) holds for integer or half-integer j (it differs from (3.23) only inthat l is replaced by j). Since (3.41) applies to all rotations through an angle α

about any axis, in general we have

χ( j)(A) = sin[( j + 1/2)A]sin(A/2)

, (3.42)

where A is the angle of rotation in radians about any axis.This result, (3.42), is not immediately obvious from the matrices given in

Table 3.6. The reason why it is not apparent is that the angle, A, is about an axis thatwill produce the same rotation as the three Euler rotations. To determine A fromα, β, and γ is a problem in spherical trigonometry that we shall not pursue here.However, the plausibility of (3.42) is evident from a few special cases. Considerthe character of D(1/2)(α, β, γ ). From Table 3.6 we have

χ(1/2)(α, β, γ ) = e−i(α+γ )/2 cos(β/2)+ ei(α+γ )/2 cos(β/2)

= 2 cos[(α + γ )/2] cos(β/2). (3.43)

If β = 0 then

χ(1/2)(α, 0, γ ) = 2 cos[(α + γ )/2] = sin(α + γ )

sin[(α + γ )/2] . (3.44)

In this case the two rotations are about the z-axis so that A = α+γ . Equation (3.44)agrees with Eq. (3.41) when A is replaced by α + γ . If α = γ = 0, then

χ(1/2)(0, β, 0) = 2 cos(β/2) = sin(β)

sin(β/2), (3.45)

which agrees with (3.42) for A = β.The matrices for half-integral j ( j = n+ 1/2, n = 0, 1, 2, . . .) are double-valued

in that D(n+1/2)(R+2π) = −D(n+1/2)(R), but D(n+1/2)(R+4π) = +D(n+1/2)(R),where R is any rotation. These matrices represent rotations of spinor objects suchas the spin of an electron or neutron or any total half-integer momentum vector j ,for which a rotation by 4π is required in order to return the object to its originalstate.

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3.8 General representations of the full rotation group 87

Table 3.6 IRs of the full rotation group. Note that the integral representations canbe expressed in terms of the angle β, while the half-integral representations aredependent on the angle β/2 and hence are double-valued.

• D(0) = 1• D(1/2)(α, β, γ )m′m = e−im′αe−imγ M (1/2)(β)m′m

M (1/2)(β)m′m =m = 1

2 m = − 12 m′

C −S 12

S C − 12

with C = cos(β/2), S = sin(β/2)

• D(1)(α, β, γ )m′m = e−im′αe−imγ M (1)(β)m′m

M (1)(β)m′m =

m = 1 m = 0 m = −1 m′

12 (1+ C) −(1/√2)S 1

2 (1− C) 1

(1/√

2)S C −(1/√2)S 0

12 (1− C) (1/

√2)S 1

2 (1+ C) −1

with C = cosβ, S = sinβ

• D(3/2)(α, β, γ )m′m = e−im′αe−imγ M (3/2)(β)m′m

M (3/2)(β)m′m =

m = 32 m = 1

2 m = − 12 m = − 3

2 m′

C3 −√3C2S√

3C S2 −S3 32√

3C2S (C3 − 2C S2) (S3 − 2C2S)√

3C S2 12√

3C S2 −(S3 − 2C2S) (C3 − 2C S2) −√3C2S − 12

S3√

3C S2√

3C2S C3 − 32

with C = cos(β/2), S = sin(β/2)

• D(2)(α, β, γ )m′m = e−im′αe−imγ M (2)(β)m′mM (2)(β)m′m =

m = 2 m = 1 m = 0 m = −1 m = −2 m′

14 (1+ C)2 − 1

2 (1+ C)S 12

√32 S2 − 1

2 (1− C)S 14 (1− C)2 2

12 (1+ C)S 1

4 [(1+ C)2 − 3S2] −√

32 C S − 1

4 [(1− C)2 − 3S2] − 12 (1− C)S 1

12

√32 S2

√32 C S 1

2 (3C2 − 1) −√

32 C S 1

2

√32 S2 0

12 (1− C)S − 1

4 [(1− C)2 − 3S2]√

32 C S 1

4 [(1+ C)2 − 3S2] − 12 (1+ C)S −1

14 (1− C)2 1

2 (1− C)S 12

√32 S2 1

2 (1+ C)S 14 (1+ C)2 −2

with C = cosβ, S = sinβ

The D(l)(α, β, γ ) matrices assume the phase convention Y |m|l = (−1)m Y−|m|l used inTable 3.1. D(l)(α, β, γ ) acting on Y m

l produces a clockwise (positive) rotation of thefunction in a fixed coordinate system.

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88 Spherical symmetry and the full rotation group

References

[3.1] E. P. Wigner, Group Theory and its Application to the Quantum Mechanics ofAtomic Spectra (New York: Academic Press, 1959).

[3.2] M. Tinkham, Group Theory and Quantum Mechanics (New York: McGraw-Hill,1969).

Exercises

3.1 Express the functions listed below in terms of dx2 and dz2 orbitals defined inTable 3.2:(a) (3x2 − r2)/r2,(b) (2y2 − x2 − z2)/r2,(c) (z2 − y2)/r2.

3.2 (a) Use Table 3.1 to show for the l = 1 spherical harmonics that

Y 01 =

√3

z

r,

Y 11 = −

√3

x + iy

r,

Y−11 =

√3

x − iy

r,√

3

x

r= − 1√

2

(Y 1

1 − Y−11

),√

3

y

r= i√

2

(Y 1

1 + Y−11

),√

3

z

r= Y 0

1 .

(b) Find the matrix elements of 120◦ rotation about an axis along the (1, 1, 1)direction for a representation based on the l = 1 spherical harmonics.

3.3 Show that the real part of Y 22 is a basis function for the Eg and that the

imaginary part is a basis function for the T2g for the Oh group.3.4 Every group has a totally symmetric IR wherein every element is repre-

sented by unity. The totally symmetric IR is usually named �A1 or simplyA1. Assume the characters are real and show that �α × �β contains the A1

representation if and only if α = β.3.5 Show that the requirement that �α×�β contain �δ is equivalent to the require-

ment that �α × �β × �δ contains the A1 representation for groups for whichthe characters are real.

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Exercises 89

3.6 Determine the degeneracies for l = 0, 1, and 2 of the spherical harmonicfunctions, Y m

l , in an environment whose symmetry is D4. The character tablefor D4 is given below.

D4 E C2 = C24 2C4 2C ′2 2C ′′2 Basis functions

A1 1 1 1 1 1 x2 + y2, z2

A2 1 1 1 −1 −1 Rz , zB1 1 1 −1 1 −1 x2 − y2

B2 1 1 −1 −1 1 xy

E 2 −2 0 0 0 (xz, yz)

{(x, y)(Rx , Ry)

3.7 Decompose the direct product of D(2) × D(1) into the IRs of D4.3.8 To what representation of Oh does the operator ∇2 = ∂2/∂x2 + ∂2/∂y2 +

∂2/∂z2 belong?3.9 (a) Use the symmetry-function-generating machine to find the three partner

functions of the T1(O) representation starting with the function (x3 +y3 + z3). The matrix elements of the T1 IR are given in Appendix E.

(b) To which IR of Oh do the partner functions belong?3.10 (a) The electric-dipole operator is a polar vector e R = e(X, Y, Z) in Carte-

sian coordinates. Find the matrix that rotates R by an angle α about thez-axis.

(b) Find the matrix that rotates R by an angle α about the z-axis and subjectsthe vector to the inversion operation.

(c) What is the trace (character) of the combined rotation–inversion matrix?(d) To what IR of Oh does R belong?

3.11 The magnetic-dipole operator is an axial vector μ ∝ R × dR/dt (vectorcross product) Since μ is a vector (a pseudo-vector) the rotation matrix ofExercise 3.10 applies.(a) What is the combined rotation–inversion matrix for μ?(b) What is the character of the matrix?(c) To what IR of Oh does μ belong?

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4

Crystal-field theory

4.1 Splitting of d-orbital degeneracy by a crystal field

As we saw in the previous chapter the s- and p-orbital degeneracies are unaf-fected when an atom or ion is placed in a site of octahedral symmetry. On theother hand, the d- and f-orbital degeneracies are changed. Transition metal ions ofTi, Fe, Ni, and Co, for example, have 3d electrons in their outer unfilled shellsand exist as positively charged ions in solids and molecular complexes. Mostfrequently, the transition metal ion is coordinated with six neighboring ligandsat a site of octahedral symmetry. The second most common situation is tetrahe-dral coordination with four neighboring ligands. Many of these transition metalsolids and molecular complexes are colored and many are magnetic. The col-ors are attributed to vibronic (electronic plus vibration) transitions between thed-orbital groups that are split in energy by the non-spherical potential of theligands. When the ligand orbitals are included in determining the splitting, the pro-cedure is called ligand-field theory. Splitting due to adjacent ligands is discussed inChapter 6.

Crystal-field theory was developed by Bethe [4.1] and Van Vleck [4.2] to explainthe optical spectra of transition metal complexes and to understand their magneticproperties. In its simplest form the crystal-field model represents the ligands sur-rounding a metal ion as point charges that interact with the transition metal iononly through an electrostatic potential. This is, of course, a very crude approxima-tion, since the orbitals of the ligands usually form electronic bonds with the metalion’s d-electrons. However, the type of splitting that occurs depends only on thesymmetry of the environment. Therefore, crystal-field theory can be applied as asemi-empirical model with the strength of the interaction taken as an adjustableparameter to be determined by experiment.

90

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4.1 Splitting of d-orbital degeneracy by a crystal field 91

4.1.1 Splitting of the d1 configuration

As an example, consider a metal ion with a single electron in its d-electron shellin a site of octahedral symmetry. The ion could, for example, be in an octahedralsite in a metal oxide such as n-type SrTiO3 or an ion in a chemical complex suchas [Ti(H2O)6]3+.

The crystal-field Hamiltonian for a d-electron in the octahedral environment is

Hoct = H0 + Voct, (4.1)

H0 = − �2

2me∇2 − e2 Z

r+ Veff(r), r = |r|, (4.2)

where H0 is the Hamiltonian for a spherically symmetric environment and Voct isthe octahedrally symmetric point-charge potential. The arrangement of the point-charges shown in Fig. 4.1(a) produces an electrostatic potential given by

Voct(r) = e2 Z ligand

6∑j=1

1

|r− Rj| , (4.3)

where eZ ligand is the effective negative point charge representing one of the ligands,e is the electron charge, and Rj is the vector from the d-ion to the j th point charge.If the potential is expanded in a Taylor’s series (for |r| < t, see Fig. 4.1(a)) thelowest-order, non-spherically symmetric term is

Voct = D

(x4 + y4 + z4 − 3

5r4

), (4.4)

D = 35

4

e2 Z ligand

t5, (4.5)

where |Rj| = t is the distance between the metal ion and any one of the equidistantligand charges as shown in Fig. 4.1(a). There are higher-order terms in the expan-sion of Voct, but they make no contribution to the energy splitting of the d-orbitals(see Exercise 4.1).

We have already established in Chapter 3 that the d-orbitals of Table 4.1 areeigenstates both of the spherically symmetric Hamiltonian, H0, and of Hoct = H0+Voct. The eigenstates are degenerate in the absence of Voct, so∫ ∫ ∫

ψ∗d H0 ψd r2 sin θ dθ dφ dr = E0, (4.6)

where E0 is the energy due to the spherically symmetric Hamiltonian and ψd isany one of the d-orbitals listed in Table 4.1. We can calculate the relative splitting

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92 Crystal-field theory

(a)

t

Zle

Zle

Zle Zle

Zle

Zle

(b) d1 or d6

E0

eg

t2g

Δc

35Δc

25Δc

(c) d4 or d9

E0

eg

t2g

Δc

25Δc

35Δc

Figure 4.1 (a) The crystal-field model. The six ligands have charge eZ ligand (writ-ten Z le in part (a)) and each is a distance t from the central ion. (b) Splittingof a d1 configuration term into two-fold-degenerate Eg states and three-fold-degenerate T2g states. The splitting is �c = 10Dq. E0 is the energy in theabsence of the crystal-field potential. (c) Splitting for the d4 or d9 configurationcorresponding to a hole in a closed-shell configuration.

of the T2g and Eg energy levels due to the octahedral potential of (4.4). For the dxz

state the crystal-field energy is given by

�(T2g) =∫ ∫ ∫

d∗xz D

(x4 + y4 + z4 − 3

5r4

)dxz r2 sin θ dθ dφ dr

= DIrad

∫ 2π

0dφ

∫ π

0sin θ dθ

15

4π(Sθ Cφ Cθ )

2

×{[

S4θ C4

φ + S4θ S4

φ + C4θ

]− 3

5

}

= DIrad

(11

21− 3

5

)= −DIrad

8

105, (4.7)

where

Irad =∫ ∞

0r4|Rnd(r)|2r2 dr = 〈r4〉avg (4.8)

is the average value of r4 for the d-orbitals. The same result is obtained for the dxy

and dyz orbitals.For the dz2 orbital we have

�(Eg) =∫ ∫ ∫

d∗z2 D

(x4 + y4 + z4 − 3

5r4

)dz2 r2 sin θ dθ dφ dr

= DIrad

∫ 2π

0dφ

∫ π

0sin θ dθ

5

16π(3C2

θ − 1)2

{[S4θ C4

φ + S4θ S4

φ + C4θ

]− 3

5

}

= DIrad

(15

21− 3

5

)= +DIrad

12

105. (4.9)

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4.1 Splitting of d-orbital degeneracy by a crystal field 93

Table 4.1 The d-orbitals, ψd , are degenerate eigenstatesof the hydrogen-like Hamiltonian, H0. The Eg states (dz2

and dx2) and the T2g states (dxy, dxz, and dyz) are alsoeigenstates of Hoct, but they have different energies.Rn2(r) is the radial part of the wavefunction for principalquantum number n and angular momentum l = 2.

Eg orbitals T2g orbitals

dz2 =√

5

16π

3z2 − r2

r2Rn2(r) dxy =

√15

xy

r2Rn2(r)

dx2 =√

15

16π

x2 − y2

r2Rn2(r) dxz =

√15

xz

r2Rn2(r)

dyz =√

15

yz

r2Rn2(r)

The same result is obtained for dx2 . The total splitting is usually defined as �c or10Dq. It is given by

10Dq = �(Eg)−�(T2g) = 10D

{Irad

2

105

}, (4.10)

from which it follows that �(Eg) = 6Dq and �(T2g) = −4Dq, with q =(2/105)Irad = (2/105)〈r4〉avg. The splitting is shown schematically in Fig. 4.1(b).The Eg states are higher in energy than the T2g states because the maximum chargedensity of the Eg orbitals, roughly speaking, “points” directly into the repulsivecrystal-field charges, whereas the maximum charge density of the T2g orbitalspoints between the crystal-field charges.

We can make a crude estimate of 10Dq for a Ti3+ ion in an octahedral crystalfield. Ti3+ has a single d-electron outside its argon closed-shell configuration. Fora hydrogen-like 3d orbital

〈r4〉avg = 336

(3a0

2Znucleus

)3

〈r〉avg, (4.11)

where a0 = 0.53 Å is the Bohr radius. Using Znucleus = 3, (4.11) gives 〈r4〉avg

approximately equal to 6.25〈r〉avg. If we take 〈r〉avg to be the ionic radius for Ti3+

(0.67 Å) then we have 〈r4〉avg approximately equal to 4.19 Å4. We assume t = 2 Å(a typical metal–oxygen distance) and, because of charge neutrality, Z ligand = 1/2(so that 6eZ ligand = 3 electron charges). Using all of these estimates we have

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94 Crystal-field theory

10Dq = 10

(2

105

)(35

4

)e2 Z ligand

t5〈r4〉avg

= 5

3(14.3942 eV Å)

1

2

4.19

32= 1.57 eV = 12,668 cm−1.

The experimentally observed absorption peak for the complex [Ti(H2O)6]3+ occursat about 20,300 cm−1. So our crude estimate is of the correct order of magnitudebut off by nearly a factor of two. Typical values of 10Dq for ions in solution rangefrom 6,000 to 30,000 cm−1.

Visible light has a wavelength ranging roughly from about 380 nm to 750 nm orenergy of 1.65 to 3.26 electron volts or 13,333 to 26,300 cm−1. The crystal-fieldsplitting of the electron levels of transition metal complexes often lies within thisrange. Therefore these complexes absorb part of the white-light spectrum, causingthe material (in a solid or in solution) to appear colored.

It should be noted that direct transitions among the d-levels are forbiddenbecause the selection rules for electric-dipole transitions require the initial and finalstates to have opposite parity. Since all d-levels are “g”, a transition between twod-levels is forbidden. The actual transitions are likely vibronic, meaning that bothelectron and vibrational states are involved. The parity of the initial or final stateis the product of the parity of the electronic state times the parity of the vibra-tional state. Since the d-levels are all “g”, this type of transition requires that thevibrational mode be “u” (antisymmetric under the inversion operation).

4.1.2 Filled and half-filled atomic shells

If each of the (2l + 1) hydrogen-like states are occupied by one electron the totalcharge density, ρ, is spherically symmetric. For example, for the d-states

ρ(d5, r, θ, φ) = e|Rn2(r)|2{|Y 02 |2 + |Y 1

2 |2 + |Y−12 |2 + |Y 2

2 |2 + |Y−22 |2}

= e|Rn2(r)|2{|dxy|2 + |dxz|2 + |dyz|2 + |dz2 |2 + |dx2 |2}= e|Rn2(r)|2 5

4π. (4.12)

A half-filled shell (d5 with one electron in each of the different d-orbitals) hasspherically symmetric charge density and so does a completely filled shell (d10

with each of the two spin states occupied). If there are four electrons occupyingfour of the five different d-orbitals, the charge density will be

ρ(d4, r, θ, φ) = ρ(d5, r, θ, φ)− e|Rn2(r)|2|dα|2, (4.13)

where dα is the empty d-orbital.

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4.2 Multi-electron systems 95

The spherically symmetric charge density, ρ(d5, r, θ, φ), can be incorporatedinto the effective potential, Veff, of the Hamiltonian. The non-spherically symmet-ric charge is the same as that of a single electron in the dα orbit except the charge is−e (positive). Therefore, under the operations of the rotation group the symmetryproperties of a half-filled shell minus one electron (d4) or a completely filled shellminus one electron (d9) are the same as those of a single electron with a positivecharge (i.e., a hole). Similarly, considering only the orbital momentum, the sym-metry properties of a half-filled shell plus one additional electron (d6) are the sameas those of a single electron.

Because the effective charges of the d4 and d9 hole states are positive, the split-ting is reversed from the d1 case. For d4 and d9 the Eg hole states lie below theT2g states, as illustrated in Fig. 4.1(c). The lowest energy state is that with thehole occupying an Eg orbital. From the electron point of view this is equivalent toremoving one electron from the upper Eg state in Fig. 4.1(b).

4.2 Multi-electron systems

The preceding analysis of crystal-field splitting applies to the states of a singled-electron or single d-hole. For a multi-electron system the electronic states arecharacterized by several schemes that specify how the orbital and spin momentashould be combined. For example, for light atoms (Z < 30 or so) the spin–orbitinteraction is small compared with the crystal-field splitting (10% or so) and theL–S coupling scheme can be employed. With the L–S scheme the individual orbitalmomenta are combined to form L and the spin momenta are combined to form S.For heavier atoms, particularly for the rare earths, where the spin–orbit interac-tion is larger, j–j coupling should be employed. In this case the orbital and spinmomenta of the kth electron are added to form jk. The total angular momentum isJ =∑

jk.For a multi-electron atom or ion the electronic configuration is specified by

indicating which types of atomic orbitals are occupied. For example, for two3d-electrons outside of closed shells the configuration would be indicated as1s2 2s2 2p6 3s2 3p6 3d2. From a group-theoretical point of view, the filled shellsare unimportant because they have spherically symmetric charge densities and areinvariant under all of the operations of the rotation group. The outer electrons deter-mine the symmetry properties and therefore the configuration is abbreviated as 3d2

or just d2 if the principal quantum number is unimportant. Given a particular con-figuration, the vector model from atomic physics can be used to determine how toadd the orbital and spin momenta of the two electrons to form different terms. WithL–S coupling of two d-electrons the orbital angular momentum, ML =∑

mi , andthe spin, S = ∑

si , are obtained by assigning the two electrons to any of the

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96 Crystal-field theory

m = 0,±1,±2 d-states consistent with the Pauli exclusion principle. In this case,ML can be 0, ±1, ±2, ±3, or ±4. There may be more than one assignment thatresults in the same value of ML . For example, m1 = 2 and m2 = 0 have the sameML as m1 = 1 and m2 = 1. The eigenstates are formed from a linear combinationof states with the same ML determined by the requirement that the eigenvalue ofM2

L = L(L + 1). The possible values of L can be determined by the decomposi-tion of the direct product: D(2) × D(2) = D(0) + D(1) + D(2) + D(3) + D(4) (seeEq. (4.18) below). Thus two d-electrons can form S, P , D, F , and G terms. Theassignment of spins to the various one-electron d-states must be done in such amanner that the total wavefunction is antisymmetric under the interchange of theelectron and spin coordinates of the two electrons. For terms with even L (0, 2, 4,. . . or S, D, G, . . .) the spatial function is symmetric under the interchange of thespatial coordinates and therefore the spin function must be antisymmetric. For twoelectrons that requires zero total spin, S = s1 + s2 = 0 (antisymmetric spin state).Conversely, for odd L (1, 3, 5, . . . or P , F , H , . . .) the spatial function is anti-symmetric, so the spin function must be symmetric. For two electrons that requiresS = 1. The convention is to label a term with a left-hand superscript that gives thespin multiplicity, 2S + 1. For the d2 configuration the terms appear as 1S, 3 P , 1 D,3 F , and 1G. A further specification of the term is given by placing the value of Jas a right-hand subscript. The complete identification is then (2S+1)L J . The valueof J can range from |ML − S| to |ML + S| in integer steps.

In the following sections we examine how the ground-state term of a d2

configuration is split by an octahedral crystal field.

4.2.1 Crystal-field splitting for a d2 configuration

In a spherically symmetric environment Y ml (r) is a basis function for the D(l)

IR. The orbitals for a two-electron wavefunction involve the products of the typeY m

l (r) Y m′l ′ (r

′). For such products there are 2l+1 functions of r and 2l ′+1 functionsof r ′ and therefore a total of (2l+1)(2l ′+1) products. These product functions formthe basis for the direct-product representation, D(l)(R)×D(l ′)(R), and therefore thecharacters of the direct-product representation are the products of the characters ofthe two representations D(l)(R) and D(l ′)(R) (see Theorem 3.2c).

However, it is important to recognize that the covering group for the Hamiltonianof a two-electron (or any many-electron) system is not the direct-product group.The Hamiltonian contains electrostatic interactions of the form e2/|r − r ′| andtherefore is invariant only under operations that are applied to both coordinates ofthe product basis functions. The Hamiltonian is invariant only under the diagonalelements of the direct-product group, PRi (r) RRi (r

′). An off-diagonal element suchas PRi (r) RR j (r

′), with i unequal to j , is an operation under which e2/|r − r ′| is

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4.2 Multi-electron systems 97

not invariant. The group of the diagonal elements is in fact just the group of thecovering operations of the Hamiltonian. The direct-product representation D(l) ×D(l ′) is a reducible representation of the group of the Hamiltonian even if it is adirect product of the one-electron IRs.

With the understanding that an operation R is applied to all of the relevant coor-dinates (r and r ′), we shall write PR f (r) g(r ′) to mean PR f (r) PR g(r ′). Thecharacter of the direct-product representation matrix is the product of the charac-

ters of the two representations, so χ(D(l)×D(l′))(R) ≡ χ(l×l ′)(R) = χ(l)(R) χ(l ′)(R).The character of a rotation by an angle A about any axis is the same as the characterof a rotation about the z-axis through an angle A and therefore

χ(l×l ′)(A) =+l∑

m=−l

e−im A+l ′∑

m′=−l ′e−im′A =

l+l ′∑k=|l−l ′|

{ +k∑M=−k

e−i M A

}

=l+l ′∑

k=|l−l ′|χ(k)(A). (4.14)

Equation (4.14) tells us that the character for a rotation through an angle of Aof the direct-product representation matrix is the sum of characters of D(k)(A)from k = |l − l ′| to (l + l ′) (see Exercise 4.5). This result suggests that the directproduct decomposes into a sum of D(l)s. In fact, the decomposition is as shown inTheorem 4.1.

Theorem 4.1 Decomposition of the direct product into the IRs of the rotationgroup:

D(l) × D(l ′) =l+l ′∑

k=|l−l ′|D(k). (GT4.1)

Some examples of the decomposition of the direct products of the IRs of therotation group are

D(0) × D(l ′) = D(l ′), (4.15)

D(1) × D(1) = D(0) + D(1) + D(2), (4.16)

D(1) × D(2) = D(1) + D(0) + D(2) + D(3), (4.17)

D(2) × D(2) = D(0) + D(1) + D(2) + D(3) + D(4). (4.18)

For integral values of l, D(l)(R) and D(l)(R)∗ have the same basis functions and thesame characters. The effect of the complex conjugation is simply to interchange themth-row and −mth-row functions. Therefore, χ(l)(R)∗ = χ(l)(R), so the characterof D(l)(R)∗ × D(l ′)(R) = χ(l)(R) χ(l ′)(R). As a consequence, the decompositionof D(l)∗ × D(l ′) is the same as that of D(l) × D(l ′).

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98 Crystal-field theory

According to Theorem 4.1 each IR on the right-hand side appears only once.Therefore, the two-electron basis functions, Y m

l Y m′l ′ , for these IRs will be eigen-

functions of a hydrogen-like Hamiltonian H(r, r ′) = H0(r)+H0(r ′), where H0 isgiven by (4.2).

4.2.2 The group-theoretical basis for the vector model of atomic physics

The results of Theorem 4.1 serve as a theoretical basis for the vector modelused in atomic physics to determine the possible spectroscopic terms. Con-sider an electron in an np-orbital and a second electron in an n′ p-orbital(n �= n′). Each orbital transforms according to D(1). The multi-electron wave-function (a Slater determinant, for example) will contain products of the typeRn1(r) Y m

1 (r, θ, φ) Rn′1(r ′) Y m1 (r ′, θ ′, φ′), where Rn1 is the radial part of the orbital.

These products serve as basis functions for the direct product D(1) × D(1). Equa-tion (GT4.1) specifies the angular-momentum terms that result when the directproduct is reduced to the IRs of the group of the Hamiltonian. We have D(1) ×D(1) = D(0) + D(1) + D(2), so the two p-electrons can form S, P , and D spec-troscopic terms. The spin states are determined by D(1/2) × D(1/2) = D(0) + D(1),which corresponds to a singlet (S = 0) and a triplet (S = 1). Therefore the possibleterms are 1S, 1 P , 1 D, 3S, 3 P , and 3 D.

The total wavefunction must be antisymmetric under the interchange of spaceand spin coordinates of the two electrons. The spin-triplet states are symmet-ric and the spin-singlet state is antisymmetric under the interchange of the spinvariables. Therefore, the 3S, 3 P , and 3 D states must have antisymmetric spa-tial wavefunctions, and the 1S, 1 P , and 1 D states must have symmetrical spatialwavefunctions.

In the case of two equivalent p electrons (n = n′) only the 3 P , 1S, and 1 D termsare permitted by the Pauli exclusion principle (see Chapter 5 for more detail).

Hund’s rules for determining the ground state

1. Of the terms arising from the electron configuration, the term with maximum spinmultiplicity, 2S + 1 (

∑si = S), lies lowest in energy.

2. Among the terms of the same spin multiplicity, the term with the largest-valueangular momentum, ML =∑

m′i (consistent with the Pauli exclusion principal),has the lowest energy.

3. For a given term, if the outer sub-shell is half-filled or less, then the level with thesmallest J lies lowest in energy. If the sub-shell is more than half-filled, then thelevel with the largest J lies lowest. (J takes on the integer values from |ML − S|to |ML + S|.)

Note: These rules apply only to the ground state and assume that the repulsionbetween the electrons is much greater than the spin–orbit interaction.

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4.2 Multi-electron systems 99

For the case of a d2 configuration we use the Russell–Saunders coupling so ML

is the sum of the m-values, and S is the sum of the spins. The possible termsresulting from two d-electrons are S, P , D, F , and G, corresponding to the D(0),D(1), D(2), D(3), and D(4) IRs on the right-hand side of (4.18). We can determinethe ground state using Hund’s rules. We have five d-orbital states, which can befilled with either an α (spin 1/2) or a β (spin −1/2) or two electrons in the samespatial orbital with spin α or β. However, the Pauli exclusion principle requiresthat no two electrons can have the same orbital and spin states. For two d-electronsoutside of a closed shell, Hund’s rules require the electrons to be assigned (a) togive the maximum value for the sum of the spins and then (b) the maximum forthe sum of the mi values. Therefore the ground state has the configuration (m = 2,s = α)(m ′ = 1, s ′ = α). With this configuration we have ML = ∑

mi = 3and S = ∑

si = 1. According to Hund’s rules the value of J to be assigned tothis configuration is |ML − S| = 3 − 1 = 2 since the d-shell is less than half-filled. Therefore the ground-state term is (2S+1)FJ = 3 F2. The F term will have(2ML + 1) = 7 states that are degenerate in a spherical environment.

We want to determine how the orbital degeneracy of this seven-fold-degenerateground state is affected when the system is in an octahedral crystal field. Theexpected type of splitting can immediately be determined by decomposing theF(D(3)) term into the IRs of the O or Oh groups. Since the basis functions ofthe direct product D(2) × D(2) are products of functions that are symmetric underinversion, the decomposition must be into “g” IRs. In the octahedral environmentD(3) = A2+ T1+ T2, or, under Oh , D(3) = A2g+ T1g+ T2g. Therefore, we see thatthe seven-fold degeneracy of the F ground state splits into a non-degenerate A2g

and two triply degenerate groups, T1g and T2g. Note that the one-electron f-orbitalsare antisymmetric under inversion, so in an octahedral environment D(3) decom-poses into A2u + T1u + T2u (Eq. (3.38)). Here the basis functions are symmetricunder inversion and hence the decomposition is into “g” states. To determine thedetails of these splittings we must first find the appropriate two-electron basis func-tions and then calculate the matrix elements of the crystal-field potential betweenthese states.

4.2.3 Basis functions for the ground state

Our task is to find the basis functions for the 3 F ground state of two electrons.The basis functions should be eigenfunctions of the operator M2

L with eigenvalue3(3 + 1) (F state with L = 3), and the eigenvalues of ML = ∑

mi are −3, −2,−1, 0, 1, 2, and 3, corresponding to the (2ML + 1) = 7 partner functions for anF(D(3)) representation.

We can use d-orbital Slater determinants to construct states that are antisymmet-ric for the interchange of space and spin variables. We define

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100 Crystal-field theory

|m, α;m ′, α′〉 =√

1

2det

[�m(r) α �m(r ′) α′

�m′(r) α �m′(r ′) α′

],

where �m(r) = Rd(r) Y m2 (r, θ, φ) and �m′(r ′) = Rd(r ′) Y m′

2 (r ′, θ ′, φ′). Expansionof the determinant gives

|m, α;m ′, α′〉 =√

1

2Rd(r) Rd(r

′)

×[Y m

2 (r, θ, φ) Y m′2 (r ′, θ ′, φ′)− Y m′

2 (r, θ, φ) Y m2 (r ′, θ ′, φ′)

]αα′

= |m,m ′〉αα′, (4.19)

where |m,m ′〉 is the determinant of the spatial functions.The two-electron state, |m,m ′〉αα′, is “g” under the inversion operation and is

antisymmetric in the interchange of r, α with r′, α′.1 The aim is now to find linearcombinations of Slater determinants with same ML that are also eigenstates of M2

L .Consider the determinantal states shown in Table 4.2. These functions have

eigenvalues of ML = 3, 2, 1, 0, −1, −2, and −3, but are not all eigenfunctionsof M2

L . To find the effect of M2L on the states in Table 4.2, we can use the following

operator result for two-electron states:2

M2L |m,m ′〉 = [l(l + 1)+ l ′(l ′ + 1)+ 2mm ′]|m,m ′〉

+ [(l − m)(l + m + 1)(l ′ + m ′)(l ′ − m ′ + 1)]1/2|m + 1,m ′ − 1〉+ [(l + m)(l − m + 1)(l ′ − m ′)(l ′ + m ′ + 1)]1/2|m − 1,m ′ + 1〉,

(4.20)

where the angular-momentum quantum numbers are l and l ′, and the magneticquantum numbers are m and m ′. For two d-electrons (l = 2 and l ′ = 2) (4.20)reduces to

M2L |m,m ′〉 = [12+ 2mm ′]|m,m ′〉

+ {[6− m(m + 1)][6− m ′(m ′ − 1)]}1/2|m + 1,m ′ − 1〉+ {[6− m(m − 1)][6− m ′(m ′ + 1)]}1/2|m − 1,m ′ + 1〉.

(4.21)

1 With three or more electrons the states can not be described by the product of a single spatial function times asingle spin function (see Exercise 4.10).

2 For the general operator formulas see Chapter 5 or references [4.3, 4.4].

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4.2 Multi-electron systems 101

Table 4.2 Slater determinant states. The first column is the Slater determinant.The second column is ML =∑

mi . The third column is the eigenvalue of M2L . If

no eigenvalue is given the state is not an eigenstate of M2L . The last column lists

the states that are coupled by the M2L operator (4.21). Linear combinations of the

coupled states must be used to construct eigenstates of M2L .

Slater determinant |m,m′〉 ML = m + m′ M2L Coupled by M2

L operator

|1, 2〉 3 3(3+ 1)|2, 0〉 2 3(3+ 1)|2,−1〉 1 |1, 0〉|1, 0〉 1 |2,−1〉|1,−1〉 0 |2,−2〉|2,−2〉 0 |1,−1〉|1,−2〉 −1 | −1, 0〉|−1, 0〉 −1 |1,−2〉|−2, 0〉 −2 3(3+ 1)|−2,−1〉 −3 3(3+ 1)

In calculating M2L |m,m ′〉 note that |m,m〉 = 0 (two electrons in identical

states) and |m ′,m〉 = −|m,m ′〉 (a property of an antisymmetric wavefunction).For example, for the state |2, 1〉 we have

M2L |2, 1〉 = (12+ 4)|2, 1〉 + 4|1, 2〉

= 16|2, 1〉 − 4|2, 1〉 = 3(3+ 1)|2, 1〉. (4.22)

Therefore |2, 1〉 is an eigenstate of M2L with eigenvalue L(L + 1) = 3(3+ 1). For

|2, 0〉 we have

M2L |2, 0〉 = (12+ 0)|2, 0〉 + 4|1, 1〉 = 12|2, 0〉 = 3(3+ 1)|2, 0〉. (4.23)

The first two states and the last two states of Table 4.2 are suitable basis functionsbecause they are eigenstates of M2

L . However, if we calculate M2L |2,−1〉, we find

that it is not an eigenstate, but is coupled to the state |1, 0〉,M2

L |2,−1〉 = 8|2,−1〉 + 2√

6|1, 0〉, (4.24)

M2L |1, 0〉 = 6|1, 0〉 + 2

√6|2,−1〉. (4.25)

In order to achieve basis functions that are eigenstates of M2L we must take linear

combinations of the coupled states. Consider the coupled states |2,−1〉 and |1, 0〉.We take the linear combination, c|2,−1〉 + d|1, 0〉, and require that

M2L[c|2,−1〉 + d|1, 0〉] = λ[c|2,−1〉 + d|1, 0〉]. (4.26)

Using (4.24) and (4.25) in (4.26), we obtain

[(8− λ)c + 2√

6d]|2,−1〉 + [2√6c + (6− λ)d]|1, 0〉 = 0, (4.27)

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102 Crystal-field theory

or, written as an eigenvalue matrix equation,(8− λ 2

√6

2√

6 6− λ

)(cd

)= 0. (4.28)

The vanishing of the determinant requires that

(8− λ)(6− λ)− 24 = 0, (4.29)

or

λ2 − 14λ+ 24 = (λ− 2)(λ− 12) = 0. (4.30)

The two eigenvalues are obviously λ = 2 = 1(1+ 1) and λ = 12 = 3(3+ 1). Thefirst eigenvalue leads to a linear combination associated with a P term. The secondeigenvalue corresponds to an F term and is the desired result. The coefficients aredetermined by substituting the eigenvalue into (4.28) and finding the ratio of c to d.This gives c/d = √6/2 and the normalized basis function

(3/5)1/2|2,−1〉 + (2/5)1/2|1, 0〉. (4.31)

For the coupled states |1,−1〉 and |2,−2〉, we have

M2L |1,−1〉 = 10|1,−1〉 + 4|2,−2〉, (4.32)

M2L |2,−2〉 = 4|2,−2〉 + 4|1,−1〉. (4.33)

On taking a linear combination of the two states we find that the state withM2

L = 3(3+ 1) is

(4/5)1/2|1,−1〉 + (1/5)1/2|2,−2〉. (4.34)

Following the above procedure for the coupled states gives us the set of sevenfunctions shown in Table 4.3.

Table 4.3 Basis functions based on linear combinations of Slater determinants.The notation �(mL) is defined in the first column.

Eigenfunctions of ML and M2L Eigenvalue of ML Eigenvalue of M2

L

�(3) = |2, 1〉 3 3(3+ 1)�(2) = |2, 0〉 2 3(3+ 1)�(1) = (2/5)1/2|1, 0〉 + (3/5)1/2|2,−1〉 1 3(3+ 1)�(0) = (4/5)1/2|1,−1〉 + (1/5)1/2|2,−2〉 0 3(3+ 1)�(−1) = (2/5)1/2|−1, 0〉 + (3/5)1/2|−2, 1〉 −1 3(3+ 1)�(−2) = |−2, 0〉 −2 3(3+ 1)�(−3) = |−2,−1〉 −3 3(3+ 1)

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4.2 Multi-electron systems 103

In Table 4.3 we have introduced a simplified notation for the states �(ML),ML = 3, 2, 1, 0, −1, −2, and −3.

Our next task is to calculate the matrix elements of the crystal-field potentialbetween the basis functions in Table 4.3. The Hamiltonian for the two-electronsystem (ignoring for the moment the electron–electron interactions) is

H(r, r ′) = H0(r)+H0(r ′)+ Voct(r)+ Voct(r′), (4.35)

where

H0 = − �2

2me∇2 − e2 Z

r+ Veff(r), (4.36)

Voct(r) = D

(x4 + y4 + z4 − 3

5r4

), (4.37)

D = 35

4

e2 Z ligand

t5. (4.38)

Veff(r) includes any effective spherical potential due to the filled-shell electrons, butexcludes the crystal-field potential and the electron–electron interactions betweenthe d-electrons. The two-electron states listed in Table 4.3 are composed of lin-ear combinations of products such as Y m

2 (r) Y m′2 (r ′) and are therefore degenerate

eigenstates of H 0(r, r ′) = H0(r)+H0(r ′). That is,

H0(r, r ′)�(ML) = 2E0 �(ML). (4.39)

As a consequence, we need only calculate the matrix elements of the crystal-fieldpotential V (r, r ′)oct = Voct(r)+ Voct(r ′) between the various �(ML) states.

4.2.4 Calculation of the matrix elements of the crystal field for d2

Since V (r, r ′)oct = Voct(r) + Voct(r ′) is a sum of one-electron operators (there areno terms involving both r and r ′), it follows that

〈m,m ′|V (r, r ′)oct|n, n′〉 = 〈m|V (r)|n〉δm′,n′ + 〈m ′|V (r ′)|n′〉δm,n

− 〈m ′|V (r)|n〉δm,n′ − 〈m|V (r)|n′〉δm′,n, (4.40)

where 〈m| is the one-electron wavefunction, Rd(r) Y m2 (r, θ, φ). Using (4.40) for

�(3), for example, gives

〈�(3)|Voct(r)+ Voct(r′)|�(3)〉 = 〈2, 1|Voct(r)+ Voct(r

′)|2, 1〉= 〈2|Voct(r)|2〉 + 〈1|Voct(r)|1〉. (4.41)

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104 Crystal-field theory

Table 4.4 One-electron integrals: 〈m|Voct(r)|m′〉 in units of Dq

m′

m 2 1 0 −1 −2

2 1 0 0 0 51 0 −4 0 0 00 0 0 6 0 0−1 0 0 0 −4 0−2 5 0 0 0 1

Written out in detail,

〈�(3)|Voct(r)+ Voct(r′)|�(3)〉 = D

∫|Rn2(r)|2r2 dr

∫dφ

∫dθ sin θ

×{|Y 22 |2 + |Y 1

2 |2}(

x4 + y4 + z4 − 3

5r4

).

(4.42)

Table 4.4 gives the results for the integrals of the crystal-field potential betweenthe different spherical harmonics. Using this table we can quickly calculate all ofthe various matrix elements.

For example, to calculate 〈�(3)|Voct(r) + Voct(r ′)|�(3)〉 = 〈2|Voct(r)|2〉 +〈1|Voct(r)|1〉 we use Table 4.4 to obtain

〈�(3)|Voct(r)+ Voct(r′)|�(3)〉 = Dq(1− 4) = −3Dq. (4.43)

For 〈�(1)|Voct(r)+ Voct(r ′)|�(1)〉, we have[√2

5〈1, 0| +

√3

5〈2,−1|

]|Voct(r)+ Voct(r

′)|[√

2

5|1, 0〉 +

√3

5|2,−1〉

]

= 2

5

[〈1|Voct(r)|1〉 + 〈0|Voct(r)|0〉

]+ 3

5

[〈2|Voct(r)|2〉 + 〈−1|Voct(r)| − 1〉

]= 2

5(−4+ 6)Dq + 3

5(1− 4)Dq =

(4

5− 9

5

)= −Dq. (4.44)

The crystal-field matrix elements of the two-electron functions of Table 4.3 areshown in Table 4.5.

Now that we have all the matrix elements of the proper basis functions we mustsolve the matrix eigenvalue equation

〈�(m)|H 0(r)+ H 0(r ′)+ Voct(r)+ Voct(r′)− E |�(m ′)〉

= 〈�(m)|{(2E0 − E)δmm′ + Voct(r)+ Voct(r′)}|�(m ′)〉 = 0. (4.45)

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4.2 Multi-electron systems 105

Table 4.5 Matrix elements of the crystal-field potential, Voct(r)+ Voct(r ′), fortwo-electron 3 F states in units of Dq. Blank spaces indicate that the integral iszero.

�(3) �(2) �(1) �(0) �(−1) �(−2) �(−3)

�(3) −3√

15

�(2) 7 5

�(1) −1√

15

�(0) −6

�(−1)√

15 −1

�(−2) 5 7

�(−3)√

15 −3

On grouping the coupled rows and columns together, we have the secularequation

det

�(3) �(−1) �(−3) �(1) �(2) �(−2) �(0)

�(3) −3Dq − λ√

15Dq

�(−1)√

15Dq −Dq − λ

�(−3) −3Dq − λ√

15Dq

�(1)√

15Dq −Dq − λ

�(2) 7Dq − λ 5Dq

�(−2) 5Dq 7Dq − λ

�(0) −6Dq − λ

= 0, (4.46)

where λ = E − 2E0. The eigenvalues of the upper two 2 × 2s are λ = 2Dq andλ = −6Dq. The eigenvalues of the lower 2 × 2 are λ = 2Dq and λ = 12Dq.The lower 1 × 1 gives λ = −6Dq. To find the eigenvectors we can treat eachof the uncoupled blocks separately. For the upper block with the eigenvalue−6Dq,the upper row requires that +3a+√15b = 0, where a and b are the amplitudes of�(3) and �(−1), respectively. This relation determines the ratio of a to b, and nor-malization yields another condition so that the coefficients a and b are completelydetermined. The normalized state is (5/8)1/2�(3)− (3/8)1/2�(−1). Proceeding inthis manner we have the terms and levels given in Table 4.6.

Since the F representation decomposes into A2g + T1g + T2g, it is clear that thenon-degenerate eigenvalue belongs to the A2g representation. However, both T1g

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106 Crystal-field theory

Table 4.6 Terms and levels for the 3 F crystal-field states

(a) T1g: A triple degeneracy States

�(0)E = 2E0 − 6Dq (5/8)1/2�(3)− (3/8)1/2�(−1)

(5/8)1/2�(−3)− (3/8)1/2�(1)

(b) T2g: A triple degeneracy States

(3/8)1/2�(3)+ (5/8)1/2�(−1)E = 2E0 + 2Dq (3/8)1/2�(−3)+ (5/8)1/2�(1)

(1/2)1/2[�(2)−�(−2)](c) A2g: A non-degenerate state State

E = 2E0 + 12Dq (1/2)1/2[�(2)+�(−2)]

and T2g are three-fold degenerate representations. In order to assign the states in(a) and (b), we must look at their symmetry properties. The functions listed in parts(a) and (b) of Table 4.6 are the bases for three-dimensional IRs. We can calculatethe trace of the matrix representing rotations using these functions. From the traceswe can identify which is T1g and which is T2g.

Consider a rotation C4 or (π/2) about the z-axis. If we operate on a Slaterdeterminant |m, n〉 we find

R(π/2)|m, n〉 = [Rn2(r) Rn2(r′)]

× 1√2

{Y m

2

(r, θ, φ + π

2

)Y n

2

(r ′, θ ′, φ′ + π

2

)

− Y m2

(r ′, θ ′, φ′ + π

2

)Y n

2

(r, θ, φ + π

2

)}= ei(m+n)π/2|m, n〉 = im+n|m, n〉. (4.47)

(Here R(π/2) operates both on r and on r′.) Using this rule, we find that

R(π/2)�(M) = i M�(M). (4.48)

For the functions in part (a) of Table 4.6 we have

R(π/2) �(0) = �(0), (4.49)

R(π/2)

[√5

8�(3)−

√3

8�(−1)

]= i3

√5

8�(3)+ i

√3

8�(−1)

= −i

[√5

8�(3)−

√3

8�(−1)

], (4.50)

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4.2 Multi-electron systems 107

R(π/2)

[√5

8�(−3)−

√3

8�(1)

]= i−3

√5

8�(−3)− i

√3

8�(1)

= +i

[√5

8�(−3)−

√3

8�(1)

]. (4.51)

Therefore, the representation matrix for R(π/2) is diagonal. The trace of the matrixrepresenting R(π/2) is the character, χ(a)(π/2), of the representation. It is just thesum of the eigenvalues: 1− i + i = +1. Repeating this procedure for group (b) inTable 4.6 yields χ(b)(π/2) = −1. From the character table for O we see that thecharacter for C4 is +1 for T1 and −1 for T2. Therefore we can identify group (a)as T1g and group (b) as T2g. The spin is S = 1 for all of these states, so the spinmultiplicity is 2S + 1 = 3, and therefore we can label the states as triplets: 3 A2g,3T1g, and 3T2g. The splitting is shown schematically in Fig. 4.2.

One might suppose that we could use the above two-electron results to deducethe behavior of the d3 or d8 states, since the d-shell would have two holes. Theelectronic (Hund’s rules) ground-state term is 4 F3/2, corresponding to S = 3/2.In the case of three d-electrons not all of the spin-orbitals can be represented bya single Slater determinant (see Exercise 4.10). Consequently, the “hole” pictureis more complex. In the most studied case (Cr3+ in ruby) the lowest crystal-fieldlevel is the 4 A2g state. For the d8 configuration (Ni2+) the lowest crystal-field levelis 3 A2g. This is what one might expect from the “two-hole” argument, since thehighest level for the d2 configuration is A2g.

We have explored the splitting of the 3 F ground state of the d2 configura-tion in detail above without regard to electron–electron interactions and the otherhigher-lying terms. In the octahedral environment higher-lying terms may havea level with the same symmetry as one of the 3 F levels. Levels with the samesymmetry will interact with one another. For example, the 3 P term has a group of

2E0

3A2g

3T2g

3T1g

2Dq

12Dq

6Dq

Figure 4.2 Crystal-field splitting of the two-d-electron 3 F term.

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108 Crystal-field theory

T1g states that will interact with the T1g states of the 3 F term. We have neglectedthe d–d-electron interactions, but in a more complete treatment [4.5, 4.6, 4.7]of the crystal-field theory the potential, e2/|r− r′|, is included explicitly in thetwo-electron Hamiltonian. We shall consider the effects of the electron–electroninteractions in Section 4.2.6 below.

4.2.5 The strong-crystal-field case

If the crystal field is very large compared with the electrostatic, orbit–orbit, andspin–orbit interactions, the eigenstates can be characterized by the IRs of thecrystal-field symmetry. For a strong octahedral crystal field, d-electron config-urations are more appropriately specified in terms of the number of electronsoccupying the T2g and Eg states. An outer-electron configuration is therefore ofthe form tmen , meaning that there are m electrons assigned to T2g orbitals and nelectrons in Eg orbitals. For the case of two d-electrons there are three possibleconfigurations: t2, e2, and te. The terms resulting from these configurations aredetermined by the direct-product decompositions.

In the case of the te states one electron resides in a t orbital and the other inan e orbital. The spins may therefore be parallel or antiparallel, so both triplets(3T1g, 3T2g) and singlets (1T1g, 1T2g) are permissible. For the e2 and t2 the termsinvolving doubly occupied orbitals can not have parallel spins, and some tripletstates are forbidden.

The symmetry-function-generating machine can be used to generate the basisfunctions for the various IRs listed in the decompositions of Table 4.7. For theone-dimensional IRs, A1g and A2g, all that is required is the character table forthe O group. (All the spatial functions must be “g” since they are constructedfrom the product of two d-functions.) As an example, we generate an e2 functionbelonging to the A1g IR using dx2(r) dx2(r ′) as the starting function (see Table 4.2for definitions of the d-orbitals). In this case the generating machine takes the form

F A1g (r, r ′) ∝∑

R

χ A1g (R) PR[dx2(r) dx2(r ′)] =∑

R

PR[dx2(r) dx2(r ′)], (4.52)

Table 4.7 Strong-crystal-field configurations and Oh terms

Configuration Direct product Decomposition into the IRs of Oh

t2 T1g × T1g A1g + Eg + T1g + T2gte T1g × Eg T1g + T2g

e2 Eg × Eg A1g + A2g + Eg

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4.2 Multi-electron systems 109

where the sum is over all of the operations of O , and the characters of the A1

representation are all equal to unity. The effect of PR on the various d-electronfunctions is given in Appendix E. This procedure gives

F A1g(r, r ′) ∝ [dx2(r) dx2(r ′)+ dz2(r ′) dz2(r)]. (4.53)

The function is symmetric upon interchange of the spatial coordinates and thereforethe spin state must be antisymmetric. The normalized spin–orbital function is

�(r, r ′, s, s ′) = 1

2[dx2(r) dx2(r ′)+ dz2(r ′) dz2(r)][α(s)β(s ′)−α(s ′)β(s)], (4.54)

where α and β are the spin +1/2 and −1/2 states, respectively.For the Eg, T1g, and T2g terms it is necessary to have the matrix elements of the

IRs. The symmetry-function-generating machine gives

F�(r, r ′) j ∝∑

R

�(R) j j PR[da(r) db(r′)], (4.55)

where F�(r, r ′) j is a function that belongs to the j th row of the � IR and �(R) j j

is the j th diagonal matrix element of the IR. In (4.55) da and db are any of thed-orbitals or linear combinations of the d-orbitals listed in Table 4.1. The IRmatrix elements, �(R)i j , for the IRs of O are given in Appendix E. Table 4.8gives the wavefunctions for the t2, e2, and te configurations calculated using (4.55).The wavefunctions are eigenstates of the crystal-field Hamiltonian in the absenceof the electrostatic potential, e2/|r− r′|. The functions F�(r, r ′) j must be sym-metrized and normalized appropriately. For symmetric spin functions (S = 1),if F�(r, r ′) j is not antisymmetric in the interchange of r and r ′ it is replaced by[F�(r, r ′) j − F�(r ′, r) j ], and then normalized. For antisymmetric spin functions(S = 0), if F�(r, r ′) j is not symmetric in the interchange of r and r ′ it is replacedby [F�(r, r ′) j + F�(r ′, r) j ], and then normalized.

4.2.6 The effect of electron–electron interactions on crystal-field splitting

In the strong-crystal-field case, the zeroth-order energy (in the strong-field limit)for a T2g-electron state is 10Dq lower than that of an Eg-electron state. It is cus-tomary to choose the zero of energy as the te configuration. With that choice, thee2 configuration’s energy is +10Dq and the t2 configuration’s energy is −10Dqas shown on the right-hand side of Fig. 4.3 on p. 115.

The states listed in Table 4.8 are the symmetry functions for the levels of thed2-configuration terms. They are eigenstates of H0(r, r ′) = H0(r) + H0(r ′) +Voct(r) + Voct(r ′). To obtain the energies in the presence of the electron–electroninteractions we must solve the eigenvalue equation (H− 2E0 − λα)�α(r, r ′) = 0,

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110 Crystal-field theory

Table 4.8 Symmetry functions for the terms of the t2, e2, and te configurations oftwo d-electrons in a strong octahedral crystal field. Each row function of thetriplets has the three spin states listed on the right. Each singlet has one spin statelisted on the right. These functions are eigenstates in the absence of thee2/|r− r′| interaction in a spherical or octahedral environment.

t2 terms3T1g , S = 1 Spatial part of the eigenstate Spin eigenstates

|t2, 3T1g, 1〉 [a(r)b(r ′)− b(r)a(r ′)]/√2 S1(1), S1(0), S1(−1)

|t2, 3T1g, 2〉 [a(r)c(r ′)− b(r)c(r ′)]/√2 S1(1), S1(0), S1(−1)

|t2, 3T1g, 3〉 [b(r)c(r ′)− b(r)c(r ′)]/√2 S1(1), S1(0), S1(−1)1T2g , S = 0

|t2, 1T2g, 1〉 [b(r)c(r ′)+ c(r)b(r ′)]/√2 S(0)

|t2, 1T2g, 2〉 [a(r)c(r ′)+ c(r)a(r ′)]/√2 S(0)

|t2, 1T2g, 3〉 [a(r)b(r ′)+ b(r)a(r ′)]/√2 S(0)1Eg , S = 0

|t2, 1Eg, 1〉 [b(r)b(r ′)− c(r)c(r ′)]/√2 S(0)

|t2, 1Eg, 2〉 [2a(r)a(r ′)− b(r)b(r ′)− c(r)c(r ′)]/√6 S(0)1 A1g , S = 0

|t2, 1 A1g, 1〉 [a(r)a(r ′)+ b(r)b(r ′)+ c(r)c(r ′)]/√3 S(0)

e2 terms1 A1g , S = 0 Spatial part of the eigenstate Spin eigenstates

|e2, 1 A1g, 1〉 [x2(r)x2(r ′)+ z2(r)z2(r ′)]/√2 S(0)3 A2g , S = 1

|e2, 3 A2g, 1〉 [x2(r)z2(r ′)− z2(r)x2(r ′)]/√2 S1(1), S1(0), S1(−1)1Eg , S = 0

|e2, 1Eg, 1〉 [x2(r)z2(r ′)+ z2(r)x2(r ′)]/√2 S(0)

|e2,1 Eg, 2〉 [x2(r)x2(r ′)− z2(r)z2(r ′)]/√2 S(0)

te terms1T1g , S = 0 Spatial part of the eigenstate Spin eigenstates

|te, 1T1g, 1〉 {c(r)[x2(r ′)+√3z2(r ′)]+ c(r ′)[x2(r)+√3z2(r)]}/√8 S(0)

|te, 1T1g, 2〉 {b(r)[x2(r ′)−√3z2(r ′)]+ b(r ′)[x2(r)−√3z2(r)]}/√8 S(0)

|te, 1T1g, 3〉 {a(r)x2(r ′)+ a(r ′)x2(r)}/√2 S(0)

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4.2 Multi-electron systems 111

Table 4.8 (cont.)

3T1g , S = 1

|te, 3T1g, 1〉 {c(r)[x2(r ′)+√3z2(r ′)]− c(r ′)[x2(r)+√3z2(r)]}/√8 S1(1), S1(0), S1(−1)

|te, 3T1g, 2〉 {b(r)[x2(r ′)−√3z2(r ′)]− b(r ′)[x2(r)−√3z2(r)]}/√8 S1(1), S1(0), S1(−1)

|te, 3T1g, 3〉 {a(r)x2(r ′)− a(r ′)x2(r)}/√2 S1(1), S1(0), S1(−1)1T2g , S = 0

|te, 1T2g, 1〉 {a(r)z2(r ′)+ a(r ′)z2(r)}/√2 S(0)

|te, 1T2g, 2〉 {b(r)[z2(r ′)+√3x2(r ′)]+ b(r ′)[z2(r)+√3x2(r)]}/√8 S(0)

|te, 1T2g, 3〉 {c(r)[z2(r ′)−√3x2(r ′)]+ c(r ′)[z2(r)−√3x2(r)]}/√8 S(0)

3T2g , S = 1

|te, 3T2g, 1〉 {a(r)z2(r ′)− a(r ′)z2(r)}/√2 S1(1), S1(0), S1(−1)

|te, 3T2g, 2〉 {b(r)[z2(r ′)+√3x2(r ′)]− b(r ′)[z2(r)+√3x2(r)]}/√8 S1(1), S1(0), S1(−1)

|te, 3T2g, 3〉 {c(r)[z2(r ′)−√3x2(r ′)]− c(r ′)[z2(r)−√3x2(r)]}/√8 S1(1), S1(0), S1(−1)

|nm, p�, q〉 : nm is the octahedral configuration (t2, te, or e2), � is the IR of Oh , q is therow of �, and p is the spin multiplicity. The total spatial eigenfunction isRn2(r) Rn2(r ′) |nm, p�, q〉. z2(r) = dz2(r), x2(r) = dx2(r), a(r) = dxy(r),b(r) = dxz(r), c(r) = dyz(r). S(0) = [α(s)β(s′)− β(s)α(s′)]/√2, where α is spin upand β is spin down, S1(1) = α(s)α(s′), and S1(0) = [α(s)β(s′)+ β(s)α(s′)]/√2, andS1(−1) = β(s)β(s′).

where H = H0(r, r ′) + e2/|r− r′|. The eigenstates �α(r, r ′) will be linear com-binations of the symmetry functions in Table 4.8 that belong to the same row ofthe same IR. For example, there are A1g states both for the t2 and for the e2 con-figurations. These two same-symmetry states will be mixed to form the final A1g

eigenstates. Therefore we need the matrix elements of e2/|r− r′| between the sym-metry functions of the same type. Note that e2/|r− r′| is invariant under all of theoperations of the group because PR rotates both r and r′ in the same way and thequantity |r− r′| is unchanged by the rotational operations. Therefore, e2/|r− r′|belongs to the totally symmetric A1g representation. Consequently, there can beno non-zero matrix elements between different IRs or between different rows of thesame IR. This feature greatly simplifies the calculations of the matrix elements.e2/|r− r′| does not involve the spins, so its matrix element between two statesvanishes unless S = S′.

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112 Crystal-field theory

When calculating the electrostatic interactions there are two important mathe-matical results that are useful. The first is

1

|r1 − r2| =∞∑

k=0

rk<

rk+1>

Pk(cos γ ), (4.56)

where r< is the smaller of r1 and r2 and r> is the larger of r1 and r2. The angle γ isthe angle between r1 and r2, and Pk is the kth Legendre polynomial.

The second result, called the “addition theorem”, is

Pk(cos γ ) = 4π

2k + 1

k∑m=−k

Y mk (θ1, φ1)

∗ Y mk (θ2, φ2). (4.57)

These two mathematical results can be used to evaluate the electrostatic interactionbetween electrons. A general interaction matrix element is of the form⟨

n1, n2, l1, l2,m1,m2

∣∣∣∣ e2

|ra − rb|∣∣∣∣ n3, n4, l3, l4,m3,m4

⟩, (4.58)

where n, l, and m are the principal, angular-momentum, and magnetic quantumnumbers, respectively.

In detail the matrix element (4.58) is∫r2

1 dr1

∫r2

2 dr2

∫dθ1 sin θ1

∫dθ2 sin θ2

∫dφ1

∫dφ2

×[

Rn1(r1) Rn2(r2) Y m1l1

(θ1, φ1) Y m2l2

(θ2, φ2)]∗

×[ ∞∑

k=0

rk<

rk+1>

2k + 1

k∑m=−k

Y mk (θ2, φ2)

∗ Y mk (θ1, φ1)

]

×[

Rn3(r1) Rn4(r2) Y m3l3

(θ1, φ1) Y m4l4

(θ2, φ2)]. (4.59)

This is a formidable-appearing result, but it can be simplified in application. Thematrix element involves the products[ ∫

dφ1

∫dθ1 sin θ1 Y m1

l1(θ1, φ1)

∗ Y mk (θ1, φ1) Y m3

l3 (θ1, φ1)]

×[ ∫

dφ2

∫dθ2 sin θ2 Y m2

l2(θ2, φ2)

∗ Y mk (θ2, φ2) Y m4

l4 (θ2, φ2)]. (4.60)

The integrals over φ1 and φ2 can be carried out immediately, with the result∫ 2π

0ei(−m1+m+m3)φ1 dφ1

∫ 2π

0ei(−m2−m+m4)φ2 dφ2

∝ δ(−m1 + m + m3)δ(−m2 − m + m4), (4.61)

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4.2 Multi-electron systems 113

which requires m1+m2 = m3+m4. Therefore the matrix element vanishes unlessthe total orbital momenta of the two states are equal. Furthermore, the direct-product selection rules tell us that the integrals vanish except for k simultaneouslyin the range of both |l1 − l2| ≤ k ≤ l1 + l2 and |l3 − l4| ≤ k ≤ l3 + l4. For twod-electrons, this requires that k ≤ 4. Thus, of the infinite number of terms in (4.59)only the first four terms of the sum over the index k contribute a non-zero result.

It is important to note that the matrix element of (4.58) is not between the sym-metry functions listed in Table 4.8. However, the matrix elements of the symmetryfunctions will consist of a sum of these types of integrals.

For our two-d-electron problem there are at most two different symmetry func-tions belonging to a given row of the same IR, as can be seen from Table 4.8. Letf α

j and gαj be two different functions both belonging to j th row of the αth IR of

Oh . The matrix elements of e2/|r1 − r2| are given by

M11 = 〈 f αj |

e2

|r1 − r2| | fαj 〉,

M22 = 〈gαj |

e2

|r1 − r2| |gαj 〉, (4.62)

M21 = M12 = 〈 f αj |

e2

|r1 − r2| |gαj 〉.

The results are independent of which row (which j) is used to evaluate the inte-gral. The matrix elements between functions belonging to different IRs or differentrow functions vanish. The elements of M for the various IRs of Oh have been com-piled by Tanabe and Sugano [4.7]. Those for the d2 symmetry functions listed inTable 4.8 are given in Table 4.9.

The constants B and C are combinations of integrals over the radial coordinatesof the hydrogen-like atomic states [4.8]. Treating these constants as adjustableparameters allows one to explore how the levels change as the ratio of thecrystal-field strength to the electron–electron-interaction strength varies.

For the V3+ ion, approximate values are Dq = 1,800 cm−1, B = 860 cm−1, andC = 3,800 cm−1. Using Table 4.9, we can calculate the energies of the interactingstates. For example, consider the 3T1g eigenvalue secular equation (the first rowof Table 4.9). The eigenvalues for these states are obtained from the 2 × 2-matrixequation, ( −5B − 10Dq − λ 6B

6B 4B − λ

), (4.63)

where λ = E − 2E0.For V3+ (d2 configuration) this gives λ(3T1g) = 4,436 cm−1 and−23,296 cm−1.

If we solve (4.63) with Dq = 0 then we obtain the energies of the terms in the

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114 Crystal-field theory

Table 4.9 Electron–electron matrix elements for the terms of the d2 configuration

IR of the statesStrong-field

configurationM11 (upper)M21 (lower)

M12 (upper)M22 (lower)

3T1g t2 −5B − 10Dq 6Bte 6B 4B

1T2g t2 B + 2C − 10Dq 2√

3Bte 2

√3B 2C

1Eg t2 B + 2C − 10Dq −2√

3Be2 −2

√3B 2C + 10Dq

1 A1g t2 10B + 5C − 10Dq√

6(2B + C)

e2√

6(2B + C) 8B + 4C + 10Dq3 A2g t2 −8B + 10Dq1T1g te 4B + 2C3T2g te −8B

weak-field limit. They are λ(3T1g) = 6,020 cm−1 and −6,880 cm−1. We can alsotrace the evolutions from the weak-field case to the strong-field case by allow-ing Dq to continuously increase from zero to its value of 1,800 cm−1. Proceedingin this manner for all of the terms listed in Table 4.9 allows us to correlate theweak-field terms with the terms for the strong crystal field. The results are shownschematically in Fig. 4.3. The eigenvalues for Dq = 0 and Dq = 1,800 cm−1 aregiven in Table 4.10.

4.2.7 Book keeping

For the free ion with two d-electrons we determine the states from D(2) × D(2) =D(0)+D(1)+D(2)+D(3)+D(4) or S+ P+D+F+G. Since all of these states areconstructed from the product of two d orbitals, they must be “g” under inversion.If we then decompose these IRs into those of Oh , we obtain

S = A1g,

P = T1g,

D = Eg + T2g, (4.64)

F = A2g + T2g + T1g,

G = Eg + T1g + T2g + A1g.

Conversely, if we start from the strong-field limit the configurations are t2, e2, andte, and the terms are derived from the direct products, T2g × T2g, Eg × Eg, andEg × T2g. These direct products are

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4.2 Multi-electron systems 115

Crystal-field theory

–20

–10

0

10

20

30

40

0 0.2 0.4 0.6 0.8 1Dq/(2.093B)−→

(E−

2E0 )

(103

cm)

−1

1S

3G

3P1D

3F

1A1g

1Eg

1A1g1T1g

1T2g

3T1g

1A1g

1T2g

1Eg

3T2g

3T1g

00.20.40.60.8←− 2.093B/(Dq)

+10Dq(e2)

0 (te)

−10Dq(t2)

1A1g

1Eg

1A1g

1T1g

3T2g

1T2g

3T1g

1Eg1A1g

1T2g

3T1g

Figure 4.3 A correlation diagram using the values of the parameters for V3+: B =860 cm−1, C/B = 4.42, and Dq = 1,800 cm−1. Terms derived from t2, e2, andte configurations asymptotically approach −10Dq, 10Dq, and 0, respectively, asB/Dq → 0.

T2g × T2g = A1g + Eg + T1g + T2g,

Eg × Eg = A1g + A2g + Eg, (4.65)

Eg × T2g = T1g + T2g + T1g + T2g.

On comparing the terms on the right-hand side of (4.64) with those of (4.65) wefind that they are exactly the same.

A schematic representation of the evolution of the terms on going from the weak-field to the strong-field case is shown in Fig. 4.3. On the extreme left the energiesare those for Dq = 0. On the extreme right the energies are for Dq →∞. In thislimit terms derived from the t2 configuration merge to−10Dq, while those derivedfrom the e2 configuration merge to +10Dq. The terms from the te configurationtend to 0 (by virtue of the choice of the origin of energy).

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116 Crystal-field theory

Table 4.10 d2 States. Energy λ = E − 2E 0. Results for Dq = 0 andDq = 1,800 cm−1 with B = 860 cm−1 and C = 3,800 cm−1. The “observed”energy levels of [V(H2O)6]3+ are given in the fourth column and the last columngives the calculated values. E1 = E(−23,296 cm−1) is the lowest 3 F state forDq = 1,800 cm−1.

Free-ionterms

Levels: Weakcrystal field

Dq = 0 cm−1

Eigenvalues Ohsymmetry

Dq = 1,800 cm−1

Observed E − E1(approximate values)

(cm−1)

CalculatedE − E1(cm−1)

1S 1 A1g 38, 640 45, 2141G 1 A1g 11, 040 4, 4661G 1Eg 11, 040 25, 8511G 1T1g 11, 040 11, 0401G 1T2g 11, 040 8, 1033 P 3T1g 6, 020 4, 436 3T1g(P)→ 3T1g(F) 25,000 27,7321 D 1T2g 5, 020 −10,0431 D 1Eg 5, 020 −9,7913 F 3 A2g −6,880 11,120 3 A2g(F)→ 3T1g(F) 37,000 34,4163 F 3T1g −6,880 −23,2963 F 3T2g −6,880 −6,880 3T2g(F)→ 3T1g(F) 17,700 16,416

4.3 Jahn–Teller effects

The Jahn–Teller distortion (or Jahn–Teller effect) refers to the tendency of amolecule whose ground state is degenerate to distort structurally. For example,the complex ion [Ti(H2O)6]3+ would be expected to consist of a Ti(d1) ion sur-rounded by an octahedron of water molecules. Since all of the six water moleculesare equivalent, there is no apparent reason why any one axis should be singled outto be different. However, experimentally the complex is observed to be compressedalong one axis (arbitrarily designated as the z-axis), resulting in D4h rather than Oh

symmetry. By contrast, [Cu(H2O)6]2+(d9) is elongated along the z-axis as shownschematically in Fig. 4.4.

Jahn and Teller [4.9] put forward the following theorem: “Any non-linear molec-ular system in a degenerate electronic state will be unstable and will undergodistortion to form a system of lower symmetry and lower energy, thereby removingthe degeneracy”.

The Jahn–Teller distortion is most often observed with transition metal ions inwhat would be expected to be an octahedral environment. The effect is found to bemost pronounced when an odd number of electrons occupy (doubly degenerate) Eg

orbitals. The effect occurs for electrons occupying degenerate T2g orbitals as well,but is much weaker.

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4.3 Jahn–Teller effects 117

(a) Oh symmetry expected

1.95 A

1.95 A

1.95 A

CuH2O

H2OH2O H2O

H2O

H2O

(b) D4h symmetry observed(elongated along z−axis)

1.95 A

2.38 A

2.38 A

CuH2O

H2OH2O H2O

H2O

H2O

Figure 4.4 The complex [Cu(H2O)6]2+: (a) the octahedral structure expected inthe absence of the Jahn–Teller effect and (b) the observed structure. The complexis elongated along the z-axis due to the Jahn–Teller effect. The resulting structurehas D4h symmetry.

To see why a distortion might reduce the ground-state energy of a complex,start with a d-ion surrounded by a perfect octahedron of ligands. A change inthe ion–ligand bond length along the z-axis changes the symmetry from Oh toD4h . For a d1 configuration the octahedral ground state is a triply degenerate T2g

level. For a single d-hole (d9) the ground state is the Eg level. The Eg and T2g

IRs of Oh are reducible representations of the D4h group. Since the d-orbitalsare symmetric under inversion, we need only consider the decomposition of Eand T2 into the IRs of D4. The character table for D4 is given in Table 4.11,together with the characters of the E(O), T2(O), and T1(O) IRs of the O group.The correspondences of the classes of the two groups are 8C3(O)→ none in D4,3C2(O)→ C2(D4)+ 2C ′2(D4), 6C2(O)→ 2C ′′2 (D4), and 6C4(O)→ 2C4(D4).

Using Table 4.11, we find that the T2(O) IR splits into a doubly degenerateE(D4) and a non-degenerate B2(D4) as shown in Fig. 4.5. The E(O) IR splits into anon-degenerate A1(D4) and a non-degenerate B1(D4) level. For a d1 configurationthe distortion produces a lower energy level for the d-electron to reside in. Forthe d9 configuration a similar argument can be made regarding the hole. The d1

electron occupies the Eg(D4h) for elongation or the B2g(D4h) for compression (e.g.,[Ti(H2O)6]3+). In either case the ground-state energy is lowered.

There are, of course, opposing reactions to a Jahn–Teller distortion, includingelastic, dipolar, and other changes in the electronic structure of the complex due todistortion. The occurrence of the Jahn–Teller distortion implies that the d-electronsplitting energy is dominant. The d9 configuration, which is typical of many Cu-containing molecules and solids, has an Eg (hole in the dx2 level) ground state, andthe Jahn–Teller effect generally results in an elongation along the z-axis.

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118 Crystal-field theory

Table 4.11 The character table for the symmetry elements of D4 and IRs of O.The upper segment is the character table for D4. The lower segment shows thedecomposition of E , T2, and T1 of O into the IRs of D4.

IR E C2 2C4 2C ′2 2C ′′2 Basis functions

A1 1 1 1 1 1 x2 + y2, z2

A2 1 1 1 −2 −2 Rz , zB1 1 1 −2 1 −2 x2 − y2

B2 1 1 −2 −2 1 xyE 2 −2 0 0 0 (xy, yz); (Rx , Ry)

Decomposition O → D4E(O) 2 2 0 2 0 A1 + B1T2(O) 3 −2 −2 −2 1 B2 + ET1(O) 3 −2 1 −2 −2 A2 + E

(a) Elongation

Eg

dx2

dz2

(B1g)

(A1g)

T2g

dxy

dxz, dyz

(B2g)

(Eg)

(b) Compression

Eg

dz2

dx2(B1g)

(A1g)

T2g

dxz, dyz

dxy (B2g)

(Eg)

z z

Figure 4.5 Jahn–Teller splitting of the d-orbital energies: (a) elongation and (b)compression.

There are two other closely related effects that can lead to structural distortion.The first is the dynamic Jahn–Teller effect. This effect may occur when there areseveral different structural distortions that are separated by small energy barriers.Quantum-mechanical tunneling between the different structures can occur with theassistance of phonons of the proper symmetry. There is experimental evidence for

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Exercises 119

the dynamic Jahn–Teller effect in molecules and solids. It is conjectured that thephenomenon may play a role in explaining high-temperature superconductivity inthe cuprate compounds.

A second effect, called the second-order Jahn–Teller effect, occurs when anexcited state of a system is very close in energy to the ground state. A distor-tion of the structure may allow the states to mix, resulting in a lower ground-stateconfiguration. This effect is believed to be involved in stereochemical non-rigidity.

References

[4.1] H. Bethe, “Termaufspaltung in Kristallen”, Ann. Phys. 3, 133–206 (1929).[4.2] J. H. Van Vleck, “The group relation between the Mulliken and Slater–Pauling

theories of valence”, J. Chem. Phys. 3, 803–807 (1935).J. H. Van Vleck, “Valence strength and the magnetism of complex salts”, J. Chem.Phys. 3, 807–814 (1935).

[4.3] H. Eyring, J. Walter, and G. E. Kimball, Quantum Chemistry (New York: Wiley,1944).

[4.4] E. U. Condon and G. H. Shortley, The Theory of Atomic Spectra (Cambridge:Cambridge University Press, 1963).

[4.5] T. M. Dunn, D. S. McClure, and R. G. Pearson, Some Aspects of Crystal Field Theory(New York: Harper & Row, 1965). (Note that there are errors in some of the formulason pages 18 and 20. These have been corrected in our discussion.)

[4.6] D. S. McClure, “Electronic spectra of molecules and ions in crystals. II. Spectra ofions in crystals”, Solid State Phys. 9, 399–525 (1958).

[4.7] Y. Tanabe and S. Sugano, “On the absorption spectra of complex ions I”, J. Phys.Soc. Japan 9, 753–766 (1954).Y. Tanabe and S. Sugano, “On the absorption spectra of complex ions II”, J. Phys.Soc. Japan 9, 766–779 (1954).

[4.8] G. Racah, “Theory of complex spectra II”, Phys. Rev. 62, 438–462 (1942).[4.9] H. A. Jahn and E. Teller, “Stability of polyatomic molecules in degenerate electronic

states. I. Orbital degeneracy”, Proc. Royal Soc. London A 161 (no. 905), 220–235(1937).

Exercises

4.1 The spherical harmonics form a complete, orthogonal set and therefore thecrystal field can be expanded in these functions:

Voct =∞∑

l=0

l∑m=−l

f ml (r) Y m

l (θ, φ)

= 7eZ ligand

3t5

[Y 0

4 +5

14

(Y 4

4 + Y−44

)]+ terms with l > 4,

where r is the magnitude of the vector r. Use the direct-product rules to showthat the terms with l > 4 make no contribution to the matrix elements of thecrystal field for d-orbitals.

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120 Crystal-field theory

4.2 An ion at the origin is surrounded by four equidistant ligands, each havingnegative charge Z ligande (Z le in Fig. 4.6).

tx

y

z

r

Zle

Zle

Zle Zle

Figure 4.6

(a) Identify the symmetry group for the three-dimensional configuration.(b) Using a Taylor’s-series expansion of the potential (for r < t) show that

the lowest-order non-spherical term is

e2 Z ligand3(x2 + y2)− 2r2

t3.

(c) From Table 4.11 identify the IR corresponding to the potential.(d) Determine the type of splitting of a d-electron due to the potential.

4.3 Consider a single d-electron and the d-states, dxy , dxz , dyz , dx2 , and dz2 .(a) Identify to which IR of D4 each d-state belongs.(b) The matrix elements of the potential in Exercise 4.2 between the d-states

can be written symbolically as 〈�a(D4)|A1(D4)|�b(D4)〉. Use direct-product arguments to show that the off-diagonal matrix elements arezero.

(c) Calculate the crystal-field splitting between the B2(D4) and E(D4) levelsin terms of the parameter γ = e2 Z ligand〈r2〉avg/t3, where

〈r2〉avg =∫ ∞

0r2[r2 Rnd(r)

2]dr.

4.4 Use Hund’s rules to find the ground states of the elements Sc(d 1), Ti(d 2),V(d 3), Cr(d 5s), Mn(d 5), Fe(d 6), Co(d 7), Ni(d 8), and Cu(d 10s).

4.5 Find the terms of the configuration f 2 and show that the Hund ground stateis 3 H4.

4.6 Show that, for the 3 H4 ground state in Exercise 4.5 the combination ofSlater determinants corresponding to ML = 3, M2

L = 5(5 + 1) is �(3) =(31/2/2)|3, 0〉 + (1/3)1/2|2, 1〉.

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Exercises 121

4.7 The D(1) representation based on spherical harmonics is equivalent to theT1 representation of the O group, meaning that D(1) is related to T1 by asimilarity transformation. Show that the decomposition of the direct productof D(1) × D(1) into the IRs of O is the same as the decomposition of T1 × T1

into the IRs of O . Indicate whether the representations of the decompositionsare “u” or “g”.

4.8 Use the formula

l∑k=0

xk = 1− x (l+1)

1− x

for x < 1 to show that

l∑k+−l

e−ikα = sin(l + 1/2)α

sin(α/2).

4.9 Prove that

χ(l)(R)χ(l ′)(R) =l+l ′∑

k=|l−l ′|χ(k)(R),

where χ(l) is the character of D(l)(R) and χ(l) = ∑lm=−l e−imθ(R), where

θ(R) is the angle associated with the operation R.4.10 Find the ground state for a d3 configuration using Hund’s rules (Exercise 4.4,

V2+(d3)).(a) Write a Slater determinant wavefunction,

�(n, l1,m1, s1z; n, l2,m2, s2z; n, l3,m3, s3z),

for three d-electrons with ML = 3 and Sz = 3/2 in terms of hydrogen-like spin-orbitals.

(b) The spin-lowering operator, S−, lowers the spin of the state it operates onby 1 (in units of �):

S−�1(l1,m1, s1z; l2,m2, s2z; . . . ; nk, lk,mk, skz; . . . ; lN ,mN , sN z)

=N∑

k=1

{(lk + mk)(lk − mk + 1)}1/2 ×

× �(n1, l1,m1, s1z; n2, l2,m2, s2z; . . . ; nk, lk,mk, (skz − 1); . . . ;lN ,mN , sN z),

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122 Crystal-field theory

where � is a Slater determinant of spin-orbitals with quantum numbersn, l, m, and sz , and S− operating on a state with skz = −1/2 gives zero.Use S− on the Sz = 3/2 state of (a) to generate the Sz = 1/2, −1/2,and −3/2 states. (Note: The spin ±1/2 states can not be separated intoa single Slater determinant of spatial functions times a product of spinfunctions.)

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5

Electron spin and angular momentum

Unlike orbital angular momentum, the total spin of a system can be integral orhalf-integral. Fermions such as electrons, positrons, neutrinos, and quarks possessintrinsic angular momentum or spin with a measurable value of ±1/2 (in units of�). Composite particles such as protons and neutrons also have measurable spin of±1/2 and atomic nuclei can have half-integral spin values (1/2, 3/2, 5/2, . . .). The“spinor” function for half-integral spin is unusual in that rotation by 2π transformsit into the negative of itself. A rotation by 4π is required in order to transform thespin function into itself. While this may at first glance seem unreasonable, thereare simple examples that display this property.

Take a strip of paper and form a Möbius strip by twisting one end 180◦ andjoining it to the other end. Start at any point on the strip and trace a line through360◦. You do not end up at the starting point, but rather on the other side of thepaper strip, as shown in Fig. 5.1. Continue tracing along the surface for another360◦ and you will return to the original starting point.

Early spectroscopic experiments on hydrogen and hydrogen-like atoms revealedthat there were twice as many states as predicted by the solutions of Schrödinger’sequation. The idea that an electron could have intrinsic angular momentum (spin)with two possible states was proposed by Kronig, Uhlenbeck, and Goudsmit [5.1]in 1925. Later, Dirac [5.2] developed a theory revealing that spin was a nat-ural consequence of requiring the electron wavefunction to be consistent withrelativity theory.

5.1 Pauli spin matrices

The spin state of an electron can be represented as a two-component spinor. ThePauli matrices [5.3] for spin 1/2 are matrices that represent the three-dimensionalspin-vector operator in terms of matrices that operate on the two-component

123

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124 Electron spin and angular momentum

B

AUpper surface

Figure 5.1 Consider a paper model of a Möbius strip. On the right-hand side ofthe paper, mark the upper surface with an A and the other side of the paper witha B. Starting from A, trace a line through 360◦ counter-clockwise on the Möbiusstrip, arriving at B on the other side of the paper. To return to A requires another360◦ around the Möbius strip.

spinors. The Pauli σ -matrices are related to the spin-vector operator by sx =(�/2)σx , sy = (�/2)σy , and sz = (�/2)σz , where

σx =(

0 11 0

), (5.1)

σy =(

0 −ii 0

), (5.2)

σz =(

1 00 −1

). (5.3)

The Pauli matrices anticommute, meaning that σiσ j + σ jσi = {σi , σ j } = 0 fori �= j . For i = j , {σi , σ j } = 2I , where I is the unit matrix.

The eigenvectors of Pauli matrices correspond to the spin states with the spinvector aligned parallel or antiparallel to the direction of the x , y, or z spatial axis.The spin eigenvalues are always ±(�/2). This is easily verified by diagonalizingthe σ -matrices. The eigenvectors and eigenvalues of sx , sy , and sz are

χ+x =1√2

(11

), eigenvalue + 1

2, spin aligned along the +x-axis, (5.4)

χ−x =1√2

(1−1

), eigenvalue − 1

2, spin aligned along the −x-axis, (5.5)

χ+y =1√2

(1i

), eigenvalue + 1

2, spin aligned along the +y-axis, (5.6)

χ−y =1√2

(1−i

), eigenvalue − 1

2, spin aligned along the −y-axis, (5.7)

χ+z =1√2

(10

), eigenvalue + 1

2, spin aligned along the +z-axis, (5.8)

χ−z =1√2

(01

), eigenvalue − 1

2, spin aligned along the −z-axis. (5.9)

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5.1 Pauli spin matrices 125

For example,

sxχ−x =

2

1√2

(0 11 0

)(1−1

)= �

2

1√2

( −11

)= −�

2χ−x . (5.10)

To determine the spin operator, se, for rotated axes we construct the dot product(scalar product) of s with a unit vector, e, in any arbitrary direction,

se = s · e = (ex sx + eysy + ezsz) = �

2

(ez ex − iey

ex + iey −ez

), (5.11)

where ex , ey , and ez are the components of e along the original x-, y-, and z-axes,respectively. Note that se is shown in bold type because it is a (scalar) operator, butit is not a vector. The eigenvectors of se, however, are two-component spinors, thatis, vectors in Pauli spin-space.

The eigenvalues of se are obtained by requiring the determinant of (se − λI) tovanish, where I is the 2× 2 unit matrix. We obtain

det

{�

2

(ez − 2λ/� ex − iey

ex + iey −ez − 2λ/�

)}=[λ2 −

(�

2

)2

(e2x + e2

y + e2z )

]= 0.

(5.12)

Since e is a unit vector, (e2x + e2

y + e2z ) = 1 and the eigenvalues are λ = ±�/2.

Therefore, the eigenvalues of the electron’s spin operator are always ±�/2 regard-less of the orientation of the axes. The eigenvectors of se can be found in the usualway by inserting the eigenvalue into the secular equation and solving for the ratioof the two amplitudes. When normalized the eigenstates are

χ+e =1√

2(1− ez)

(ex − iey

1− ez

), (5.13)

χ−e =1√

2(1+ ez)

(ex − iey

−(1+ ez)

). (5.14)

Equations (5.13) and (5.14) are indeterminate when ez = ±1 (along the±z-axis). For these two exceptional cases we can simply use χ+z and χ−z givenby (5.8) and (5.9), respectively.

To determine the eigenvalue of s2 for a single electron, we note that s2x = s2

y =s2

z = (�/2)2 I, where I is a 2× 2 unit matrix. Therefore,

s2 =(

2

)2

(σ 2x + σ 2

y + σ 2z ) = 3

(�

2

)2

I = �2 1

2

(1

2+ 1

)I.

It follows that s2 χ = �2(1/2)(1/2+ 1)χ for every one-electron spin state.

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126 Electron spin and angular momentum

5.2 Measurement of spin

According to the generally accepted postulates of quantum theory, performing ameasurement of the spin along a given direction causes the spin function to “col-lapse” into one of the two corresponding eigenstates for that direction. Supposethe spin was initially in the eigenstate χ s

I . After a measurement along the directionspecified by the unit vector e the probability of the state being χ+e plus the proba-bility of the state being χ−e is unity. According to this postulate, the measurementprocess destroys the initial state and puts the spin into either χ+e or χ−e . In gen-eral, the individual probabilities for χ+f and χ−f depend upon the initial eigenstateand the direction of measurement, but the sum of the probability of +�/2 plus theprobability of −�/2 along e is always unity.

The probability that the spin of an electron will be ±�/2 along the e-axis (anarbitrary axis) is given by the absolute value of the square of the projection of thespin-eigenstate state χI onto the eigenstate χ±e . The probability of measuring±�/2along the e-direction when the spin was initially in the state χI is

|〈χI |χ+e 〉|2 + |〈χI |χ−e 〉|2 = 1, (5.15)

where

|〈χI |χ+e 〉|2 = Probability of the spin being parallel to e, (5.16)

|〈χI |χ−e 〉|2 = Probability of the spin being antiparallel to e. (5.17)

In equations (5.15)–(5.17) |〈χI |χ+e 〉|2 = (χI 1 ·χ+e1)2+ (χI 2 ·χ+e2)

2. For example,suppose χI = χ+z and we want the probability for the spin states along the direc-tion e = (1/3)1/2(1, 1, 1). According to (5.13) the eigenvector for the spin in thepositive e-direction is

χ+e =1√

2(1− ez)

(ex − iey

1− ez

)= 1/

√3√

2(1− 1/√

3)

(1− i√3− 1

). (5.18)

The probability amplitude is given by

〈χ+z |χ+e 〉 =1/√

3√2(1− 1/

√3)

(1 0)

(1− i√3− 1

)= 1/

√3√

2(1− 1/√

3)(1− i)

(5.19)and the probability of measuring +�/2 is

|〈χ+z |χ+e 〉|2 =2/3

2(1− 1/√

3)= 0.788 675. (5.20)

A similar calculation shows that |〈χ+z |χ−e 〉|2 = (2/3)/(2+ 2/√

3) = 0.211 325, sothe probability of measuring either +�/2 or −�/2 along the (1,1,1) axis is unity.

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5.3 Irreducible representations of half-integer angular momentum 127

In general the probability of measuring either +�/2 or −�/2 along any directionis unity.

It is evident that the probabilities are reciprocal since |〈χ+z |χ+e 〉|2 = |〈χ+e |χ+z 〉|2.The probability of measuring+�/2 along the+z-direction if the state was initiallyalong the positive (1,1,1) direction is the same as the probability of measuring+�/2 along the (1,1,1) direction if the state was initially along the +z-axis.

If the spin is initially in an eigenstate along the direction e, the probability ofmeasuring ±�/2 in a direction perpendicular to e is always 1/2. For example, ifthe spin was initially in the state χ+y the probability of measuring −�/2 along thex-axis is 1/2:

〈χ+y |χ−x 〉 =1

2(1 i)

(1−1

)= 1

2(1− i), (5.21)

|〈χ+y |χ−x 〉|2 =1

2. (5.22)

5.3 Irreducible representations of half-integer angular momentum

In Chapter 3 we discussed the D(l) irreducible representations for the rotationgroup. The matrix elements of the IR for orbital momentum l are given byTheorem 3.3. In that discussion l was an integer and the basis functions werethe spherical harmonic functions. The basis functions for half-integer momen-tum are the spinor functions. For spin-1/2 particles the basis functions are thetwo-component Pauli spinors. Even though the basis functions are not sphericalharmonic functions, the formula for D(l) remains valid, where now l is allowedto be integral or half-integral. For the spin J the rotation of the spin vector isachieved by applying the matrix D(J )(α, β, γ ) to the spin function. For spin 1/2,the IR matrix is

D(1/2)(α, β, γ ) =(

eiα/2 cos(β/2) eiγ /2 eiα/2 sin(β/2) e−iγ /2

−e−iα/2 sin(β/2) eiγ /2 e−iα/2 cos(β/2) e−iγ /2

).

(5.23)

(Note that the labeling of the matrix is D(1/2)11 = D(1/2)

+1/2,+1/2, D(1/2)12 = D(1/2)

+1/2,−1/2,

D(1/2)21 = D(1/2)

−1/2,+1/2, and D(1/2)22 = D(1/2)

−1/2,−1/2. The matrix is often written with therows and columns interchanged from this choice. The matrix of (5.23) rotates thefunction clockwise. Other texts give the matrix for a “positive” rotation, namely acounter-clockwise rotation.) We define the operator OR = PR Q R , where PR is ourprevious operator that operates on the spatial coordinates and Q R operates on thespin vector. A rotation of the state φ(r) χ(s) is given by

OR φ(r) χ(s) = PR φ(r) Q R χ(s) = φ(R−1r) D(1/2)(R) χ(s). (5.24)

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128 Electron spin and angular momentum

Consider a rotation by β about the y-axis. The result is

P(0,β,0) φ(r) D(1/2)(0, β, 0) χ(s) = φ(R(β)−1r) D(1/2)(0, β, 0) χ(s), (5.25)

where R(β)−1 is a rotation about the y-axis by an angle β in a clockwise sense.The transformation of an arbitrary spin state due a rotation by β is

D(1/2)(0, β, 0) χ(s) =(

cos(β/2) sin(β/2)−sin(β/2) cos(β/2)

)(ab

)

=(

a cos(β/2)+ b sin(β/2)−a sin(β/2)+ b cos(β/2)

), (5.26)

where |a|2 + |b|2 = 1. For the spin state χ+z , a = 1 and b = 0, so the rotated state,χβ , is

χβ =(

cos(β/2)−sin(β/2)

). (5.27)

The probability of measuring +�/2 along the rotated vector direction is

|〈χ+z |χβ〉|2 = cos2

2

), (5.28)

and the probability of measuring −�/2 along the rotated vector is sin2(β/2). Forthe state χ−z we have a = 0 and b = −1, and the rotated vector is

χβ =(−sin(β/2)

cos(β/2)

), (5.29)

|〈χ+z |χβ〉|2 = sin2

2

), (5.30)

|〈χ−z |χβ〉|2 = cos2

2

). (5.31)

In general, we can write |〈χ±|D(1/2)(R)|χ±〉|2 as the probability of measuring±�/2 along the direction of the rotated vector, D(1/2)(R)χ±, if the spin eigen-state was χ± before rotation, where R represents the rotation in terms of the Eulerangles, α, β, and γ . In some cases, it is easier to specify the rotation in terms ofthe components of e rather than in terms of the Euler angles.

The combination of the Pauli matrices (σx + iσy)/2 ≡ σ+ is a spin-raising oper-ator. When acting on χ−z it produces the state χ+z . When acting on χ+z it producesa null state,

σ+ =(

0 10 0

), (5.32)

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5.4 Multi-electron spin–orbital states 129

and (0 10 0

)(01

)=(

10

), (5.33)

(0 10 0

)(10

)=(

00

). (5.34)

Similarly, σ− ≡ (σx − iσy)/2 lowers the spin. When operating on χ+z it producesχ−z . When operating on χ−z it produces a null state,

σ− =(

0 01 0

), (5.35)

and (0 01 0

)(10

)=(

01

), (5.36)

(0 01 0

)(01

)=(

00

). (5.37)

5.4 Multi-electron spin–orbital states

For the Russell–Saunders or L–S coupling scheme [5.4] the spins combine to forma total spin vector, S = ∑

si , and the orbital angular momenta combine to forma total orbital angular momentum, L = ∑

li . The basis states for a multi-electronsystem can be built up from products of one-electron orbitals and one-spin func-tions. Usually the spatial orbitals used are the hydrogen-like orbitals (sphericalharmonic functions) and the spinors are the Pauli two-state functions discussedin Section 5.1. For the L–S scheme the built-up functions for the angular and spinvariables are constructed so that they are eigenfunctions of the operators Lz , L2, Sz ,and S2. A multi-electron state is labeled by the eigenvalues of these operators. Suchfunctions are eigenstates of the hydrogen-like Hamiltonian, H0. As a consequence,there are no non-zero matrix elements of H0 between two states that differ in anyof the eigenvalue labels. All angular-momentum operators have units of � and thesquared angular-momentum operator has units of �2. It is conventional to use a setof dimensionless operators defined as follows for the total angular momenta:

Ms = 1

�Sz, M2

s =1

�2S2, (5.38)

ML = 1

�Lz, M2

L =1

�2L2, (5.39)

MJ = 1

�Jz, M2

J =1

�2J2. (5.40)

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130 Electron spin and angular momentum

Table 5.1 Notation for the operators and their eigenvalues used in thischapter. Bold symbols indicate operators. Non-bold symbols are numbers.Capitalized symbols refer to multi-electron operators and uncapitalizedsymbols to single electrons.

Operator Eigenvalue

M2L (L2/�2) L(L + 1), where L is a positive integer

MLz (Lz/�) ML , a positive or negative integerm2

l (l2/�2) l(l + 1), where l is a positive integermlz (lz/�) ml , a positive or negative integerM2

S (S2/�2) S(S + 1), S is a positive integer or half-integerMSz (Sz/�) MS , a positive or negative integer or half-integerms2 (s2/�2) s(s + 1), where s =1/2msz (lz/�) ms,±1/2M2

J (J2/�2) J (J + 1), J is a positive integer or half-integerMJz (Jz/�) MJ , a positive or negative integer or half-integerm2

j (j2/�2) j ( j + 1), j is a positive integer or half-integerm jz (jz/�) m j , a positive or negative integer or half-integer

The notation for the single-electron operators is

ms = 1

�sz, m2

s =1

�2s2, (5.41)

ml = 1

�lz, m2

l =1

�2l2, (5.42)

m j = 1

�jz, m2

j =1

�2j2. (5.43)

The notation for the operators and their eigenvalues is given in Table 5.1.

5.5 The L–S-coupling scheme

For the L−S or Russel–Saunders coupling scheme [5.4], the multi-electron statesare eigenstates of M2

L ,MLz ,M2S , and MSz . The multi-electron states are built up

from one-electron eigenstates of the operators, m2l ,mlz ,m2

s , and ms . These one-electron spin-orbitals are products of a hydrogen-like orbital function times a Paulispin function.

Consider the product of N , one-electron spin-orbitals:

� = φ(r1, s1; n1,ml1,ms1) φ(r2, s2; n2,ml2,ms2) . . . φ(rN , sN ; nN ,mlN ,msN ),

(5.44)where n is the radial or principal quantum number, φ(r, s; n,ml,ms) =Rnl Y ml

l χz(ms). The eigenvalue of χz(ms) is ms . That is, msχz(ms) = msχz(ms)

and ms can have the values ±1/2.

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5.5 The L – S-coupling scheme 131

For our discussions in this section we can drop some of the labels and write theproduct of the N one-electron states as

� = φ(ml1,ms1;ml2,ms2;ml3,ms3; . . . ;mlN ,msN ). (5.45)

As defined, � is an eigenstate of the operators, ml,ms,ML , and MS ,

ML� = ML� =(∑

mlk

)� (k = 1, 2, . . . , N ), (5.46)

MS� = MS� =(∑

msk

)� (k = 1, 2, . . . , N ). (5.47)

Usually, � is not an eigenstate of M2L or M2

S and it is necessary to construct a linearcombination of such product states in order to obtain eigenstates.

A physically allowed multi-electron state must be antisymmetric in the inter-change of any two different pairs of space–spin variables. One method of formingsuch an antisymmetric state is to use a Slater determinant,

det{φ(ml1,ms1;ml2,ms2;ml3,ms3; . . . ;mlN ,msN )} =1√N ! ×∣∣∣∣∣∣∣∣∣

φ(r1, s1; n1,ml1,ms1) φ(r1, s1; n2,ml2,ms2) . . . φ(r1, s1; nN ,mlN ,msN )

φ(r2, s2; n1,ml1,ms1) φ(r2, s2; n2,ml2,ms2) . . . φ(r2, s2; nN ,mlN ,msN )...

.... . .

...

φ(rN , sN ; n1,ml1,ms1) φ(rN , sN ; n2,ml2,ms2) . . . φ(rN , sN ; nN ,mlN ,msN )

∣∣∣∣∣∣∣∣∣.

(5.48)

The coordinate variables r and s are constant across the rows, and the state labelsare constant down the columns.

It is useful to introduce antisymmetrization operator, A, which is defined by itsaction on a product function:

A φ(ml1,ms1;ml2,ms2;ml3,ms3; . . . ;mlN ,msN )

= 1√N ! det{φ(ml1,ms1;ml2,ms2;ml3,ms3; . . . ;mlN ,msN )}. (5.49)

A commutes with all of the angular-momentum operators and this allows a greatsimplification. We can apply any of the angular-momentum operators to a productfunction and then apply A to the result instead of having to work with the N ! termsof the determinantal state. For example,

M2L det{φ(ml1,ms1;ml2,ms2;ml3,ms3; . . . ;mlN ,msN )}=M2

L A φ(ml1,ms1;ml2,ms2;ml3,ms3; . . . ;mlN ,msN )

= A{M2Lφ(ml1,ms1;ml2,ms2;ml3,ms3; . . . ;mlN ,msN )}

= 1√N

det{M2L φ(ml1,ms1;ml2,ms2;ml3,ms3; . . . ;mlN ,msN )}.

(5.50)

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132 Electron spin and angular momentum

Another method used to construct antisymmetrized wavefunctions is to use theproduct of an antisymmetric spatial state and a symmetric spin state or an antisym-metric spin state and a symmetric space state. In many instances the determinantalstate can be factored into a product of a space times a spin function. A deter-minantal state with all spins parallel can be factored into an antisymmetricspace function, det[φ1(ml1)φ2(ml2) . . . φN (mlN )], and a symmetric spin product,σ1σ2 . . . σN , where all the σ j are spin up or all are spin down. The two-electronstate, (1/2)[φa(1)φa(2)±φb(1)φb(2)](α1β2−β1α2), is the product of a symmetri-cal space function times an antisymmetric spin function. This state can be obtainedfrom two Slater determinants,

1

2[φa(1)φa(2)± φb(1)φb(2)](α1β2 − β1α2)

= 1

2{det[φaα, φaβ] ± det[φbα, φbβ]}. (5.51)

Spin orientations in real space

In the discussions we have referred loosely to “spin-up” and “spin-down” statesof an electron. However, it is worth noting that the α (spin up) and β (spin down)eigenstates (1,0) and (0,1) do not correspond to spin vectors in real space that arealigned along the z-axis. For the α state (1,0) is the eigenstate of the operator sz

with eigenvalue �/2, and also the eigenstate of the operator s2 with eigenvalue(1/2)(1/2 + 1)�2 = (3/4)�2. (1,0) is not an eigenstate of the operator s. Themagnitude |s| = (3/4)1/2� and the magnitude of the component along the z-axis is�/2. Therefore the spin vector in real space makes an angle with the z-axis givenby (3/4)1/2 cos θ = 1/2 or θ = cos−1(1/

√3) = 54.74◦.

We have spoken also of “parallel” and “antiparallel” two-electron spin states,for example, α1α2 or α1β2. For two L−S-coupled spins, the state α1α2 does notcorrespond to the spin vectors being parallel in real space. We have S = s1 + s2

and S2 = s21 + s2

2 + 2s1 · s2 = s21 + s2

2 + 2|s1||s2| cosϕ, where ϕ is the anglebetween the real-space vectors s1 and s2. Thus cosϕ = [S2 − (s2

1 + s22)]/(2|s1||s2|)

= 1/3. This result gives ϕ = 70.53◦. On the other hand, for “antiparallel” spins,S2 = 0, so cosϕ = −1 or ϕ = π . Therefore the real-space spins are actuallyantiparallel.

5.6 Generating angular-momentum eigenstates

We shall use the notation (ml1,ms1; ml2,ms2; ml3,ms3; . . . ; ml N ,ms N ) to indi-cate the product state, �p φ(mlp,msp) = φ(ml1,ms1; ml2,ms2; ml3,ms3; . . . ;ml N ,ms N ), and |ml1,ms1; ml2,ms2; ml3,ms3; . . . ; ml N ,ms N 〉 to represent thedeterminantal state. By construction, both the product states and the determinantalstates are eigenstates of MSz and MLz ,

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5.6 Generating angular-momentum eigenstates 133

Table 5.2 Raising and lowering operators

Operator Effect of operator

m±sk = (sxk ± isyk)/� m±sk φ(mlk,msk) = φ(mlk,msk + 1) δ(msk,∓1/2)

M±S =∑

k m±sk M±S �p φ(mlp,msp) =∑k �p �=k φ(mlp,msp)×φ(mlk,msk + 1) δ(msk,∓1/2)

Also M±S �(L , S, ML , MS) = √(S ∓ MS)(S ± MS + 1)×�(L , S, ML , MS ± 1)

m±lk = (lxk ± i lyk)/� m±lk φ(mlk,msk) = √(l ∓ mlk)(l ± mlk + 1) φ(mlk ± 1,msk)

M±L =∑

k m±lk M±L �p φ(mlp,msp) =∑k �p �=k φ(mlp,msp)

×√(l ∓ mlk)(l ± mlk + 1)×φ(mlk ± 1,msk)

Also M±L �(L , S, ML , MS) = √(L ∓ ML)(L ± ML + 1)×�(L , S, ML ± 1, MS)

m±jk = (jxk ± ijyk)/� m±jk φ(mlk,msk) = (m±lk +m±sk) φ(mlk,msk)

M±J =∑

k m±jk M±J �p φ(mlp,msp) =∑k(m±lk +m±sk) �p �=k φ(mlp,msp)

Also M±J �(J, MJ ) = √(J ∓ MJ )(J ± MJ + 1) �(J, MJ ± 1)

�p φ(mlp,msp) = φ(ml1,ms1;ml1,ms1; . . . ;ml N ,ms N )= φ(ml1,ms1) φ(ml2,ms2) . . . φ(ml N ,ms N )

�(L , S, ML , MS) is a properly normalized L S eigenstate�(J, MJ ) is a properly normalized J M eigenstate

MSz� ={∑

k

msk

}� =

{∑k

msk

}� = Ms�, (5.52)

MLz� ={∑

k

mlk

}� =

{∑k

mlk

}� = Ml�, (5.53)

where � is the product state φ(ml1,ms1; ml2,ms2; ml3,ms3; . . . ; ml N ,ms N ) or thedeterminantal state |ml1,ms1; ml2,ms2; ml3,ms3; . . . ; ml N ,ms N 〉.

Raising and lowering operators

The angular-momentum-raising and -lowering operators are defined in Table 5.2.We have

m±sk(mlk,msk) = (mlk, (msk ± 1)), (5.54)

with the understanding that a state with spin > 1/2 or spin < −1/2 is a null state.

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134 Electron spin and angular momentum

M2s can be expressed in terms of the one-electron raising and lowering operators

M2s =

(∑k

msk

)·(∑

k′msk′

)

=∑

k

(msk)2 +

∑k

∑k′ �=k

{mskmsk′ + 1

2(m+skm−sk′ +m−skm+sk′)

},

(5.55)

where the sum is over the N electrons, k = 1 to N and k ′ = 1 to N for the doublesummation.

With the aid of (5.55) we can operate on a determinantal state with M2s ,

M2s |ml1,ms1;ml2,ms2; . . . ;ml N ,ms N 〉

={

N

(3

4

)+∑

k′

∑k>k′

2mskmsk′

}|ml1,ms1;ml2,ms2; . . . ;ml N ,ms N 〉

+∑

k,k′<k

|ml1,ms1;ml2,ms2; . . . ;mlk,msk+1; . . . ;mlk′,msk′−1; . . . ;ml N ,ms N 〉

+∑

k,k′<k

|ml1,ms1;ml2,ms2; . . . ;mlk,msk−1; . . . ;mlk′,msk′+1; . . . ;ml N ,ms N 〉,

(5.56)

where (mlk,msk ± 1) indicates that the spin is raised (lowered) by one unit.For two electrons,

M2s = 2

(3

4

)+ 2ms1ms2 + (m+s1m−s2 +m−s1m+s2) (5.57)

and

M2s |ml1,ms1;ml2,ms2〉 =

(3

2+ 2ms1ms2

)|ml1,ms1;ml2,ms2〉

+ |ml1, (ms1 + 1);ml2, (ms2 − 1)〉+ |ml1, (ms1 − 1);ml2, (ms2 + 1)〉. (5.58)

For three electrons the operator is

M2s = 3

(3

4

)+ (2ms1ms2 + 2ms1ms3 + 2ms2ms3)

+ (m+s1m−s2 +m−s1m+s2)+ (m+s1m−s3 +m−s1m+s3)

+ (m+s2m−s3 +m−s2m+s3). (5.59)

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5.6 Generating angular-momentum eigenstates 135

In a similar fashion we can express the operator M2L in terms of the one-electron

operators. We define the orbital raising and lowering operators by

m±lk =lx ± i ly

�. (5.60)

Their action on a one-electron state according to Table 5.2 is

m−lk(mlk,msk) =√(lk + mlk)(lk − mlk + 1) ((mlk − 1),msk), (5.61)

m+lk(mlk,msk) =√(lk − mlk)(lk + mlk + 1) ((mlk + 1),msk). (5.62)

Also,

m2lk(mlk,msk) = lk(lk + 1)(mlk,msk). (5.63)

The value of lk appearing on the right-hand side of (5.60)–(5.62) is the value of theangular-momentum quantum number of Y m

l from which (mlk,msk) is constructed.For example, if the one-electron orbital, φk , is for a d-electron state, then lk = 2.Using these one-electron operators, we find that

M2L =

(∑k

mlk

)·(∑

k′mlk′

)

=∑

k

(mlk)2 +

∑k

∑k′ �=k

{2mlkmlk′ + 1

2(m+lkm−lk′ +m−lkm+lk′)

}.

(5.64)

For two electrons (5.64) gives

M2L |ml1,ms1;ml2,ms2〉= {l1(l1 + 1)+ l2(l2 + 1)+ 2(ml1ml2)}|ml1,ms1;ml2,ms2〉+√(l1 − ml1)(l1 + ml1 + 1) |(ml1 + 1),ms1; (ml2 − 1),ms2〉+√(l2 + ml2)(l2 − ml2 + 1) |(ml1 − 1),ms1; (ml2 + 1),ms2〉.

(5.65)

5.6.1 Russell–Saunders (L–S) states

For this coupling scheme, the orbital momenta of the electrons combine to forma total L = ∑

lk and the spins combine to form a total S = ∑sk . Table 5.2

defines the raising and lowering operators that operate on the product of states ordeterminantal states,

M−S =∑

k

m−sk, (5.66)

M−L =∑

k

m−lk . (5.67)

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136 Electron spin and angular momentum

Let �(L , S, ML , MS) be an eigenstate of the operators L2, S2, Lz , and Sz , witheigenvalues L(L + 1), S(S+ 1), ML , and MS , respectively. �(L , S, ML , MS) maybe a determinantal state or a linear combination of determinantal states. Accordingto Table 5.2, we have

M−S �(L , S, ML , MS) =√(S + MS)(S − MS + 1) �(L , S, ML , MS − 1),

(5.68)where �(L , S, ML , MS − 1) is a properly normalized eigenstate whose MS islowered by 1. If we operate with M−L ,

M−L �(L , S, ML , MS) =√(L + ML)(L − ML + 1) �(L , S, ML − 1, MS),

(5.69)where �(L , S, ML − 1, MS) is a properly normalized eigenstate with the ML low-ered by one unit. The operators M−S and M−L allow us to generate all of the levelsof a given term. For example, consider a set of three equivalent d-electrons. Thehighest spin state with the highest ML will have an ml1 = 2, ms1 = +1/2, ml2 = 1,ms2 = +1/2, ml3 = 0, ms3 = +1/2 state. We simplify our notation so that|2+1+0+〉 represents the corresponding determinantal state, where the numbers arethe ml values and ± indicates the spin state, ±�/2. Using the operators in theprevious section, we find that |2+1+0+〉 is an eigenstate with L = 3, S = 3/2,ML = 3, and MS = 3/2. Using the notation 4 F(L , ML , S, MS), this particularstate is 4 F(3, 3, 3/2, 3/2).1

To generate the 2S + 1 spin states (all with L = 3 and S = 3/2), we operate on|2+1+0+〉 with M−S . Using Table 5.2, we obtain

M−S [4 F(3, 3, 3/2, 3/2)] = √3 4 F(3, 3, 3/2, 1/2) =∑

k

m−sk |2+1+0+〉

= [|2−1+0+〉 + |2+1−0+〉 + |2+1+0−〉]. (5.70)

Hence,

4 F(3, 3, 3/2, 1/2) = 1√3

[|2−1+0+〉 + |2+1−0+〉 + |2+1+0−〉

]. (5.71)

Each application of M−S reduces MS by 1:

M−S [4 F(3, 3, 3/2, 1/2)] = 2 [4 F(3, 3, 3/2,−1/2)].Also,

M−S [4 F(3, 3, 3/2, 1/2)] =∑

k

m−sk

1√3

{|2−1+0+〉 + |2+1−0+〉 + |2+1+0−〉

}

1 Note that the order of L , ML , S, and Ms in 4 F(L , ML , S, Ms ) is different from that used for �(L , S, ML , Ms ).

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5.6 Generating angular-momentum eigenstates 137

= 1√3

{2|2−1−0+〉 + 2|2−1+0−〉 + 2|2+1−0−〉

};(5.72)

therefore

4 F(3, 3, 3/2,−1/2) = 1√3

{|2−1−0+〉 + |2−1+0−〉 + |2+1−0−〉

}. (5.73)

Then

M−S [4 F(3, 3, 3/2,−1/2)] = 3[4 F(3, 3, 3/2,−3/2)]=∑

k

m−sk

1√3

{|2−1−0+〉 + |2−1+0−〉 + |2+1−0−〉

}

= 1√3{3|2−1−0−〉}, (5.74)

and therefore

4 F(3, 3, 3/2,−3/2) = |2−1−0−〉. (5.75)

To generate the 2L + 1 orbital states (all with L = 3 and S = 3/2) we apply M−Lto the eigenstate |2+1+0+〉:

M−L [4 F(3, 3, 3/2, 3/2)] = √6[4 F(3, 2, 3/2, 3/2)] =∑

k

m−lk |2+1+0+〉

= {2|1+1+0+〉 + √6|2+0+0+〉 + √6|2+1+−1+〉}.(5.76)

The determinantal states |1+1+0+〉 and |2+0+0+〉 vanish, since each has twoidentical rows; therefore

4 F(3, 2, 3/2, 3/2) = |2+1+−1+〉. (5.77)

A second application of M−L gives

M−L [4 F(3, 2, 3/2, 3/2)] = √10[4 F(3, 1, 3/2, 3/2)] =∑

k

m−lk |2+1+−1+〉

={

2|2+1+−2+〉 + √6|2+0+−1+〉}, (5.78)

so that

4 F(3, 1, 3/2, 3/2) ={√

3

5|2+0+−1+〉 +

√2

5|2+1+−2+〉

}(5.79)

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138 Electron spin and angular momentum

Continuing this process gives

4 F(3, 0, 3/2, 3/2) = 1√5|1+0+−1+〉 +

√4

5|2+0+−2+〉, (5.80)

4 F(3,−1, 3/2, 3/2) =√

3

5|1+0+−2+〉 +

√2

5|2+−1+−2+〉, (5.81)

4 F(3,−2, 3/2, 3/2) =(√

18

50+√

8

50

)|1+−1+−2+〉

= |1+−1+−2+〉, (5.82)4 F(3,−3, 3/2, 3/2) = |0+−1+−2+〉. (5.83)

All (2S + 1)(2L + 1) = 4 × 7 = 28 states of the 4 F term can be generated inthis manner by operating with both M−L and M−S . For example, to find the state4 F(3,−1, 3/2, 1/2) we can operate with M−S on 4 F(3,−1, 3/2, 3/2) or operatewith M−S M−L on 4 F(3, 0, 3/2, 3/2) to obtain

4 F(3,−1, 3/2, 1/2) =√

3

15

{|1−0+−2+〉 + |1+0− − 2+〉 + |1+0+−2−〉

}

+√

2

15

{|2− − 1+−2+〉 + |2+−1− − 2+〉+ |2+−1+−2−〉

}. (5.84)

5.7 Spin–orbit interaction

The spin–orbit effect in atomic physics arises from the interaction between themagnetic moment of the electron and the charge of the nucleus. A simple clas-sical model serves to show how this interaction comes about. If one “sits” onthe electron, the nuclear charge appears to be circling around the electron. Themoving nuclear charge creates an electrical current that produces a magnetic fieldB = −v × E/c2, where v is the velocity of the electron and E is the electricfield due to the nuclear charge. E = −∇r V (r) = [−∂V (r)/∂r ]r/r , with V (r)being the electrostatic potential energy experienced by the electron due to thenuclear charge. In the case of a hydrogenic atom, V (r) = −Ze2/r , where +Zeis the effective nuclear charge. For this case, [−∂V (r)/∂r ]r/r = (−Ze2/r2)r/r .The angular momentum, L, of the electron is given by L = r × p = mer × v,so B = 1/(mec2)[(1/r)(∂V (r)/∂r)]L. The magnetic moment of the electron isμs = −(gsμB/�)S, where the g-factor, gs, is approximately 2, and μB is the

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5.7 Spin–orbit interaction 139

Bohr magneton. The interaction energy between the spin magnetic moment andthe orbital magnetic field is

− μs · B = μBgs

mec2�

1

r

∂V (r)

∂rL · S. (5.85)

This classical result differs from the correct result by a factor of 1/2. According toDirac’s relativistic quantum theory, the actual coefficient of L · S is half that givenin (5.85). For the N -electron case each electron spin interacts with its own orbit, so

Hso =N∑

k=1

ξk(nl)(r)lk · sk, (5.86)

ξk(r) = 1

2μB

gs

mec2�

1

r

∂Vk(r)

∂rk. (5.87)

The result (5.86) includes only the interaction between an electron’s spin and itsown orbit. There are also interactions between a spin and another electron’s orbitalmomentum. In most cases these “spin–other-orbit” interactions are much smallerthan the “spin–same-orbit” interactions.

When the spin–orbit interaction is large, as it is with heavier atoms, the kthelectron’s spin couples with its own orbital momentum to form jk = lk + sk . Fora multi-electron system, J = ∑

jk . As we shall show below, we can find linearcombinations of one-electron spin-orbitals that are eigenstates of j2, l2, s2, and jz .For these states the spin–orbit interaction is diagonal:

lk · sk = �2

2[ jk( jk + 1)− lk(lk + 1)− sk(sk + 1)]. (5.88)

The energy associated with the spin–orbit interaction for a single-electron spin-orbital that is an eigenstate of j2, l2, and s2 is

〈Hso〉 = �2

2〈ξnl〉{ j ( j + 1)− l(l + 1)− s(s + 1)}, (5.89)

where

〈ξnl〉 =∫|Rnl(r)|2ξ(r)r2 dr. (5.90)

The matrix elements of l · s between states differing in any of the eigenvalues ofj2, l2, s2, and jz vanish. From (5.89) we see that 〈Hso〉 is independent of the valuesof jz , lz , and sz . The factor 〈ξnl〉 depends on the principal quantum number n, andon l.

For an N -electron system the spin–orbit interaction is

〈Hso〉 =∑

k

〈ξk(nl)〉lk · sk = 1

2

∑k

〈ξk(nl)〉[j2k − l2k − s2

k], (5.91)

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140 Electron spin and angular momentum

and, for states for which j , l, and s are “good” quantum numbers, the energy is

〈Hso〉 = 1

2�2

∑k

〈ξk(nl)〉{ jk( jk + 1)− lk(lk + 1)− sk(sk + 1)}. (5.92)

5.7.1 Spin–orbit splitting of one-electron states

The eigenvalue of j2 is j ( j+1). For a single electron with orbital momentum l, thebasis functions transform according to D(l) and the spinor functions according toD(1/2); therefore the spin-orbitals are the basis for the direct product D(l)×D(1/2) =D(l−1/2)+D(l+1/2). As a result, j can have two possible values, namely j = l+1/2and j = l − 1/2, corresponding to s parallel and antiparallel to l. The spin–orbitenergy, 〈Hso〉, has two possible values, one for each eigenvalue of j2:

[〈Hso〉]± = �2

2〈ξnl〉(l ± 1/2){(l ± 1/2+ 1)− l(l + 1)− 3/4}

= �2

2〈ξnl〉[(l − 1/2)± 1/2]. (5.93)

For l = 0 the spin–orbit interaction is zero, as may be seen from (5.88). Thereforeeach one-electron state with l > 0 is split into two states by the spin–orbit interac-tion. This splitting is observed spectroscopically and is referred to as fine structure.For a hydrogenic atom the spin–orbit energy can be evaluated analytically. Theresult is that

〈Hso〉 = A(n, l)Z4[ j ( j + 1)− l(l + 1)− s(s + 1)], (5.94)

where

A(n, l) = μ0

gsμ2B

n3a30 l(l + 1/2)(l + 1)

, (5.95)

n is the principal quantum number, l > 0 is the angular-momentum quantum num-ber, μ0 is the permeability of vacuum (μ0/(4π) = 10−7 N/A2), and a0 is the Bohrradius. The 2p level (l = 1) of hydrogen splits into j = 3/2 and j = 1/2 stateswith an energy separation of about 4.5× 10−5 eV.

5.7.2 j–m j spin-orbitals

The one-electron spin-orbitals can be characterized by n, l, ml , and ms . We canalso choose the quantum numbers n, l, j , and m j . Multi-electron states can be builtup from the latter spin-orbitals with the quantum numbers J and MJ (instead of Land ML). We have that M2

J = (∑

m jk)2 and MJ z = ∑

m j z = ∑(mlkz +mskz),

and, according to Table 5.1, the eigenvalues of M2J and MJ z are J (J + 1) and MJ .

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5.7 Spin–orbit interaction 141

Our first task is to find one-electron states that are eigenstates of j2. Thehydrogen-like spin-orbital φ(n, l,ml,ms) = Rnl(r) Y ml

l (θ, φ) γ (s), where n isthe principal quantum number and l is the angular-momentum quantum num-ber. The spinor function γ (s) is either α (spin up or +) or β (spin down or −).φ(n, l,ml,ms) is an eigenstate of jz but not of j2. The problem then comes downto finding linear combinations of these spin-orbitals that are eigenfunctions of j2.To simplify the notation in this section, we suppress the n and l quantum numbersand write φ(n, l,ml,ms) (for a fixed n and l) as φ(ml,ms).

Since m j = ml +ms , we have

m2j = m2

l +m2s + 2ml · ms

= l(l + 1)+ s(s + 1)+ 2mlms +m+l m−s +m−l m+s . (5.96)

Therefore

m2j φ(ml,ms) = [l(l + 1)+ s(s + 1)+ 2mlms] φ(ml,ms)

+√(l − ml)(l + ml + 1) φ(ml + 1,ms − 1)

+√(l + ml)(l − ml + 1) φ(ml − 1,ms + 1). (5.97)

Consider the spin-orbital φ(l, 1/2). This function has the maximum ml (ml = l)and maximum ms (ms = 1/2). By operating on this state with m2

j we obtain

m2j φ(l, 1/2) = [l(l + 1)+ 3/4+ l] φ(l, 1/2)

= [(l + 1/2)(l + 1/2+ 1)] φ(l, 1/2). (5.98)

Therefore this state is an eigenstate of jz and j2 with eigenvalues m j = l+ 1/2 andj = l + 1/2. However, if we operate on the state φ(l,−1/2) we obtain

m2j φ(l,−1/2) = [l(l + 1)+ 3/4− l] φ(l,−1/2)+√2l φ(l − 1, 1/2). (5.99)

Similarly,

m2j φ(l−1, 1/2) = [l(l+1)+3/4+l−1] φ(l−1, 1/2)+√2l φ(l,−1/2). (5.100)

Equations (5.99) and (5.100) show that neither φ(l,−1/2) nor φ(l − 1, 1/2) isan eigenstate of j2. However, these two states are coupled only to each other, andtherefore we can use a linear combination to form an eigenstate of j2. We take� = A φ(l,−1/2)+B φ(l−1, 1/2), where A and B are constants to be determinedby requiring that j2� = λ�. The resulting secular equation is(

E1 − l − λ√

2l√2l E1 + l − 1− λ

)(AB

)= 0, (5.101)

where E1 = l(l+1)+3/4. The two solutions are λ1 = l2+2l+3/4 = (l+1/2)(l+1/2+1), corresponding to j = l+1/2, and λ2 = l2−1/4 = (l−1/2)(l−1/2+1),

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142 Electron spin and angular momentum

corresponding to j = l − 1/2. The eigenvalue λ1 corresponds to s parallel to l andλ2 corresponds to s antiparallel to l. By substituting the eigenvalues for λ into(5.101) the ratio A/B can be determined, and hence the eigenvectors. This processyields

�(l + 1/2, l − 1/2) =√

2l φ(l − 1, 1/2)+ φ(l,−1/2)√2l + 1

, (5.102)

�(l − 1/2, l − 1/2) = −√

2l φ(l,−1/2)+ φ(l − 1, 1/2)√2l + 1

. (5.103)

Now that we have the two eigenstates corresponding to j = l+1/2 and j = l−1/2,we can generate all of the other m j states by applying the operator m−j = m−l +m−sto these two eigenstates. Using Table 5.2, we obtain

m−j �( j,m J ) =√( j + m J )( j − m J + 1) �( j,m J − 1). (5.104)

Applying this to (5.102) gives

m−j �(l + 1/2, l − 1/2) = √4l �(l + 1/2, l − 3/2). (5.105)

We can also write

m−j �(l + 1/2, l − 1/2) = (m−l +m−s )√

2l φ(l − 1, 1/2)+ φ(l,−1/2)√2l + 1

=√

2l[√4l − 2 φ(l − 2, 1/2)+ 2φ(l − 1,−1/2)]√2l + 1

,

(5.106)

so that

�(l + 1/2, l − 3/2) =√

2l[√4l − 2 φ(l − 2, 1/2)+ 2φ(l − 1,−1/2)]√4l√

2l + 1.

(5.107)

Similarly,

m−j �(l − 1/2, l − 1/2) = √2l − 1 φ(l − 1/2, l − 3/2), (5.108)

and

(m−l +m−s )−√2l φ(l,−1/2)+ φ(l − 1, 1/2)√

2l + 1

= −(2l − 1)φ(l − 1,−1/2)+√4l − 2 φ(l − 2, 1/2)√2l + 1

, (5.109)

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5.7 Spin–orbit interaction 143

Table 5.3 �( j,m j ) functions in terms of φ(ml,ms) for l = 1 (p-electron)

j m j �( j,m j ) in terms of φ(ml ,ms)

3/2 3/2 (1+)3/2 1/2 [√2(0+)+ (1−)]/√33/2 −1/2 [(−1+)+√2(0−)]/√33/2 −3/2 (−1−)1/2 1/2 [(0+)−√2(1−)]/√31/2 −1/2 [√2(−1+)− (0−)]/√3

Notation: (m±l ) = φ(l = 1,ml ,ms = ±1/2)

Table 5.4 �( j,m j ) functions in terms of φ(ml,ms) for l = 2 (d-electron)

j m j �( j,m j ) in terms of φ(ml ,ms)

5/2 5/2 (2+)5/2 3/2 [2(1+)+ (2−)]/√55/2 1/2 [√3(0+)+√2(1−)]/√55/2 −1/2 [√2(−1+)+√3(0−)]/√55/2 −3/2 [(−2+)+ 2(−1−)]/√55/2 −5/2 (−2−)3/2 3/2 [(1+)− 2(2−)]/√53/2 1/2 [√2(0+)−√3(1−)]/√53/2 −1/2 [√3(−1+)−√2(0−)]/√53/2 −3/2 [−2(−2+)+ (−1−)]/√5

Notation: (m±l ) = φ(l = 2,ml ,ms = ±1/2)

from which it follows that

�(l − 1/2, l − 3/2) = −(2l − 1)φ(l − 1,−1/2)+√4l − 2 φ(l − 2, 1/2)√2l + 1

√2l − 1

.

(5.110)

Using the procedure outlined above, we can easily find the �( j,m j ) states for anyvalue of the orbital momentum, l. Results for l = 1, 2, and 3 are summarized inTables 5.3–5.5.

5.7.3 Two-electron j– j spin-orbitals

Many-electron states characterized by j1, j2, J , and MJ can be built up for the one-electron spin-orbitals of the type in Tables 5.3–5.5 which are characterized by theeigenvalues j and m j . In general, the eigenstates of J2 involve a linear combinationof products of these states with j1 and j2 fixed, but with various values of m j . The

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144 Electron spin and angular momentum

Table 5.5 �( j,m j ) functions in terms of φ(ml,ms) for l = 3 ( f -electron)

j m j �( j,m j ) in terms of φ(ml ,ms)

7/2 7/2 (3+)

7/2 5/2 [√6(2+)+ (3−)]/√7

7/2 3/2 [√5(1+)+√2(2−)]/√7

7/2 1/2 [2(0+)+√3(1−)]/√7

7/2 −1/2 [√3(−1+)+ 2(0−)]/√7

7/2 −3/2 [√2(−2+)+√5(−1−)]/√7

7/2 −5/2 [(−3+)+√6(−2−)]/√7

7/2 −7/2 (−3−)

5/2 5/2 [(2+)−√6(3−)]/√7

5/2 3/2 [√2(1+)−√5(2−)]/√7

5/2 1/2 [√3(0+)− 2(1−)]/√7

5/2 −1/2 [2(−1+)−√3(0−)]/√7

5/2 −3/2 [√5(−2+)−√2(−1−)]/√7

5/2 −5/2 [√6(−3+)− (−2−)]/√7

Notation: (m±l ) = φ(l = 3,ml ,ms = ±1/2)

reason for this is that J2, J−, or J+ operating on �( j1,m j1)�( j2,m j2) does notchange j1 or j2. Because of this feature, the spin–orbit interaction energy can easilybe evaluated.

A one-electron spin-orbital characterized by j will transform according to D( j),where j can have the values l ± 1/2. A two-electron state constructed from aproduct of two one-electron states will transform according to the direct productD( j1) × D( j2) = D( j1+ j2) + D( j1+ j2−1) + D( j1+ j2−2) + · · · + D| j1− j2|. As an exampleconsider the configuration npn′ p with n unequal to n′. There are several possiblepairs of j1 and j2: (3/2, 3/2), (3/2, 1/2), (1/2, 3/2), and (1/2, 1/2). The followingdecompositions apply:

D(3/2)n × D(3/2)

n′ = D(3) + D(2) + D(1) + D(0), (5.111)

D(3/2)n × D(1/2)

n′ = D(2) + D(1), (5.112)

D(1/2)n′ × D(3/2)

n = D(2) + D(1), (5.113)

D(1/2)n × D(1/2)

n′ = D(1) + D(0), (5.114)

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5.7 Spin–orbit interaction 145

where the subscripts n and n′ are reminders that the principal quantum numbers nand n′ are not equal. Thus the combination of two inequivalent p-electrons leadsto terms with J = 3, 2, 1, and 0; J = 2, 1; J = 2, 1; and J = 1, 0. Each levelis (2J + 1)-fold degenerate, and therefore there is a total of 36 states. In the L−Sscheme we have six np spin-orbitals and six n′ p spin-orbitals. The first electroncan go into any of the six np states and the second into any of the six n′ p states.Therefore the total number of states for npn′ p is 6× 6 = 36.

In the case of two equivalent electrons, (np)2, the Pauli exclusion principle for-bids two electrons occupying the same spin-orbital. The exclusion principle is notbuilt into the results of (5.111)–(5.114). The effect is to forbid D(3) and D(1) inthe D(3/2) × D(3/2) decomposition and D(1) in the D(1/2) × D(1/2) decomposition.In addition, the terms in (5.112) and (5.113) are indistinguishable. Therefore, thepermitted terms are J = 0 and J = 2 for D(3/2) × D(3/2), J = 1 and J = 0 forD(1/2)× D(3/2) and J = 0 for D(1/2)× D(1/2). These terms give a total of 15 states.This result is clear in the L–S scheme. The first spin-orbital can be chosen in sixways but the second can be chosen in only five ways. This reduces the number ofstates from 6 × 6 = 36 to 6 × 5 = 30. The states (ml1 , ms1 , ml2 , ms2 ) are indis-tinguishable from the states (ml2 , ms2 , ml1 , ms1 ), and this reduces the number ofremaining states by a factor of 2. Therefore, the total number of distinct states fortwo equivalent p-electrons is 15.

It is easy to see why the J = 3 for D(3/2) × D(3/2) and the J = 1 forD(1/2) × D(1/2) are present for two inequivalent electrons but absent for the twoequivalent electrons. The state

√2|nl+, n′l+〉 (l = 1, n �= n′) is an eigenstate of

j21, j2

2, and J2 with eigenvalues j1 = 3/2, j2 = 3/2, and J = 3. For equivalentelectrons

√2|nl+, nl+〉 (l = 1, n = n′) is clearly a null state and forbidden by

the exclusion principle. Similarly, the state√

2|nl−, n′l−〉 is an eigenstate of j21, j2

2,and J2 with eigenvalues j1 = 1/2, j2 = 1/2, and J = 1. For equivalent electrons,√

2|nl−, nl−〉 (l = 1, n = n′) is a null state.For two electrons the determinantal state can be written as a product of a spin

and a space function. The decomposition of the direct product for the spatial rep-resentations, D(1) × D(1) = D(2) + D(1) + D(0), where D(0) and D(2) correspondto symmetric spatial functions and D(1) corresponds to an antisymmetric spatialfunction. Therefore the S and D terms are singlets, while the P term is a triplet.The spectroscopic terms are two 1S0, one 3 P1, and two 1 D2.

As an example of how the j–j states can be constructed, consider the case of twoequivalent p-electrons. Our aim is to find the 15 eigenstates of the operators j1, j2,J2, and MJ . If we use products of the spin-orbital in Table 5.3 we are assured thatthe basis states are eigenstates of j1, j2, m j1 , and m j2 . Furthermore, we know thatthe eigenstates of J2 must be linear combinations of states having the same value

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146 Electron spin and angular momentum

Table 5.6 j– j-Coupled states for J = 2 for two equivalent p-electrons

J = 2 ( j1, j2) = (1/2, 1/2)

MJ |�( j1,m j1) �( j2,m j2)〉2 |�(3/2, 3/2) �(1/2, 1/2)〉1 {√3|�(3/2, 1/2) �(1/2, 1/2)〉 + |�(3/2, 3/2) �(1/2,−1/2)〉}/2

0 {|�(3/2,−1/2) �(1/2, 1/2)〉 + |�(3/2, 1/2) �(1/2,−1/2)〉}/√2

−1 {√3|�(3/2,−1/2) �(1/2,−1/2)〉 + |�(3/2,−3/2) �(1/2, 1/2)〉}/2

−2 |�(3/2,−3/2) �(1/2,−1/2)〉Notation: |�( j1,m j1) �( j2,m j2)〉 = det{�( j1,m j1) �( j2,m j2)}/

√2

�( j,m j ) is defined in Table 5.3.

for m j1+m j2 = MJ . This is true because J2 applied to any of the spin-orbitals mayresult in a sum of orbital functions, but they all have the same MJ .

As mentioned earlier, for two equivalent p-electrons J = 3 does not occur. Togenerate the J = 2 level, we start with any product function �( j1,m j1) �( j2,m j2)

with m j1 + m j2 = MJ = 2, e.g., the product �(3/2, 3/2) �(1/2, 1/2), where the�( j,m j ) functions are listed in Table 5.3. We operate with J2 on the state, andobtain

J2|�(3/2, 3/2) �(1/2, 1/2)〉= { j2

1 + j22 + 2(m j1m j2)+ j+1 j−2 + j−1 j+2 }|�(3/2, 3/2) �(1/2, 1/2)〉

= 6|�(3/2, 3/2) �(1/2, 1/2)〉. (5.115)

Therefore, �(3/2, 3/2) �(1/2, 1/2) is an eigenstate of J2 arising from D(3/2) ×D(1/2), with J = 2 and MJ = 2. We can generate all of the states belonging to thisterm by operating with M−J . The results are shown in Table 5.6.

Another product that yields MJ = 2 is |�(3/2, 3/2) �(3/2, 1/2)〉, and

J2|�(3/2, 3/2) �(3/2, 1/2)〉= 9|�(3/2, 3/2) �(3/2, 1/2)〉 + 3 |�(3/2, 1/2) �(3/2, 3/2)〉= 6|�(3/2, 3/2) �(3/2, 1/2)〉= 2(2+ 1)|�(3/2, 3/2) �(3/2, 1/2)〉. (5.116)

Therefore this state is an eigenstate with J = 2 and MJ = 2 arising fromD(3/2)×D(3/2). Operating repeatedly on this state with M−J gives the results shownin Table 5.7.

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5.7 Spin–orbit interaction 147

Table 5.7 j– j-Coupled states for J = 2 for two equivalent p-electrons

J = 2 ( j1, j2) = (3/2, 3/2)

MJ |�( j1,m j1) �( j2,m j2)〉2 |�(3/2, 3/2) �(3/2, 1/2)〉1 |�(3/2, 3/2) �(3/2,−1/2)〉0 {|�(3/2, 1/2) �(1/2,−1/2)〉 + |�(3/2, 3/2) �(3/2,−3/2)〉}/√2

−1 |�(3/2, 1/2) �(3/2,−3/2)〉−2 |�(3/2,−1/2) �(3/2,−3/2)〉Notation: |�( j1,m j1) �( j2,m j2)〉 = det{�( j1,m j1) �( j2,m j2)}/

√2

�( j,m j ) is defined in Table 5.3.

For the J = 1 states we select the product |�(3/2, 3/2) �(1/2,−1/2)〉 forwhich MJ = 1. We find that

M2J |�(3/2, 3/2) �(1/2,−1/2)〉 = 3|�(3/2, 3/2) �(1/2,−1/2)〉

+√3|�(3/2, 1/2) �(1/2, 1/2)〉(5.117)

and

M2J |�(3/2, 1/2) �(1/2, 1/2)〉 = 5|�(3/2, 3/2) �(1/2,−1/2)〉

+√3|�(3/2, 3/2) �(1/2,−1/2)〉.(5.118)

To form an eigenstate of M2J we take a linear combination of the two coupled states:

� = A|�(3/2, 3/2) �(1/2,−1/2)〉 + B|�(3/2, 1/2) �(1/2, 1/2)〉, (5.119)

where A and B are coefficients to be selected so that M2J� = λ�. The eigenvalue

equation is (3− λ

√3√

3 5− λ

)(AB

)= 0. (5.120)

The vanishing of the determinant give the roots λ = 4 ± 2. The lower sign givesλ = 2 = 1(1 + 1) and corresponds to J = 1. Substituting λ = 2 into the secularequation determines the ratio of A to B, and the normalized state arising fromD(3/2) × D(1/2) is

� = −√3|�(3/2, 3/2) �(1/2,−1/2)〉 + |�(3/2, 1/2) �(1/2, 1/2)〉. (5.121)

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148 Electron spin and angular momentum

Table 5.8 j– j-Coupled states for J = 1 for two equivalent p-electrons

J = 1 ( j1, j2) = (3/2, 1/2)

MJ |�( j1,m j1) �( j2,m j2)〉1 {|�(3/2, 1/2) �(1/2, 1/2)〉 − √3|�(3/2, 3/2) �(1/2,−1/2)〉}/2

0 {|�(3/2,−1/2) �(1/2, 1/2)〉 − |�(3/2, 1/2) �(1/2,−1/2)〉}/√2

−1 {|�(3/2,−1/2) �(1/2,−1/2)〉 − √3|�(3/2,−3/2) �(1/2, 1/2)〉}/2

Notation: |�( j1,m j1) �( j2,m j2)〉 = det{�( j1,m j1) �( j2,m j2)}/√

2

�( j,m j ) is defined in Table 5.3.

The other states for J = 1 generated by use of M−J are shown in Table 5.8.So far we have found 13 of the 15 eigenstates. The remaining two have J =

0. As a generating function we choose |�(3/2, 3/2) �(3/2,−3/2)〉, which hasMJ = 0. We have

M2J |�(3/2, 3/2) �(3/2,−3/2)〉 = 3|�(3/2, 3/2) �(3/2,−3/2)〉

+ 3|�(3/2, 1/2) �(3/2,−1/2)〉(5.122)

and

M2J |�(3/2, 1/2) �(3/2,−1/2)〉 = 3|�(3/2, 1/2) �(3/2,−1/2)〉

+ 3|�(3/2, 3/2) �(3/2,−3/2)〉.(5.123)

The two states |�(3/2, 1/2)�(3/2,−1/2)〉 and |�(3/2, 3/2)�(3/2,−3/2)〉 arecoupled. On solving the secular equation we find that the J = 0 state arising fromD(3/2) × D(3/2) is

1√2

{|�(3/2, 3/2) �(3/2,−3/2)〉 − |�(3/2, 3/2) �(3/2,−3/2)〉

}. (5.124)

The other J = 0 state, namely that arising from D(1/2) × D(1/2), is

|�(1/2, 1/2) �(1/2,−1/2)〉. (5.125)

Thus, by selecting product functions having the desired MJ value and using theoperators M2

J and M−j , we are able to generate all of the j–j-coupled eigenstates.These j–j-coupled states can be expressed in terms of the L–S-coupled states. Ingeneral the transformation of the j–j-coupled states to or from the L–S-coupledstates is a tedious task when the number of electrons is more than two.

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5.7 Spin–orbit interaction 149

Table 5.9 Spin–orbit energies of j– j-coupled states with n = 1 and J = 2 fortwo equivalent p-electrons

j1 j212

∑{ jk( jk + 1)− lk(lk + 1)− sk(sk + 1)}ξknl Degeneracy

3/2 3/2 ξknl 63/2 1/2 −ξknl/2 81/2 1/2 −2ξknl 1

With the j– j coupling the spin–orbit energy is easily evaluated. For the twoequivalent p-electrons lk = 1 and sk = 1/2, and the values of jk can be readdirectly from the eigenstate labels. The spin–orbit energy does not depend on MJ

or m j , and hence is constant for the various levels of a given J term.For two equivalent p-electrons the spin–orbit energies are shown in Table 5.9.

The column labeled “degeneracy” lists the number of states that have the pair j1 andj2. We note that the (np)2 configuration is split into three equally spaced groups bythe spin–orbit interaction. In addition, the sum of the spin–orbit energies times theirdegeneracies is equal to zero. This is a general feature of spin–orbit splitting. The“center of gravity” of the energy levels is unaffected by the spin–orbit interaction.

5.7.4 The nuclear-shell model

The neutrons and protons of a nucleus are held together by very strong nuclearinteractions. The binding energy or stability of a nucleon depends on the numberof protons (Z ) and the number of neutrons (N ) comprising the nucleus. Imaginebuilding a nucleus by successively adding neutrons and protons. It is experimen-tally observed that the binding energy of the last added nucleon is particularlylarge for special numbers called “magic numbers”. The magic numbers for neu-trons or protons are 2, 8, 20, 28, 50, 82, and 126. In some cases both Z andN are magic numbers (not necessarily the same number), in which case thenucleus is “doubly magic” and has exceptional stability. The calcium isotopeshaving Z = 20 and N = 20 and 28 are doubly magic nuclei. These twoisotopes are much more stable than the other isotopes of calcium. The occur-rence of magic numbers suggests that the nucleus may have a shell structureanalogous to the atomic shells for electrons. Other evidence includes the obser-vations that (1) the stable isotopes of radioactive series all have magic numbersfor N and Z , and (2) the neutron-absorption cross-section is lower for magic-numbered nuclei than for other isotopes. An empirical nuclear-shell model similarto the electronic-shell model was developed by E. P. Wigner, M. Goeppert-Mayer,and J. H. D. Jensen, who shared the Nobel prize in 1963 for their contributions.

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150 Electron spin and angular momentum

The model accounts for the magic numbers and also predicts the total spin ofnuclei [5.5]. Although there are strong similarities between the nuclear-shell modeland the electronic-shell model, protons and neutrons are different particles, andeach species has its own shells. Second, there is no physical, central core aboutwhich the nucleons circulate. Nevertheless, the nuclear-shell model supposes thateach nucleon moves in a spherically symmetric potential due to its interactions withthe other nucleons. Various empirical potentials that decay exponentially beyondthe nuclear radius have been suggested, but all depend only on the magnitude ofr . This assumption is essential because it allows the Schrödinger equation for anucleon to be separated into radial and angular functions. The angular functionsare the same spherical harmonics as those used for atomic states. The one-nucleoneigenstates are then characterized by angular momentum l, magnetic quantumnumber ml , spin s, and ms . Unlike in the atomic case, l is not limited by n (inthe atomic case l can not exceed n.) The magic numbers are then associatedwith the filling of a proton or neutron shell. To obtain the correct magic num-bers, a strong spin–orbit interaction must be included in the model. This meansthat the states must be characterized by j , m j , J , M j , and parity (“g” or “u”states). The nucleons are Fermi particles, and therefore obey the Pauli exclu-sion principle. They can be assigned to the various shells just as one does forelectrons. The filled shells have zero total J . Therefore the nucleons occupyingthe unfilled shells determine the value of the total angular momentum and thevalue of the total spin. The shell model correctly predicts the total J and spin ofnuclei [5.5].

5.8 Crystal double groups

For the rotation group the formula for the character of a rotation through an angleα given by (3.20),

χ l(α) = sin[(l + 1/2)α]sin(α/2)

, (5.126)

was derived from the properties of the spherical harmonic functions for integerangular momentum, l. The spinor functions that are the basis for half-integermomentum are not spherical harmonic functions; however, the same formulaapplies. The formula for the character can be generalized to read

χ J (α) = sin[(J + 1/2)α]sin(α/2)

, (5.127)

where J represents integer or half-integer angular momentum. If we now considera rotation through an angle α + 2π we obtain the result that

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5.8 Crystal double groups 151

χ J (α + 2π)

= sin[(J + 1/2)(α + 2π)]sin(α/2+ π)

= sin[(J + 1/2)α] cos[2π(J + 1/2)] + cos[(J + 1/2)α] sin[2π(J + 1/2)]sin(α/2) cos(π)+ cos(α/2) sin(π)

= sin[(J + 1/2)α] cos[2π(J + 1/2)]sin(α/2) cos(π)

= χ J (α) cos(2π J )

= (−1)2Jχ J (α). (5.128)

When J is a half-integer (5.128) shows that a rotation by 2π has the character−1 and that χ J (α+ 2π) is the negative of χ J (α). However, χ J (α+ 4π) = χ J (α)

when J is an integer or half-integer. Since any rotation that is a multiple of 2πhas been assumed to be equivalent to E , the identity, we see that the character ofa rotation for half-integer momentum is not single-valued and, in fact, the D(J )(R)

representation matrices for half-integer J are determined only up to a ± sign.A method for handling this sign ambiguity for half-integer J has been developed.

The approach introduces an element, R, that corresponds to a rotation by 2π . Ecorresponds to a rotation by 4π . A new group, G∗ (the crystal double group), canbe constructed from the original group, G, using as elements C and RC , where Cis any element of G. This results in a group that has twice as many elements andtherefore additional IRs. Although there are twice as many elements, there need notbe twice as many classes, and hence there need not be twice as many IRs. However,if � is an IR of G with characters χ(Ci ), then an IR of the double group has thecharacters χ(RCi ) = χ(Ci ). Thus the characters of Nκ IRs of G∗ are immediatelydetermined, where Nκ is the number of IRs of G.

The rules for determining the class structure, originally due to Opechowski [5.6],are summarized below.

Let G be the original group and {Cα} a class in G, then the sets {Cα} and {RCα} aredifferent classes in G∗. An exception to this rule occurs if and only if Cαi is a rotation byπ and there is another C2 rotation about an axis perpendicular to the symmetry axis of Cαi .In such a case, Cαi and RCαi are in the same class.

In atomic theory the basis functions for the IRs of the rotation group are spher-ical harmonic functions when J is an integer [5.6]. For half-integer J the basisfunctions are linear combinations of the spin-orbitals we have been discussing inthis chapter. The double group provides the IRs needed in order to decompose therepresentations based on spin-orbitals in exactly the same manner as that in whichthe representations based on spatial orbitals are decomposed. The complexities ofconstructing the IRs of the double group can be found elsewhere [5.7] and will notbe discussed here. As an example of a double group, the character table for the

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152 Electron spin and angular momentum

Table 5.10 O ′: The crystal double group for octahedral symmetry. The Roperator is a rotation by 2π and E is a rotation by 4π . The IRs of the O groupare shown on the far right. The new IRs, namely �6, �7, and �8, are thoseresulting from extending the O group.

O ′ E R 8C3 8RC3 3C2 + 3RC2 6C2 + 6RC2 6C4 6RC4 O

�1 1 1 1 1 1 1 1 1 A1

�2 1 1 1 1 1 −1 −1 −1 A2

�3 2 2 −1 −1 2 0 0 0 E

�4 3 3 0 0 −1 −1 1 1 T1

�5 3 3 0 0 −1 1 −1 −1 T2

�6 2 −2 1 −1 0 0√

2 −√2 New

�7 2 −2 1 −1 0 0 −√2√

2 IRs

�8 4 −4 −1 1 0 0 0 0

extended O group, denoted as O ′, is presented in Table 5.10. We shall refer to theupper block (�1 through �5) as the “old IRs” and the lower block (�6 through �8)as the “new IRs”.

For integral values of J , the representation D(J ) has χ J (RC) = χ J (C), whilethe characters of the new IRs satisfy χ�(new)(RC) = −χ�(new)(C). Therefore thedecomposition of D(J ) into the new IRs is

n�(new IR) = 1

48

∑{χ�(new)(C)}∗χ J (C)+ {χ�(new)(RC)}∗χ J (RC)

= 1

48

∑{χ�(new)(C)}∗χ J (C)− {χ�(new)(C)}∗χ J (C) = 0,

(5.129)

where the sum is over all of the 24 operations of the O group. Equation (5.129)shows that the representations corresponding to integral J contain none of thenew IRs and therefore decompose entirely into the old IRs of O ′. Conversely,for half-integer J , χ(RC) = −χ(C), while the characters of the old IRs satisfyχ�(RC) = +χ�(C). Consequently, the decomposition for a half-integer represen-tation is entirely into the new IRs. The decomposition of D(J ) into the IRs of thedouble group corresponding to octahedral symmetry is given for several values ofJ in Table 5.11.

It follows from these decompositions that the J = 1/2 and J = 3/2 levelsare not split by an octahedral field. A “p” spin-orbital will have j = 3/2 and

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5.9 The Zeeman effect (weak-magnetic-field case) 153

Table 5.11 Decomposition of D(J ) into the IRs of O ′

D(1/2) = �6 D(7/2) = �6 + �7 + �8

D(3/2) = �8 D(9/2) = �6 + 2�8

D(5/2) = �7 + �8 D(11/2) = �6 + �7 + 2�8

D(0) = �1 D(3) = �2 + �4 + �5

D(1) = �4 D(4) = �1 + �3 + �4 + �5

D(2) = �3 + �5 D(5) = �3 + 2�4 + �5

4F

4F9/2

4F7/2

4F5/2

4F3/2

Γ8

Γ6

Γ8

Γ7

Γ8

Γ6

Γ8

Γ7

Γ8

Free ion Octahedralcrystal field

Figure 5.2 A schematic representation of the splitting of the 4 F term in anoctahedral environment.

j = 1/2 levels, neither of which will be split by an octahedral crystal field. A “d”spin-orbital will have j = 5/2 and j = 3/2 levels. The 5/2 level splits into adoublet (�7) and a quartet (�8), while the 3/2 level remains unsplit.

The splitting of a multi-electron term can also be easily deduced now. For exam-ple, for a d3 configuration (e.g., V2+) the 4 F term has L = 3, S = 3/2. J can haveinteger values ranging from L + S to |L − S|. That is, J can have the values 9/2,7/2, 5/2, and 3/2. From Table 5.11 we see that in an octahedral environment theJ levels are split as shown in Fig. 5.2.

5.9 The Zeeman effect (weak-magnetic-field case)

In the presence of an external magnetic field, the orbital and spin magneticmoments interact with the field. The potential energy due to this magneticinteraction is

HM = −μ · B = μB

∑k

(lk + gssk) · B = μB

∑k

(mlk + gsmsk)B, (5.130)

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154 Electron spin and angular momentum

JL

S

J = L + S

L

S

Figure 5.3 A schematic representation of the classical picture of L and S pre-cessing about J as J precesses about the z-axis. The projection of J onto thez-axis is m J . Since L and S precess about J , their projection onto the z-axis isnot constant.

where μB is the Bohr magnetron, B is the external magnetic field (also knownas the induction, field density, or flux density). The g-factor for the spin, gs =2.002 319, will for simplicity be taken as 2 in our discussion. We may assumethat B is oriented along the z-axis, so for states described by the L–S-couplingscheme

HM = μB(Lz + 2Sz)B = μB(ML + 2MS)B. (5.131)

For a system with spherically symmetric electron potentials the (2L + 1) leveleigenstates of H0 are degenerate as are the (2S + 1) states of each level. Theperturbation, HM , splits the degeneracy of the L–S states.

However, often the spin–orbit interaction is larger than HM . In that case thestates are best described in terms of J and MJ rather than L and S. The magneticinteraction will remove the (2J + 1) degeneracy of the J terms, but the calculationof the splitting is complicated because ML and MS are no longer appropriate quan-tum numbers. The classical picture is that the vectors L and S precess about J asshown in Fig. 5.3.

Equation (5.130) is not useful because L and S are no longer “good” quantumnumbers. Since L is precessing about J , its time-averaged value is just the projec-tion of L onto J . The components perpendicular to J average to zero. The sameargument applies to S. Therefore,

〈L〉avg = (L · J)JJ 2

, (5.132)

〈S〉avg = (S · J)JJ 2

, (5.133)

HM ≈ μB

(L · J)J+ 2(S · J)JJ 2

J · B,

= μB MJ B(L · J)J+ 2(S · J)J

J 2. (5.134)

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References 155

The terms (L · J) and (S · J) can be eliminated by use of the relations

(L · J) = 1

2(J 2 + L2 − S2) = �2

2[J (J + 1)+ L(L + 1)− S(S + 1)], (5.135)

(S · J) = 1

2(J 2 − L2 + S2) = �2

2[J (J + 1)− L(L + 1)+ S(S + 1)], (5.136)

J · B = Jz Bz = � MJ B. (5.137)

On substituting (5.135)–(5.137) into (5.134) we obtain

〈HM〉avg = μB MJ B

× [J (J + 1)+ L(L + 1)− S(S + 1)] + 2[J (J + 1)− L(L + 1)+ S(S + 1)]2J (J + 1)

= μB MJ B

{3

2+ S(S + 1)− L(L + 1)

2J (J + 1)

}= μBgL MJ B, (5.138)

where gL, the factor enclosed in curly brackets, is called the Landé g-factor. Itsvalue depends on L and S as well as on J .

Equation (5.138) is useful in evaluating 〈HM〉avg for a J level belonging to anL–S-coupling term. For example, the ground state of V with the d3 configurationis (2S+1)FJ = 4 F3/2, so J = 3/2, L = 3, and S = 3/2. In a spherically symmet-ric environment the (2J + 1) values of MJ are degenerate. In the presence of amagnetic field these levels are split. According to (5.138),

〈HM〉avg = μB MJ B

{3

2+ 15/4− 12

15/2

}= 2

5MJμB B. (5.139)

Therefore, for MJ = ±3/2 and ±1/2, 〈HM〉avg = ±3/5 and ±1/5, respectively,in units of μB B.

References

[5.1] S. A. Goudsmit and G. E. Uhlenbeck, “Ersetzung der Hypothese vom unmechanis-chen Zwang durch eine Forderung bezüglich des inneren Verhaltens jedes einzelnenElektrons”, Naturwissenschaften 47, 953–954 (1925).S. A. Goudsmit and G. E. Uhlenbeck, “Over het roteerende electron en de structuurder spectra”, Physica 6, 273–290 (1926).

[5.2] P. A. M. Dirac, “The quantum theory of the electron”, Proc. Roy. Soc. A 117, 610–624 (1928).

[5.3] W. Pauli, “Zur Quantenmechanik des magnetischen Elektrons”, Z. Phys. 43, 601–623 (1927).

[5.4] E. U. Condon and G. H. Shortley, The Theory of Atomic Spectra (Cambridge:Cambridge University Press, 1963).

[5.5] M. Goeppert-Mayer, Elementary Theory of Nuclear Shell Structure (New York:Wiley, 1960). M. Goeppert-Mayer, “On closed shells in nuclei, II”, Phys. Rev. 75,1969–1970 (1949).

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156 Electron spin and angular momentum

[5.6] W. Opechowski, “Sur les groupes cristallographiques ‘doubles’ ”, Physica 7, 552(1940).N. B. Backhouse, “Projective character tables and Opechowski’s theorem”, Physica70, 505 (1973).

[5.7] V. Heine, Group Theory in Quantum Mechanics (New York: Pergamon Press,1960).

Exercises

5.1 A spin is in the eigenstate χ+x and a measurement is made with the z′-axisoriented along the (1, 1, 0) direction.(a) Find the probability that the measurement result is +(1/2)�.(b) Find the probability of measuring −(1/2)�.(c) Show that the sum of the probabilities is unity.

5.2 (a) Find the spin eigenstate χ+e for the z′-axis along the (1, 0, 1) directionusing Eq. (5.13).

(b) Using the IR-matrix, D(1/2), rotate χ+z so that it lies along the (1, 0, 1)direction.

(c) Show that the vectors of (a) and (b) are equal.5.3 Consider the d2 state, |2+, 1+〉.

(a) Using the operators M2L and M2

S show that L = 3 and S = 1 (in unitsof �).

(b) Use the operator L− to generate the remaining 2L states. (Compare yourresults with those derived in Chapter 4, in Table 4.4.)

(c) Show that 3 F(3, 3, 1) = |2+, 1+〉 can be separated into the product of anantisymmetric spatial function times a symmetric spinor.

(d) Use M−S on the spinor to find the three spin states.5.4 Consider two equivalent d-electrons.

(a) Use Table 5.3 to show that the determinantal j– j-coupled state|�(5/2, 5/2) �(3/2, 3/2)〉 = 1/

√5|2+, 1+〉 − 4/

√5|2+2−〉 in L–S

notation.(b) Show that the state is an eigenstate of m2

j1, m2j2, MJ , and M2

J with eigen-values j1 = 5/2, j2 = 3/2, MJ = 4, and J = 4. Use m jk = mlk+msk ,MJ =∑

(m jk)z , and M2J = (

∑m jk)

2.(c) What is the spin–orbit energy of this level?

5.5 (a) Find one of the eigenstates for the J = 7/2 and J = 5/2 levels for anf -electron (l = 3) in terms of L–S-coupled states.

(b) What is the spin–orbit energy of each level?5.6 (a) For a pair of f -electrons find the J = 5 and J = 6 eigenstates in the

j–j scheme for the pair j1 = 7/2 and j2 = 5/2.

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Exercises 157

Hint: Operate on the state |�(7/2, 5/2) �(5/2, 5/2)〉 with M2J .

(b) What is the spin–orbit energy of these states?5.7 Consider an f 3 outer electron configuration.

(a) Show that the L–S state |3+, 2+, 1+〉 is an eigenstate of M2L and M2

S andalso of M2

J .(b) How would this spectroscopic term be written?

5.8 (a) Use the direct-product rules to determine the allowed J values for a pdelectronic configuration.

(b) Show that the total number of states is 60.(c) Find the J = 4, MJ = 3 state expressed as a linear combination of

L–S-coupled states.5.9 Find the Zeeman level splittings of the 1S1/2, 2 P1/2, and 2 P3/2 terms of

hydrogen (in units of μB B).

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6

Molecular electronic structure: The LCAO model

Much of what we understand about the chemistry and optical properties ofmolecules has come from theoretical studies of very simple, empirical models.In most cases the theoretical models employ such drastic approximations that onemay wonder why the results have any relevance at all to actual molecular systems.The success of these models may be attributed almost entirely to their use of group-theoretical concepts. In many cases symmetry is the dominant factor determiningthe electronic structure of a molecule. While the models are crude approximations,the general structure imposed by symmetry is usually exact and often independentof the details of the model employed. As a result many of the features have a muchdeeper truth than the model from which they are derived.

In this chapter we discuss the use of the LCAO method (linear combinations ofatomic orbitals) to analyze the electronic structure of molecules. The term “atomicorbitals” is used loosely to mean one-electron orbitals whose angular functions arethe spherical harmonics. The precise specifications of the radial parts of the orbitalsare not needed for our discussion.

6.1 N-electron systems

It is generally assumed that the electronic states of a molecule or solid can becalculated for fixed positions of the nuclei. The electron’s velocity is very largecompared with the speed of vibratory motion, so that in effect the electronsinstantly readjust to any motion of the nuclei. This assumption is referred to asthe Born–Oppenheimer approximation. With that assumption the Hamiltonian forelectrons of the molecule is

Hmol =∑

i

−�2

2m∇2

i +∑

i

∑j<i

e2

|ri − r j | −∑RA

∑i

Z Ae2

|ri − RA| , (6.1)

158

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6.1 N-electron systems 159

where ri is the position vector of the i th electron, RA is the position vector of theAth nucleus, and e is the magnitude of the electron’s charge. We seek solutions ofthe Schrödinger equation,

Hmol �(r1, s1, r2, s2, . . . , rN , sN ) = E �(r1, s1, r2, s2, . . . , rN , sN ), (6.2)

where � is an antisymmetrized determinantal wavefunction. For a molecule withN > 2 the solutions of (6.1) are essentially impossible to find analytically. Indeed,even with the use of high-speed computers, numerical solutions of (6.2) withoutapproximations are virtually impossible. To derive useful information it is neces-sary to make a myriad of approximations. A large number of standard computerprograms that employ various approximations and limited basis sets can producereasonable solutions for the molecular states of small molecules. However, sim-pler models that employ more drastic approximations are susceptible to theoreticalanalysis and provide substantial insight into the electronic structure of molecules.

6.1.1 Hartree–Fock equations

The Hartree–Fock approach is the simplest method of obtaining approximate solu-tions for an N -electron molecule consistent with the Pauli principle. This methodreplaces the electron–electron interactions with an average or mean-field potentialbut neglects actual electron–electron correlations. Improved results can be obtainedby adding an empirical correlation potential that is based on the local electrondensity. The wavefunctions are Slater determinantal states constructed from theone-electron orbitals which are determined by the variational principle.

We begin with an orthogonal set of N (one-electron) wavefunctions ψk(τ ) =�k(r) χk(s), k = 1, 2, . . ., N , where τ represents r and s, and χk(s) is the spinoreigenstate (χ+z or χ−z ) for the kth state. The initial wavefunction is the Slaterdeterminant,

� = 1√N ! det{ψ1(τ1) ψ2(τ2) ψ3(τ3) . . . ψN (τN )}

= |ψ1(τ1) ψ2(τ2) ψ3(τ3) . . . ψN (τN )〉. (6.3)

The expectation value for the energy of the system is

〈E〉 = 〈ψ1(τ1)ψ2(τ2)ψ3(τ3) . . . ψN (τN )|H|ψ1(τ1)ψ2(τ2)ψ3(τ3) . . . ψN (τN )〉.(6.4)

According to the variational principle, the best choice of the spin orbitals (theψi (τi )) is determined by minimizing 〈E〉 with respect to variations in the orbitals.The variations are subject to the constraint that the orbitals remain orthogonalto one another. This constraint is imposed by the use of Lagrange multipliers.

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160 Molecular electronic structure: The LCAO model

The details of this calculation are available elsewhere [6.1, 6.2, 6.3] and will not begiven here. The resulting Hartree–Fock equations for the one-electron orbitals are{

− �2

2m∇2 + VN(r)+ VC(r)− Vex(r)

}�k(r) = εk �k(r), (6.5)

where

VN(r) = −∑RA

Z Ae2

|r− RA| , (6.6)

VC(r) =∑

j

∫dr′ �∗j (r

′)e2

|r− r′| � j (r′), (6.7)

V (k)ex (r) =

∑j

∫dr′ �∗j (r

′)e2

|r− r′|� j (r)�k(r)

�k(r′)〈χ(sk)|χ(s j )〉. (6.8)

VN(r) is the potential due to the nucleus at RA whose charge is eZ A, VC(r) is theCoulomb (electrostatic) interaction of the kth electron with all of the other elec-trons of the system, and V (k)

ex (r) is the exchange interaction with all of the otherelectrons. The exchange interaction between electrons vanishes unless the spinsare parallel. That is, the term 〈χ(sk)|χ(s j )〉 vanishes unless both electrons are inthe same spin state. It should also be noted that the contribution to VC for j = k isexactly canceled out by a contribution to Vex for j = k so that there is no interactionof the electron with itself. Equation (6.5) has the form of a Schrödinger equa-tion for a one-electron system. The eigenvalue, εk , of the Hartree–Fock equationfor �k(r) plays the role of a one-electron energy. However, εk is not the ioniza-tion energy of an electron in the state �k(r); rather it is the energy differencebetween an N -electron determinantal state and an (N − 1)-electron determinan-tal state obtained by removing the kth spin-orbital (Koopman’s theorem). For largeN , εk approaches the ionization energy, but for small N the difference between thetwo can be substantial.

An approximate form of the exchange potential is often employed. One approx-imation, known as the “Xα approximation”, is based on the Fermi–Thomas modelfor which the exchange interaction is found to be proportional to the cube root ofthe charge density. The Xα exchange potential is

VXα(r) = 9

(3ρ(r)

)1/3

, (6.9)

where ρ(r) is the charge density at r,

ρ(r) = e∑

j

φ j (r)∗φ j (r). (6.10)

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6.1 N-electron systems 161

The empirical parameter, α, is usually chosen to be in the range between 2/3 and 1.The Hartree–Fock equations can not be solved directly since the potentials

depend on a-priori knowledge of the orbitals being sought. The equations mustbe solved by use of an iteration scheme. An initial set of orbitals is chosen andused to calculate the potentials. A new set of orbitals can be obtained from theHartree–Fock equations. These new orbitals can then be used to determine newpotentials. The process is iterated until the results are self-consistent, meaning thatthe input potentials do not differ significantly from the output potentials.

6.1.2 The LCAO method

The Hartree–Fock equations are the basis for numerical calculations of molecularand solid-state electronic structures. The LCAO method employs a finite set ofatomic or atomic-like orbitals as the starting set of basis states. The advantage ofthis approach is that it provides a very simple and intuitive interpretation of theelectronic structure.

For the LCAO method we assume the orbitals comprising the Slater determinantare linear combinations of hydrogen-like orbitals (states whose angular functionsare spherical harmonics),

�k(r) =∑

n

∑μ

C (k)nμ ϕμ(r− Rn), (6.11)

where ϕμ(r − Rn) is a hydrogen-like orbital for an atom centered at Rn . Theindex μ labels the different angular symmetry types μ = s, px , py , pz , dxy ,dxz , . . . The coefficients C (k)

nμ specify the amplitudes of the hydrogen-like orbitalscomprising �k .

If (6.11) is used in the Hartree–Fock equations (6.5), an eigenvalue equation forthe coefficients is obtained, namely∑

n

∑α

{Hmβ,nα − εk Smβ,nα}C (k)nα = 0, (6.12)

where

Hmβ,nα =∫

dr ϕβ(r− Rm)∗{−�2

2m∇2 + VN (r)+ VC(r)− V (k)

ex (r)}

×ϕα(r− Rn). (6.13)

Hmβ,nα is the matrix element of the Hartree–Fock Hamiltonian between hydrogenicstates centered at Rm and Rn with symmetries β and α, respectively, and

Smβ,nα =∫

dr ϕβ(r− Rm)∗ ϕα(r− Rn) (6.14)

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162 Molecular electronic structure: The LCAO model

is the overlap integral between the states. It should be noted that different hydrogen-like orbitals located on the same site are orthogonal to one another,

Smβ,mα = δαβ,

but orbitals located on different sites are not. The overlap-matrix elements, Smβ,nα,have magnitudes less than or equal to 1. From (6.14) it follows that Smβ,nα =(Snα,mβ)

∗. Since the overlap matrices are real, Smβ,nα = Snα,mβ .As mentioned before, the equations must be solved self-consistently. In princi-

ple, an infinite set of hydrogen-like orbitals is required. However, in practice theset is finite, and this leads to what are called “truncation” errors.

6.2 Empirical LCAO models

Equation (6.12) is the basis for empirical LCAO models that employ a smallnumber of atomic basis states and treat the matrix elements of the Hartree–FockHamiltonian and the overlap integrals as adjustable parameters to be determinedby experiment or more accurate numerical calculations. Such empirical modelsprovide conceptual understanding of the electronic structures of molecules.

6.2.1 LCAO matrix elements

If the Hartree–Fock Hamiltonian is approximated as a sum of spherically symmet-ric potentials centered on the atomic sites,

H(r) =∑

t

Ha(r− Rt), (6.15)

then

Hnα,mβ =∑

t

∫ϕα(r− Rn)

∗ Ha(r− Rt) ϕβ(r− Rm) dr. (6.16)

Equation (6.16) shows that there are several different types of matrix elements.

1. One-center integrals:∫ϕα(r− Rm)

∗ Ha(r− Rm) ϕα(r− Rm) dr (t = m = n).2. Two-center integrals:∫

ϕα(r− Rn)∗ Ha(r− Rt) ϕβ(r− Rm) dr (t = n or t = m).

3. Three-center integrals:∫ϕα(r− Rn)

∗ Ha(r− Rt) ϕβ(r− Rm) dr (n, t, m; all different).

In applying the LCAO method to large molecules or to solids it is necessaryto impose some cut-off distance Rc beyond which interactions and overlaps are

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6.3 Parameterized LCAO models 163

neglected. That is, interactions for which |Rn − Rm | > Rc, |Rt − Rm | > Rc, or|Rt − Rn| > Rc are neglected.

Since atomic wavefunctions decrease exponentially with distance, it is physi-cally reasonable to expect that the one-center integrals are larger than the two-center integrals and that the two-center integrals are larger than the three-centerintegrals. For a molecule with N atoms the numbers of one-center, two-center,and three-center integrals increase approximately as N , N 2, and N 3, respectively.Therefore, three-center integrals can only be neglected if N times the magni-tude of three-center integrals is small compared with the magnitude of two-centerintegrals.

6.3 Parameterized LCAO models

A simple, but very useful, approach to understanding the electronic structure ofmolecules is obtained by keeping only the one- and two-center integrals. Considertwo atoms, A and B, of a molecule separated by the vector RAB with hydrogen-like orbitals ϕα(r − RA) and ϕβ(r − RB). We define the matrix elements of theHamiltonian as Hα,β(RAB) and the overlap integral as Sα,β(RAB), and regard thesematrix elements as parameters. It is further assumed that the potentials are spheri-cally symmetric about each atomic site. This is a rather drastic assumption since thenon-spherical charge density between atoms is responsible for molecular bonding.However, the use of this assumption does not mean that the resulting wavefunc-tions will produce a spherical charge density. In fact, the resulting wavefunctionslead to accumulated density between bonded atoms just as one would expect. Theway to view this spherical-potential assumption is that it is the first step of a self-consistent procedure. The density computed from the wavefunctions is the secondstep in a self-consistent procedure.

With these assumptions the two-center matrix elements and overlap integralsvanish unless the two orbitals have the same symmetry about the internuclear axis,RAB . As a result matrix elements of the Hamiltonian have the same symmetryproperties as the overlap integrals.

The type of overlap between two orbitals centered on different atoms is classi-fied according to the symmetry about the line, RAB , joining the centers of the twoorbitals. If the overlap is axially symmetric about RAB it is referred to as a “sigma”(σ ) overlap. If the overlap has a nodal plane containing RAB it is called a “pi”(π ) overlap. If there are two different nodal planes containing RAB the overlap iscalled a “delta” (δ) overlap. We can then define the two-center matrix elements ofthe Hamiltonian using the same convention as for the overlap integrals. We shallwrite the overlap integrals as

[αβη]AB =∫

dr ϕα(r− RA)∗ ϕβ(r− RB), (6.17)

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164 Molecular electronic structure: The LCAO model

ϕα ϕβ

s s

(ssσ)

zH

pz s

(spσ)

z

H

z

y

(ppπ)

py py

Hz

pz pz

(ppσ)

H

Figure 6.1 Fundamental Hamiltonian matrix elements for s- and p-orbitals. Theshaded regions have positive phase and the unshaded regions indicate negativephase. The overlap integrals are defined in the same way (without the H).

where η indicates the overlap type, i.e., σ , π , δ, . . . For the Hamiltonian matrixelements we write

(αβη)AB =∫

dr ϕα(r− RA)∗ H(r) ϕβ(r− RB). (6.18)

With this convention, brackets indicate an overlap integral and parentheses indi-cate a one- or two-center matrix element of the Hamiltonian. Figure 6.1 defines thefundamental Hamiltonian matrix elements for s- and p-orbitals. The same defini-tions can be used for the two-center overlaps. The phases of the orbitals shown inFig. 6.1 are part of the definitions. A reversal of the phase of one orbital changesthe sign of the overlap or matrix element. For example, if we reverse the + and −lobes of pz , the [spσ ] overlap becomes −[spσ ]. If we reverse the phase of bothfunctions [spσ ] remains unchanged.

Given ϕα(r − Ri ) and ϕβ(r − R j ), on arbitrary centers the overlap can beexpressed in terms of the fundamental overlaps. To accomplish this the orbitalsare expressed as linear combinations of orbitals parallel and perpendicular to theinternuclear axis. The interactions can then be reduced to the fundamental matrixelements and overlap integrals. The fundamental integrals for the s- and p-orbitalsare listed in Table 6.1.

Consider two atomic sites for which Ri − R j = d(−1, 1, 0). The directioncosines are −l = m = 1/

√2. The interaction between a px orbital at the origin

and a py orbital at d(−1, 1, 0) according to Table 6.1 is Sxy = − 12 [ppσ ]+ 1

2 [ppπ ]and the Hamiltonian matrix element is Hxy = − 1

2(ppσ) + 12(ppπ). To distin-

guish between interactions at different distances we shall affix a subscript on the

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6.3 Parameterized LCAO models 165

Table 6.1 Overlap and matrix elements of the Hamiltonian betweenϕα(r− Ri ) and ϕβ(r− R j ). Here l, m, and n are the directioncosines of the vector R j − Ri .

Overlap Hamiltonian matrix element

Sss = [ssσ ] Hss = (ssσ)Ssx = l[spσ ] Hsx = l(spσ)Ssy = m[spσ ] Hsy = m[spσ ]Ssz = n[spσ ] Hsz = n[spσ ]Sxx = l2[ppσ ] − (1− l2)[ppπ ] Hxx = l2(ppσ)− (1− l2)(ppπ)Sxy = lm[ppσ ] − lm[ppπ ] Hxy = lm(ppσ)− lm(ppπ)Sxz = ln[ppσ ] − ln[ppπ ] Hxz = ln(ppσ)− ln(ppπ)For Ri = R j :Sαβ = δαβ Hαβ = εαδαβ

parentheses and brackets. For example, a nearest-neighbor interaction is [αβη]1 or(αβη)1, and a second-nearest-neighbor interaction is [αβη]2 or (αβη)2.

For the s- and p-orbital interactions a graphical method is efficient. The p-orbitals transform as ordinary vectors, and each p-orbital may be represented as anarrow, with the tip corresponding to positive phase. The s-orbital of positive phasecan be represented by a circle. The fundamental interactions are then graphicallyrepresented as [ssσ ] = ©©, (ssσ) = ©h©, [spσ ] = ©←, (spσ) = ©h←,[ppσ ] =←←, (ppσ) =←h←, [ppπ ] =↑↑, and (ppπ) =↑h↑. From symmetryconsiderations we see that©↑= 0, ↑→= 0,©h↑= 0, and ↑h→= 0. Figure 6.2illustrates how to calculate an s–p and a p–p overlap when the two atomic sitesare on the diagonal of a square. The same resolution applies to the Hamiltonianmatrix elements.

6.3.1 Orthogonalized basis functions

The matrix equation (6.12) has the same form as the molecular-vibration problemconsidered in Chapter 1. The group-theoretical methods employed in Chapter 1 canbe used to find the eigenvalues and eigenvectors for the electronic states. We cantransform the eigenvalue equation to standard form by a similarity transformation.Let C (k) to be the column vector whose elements are C (k)

nμ , and define a new vectorand Hamiltonian by

H′ = S−1/2 H S−1/2, (6.19)

D(k) = S1/2 C (k). (6.20)

Then the eigenvalue equation becomes

H′ D(k) = εk D(k). (6.21)

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166 Molecular electronic structure: The LCAO model

ϕs(r − Ri)∗ ϕpx(r − Rj) dr Ri = (0, 0, 0),Rj = (−1, 1, 0)

=

1√2

1√2

= 1√2

← + 1√2

= 1√2[spσ]

ϕpx(r − Ri)∗ ϕpx

(r − Rj) dr

== 1√

21√2

← ← + 1√2

1√2

↓ ↓

= 12[ppσ] + 1

2[ppπ]

ϕpx(r − Ri)∗ ϕpy

(r − Rj) dr

== 1√

21√2

← → + 1√2

1√2

↓ ↓

= −12[ppσ] + 1

2[ppπ]

Figure 6.2 A graphical method for calculating the s–p overlap and Hamiltonianmatrix elements.

The Hamiltonian H′ is a unitary, N × N matrix, where N is the total numberof hydrogen-like orbitals included in the analysis. C (k) is a column vector withN components, C (k)

nμ . Equation (6.21) is in standard eigenvalue form and thereforethe eigenvectors D(k) (k = 1 to N ) are orthogonal to one another. The new basisorbitals corresponding to (6.20) are

ξμ(r− Rn) =∑

m

∑ν

(S−1/2)mν,nμ ϕν(r− Rm). (6.22)

These functions, known as Löwdin orbitals, are more extended than thehydrogen-like orbitals, but they transform under the operations of the coveringgroup in exactly the same way as do the hydrogen-like orbitals. As a result, thesymmetry types of the matrix elements are identical to those of the hydrogen-likeorbitals. The advantage of the Löwdin orbitals is that they form an orthogonal set,and the overlaps between different sites are zero:∫

dr ξμ(r− Rn)∗ ξν(r− Rm)

=∑

j

∑α

∑k

∑β

S−1/2jα,nμ

∫dr ϕα(r− R j )

∗ ϕβ(r− Rk) S−1/2kβ,mν

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6.3 Parameterized LCAO models 167

=∑

j

∑α

∑k

∑β

S−1/2jα,nμ S jα,kβ S

−1/2kβ,mν

= (S−1/2 S S−1/2)nμ,mν = Inμ,mν = δnmδμν. (6.23)

In arriving at the final result in (6.23) we have used the fact that S−1/2jα,nμ = S

−1/2nμ, jα.

As a very simple example consider a diatomic molecule, AB, with a singleorbital on each atom. The hydrogen-like orbitals are ϕα(r− RA) and ϕβ(r− RB),and the overlap matrix and its inverse are

S =(

1 SS 1

), (6.24)

S−1/2 =(

a bb a

), (6.25)

where

a = 1

2

{1√

1+ S+ 1√

1− S

},

b = 1

2

{1√

1+ S− 1√

1− S

}. (6.26)

The Löwdin orbitals are then

ξα(r− RA) = a ϕα(r− RA)+ b ϕβ(r− RB), (6.27)

ξβ(r− RB) = a ϕβ(r− RB)+ b ϕα(r− RA). (6.28)

As an illustration, assume an overlap of S = 0.2, then a = 1.015 452 andb = −0.102 582. The Löwdin orbitals ξα(r − RA) and ξβ(r − RB) are orthogo-nal to one another since (a2 + b2)S + 2ab = 0. They are also properly normalizedsince a2 + b2 + 2abS = 1.

Empirical models using Löwdin orbitals are much simpler to deal with becausethe basis states form an orthogonal set. The absence of overlap integrals reducesthe number of parameters by roughly a factor of two. The question of whether thehydrogen-like-orbital model with overlap integrals has more adjustable parame-ters than an empirical model that is based on the Löwdin orbitals then arises. Wecould have selected a set of overlap and interaction parameters (e.g., by fitting theresults to data) for a hydrogen-like model. If instead we use the Löwdin orbitalsthe number of apparent empirical parameters is greatly reduced. If we now selectthe parameters to fit data, the parameters will not be the same as in the previousmethod because we are identifying the eigenvalues of different basis orbitals withthe same physical data.

Often chemists employ what are called hybrid orbitals as the basis functions.These are linear combinations of orbitals that have the same symmetry about the

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168 Molecular electronic structure: The LCAO model

internuclear axis joining two atomic sites. The s–p hybrid orbitals, for example,are formed by the sum and difference of s- and p-orbitals with the same principalquantum number on the same site. The idea is to form hybrid functions that lead toan accumulation of charge between the atoms that bond in the molecule.

6.4 An example: The electronic structure of squarene

As a tutorial example we shall analyze the electronic structure of the fictitioussquarene molecule. Recall that squarene is a square molecule with identical atomson each corner. For this example let us assume that the outer electrons on thesquarene atoms possess 2s and 2p orbitals. The 1s orbitals are tightly bound totheir nuclei. These “core states” are far removed in energy from the outer 2p and 2selectrons’ levels and may be omitted since the molecular binding is due principallyto the interactions among the 2s and 2p electrons.

In Chapter 1 we restricted our analysis to the plane of the molecule. Here wewant to consider all three dimensions, therefore the point group of the coveringoperations is D4h (D4 × i) rather than C4v. However, as we shall see, we can stillderive all of the information we need using the simpler C4v group. The behavior ofthe symmetry functions under inversion will be obvious. The coordinates and basisfunctions to be used in our analysis are shown schematically in Fig. 6.3.

(a)

z1

z3

z2

z4

y1

y3

y2

y4

x1

x3

x2

x4

(c)

px4

px3

px2

px1

(b)

s1 s2

s3s4

(d)py4 py3

py2py1

(e)

pz4 pz3

pz2pz1

Figure 6.3 The coordinate system and basis functions for squarene: (a) coordi-nates, (b) 2s orbitals, (c) 2px orbitals, (d) 2py orbitals, and (e) 2pz orbitals.

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6.4 An example: The electronic structure of squarene 169

Table 6.2 The action table for the 2s and 2p orbitals ofsquarene. Here s = ϕ2s(r− R) and pκ = ϕpκ(r− R), whereκ = x, y, or z.

E C4 C2 C34 σv1 σv2 σd1 σd2

1 1 4 3 2 1 3 2 42 2 1 4 3 4 2 1 33 3 2 1 4 3 1 4 24 4 3 2 1 2 4 3 1px px −py −px py −py py px −pxpy py px −py −px −px px −py pypz pz pz pz pz pz pz pz pzs s s s s s s s s

(a)

1 2

34

E

(b)

C4, C2, C34

(c) σd1 (d)

σd2

(e)

σv1

(f)

σv2

Figure 6.4 Symmetry operations of the C4v group. (a) E is the identity. (b) Thereis a four-fold rotation axis perpendicular to the plane of the molecule. Rotationsabout this axis include C2, C4, and C3

4 . In (c) and (d) we show σd1 and σd2:reflections in a line bisecting the sides of the square. In (e) and (f) we show σv1and σv2: reflections in a diagonal line passing through diagonally opposite cornersof the square.

The operations of the group are to be applied to the basis functions while thecoordinate system remains fixed in space. The operations of the C4v group aregiven in Chapter 1, but are repeated here for convenience in Fig. 6.4.

The action table for the s- and p-orbitals is shown in Table 6.2. This tableallows us to determine the effect of any operation on any s or p basis function.For example, under the operation C4, a pz orbital at corner 1 is rotated into a

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170 Molecular electronic structure: The LCAO model

pz orbital on corner 4. That is, P(C4) ϕpz (r − R1) = ϕpz (r − R4). Similarly,P(σv2) ϕpx (r− R1) = ϕpy (r− R3).

Since ϕpx (r−Ri ), ϕpy (r−Ri ), and ϕpz (r−Ri ) transform under the group oper-ations in the same way as the displacement vectors rxi , ryi , and rzi , the symmetryproperties of the p-orbital system are identical to the symmetry properties of thenuclear displacements considered in Chapter 1. The symmetry functions for thep-orbitals are completely analogous to those of the vibrational problem. However,for the p-orbital system there are no analogs to the body modes of translation androtation. As a result the eigenvectors (linear combinations of symmetry functions)will in general be different.

There are in total 16 basis functions for the matrix representations. These arefour 2s orbitals, four pz orbitals, four px orbitals, and four py orbitals. However, nosymmetry element can transform a 2s orbital into a p-orbital or a pz orbital into apx or py orbital. The 2s and also the pz orbitals transform only among themselves.However, the px and py orbitals are transformed into one another. As a result ofthis, the determination of the symmetry functions can be carried out separately forthe 2s, pz , and (px , py) orbitals.

6.4.1 s- and pz-orbital symmetry functions

In order to find the orbital symmetry functions, we need the characters for the basis-function representations of the operations. These may be obtained from the actiontable by simply adding the number of basis functions that are unchanged by anoperation minus the number transformed into the negative of themselves. Considerthe pz orbitals first. From the action table it is clear that only the operations E ,σv1, and σv2 leave all or some of orbitals in their original positions. On lookingat the column for σv1 we see that pz1 and pz3 are transformed into themselves,and therefore the trace of the representation will be 2. For σv2, pz2 and pz4 aretransformed into themselves, so the representation character is also 2 (as it mustbe, since σv1 and σv2 are in the same class). Clearly the character of the matrixrepresenting E is 4, the number of pz orbitals. The same characters are obtainedfor the 2s orbitals. We designate the representation based on the 2s orbitals as �2s

and the representation based on the pz orbitals as �z . The character table for C4v

is shown in Table 6.3. Characters for the representation matrices of �z and �2s areshown on the last two lines of the character table.

The decomposition of a reducible representation, �, into the IRs of a group isgiven by

nα = 1

h

∑R

χ(R)∗ χα(R), (6.29)

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6.4 An example: The electronic structure of squarene 171

Table 6.3 The character table for C4v (top) and thecharacters for the representations �xy, �z, and �2s

C4v E C2 2C4 2σv1 2σd1

A1 1 1 1 1 1A2 1 1 1 −1 −1B1 1 1 −1 1 −1B2 1 1 −1 −1 1E 2 −2 0 0 0�xy 8 0 0 0 0�z 4 0 0 2 0�2s 4 0 0 2 0

where h is the order of the group (h = 8 in this case), χ(R) is the character of theoperation for the reducible representation, χα(R) is the character of the αth IR ofthe group (C4v in this case), and nα is the number of times �α is contained in �.The index R runs over the operations of the group. For the 2s and pz orbitals thedecomposition gives

�2s = a1 + b1 + e, (6.30)

�z = a1 + b1 + e. (6.31)

In (6.30) and (6.31) the labels for the IRs are shown as lower-case letters. Inthis chapter we shall use lower-case letters for the IRs of one-electron functionsand upper-case letters such as A1 or B1 to characterize a many-electron molecularstate.

To generate the symmetry coordinates for squarene we proceed in the usual way.For the one-dimensional IRs,

f α ∝∑

R

χα(R) PR f, (6.32)

where f is an arbitrary combination of orbitals. For a representation of higherdimensionality,

f αi ∝

∑R

Dα(R)i i PR f, (6.33)

where Dα(R)i i is a diagonal element of the α IR matrix for the R operation. For thee representation the matrix elements were obtained in Chapter 1, in Table 1.7. Thenon-zero, diagonal matrix elements for row 1 are �E(E)11 = 1, �E(C2)11 = −1,�E(σd1)11 = −1, and �E(σd2)11 = 1. For row 2 the non-zero diagonal matrixelements are �E(E)11 = 1, �E(C2)11 = −1, �E(σd1)11 = 1, and �E(σd2)11 = −1.

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172 Molecular electronic structure: The LCAO model

(These matrix elements can be deduced from the action table since px and py

are basis functions for the e IR.) Using s1 as f , we find (after normalizing thefunctions) that

f 2s(a1) = 1

2(s1 + s2 + s3 + s4), (6.34)

f 2s(b1) = 1

2(s1 − s2 + s3 − s4), (6.35)

f 2s1(e) = 1

2(s1 − s2 − s3 + s4) (row 1), (6.36)

f 2s2(e) = 1

2(s1 − s2 + s3 − s4) (row 2). (6.37)

Using f = pz1 yields the symmetry coordinates,

f z(a1) = 1

2(pz1 + pz2 + pz3 + pz4), (6.38)

f z(b1) = 1

2(pz1 − pz2 + pz3 − pz4), (6.39)

f z1 (e) =

1

2(pz1 − pz2 − pz3 + pz4) (row 1), (6.40)

f z2 (e) =

1

2(pz1 + pz2 − pz3 − pz4) (row 2). (6.41)

The 2s and 2pz symmetry functions are shown schematically in Fig. 6.5. For the2s symmetry functions it can be seen that the a1 and b1 functions are invariant underinversion and therefore they can be labeled as “g” functions. The 2s e functionsare antisymmetric under inversion and can be labeled as “u” functions. For the pz

functions the converse is true. The a1 and b1 are antisymmetric under inversion,while the e symmetry functions are symmetric.

Therefore, the decompositions of �2s and �z can be written as

�2s = a1g + b1g + eu, �z = a1u + b1u + eg. (6.42)

6.4.2 px−py symmetry functions

The orbitals px and py are transformed into one another by some of the symmetryoperations. Therefore they must be considered together. There are four px and fourpy orbitals, so the representations based on these functions are 8 × 8 matrices. Itcan be seen from the action table that the only operation with a non-zero trace isthe identity, for which χ xy(E) = 8.

Therefore the decomposition into the IRs of C4v is

�xy = a1 + a2 + b1 + b2 + 2e. (6.43)

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6.4 An example: The electronic structure of squarene 173

Table 6.4 Symmetry functions for the px and py basis orbitals. There are two erepresentations. The first is labeled “a” with partners 1a (row 1) and 2a (row 2);the second is labeled “b” with partners 1b and 2b. The “g” and “u” labels areassigned according the symmetry of the function under inversion.

f xy(a1g) = (1/√

8)(px1 + px2 − px3 − px4 − py1 + py2 + py3 − py4)

f xy(a2g) = (1/√

8)(px1 − px2 − px3 + px4 + py1 + py2 − py3 − py4)

f xy(b1g) = (1/√

8)(px1 − px2 − px3 + px4 − py1 − py2 + py3 + py4)

f xy(b2g) = (1/√

8)(px1 + px2 − px3 − px4 + py1 − py2 − py3 + py4)

f xy1a (eu) = 1

2 (px1 − px2 + px3 − px4)

f xy2a (eu) = 1

2 (px1 + px2 + px3 + px4)

f xy1b (eu) = 1

2 (py1 + py2 + py3 + py4)

f xy2b (eu) = 1

2 (py1 − py2 + py3 − py4)

(a)

a1g b1g

eu row 1 eu row 2

(b)

a1u b1u

eg row 1 eg row 2

Figure 6.5 Symmetry functions for 2s and 2pz orbitals. The g and u labels havebeen added to indicate the behavior under the inversion operation.

The px –py symmetry functions given in Table 6.4 are shown schematically inFig. 6.6.

From Fig. 6.6 we can see that the a1, a2, b1, and b2 symmetry functions aresymmetric under inversion. The e symmetry functions are antisymmetric underinversion. Therefore, the decomposition can be written as

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174 Molecular electronic structure: The LCAO model

a1g a2g

b1g b2g

eu row 1a eu row 2a

eu row 1b eu row 2b

Figure 6.6 px –py symmetry functions. The labels u and g have been added toindicate the behavior of the functions under the inversion operation.

�xy = a1g + a2g + b1g + b2g + 2eu . (6.44)

6.4.3 Analysis of the decompositions

The 16 × 16 representation for the molecule, �(molecule), based on all of theorbitals is

�(molecule) = �2s + �z + �xy

= 2a1g + a2g + 2b1g + b2g + eg + a1u + b1u + 3eu .

We can now determine the blocks of the secular equation (6.12) when it is trans-formed to symmetry-function coordinates. First we note that �z has no IRs incommon with either �xy or �2s . Therefore the eigenvectors of the �z states willhave no mixing with the 2s, px , or py orbitals. Since the pz orbitals interact witheach other only by means of the π overlaps and π matrix elements, the pz orbitalsform pure “π states”. Second, the one-dimensional IRs, a2g, b2g, a1u , and b1u , eachoccur only once in �(molecule), and therefore the symmetry functions for theserepresentations must also be eigenvectors.

In the symmetry-function representation the secular equation has four 1 × 1blocks (a2g, b2g, a1u , and b1u). Furthermore, since the Hamiltonian matrix elementsand overlaps between different rows of the same representation vanish, the twopartner functions for the eg representation are also eigenvectors. That is, the 2× 2

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6.4 An example: The electronic structure of squarene 175

a2g

b2g

a1u

b1u

a1g

⎧⎪⎪⎪⎨⎪⎪⎪⎩

× ×× ×

eg

⎧⎪⎪⎪⎨⎪⎪⎪⎩

× 0 (row 1)

0 × (row 2)

b1g

⎧⎪⎪⎪⎨⎪⎪⎪⎩

× ×× ×

× × ×eu (row 1)

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

× × ×× × ×

× × ×eu (row 2) × × ×

× × ×

Figure 6.7 The block-diagonal form of the secular matrix when transformed tosymmetry-function coordinates.

eg block must be diagonal. The a2g and b2g symmetry functions are unique to �xy ,and therefore the eigenvectors involve only px and py orbitals. Finally, eu occursthree times and that leads to two 3 × 3 blocks; one 3 × 3 block for the first-rowfunctions and one 3 × 3 block for the second-row functions. The eigenfunctionswill be admixtures of 2s, 2px , and 2py orbitals. The block-diagonal form of theHamiltonian matrix in the symmetry-function representation is shown in Fig. 6.7.

To find the eigenvalues and eigenvectors of squarene, we can transform thesecular matrix from the 2s and 2p representation to the symmetry-function repre-sentation. The matrix U which accomplishes this is constructed from the symmetry

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176 Molecular electronic structure: The LCAO model

functions. The columns of the (16 × 16) U matrix are 16-component columnvectors whose components are the coefficients of the symmetry functions. Thetransformation

U−1(H− ε S)U (6.45)

casts the eigenvalue problem into the block-diagonal form shown in Fig. 6.7.Alternately, we can compute the overlap- and interaction-matrix elements

between the symmetry functions belonging to the same IRs. In Chapter 1 wecarried out the U transformation and then solved the secular equations for thesmaller blocks. Here we shall use the alternate method of calculating the matrixelements between symmetry functions, a procedure mathematically equivalent totransforming the H− ε S matrix.

Since all of the �z symmetry functions are eigenstates, we need only calcu-late the diagonal matrix elements and overlap for each �z symmetry function(Table 6.4) to obtain the eigenvalues.

For these functions,

ε =∫

f z∗ H f z dr∫f z∗ f z dr

. (6.46)

The overlap for f z(a1u) is∫f z(a1u)

∗ f z(a1u) dr

= 1

4

∫dr(pz1 + pz2 + pz3 + pz4)

∗(pz1 + pz2 + pz3 + pz4)

= 1

4

[ ∫dr(p2

z1 + p2z2 + p2

z3 + p2z4)

+ 2{pz1(pz2 + pz3 + pz4)+ pz2(pz3 + pz4)+ pz3 pz4}]

=∫

dr{(p∗z1 pz1)+ [p∗z1(pz2 + pz4)] + (p∗z1 pz3)}= 1+ 2 ↑↑1 + ↑↑2

= 1+ 2[ppπ ]1 + [ppπ ]2, (6.47)

where [ppπ ]1 is the overlap between pz orbitals on adjacent sites and [ppπ ]2 is theoverlap between pz orbitals on diagonally separated sites. For the matrix elementsof the Hamiltonian,∫

f z(a1u)∗H f z(a1u) dr =

∫pz1 + pz2 + pz3 + pz4

2H

pz1 + pz2 + pz3 + pz4

2dr

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6.4 An example: The electronic structure of squarene 177

= 4∫

1

4{(p∗z1H pz1)+ [p∗z1 H (pz2 + pz4)]+ (p∗z1 H pz3)}dr

= εp + 2(ppπ)1 + (ppπ)2, (6.48)

where (ppπ)1 is the Hamiltonian matrix element between adjacent pz functions onadjacent sites and (ppπ)2 is the matrix element between pz functions on diagonallyseparated sites. The diagonal interaction, εp, is roughly the energy of the isolatedp-orbital without any 2s orbital mixing.

From (6.47) and (6.48) we obtain the eigenvalue for the a1u state,

ε(a1u) = εpz + 2(ppπ)1 + (ppπ)2

1+ 2[ppπ ]1 + [ppπ ]2 . (6.49)

The result for ε(a1u) involves only π -type overlaps and interactions, and thereforeit is sometimes labeled by the symbol π(a1u).

As a second example we consider the b2g state constructed from px and py

orbitals. The analysis is easily carried out using either the graphical method orTable 6.1. To simplify the notation, we use xi and yi to represent the pxi and pyi

orbitals.For the b2g state we need to calculate∫

dr f xy(b2g)∗ H f xy(b2g)∫

dr| f xy(b2g)|2

with

f xy(b2g) = 1√8{x1 + x2 − x3 − x4 + y1 − y2 − y3 + y4}. (6.50)

Consider one of the products,

∫dr

(x1√

8

)∗H f xy(b2g)

=∫

dr1

8(x1)

∗H{x1 + x2 − x3 − x4 + y1 − y2 − y3 + y4}

= 1

8

[εp+ ↓↓1 −

{1

2↓↓2 +1

2←←2

}−→→1

+ ↓→0 + ↓→1 −{−1

2→→2 +1

2↓↓2

}+ ↓→1

]

= 1

8{εpx + (ppπ)1 − (ppσ)1 − (ppπ)2}. (6.51)

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178 Molecular electronic structure: The LCAO model

It is easily verified that∫

dr (φ/√

8)∗ H f xy(b2g) gives the same result, where φ

is −x2, −x3, x4, −y1, −y2, y3, or y4, so∫dr f xy(b2g)

∗ H f xy(b2g) = {εpx + (ppπ)1 − (ppσ)1 − (ppπ)2}. (6.52)

A similar calculation gives the overlap as∫dr f xy(b2g)

∗ f xy(b2g) = 1+ [ppπ ]1 − [ppσ ]1 − [ppπ ]2. (6.53)

Therefore, the eigenvalue is

ε(b2g) = εpx + (ppπ)1 − (ppσ)1 − (ppπ)2

1+ [ppπ ]1 − [ppσ ]1 − [ppπ ]2 . (6.54)

The diagonal parameters εpx and εpy are necessarily equal, since they representintegrals that are symmetrically equivalent; however, εpz will be different becausethe lobes of the pz orbital point into a different environment. For simplicity wehave taken all of them to be equal to εp.

As a last example we consider the 2 × 2 block for the a1g states. These statesinvolve f xy(a1g) and f 2s(a1g) symmetry coordinates.

The 2× 2 secular equation that mixes these two symmetry functions is∣∣∣∣ M11 − εS11 M12 − εS12

M12 − εS12 M22 − εS22

∣∣∣∣ = 0, (6.55)

where

M11 = 1

4

∫dr(s1 + s2 + s3 + s4)H(s1 + s2 + s3 + s4)

= εs + 2(ssσ)1 + (ssσ)2, (6.56)

M12 = 1√32

∫dr(s1 + s2 + s3 + s4)

× H(x1 + x2 − x3 − x4 − y1 + y2 + y3 − y4)

= −√2(ssσ)1 − (ssσ)2, (6.57)

M22 = 1

8

∫dr(x1 + x2 − x3 − x4 − y1 + y2 + y3 − y4)

× H(x1 + x2 − x3 − x4 − y1 + y2 + y3 − y4)

= εx + (ppπ)1 − (ppσ)1 − (ppσ)2, (6.58)

S11 = 1+ 2[ssσ ]1 + [ssσ ]2, (6.59)

S12 = −√

2[ssσ ]1 − [ssσ ]2, (6.60)

S22 = 1+ [ppπ ]1 − [ppσ ]1 − [ppσ ]2. (6.61)

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6.4 An example: The electronic structure of squarene 179

The eigenvalue equation is of the form ε2 − 2Aε − B = 0, where

A = S11 M22 + S22 M22 − 2M12S12

2(1− S212)

, (6.62)

B = M212 − M11 M22

1− S2sp

. (6.63)

The eigenvalues and eigenvectors for any matrix of the form of (6.55) are

ε± = A ±√

A2 + B, (6.64)

f ± = −a± f 1 + b± f 2√a±2 + b±2 + 2a±b±S12

, (6.65)

where

a± = M12 − ε±S12, (6.66)

b± = M11 − ε±S11. (6.67)

For our problem here, ε± = ε±(a1g), f 1 = f 1s(a1g), and f 2 = f xy(a1g).A schematic representation of the eigenvectors is shown in Fig. 6.8. The asym-

metrical contours represent the admixtures of s- and p-orbitals that make up theeigenfunctions. It can be seen that for the f −(a1g) state there will be a build upof electron density between the atoms. This internuclear charge enhancement rep-resents the formation of molecular bonds. This state is called a “bonding” state.Conversely, for the f +(a1g) state the build up of electron density will be pointedaway from the internuclear axes. This type of state is called an “antibonding”state. The bonding state is pushed below the atomic 2s level in energy, and theantibonding state is raised above the 2p level.

A correlation diagram showing how all of the molecular states evolve from the2s and 2p states is presented in Fig. 6.9. The actual ordering of the levels dependson the values of matrix elements and overlap integrals.

For our squarene example let us assume that the outer atomic configuration ofeach atom of squarene is 2s22p2. There will be a total of 16 electrons to distributeamong the one-electron states of Fig. 6.9. If we assign paired electrons (one spin upand one spin down) to the lowest states (1a1g, 1b1g, 1eu , b2g, a2g, 2b1g, and 2a1g),each level will be doubly occupied.

The total wavefunction for the molecule is the determinantal state, �,

� = 1√16! det

{(1a1g ↑)(1a1g ↓)(1b1g ↑)(1b1g ↓)(1eu ↑)(1eu ↓)× (1eu ↑)(1eu ↓)(b2g ↑)(b2g ↓)(a2g ↑)(a2g ↓)× (2b1g ↑)(2b1g ↓)(2a1g ↑)(2a1g ↓)

}, (6.68)

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180 Molecular electronic structure: The LCAO model

(a)

+f−(a1g)Bonding

+f+(a1g)Antibonding

(b)

a1g

2p(x, y)

f+(a1g) Antibonding

2s

f−(a1g) Bonding

Figure 6.8 One-electron molecular states of squarene. (a) The a1g eigenstatesare a combination of s- and p-orbitals. f −(a1g) is a bonding state with chargeaccumulating between the atoms. f +(a1g) is an antibonding state with chargedepleted between the atoms. (b) A correlation diagram showing how the a1g statesevolve from s and p levels.

or, in abbreviated notation,

� = (1a1g)2(1b1g)

2(1eu)2(1eu)

2(b2g)2(a2g)

2(2b1g)2(2a1g)

2.

In (6.68) the spin states are indicated by the ↑ and ↓ arrows. � is symmetric underinversion since the product of four eu functions is a g function. The total spin iszero, and therefore the molecular state is a singlet. The total wavefunction, �, islabeled by the IR according to which it transforms. To determine the IR of thetotal molecular state, we must consider the transformation properties of the prod-uct of all 16 occupied states. This is usually not a difficult thing to do becausea doubly occupied one-electron state belonging to a one-dimensional represen-tation will transform as the totally symmetric �a1g representation. Furthermore,�α × �a1g = �α. For our example above, the total molecular state transformsas (1eu ↑)(1eu ↓)(1eu ↑)(1eu ↓). These four states consist of two row-1 eu

eigenfunctions and two row-2 eu eigenfunctions. The product of two row-1 (or

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6.4 An example: The electronic structure of squarene 181

2

22

1

1

2

1

2

1

3

2

2 1

2 1

2

2

11 1

1 1

1

1

2

Bondingstates

Antibondingstates

Non-bondingstates

states

Figure 6.9 A correlation diagram, showing schematically the one-electron energylevels. The px and py states interact to split as shown. The addition of the 2sinteractions lead to the formation of the molecular levels shown at the center ofthe diagram. The bonding states are pushed down in energy (become more stable)compared with the 2s energy. The antibonding states are pushed up in energy(become less stable) compared with the 2p(x, y) energy. The b2g and a2g statesare non-bonding. The states labeled π states are constructed from 2pz orbitalsperpendicular to the plane of the molecule.

two row-2) eu functions transforms as �A1g . (This may be verified by applying thesymmetry-function-generating machine to the product of two such functions.) Asa result of the above reasoning we conclude that the total wavefunction (6.68) is an1 A1g state. The left-hand superscript indicates the spin multiplicity and the (uppercase) A indicates that it is a many-electron, totally symmetric, molecular state.

A different assignment of the electrons results in a different molecular state. Forexample, if we place only one electron in the 2a1g and add the remaining electronto the 2eu , the new molecular state is antisymmetric under inversion. If the 2a1g

and 2eu electrons have opposite spin states then the total molecular spin is stillzero, S = 0. However, if the 2a1g and 2eu electrons have parallel spins, the totalmolecular spin is S = 1. In the first case, with S = 0, the molecular state is 1Eu

and would be doubly degenerate, since there are two possible degenerate eu spatial

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182 Molecular electronic structure: The LCAO model

functions. In the second case, S = 1, the molecular state is a triplet, namely 3Eu ,and doubly degenerate.

The determinantal state |(1a1g)2(1b1g)

2(1eu)2(1eu)

2(b2g)2(a2g)

2(2b1g)2 (2a1g ↑)

(2eu ↑)〉 corresponds to S = 1, Ms = 1. In fact, only the S = 1, Ms = 1 (orMs = −1) state is described by a single Slater determinant. However, starting withthe Ms = 1 state, the Ms = 0 and Ms = −1 can be obtained by applying thespin-lowering operator, S−, to the Ms = 1 state. Since S− gives a non-zero resultonly when acting on unpaired spins (singly occupied states), only the one-electronstates 2a1g ↑ and 2eu ↑ are affected. Applying the lowering operator to the productof these states gives

S−�(2a1g ↑)�(2eu ↑) = �(2a1g ↓)�(2eu ↑)+�(2a1g ↑)�(2eu ↓), (6.69)

and

S−{�(2a1g ↓) �(2eu ↑)+�(2a1g ↑) �(2eu ↓)

}= 2�(2a1g ↓) �(2eu ↓).

(6.70)Therefore the normalized, determinantal states are

3E(Ms = 1) = 1√16! det

{(1a1g ↑)(1a1g ↓) . . . (2a1g ↑)(2eu ↑)

}= (1a1g)

2(1b1g)2 . . . (2a1g ↑)(2eu ↑), (6.71)

3E(Ms = 0) = 1√16!

1√2

[det

{(1a1g ↑)(1a1g ↓) . . . (2a1g ↑)(2eu ↓)

}+ det

{(1a1g ↑)(1a1g ↓) . . . (2a1g ↓)(2eu ↑)

}]= 1√

2

{(1a1g)

2(1b1g)2 . . . (2a1g ↑)(2eu ↓)

+ (1a1g)2(1b1g)

2 . . . (2a1g ↓)(2eu ↑)}, (6.72)

3E(Ms = −1) = 1√16! det

{(1a1g ↑)(1a1g ↓) . . . (2a1g ↓)(2eu ↓)

}= (1a1g)

2(1b1g)2 . . . (2a1g ↓)(2eu ↓). (6.73)

6.5 The electronic structure of H2O

As an application to a real molecule, we analyze the electronic structure of H2Ousing molecular-orbital theory. The group for the molecule is C2v. The coordinatesystem and symmetry operations are illustrated in Fig. 6.10. We employ the sameright-handed coordinate system as we used in Chapter 2. The operations includethe identity, a 180◦ (C2) rotation about the z-axis, a reflection in the plane of themolecule (σ ′v(yz)), and a reflection in a plane perpendicular to the plane of the

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6.5 The electronic structure of H2O 183

θ

C2σv(yz)

σv(xz)

x y

z

O

H1

H2

C2v(2mm) E C2 σv(xz) σ′v(yz)

x2, y2, z2 z A1 1 1 1 1

xy Rz A2 1 1 −1 −1

xz Ry, x B1 1 −1 1 −1

yz Rx, y B2 1 −1 −1 1

Figure 6.10 Coordinates and symmetry operations for H2O. C2 is a clockwiserotation about the axis through O in the molecular plane, σ ′v(yz) is a reflection inthe plane of the molecule, and σv(xz) is a reflection in the plane perpendicularto the molecular plane. The hydrogen atoms are labeled 1 and 2. The symmetryelements operate on the orbitals, not the coordinates.

molecule (σv(xz)). The reader should note that some authors label the reflection inthe plane of the molecule as σv rather than σ ′v. This interchanges the role of x andy, and is not compatible with the character table in Fig. 6.10.

The oxygen and hydrogen atoms have outer electron configurations 2s22p4 and1s, respectively. As a simple model we shall use as basis functions 2s(O) = 2s,2px(O) = x , 2py(O) = y, 2pz(O) = z, 1s(H1) = s1, and 1s(H2) = s2 for atotal of six orbitals. The representation based on these six functions is denoted by�(molecule).

6.5.1 Decomposition of the �(molecule) representation

The action table, Table 6.5, shows that the basis functions (s1, s2), 2s, px , py , andpz transform respectively only among themselves, and therefore can be consideredseparately in the symmetry analysis. We can write �(molecule) = �1s + �2s +�x + �y + �z .

The characters of the operations can be obtained from the action table. For ex-ample, for the two 1s orbitals, χ(E) = χ(σ ′v) = 2, while χ(C2) = χ(σv) = 0.The decomposition is �1s = a1 + b2. For the other orbitals, �2s = a1, �x = b1,�y = b2, and �z = a1. Therefore,

�mol = 3a1 + 2b2 + b1. (6.74)

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184 Molecular electronic structure: The LCAO model

Table 6.5 The action table for the 2s, 2p, 1s1, and 1s2 orbitals of H2O. The 1sorbitals of the two hydrogen atoms are labeled s1 and s2. 2s indicates the oxygen2s orbital. The 2p orbitals are labeled as 2px = x, 2py = y, and 2pz = z.

C2v E C2 σv(xz) σ ′v(yz)

s1 s1 s2 s2 s1s2 s2 s1 s1 s2x x −x x −xy y −y −y yz z z z z2s 2s 2s 2s 2s

The symmetry functions are

f 1s(a1) = 1√2[1s1 + 1s2], (6.75)

f 1s(b2) = 1√2[1s1 − 1s2], (6.76)

f 2s(a1) = 2s, (6.77)

f x(b1) = px , (6.78)

f y(b2) = py, (6.79)

f z(a1) = pz. (6.80)

From the decomposition results it is clear that the secular equation will be block-diagonalized into a 1 × 1(b1), a 2 × 2(b2), and a 3 × 3(a1) when transformed tothe symmetry-function representation. One immediate result is that f x(b1) is aneigenvector. That is, the oxygen 2px (out-of-plane) orbital does not interact withany of the other orbitals. This type of state is called “non-bonding” because itcontributes nothing to the hydrogen–oxygen bond. In the ground state the b1 stateis occupied by a pair of electrons called a “lone pair”.1

Since b1 occurs only once, px is an eigenvector with eigenvalue εp (roughly theenergy of an isolated 2p oxygen orbital).

The b2 block involves the interaction between f y(b2) and f 1s(b2) symmetryfunctions. The secular-equation matrix for the b2 states is∣∣∣∣ M11 − εS11 M12 − εS12

M12 − εS12 M22 − εS22

∣∣∣∣ = 0, (6.81)

1 Some texts use a different system and have the 2pz as the non-interacting orbital. Whatever system used, it isthe 2p orbital perpendicular to the plane of the molecule which is the non-interacting orbital.

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6.5 The electronic structure of H2O 185

where

M11 = ε1s − (s1s2σ), (6.82)

M12 = − 1√2

sin

2

)(1spσ), (6.83)

M22 = εp, (6.84)

S11 = 1− [1spσ ], (6.85)

S12 = − 1√2

sin

2

)[1spσ ], (6.86)

S22 = 1, (6.87)

where the angle θ is the H–O–H angle, about 105◦, and εpy = εp is the diagonalenergy of the 2py orbital. It should be noted that εpx , εpy , and εpz should all bedifferent because no two are in the same symmetry environment. For simplicitywe shall take εpx = εpy = εpz = εp. The diagonal (on the same site) 2s–2pinteractions vanish by symmetry. The eigenvalues and eigenvectors for the secularequation, (6.81), can be found in the usual manner. As expected, the eigenstates ofthe 2× 2 form bonding and antibonding states.

The remaining block is a 3 × 3 for the a1 molecular states involving mixing off 1s(a1), f 2s(a1), and f z(a1). The secular-equation matrix elements are∣∣∣∣∣∣

M1s,1s − εS1s,1s M1s,2s − εS1s,2s M1s,z − εS1s,z

M1s,2s − εS1s,2s M2s,2s − ε 0M1s,z − εS1s,z 0 Mz,z − ε

∣∣∣∣∣∣ = 0, (6.88)

where

M1s,1s = ε1s + (s1s2σ), S1s,1s = +[s1s2σ ], (6.89)

M1s,2s =√

2(s1, 2s, σ ), S1s,2s =√

2[s1, 2s, σ ], (6.90)

M1s,z =√

2 cos

2

)(s1 pσ), S1s,z =

√2 cos

2

)[s1 pσ ], (6.91)

M2s,2s = ε2s, (6.92)

Mz,z = ε2p. (6.93)

Equation (6.88) leads to a cubic equation in ε with all roots being real. The eigen-values can be found from the standard formulas for the roots of a cubic equation,but will not be presented here.

Figure 6.11 shows schematically the molecular levels of H2O. For the free atoms,the oxygen 1s energy, ε1s = −20.6 au (1 au = 27.2 eV), is far removed in energyfrom the 2s and 2p orbital energies, and therefore has little effect on the electronicstructure of the valence states. The oxygen 2s energy is ε2s = −1.24 au, the 2penergy is εp = −0.63 au, and the energy of the hydrogen 1s is −0.50 au. In

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186 Molecular electronic structure: The LCAO model

O(1s)

2s

2p

H(1s)

2b2

4a1

1b1

3a1

1b2

2a1

1a1

↑↓↑↓

↑↓

↑↓

↑↓

⎫⎪⎪⎬⎪⎪⎭

σ∗ (antibonding)

⎫⎬⎭ nb (non-bonding)

⎫⎪⎪⎪⎪⎪⎪⎪⎪⎬⎪⎪⎪⎪⎪⎪⎪⎪⎭

σ (bonding)

Core electrons

One-electronstates

(a) (b)

2 1A1

3A1

1B1

3B1

1 1A1

Molecularstates

ν=

6.0

×10

4cm

−1

ν=

7.8

×10

4cm

−1

Figure 6.11 The electronic structure of H2O. (a) One-electron levels. (b) Electric-dipole transitions between molecular states.

Fig. 6.11 the oxygen 1s state is shown as 1a1. The next two levels (2a1 and 1b2)are labeled as “bonding” or “σ” because they are principally responsible for thehydrogen–oxygen molecular bonds. The middle levels (1b1 and 3a1) contribute lit-tle or nothing to the molecular bonding, and are labeled “nb” for non-bonding. Theupper two levels (4a1 and 2b2) are called “antibonding” or “σ ∗” because occupyingthese levels with electrons reduces the stability of the molecule and because thewavefunctions have depleted rather than enhanced charge between the hydrogenand oxygen atoms.

The configuration of the valence electrons for the oxygen atom is 2s22p4, andthat for each of the two hydrogen atoms is 1s. Thus there are eight electrons to bedistributed among the one-electron valence states (ten including two in the oxygen1s level) The ground state is constructed by filling the lowest one-electron stateswith paired electrons as shown in Fig. 6.11. The stability or binding energy ofthe H2O molecule relative to the free atoms is the reduction of the total energy

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6.5 The electronic structure of H2O 187

z

y

z

x

z

y

2a1 1b2

OH H

OH H

3a1 1b1

OH

OH

4a1 2b2

OH H O

H H

Figure 6.12 Charge distributions for the various one-electron states of H2O.

achieved by filling the bonding, one-electron molecular states compared with thetotal energy of the isolated atoms,

E(binding) = {2ε2s + 4ε2p + 2ε1s} − {2ε(2a1)+ 2ε(1b2)+ 2ε(3a1)+ 2ε(1b1)},(6.94)

where ε2s , ε2p, and ε1s are the free-atom energies. A positive value for E(binding)means that the molecule is stable. The b1 energy level is nearly unshifted fromthe free-atom 2p energy. Clearly the binding energy is positive principally due tothe 2a1 and 1b2 bonding states. Sketches of the charge densities of the variousone-electron states of H2O are shown in Fig. 6.12.

For the one-dimensional representations, we have � × � = a1 andtherefore all of the doubly occupied one-electron states belong to a1. Thetotal spin is zero, so the molecular state, �, is 1 A1. In detail �(1 A1) =(1/10!)1/2 det{(1a1)

2(2a1)2(1b2)

2(3a1)2(1b1)

2} is usually abbreviated as

�(1 A1) = (1a1)2(2a1)

2(1b2)2(3a1)

2(1b1)2,

where (�)2 = � ↑ � ↓ means paired electrons.From the character table for C2v we see that z transforms as a1, x as b1, and y as

b2; therefore, for the electric-dipole operator μe, the z-component belongs to a1, thex-component to b1, and the y-component to b2. Optical transitions from an initialstate �(�i ) to a final state �(� f ) are not forbidden if �i × � f contains a1, b1,or b2. However, since the dipole-moment vector lies in the plane of the molecule,

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188 Molecular electronic structure: The LCAO model

the x-component (perpendicular to the H2O plane) of μe is zero, so the transition1b1 ⇒ 4a1 observed both in optical and in electron inelastic scattering experimentsis not a simple electric-dipole transition. The transition 3a1 ⇒ 4a1 is an allowedelectric-dipole transition since a1×a1 = a1. The radiation must be polarized alongthe z-axis, in the plane of the molecule.

The excited molecular state created by the transition 3a1 ⇒ 4a1 with no changein spin is a 1 A1. This state is shown on the right-hand side in Fig. 6.11. The molec-ular transition is 1 1 A1 ⇒ 2 1 A1. The observed peaks in the optical and electronscattering corresponding to this transition occurs near ν = 7.8× 104 cm−1. For the1b1 ⇒ 4a1(1 1 A1 ⇒ 1 B1) transition, ν = 6× 104 cm−1.

The determinantal wavefunctions for the 1 B1 and 2 1 A1 molecular states are (inabbreviated form)

�(1 B1) = (1a1)2(2a1)

2(1b2)2(3a1)

2(1b1σ)(4a1σ ), (6.95)

�(2 1 A1) = (1a1)2(2a1)

2(1b2)2(3a1σ)(1b1)

2(4a1σ ), (6.96)

where σ =↑ or ↓ and σ = −σ .Also shown in Fig. 6.11 are the triplet molecular states. However, these states

are not accessible from the ground state by electric-dipole transitions because achange in total spin would be required. Though well studied, the spectra observedfor water are complex and as yet not well understood [6.4]. However, there isgeneral agreement that the electronic structure is as outlined here.

References

[6.1] R. M. Martin, Electronic Structure: Basic Theory and Practical Methods, Vol. 1(Cambridge: Cambridge University Press, 2008).

[6.2] T. Wolfram and S. Ellialtıoglu, Electronic and Optical Properties of d-BandPerovskites (Cambridge: Cambridge University Press, 2006), pp. 27–39.

[6.3] M. Tinkham, Group Theory and Quantum Mechanics (New York: McGraw-HillBook Company, 1964).

[6.4] M. B. Robin, Higher Excited States of Polyatomic Molecules, Vol. 1, (New York:Academic Press, 1974).

Other suggested reading

C. J. Ballhausen and H. B. Gray, Molecular Orbital Theory (New York: W. A.Benjamin, Inc. 1964).F. A. Cotton, Chemical Applications of Group Theory (New York: Wiley–Interscience, 2nd edn., 1971).D. C. Harris and M. D. Bertolucci, Symmetry and Spectroscopy: An Introduc-tion to Vibrational and Electronic Spectroscopy (New York: Dover Publications,Inc., 1989).

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Exercises 189

Exercises

6.1 Consider two sites, A and B, with φ(px), φ(py), and φ(pz) orbitals on A anda φ(1s) orbital on B (Fig. 6.13).

x

y

z

spz

px

py

Figure 6.13

(a) Find the LCAO overlap and Hamiltonian matrices, S and H. Arrange therows and columns in the order s, pz , px , and py . Use εs and εp for thediagonal Hamiltonian matrix elements and S for the overlap between thes and pz orbitals.

(b) Show that the eigenvalues of H− ε S are given by

ε = εp (twice),

ε± = A ± (A2 + B)1/2,

A = εs + εp − 2S(spσ)

2(1− S2),

B = (spσ)2 − εsεp

1− S2.

(c) Show that the normalized eigenvectors are φ(px), φ(py), and

φ±(1s, pz)

= −{(spσ)− ε±S}φ(1s)+ (εs − ε±) φ(pz)√{(spσ)− ε±S}2 + (εs − ε±)2 − 2S{(spσ)− ε±S}(εs − ε±).

6.2 (a) For the molecule in Exercise 6.1, show that the Löwdin orbitals areξ(px) = φ(px), ξ(py) = φ(py), ξ 1s = a φ(1s) + b φ(pz) and ξ pz =a φ(pz) + b φ(1s), where a = (1/2){(1 + S)−1/2 + (1 − S)−1/2} andb = (1/2){(1 + S)−1/2 − (1 − S)−1/2}, where S is the (1s, pz) overlapintegral.

(b) Prove that ξ 1s is orthogonal to ξ pz .(c) Find the coefficients, a and b, for S = 0.25, and sketch the two Löwdin

orbitals, ξ 1s and ξ pz .

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190 Molecular electronic structure: The LCAO model

6.3 (a) For the system in Exercise 6.2, show that

(spσ)L = (a2 + b2)(spσ)+ ab (ε1s + εp) = (spσ)− S(ε1s + εp)/2

(1− S2),

(ε1s)L = a2(ε1s)+ b2(εp)+ 2(spσ) ab,

(εp)L = b2(ε1s)+ a2(εp)+ 2(spσ) ab,

where the superscript “L” means for Löwdin orbitals, and a and b aregiven in Exercise 6.2. The Löwdin parameters are defined by

(spσ)L =∫

dr (ξ 1s)∗ H ξ pz ,

(ε1s)L =

∫dr (ξ 1s)∗ H ξ 1s,

(εp)L =

∫dr (ξ pz )∗ H ξ pz .

(b) For ε1s = −13 eV, εp = −15 eV, S = 0.25, and (spσ) = 3 eV, calculate(spσ)L, (ε1s)

L, and (εp)L.

6.4 Suppose an electron in the 1b2 state of H2O is removed, leaving an H2O+

molecule in an excited state.(a) Give the two possible determinantal states of the molecule.(b) What are the molecular states expressed in the notation (2s+1)�α?(c) What electric-dipole transitions can occur from initial states 1b1 and 2a1

to the final state 2b2 for H2O+, and how must light be polarized in orderto achieve the transition? Specify the transition between the molecularstate labels (i.e., (2s+1)�α → (2s+1)�α′).

(d) What is the frequency of the light required for each transition?6.5 Consider a dxy orbital located at the center of a square in the x–y plane as

shown in (a) in Fig. 6.14. Each corner of the square has px and py orbitals.The fundamental definitions of (pdπ) and [pdπ ] are shown in (b) at the bot-tom left of the figure. The covering group is C4v and the symmetry elementsare shown in (c) at the bottom right of the figure.(a) Using the geometry and symmetry elements as shown in Fig. 6.14,

construct the action table for the corner numbers, px , py , and the dxy

orbitals.(b) Using the C4v character table and the action table, find the characters for

�(dxy) = �d and �(px , py) = �xy , and decompose the representationsinto the IRs of C4v.

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Exercises 191

(c) Determine the dimensions of the sub-blocks of the secular equation in thesymmetry-function representation and label each according to its IR.

6.6 For the p–d system in Exercise 6.5, generate the symmetry functions foreach sub-block. For the e functions use px1 , px2 , py1 , and py2 as the startingfunctions for the symmetry-function generator. List the symmetry functionsthat are also eigenstates and determine their eigenvalues.

H

1 11

22

2

3 3 3

44

+ 4+

(a)

(b) (c)

dxy orbital px orbitals

(pdπ) [pdπ]

1

2

3

4

py orbitals

σd2

σv1

σv2

σd1

x

y

C4v symmetry elementsDefinition of (pdπ)

Figure 6.14

6.7 For the p–d system in Exercise 6.5, determine the 2 × 2 secular equationfor the b2 states and find expressions for the eigenvalues and eigenvectorsaccording to the LCAO model including overlap integrals.

6.8 (a) For Exercise 6.5 sketch the two row-1 e symmetry functions and the tworow-2 e symmetry functions.

(b) Find the 2× 2 secular equations for row-1 and row-2 functions.(c) Prove that the eigenvalues of the 2 × 2 row-1 secular equation must be

the same as those of the 2× 2 row-2 secular equation.6.9 (a) Calculate the energies of all of the states for Exercise 6.5 using the

following parameters (energies in eV, overlaps are dimensionless):

ed = −6, (pdπ) = 1, (ppσ)1 = 0.5, (ppπ)1 = 0.2,(ppσ)2 = 0.05, (ppπ)2 = 0.025,

ep = −10, [pdπ ] = 0.2, [ppσ ]1 = 0.1, [ppπ ]1 = 0.1,[ppσ ]2 = 0.03, [ppπ ]2 = 0.01.

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192 Molecular electronic structure: The LCAO model

(b) Make a sketch of the correlation diagram showing how the isolated p andd levels correlate with the one-electron molecular levels. Label the statesas “g” or “u”.

6.10 Suppose the isolated corner atoms in the system in Exercise 6.5 each havetwo electrons and the center atom has one.(a) Specify the molecular ground-state configuration and its symbol.(b) Which one-electron electric-dipole transitions to the 1a2g, 2b2g, and 2eu

are forbidden and which are not forbidden? What is the polarization ofthe light for non-forbidden transitions?

(c) Give the configurations and the molecular symbols for the final states thatare not forbidden.

(d) Using the results from Exercise 6.9, calculate the ground-state molecularbinding energy, assuming that εp and εd are the energies of the isolatedp- and d-orbitals.

(e) Using the results of Exercise 6.9, calculate the energies of the photonsrequired for the 1b1g → 2eu and the 1b2g → 2eu transitions.

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7

Electronic states of diatomic molecules

7.1 Bonding and antibonding states: Symmetry functions

In Chapter 6 we discussed the general approach to molecular-orbital theory formolecules. The methods employed are, of course, equally applicable to diatomicmolecules. However, diatomic (or linear) molecules have some special propertiesthat make them worth discussing separately. One special property is the existenceof a continuous symmetry operation. For a molecule A–A or A–B a rotation by anyangle about the internuclear axis is a symmetry element. Thus there is an infinitenumber of operations. A second feature is the special notation used to specify themolecular states.

There are two types of diatomic molecules: homonuclear A–A and heteronuclearA–B. The covering group for homonuclear diatomic molecules is D∞h . The “infin-ity” refers to the infinity of rotations about the internuclear axis. The coveringgroup for A–B molecules is C∞v. The symmetry elements and character tables forthese two groups are shown in Table 7.1.

The C∞v group has as elements the identity E , rotation Cφ (and C−φ) by anyangle φ (or −φ) about the internuclear axis (the z-axis), and a reflection in the σv

plane. The group D∞h is the direct product D∞h = i × C∞v.1

Consider a homonuclear, diatomic molecule with an arbitrary hydrogen-likeorbital, φα, centered on atoms 1 and 2, and denote them as φα1 and φα2, whereα specifies the orbital symmetry type (s, p, d, . . .). The LCAO secular equation forthe interaction of these two orbitals is

∣∣∣∣ ε − λ H12 − λS12

H12 − λS12 ε − λ

∣∣∣∣ = 0, (7.1)

1 Other choices of the elements can be made for D∞h . For example, σv can be omitted and a two-fold rotationadded.

193

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194 Electronic states of diatomic molecules

Table 7.1 (a) The D∞h character table and symmetry elements for ahomonuclear, A−A, molecule (c.m. is the center of mass). (b) The C∞v charactertable and symmetry elements for a heteronuclear, A−B, molecule.

(a)

D∞h(∞/mm) E 2Cφ σv i 2iCφ iσv

x2 + y2, z2 A1g(�+g ) 1 1 1 1 1 1

A1u(�−u ) 1 1 1 −1 −1 −1

Rz A2g(�−g ) 1 1 −1 1 1 −1

z A2u(�+u ) 1 1 −1 −1 −1 1

(xz, yz) (Rx , Ry) E1g(�g) 2 2 cosφ 0 2 2 cosφ 0(x , y) E1u(�u) 2 2 cosφ 0 −2 −2 cosφ 0

(x2 − y2, xy) E2g(�g) 2 2 cos 2φ 0 2 2 cos(2φ) 0E2u(�u) 2 2 cos 2φ 0 −2 −2 cos(2φ) 0

. . . . . . . . . . . . . . . . . . . . .

AA c.m.

σvx

z

y

C2

(b)

C∞v(∞m) E 2Cφ σv

x2 + y2, z2 z A1(�+) 1 1 1

Rz A2(�−) 1 1 −1

(xz, yz) (x , y) E1(�) 2 2 cosφ 0(Rx , Ry)

(x2 − y2, xy) E2(�) 2 2 cos(2φ) 0. . . . . . . . . . . .

AB

σvx

z

y

where ε is the diagonal energy, H11. H12 is the Hamiltonian matrix element, andS12 is the overlap of the two orbitals. The eigenvalues and eigenvectors of (7.1) are

λ+(α) = ε + H12

1+ S12, ψ+(α) = φα1 + φα2√

2(1+ S12), (7.2)

λ−(α) = ε − H12

1− S12, ψ−(α) = φα1 − φα2√

2(1− S12). (7.3)

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7.1 Bonding and antibonding states: Symmetry functions 195

Table 7.2 The action table for s- and pz orbitals.ψb(s) = [φs1 + φs2] /√2+ [ssσ ], ψa(s) = [φs1 − φs2]/√2− [ssσ ],ψb(pz) = [φpz1 −φpz2] /

√2− [spσ ], and ψa(pz) = [φpz1 +φpz2] /

√2+ [spσ ].

The superscripts “b” and “a” denote bonding and antibonding combinations. Anentry in the table of “1” means that the function goes into itself and “−1” meansthat it goes into the negative of itself.

Atom position E Cφ C ′2 i iCφ iC ′21 1 1 1 2 2 22 2 2 2 1 1 1

Functions IR basis

ψb(s) 1 1 1 1 1 1 σ+gψa(s) 1 1 −1 −1 −1 1 σ+uψb(pz) 1 1 1 1 1 1 σ+gψa(pz) 1 1 −1 −1 −1 1 σ+u

The bonding/antibonding symmetry of ψ±(α) depends on α, the symmetryindex. For an s-state combination ψ+(α) is the bonding state and ψ−(α) is theantibonding state. The same is true for π orbitals (px or py). For the pz-orbitalscombination the opposite is true; ψ−(pz) is the bonding state and ψ+(pz) is theantibonding state. As we shall see later, the states of (7.2) and (7.3) are basisfunctions for the IRs of D∞h .

Let us see how these combinations, ψ±, transform under the symmetry oper-ations of D∞h . The results depend on the symmetry of the orbitals with respectto the internuclear axis. The action table for s-orbitals and pz orbitals is shownin Table 7.2. The superscripts “b” and “a” denote bonding and antibondingcombinations.

Comparing the results of Table 7.2 with the character table for D∞h shows thatψb(s) and ψb(pz) transform as A1g or �+g . The notation �+g indicates a functionwhose symmetry with respect to the internuclear axis is “sigma” (axially symmet-ric), and is symmetric, “g”, under inversion. The “+” indicates that the functionis symmetric under reflection in the σv plane. We shall maintain our rule of usinglower-case letters for the one-electron states and an upper-case letter for the totalmolecular state. Therefore we shall use the notation σ+g for a one-electron functionand �+g for a multi-electron, molecular state.

According to the character table for D∞h , the functions ψa(s) and ψa(pz) belongto σ+u (A2u), meaning that they are “sigma” with respect to the internuclear axis,symmetric under σv, and antisymmetric under inversion.

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196 Electronic states of diatomic molecules

The pπ orbitals perpendicular to the A–A axis transform into one another undersome operations of the group and therefore will form a 2×2 representation. Underthe operation of σv, py is antisymmetric (−) and px is symmetric (+). We can formthe bonding and antibonding orbitals,

ψb(pt) = pt1 + pt2√2(1+ S12)

, ψa(pt) = pt1 − pt2√2(1− S12)

(t = x and y),

where t = x and y indicate px and py orbitals, respectively. Unlike the sigmastates, ψb(pt) is antisymmetric and ψa(pt) is symmetric under inversion.

If we rotate px by φ clockwise px → px cosφ − py sinφ. The function py →px sinφ + py cosφ. Therefore, the 2 × 2 representation matrix for the operatorPR = PCφ is

PCφ=(

cosφ −sinφ

sinφ cosφ

). (7.4)

For the inversion and σv operations,

Pi =( −1 0

0 −1

), (7.5)

Pσv =( −1 0

0 1

). (7.6)

From these properties we can calculate the characters for ψb(px), ψa(px), ψb(py),and ψa(py). For example,

PCφψb(px) = PCφ

px1 + px2√2(1+ S12)

= cosφ ψb(px)− sinφ ψb(py), (7.7)

PCφψb(py) = PCφ

py1 + py2√2(1+ S12)

= cosφ ψb(py)+ sinφ ψb(px), (7.8)

On the other hand, for the inversion operation,

Pi ψb(px) = Pi

px1 + px2√2(1+ S12)

= −px2 − px1√2(1+ S12)

= −ψb(px), (7.9)

Pi ψa(px) = Pi

px1 − px2√2(1− S12)

= −px2 + px1√2(1− S12)

= +ψa(px). (7.10)

The action table for the p-orbitals perpendicular to the internuclear axis is givenin Table 7.3(a). Using this table one can calculate the characters of the 2 × 2representations based on the pairs {ψb(px), ψ

b(py)} and {ψa(px), ψa(py)}. The

character table is shown in Table 7.3(b).On comparing the results of Table 7.3(b) with the character table of Table 7.1(b)

it is seen that the sets {ψb(px), ψb(py)} and {ψa(px), ψ

a(py)} are basis functionsfor the πu(E1u) and πg(E1g) IRs, respectively. The πu and πg are each doubly

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7.1 Bonding and antibonding states: Symmetry functions 197

Table 7.3 Symmetry properties of the px and py orbitals. (a) The action table.(b) The character table for the 2× 2 representations with basis functions{ψb(px ), ψ

b(py)} and {ψa(px ), ψa(py)}. (c) The πu(E1u) IR matrices for the

basis functions {ψb(px ), ψb(py)}. (d) The πg(E1g) IR matrices for the basis

functions {ψa(px ), ψa(py)}.

(a)

Atom E 2Cφ σv i 2iCφ iσv

1 1 1 1 2 2 22 2 2 2 1 1 1

Orbital

px px px cosφ −px −px −px cosφ px− py sinφ + py sinφ

py py py cosφ +py −py −px cosφ −py+ px sinφ − px sinφ

(b) Character table for the π states

Functions IR χ(E) 2χ(Cφ) χ(σv) χ(i) 2χ(iCφ) χ(iσv)

{ψb(px ), ψb(py)} πu(E1u) 2 2 cosφ 0 −2 −2 cosφ 0

{ψa(px ), ψa(py)} πg(E1g) 2 2 cosφ 0 2 2 cosφ 0

ψb(pt) = pt1 + pt2√2(1+ S12)

ψa(pt) = pt1 − pt2√2(1− S12)

(t = x and y)

(c) πu(E1u) IR matrices for the basis functions {ψb(px), ψb(py)}

PE =(

1 00 1

), PCφ

=(

cosφ −sinφ

sinφ cosφ

), Pσv =

( −1 00 1

),

Pi =( −1 0

0 −1

),

PiCφ= Pi PCφ

,

Piσv = Pi Pσv .

(d) πg(E1g) IR matrices for the basis functions {ψa(px), ψa(py)}

PE =(

1 00 1

), PCφ

=(

cosφ −sinφ

sinφ cosφ

), Pσv =

( −1 00 1

),

Pi =(

1 00 1

),

PiCφ= Pi PCφ

,

Piσv = Pi Pσv .

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198 Electronic states of diatomic molecules

degenerate and therefore each can contain two sets of paired electrons. The IRmatrices are shown in Tables 7.3(c) and 7.3(d).

The analysis can be continued for d-, f -, . . . orbitals. For example, the bondingand antibonding combinations of dxz and dyz are

ψa(dxz) = dxz1 + dxz2√2+ 2[ddπ ] , ψb(dxz) = dxz1 − dxz2√

2− 2[ddπ ] , (7.11)

ψa(dyz) = dyz1 + dyz2√2+ 2[ddπ ] , ψb(dyz) = dyz1 − dyz2√

2− 2[ddπ ] . (7.12)

The formation of bonding and antibonding combinations for hydrogen-likefunctions on the two atoms that have different symmetries with respect to the inter-nuclear axis need not be considered because the Hamiltonian matrix element andthe overlap of functions vanish.

What we have accomplished in the above analysis is a simple method for find-ing the symmetry functions for any type of orbital centered on two atomic sites.Symmetry functions belonging to the same IR will interact, and those belong-ing to different IRs or different rows of the same IR will not. Thus the blockdiagonalization of the secular equation can easily be determined.

7.2 The “building-up” of molecular orbitals for diatomic molecules

The bonding and antibonding molecular orbitals serve as the first-order approxima-tion to the molecular eigenstates, similar to the manner in which the hydrogen-likeorbitals are the first-order approximation to the atomic states. In fact, the labelingof the molecular orbitals, �, �, �, . . ., σ , π , δ, . . . is just the Greek-letter equiva-lent of the S, P , D, . . . or s, p, d, . . . for atomic labels. However, the degeneraciesare not the same. For example, while P and D are three- and five-dimensional,respectively, � and � are both two-dimensional.

The bonding and antibonding molecular orbitals are not eigenfunctions of themolecular Hamiltonian; they are symmetry functions. The eigenstates are linearcombinations of the symmetry functions that transform according to the same rowof the same IR of the covering group. Nevertheless, we can specify the molec-ular electronic configurations in terms of these symmetry functions and deducequalitative results.

7.2.1 Quantum numbers for homonuclear diatomic molecules

Since A–A diatomic molecules have two centers, the Hamiltonian can not havespherical symmetry. As a result L2 is not a “good” quantum number. However, Lz ,the angular momentum about the z-axis, is still appropriate. For A–A molecules the

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7.2 The “building-up” of molecular orbitals for diatomic molecules 199

notation � is customarily used instead of Lz to specify the total angular momentumabout the z-axis, where � = |Lz|. The angular momentum for the i th one-electronmolecular state is labeled as λi , and � = ∑

λi . The spin of the i th one-electronmolecular state is siz , and, for the total molecular spin, Sz = ∑

siz . The values ofλ are 0, 1, and 2 for the molecular orbitals σ , π , and δ, or � = 0, 1, and 2 for themolecular states �, �, and �, respectively.

The one-electron molecular orbitals belonging to a given irreducible representa-tion play a role similar to that of the orbitals of a given atomic shell. A σ+g state canbe occupied by two electrons of opposite spin, denoted by (σ+g )2. This constitutesa closed shell in the same way that (ns)2 is a closed shell. Closed shells trans-form as �+g (A1g). The same may be said for (σ+u )2, (σ−g )2, and (σ−u )2, which alsotransform as �+g (A1g). However, the pi states are different. They are derived fromcombinations of px and py orbitals and transform according to the two-dimensionalIRs of D∞h labeled πu or πg. Therefore each pi shell can accommodate four elec-trons, namely two pairs of electrons with oppositely directed spins. As a result(πu)

4 and (πg)4 are the closed molecular-shell configurations that transform as

�+g (A1g).In the absence of spin–orbit effects the symbol specifying the total molecular

state has the form

(2S+1)�(±)(g/u), (7.13)

where (2S + 1) is the spin multiplicity, � is the total orbital momentum about thez-axis, +/− is the reflection symmetry in a plane containing the A–A axis, andg/u is the parity (the inversion symmetry of the spatial part of the function).

To specify the molecular configurations, we need to know the energy orderingof the various bonding and antibonding states. In general the atomic levels areenergetically ordered according to 1s < 2s < 2p. For the molecular orbitals,σ+g < σ+u and πu < πg. With these rules we can construct two different energy-level arrangements depending upon whether, for the 2p electrons, σ+g < πu orπu < σ+g . These two possibilities are shown schematically in Fig. 7.1.

The difference between the orderings in Figs. 7.1(a) and (b) is that for level-ordering 1 (LO1) 1πu < 3σ+g and in (b) the opposite is true. It is generally believedthat LO1 holds for the lightest diatomic molecules, H2, He2, Li2, B2, N+2 , and C2.For O2 and F2, LO2 is appropriate. The explanation for this switch is that the split-ting of the bonding and antibonding levels decreases as the nuclear charge (Z )increases. Heavier diatoms tend to have greater bond lengths and smaller overlap.Starting from F2 and moving toward lighter molecules the πg–πu splitting increasesfaster than the σ+u −σ+g splitting. The πu level in LO2 drops below the σ+g level,resulting in the LO1 scheme. The crossover is shown schematically in Fig. 7.2.

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200 Electronic states of diatomic molecules

2p 2p

3σ+u

1πg

3σ+g

1πu

2s 2s2σ+

u

2σ+g

1σ+u

1σ+g

1s 1s

LO1

(a)

2p 2p

3σ+u

1πg

1πu

3σ+g

2s 2s2σ+

u

2σ+g

1σ+u

1σ+g

1s 1s

LO2

(b)

Figure 7.1 A schematic representation of the ordering of the levels for homo-nuclear, diatomic molecules. (a) Level-ordering 1 (LO1), 1πu < 3σ+g . (b) Level-ordering 2 (LO2), 3σ+g < 1πu .

7.2.2 The united-atom concept

A qualitative explanation of the πu–σ+g crossover can be obtained from the “united-atom” concept. Consider decreasing the distance between the two atoms of themolecule. When the distance approaches zero we have a single “united atom” withtwice the nuclear charge and twice as many electrons. In this limit the orbitals mustbecome the atomic states 1s, 2s, 2p, . . . Since the nuclear charge is twice that ofa single nucleus, these atomic states must lie at much lower energies than those ofa single atom. Increasing the distance between the nuclei eventually produces twonon-interacting atoms. Again, in the limit of “separated atoms”, the states mustbe two identical sets of 1s, 2s, 2p, . . . states for an atom with a single nuclearcharge. These states will be at higher energies than for the corresponding united-atom limit. The internuclear distance changes, but the covering group does not.Group-theoretical considerations require that a molecular orbital evolving towardthe limit of the united atom or the limit of the separated atoms not change itssymmetry. The σ+g molecular level arises from the in-phase, linear combinationof two 1s states of the separated atoms. As the united-atom limit is approachedthe two nearly overlapping 1s orbitals have the symmetry of an s-orbital. Thusσ+g ⇒ 1s (united atom). On the other hand, σ+u , constructed from two out-of-phase1s orbitals, has the symmetry of a p-orbital. Therefore σ+u ⇒ 2p (united atom).A πg state has the same symmetry as a dxz or dyz orbital, so φ1(px) − φ2(px) ⇒πg ⇒ dxz (united atom), while φ1(px)+φ2(px)⇒ πu ⇒ px (united atom). These

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7.2 The “building-up” of molecular orbitals for diatomic molecules 201

4p

3d

3p

3s

2p

3σ+u

1πg

1πu

3σ+g

2p

2s

2s2σ+

u

2σ+g

1σ+u

1σ+g

1s

1s

O2

N2

Crossing

LO1 LO2

(Li2, Be2, B2, C2, N2) (O2, F2, Ne2)

Separated atomUnited atom

Figure 7.2 A schematic correlation diagram showing the evolution of the stateson going from separated atoms to the united atom. The 1πu level drops below the3σ+g on going from O+2 to N2 because it correlates with the 2p (united atom),while the 3σ+g correlates with the 3s (united atom).

rules allow us to construct the qualitative correlation diagram in Fig. 7.2 showinghow the molecular states evolve as a function of the internuclear distance.

7.2.3 Molecular configurations

We can now build up a picture of the electronic configurations of various diatomsby filling the molecular levels with electrons. For example, for N2, each nitrogenatom has 7 electrons, so there are 14 electrons to distribute among the N2 molecularorbitals. The configuration is (1σ+g )2(1σ+u )2(2σ+g )2(2σ+u )2(1πu)

4(3σ+g )2 (using theLO1 scheme). The label for the total molecular state is the IR of D∞h according towhich the product of all of the occupied states transforms. For N2 the ground state

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202 Electronic states of diatomic molecules

is �+g , since all molecular shells are closed. Some typical configurations are shownin Table 7.4 on p. 204.

A more interesting example is that of the O2 molecule. The electronic configura-tion is (1σ+g )2(1σ+u )2(2σ+g )2(2σ+u )2(3σ+g )2(1πu)

4(1πg)2 = (closed shells) (1πg)

2.The total molecule symmetry is determined by the partially filled 1πg shell. Thereare two electrons in this shell and the possible choices for the spins are S = 0 for↑↓ and S = 1 for ↑↑ or ↓↓. The first choice will produce a singlet state and thesecond a triplet state. There are two choices for the molecular orbitals: ψa(px) andψa(py) (see Table 7.3(b)). The product of two πg states is a “g” state, The orbitalsymmetries of the possible states are determined by the direct product, πg × πg.From Table 7.5 on p. 205 we conclude that the possible total molecular states are�+g , �−g , and �g. The spin multiplicity of these states is discussed below.

When a configuration results in several terms, Hund’s rules (discussed inChapter 3) state that the ground state is the term with the maximum spin andmaximum angular momentum. There are exceptions to Hund’s rules but in a largemajority of cases they are valid.

It is useful to introduce a simplified notation for the determinantal state. We shalluse 〈ψa(pxσ)ψ

a(pyσ′)| as an abbreviation for the full determinantal state,

〈ψa(pxσ)ψa(pyσ

′)| = 1√16! det{(1σ+g )2(1σ+u )2(2σ+g )2(2σ+u )2(3σ+g )2

(1πu)4ψa(pxσ)ψ

a(pyσ′)}. (7.14)

The determinantal states that can be formed with S = 1 are

(1) 〈ψa(px ↑) ψa(px ↑)| = 0 (two states are identical), (7.15)

(2) 〈ψa(px ↑) ψa(py ↑)|, (7.16)

(3) 〈ψa(py ↑) ψa(py ↑)| = 0 (two states are identical), (7.17)

(4) 〈ψa(py ↑) ψa(px ↑)| = −〈ψa(px ↑) ψa(py ↑)|, (7.18)

and, for S = 0,

(5) 〈ψa(px ↑) ψa(px ↓)|, (7.19)

(6) 〈ψa(py ↓) ψa(py ↑)|, (7.20)

(7) 〈ψa(px ↑)ψa(py ↓)|, (7.21)

(8) 〈ψa(px ↓) ψa(py ↑)|. (7.22)

There are in total five independent, non-zero, determinantal states. From the directproduct we know that the total molecular states are �+g +�−g +�g. There is onlyone independent S = 1 determinant, 〈ψa(px ↑) ψa(py ↑)|. This state must belongto a triplet term since S = 1. Therefore there must be two singlets and one triplet

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7.2 The “building-up” of molecular orbitals for diatomic molecules 203

for a total of five states. To identify which state is the triplet and which states aresinglets we must consider how the states transform under the operations of thegroup.

We first consider the product, ψa(px ↑) ψa(py ↑). Making use of equations (7.7)and (7.8) gives

PCφ ψa(px(r) ↑) ψa(py(r′) ↑) = PCφ ψ

a(px(r) ↑)PCφ ψa(py(r′) ↑)

= [cosφ ψa(px(r) ↑)− sinφ ψa(py(r) ↑)]× [cosφ ψa(py(r′) ↑)+ sinφ ψa(px(r′) ↑)]= cos2 φ[ψa(px(r) ↑) ψa(py(r′) ↑)]− sin2 φ[ψa(py(r) ↑) ψa(px(r′) ↑)]+ cosφ sinφ[ψa(px(r) ↑) ψa(px(r′) ↑)]− cosφ sinφ[ψa(py(r) ↑) ψa(py(r′) ↑)].

(7.23)

We have added the variables r and r′ as a reminder that the states are functions oftwo different electron coordinates. The determinantal state satisfies the relation

PCφ〈ψa(px ↑) ψa(py ↑)| = cos2 φ〈ψa(px ↑) ψa(py ↑)|− sin2 φ〈ψa(py ↑) ψa(px ↑)|+ cosφ sinφ〈ψa(px ↑) ψa(px ↑)|− cosφ sinφ〈ψa(py ↑) ψa(py ↑)|. (7.24)

The third and fourth determinants of (7.24) vanish because the last two columnsare identical. In addition, the second determinant is the negative of the firstdeterminant, so finally we have

PCφ〈ψa(px ↑) ψa(py ↑)| = +1〈ψa(px ↑) ψa(py ↑)| ; (7.25)

that is, the original determinant is invariant under PCφ .It should be noted that the spin is oriented along the z-axis and therefore it is

invariant under Cφ and σv. It is also invariant under inversion since it is an angular-momentum vector: Pi [r×dr/dt] = [−r×d(−r)/dt] = [r×dr/dt]. Consequentlythe spin is invariant under all of the operations of the group.

For the reflection σv, we have

Pσv ψa(px ↑) ψa(py ↑) = Pσv [px1 − px2][py1 − py2]

Nx Ny

= −[px1 − px2][py1 − py2]Nx Ny

= −ψa(px ↑) ψa(py ↑), (7.26)

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204 Electronic states of diatomic molecules

Table 7.4 Molecular-orbital configurations for some homomolecular diatomicmolecules (Mol., molecule)

Mol. Electronic configuration# ofes

Molecularsymbol

Bondorder

H+2 (1σ+g ) 1 2�+g 1/2H2 (1σ+g )2 2 1�+g 1He+ (1σ+g )2(1σ+u )1 3 2�+u 1/2He2 (1σ+g )2(1σ+u )2 4 1�+g 0Li2 (1σ+g )2(1σ+u )2(2σ+g )2 6 1�+g 1Be2 (1σ+g )2(1σ+u )2(2σ+g )2(2σ+u )2 8 1�+g 0N2 (1σ+g )2(1σ+u )2(2σ+g )2(2σ+u )2(1πu)

4(3σ+g )2 14 1�+g 3O+2 (1σ+g )2(1σ+u )2(2σ+g )2(2σ+u )2(3σ+g )2(1πu)

4(1πg)1 15 2 Ag 5/2

O2 (1σ+g )2(1σ+u )2(2σ+g )2(2σ+u )2(3σ+g )2(1πu)4(1πg)

2 16 1�+g , 3�−g , 1�g 2F2 (1σ+g )2(1σ+u )2(2σ+g )2(2σ+u )2(3σ+g )2(1πu)

4(1πg)4 18 1�+g 1

where Nx and Ny are the normalization factors. Under inversion,

Pi ψa(px ↑) ψa(py ↑) = −ψa(px ↑)×−ψa(py ↑) = +ψa(px ↑) ψa(py ↑).

(7.27)From these symmetry properties and Table 7.1(a) we can identify 〈ψa(px ↑)ψa(py ↑)| as 3�−g with Ms = +1. The other two spin states, Ms = 0 and Ms = −1,can be generated by operating on 〈ψa(px ↑) ψa(py ↑)| with the spin-loweringoperator.

In summary, the O2 configuration, (1σ+g )2(1σ+u )2(2σ+g )2(2σ+u )2(3σ+g )2 (1πu)4

(1πg)2, yields three terms, 1�+g , 3�−g , and 1�g. According to Hund’s rules, 3�−g

is the ground state. The lowest excited state is 1�g and the next higher energystate is 1�+g . The two excited states are involved in the absorption of light by theatmosphere.

The last column of Table 7.4 specifies the “bond order”. The bond order isdefined as the number of electrons occupying bonding states minus the numberoccupying antibonding states divided by 2. For example, for N2 the configura-tion of the ground state includes 10 bonding states and 4 antibonding states, sothe bond order is 6/2 = 3. The bond order is a useful qualitative measure of thestrength of the bond holding the molecule together. A number of physical prop-erties of a molecule can be correlated with the bond order. The chemical bondenthalpy, D0, increases with the bond order. The bond length also correlates withbond order, shorter bond length corresponding to larger bond order for a particularseries such as the series B2, C2, N2, O2, and F2 or the series Si2, P2, S2, and Cl2. Inaddition, the vibrational stretching frequency is roughly proportional to the bond

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7.2 The “building-up” of molecular orbitals for diatomic molecules 205

Table 7.5 Direct products for C∞v and D∞h. For D∞h the “g” and “u” labelsare omitted. They may be added according to the rule g × g = g, g × u = u, andu × u = g. The column labeled � gives the angular momentum about theinteratomic axis.

C∞v , D∞hIR � �+ �− � � �

�+ 0 �+ �− � � �

�− 0 �− �+ � � �

� 1 � � �+ +�− +� �+� �+ �

� 2 � � �+� �+ +�− + � �+ H

order. Molecules with zero bond order (e.g., He2 and Be2) are considered unstableand to date have not been experimentally detected.

7.2.4 Electric-dipole transitions

The electric-dipole transition-matrix element is∫

dr�∗i μ�f, where μ is theelectric-dipole moment operator, �i is the initial molecular state, and �f is the finalstate. From the character table for D∞h we see that the z-component of the electric-dipole operator belongs to �+u (A2u) and the x- and y-components belong to�u(E1u). Therefore the product �∗i �f must transform as �+u or �u or else thetransition is forbidden. Alternately we can require that �μ × �i contains �f, that�μ × �f contains �i, or that �i × �f × �μ contains �+g . In any case, �i and �f

must have different parities since μ is odd under inversion. For z-polarized lightselection rules for symmetry-allowed transitions are easily worked out from thedirect-product rule that �i × �+u must contain �f. Using the character table, wefind that the allowed transitions are �+u ⇔ �+g , �−u ⇔ �−g , and �u ⇔ �g.

For light polarized in the x–y plane (electric dipole belonging to �u) the directproduct of two representations of dimensions higher than one can not be obtainedfrom the character table alone. The direct products of some of the IRs are listed inTable 7.5.

If we look at the results in Table 7.5 for � × � we see that the direct productcontains the IRs ��+1 and ��−1 (recall that � is the absolute value of the angularmomentum about the z-axis.) For light polarized in the x–y plane, the electric-dipole operator transforms as �u . If ��

g is the initial state then the final state mustbe ��+1

u or ��−1u , or else the transition is forbidden. Conversely, if ��

u is the initialstate then the final state must be ��+1

g or ��−1g , or else the transition is forbidden.

The selection rule requires �� = ±1. Some of the symmetry-allowed transitions

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206 Electronic states of diatomic molecules

for radiation polarized in the x–y plane are �±g ⇔ �u , �±u ⇔ �g, �g ⇔ �u , and�u ⇔ �g.

7.3 Heteronuclear diatomic molecules

The covering group for A–B molecules, C∞v, is given in Table 7.1(b). It consistsof only the operations E , Cφ , and σv, since there is no inversion center. We canproceed with the analysis much as we did with the A–A molecules, forming bond-ing and antibonding combinations. Let φA(r − RA) and φB(r − RB) be orbitalscentered on the atoms A and B. For these two orbitals the secular equation is∣∣∣∣ εA − λ HAB − SABλ

HAB − SABλ εB − λ

∣∣∣∣ = 0, (7.28)

where εA and εB are the diagonal Hamiltonian matrix elements. The element, HAB

is the matrix element∫

dr φ∗A H φB , and SAB is the corresponding overlap betweenφA and φB . We can assume that the two orbitals have the same symmetry withrespect to the internuclear axis (if not, HAB and SAB vanish). The eigenvalues for(7.28) are

λ+ = C − D, (7.29)

λ− = C + D, (7.30)

where

C = εA + εB − 2HAB SAB

2(1− S2AB)

, (7.31)

D ={

C2 − εAεB − H 2AB

1− S2AB

}1/2

. (7.32)

The eigenvectors are

ψ− = 1√NA

(HAB − SABλ−)φB − (εB − λ−)φA, (7.33)

ψ+ = 1√NB

(HAB − SABλ+)φA − (εA − λ+)φB, (7.34)

NA/B = {(HAB − SABλ±)2 + (εA/B − λ±)2

− 2(HAB − SABλ±)(εA/B − λ±)SAB}1/2. (7.35)

The bonding/antibonding combinations depend on the symmetry of the individ-ual orbitals. For example, for sA and sB orbitals the bonding state corresponds tothe plus sign. For pz1 and s2 orbitals the minus sign gives the bonding state. Ingeneral, for the bonding state, the sign is such that the phases of the orbitals are thesame in the overlap region between atoms.

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7.3 Heteronuclear diatomic molecules 207

Table 7.6 AB molecule bonding and antibonding orbital combinations.ψb/a = Ab/aφA + Bb/aφB, where the coefficients Ab/a and Bb/a are given inEqs. (7.33) and (7.34).

Function φA φB χ(E) χ(Cφ) χ(σv) IR

ψb(s, s) s s 1 1 1 σ+ψa(s, s) s s 1 1 1 σ+ψb(s, pz) s pz 1 1 1 σ+ψa(s, pz) s pz 1 1 1 σ+ψb(pz, pz) pz pz 1 1 1 σ+ψa(pz, pz) pz pz 1 1 1 σ+{ψb(px , px ), ψ

b (py, py)} (2× 2 IR) 2 2 cosφ 0 π{ψa(px , px ), ψ

a(py, py)} (2× 2 IR) 2 2 cosφ 0 π

ψb/a(px , px ) px px

ψb/a(py, py) py py

(a)

A

(λa − A)

λa

λb

B

( B − λb)

(b)

ψb

ψa

Figure 7.3 (a) A schematic representation of the energy levels for the ABmolecule. (b) A schematic representation of the eigenfunctions when φA is ap-orbital and φB is an s-orbital.

The energy levels resulting from the interaction of φA and φB are shownschematically in Fig. 7.3, where λb and λa are the bonding and antibonding ener-gies, respectively. The lower level, εB , is pushed down by an amount (εB −λb) andthe upper level is pushed up by an amount (λa−εA). The eigenvectors are illustratedschematically in Fig. 7.4(b), where φA is a p-orbital and φB is an s-orbital.

7.3.1 Bonding and antibonding combinations for the AB molecule

The action table for the bonding and antibonding combinations is given inTable 7.6. The orbitals are of the form AφA + BφB . The φA and φB orbitals musthave the same symmetry about the internuclear axis but need not be the same type

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208 Electronic states of diatomic molecules

2p

2p

6σ+

5σ+

2s

2s

4σ+

3σ+

2σ+

1σ+

1s1s

A B

(a)

1s 2p

4σ+

↑↓↑↓3σ+↑↓2σ+

↑↓1σ+

↑↓

2s

1s

H F

(b)

Figure 7.4 Level ordering for A–B diatomic molecules. (a) Ordering formolecules with not too dissimilar atoms such as CO. (b) Ordering for HF. TheH(1s) orbital interacts strongly only with the σ(2p) orbitals. The 1π is a two-folddegenerate, non-bonding level that can accommodate up to four electrons (twosets of paired electrons). The HF ground-state configuration is (1σ+)2(2σ+)2,(3σ+)2(1π)4. The occupied states are indicated by the spin arrows.

of orbital. For example, we could use an s-orbital on atom A and a pz orbital onatom B. The same is true of the A–A molecule, but in that case we know that thesame orbitals have the same energy and interact most strongly. Therefore, as a firstapproximation, we can ignore the hybrid combinations such as a pz orbital withan s-orbital. For the A–B molecule, it is possible that the two orbitals closest inenergy are different types of orbitals. This is the case, for example, for HF, wherethe H(1s) (hydrogen 1s orbital) interacts with the F(2pz) (fluorine 2pz orbital). TheF(1s) and F(2s) are energetically too far removed from the H(1s) to form a bondof any consequence. The level ordering for HF is shown in Fig. 7.4(b). The F(px)

and F(py) orbitals do not interact with the H(1s) orbitals. They form the 1π level, atwo-fold degenerate, non-bonding level that can accommodate up to four electrons(two sets of paired electrons). For H–B molecules with B less electronegative thanF the B(2s) may form a sigma bond with the H(1s) if the energy separation is nottoo large. The possible choices for s- and p-orbitals are given in Table 7.6.

The first column of Table 7.6 specifies ψb/a, the second and third columnsspecify the two orbitals used in constructing ψb/a. Columns 4, 5, and 6 givethe characters under the operations of the group. The functions ψb(px , px) andψb(py, py) are the bases for a two-dimensional representation, as are the functionsψa(px , px) and ψa(py, py). The characters are the sum of the diagonals of the 2×2

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Exercises 209

representation matrices. Each function is a basis function for an IR (or row of anIR) of C∞v. The IRs for the functions are given in the last column of Table 7.6.

Figure 7.4(a) shows the level ordering for diatomic molecules with not too dis-similar atoms. On comparing Fig. 7.4(a) with Fig. 7.4(b) it is clear that the actuallevel ordering depends on the atoms that form the molecule. Experimental datais often required in order to determine the exact ordering. The molecular orbitalstates we have been discussing are analogous to the 1s, 2s, 2p, . . . states of thehydrogenic atoms. They provide a starting point for understanding the electronicstructure, but are not the molecular eigenstates. Levels having the same symme-try will interact, and the effects of various Hamiltonian perturbations (exchangeenergy, spin–orbit effects, . . .) must be added to the theory in order to find goodapproximations to the eigenstates. Nevertheless, a great deal of useful informa-tion is obtained using this simple, molecular “building-up” principle. Whatever theactual eigenstates are, we can be assured that they possess the same symmetries asthose predicted by molecular-orbital theory.

Exercises

7.1 (a) NO and O+2 are similar molecules. O+2 has an additional proton (and neu-tron) compared with NO, but they have the same number of electrons.Assume the level ordering (LO2) for the NO molecule is the same as O+2 .What is the molecular configuration for NO? (Use Table 7.4. Drop theu/g designation and renumber the levels.)

(b) What is the symmetry of the total molecular state of NO?(c) What is the bond order of NO?(d) Assume the stretching vibration frequency is proportional to the bond

order. Calculate the frequencies in cm−1 for NO and O2, given that thefrequency for N2 is 2,331 cm−1. (The experimentally observed values are1,876 cm−1 and 1,555 cm−1.)

7.2 Find the O2 determinantal state for the 3�−g term with S = 1 and Ms = 0.7.3 Show that the determinantal state, 1�+g , for O2 is

1√2

[〈ψa(px ↑)ψa(px ↓)| + 〈ψa(py ↑)ψa(py ↓)|].

7.4 If � = (�r +�s), show that

det(�, φ1, φ2, φ3, . . . , φN ) = det(�r , φ1, φ2, φ3, . . . , φN )

+ det(�s, φ1, φ2, φ3, . . . , φN ).

7.5 Use the results of Exercise 7.3 to derive Eq. (7.24). Start from Eq. (7.23).

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210 Electronic states of diatomic molecules

7.6 For the molecule C2, answer the following questions.(a) What is the (LO1) molecular configuration and the symmetry?(b) What is the configuration and symmetry of the lowest-lying excited state?(c) What are the possible values of S?(d) Is an electronic dipole transition from the ground state to the lowest

excited state allowed or forbidden?7.7 The configuration (πu)

3 is equivalent to a (πu) hole state. A (πu)3 determinan-

tal state can be formed in two ways. The first is �1 = 〈ψb(py) ↑ ψb(py) ↓ψb(px) ↑ | and the second is �2 = 〈ψa(px) ↑ ψa(px) ↓ ψa(py) ↑ |.Show that �1 and �2 are basis functions for the IR πu , and therefore (πu)

3

transforms according to πu IR.7.8 (a) Draw a schematic diagram of the molecular levels for the (OH)+

molecule assuming that the H(1s) state is nearest in energy to the O(2p)state.

(b) What is the molecular electronic configuration? Indicate the occupiedmolecular orbitals. (Assume Hund’s rule for spin is obeyed.)

(c) What are the molecular terms that can be formed from the configuration?(d) Which state would be the ground state according to Hund’s rules?

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8

Transition-metal complexes

In Chapter 3 we discussed crystal-field theory in which the splitting of thed-electron levels was attributed to the electrostatic interactions due to the surround-ing ligands. In this chapter we extend the theory to include the molecular orbitalsrepresenting the ligands.

8.1 An octahedral complex

We consider a complex consisting of a transition metal ion, M , surrounded by anoctahedron of ligands (L1 through L6) as illustrated in Fig. 8.1. The covering groupis Oh and its character table is given in Table 8.1.

As a starting point we assume that each ligand has s- and p-electrons outside aclosed shell. The metal ion is assumed to have s-, p-, and d-electrons outside itsclosed shell.

No operation of the group can transform a ligand orbital into a metal orbital(or vice versa), and no operation can transform an s-orbital into a p- or d-orbital.Therefore we can express the representation for the complex as a sum of represen-tations,

�C = �s(M)+ � p(M)+ �d(M)+ �s(L)+ � p(L), (8.1)

where �C is the representation for the total complex, �α(M) is the α-orbital rep-resentation for the metal ion, and �β(L) is the β-orbital representation for theligands.

In Chapter 3 we discussed the Oh group and its operations. We found that thed-orbitals dxy , dxz , and dyz were basis functions for the t2g IR, and that dz2 and dx2

were basis functions for the eg IR of the Oh group. Therefore we have

�d(M) = t2g + eg. (8.2)

211

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212 Transition-metal complexes

Table 8.1 The character table for O. Oh = O × i

O(432) E 8C3 3C2 6C2 6C4

A1 1 1 1 1 1A2 1 1 1 −1 −1

(x2 − y2, 3z2 − r2) E 2 −1 2 0 0(Rx , Ry , Rz) (x , y, z) T1 3 0 −1 −1 1(xy, yz, xz) T2 3 0 −1 1 −1

x

y

z

ML4

L2

L1L3

L6

L5

Figure 8.1 A transition metal ion surrounded by an octahedron of ligands. Theligands are numbered L1 through L6. In terms of distance of separation, M andLi are nearest neighbors. Ligands adjacent to one another are second-nearestneighbors. Ligands on opposite sides of the metal ion are third-nearest neighbors.

Without specifying the ligand orbitals, we can already draw a conclusion aboutthe nature of the molecular states. The d-states are all “g” states; hence only ligand“g” states can interact with them to form metal–ligand bonds. The “u” ligand stateswill not interact with the d-states. If only d-electrons are considered for the metalion, the “u” ligand states will be non-bonding. Now suppose we add an s-orbitalto the collection of metal-ion orbitals. We know that �s(M) belongs to the a1g IRof Oh and these states interact only with “g” ligand states. However, when we addp-electrons to the metal orbitals the situation changes. Since � p(M) belongs tot1u , the p-orbitals will interact only with the “u” ligand states. In summary, the s-and d-orbitals of the metal ion form only “g” bonds and the p-orbitals form only“u” bonds.

Whatever types of orbitals are used for the ligands’ states, we know that theymust be basis functions for a representation of the group and the representation

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8.1 An octahedral complex 213

must decompose into the IRs of the Oh . For a bond to be formed with the metal ion,the decomposition of the ligand representation must contain eg and/or t2g and/ort1u . All other IRs in the decomposition are non-bonding with respect to the metal–ligand interactions.1

Representations and symmetry functions

The characters for the representations based on various hydrogen-like atomicorbitals can be deduced from the action table, Table 8.2.

Under the inversion operation the ligand positions change as follows:

Pi (L1)⇒ L3, (8.3a)

Pi (L2)⇒ L4, (8.3b)

Pi (L3)⇒ L1, (8.3c)

Pi (L4)⇒ L2, (8.3d)

Pi (L5)⇒ L6, (8.3e)

Pi (L6)⇒ L5. (8.3f)

Equations (8.3) together with the action table Table 8.2 can be employed to find thecharacters for the inversion operations. The characters for representations based onthe various orbitals are given in Table 8.3.

The decompositions of the representations are

�s(M) = a1g, (8.4a)

� p(M) = t1u, (8.4b)

�d(M) = eg + t2g, (8.4c)

�s(L) = a1g + eg + t1u, (8.4d)

� p(L) = a1g + eg + t1g + t2g + 2t1u + t2u, (8.4e)

�C = 3a1g + 3eg + t1g + 2t2g + 4t1u + t2u. (8.4f)

From Eq. (8.4f) we see that t1g and t2u occur only once in �C . These representa-tions correspond to non-bonding combinations of p-orbitals. The secular equation,det{H− S ε} = 0, for the molecular states will block-diagonalize as follows.

a1g: One 3× 3 block involving the s(M), s(L), and p(L) orbitals.eg: Two 3× 3 blocks involving d(M), s(L), and p(L) orbitals.t1g: Three 1 × 1 blocks involving only p(L). These states are non-bonding and

the symmetry functions are eigenstates.

1 The ligand orbitals that do interact with the metal orbitals can in some cases also produce non-bonding statesas well as bonding and antibonding states.

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Table 8.2 The action table for the O group. Functions in bold face go into a multiple of themselves under an operation of thegroup. A complete discussion of the operations of the group and the irreducible representations is given in Appendix E.

E C13 C1

32

C23 C2

32

C33 C3

32

C43 C4

32

C12 C2

2 C32 C1

2 C22 C3

2 C42 C5

2 C62 C1

4 C14

3C2

4 C24

3C3

4 C34

3

X −Y −Z −Z Y −Y Z Z Y X −X −X −X −X −Z Z −Y Y X X Z −Z −Y Y

Y Z −X X −Z −Z −X X Z −Y Y −Y Z −Z −Y −Y −X X −Z Z Y Y X −X

Z −X Y −Y −X X −Y Y X −Z −Z Z Y −Y −X X −Z −Z Y −Y −X X Z Z

1 6 2 4 5 5 2 4 6 3 1 3 6 5 3 3 2 4 5 6 1 1 4 2

2 1 6 6 3 1 5 5 3 2 4 4 4 4 6 5 1 3 2 2 5 6 1 3

3 5 4 2 6 6 4 2 5 1 3 1 5 6 1 1 4 2 6 5 3 3 2 4

4 3 5 5 1 3 6 6 1 4 2 2 2 2 5 6 3 1 4 4 6 5 3 1

5 4 3 1 4 2 1 3 2 6 6 5 3 1 4 2 6 6 3 1 4 2 5 5

6 2 1 3 2 4 3 1 4 5 5 6 1 3 2 4 5 5 1 3 2 4 6 6

XY −YZ XZ −XZ −YZ YZ −XZ XZ YZ −XY −XY XY −XZ XZ YZ −YZ XY XY −XZ XZ YZ −YZ −XY −XY

XZ XY −YZ YZ −XY −XY −YZ YZ XY −XZ XZ −XZ −XY XY XZ XZ YZ −YZ XY −XY −XZ −XZ −YZ YZ

YZ −XZ −XY −XY XZ −XZ XY XY XZ YZ −YZ −YZ YZ YZ XY −XY XZ −XZ −YZ −YZ −XY XY XZ −XZ

3Z

2−r

2

2√3=

bX

2−Y

22=

a

−1 2

b+√ 3 2

a−

1 2a−√ 3 2

b

−1 2

b−√ 3 2

a−

1 2a+√ 3 2

b

−1 2

b−√ 3 2

a−

1 2a+√ 3 2

b

−1 2

b+√ 3 2

a−

1 2a−√ 3 2

b

−1 2

b+√ 3 2

a−

1 2a−√ 3 2

b

−1 2

b−√ 3 2

a−

1 2a+√ 3 2

b

−1 2

b−√ 3 2

a−

1 2a+√ 3 2

b

−1 2

b+√ 3 2

a−

1 2a−√ 3 2

b

ba

ba

ba

−1 2

b−√ 3 2

a1 2

a−√ 3 2

b

−1 2

b−√ 3 2

a1 2

a−√ 3 2

b

−1 2

b+√ 3 2

a1 2

a+√ 3 2

b

−1 2

b+√ 3 2

a1 2

a+√ 3 2

b

b−a

b−a

−1 2

b−√ 3 2

a1 2

a−√ 3 2

b

−1 2

b−√ 3 2

a1 2

a−√ 3 2

b

−1 2

b+√ 3 2

a1 2

a+√ 3 2

b

−1 2

b+√ 3 2

a1 2

a+√ 3 2

b

b−a

b−a

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8.1 An octahedral complex 215

Table 8.3 The character table for the different types of orbital representations

Orbital E 8C3 3C2 6C2 6C4 i 8iC3 3iC2 6C2 6C4

�s(M) 1 1 1 1 1 1 1 1 1 1� p(M) 3 −1 1 1 −1 5 −1 1 1 −1�d(M) 5 0 −1 −1 1 −3 0 1 1 −1�s(L) 6 0 2 0 2 0 0 4 2 0� p(L) 18 0 −2 0 2 0 0 4 2 0�C 33 0 1 1 5 3 0 11 7 −1

t2g: Three 2× 2 blocks involving d(M) and p(L) orbitals.t1u: Three 4× 4 blocks involving p(M), s(L), and p(L) orbitals.t2u: Three 1 × 1 blocks involving only p(L). These states are non-bonding and

the symmetry functions are eigenstates.

The symmetry functions for the representations can be found using thesymmetry-function-generating machine,

f αi ∝

∑R

Dαi i (R)∗ PR f, (8.5)

where f αi is a symmetry function that transforms according to the i th row of the

�α IR, Dαi i (R) is the diagonal element of the IR matrix for the R operation of the

group, and f is an arbitrary linear combination of orbitals. The matrix elements forthe different IRs of Oh are given in Appendix E, in Section E.4. The results for thesymmetry functions are listed in Table 8.4.

8.1.1 Energy levels of an octahedral complex

The eigenvectors and eigenvalues of an octahedral complex can be calculated interms of the LCAO overlap and interaction integrals. It is instructive to begin withthe model for which the metal s-, p-, and d-orbitals interact only with the ligandp-orbitals. In many cases the ligand s-orbitals are far removed in energy from themetal-ion orbitals. For example, for F the 2s ionization energy is 46.4 eV andthe 2p ionization energy is 18.7 eV, while the ionization energy for a d-electron(3dn−14s ⇒ 3dn−24s) is 5.6 and 7.2 eV for Ti and Cr, respectively. Therefore,omission of the 2s ligand orbitals is not a severe approximation.

The ligand p-orbitals divide into two classes. The first class consists of p-orbitals that can form a sigma interaction with the metal ion. These “p-sigma”orbitals are aligned along the x-, y-, and z-axes. They are labeled y1, x2, y3, x4, z5,and z6, where xi , yi , and zi represent the p-orbitals, px , py , and pz , respectively.

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216 Transition-metal complexes

Table 8.4 Symmetry functions for the octahedral complex. The first column liststhe IR of Oh. The second column lists the row of the IR for multidimensional IRs.The third column lists the orbital type: (L) indicates ligand and (M) indicatescentral metal ion. The fourth column lists the symmetry functions for each row ofthe IR. Notation: si and xi , yi , and zi (i = 1, 2, . . . , 6) indicate the ligand s- andp-orbitals, respectively. The last column lists the type of interaction the metalorbitals have with the ligand orbitals of the same symmetry type, where “NB”indicates non-bonding with the metal orbitals.

IR Row Orbital type Symmetry function Type

a1g s(M) s(M)

s(L) (s1 + s2 + s3 + s4 + s5 + s6)/√

6 Sigmap(L) (y1 − x2 − y3 + x4 − z5 + z6)/

√6 Sigma

eg 1 d(M) dx2(M)

2 dz2(M)

1 s(L) (s1 − s2 + s3 − s4)/2 Sigma2 (s1 + s2 + s3 + s4 − 2s5 − 2s6)/

√12 Sigma

1 p(L) (y1 + x2 − y3 − x4)/2 Sigma2 (y1 − x2 − y3 + x4 + 2z5 − 2z6)/

√12 Sigma

t1g 1 p(L) (z1 − z3 + y5 − y6)/2 NB pi2 (x5 − x6 + z4 − z2)/2 NB pi3 (x1 − x3 + y2 − y4)/2 NB pi

t2g 1 d(M) dxy(M)

2 dxz(M)

3 dyz(M)

1 p(L) (x1 − y2 − x3 + y4)/2 Pi2 (z2 − z4 − x6 + x5)/2 Pi3 (z1 − z3 + y6 − y5)/2 Pi

t1u 1 p(M) x(M)

2 y(M)

3 z(M)

1 s(L) (s2 − s4)/√

2 Sigma2 (s1 − s3)/

√2 Sigma

3 (s5 − s6)/√

2 Sigma1 p(L) (x1 + x3 + x5 + x6)/2 Pi2 (y2 + y4 + y5 + y6)/2 Pi3 (z1 + z2 + z3 + z4)/2 Pi1 p(L) (x2 + x4)/

√2 Sigma

2 (y1 + y3)/√

2 Sigma3 (z5 + z6)/

√2 Sigma

t2u 1 p(L) (z1 + z3 − z2 − z4)/2 NB pi2 (y2 + y4 − y5 − y6)/2 NB pi3 (x1 + x3 − x5 − x6)/2 NB pi

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8.1 An octahedral complex 217

The second class of ligand p-orbitals consists of those that can form pi bonds withthe metal orbitals, the “p-pi” orbitals. The p-pi orbitals are oriented perpendicularto the x-, y-, and z-axes. They include x1, z1, y2, z2, x3, z3, y4, z4, x5, y5, x6, and y6.The class of p-orbitals involved in an eigenstate is indicated in the last column inTable 8.4. All of the eigenstates involving ligand–d-orbital interactions are eitherpi or sigma. The t1u states are different. They involve ligand orbitals and metalp-orbitals, and have both sigma and pi ligand–metal interactions.

In many chemistry texts s–p hybrid orbitals are used for the ligand sigmaorbitals. The use of hybrid orbitals is not necessary if both the s and the p ligandorbitals are included in the set of basis functions. The formation of s–p hybrid-likebonds and anti-bonds occurs automatically. The use of the s–p hybrid orbitals hasno effect on the group-theoretical analysis. The hybrid orbitals transform under thegroup operations exactly as the p-sigma orbitals.

The orbitals for the eg and t2g symmetries are given in Table 8.4. The secularequation for the t2g states consists of three symmetry-equivalent 2 × 2 matrices.For simplicity we shall assume the orbitals are orthogonal Löwdin functions sothat we do not have to deal with overlap integrals. The LCAO interactions for s-,p-, and d-orbitals are given in Table 8.5 [8.1].

The secular equation for the interaction of the two row-1 t2g symmetry functionsis then ∣∣∣∣ εd − ε −2(pdπ)1

−2(pdπ)1 ε′p − ε

∣∣∣∣ = 0, (8.6)

where

εd =∫

d∗xy H dxy dr,

ε′p =∫

1

2(px1 − py2 − px3 + py4)

∗H1

2(px1 − py2 − px3 + py4)dr

= εp + (pL pLπ)2 − (pL pLσ)2 − (pL pLπ)3,

εp =∫

p∗α j H pα j dr (α = x, y, or z, and i = 1, 2, . . . , 6),

and ∫d∗xy H

1

2(px1 − py2 − px3 + py4)dr = −2(pdπ).

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218 Transition-metal complexes

Table 8.5 LCAO two-center integrals [8.1] for orbitals centered at �Ri and �R j .The variables , m, and n are the direction cosines of the vector ( �R j − �Ri ).

Es,s (ssσ)

Es,x (spσ)

Ex,x 2(ppσ)+ (1− 2)(ppπ)

Ex,y m(ppσ)− m(ppπ)

Ex,z n(ppσ)− n(ppπ)

Es,xy√

3 m(sdσ)

Es,x2−y212

√3( 2 − m2)(sdσ)

Es,3z2−r2 [n2 − 12 (

2 + m2)](sdσ)

Ex,xy√

3 2m(pdσ)+ m(1− 2 2)(pdπ)

Ex,yz√

3 mn(pdσ)− 2 mn(pdπ)

Ex,zx√

3 2n(pdσ)+ n(1− 2 2)(pdπ)

Ex,x2−y212

√3 ( 2 − m2)(pdσ)+ (1− 2 + m2)(pdπ)

Ey,x2−y212

√3m( 2 − m2)(pdσ)− m(1+ 2 − m2)(pdπ)

Ez,x2−y212

√3n( 2 − m2)(pdσ)− n( 2 − m2)(pdπ)

Ex,3z2−r2 [n2 − 12 (

2 + m2)](pdσ)−√3 n2(pdπ)

Ey,3z2−r2 m[n2 − 12 (

2 + m2)](pdσ)−√3mn2(pdπ)

Ez,3z2−r2 n[n2 − 12 (

2 + m2)](pdσ)+√3n( 2 + m2)(pdπ)

Exy,xy 3 2m2(ddσ)+ ( 2 + m2 − 4 2m2)(ddπ)+ (n2 + 2m2)(ddδ)

Exy,yz 3 m2n(ddσ)+ n(1− 4m2)(ddπ)+ n(m2 − 1)(ddδ)

Exy,xz 3 2mn(ddσ)+ mn(1− 4 2)(ddπ)+ mn( 2 − 1)(ddδ)

Exy,x2−y232 m( 2 − m2)(ddσ)+ 2 m(m2 − 2)(ddπ)+ 1

2 m( 2 − m2)(ddδ)

Eyz,x2−y232 mn( 2 − m2)(ddσ)− mn[1+ 2( 2 − m2)](ddπ)

+mn[1+ 12 (

2 − m2)](ddδ)

Ezx,x2−y232 n ( 2 − m2)(ddσ)+ n [1− 2( 2 − m2)](ddπ)

− n [1− 12 (

2 − m2)](ddδ)

Exy,3z2−r2

√3 m[n2 − 1

2 ( 2 + m2)](ddσ)− 2

√3 mn2(ddπ)

+ 12

√3 m(1+ n2)(ddδ)

Eyz,3z2−r2

√3mn[n2 − 1

2 ( 2 + m2)](ddσ)+√3mn( 2 + m2 − n2)(ddπ)

− 12

√3mn( 2 + m2)(ddδ)

Ezx,3z2−r2

√3 n[n2 − 1

2 ( 2 + m2)](ddσ)− 2

√3 n( 2 + m2 − n2)(ddπ)

+ 12

√3 n( 2 + m2)(ddδ)

Ex2−y2,x2−y234 (

2 − m2)2(ddσ)+ [ 2 + m2 − ( 2 − m2)2](ddπ)

+[n2 + 14 (

2 − m2)2](ddδ)

Ex2−y2,3z2−r212

√3( 2 − m2)[n2 − 1

2 ( 2 + m2)](ddσ)+ √3n2(m2 − 2)(ddπ)

+ 14

√3(1+ n2)( 2 − m2)(ddδ)

E3z2−r2,3z2−r2 [n2 − 12 (

2 + m2)](ddσ)+ 3n2( 2 + m2)(ddπ)+ 34 (

2 + m2)2(ddδ)

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8.1 An octahedral complex 219

The subscripts 2 and 3 on the (pLπpLπ) parameters indicate interactions betweenorbitals that are second-nearest and third-nearest neighbors as indicated in thecaption of Fig. 8.1. The eigenvalues are

ε±(t2g) = 1

2(εd + ε′p)±

√[1

2(εd − ε′p)

]2

+ 4(pdπ)2. (8.7)

The plus sign in (8.7) corresponds to an antibonding level, 2t2g, and the minus signgives the 1t2g bonding level. The other two rows of the t2g symmetry functionsyield the same result, therefore the eigenvalues are three-fold degenerate. For theeg states we have that the secular equation (ignoring the ligand s-orbitals) is∣∣∣∣ εd − ε −√3(pdσ)1

−√3(pdσ)1 ε′′p − ε

∣∣∣∣ = 0. (8.8)

The eigenvalues are

ε±(eg) = 1

2(εd + ε′′p)±

√[1

2(εd − ε′′p)

]2

+ 3(pdσ)2, (8.9)

ε′′p = εp + (pL pLσ)2 − (pL pLπ)2 − (pL pLσ)3. (8.10)

The plus sign gives the energy of the antibonding state, 2eg, and the minus signgives the energy of the 1eg bonding state. The row-2 eg-symmetry functions givethe same result, so the eg states are each two-fold degenerate. The splitting of thed-orbital antibonding levels is

� = ε+(eg)− ε+(t2g)

= (pL pLπ)2 + (pL pLσ)2 − 1

2(pL pLσ)3 − (pL pLπ)3

+√[

1

2(εd − ε′′p)

]2

+ 3(pLdMσ)21

−√[

1

2(εd − ε′p)

]2

+ 4(pLdMπ)21, (8.11)

where the subscripts L and M indicate ligand and metal, respectively.The energy difference � is the molecular-orbital equivalent of “10Dq” in

crystal-field theory (Chapter 3). Here, however, the splitting derives from cova-lent mixing of the d-ion orbitals with the ligand orbitals, whereas in crystal-fieldtheory the splitting is due to the electrostatic field of the ligands. The covalent mix-ing of the eg states is larger than that of t2g states, so � is positive. With this typeof simple model the interaction parameters are usually determined by fitting the

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220 Transition-metal complexes

results to experimental data or more accurate numerical calculations. The electric-dipole transition t2g ⇒ eg is forbidden unless accompanied by “u” excitation (e.g.,a molecular vibration transition). A weak absorption band observed for [TiF6]3−indicates that � is about 17,000 cm−1 (2.11 eV).

Usually |(pdσ)| is in the range of 2–5 eV in magnitude. It is two to five timeslarger than |(pdπ)|. The pL−pL interactions |(pL pLσ)2| and |(pL pLπ)2| areroughly a factor of 10 smaller than |(pdσ)|. The two quantities |(pL pLσ)3| and|(pL pLπ)3| are often small enough to neglect.

The empirical LCAO model can not be used to carry out accurate calculationsof the energies or the wavefunctions. However, the symmetries of the molecularstates derived from the LCAO model are correct. The principal value of the modelis conceptual. It provides a scheme for understanding the physics, chemistry, andoptical properties of transition metal complexes.

The energies of the secular equations for the various molecular levels aresummarized in Table 8.6.

Figure 8.2 is a schematic representation of the molecular levels resulting fromM−L and L−L interactions and overlaps.

As examples, we consider the electronic structures of two complexes, [TiF6]3−and [CrF6]3−. The electronic configurations of the atoms are Ti = [Ar]3d24s2,Cr = [Ar]3d54s, and F = [He]2s22p5. The number of electrons outside the closed[Ar] shell is four for Ti, and for F the number of electrons outside the [He]2s2

closed shells is five. The total number of electrons to be assigned to the molecularorbitals for the [TiF6]3− complex is therefore 4 + 6(5) + 3 = 37, where the lastnumber of 3 is due to the charge on the complex. For [CrF6]3− the number ofelectrons to be assigned is 6+ 6(5)+ 3 = 39.

If we fill the lowest energy levels in Fig. 8.2, then the configuration for[TiF6]3− is

(1a1g)2(1eg)

4(1t2g)6(1t1u)

6(2t1u)6(1t2g)

6(1t1g)6(2t2g)

1. (8.12)

Since there is a single 2t2g electron outside of closed molecular shells, the totalmolecular state is T2g; and since there is a single unpaired spin, the state is a spindoublet, 2T2g. For [CrF6]3− we must assign two more electrons to the 2t2g level, sothe molecular configuration is

(1a1g)2(1eg)

4(1t2g)6(1t1u)

6(2t1u)6(1t2g)

6(1t1g)6(2t2g)

3. (8.13)

The three electrons in the 2t2g molecular orbitals produce several terms. InTable 8.4 there are two t2g symmetry functions for each of the three rows, giv-ing a total of six symmetry functions. The eigenstates will be linear combinationsof the functions from the same rows. Three of the six eigenstates correspond to the1t2g bonding states, while the other three are the 2t2g antibonding states. We have

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8.1 An octahedral complex 221

Table 8.6 LCAO molecular energy levels for an octahedral complex usingorthogonal Löwdin orbitals. Included are s, p, and d metal orbitals and sixligand p-orbitals. Here (pLdMσ)1 = (pdσ) and (pLdMπ)1 = (pdπ), withsubscripts L for ligand and M for metal. “NB” denotes non-bonding. The lastcolumn gives the degeneracy (D) of the molecular levels

IR Energy Type D

a1g ε±(a1g) = 12 (εs(M)− ε′p)±

√[ 12 (εs(M)− ε′p)

]2 + 6(sM pLσ)21 Sigma 1

ε′p = εp + 2[(pL pLπ)2 − (pL pLσ)2] − (pL pLσ)3

eg ε±(eg) = 12 (εd + ε′′p)±

√[ 12 (εd − ε′′p)

]2 + 3(pdσ)2 Sigma 2

ε′′p = εp + (pL pLσ)2 − (pL pLπ)2 + (pL pLσ)3

t1g ε(t1g) = εp + (pL pLσ)2 − (pL pLπ)2 − (pL pLσ)3 NB 3

t2g ε±(t2g) = 12 (εd + ε′p)±

√[ 12 (εd − ε′p)

]2 + 4(pdπ)2 Pi 3

ε′p = εp + (pL pLπ)2 − (pL pLσ)2 − (pL pLπ)3

t1u

∣∣∣∣∣∣∣∣εd − ε

√2(pM pLπ)1

√2(pM pLσ)1√

2(pM pLπ)1 ε′p − ε [(σ )2 + (π)2]/√

2√2(pM pLσ)1 [(σ )2 + (π)2]/

√2 ε′′p − ε

∣∣∣∣∣∣∣∣= 0

ε′p = εp + 2(pL pLπ)2 + (pL pLπ)3 Pi

ε′′p = εp + (pL pLσ)3 and 3

[(σ )2 + (π)2] = (pM pLσ)2 + (pM pLπ)2 sigma

t2u ε(t2u) = εp − 2(pL pLπ)2 + (pL pLπ)3 NB 3

three electrons outside of the closed molecular shells to be distributed among thethree 2t2g states.

Let us label the three 2t2g molecular eigenstates as “a”, “b”, and “c”, where “a”is the row-1 eigenstate, “b” is the row-2 eigenstate, and “c” is the row-3 eigenstate.There are 10 independent spatial determinantal states that can be formed with threeelectrons distributed among the three eigenstates. They are

(group I) 〈abc|, (8.14)

(group II) 〈aaa|, 〈bbb|, 〈ccc|, (8.15)

(group III) 〈aab|, 〈aac|, 〈bba|, 〈bbc|, 〈cca|, 〈ccb|. (8.16)

(Permutations of a, b, and c do not produce a new state; they simply produce thesame or the negative of the same determinant.) The group II states are forbid-den by the Pauli exclusion principle. Therefore we are left with seven independent

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222 Transition-metal complexes

Δ

4p

4s

3d

3t1u

2a1g

2eg

2t2g

1t1g

1t2u

2t1u

1t1u

1t2g

1eg

1a1g

2p

Metalorbitals

Molecularorbitals

Ligandp-orbitals

2p (sigma)1t1g

1t2u

2t1u

1t1u

1t2g

1eg

1a1g

2p (pi)

Molecularorbitals

Figure 8.2 Molecular-orbital energy levels. Typical ordering of molecular lev-els for a transition metal ion surrounded by an octahedron of ligands such as[CrF6]3− or [TiF6]3−. The energy difference between the 2t2g and 2eg levels(labeled �) is the molecular-orbital equivalent of 10Dq in crystal-field theory.The inset shows how the ligand p-sigma and p-pi orbitals are correlated with themolecular bonding levels. Note that the t1u levels include both the sigma and thepi ligand orbitals.

determinantal states. Our task is to find the linear combinations of these states thatare basis functions for the IRs of the Oh group.

The molecular state “a” must transform under the group operations exactly asthe function dxy . Similarly, “b” must transform as dxz and “c” as dyz .

The group I state is the only determinant that allows all of the spins to be parallel.The lowest energy state, according to Hund’s rules, is the determinantal state, 〈a ↑b ↑ c ↑| = 〈abc|α1α2α3. This state has S = 3/2 (also Sz = 3/2), and is therefore aspin quartet. Using the action table, we can determine the characters of the productabc. Since abc must transform as (xy)(xz)(yz), we have

P(E) abc ⇒ P(E) (xy)(xz)(yz) = (xy)(xz)(yz)⇒ abc, (8.17)

P(C13) abc⇒ P(C1

3) (xy)(xz)(yz) = (−yz)(xy)(−xz)⇒ abc, (8.18)

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8.1 An octahedral complex 223

P(C12) abc⇒ P(C1

2) (xy)(xz)(yz) = (−xy)(−xz)(yz)⇒ abc, (8.19)

P(C12) abc⇒ P(C1

2) (xy)(xz)(yz) = (−xz)(−xy)(yz)⇒ abc, (8.20)

P(C14) abc⇒ P(C1

4) (xy)(xz)(yz) = (−xz)(xy)(−yz)⇒ abc, (8.21)

and χ(E) = χ(C3) = χ(C2) = χ(C2) = χ(C4) = 1.The arrows in Eqs. (8.17)–(8.21) indicate symmetry-equivalent functions. From

the characters we conclude that abc belongs to the a1g representation. From theabove information we see that the ground state of [CrF6]3− is, according to Hund’srules, 4 A1g. The other spin states, Sz = 1/2, −1/2, and −3/2, can be obtained inthe usual way by operating with the spin-lowering operator. For example,

S−〈abc|α1α2α3

= 〈abc| 1√3{β1α2α3 + β2α1α3 + β3α1α2}

(Sz = 1

2

), (8.22)

S−〈abc|{β1α2α3 + β2α1α3 + β3α1α2}= 〈abc| 1√

3{α1β2β3 + α2β1β3 + α3β1β3}

(Sz = −1

2

), (8.23)

S−〈abc| 1√3{α1β2β3 + α2β1β3 + α3β1β3} = 〈abc|{β1β2β3}

(Sz = −3

2

).

(8.24)

The molecular-orbital states of the other IRs (terms) can be obtained using thesymmetry-function-generating machine. These calculations are simple, but can betedious.

Let us consider the functions from group III. The six symmetry-equivalent func-tions 〈aab|, 〈ccb|, 〈bba|, 〈bbc|, 〈cca|, and 〈ccb| are all “g” functions. Under theoperations of the group, aab must transform as (xy)(xy)(xz), ccb must transformas (yz)(yz)(xz), and so forth. From the action table we find, for example,

P(C13) |aab〉 ⇒ P(C1

3) (xy)(xy)(xz) = (−yz)(−yz)(xy)⇒ 〈cca|,where the arrows indicate symmetry-equivalent functions. Table 8.7 summarizesthe results for all of the operations of the group acting on 〈aab|.

If we apply the symmetry-function-generating machine, (8.5), to the results inTable 8.7, using the matrix elements for T1 in Appendix E, we find that

f (t1)1 ∝

∑R

Dt111(R)∗ PR〈aab| = 0, (8.25)

f (t1)2 ∝

∑R

Dt122(R)∗ PR〈aab| = 4〈aab| − 4〈ccb|, (8.26)

f (t1)3 ∝

∑R

Dt133(R)∗ PR〈aab| = 0. (8.27)

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224 Transition-metal complexes

Table 8.7 Results of operating on 〈aab| with the operations of the Oh group

E C (1)3 C (1)2

3 C (2)3 C (2)2

3 C (3)3 C (3)2

3 C (4)3〈aab| 〈cca| −〈bbc| 〈bbc| −〈cca| −〈cca| −〈bbc| 〈bbc|

C(4)2

3 C (1)2 C (2)

2 C (3)2 C (1)

2 C (2)2 C (3)

2 C (4)2〈cca| −〈abb| 〈abb| −〈aab| −〈bba| 〈bba| 〈ccb| 〈ccb|

C (5)2 C (6)

2 C (1)4 C (1)3

4 C (2)4 C (2)3

4 C (3)4 C (3)3

4〈aac| −〈aac| 〈bba| −〈bba| −〈ccb| −〈ccb| −〈aac| 〈aac|

Thus, we can conclude that the function [〈aab| − 〈ccb|] transforms accordingto the second row of the t1g IR of Oh . Similar calculations show that the three t1g

(normalized) functions are

t1g(row 1) = 1√2

[〈aac| − 〈bbc|

], (8.28)

t1g(row 2) = 1√2

[〈aab| − 〈ccb|

], (8.29)

t1g(row 3) = 1√2

[〈bba| − 〈cca|

]. (8.30)

Since each of the functions has one doubly occupied orbital, it follows that S =1/2. Thus we see that the molecular state is 2T1g. Including spin, the determinantalstates are

t1g(row 1) = 1√2

[〈(aα)(aβ)(cα)| − 〈(bα)(bβ)(cα)|

], (8.31)

t1g(row 2) = 1√2

[〈(aα)(aβ)(bα)| − 〈(cα)(cβ)(bα)|

], (8.32)

t1g(row 3) = 1√2

[〈(bα)(bβ)(aα)| − 〈(cα)(cβ)(aα)|

]. (8.33)

A similar calculation using the diagonal matrix elements of the T2g IR gives thedeterminantal states

t2g(row 1) = 1√2

[〈(bα)(bβ)(aα)| + 〈(cα)(cβ)(aα)|

], (8.34)

t2g(row 2) = 1√2

[〈(aα)(aβ)(bα)| + 〈(cα)(cβ)(bα)|

], (8.35)

t2g(row 3) = 1√2

[〈(aα)(aβ)(cα)| + 〈(bα)(bβ)(cα)|

]. (8.36)

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8.1 An octahedral complex 225

(2t2g)3

2T1g

2T2g

4A1g

Figure 8.3 Molecular terms for three electrons in 2t2g orbitals. The ground stateis 4 A1g according to Hund’s rules.

Figure 8.3 shows schematically the (total) molecular states arising from threeelectrons in the three 2t2g level. The ordering of the states is not determined bygroup theory.

There are many ways to form excited states. For example, we could assign twoelectrons to the 2t2g and one to the 2eg. States of this type, considering only themetal orbitals, were discussed in Chapter 3.

The molecular eigenstates are linear combinations of metal and ligand orbitals,and, for Löwdin orbitals, are of the form

� = 1√N

{∑i

ai (M)φi (M)+∑

j

b j (L)φ j (L)}, (8.37)

N =∑

i

a2i +

∑j

b2j , (8.38)

where ai and b j are the amplitudes for the metal and ligand orbitals, respectively,and√

N is the normalization factor. In a very approximate way, � distributes theelectron’s charge between the metal and ligand ions,

q(M) = e

N

∑i

a2i , (8.39)

q(L) = e

N

∑j

b2j , (8.40)

where q(M) is the charge residing in metal orbitals and q(L) is the charge residingon ligand orbitals. Generally, the bonding and non-bonding eigenstates have morecharge in the ligand orbitals than in the metal orbitals, q(L) > q(M). For theantibonding eigenstates the converse is true, q(M) > q(L).

Therefore, an electronic transition from a bonding or non-bonding eigenstate toan antibonding eigenstate has the effect of moving some charge from the ligands tothe metal ion. It is also possible for transitions to transfer charge from the metal ionto the ligands. These types of transitions are called “charge-transfer” transitions.

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226 Transition-metal complexes

An example of an L ⇒ M charge transfer is 1t1g ⇒ 2t2g+ “u vibration” (seeFig. 8.2). The 1t1g states are non-bonding, so for the initial state there is littlecharge residing in the metal orbitals. The final state is an antibonding state, andtherefore most of the charge resides in metal orbitals.

8.1.2 High- and low-spin states

For transition metal molecules or complexes the highest occupied levels are usuallypartially filled t2g and/or eg d-electron antibonding levels. The magnetic propertiesof a complex are dependent upon the electron occupation of these d-electron states.If the net spin is non-zero, the complex is paramagnetic, whereas if the net spin iszero it is diamagnetic.

In the simple model that neglects spin–orbit interactions there are two princi-pal energies to be considered. The first is the “pairing energy”, P . The repulsionbetween two electrons is less for parallel spins than for antiparallel spins due to theexchange energy. Also the repulsion between electrons occupying the same orbitalis greater than the repulsion between electrons occupying different orbitals. Thepairing energy is a measure of the extra energy required for electrons with pairedspins (↑↓) compared with the case of electrons with unpaired spins (↑↑) or (↓↓).The second important energy, �, is the splitting of the eg and t2g levels.

For complexes with three or fewer electrons in d-levels, Hund’s rules require thatthe electron be assigned to the t2 states with parallel spins. These configurations areindicated below:

For four d-electrons the lowest energy state depends upon whether � is greaterthan P or less than P . If P > �, it is energetically favorable to place the fourthelectron in the eg level with its spin up. However, if P < �, the fourth electronwill occupy one of the three t2g levels with spin down. These two cases are shownbelow:

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8.2 A tetrahedral complex 227

The “high-spin configuration” results when the pairing energy is greater thanthe eg − t2g splitting. When the number of d-electrons lies between four and seventhere will be a high- or low-spin configuration for P > � or P < �, respectively.For five d-electrons the high-spin configuration has S = 5/2, and the low-spinconfiguration has S = 1/2. For six d-electrons, S = 2 and S = 0; and for sevend-electrons, S = 3/2 and S = 1/2. For 8, 9, and 10 d-electrons the spin of thelowest energy state is the same for P > � and P < �, since the t2g levels are fullyoccupied by electron pairs.

The pairing energy is in the range 12,000–25,000 cm−1, and � is in the range6,000–30,000 cm−1. Therefore, these two energies are comparable, and which islarger depends upon the transition metal ion and the ligands of the complex. As anexample, Fe3+ has five d-electrons. The complex [Fe(NO2)]3− has � > P , andexists as a low-spin complex with S = 1/2. However, Fe3+ in [Fe(Br)]3− is in ahigh-spin configuration with S = 5/2.

Some examples of transition metal ions with high/low-spin states are givenbelow:

Configuration Ion S high spin S low spin

d4 Cr2+ 2 1d5 Fe3+ 5

212

d6 Fe2+ 2 0d7 Co2+ 3

212

d8 Ni2+ 1 —

8.2 A tetrahedral complex

The geometry and coordinate system for a tetrahedral molecular complex is shownin Fig. 8.4 for the molecular complex M L4 (inscribed in a cube), where M is themetal ion and L is a ligand. The metal ion is at the center surrounded by fourligands. The group of covering operations is Td . The tetrahedron does not have acenter of inversion, and therefore the “g” and “u” designations are inapplicable.

8.2.1 Elements of Td

The Td group and its IRs are discussed in detail in Appendix F. It consists of the 24operations, namely rotations and reflections, that leave a tetrahedron unchanged.For visual purposes it is useful to show the tetrahedron inscribed within a cubeas shown in Fig. 8.4. The symmetry elements in addition to the identity arelisted below.

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228 Transition-metal complexes

x4 y4

z4

x3y3

z3

x2y2

z2

x1 y1

z1

x0 y0

z0

(a) (b)

M

L3

L2

L1

L4

Figure 8.4 A tetrahedral molecular complex consisting of five atoms. (a) Thecoordinate system. (b) The tetrahedron for M L4, where M is the metal ion and Lrepresents a ligand. The ligands are labeled 1 through 4, and the M ion is at thecenter of the tetrahedron.

(1) C3. Four three-fold axes: 120- and 240-degree, clockwise rotations about thecube body diagonals.

(2) C2. Three two-fold axes: 180-degree, clockwise rotations about the x-, y-, andz-axes.

(3) σd . Six reflections. The reflection planes are defined by a pair of parallel facediagonals on opposite sides of the circumscribed cube.

(4) S4. Six improper rotations: Clockwise rotations by 90 degrees and 270 degreesabout one of the three C2 axes (x-, y-, and z-axes through the center atom),followed by a reflection in the plane perpendicular to the axis of rotation.

The symmetry axes and reflection planes are shown in Appendix F, in Fig. F.1.

8.2.2 Symmetry functions for the ML4 complex

We consider a complex, M L4, that has s-, p-, and d-orbitals on the metal (M) ion,and s- and p-orbitals on each of the four equivalent ligands (L4). We have in mindhere a transition metal ion with 3d, 4p, and 4s orbitals interacting with ligandsthat have 2s and 2p or 3s and 3p valence orbitals. To analyze the electronic statesof the M L4 complex, our first task is to determine the number and types of IRscontained in the orbital representation. To do this we make use of the charactertable, Table 8.8, and the action table, Table 8.9. The representation for the tetrahe-dral complex, �C , using the s(M), p(M), d(M), s(L), and p(L) orbitals, may bewritten as

�C = �d(M) + �s(M) + �s(L) + � p(M) + � p(L). (8.41)

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8.2 A tetrahedral complex 229

Table 8.8 The character table for the Td group. Basis functions are listed in orderof rows. Note the order for T2.

A1 1 1 1 1 1A2 1 1 1 −1 −1

(x2 − y2, 3z2 − r2) E 2 −1 2 0 0[x(z2 − y2), y(z2 − x2), z(x2 − y2)] T1 3 0 −1 −1 1(x , y, z); (yz, xz, xy) T2 3 0 −1 1 −1

s-orbitals on metal ion �s(M) 1 1 1 1 1s-orbitals on ligands �s(L) 4 1 0 2 0p-orbitals on metal ion � p(M) 3 0 −1 1 −1p-orbitals on ligands � p(L) 12 0 0 2 0d-orbitals on metal ion �d(M) 5 −1 1 1 −1all orbitals on molecule �C 25 1 1 7 −1

We can find the characters for each of the orbital representations on the right-handside of (8.41). The representations may then be decomposed into the IRs of Td

using the character table (Table 8.8). We find

�d(M) = e + t2, (8.42)

�s(M) = a1, (8.43)

�s(L) = a1 + t2, (8.44)

� p(M) = t2, (8.45)

� p(L) = a1 + e + t1 + 2t2, (8.46)

�C = 3a1 + 2e + t1 + 5t2. (8.47)

Analysis of (8.47) yields some useful information. The d(M) orbitals belongto the e and t2 IRs. Therefore we can conclude that there are in total five triplydegenerate t2 states (15 states) and two doubly degenerate e states (4 states) thatcould possibly form bonding and antibonding states with metal d-orbitals. Thethree states derived from the a1 IRs form bonding and antibonding states withthe metal s-orbital. The t1 states are non-bonding, and, since t1 occurs only oncein (8.47), the three t1 symmetry functions are also the eigenstates. The secularequation for the complex is a 25× 25 matrix that will block-diagonalize into

one 3× 3 matrix for the a1 states,two 2× 2 matrices for the e states,three 1× 1 matrices for the t1 states,three 5× 5 matrices for the t2 states.

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Table 8.9 The action table for an M L4 complex with Td symmetry. The first row gives the symmetry elements. The first columnlists the functions and ligand numbers. The other rows give the result of operating on the function with the symmetry operationat the top of the column.

E C13 C1

32

C23 C2

32

C33 C3

32

C43 C4

32

C12 C2

2 C32 σ1

d σ2d σ3

d σ4d σ5

d σ6d S1

4 S14

3S2

4 S24

3S3

4 S34

3

X Z Y −Z Y Z −Y −Z −Y X −X −X X X −Z Z Y −Y −X −X Z −Z −Y Y

Y X Z X −Z −X −Z −X Z −Y Y −Y Z −Z Y Y X −X −Z Z −Y −Y X −X

Z Y X −Y −X −Y X Y −X −Z −Z Z Y −Y −X X Z Z Y −Y −X X −Z −Z

1 1 1 4 3 2 4 3 2 4 3 2 1 4 3 1 1 2 3 2 2 4 4 3

2 3 4 2 2 4 1 1 3 3 4 1 3 2 2 4 2 1 1 4 3 1 3 4

3 4 2 1 4 3 3 2 1 2 1 4 2 3 1 3 4 3 4 1 4 2 1 2

4 2 3 3 1 1 2 4 4 1 2 3 4 1 4 2 3 4 2 3 1 3 2 1

XY X Z Y Z −X Z −Y Z −X Z Y Z X Z Y Z −XY −XY XY X Z −X Z −Y Z Y Z XY XY X Z −X Z −Y Z Y Z −XY −XY

X Z Y Z XY Y Z −XY −Y Z −XY Y Z XY −X Z X Z −X Z XY −XY X Z X Z Y Z −Y Z −XY XY −X Z −X Z Y Z −Y Z

Y Z XY X Z −XY X Z XY −X Z −XY −X Z Y Z −Y Z −Y Z Y Z Y Z −XY XY X Z −X Z −Y Z −Y Z XY −XY −X Z X Z

3Z

2−r

2

2√3=

bX

2−Y

22=

a

−1 2

b+√ 3 2

a−

1 2a−√ 3 2

b

−1 2

b−√ 3 2

a−

1 2a+√ 3 2

b

−1 2

b−√ 3 2

a−

1 2a+√ 3 2

b

−1 2

b+√ 3 2

a−

1 2a−√ 3 2

b

−1 2

b−√ 3 2

a−

1 2a+√ 3 2

b

−1 2

b+√ 3 2

a−

1 2a−√ 3 2

b

−1 2

b−√ 3 2

a−

1 2a+√ 3 2

b

−1 2

b+√ 3 2

a−

1 2a−√ 3 2

b

ba

ba

ba

−1 2

b−√ 3 2

a1 2

a−√ 3 2

b

−1 2

b−√ 3 2

a1 2

a−√ 3 2

b

−1 2

b+√ 3 2

a1 2

a+√ 3 2

b

−1 2

b+√ 3 2

a1 2

a+√ 3 2

b

b−a

b−a

−1 2

b−√ 3 2

a1 2

a−√ 3 2

b

−1 2

b−√ 3 2

a1 2

a−√ 3 2

b

−1 2

b+√ 3 2

a1 2

a+√ 3 2

b

−1 2

b+√ 3 2

a1 2

a+√ 3 2

b

b−a

b−a

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8.2 A tetrahedral complex 231

Table 8.10 Symmetry functions for the tetrahedral complex M L4. Here xi , yi ,and zi are px , py, and pz orbitals centered on the ith ligand site. Metal (M)orbitals are centered at the origin.

IR Row Orbital Symmetry function Bonding

a1 s(M) s(M) Sigmas(L) (s1 + s2 + s3 + s4)/2p(L) (x1 + y1 + z1 − x2 − y2 + z2 − x3 + y3

− z3 + x4 − y4 − z4)/√

12e 1 d(M) dx2(M)

2 dz2(M)

1 p(L) (x1 − y1 − x2 + y2 − x3 − y3 + x4 + y4)/√

8 Pi2 (x1 + y1 − 2z1 − x2 − y2 − 2z2 − x3 + y3

− 2z3 + x4 − y4 + 2z4)/√

24t1 1 p(L) (x1 − z1 + x2 + z2 − x3 + z3 − x4 − z4)/

√8 NB

2 (y1 − z1 + y2 + z2 − y3 − z3 − y4 + z4)/√

83 (x1 − y1 − x2 + y2 + x3 + y3 − x4 − y4)/

√8

t2 1 p(M) px (M)

2 py(M)

3 pz(M)

1 d(M) dyz(M)

2 dxz(M)

3 dxy(M)

1 s(L)1 (s1 − s2 − s3 + s4)/2 Sigma2 s(L)2 (s1 − s2 + s3 − s4)/23 s(L)3 (s1 + s2 − s3 − s4)/2

1 p(L)1a (x1 + x2 + x3 + x4)/2 Pi+ Sigma2 p(L)2a (y1 + y2 + y3 + y4)/23 p(L)3a (z1 + z2 + z3 + z4)/2

1 p(L)1b (y1 + z1 + y2 − z2 − y3 + z3 − y4 − z4)/√

8 Pi+ Sigma2 p(L)2b (x1 + z1 + x2 − z2 − x3 − z3 − x4 + z4)/

√8

3 p(L)3b (x1 + y1 − x2 − y2 + x3 − y3 − x4 + y4)/√

8

Equation (8.5) can be used to find the required symmetry functions. The results aregiven in Table 8.10.

8.2.3 t2 symmetry functions

The symmetry functions listed in Table 8.10 are not unique when a representation iscontained more than once in the decomposition. The symmetry function generatedby the generating machine depends upon the choice of the “arbitrary” function.

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232 Transition-metal complexes

Consider the a1 ligand functions generated by the arbitrarily chosen functionsx1(px1(L)) and s1(L), respectively. If we had chosen s1(L)± x1 as the generatingfunction, we would have obtained the symmetry functions f +(a1) and f −(a1),

f ±(a1) = (s1 + s2 + s3 + s4)

± (x1 + y1 + z1 − x2 − y2 + z2 − x3 + y3 − z3 + x4 − y4 − z4).

(8.48)

The combinations of the p-orbitals of f ±(a1) are aligned parallel or antiparallel tothe metal–ligand intra-atomic axes. When these are combined with the s-orbitals,“sp3” hybrid orbitals are obtained, as shown schematically in Fig. 8.5.

Functions oriented along the internuclear axes are often called sigma symmetryfunctions. For the tetrahedron these functions do have sigma overlap with the t2metal s-orbitals but not with the t2 metal p and d symmetry functions. The LCAOinteractions between the p (or sp3) functions and the p and d metal functionsof t2 involve both sigma and pi bonding. Regardless of what linear combinationsare taken for the symmetry functions, the eigenvectors and eigenvalues obtainedfrom the secular equations are the same. The hybrid orbitals are conceptually use-ful, but there is no particular mathematical advantage to choosing axes alignedalong and perpendicular to the internuclear axes. The formation of hybrid orbitalcombinations occurs automatically when the secular equation is diagonalized. Thatis, the eigenvectors will contain the appropriate linear combinations of symmetryfunctions.

8.2.4 Eigenvalues for a tetrahedral complex

The LCAO interactions for the tetrahedral complex are different from those of theoctahedral complex. For example, in the octahedral case the metal orbitals dxy ,dxz , and dyz form pi bonds, while the metal orbitals dx2 and dz2 form sigma bonds

(a) (b)

Figure 8.5 The sp3 hybrid orbitals. (a) f −(a1), bonding hybrid orbitals.(b) f +(a1), antibonding hybrid orbitals.

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8.2 A tetrahedral complex 233

with the p-orbitals of the ligands. For the tetrahedral complex dxy , dxz , and dyz

form both sigma and pi metal–ligand bonds, while the dz2 and dx2 orbitals formonly pi metal–ligand bonds. Also, if we choose x, y, and z as the basis functionsfor the first, second, and third rows of the t2 IR, then the dyz orbital is a first-rowfunction. The d-orbitals dxz and dxy are second- and third-row symmetry functions,respectively. The d-orbitals split into the e and t2 groups as in the octahedral case;however, for the tetrahedron the e level lies lower in energy than the t2 level. Thisis explained by noting that the lobes of the dxy , dxz , and dyz are directed moretoward the ligands than the lobes of dx2 and dz2 , resulting in greater electrostaticrepulsion.

The eigenvalues and eigenvectors for the various molecular levels can be foundin terms of the LCAO two-center interactions and overlap integrals. The t1 and eeigenvalues can easily be obtained; however, the remaining levels require solutionof 3× 3 and 5× 5 secular equations.

As an example we shall find the LCAO eigenvalues for the e states using Löwdinorbitals. We need only find the energy for the row-1 functions. The eigenvalues ofthe row-2 functions are the same. The following definitions are used:

εd =∫

dr d∗x2 H dx2 =∫

dr d∗z2 H dz2, (8.49)

εp =∫

dr p∗k H pk (k = x, y, and z), (8.50)

εpL =∫

dr1√8(x1 − y1 − x2 + y2 − x3 − y3 + x4 + y4)

×H1√8(x1 − y1 − x2 + y2 − x3 − y3 + x4 + y4), (8.51)

M12 =∫

dr d∗x2 H1√8(x1 − y1 − x2 + y2 − x3 − y3 + x4 + y4) = M21.

(8.52)

We can evaluate εpL and M12 in terms of the LCAO parameters. Consider the firstterm of (8.51),

1

8

∫dr x1 H(x1 − y1 − x2 + y2 − x3 − y3 + x4 + y4)

= 1

8

[εp − Hx1,x2 − Hx1,x3 + Hx1,x4 − Hx1,y1 + Hx1,y2 − Hx1,y3 + Hx1,y4

].

(8.53)

From Table 8.5 we have

Hx,x = l2(ppσ)+ (1− l2)(ppπ) (8.54)

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234 Transition-metal complexes

and

Hx,y = lm(ppσ)− lm(ppπ), (8.55)

where l and m are the direction cosines for the vector joining the two ligand sites.For our tetrahedron,

l2 m2 lmR12 1/2 1/2 1/2R13 1/2 0 0R14 0 1/2 0

where Ri j is the vector joining ligand sites i and j . Using these results in (8.49)and (8.50) gives

Hx1,x2 = Hx1,x3 =1

2[(ppσ)+ (ppπ)], (8.56)

Hx1,x4 = (ppπ), (8.57)

Hx1,y1 = Hx1,y3 = Hx1,y4 = 0, (8.58)

Hx1,y2 =1

2[(ppσ)− (ppπ)], (8.59)

1

8[εp − Hx1,x2 − Hx1,x3 + Hx1,x4 + Hx1,y1 + Hx1,y2 − Hx1,y3 + Hx1,y4]

= 1

8

{εp − 1

2[(ppσ)+ (ppπ)]

}. (8.60)

All of the other terms of the product in (8.51) yield the same result, so

εpL = εp − 1

2[(ppσ)+ (ppπ)]. (8.61)

For M12 we have∫dr d∗x2 H

1√8(x1 − y1 − x2 + y2 − x3 − y3 + x4 + y4)

= Hx2,x1− Hx2,y1

− Hx2,x2+ Hx2,y2

− Hx2,x3− Hx2,y3

+ Hx2,x4+ Hx2,y4

.

(8.62)

The direction cosines for the vectors from the metal ion to the ligands are

l m nR01 1/

√3 1/

√3 1/

√3

R02 −1/√

3 −1/√

3 1/√

3R03 −1/

√3 1/

√3 −1/

√3

R04 1/√

3 −1/√

3 −1/√

3

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8.2 A tetrahedral complex 235

where R0i is the vector from the metal ion to the i th p-ligand site.From Table 8.5 we have

Hx2,x =√

3

2l(l2 − m2)(pdσ)+ l(1− l2 + m2)(pdπ) = l(pdπ), (8.63)

Hx2,y =√

3

2l(l2 − m2)(pdσ)− m(1+ l2 − m2)(pdπ) = −m(pdπ), (8.64)

Hx2,x1= Hx2,y2

= Hx2,y4= Hx2,x4

= 1√3(pdπ), (8.65)

Hx2,x2= Hx2,x3

= Hx2,y3= Hx2,y1

= − 1√3(pdπ). (8.66)

Using these results we find that

M12 =√

8

3(pdπ). (8.67)

The eigenvalues of the e states are determined by the condition that

det

⎛⎝ εd − ε

√83(pdπ)√

83(pdπ) εpL − ε

⎞⎠ = 0, (8.68)

where εpL = εp − 12 [(ppσ)+ (ppπ)].

The eigenvalues are

ε(e)± = 1

2(εd + εpL )±

√[1

2(εd − εpL )

]2

+ 8

3(pdπ)2. (8.69)

The “+” sign goes with the antibonding e state and the “−” sign with the bondinge state.

Figure 8.6 shows a schematic representation of the molecular energy levels forthe tetrahedral complex. The ordering of the levels depends upon the particularM L4 complex. The ordering shown is approximately what is expected for the firsttransition series with oxygen or chlorine ligands. For tetrahedral oxide anions suchas TiO−4 , VO3−

4 , CrO2−4 , MnO−4 , and FeO4 the ordering of the lower levels has been

suggested to be [8.2] 1a1, 1t2, 1e, 2t2, 1t1, 3t2, 2e, 4t2, 2a1.

8.2.5 High- and low-spin states for a tetrahedral complex

As in the case of the octahedral complex high- and low-spin configurations arepossible. Assuming that the antibonding t2 and e molecular orbitals are partiallyoccupied, the spin configuration depends on the size of the paring energy, P ,compared with the e–t2 splitting, �tet. The spin configurations appropriate for thetetragonal complex are different from those of the octahedral complex because

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236 Transition-metal complexes

⎧⎪⎨⎪⎩

⎫⎪⎬⎪⎭

4pt2

4sa1

3de, t2

5t2

3a1

4t2

2e

1t12a13t21e2t2

1t2

1a1

2pa, e, t1, t2

2sa1, t2

Metalorbitals

Molecularorbitals

Ligandp-orbitals

Figure 8.6 A schematic representation of the molecular levels of a tetrahedrallycoordinated transition metal ion. The order of the levels 1e, 2a1, 2t2, 1t1, and 3t2depends on the particular M L4 molecule or complex.

the e antibonding level lies below t2. (In Fig. 8.6 the two d-levels are labeled 2eand 4t2.) For three to five d-electrons both high- and low-spin states are possible,depending upon whether P > �tet or P < �tet. The spin configurations for threeto six d-electrons are shown schematically below.

High-spin states (tetrahedral)

Low-spin states (tetrahedral)

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Exercises 237

For more than six d-electrons, the spin configuration for the lowest energy statesdoes not depend on the magnitude of P compared with �tet.

References

[8.1] J. C. Slater and G. F. Koster, “Simplified LCAO method for the periodic potentialproblem”, Phys. Rev. 94, 1498–1524 (1954).

[8.2] G. V. Ionova and M. E. Dyatkina, “Molecular orbitals of tetrahedral oxide anions oftransition metals communication IV. FeO4 molecule”, J. Struct. Chem. 6, 764–765(1966).

Exercises

8.1 Make schematic sketches of the symmetry functions for the a1g, eg, and t2g

states of an octahedral complex.8.2 Calculate the LCAO matrix elements for the secular equation, det(H−εS)=0,

for the a1g levels for an octahedral complex. Include all three atomic orbitals,s(M), s(L), and p(L), and overlap integrals. Table 8.5 can also be used todetermine the overlap integrals. Take HsM sM = εs(M), HsL sL = εs(L), andHxx = Hyy = Hzz = εp(L).

8.3 (a) Calculate the row-1 t1g energy for hydrogen-like orbitals in terms ofLCAO parameters including the overlap.

(b) Show that the energy calculated using Löwdin orbitals is equal to theenergy calculated using hydrogen-like orbitals.

8.4 Consider an octahedral complex with 38 electrons. Assign the electrons tothe levels in Fig. 8.2.(a) What is the configuration for the lowest energy state according to Hund’s

rules?(b) What is the symmetry symbol for the total molecular ground state?

8.5 (a) For the molecular state in Exercise 8.4, what is the lowest-energy electric-dipole transition where the final state is 2t2g (according to Fig. 8.2)?

(b) Would it be correct to call the transition in (a) a charge-transfer transition?(c) Show that the excited molecular state with S = 3/2 is 4T2u .

8.6 (a) For an octahedral complex with 8, 9, or 10 d-electrons Hund’s ground-state spin configuration (high spin/low spin) does not depend on the ratioP/�oct. Explain why.

(b) For a tetrahedral complex with more than six d-electrons, Hund’s ground-state spin configuration does not depend on the ratio of P/�tet. Why dothe tetrahedral and octahedral cases differ?

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238 Transition-metal complexes

8.7 Methane, CH4, is a tetrahedral molecule. Using 2s and 2p orbitals for the Catom and 1s orbitals for the H atoms, do the following:(a) Find the IRs of Td contained in the orbital representations.(b) Describe the block-diagonalized secular equation: the symmetry, the

dimensions of the secular equations, and what orbitals are involved.(c) Find the symmetry functions.(e) What is the electronic configuration of the ground state in terms of the

one-electron molecular states? What is the symmetry label for the totalmolecule ground state?

8.8 (a) Calculate the energies of the molecular levels in terms of the LCAOparameters for the methane molecule in Exercise 8.7, including overlapintegrals.

(b) Construct a schematic representation of the molecular levels and labelthem according to their IRs. Assume ε2s < ε1s < ε2p and that CH4 isdiamagnetic.

(c) What is the symmetry symbol for the ground state of the molecule?(d) Photoelectron spectroscopy shows ionization of electrons corresponding

to 14 eV and 23 eV. How would you interpret these results?8.9 Using the action table and the matrix elements of the t2 IR for the Td , show

that dyz (not dxy) is a row-1 basis function for the t2 IR.8.10 For an octahedral complex, Hund’s ground state for (t2)3 has S = 3/2

and each of the t2 d-orbitals is singly occupied. The determinantal stateis 〈dxy↑dxz↑dyz↑|. Show that the product dxydxzdyz transforms as a1 (i.e., isinvariant under all of the operations of the group).

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9

Space groups and crystalline solids

9.1 Definitions

A crystal or crystalline solid is an ordered array of atoms, molecules, or ionswhose pattern or lattice is repeated periodically. A “single crystal” is perfectlyordered. Most crystalline solids are polycrystalline, meaning that they are com-posed of many small single crystals with defective, bounding surfaces betweenthem. Large single crystals occur in nature, but often it is necessary to prepare themby special crystal-growth methods in laboratories. Single crystals are the preferredform for studying the intrinsic properties of a crystalline material.

Theoretical analysis of single-crystal properties usually assumes the crystallinestructure is infinite or imposes periodic boundary conditions requiring the wave-function to repeat itself after a sufficiently large number of chemical units. This isa reasonable approximation, since a cubic-centimeter crystal contains on the orderof 1022 to 1024 repeated units.

To sharpen our description of a lattice some definitions are useful.

● Bravais lattice. A Bravais lattice is a space-filling array of points generated bythree primitive lattice vectors, a, b, and c. The vectors of the infinite set ofvectors {R(l,m, n)} terminate on the Bravais lattice points. These vectors areR(l,m, n) = la + mb + nc, where l, m, and n are positive or negative integersor zero. Instead of the vectors a, b, and c, a Bravais lattice can be specified bythe magnitudes of the primitive vectors, a = |a|, b = |b|, and c = |c|, and theangles between them, α, β, and γ . Here α is the angle between a and c, β is theangle between b and c, and γ is the angle between a and b. The numbers a, b,and c are called the lattice constants. The 14 distinct Bravais lattices are shownin Fig. 9.1.

● Unit cell. The parallelepiped defined by the vectors a, b, and c containing aformula unit (basis) of the material is called the unit cell. The volume of the unitcell is V = a · (b × c). The atoms or molecules of a crystal may be located

239

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240 Space groups and crystalline solids

Crystal systems

triclinic

monoclinic

orthorhombic

tetragonal

hexagonal

cubic

rhombohedral(trigonal)

α, β, γ ≠ 90˚

α, β, γ ≠ 90˚

γ

γ

β

βα

α

α ≠ 90˚β, γ ≠ 90˚

a ≠ b ≠ c

a

b

c

γ

βα

α ≠ 90˚β, γ ≠ 90˚

a ≠ b ≠ c

a

b

c

a ≠ b ≠ c

a

b

c

a ≠ b ≠ c

a ≠ c

a ≠ c

a

b

aa

a aa

c

a

a

aa

a

γ

β

c

c

a ≠ c

aa

c

a

aa

a

aa

Figure 9.1 The 14 distinct Bravais lattices.

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9.1 Definitions 241

c

b

a

c

b

a

(a) (b)

Figure 9.2 (a) Cubic Bravais-lattice primitive vectors. a = b = c; α = β =γ = 90 degrees. (b) The unit cell with ions for the ABC3 cubic perovskite struc-ture. The B ion is at (0, 0, 0), the C ions are located at (a/2, 0, 0), (0, a/2, 0),(0, 0, a/2), and the A ion is at (a/2, a/2, a/2), where a is the lattice spacing.Translation of the unit cell by the set of vectors {R(l,m, n)} leads to the perovskitestructure with B atoms at a(l,m, n), C atoms at a(l+1/2,m, n), a(l,m+1/2, n),and a(l,m, n + 1/2), and A atoms at a(l + 1/2,m + 1/2, n + 1/2).

anywhere on or within the unit cell, but are located at the same relative positionsin every unit cell of the crystal. The chemical composition of the material isrepresented by the atoms belonging to a unit cell. For example, the unit cell ofan ABC3 crystal (SrTiO3 for example) contains one A atom, one B atom, andthree C atoms as shown in Fig. 9.2.

● Translational invariance. A Bravais lattice is invariant under a translation byany of the vectors {R(l,m, n)}. The smallest non-zero translations under whichthe lattice is invariant are the primitive vectors a, b, and c. The infinite, single-crystalline solid is invariant under the same set of translations as its Bravaislattice.

Before launching into a formal discussion of space groups it is worthwhileto look at a couple of very simple examples of wave solutions in a crystal.We consider a two-dimensional square lattice with orthogonal Löwdin s-orbitals,�[r − R(m, n)], on the Bravais lattice points with lattice vectors aex and aey , asshown in Fig. 9.3. The LCAO wavefunction for the electron states is

ψ(r) =∑m,n

Amn �[r− R(m, n)], (9.1)

with ∫�[r− R(m, n)]∗ �[r− R(m ′, n′)] dr = δmm′δnn′, (9.2)

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242 Space groups and crystalline solids

a ex

a ey

Figure 9.3 The square Bravais lattice with s-orbitals on each lattice point. Thenearest-neighbor LCAO interaction parameter is (ssσ).

where �[r−R(m, n)] is a Löwdin s-orbital located at the lattice point R(m, n) =a(mex + ney) and Amn is the amplitude of the orbital at that site. We want to findthe solutions to the eigenvalue equation

H ψ(r) = E ψ(r). (9.3)

For simplicity assume only nearest-neighbor atoms interact. Substitute (9.1) into(9.3), multiply both sides of the equation by �[r−R(m ′, n′)]∗, and integrate over r.This procedure leads to an eigenvalue matrix equation,

∑m′,n′{(ε − E)δmm′ δnn′ + (ssσ)δm±1,m′δnn′ + (ssσ)δmm′δm,n±1}Am′n′ = 0,

(9.4)

where

ε =∫

�[r− R(m, n)]∗ H �[r− R(m, n)] dr, (9.5)

(ssσ) =∫

�[r− R(m, n)]∗ H �[r− R(m ± 1, n)] dr

=∫

�[r− R(m, n)]∗ H �[r− R(m, n ± 1)] dr. (9.6)

To obtain solutions to (9.4), we write

Amn = A(k)eik·R(m,n), (9.7)

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9.1 Definitions 243

where k = kx ex +kyey and A(k) is a constant for fixed k. Substitution of (9.7) into(9.4) gives {

(ε − Ek)+ 2(ssσ)[cos(kxa)+ cos(kya)]}

A(k) = 0. (9.8)

If A(k) is not zero, the eigenenergy is

Ek = ε + 2(ssσ)[cos(kxa)+ cos(kya)]. (9.9)

As the k-vector varies, Ek describes a band of energies that lie between ε−4(ssσ)and ε + 4(ssσ). For this reason, a graph of Ek versus k is called an energy band.

The energy of a particular state is determined by the k-vector. Therefore we shallwrite the solutions of (9.3) as ψk(r).

Since cos(kxa) and cos(kya) are periodic functions, the energy of ψk(r) is thesame as the energies of the states ψ(k+K)(r), where

Ka = 2π(sex + tey) (s, t = 0, or any positive or negative integer). (9.10)

Therefore we can restrict kxa and kya to the ranges

0 ≤ kx ≤ 2π

aand 0 ≤ ky ≤ 2π

a. (9.11)

A different, but equivalent, choice is

−π ≤ kxa ≤ π and −π ≤ kya ≤ π. (9.12)

For our two-dimensional square Bravais lattice, the area described by (9.12) is thefirst Brillouin zone and the K-vectors of (9.10) are reciprocal-lattice vectors. Ageneral description of the reciprocal-lattice vectors and the first Brillouin zone forany Bravais lattice is given later in this chapter.

There are some features of Ek worth mentioning. If α is an operation that leavesthe Bravais lattice invariant, then Ek = Eαk. For our example above, a counter-clockwise rotation α = π/2 applied to the k-vector takes kx → ky and ky →−kx ,so that Eαk = ε + 2(ssσ)[cos(kya)+ cos(−kxa)] = Ek. Similarly, a reflection, σ ,through the x-axis takes kx → kx and ky →−ky , and Eσk = Ek. In general, everyoperation, R, that leaves the Bravais lattice invariant produces a new wavevector,k′ = Rk, such that ERk = Ek. In addition, because of the periodic nature of ψk,Ek = Ek+K, where K is any reciprocal-lattice vector.

In the case that there are n different atomic orbitals associated with each lat-tice point, there will be n different energy bands. When there are several atoms inthe unit cell, the number of energy bands that result is equal to the total numberof atomic orbitals associated with all of the atoms of the unit cell. If we denotethese energy bands as Enk, where n indexes the various bands, then each Enk

possesses the properties discussed in the previous paragraph: Enk = Enαk and

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244 Space groups and crystalline solids

En(k+K) = Enk. To illuminate this feature, for the model just discussed, let usassign two orbitals to each lattice point. For simplicity, suppose there is a 1s and a2s orbital on each site, and again assume nearest-neighbor interactions. In this casethe LCAO wavefunction is a linear combination of both 1s and 2s orbitals,

ψ(r) =∑m,n

{Amn �1s[r− R(m, n)] + Bmn �2s[r− R(m, n)]}, (9.13)

where Amn is the amplitude of the 1s orbital and Bmn is the amplitude of the 2sorbital at R(m, n). The LCAO parameters involved are ε1, ε2, (s1s1σ), (s1s2σ), and(s2s2σ). The subscripts “1” and “2” refer to the 1s and 2s orbitals, respectively.Proceeding as before, we take

Amn = A(k)eik·R(m,n), (9.14)

Bmn = B(k)eik·R(m,n), (9.15)

and obtain a 2× 2 matrix eigenvalue equation,(ε1 + 2(s1s1σ)C(k)− Ek 2(s1s2σ)C(k)

2(s1s2σ)C(k) ε2 + 2(s2s2σ)C(k)− Ek

)(A(k)B(k)

)= 0,

(9.16)

where C(k) = cos(kxa)+cos(kya), and ε1 and ε2 are the diagonal matrix elementsof the Hamiltonian for the 1s and 2s orbitals, respectively.

There are two orbitals per unit cell and two energy bands,

E±k =1

2{ε1 + ε2 + 2[(s1s1σ)+ (s2s2σ)]C(k)}

±√{

1

2{ε1 − ε2 + 2[(s1s1σ)− (s2s2σ)]C(k)}

}2

+ 4(s1s2σ)2C(k)2.

(9.17)

Each of these energy bands satisfies the conditions that E±k = E±k+K andE±k = E±αk, where K is a reciprocal-lattice vector and αk = Pα−1k.

The properties of the eigenfunctions and eigenvalues derived above for thesquare lattice with nearest-neighbor interactions are true in general for any lat-tice and any Hamiltonian with a periodic potential, irrespective of the range of theinteractions. The proof lies in the mathematics of space groups.

9.2 Space groups

An infinite single crystal is an array of repeated unit cells. The covering group ofthe Bravais lattice is a space group, G. The elements of G consist of combinations

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9.2 Space groups 245

of point-group elements (rotations, reflections, inversions) and translations underwhich an infinite, single-crystal structure is invariant. Alternately, we can considera finite crystal subject to periodic boundary conditions that require the physicalfunctions to repeat after a significantly large number of translations. After the anal-ysis is complete we may then let the large number of translations tend to infinity.In either case the electronic Hamiltonian of a crystal must also be invariant underany of the operations of the space group.

The convention is to write a space-group symmetry element in the form

{α|T}, (9.18)

where α is a point-group element (rotation, reflection, or inversion) performed onthe coordinate system with one point (the origin) fixed and T is a translation vector.A point-group element is {α|0} and a pure translation is {E |T}, where E is theidentity. A space-group operation applied to a vector r is a rotation, reflection, orinversion followed by a translation,

{α|T}r = αr+ T, (9.19)

where on the right-hand side of (9.19) αr represents a rotation of the coordinatesystem and T is a translation of the origin of coordinates. In some instances, Tneed not be a translation by a lattice vector, as will be explained shortly. Groupmultiplication of elements is defined as

{α|T1}{β|T2} = {αβ|αT2 + T1}. (9.20)

To constitute a space group the elements must meet the same requirements as thoseimposed on point groups, namely the following.

● The product of any two elements in the space group is an element of the spacegroup.

● There is an identity element, {E |0}, in the space group.● Every element, {α|T1}, in the space group has an inverse, {α|T1}−1, such that{α|T1}{α|T1}−1 = {α|T1}−1{α|T1} = {E |0}.

From the definition of multiplication (9.20) it follows that

{α|T}−1 = {α−1| − α−1T}, (9.21)

since

{α|T}−1{α|T} = {α−1| − α−1T}{α|T} = {α−1α|α−1T− α−1T} = {E |0}(9.22)

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246 Space groups and crystalline solids

and

{α|T}{α|T}−1 = {α|T} {α−1| − α−1T} = {αα−1| − αα−1T+ T} = {E |0}.(9.23)

Pure translation elements commute and therefore every element of the translationsubgroup is in a class by itself. However, in general space-group operations do notcommute. For example,

{α|T1}{β|T2} = {αβ|αT2 + T1}, (9.24)

but

{β|T2}{α|T1} = {βα|βT1 + T2}. (9.25)

For these two products to be the same one requires that {αβ|αT2 + T1} ={βα|βT1 + T2}, a condition that is generally not true.

The space group Gs contains operations of the form {E |T}. One might sup-pose that the translation, T, in (9.18) must be one of the Bravais lattice vectors,R(l,m, n); however, this is not the case. Some crystal space groups possesscompound symmetry elements that require both a point-group operation and anon-primitive translation. Such elements are not of the form {α|R(l,m, n)}, whereR(l,m, n) is a lattice vector. There are two types of such compound symmetry ele-ments, screw axes and glide planes. As the name implies, a screw axis consists ofa rotation about an axis accompanied by non-primitive translation along the sameaxis. A glide-plane operation consists of a translation parallel to a given plane plusa reflection in the same plane.

We can express a general space-group operation as

{α|T} = {α|R(l,m, n)+ tα}, (9.26)

where R(l,m, n) is a lattice vector and tα is a translation vector whose magnitudeis less than that of a primitive lattice vector. If all of the elements of G have tα = 0,the space group is symmorphic and the set {α|T} is equal to the set {α|T}|T=0. Ifany tα is unequal to zero, G is non-symmorphic and the set {α|0} is not the same as{α|T}|T=0 (distinct elements obtained by setting T = 0). In three dimensions thereare 230 distinct space groups, of which 73 are symmorphic.

9.3 The reciprocal lattice

The primitive vectors that define the Bravais (direct) lattice, namely a, b, and c,can be used to construct what is called the reciprocal lattice. The primitive vectorsfor the reciprocal lattice, denoted as a∗, b∗, and c∗, are defined by

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9.4 Brillouin zones 247

a∗ = 2π

V(b× c), b∗ = 2π

V(c× a), c∗ = 2π

V(a× b), (9.27)

where V = a · (b × c) is the volume of the unit cell. The reciprocal lattice isgenerated from the primitive reciprocal-lattice vectors in the same way as a Bravaislattice is generated from direct-lattice vectors. The vectors in the set of vectorsK(s, t, u) = sa∗ + tb∗ + uc∗, where s, t , and u are zero, positive or negativeintegers, terminate on the points of the reciprocal lattice. The reciprocal-latticevectors have the important property that

eiK(s,t,u)·R(l,m,n) = 1 for all r, s, t and all l,m, n.

Thus

K(s, t, u) · R(l,m, n) = a∗ · a(sl)+ b∗ · b(tm)+ c∗ · c(un)

= 2π

V[a · (b× c)](sl + tm + un)

= 2π(integer), (9.28)

and hence

eiK(s,t,u)·R(l,m,n) = e2π i(integer) = 1. (9.29)

The reciprocal lattice for a simple-cubic Bravais lattice is also a simple-cubicBravais lattice. The reciprocal lattice for a face-centered cubic lattice is a body-centered cubic lattice. The reciprocal lattice for a body-centered cubic lattice is aface-centered cubic lattice. Finally, the reciprocal lattice for a simple hexagonallattice is a simple hexagonal lattice as well, but with a rotation of 30◦.

9.4 Brillouin zones

Consider a reciprocal lattice and select any lattice point to be the origin. TheBrillouin zones are constructed by erecting perpendicular planes1 that bisect thereciprocal-lattice vectors from the origin to any of the other reciprocal-latticepoints. The collection of points that can be connected to the origin without crossingany of the erected planes comprises a volume called the first Brillouin zone. Thesecond Brillouin zone is the volume whose points can be connected to the origin bya line that crosses one and only one of the erected planes. The nth Brillouin zoneis the volume whose points can be connected to the origin by a line that crossesn−1 and only n−1 erected planes. The volume in reciprocal-lattice space of eachof the Brillouin zones is the same. The construction of the first Brillouin zone fora two-dimensional hexagonal lattice is shown in Fig. 9.4.

1 Each of these planes is a Bragg plane.

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248 Space groups and crystalline solids

a∗b∗

2π/a

(a)

(b)

b a

a

(c)a∗b∗

(d)

Figure 9.4 (a) A two-dimensional, hexagonal reciprocal lattice. The centerhexagon shows the construction of the first Brillouin zone. The dashed lines arethe bisectors of the nearest-neighbor reciprocal vectors and the darker shaded areais the first Brillouin zone. The area outside the first Brillouin zone within the six-pointed star formed by the bisectors is the second Brillouin zone (pale shaded).(b) The primitive lattice vectors, a and b, in direct space. (c) The first Bril-louin zone, showing the primitive reciprocal-lattice vectors, a∗ and b∗. (d) Foldedsecond Brillouin zone.

X

X

R

Σ

Λ

MZ

S

X

TΔΓ

kx

ky

kz

ΓΛ

LU

S

XZW

Q

Δ

kx

ky

kz

Γ Λ

PF

HGN

Σ

D Δ

kx

ky

kz

(a) (b) (c)

Figure 9.5 Brillouin zones for the (a) simple-cubic, (b) face-centered cubic, and(c) body-centered cubic Bravais lattices. The heavy lines define the irreduciblezones (1/48th of the whole), and letters label special symmetry features.

The first Brillouin zones for the simple-cubic, face-centered cubic, and body-centered cubic lattices are illustrated in Fig. 9.5. The letters label special high-symmetry lines and points. A point on the interior of the Brillouin zone with nospecial symmetry properties is referred to as a general point. The primitive direct-lattice and reciprocal-lattice vectors are given in Table 9.1.

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9.5 Bloch waves and symmorphic groups 249

Table 9.1 Direct- and reciprocal-lattice vectors for the simple-cubic,face-centered cubic, body-centered cubic, and hexagonal lattices

Lattice type Primitive direct-lattice vectors Primitive reciprocal-lattice vectors

Simple cubic a = aex a∗ = (2π/a)exb = aey b∗ = (2π/a)eyc = aez c∗ = (2π/a)ez

Face-centered a = (a/2)(ey + ez) a∗ = (2π/a)(−ex + ey + ez)cubic b = (a/2)(ex + ez) b∗ = (2π/a)(ex − ey + ez)

c = (a/2)(ex + ey) c∗ = (2π/a)(ex + ey − ez)Body-centered a = (a/2)(−ex + ey + ez) a∗ = (2π/a)(ey + ez)cubic b = (a/2)(ex − ey + ez) b∗ = (2π/a)(ex + ez)

c = (a/2)(ex + ey − ez) c∗ = (2π/a)(ex + ey)

Hexagonal a = a((√

3/2)ex + 12 ey) a∗ = (2π/a)((1/

√3)ex + ey)

b = a(−(√3/2)ex + 12 ey) b∗ = (2π/a)(−(1/√3)ex + ey)

c = cez c∗ = (2π/c)ez

9.5 Bloch waves and symmorphic groups

Until stated otherwise, we shall limit our discussion to symmorphic space groups.For these groups the allowed translations, R(l,m, n), are Bravais-lattice vectors.

The Hamiltonian, H(r), describing electrons or phonons or any wave-like exci-tation in a crystalline solid must be invariant under a translation R(l,m, n). Thequantum eigenvalue equation for a wave excitation is of the form

H(r) ψ(r) = λ ψ(r), (9.30)

where ψ(r) is the wavefunction and λ is the eigenvalue. According to Bloch’stheorem, Theorem 9.1, the wavefunction is of the form ψnk(r) = exp(ik ·r)unk(r),where k belongs to the first Brillouin zone. The function unk(r) is periodic, unk(r+R(l,m, n)) = unk(r), with R any lattice vector. The subscripts n and k characterizethe solutions of wavefunctions (label the IRs) for a periodic structure. The meaningof n, the band index, will be made clear later in this chapter.

Theorem 9.1 (Bloch waves) The eigenfunctions of the equation

H(r) ψ(r) = λ ψ(r),

where H(r) is invariant under translation by any lattice vector, are of the form

ψnk(r) = eik·runk(r), (GT9.1)

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250 Space groups and crystalline solids

where n labels the different bands, k is a vector belonging to the first Brillouinzone, and unk(r) is a function with the same translation symmetry (periodicity) asH(r) so that unk(r+ R(l,m, n)) = unk(r), where R is any lattice vector.

Bloch’s theorem holds for the wave-like eigenfunctions of a Hamiltonian with aperiodic potential, including, but not limited to, electrons, phonons, magnons, andelectromagnetic waves inside a periodic crystal.

Let P{α|R} be the operator for the point-group element, α, and translation, R.(For a symmorphic group, R is a lattice vector.) Consistently with our conventionthat P operates on a function, rather than on the coordinate system, we have that

P{α|R} f (r) = f [{α|R}−1r] = f (α−1r− α−1R) = f [α−1(r− R)], (9.31)

so that for the pure point-group elements

P{α|0} f (r) = f (α−1r), (9.32)

a result that is consistent with our definition of the action of the operator for a point-group operation α as defined in Chapter 3, in Eq. (3.2). For a pure translation, (9.31)gives

P{E |R} f (r) = f (r− R). (9.33)

However, some texts [9.1] define P{E |R} f (r) = f (r+ R).We can establish that the Bloch functions,

ψnk(r) = exp(ik · r)unk(r), (9.34)

are basis functions for a diagonal representation of the translation subgroup:

P{E |R(l,m,n)}ψnk(r) = eik·(r−R(l,m,n))unk(r− R(l,m, n))

= e−ik·R(l,m,n)ψnk(r). (9.35)

Equation (9.35) shows that the operator P{E |R(l,m,n)} transforms ψnk(r) into a mul-tiple of itself. Therefore the representation matrix is one-dimensional and thecharacter for the operator P{E |R(l,m,n)} is exp(−ik · R(l,m, n)).

To understand why the k-vector in Bloch’s theorem is restricted to the first Bril-louin zone, consider a finite cubic crystal with N unit cells in the three primitivelattice directions. The operator for any lattice translation can be written as theproduct of three translation operators:

P{E |R(l,m,n)} = P{E |R(l,0,0)}P{E |R(0,m,0)}P{E |R(0,0,n)}. (9.36)

The set of translations along the a-direction, R(l, 0, 0) (l = 1, 2, 3, . . . , N ), formsa subgroup of translations. So do the sets R(0,m, 0) (m = 1, 2, 3, . . . , N ) and

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9.5 Bloch waves and symmorphic groups 251

R(0, 0, n) (n = 1, 2, 3, . . . , N ). We impose periodic boundary conditions on thesystem and require that

P{E |R(N ,0,0)} = P N{E |R(1,0,0)} = P{E |0}, (9.37)

P{E |R(0,N ,0)} = P N{E |R(0,1,0)} = P{E |0}, (9.38)

P{E |R(0,0,N )} = P N{E |R(0,0,1)} = P{E |0}. (9.39)

With these periodic boundary conditions it may be seen that the subgroup withoperators P{E |R(l,0,0)} (l = 1, 2, 3, . . . , N ) is a cyclic group with period N , sinceP{E |R(l,0,0)P{E |R(l ′,0,0)} = P{E |R(l+l ′,0,0)} and P N

{E |R(1,0,0)} = P{E |0}. The group is alsoan Abelian group because the translation operations commute with one another.The same is true for the subgroups for translations along the b- and c-directions.

Theorem 9.2 (Cyclic groups) A group, all of whose elements can be generatedfrom a single element, g, as powers of g is a cyclic group. A cyclic group of periodN consists of the elements (g, g2, g3, . . . , gN ) with gN = E. The element g is calledthe generator. Cyclic groups are Abelian, meaning that the elements commute withone another. Because the elements commute, each element is in a class by itself.Therefore there are N elements, N classes, and N irreducible representations.

The characters of g for the IRs of a cyclic group of elements of period N are theN roots of unity, exp(iq2π/N ), q = 1, 2, 3, . . . , N . The character of the element,gm, for the qth IR is [exp(2π iq/N )]m. There is an IR for each value of q, so thevalues of q can be used as unique labels for the IRs of a cyclic group.

According to Theorem 9.2 the character of the generating element, g ={E |R(1, 0, 0)}, for the qth IR of the subgroup associated with translation along thea-direction is exp(2π iqa/N ). For the operator, Pg = P{E |R(1,0,0)}, the character is

χq(Pg) = e−2π iqa/N . (9.40)

According to (9.35) (with l = 1, m = 0, and n = 0) the character for a Blochwave is

χq(Pg) = e−ik·R(l,0,0) = e−ik·R(1,0,0) = e−ikaa, (9.41)

since unk(r) = unk(r− R). In (9.41), a is the length of the primitive lattice vectora and ka is the component of the k-vector along the a-direction.

Equating the characters from (9.35) and (9.41) defines ka for the qa IR:

e−ikaa = e−2π iqa/N , (9.42)

ka = 2πqa

Na, (9.43)

qa = 1, 2, 3, . . . , N . (9.44)

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252 Space groups and crystalline solids

Each value of qa specifies one of the N IRs of the cyclic group. We can alsoconsider qa > N or qa < 0. In these cases qa can be written as q ′ + nN , wheren is an appropriate positive or negative integer and 0 ≤ q ′ ≤ N . The character ofsuch a qa is

e2π iqa/N = e2π iq ′/N ei2nπ = e2π iq ′/N . (9.45)

In other words, any qa > N is equivalent to some IR with 0 < qa < N .From (9.36) it follows that a k-vector can be defined to uniquely specify the IRs

of the three-dimensional subgroup of translation operators,

k(q) = 2π

N

[(qa

a

)e∗a +

(qb

b

)e∗b +

(qc

c

)e∗c

], (9.46)

where e∗a , e∗b, and e∗c are unit vectors along the a∗-, b∗-, and c∗-directions, and a, b,and c are the magnitudes of a, b, and c, respectively. Each component of k(q) takeson the values between 2π/(Nd) and 2π/d, where d = a, b, or c depending on thecomponent chosen. As N →∞, k(q) approaches a continuous variable with eachcomponent restricted to the reciprocal-lattice space from 0 to 2π/d. An alternatebut equivalent scheme is to restrict each component to be in the range from−π/d to+π/d. For this choice, the k-vector is restricted to the first Brillouin zone. A vectork that is outside of the first Brillouin zone can be written as k = k′ + K(s, t, u),where k′ is in the first Brillouin zone, and s, t , and u are appropriate integers.Adding a reciprocal-lattice vector to k′ is analogous to adding multiples of N toqa , qb, or qc. Thus k = k′ +K belongs to the same IR as k′.

9.6 Point-group symmetry of Bloch waves

We confine our discussion to symmorphic space groups for which {α|R} can befactored into the product of a point-group element and a pure translation element{α|R} = {E |R}{α|0}. Note that {α|0} and {E |R} do not commute, and neither dothe operators P{α|0} and P{E |R}. On the other hand, the set {g} consisting of all ofthe operations of the space group of the form P{α|0} and the set {gR} consisting ofall of the operations of the space group of the form P{E |R} do commute. That is,{g}{gR} contains the same set of operators as {gR}{g} (but the order in which theoperators appear is different). The entire symmorphic group, G, can be obtainedfrom the direct product of the two subgroups: G = g × gR. This suggests that theIRs of G are the direct products of the IRs of g with those of gR. This conjecture istrue, but the proof will not be presented here.

It is clear from the form of the Bloch wave, ψk(r) = exp(ik · r)unk(r), that itdoes not have the same periodic nature as the crystal (excluding the cases k = 0or K). However, since unk(r) is periodic, the symmetry properties of a Bloch wave

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9.6 Point-group symmetry of Bloch waves 253

are determined by the modulating exponential factor. The k-vector appears in theexponential only as the scalar product, k · r. Let us consider the effect of a point-group operation on the scalar product,

(k · P{α|0}r) = k · [(α−1r)]. (9.47)

The scalar product on the right-hand side of (9.47) must be unchanged if both kand α−1r are rotated through the same angle. Therefore, if we apply P−1

{α|0} to bothvectors, the scalar product must be unchanged:

[P−1{α|0}k] · [P−1

{α|0}(α−1r)] = (αk) · (αα−1r) = (αk) · r

= (P−1{α|0}k) · r = (k · P{α|0}r). (9.48)

Equation (9.48) states that the effect of P{α|0} operating on r in the scalar productis the same as applying P−1

{α|0} to k.From (9.48) it also follows that, for a Bloch wave,

P{α|0}ψnk(r) = ψnk(α−1r) = eik·α−1runk(α

−1r) = ei(αk)·ru′n(αk)(r). (9.49)

The function u′n(αk)(r) need not be equal to un(αk)(r), but it must have the same peri-odicity. If the energy associated with the Bloch wave, Enk, is non-degenerate, thetwo functions can differ at most by a phase factor, eiφ(α). However, if Enk is degen-erate, u′n(αk)(r) may be a linear combination of the u-functions of the degeneratestates.

9.6.1 The group of the k-vector

If R is a lattice vector, then P{α|0}R = R′ is also a lattice vector. Similarly, if Kis a reciprocal-lattice vector, then P{α|0}K = K′ is also a reciprocal-lattice vec-tor. A Bravais lattice and its reciprocal lattice are invariant under the point-groupoperations of the same space group.

If k and k′ differ by a reciprocal-lattice vector (k = k′ +K), they are symmetry-equivalent or simply “equivalent”, and the two Bloch waves, ψnk and ψnk′ , belongto the same IR. Here, K is a reciprocal-lattice vector and k and k′ are vectors in thefirst Brillouin zone.

The collection of all the point-group operations that transform k into itself oran equivalent vector forms a subgroup of the space group called the group ofthe k-vector or the group of the wavevector. We shall denote this group by thesymbol gk.

We can demonstrate that gk is, in fact, a group. Let P{α|0} and P{β|0} be point-group operations in gk so that

P{α|0}k = k+K, (9.50)

P{β|0}k = k+K′. (9.51)

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254 Space groups and crystalline solids

Then,

P{β|0}P{α|0}k = P{β|0}[k+K] = k+K′ + P{β|0}K. (9.52)

P{β|0}K is a reciprocal-lattice vector. Since K′ + P{β|0}K is the sum of tworeciprocal-lattice vectors, it also is a reciprocal-lattice vector,

K′ + P{β|0}K = K′′. (9.53)

Therefore,

P{β|0}P{α|0}k = P{βα|0}k = k+K′′. (9.54)

Equation (9.54) shows that, if P{β|0} and P{α|0} belong to gk, then P{βα|0} alsobelongs to gk. The set of operators gk always has the identity P{E |0}, therefore allthat remains is to show that every operator in gk has an inverse that is also in gk.

P{α|0} = {α−1|0} and its inverse P{α−1|0} = {α|0}. We need to show that, if α−1kis equivalent to k, then so is αk. We have that

α−1k = k+K. (9.55)

Multiplying both sides of (9.55) by α gives

αα−1k = k = α(k+K) = αk+K′, (9.56)

where K′ is a reciprocal-lattice vector. From (9.56) we obtain

αk = k+ (−K′), (9.57)

showing that αk is equivalent to k. Thus the operators of gk satisfy the requirementsof a group.

All of the set of Bloch waves

ψn(αk)(r) = eiαk·r un(αk)(r) (for all α in gk including E) (9.58)

have the same energies (are degenerate), Enk = Enαk , since the wavefunctions arerelated to one another by a symmetry operation. More than one of the operationsof gk may produce the same (identical, not just equivalent) k-vector. The proper setof degenerate states consists of those that are distinct. We know that the distinct,degenerate eigenstates of a Hamiltonian form the basis for a representation of thegroup, hence the (distinct) degenerate Bloch waves of (9.58) form a basis for arepresentation of the subgroup gk.

It is useful to distinguish two different types of degeneracies. The first occursas a result of symmetry requirements. The second, called “accidental” degeneracy,is not a symmetry requirement but rather a fortuitous result dependent upon thedetails of the Hamiltonian. For energy bands we can distinguish degeneracies ofthe form Enk = En(αk) and Enk = En′k. For the latter case the energies of two

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9.6 Point-group symmetry of Bloch waves 255

Bloch waves from different bands, n and n′, coincide for the same k-vector. Thefirst condition, Enk = En(αk), is a result of symmetry and therefore is an essentialdegeneracy. The condition Enk = En′k can be a result of symmetry or an accidentaldegeneracy.

Basis functions for representations (symmorphic space groups)

The periodic parts of the Bloch-wave functions, un(αk)(r) (where α is in gk), arebasis functions for a representation of gk:

P{α|0} unk(r) =∑

k′D(α)k′k uk′(r), (9.59)

where k′ runs over the distinct, equivalent k-vectors generated by the operators ingk, and D(α)k′k are the representation-matrix elements. It should be noted that Dmay be reducible.

The exponential functions, eik·r, are basis functions for the translation subgroup,gR, whose operators are P{E |R}, where R = R(l,m, n) is any lattice vector. Thematrix representation is diagonal:

P{E |R}eik·r = e−ik·Reik·r. (9.60)

Since this representation is one-dimensional, it is irreducible.

9.6.2 Subgroups and k-vectors

The notation to be used for the various operators and groups is summarized inBox 9.1.

Box 9.1 Notation for groups and subgroups of a symmorphic space group

Symbol DefinitionG Space group with operators P{α|R}.g Subgroup of G consisting of all pure point-group operators, P{α|0}.

(Sometimes called the big group.)h(g) Number of operators in g.Pα Operator belonging to g. Pα = P{α|0}.gk Group of the wavevector. Subgroup of G and g consisting of all pure

point-group operators of G for which P{α|0}k = k+K. (Sometimescalled the little group.)

h(k) Number of operators in gk.P(k)α Operator in the group gk.gR Translation subgroup. Subgroup of G consisting of pure translation

operators, P{E |R}.PR Operator in gR.

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256 Space groups and crystalline solids

gsk The space group of the k-vector (or wavevector), gk × gR.

s(k) The star of k. The set of distinct, mutually inequivalent wavevectorsproduced by the operators of g acting on k.

h(star) Number of distinct wavevectors in s(k).Array of k The set of wavevectors produced by the operators of g acting on k.

The pure point-group operations form a subgroup of the space group, G. Wedenote this subgroup by the symbol, g, and its operations by Pα, and use h(g) forthe number of operators in g.

Given a wavevector, k, we can form h(g) wavevectors by applying the operatorsof g to k,

Pαk = α−1k, (9.61)

where α runs over all h(g) operations of g. We shall refer to the collection of thesevectors as the array of the k-vector. The h(g) wavevectors of the array are notnecessarily distinct. For a general point k in the first Brillouin zone, all the vectorsare distinct, but when k = 0 or k corresponds to a point or line of high symmetryor to the surface of the Brillouin zone the number of distinct k-vectors producedwill usually be less than h(g).

The array of k-vectors can be separated into two sets: (a) those vectors Pαk thatare equivalent to k and (b) those that are inequivalent to k. The set of operationsof g that produces the mutually equivalent vectors forms the group of the k-vectoror the group of the wavevector, gk. We represent the number of elements in gk byh(k). The set of distinct, mutually inequivalent vectors forms what is called thestar of the wavevector.

The star, denoted as s(k), is defined to include the original k. For a k havingno special symmetry properties (general point), s(k) contains all of the distinct k-vectors generated by all of the Pαk. In this case, gk has only the identity operation.If, however, the k-vector is a special symmetry point, the number of k-vectors inthe star will be smaller than in the case of the general point and the operatorsbelonging to gk form a larger group.

If we select any of the vectors of the star of k, say k′, and apply the operations ofg to it, the same array of k-vectors is produced again. The operators that produceequivalent k′-vectors need not be the same as those that produce equivalent k-vectors, but they form a group that is isomorphic to gk. In general, if the number ofoperators of gk is h(k) and the number of distinct vectors of the star is h(star), then

h(g) = h(k)h(star). (9.62)

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9.6 Point-group symmetry of Bloch waves 257

Pσv3

Pσv2

Pσh1

Pσv1

Pσh2

Pσh3

PC3 , PC2, PC−16

, PC−13

, PC6

(b)(a) (c)

Figure 9.6 Symmetry operations for C6v: (a) six-fold axis, (b) three σv reflectionlines, and (c) three σh reflection lines.

As examples of the star and the group of the wavevector, consider a simplehexagonal reciprocal lattice in two dimensions. The lattice is formed by the primi-tive reciprocal-lattice vectors a∗ and b∗ (given in Table 9.1). As shown in Fig. 9.4,the reciprocal lattice is hexagonal and so is the first Brillouin zone.

The elements consist of the identity, rotations about a six-fold axis, and six mir-ror planes perpendicular to the plane of the paper. By class, the rotations include(a) E , (b) 2C6 (±60◦ rotations), (c) 2C3 (±120◦ rotations), and (d) C2 (180◦ rota-tion). By class the reflection planes are 3σv (planes passing though the corners ofthe hexagon) and 3σh (planes bisecting the edges of the hexagon). The symmetryoperations are shown in Fig. 9.6.

For k = 0 (� in the Brillouin zone) all of the vectors Pαk = 0, and therefore gk

= g = C6v. The star of the k-vector is just k. Now consider a non-zero k-vector fromthe origin (�) to an arbitrary, general point within the first Brillouin zone. For thetwo-dimensional hexagon the point-group operations of the space group applied tothe k-vector produces 12 inequivalent k-vectors as illustrated in Fig. 9.7(b). Themutually inequivalent k-vectors form the star of the k-vector. The only operationthat transforms k into itself or into (k + K) is the identity operation. Thereforethe group of the k-vector, gk, is just the identity. The star of the k-vector has 12inequivalent vectors. Next consider a k-vector extending from � to some pointon the � line, but not extending to M, as shown in Fig. 9.7(c). The point-groupoperations that transform k into itself or (k + K) are PE and Pσh2. In this casethese two operations form gk, the group of the k-vector. The six vectors shown arethe star of the k-vector. Figure 9.7(d) shows a k-vector extending from � to M.In this case there are four operations in gk, namely E , PC2, Pσv3, and Pσh2. PC2

and Pσv3 transform k into k + K, where K = a∗ + b∗. The star of the k-vectorconsists of k and the two dashed vectors in Fig. 9.7(d). Finally, Fig. 9.7(f) showsthe vector extending from � to K. In this instance all of the point-group operations

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258 Space groups and crystalline solids

MΓΣ

Λ K

a∗

12

3

4

56 k

k k

k k

b∗

PC2k

(a)

(f)(e)(d)

(c)(b)

Figure 9.7 (a) The first Brillouin zone, showing the high-symmetry points andlines. (b) An arbitrary, general k-vector within the first Brillouin zone. Only theidentity transforms k into an equivalent vector. The 12 operations of C6v appliedto k produce 12 inequivalent k-vectors that form the star of the wavevector. (c) kalong the line � in the first Brillouin zone. The operations E and σh2 form gkin this case. The star of the wavevector has six mutually inequivalent vectors.(d) k extends from � to the point M on the edge of the Brillouin zone. The gkgroup consists of PE , Pσh2 , PC2 and Pσv3 . The last two operations transform kinto k+K, where K is a reciprocal-lattice vector a∗ +b∗ as illustrated in (e). Thestar consists of the three, distinct, non-equivalent vectors labeled 1, 2, and 3. Thedashed vector labeled 4 is equivalent to 3 (k4 = k3 + a∗), and the dashed vector5 is equivalent to 2 (k5 = k2 + b∗). (f) k extends from � to K. All six vectors(each generated twice) are equivalent, so the star has only the vector k. The groupof the wavevector is g(C6v). For k = 0, gk is also C6v .

of the space group transform k into k or a (k + K)-vector, and gk = g is the fullpoint group, C6v. The star consists of k alone.

9.7 The space group of the k-vector, gsk

To simplify the notation we shall denote a pure point-group operation, P{α|0}, byPα, and a pure translation, P{E |R}, by PR. Also we shall use gR to represent thesubgroup of pure lattice-vector translations.

The group of the wavevector, gk, consists of operators, Pα, for which k and Pαkare equivalent. To denote when Pα is an operation belonging to gk we shall usethe notation P(k)α. Thus Pα is a member of the space group for any α, but P(k)αincludes only the Pα that belong to gk.

The space group of the k-vector, gsk, includes both point-group operations and

translations. For a symmorphic group the operators of gsk are of the type P(k)α PR,

where R is any lattice vector. The space group gsk consists of products of P(k)α

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9.8 Irreducible representations of gsk 259

and PR. The operators P(k)α and PR do not commute, but the set of all pure trans-lations, {PR}, and the set of all pure point-group operations, {P(k)α}, do commute(see Appendix B for a discussion of a factor group). The space group gs

k is thedirect product of gk and gR,

gsk = gk × gR. (9.63)

gsk is a group of importance when analyzing the energy bands or wave-like

phenomena of crystals.

9.8 Irreducible representations of gsk

The IRs of gsk are the direct products of the IRs of gk and gR. We have already

shown that the Bloch waves ψnk(r) are basis functions for the IRs of gR. SincePRψnk(r) = exp(−ik · R)ψnk(r) the IRs are diagonal matrices,

D(PR)k′k = e−ik·Rδk′k. (9.64)

The group gk is a set of point-group operations, isomorphic to one of the 32crystallographic point groups described in Appendix B. As mentioned above, thedistinct functions produced by P(k)αun(k) = un(αk) are basis functions for a rep-resentation of gk. If there are, in number, n(gk) distinct un(αk) functions they formthe basis for an n(gk)-dimensional representation of gk. For notational simplicity,let us designate these distinct u-functions as ui

nk, where i = 1, 2, 3, . . . , n(gk).Since the u-functions are a basis for a representation, we have that

P(k)α uink =

∑ν

[D(α)�]νi uνnk, (9.65)

where � is the representation based on the unκ functions and [D(α)�]νi is the νimatrix element. The representation is usually reducible and can be decomposedinto the IRs of gk by the usual methods. The group gk is sometimes called the littlegroup, and an IR of gk is called a small representation. The above discussion ofrepresentations may seem a bit confusing at first. However, an example may serveto clarify the situation.

As noted in Section 9.5.1, for the two-dimensional hexagonal lattice, g is C6v

and gk is C6v or a subgroup of C6v depending on k. For a simple cubic Bravaislattice, g will be Oh and gk will be Oh or a subgroup of the Oh . As a result, the IRsof gk can be obtained in the same way as for point groups. Let [D j (α)k]κν be thematrix element of the j th IR of gk; then the matrix elements of the IR for gs

k are

D j (α,R)κν = e−ik·R [D j (α)]κν. (9.66)

Thus, for symmorphic space groups, the IRs for a given k are determined by thepoint-group IRs of gk and a phase factor corresponding to the translation, R.

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260 Space groups and crystalline solids

R

Σ

Λ

MZ

S

X

TΔΓ

X

X

kx

ky

kz

Figure 9.8 The Brillouin zone for a simple cubic Bravais lattice. The reducedBrillouin zone is the tetrahedron bounded by the lines �–X, X–M, M–�, �–R,R–X, and R–M, which represents 1/48 of the total Brillouin zone.

9.9 Compatibility of the irreducible representations of gk

At �(k = 0) the group of the k-vector, gk, is the full point group of the Bravaislattice, g. The matrix elements of the IRs are the same as those for the point group.As k is extended away from � to k′ the symmetry is usually reduced and gk′ corre-sponds to a subgroup of g. At a general point in the Brillouin zone, the symmetrymay be reduced to a group containing only the identity. As a result, in most casesthe number of degenerate Bloch-wave solutions will be reduced, and therefore thedimensions of the IRs will diminish. In general, the IRs of gk can be decomposedinto the IRs of the lower symmetry group, gk′ . The process is analogous to thesplitting of degenerate atomic wavefunctions by a crystal field. However, in thisinstance the symmetry-lowering factor is the k-vector of the Bloch wave. If �k isan IR of gk, and �k′ is an IR of gk′ , the two IRs are said to be compatible if andonly if �k′ is contained in the decomposition of �k. The concept of compatible IRsprovides a method of connecting energy bands as the wave-vector traverses somecircuit in the Brillouin zone. We shall now present a few examples for a cubiclattice to illustrate the principles outlined above.

The first Brillouin zone and special k-vector points for a simple cubic solid areillustrated in Fig. 9.8. (A discussion of the Oh group may be found in AppendixE.) The volume bounded by the lines �–X, X–M, M–�, �–R, R–X, and R–M is1/48 of the total volume. If this volume is subjected to all 48 operations of Oh ,the entire volume of the first Brillouin zone is mapped out.2 Therefore, all of thepossible energies, Enk, and the IRs of gk can be obtained from this segment ofthe first Brillouin zone. The segment is called the irreducible Brillouin zone, thereduced Brillouin zone, or just the reduced zone. Any k in the first Brillouin zone

2 The statistical weights of points on the bounding surface of the reduced zone are 1/ns, where ns is the numberof segments that share the point.

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9.9 Compatibility of the irreducible representations of gk 261

Table 9.2 The character table for gk = Oh at � (k = 0). The labels for the IRsare listed in the first three columns. The first column is the label used inpoint-group theory. The second and third columns are labels used in solid-statephysics; in the second the ± superscripts indicate the parity of a basis functioncorresponding to the IR

IR E 8C3 3C34 6C4 6C2 i 8iC3 3iC3

4 6iC4 6i C2

A1g �1(�+1 ) �1g 1 1 1 1 1 1 1 1 1 1

A2g �2(�+2 ) �2g 1 1 1 −1 −1 1 1 1 −1 −1

Eg �12(�+12) �3g 2 −1 2 0 0 2 −1 2 0 0

T1u �15(�−15) �4u 3 0 −1 1 −1 −3 0 1 −1 1

T2u �25(�−25) �5u 3 0 −1 −1 1 −3 0 1 1 −1

A1u �′1(�−1 ) �1u 1 1 1 1 1 −1 −1 −1 −1 −1

A2u �′2(�−2 ) �2u 1 1 1 −1 −1 −1 −1 −1 1 1

Eu �′12(�−12) �3u 2 −1 2 0 0 −2 1 −2 0 0

T1g �′15(�+15) �4g 3 0 −1 1 −1 3 0 −1 1 −1

T2g �′25(�+25) �5g 3 0 −1 −1 1 3 0 −1 −1 1

that does not belong to the reduced zone can be obtained from some k1 that doesbelong to the reduced zone by the relation Pαk1 = k, and therefore Ek1 = Ek.Although the star of k and that of k1 may have different vectors, the group of gk1

is isomorphic to the group of gk, and hence they have the same IRs. The charactertable and IRs for gk = Oh are given in Table 9.2. For any model, an energy bandstate must belong to one of these IRs of gk. Different bands (different n values ofthe unk(r) functions) may belong to different IRs or may serve as different rows ofthe same IR. The question of what band belongs to which IR depends on the basisfunctions selected for the energy band model.

Consider a k-vector extending from � to some point on the �-line but short ofR. The operations of g which transform k into an equivalent vector are E , 2C3, and3iC2. (The precise elements can be determined from the operations discussed inAppendix E. They are E , C3 about axis 4, C2

3 about axis 4, i C2 about axis 2, i C2

about axis 3, and i C2 about axis 5.) These operations form a group that is isomor-phic to C3v. Therefore, for a k-vector along the �-line, gk abruptly changes fromOh for k = 0 to C3v. However, if k extends to the R-point, gk becomes Oh again.The character tables for C3v and C4v are given in Table 9.3 and Table 9.4, respec-tively. The labels of the IRs are given for the atomic and solid-state conventions.The IRs at � must decompose into IRs of C3v for k along the �-line. It may beseen that the dimensions of the IRs for C3v are 1 for �1(A1), 1 for �2(A2), and2 for �3(E). Therefore any three-dimensional IR at � must split into at least two

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262 Space groups and crystalline solids

Table 9.3 The character table for C3v . The operator classes shownare those from Appendix E. The specific operations within theclasses are also indicated.

C3v 2C3 3i C2

IRs E C (4)3 , (C (4)

3 )2 i C (2)2 , i C (3)

2 , i C (5)2

�1(A1) 1 1 1�2(A2) 1 1 −1�3(E) 2 −1 0

Table 9.4 The character table for C4v . The operations are those from AppendixE. The equivalent reflections, σd and σv for C4v , are also indicated.

C4v 2C4 C2 2iC2 = 2σv 2i C2 = 2σd

IRs E C (1)4 , (C (1)

4 )3 C (1)2 iC (2)

2 , iC (3)2 i C (1)

2 , i C (2)2

�1(A1) 1 1 1 1 1�1′(A2) 1 1 1 −1 −1�2(B1) 1 −1 1 1 −1�2′(B2) 1 −1 1 −1 1�5(E) 2 0 −2 0 0

IRs of C3v for k along the �-line (excluding R). Table 9.5 shows how the Oh IRsdecompose into those of C3v.

Similarly, for k along the y-axis, the �-line, gk is isomorphic to C4v. Thedecompositions of the IRs of Oh into those of C4v are also shown in Table 9.5.

As an example, consider the energy bands for a simple-cubic lattice with p-orbitals on each site, with on-site and nearest-neighbor interactions. There are threeorbitals in each unit cell, so there will be three energy bands. For simplicity we canuse orthogonal Löwdin orbitals and assume solutions of the form

ψk(r) =∑

R

∑α

Aα(R) pα(r− R), (9.67)

where the sum over R is over all lattice vectors. Owing to the translationalsymmetry we may take Aα(R) = exp(ik · R) Aα(0), with α = x , y, or z.Consequently,

ψnk(r) =∑

R

∑α

Aα(0) eik·R pα(r− R) (α = x, y, or z). (9.68)

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9.9 Compatibility of the irreducible representations of gk 263

Table 9.5 Decomposition of the IRs of Oh into the IRsof C3v and C4v . The first column is the IR of Oh. Thesecond column is the decomposition into the IRs ofC3v . The numbers in parentheses are the dimensionsof the representations. The third column is thedecomposition into the IRs of C4v .

�(l) IRs of C3v IRs of C4v

�1(1) �1(1) �1(1)�2(1) �2(1) �2(1)�12(2) �3(2) �1(1)+�2(1)�15(3) �1(1)+�3(2) �1(1)+�5(2)�25(3) �2(1)+�3(2) �2(1)+�5

�′1(1) �1(2) �′1(1)�′2(1) �1(1) �′2(1)�′12(2) �3(2) �′1(1)+�′2(1)�′15(3) �2(1)+�3(2) �′1(1)+�5(2)

�′25(3) �1(1)+�3(2) �′2(1)+�5(2)

Note that the wavefunction of (9.68) can be cast in the Bloch-wave form, ψk(r) =uk(r) exp(ik · r), where uk(r) = ∑

R

∑α Aα(0) e−ik·(r−R) pα(r− R) is clearly

invariant under any lattice-vector translation, r→ r+ R′.The energies of the three bands are given by

Eαβγ (k) = εp + 2(ppσ) cos(kαa)+ 2(ppπ)[cos(kβa)+ cos(kγ a)], (9.69)

where α, β, and γ are any of the three combinations kx , ky , kz; ky , kx , kz; or kz , kx ,ky , respectively. At �, gk is Oh . The three bands are degenerate, with Eαβγ (�) =εp + 2(ppσ)+ 4(ppπ), and belong to the �15(T1u) IR.

For k = (kx , 0, 0), k = (0, ky, 0), or k = (0, 0, kz), gk is isomorphic to thegroup C4v, and the energies are

E xyz(�1) = εp + 2(ppσ) cos(kxa)+ 4(ppπ), (9.70)

E yxz(�5) = Ezxy(�5) = εp + 2(ppσ)+ 2(ppπ)[cos(kxa)+ 1]. (9.71)

Along the �-line, the three-fold degeneracy present at � evolves into a non-degenerate �1 energy band and a two-fold degenerate �5 energy band as indicatedin Table 9.5. The wavefunctions are

ψ�1(r) =1

N 3/2

∑R

eikx Rx px(r− R), (9.72)

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264 Space groups and crystalline solids

ψ(1)�5

(r) = 1

N 3/2

∑R

eikx Rx py(r− R), (9.73)

ψ(2)�5

(r) = 1

N 3/2

∑R

eikx Rx pz(r− R), (9.74)

where we assume periodic boundary conditions of order N along the x-, y-, andz-axes. The superscripts (1) and (2) indicate row-1 and row-2 partner functions forthe �5 IR, respectively.

Along the �-line k = kx = ky = kz . The eigenvectors transform as �1(A1) and�3(E). The energies of the �1 and �3 bands are accidentally degenerate, with

E = εp + [2(ppσ)+ 4(ppπ)] cos(ka). (9.75)

The wavefunctions can be generated using the symmetry-function-generatingmachine. We find that

ψ�1(r) =1√N 3

∑R

ei(kx Rx+ky Ry+kz Rz)1√3

∑α

pα(r− R), (9.76)

ψ(1)�5

(r) = 1√N 3

∑R

ei(kx Rx+ky Ry+kz Rz)1√2[px(r− R)− py(r− R)],

(9.77)

ψ(2)�5

(r) = 1√N 3

∑R

ei(kx Rx+ky Ry+kz Rz)

× 1√6[px(r− R)+ py(r− R)− 2pz(r− R)]. (9.78)

9.9.1 Representations of non-symmorphic space groups

A non-symmorphic space group, G, contains operations {α|R+ t}, where t is not alattice vector. Such an element can not be separated into a product of operators ofG of the form {E |R+t}{α|0}, since {E |R+t} is not an operation of G. This featuredoes not change the procedure for obtaining the IRs of the group of the wavevectorprovided that k does not lie on the boundary of the Brillouin zone. When k lies onthe zone boundary, special theoretical methods involving “multiplier groups” arerequired in order to effectively separate the representations of point-group oper-ations from the translation representations. These methods will not be discussedhere, but may be found in the literature [9.2, 9.3].

In any case, we note that the Bloch waves still form the basis for the repre-sentations and the space-group IRs are still derived from the usual point-grouprepresentations for the group of the k-vector. However, the characters for opera-tions involving t, a non-lattice vector, may acquire an additional phase factor of theform exp(ik · t).

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9.10 Energy bands in the plane-wave approximation 265

9.10 Energy bands in the plane-wave approximation

We consider the wavefunctions for a single electron in a crystal with a periodicpotential V (r) that satisfy the Schrödinger eigenvalue equation,

H ψnk =[− �2

2m∇2 + V (r)

]ψnk = Enk ψnk, (9.79)

where ψnk is a normalized Bloch wave,

ψnk = 1√!

eik·r unk(r), (9.80)

with !, the normalization factor, equal to the volume of the crystal.Substitution of (9.80) into Schrödinger’s equation yields an equation for unk,

− �2

2m

[∇2 − k2 + 2ik · ∇

]unk + V (r) unk = Enk unk, (9.81)

or

− �2

2m∇2unk +

[V (r)+ �2

mik · ∇ + �2k2

2m

]unk = Enk unk. (9.82)

The function unk(r) can be expressed as a Fourier series in reciprocal-lattice space,

unk(r) =∑

K

Ank(K) eiK·r, (9.83)

where Ank(K) are the expansion coefficients and K runs over all of the reciprocal-lattice vectors. We see that the functions of (9.83) have the proper periodicity, since

unk[r+ R(l,m, n)] =∑

K

Ank(K) eiK·reiK·R(l,m,n) = unk(r), (9.84)

where we have used the fact that exp[iK · R(l,m, n)] = 1.Using (9.83) in (9.82) gives

∑K

{�2

2m[K2 + k2 + 2k · K] + V (r)− Enk

}Ank(K) eiK·r = 0. (9.85)

Multiply (9.85) by (1/!)1/2 exp(−iK′ · r) and integrate over all space to get[�2

2m(K′ + k)2 − Enk

]Ank(K′) = −

∑K

VKK′ Ank(K), (9.86)

where

VKK′ = 1

!

∫dr V (r)ei(K−K′)·r. (9.87)

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266 Space groups and crystalline solids

In obtaining (9.86) we have used the orthogonality of the plane waves, namely

1

!

∫dr ei(K−K′)·r = δ(K−K′). (9.88)

From (9.87) it can be seen that VKK′ depends only on the vector difference K−K′

and the diagonal matrix element, when K = K′, is a constant, and does not dependupon K or K′. VKK is the integral of V (r) over the entire crystal divided by thecrystal volume. Therefore VKK is the average of V (r) over a unit cell. Since VKK isindependent of K, we shall denote it by the symbol V0.

The off-diagonal term, VKK′ = V(K−K′) = VK′′ , is the Fourier transform ofV (r) for the component K′′ = K − K′. Thus increasing values of |K′′| measureincreasingly rapid spatial variations of V (r) within the unit cell.

Equation (9.86) defines a matrix eigenvalue problem for the determination ofthe energy eigenvalues, Enk, and the Fourier expansion coefficients (eigenvectors),Ank(K). The matrix equation is of degree N (of the order of the number of atomsin the crystal) and therefore actual solutions of (9.86) are not possible. In practiceapproximate solutions are found by truncating the equations for |K′′| > K0, whereK0 is an arbitrarily chosen number.

9.10.1 Constant potential

It is useful to investigate the energy bands for a constant potential, V (r) = V0. Forthis case the eigenvalue equation (9.86) reduces to[

�2

2m(Kn + k)2 + V0 − Enk

]Ank = 0. (9.89)

Kn is uncoupled to any other reciprocal-lattice vector. That is, the eigenvaluematrix is diagonal and the eigenvalues are

Enk = �2

2m(Kn + k)2 + V0. (9.90)

Each reciprocal-lattice vector Kn has associated with it an energy band whoseenergy as a function of k is given by (9.90). Here k is a wavevector restrictedto the first Brillouin zone. From (9.90) we can see that the energy bands for a non-zero V0 and for V = 0 (called the empty-lattice model) are the same except for aconstant shift in the zero of energy.

As an example, consider the energy bands for a simple cubic lattice and write

Kn = 2π

a(nx , ny, nz), (9.91)

k = 2π

a(κx , κy, κz), (9.92)

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9.10 Energy bands in the plane-wave approximation 267

where the components of n = (nx , ny, nz) are integers that specify the reciprocal-lattice vector, Kn . The energy bands (9.90) can then be written in dimensionlessform,

εnκ =(

a

)2 (2m

�2

)(Enκ − V0)

= (nx + κx)2 + (ny + κy)

2 + (nz + κz)2. (9.93)

For the first energy band, n = (0, 0, 0) and ε(0,0,0)κ = κ2. For the next six lowestbands, having n = (±1, 0, 0), (0,±1, 0), and (0, 0,±1), the energies are given by

ε(±1,0,0)κ = (1± κx)2 + κ2

y + κ2z , (9.94)

ε(0,±1,0)κ = κ2x + (κy ± 1)2 + κ2

z , (9.95)

ε(0,0,±1)κ = κ2x + κ2

y + (κz ± 1)2. (9.96)

The k-vector is restricted to the first Brillouin zone, so for the simple cubic system−1/2 ≤ κx , κy, κz ≤ 1/2. The next 12 lowest bands are for n = (±1,±1, 0),(±1, 0,±1), and (0,±1,±1) (the plus and minus signs on the components may bechosen independently), and the energies are

ε(±1,±1,0)κ = (1± κx)2 + (1± κy)

2 + κ2z , (9.97)

ε(±1,0,±1)κ = (1± κx)2 + κ2

y + (1± κz)2, (9.98)

ε(0,±1,±1)κ = κ2x + (1± κy)

2 + (1± κz)2. (9.99)

For the cases of κ = (t, 0, 0) and κ = (t, t, 0) with t varying from 0 to 1/2 (i.e., kvaries from � to X and � to M) the lowest 19 energy bands are given in Table 9.6and graphs are shown in Fig. 9.9.

The statement Enk = En(k+K) can lead to some confusion, particularly whendealing with the plane-wave approximation. Clearly the energy expressed in (9.90),Enk = (�2/(2m))(Kn + k)2+ V0, is not equal to En(k+K) = (�2/(2m))(Kn + (k+K))2+ V0, unless K = 0. The problem is that the expressions for the energies holdonly for k restricted to the first Brillouin zone and for the same band (same Kn).The energy given above for En(k+K) is actually the energy for the band En′k, wheren′ corresponds to the band whose reciprocal-lattice vector is Kn + K, that is, theenergy for a different band. If we let k go beyond the first Brillouin zone, then theenergy bands simply repeat themselves, as shown in Fig. 9.10 for two bands alongthe �–X direction.

9.10.2 Reciprocal-lattice-vector and plane-wave representations of gk

The Bravais-lattice vectors and also the reciprocal-lattice vectors are bases forrepresentations of the space group, G. The group of the k-vector is a subgroup

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268 Space groups and crystalline solids

Table 9.6 Dimensionless energies and irreducible representations (a) forκ = (t, 0, 0), with k along the [100] direction, and (b) for κ = (t, t, 0), with kalong the [110] direction. The degeneracy (D) of a particular band is shown incolumn 4. The numerical band label (arbitrary) given in column 5 (BL) is used asidentification in Fig. 9.9.

(a) κ = (t, 0, 0) εnκ κ+ n = (t + l,m, n) D BL

�1(A1) t2 (t, 0, 0) 1 1�1(A1) (t + 1)2 (t + 1, 0, 0) 1 4�1(A1) (t − 1)2 (t − 1, 0, 0) 1 2�1 +�2 +�5 t2 + 1 (t, 1, 0), (t, 0, 1), 4 3

(t,−1, 0), (t, 0,−1)�1 +�2 +�5 (t + 1)2 + 1 (t + 1, 1, 0), (t + 1,−1, 0), 4 7

(t + 1, 0, 1), (t + 1, 0,−1)�1 +�2 +�5 (t − 1)2 + 1 (t − 1, 0, 1), (t − 1, 1, 0), 4 5

(t − 1,−1, 0), (t − 1, 0,−1)�1 +�2 +�5 t2 + 2 (t, 1, 1), (t,−1, 1), 4 6

(t, 1,−1), (t,−1,−1)

(b) κ = (t, t, 0) εnκ κ+ n = (t + l, t + m, n) D BL

�1(A1) 2t2 (t, t, 0) 1 1�1(A1)+�2(A2) (t + 1)2 + t2 (t + 1, t, 0), (t, t + 1, 0) 2 4�1(A1)+�2(A2) (t − 1)2 + t2 (t − 1, t, 0), (t − 1, 0, t) 2 2�1(A1)+�3(B1) 2t2 + 1 (t, t, 1), (t, t,−1) 2 3�1(A1) 2(t + 1)2 (t + 1, t + 1, 0) 1 9�1(A1) 2(t − 1)2 (t − 1, t − 1, 0) 1 5�1(A1)+�3(B1) 2(t2 + 1) (t + 1, t − 1, 0), (t − 1, t + 1, 0) 2 7�1 +�2 +�3 +�4 (t + 1)2 + t2 + 1 (t, t + 1,±1), (t + 1, t,±1) 4 8�1 +�2 +�3 +�4 (t − 1)2 + t2 + 1 (t, t − 1,±1), (t − 1, t,±1) 4 6

of G, and therefore the dimensionless reciprocal-lattice vectors, n, are also basesfor representations of gk. The dimensionless reciprocal-lattice vectors n + κ alsoare bases for representations of gk. Finally, of course, the Bloch waves are basisfunctions for a representation of the space group gs

k. In general, however, theserepresentations are reducible. The energy bands are labeled by the IRs of thegroup of the k-vector. We can find which energy bands are labeled by whichIRs by decomposing the reducible representations into the IRs of gk. Since nopoint-group operation of gk changes the length of a vector, a representation of gk

in terms of reciprocal-lattice vectors will involve only vectors having the samelength. For example, the set of six vectors (±1, 0, 0), (0,±1, 0), (0, 0,±1) pro-vides bases for a six-dimensional reducible representation. However, if, instead,

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9.10 Energy bands in the plane-wave approximation 269

0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

1

2

3

4

5

6

7

Δ1 + Δ3 + Δ5

Δ1

Δ1

Γ (t, 0

(a) (b)

, 0) X0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

1

2

3

4

5

6

7

8

9

Γ (t, t, 0) M

Figure 9.9 Energy bands for the simple-cubic lattice. The IRs and associatedreciprocal-lattice-vector degeneracies for each band are shown in Table 9.6.(a) Energy bands �–X for κ = (t, 0, 0). (b) Energy bands �–M for κ = (t, t, 0).

we use the (K+ k)-vectors, we see that along the �–X line there are two differentlengths, |(κx + 1, 0, 0)| and |(κx − 1, 0, 0)|, and four equal lengths |(κx ,±1, 0)|and |(κx , 0,±1)|. Therefore, the (K+ k)-based representation is useful because itis already partially decomposed. The two representations based on (κx + 1, 0, 0)and (κx − 1, 0, 0) are one-dimensional, so they must be irreducible. The remainingfour vectors are bases for a four-dimensional (reducible) representation. The Blochwaves exp[i(K+k)·r] form a reducible representation that is the same as that asso-ciated with the (K+ k)-vectors. This follows because, if Pα(K+ k) = α(K+ k),then Pα exp{i(K+ k) · r} = exp{iα(K+ k) · r}. The decomposition of a reduciblerepresentation is accomplished in the usual manner by calculating the trace of thematrices or by using the symmetry-function-generating machine to find the linearcombinations of the vectors which belong to a particular row of an IR of gk.

As an example let us find the IRs corresponding to the energy bands of Table 9.5shown in Fig. 9.9. We start with the energy band corresponding to κ = (κx , 0, 0)

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270 Space groups and crystalline solids

0.0

0.5

1.0

1.5

2.0

2.5

0.0Γ

κ 0.5X

1.0Γ

1.5X

2.0Γ

n = (1, 0, 0)

n = (1, 0, 0)

(0,0,0)κ

(1,0,0)κ

(0,0,0)(κ+n)

(1,0,0)(κ+n)

Figure 9.10 The extended Brillouin zone showing εnκ = εn(κ+n) (for band 1 andband 4 of Fig. 9.9) for a simple-cubic crystal along the �–X direction.

along the line �–X. The group of the k-vector is C4v and its operations are givenin Table 9.2. The vector k = (kx , 0, 0) is invariant under all of the operationsof gk. The vectors K + k = (2π/a)(1 ± nx , 0, 0) are also invariant under all ofthe operations of gk and therefore these IRs must belong to the totally symmet-ric IR, �1(A1). Each of the corresponding Bloch waves, exp[i(2π/a)nx x] andexp[i(2π/a)(1± nx)x], is a symmetry function for a �1 IR.

The next set of vectors, (κ,±1, 0) and (κ, 0,±1), is associated with degeneratestates, and therefore these vectors are bases for a four-dimensional representa-tion. To decompose the representation based on these vectors we need to findtheir behavior under the operations of C4v. This can be conveniently accomplishedusing the action table in Appendix E, Table E.1. To employ the table we needto recognize that a reciprocal-lattice vector, (nx , ny, nz), transforms as (x, y, z)under the operations of the group. To simplify the notation, define (κ, 1, 0) = A,(κ,−1, 0) = B, (κ, 0, 1) = C, and (κ, 0,−1) = D. The action of P(i C (1)

2 ) on A,for example, is, according to Table E.1, x → x , y → −z, and z → −y. Therefore(1, 1, 0) → (1, 0,−1), that is, A → D. Table 9.7 shows the results of operatingwith the elements of C4v on the four basis functions.

The characters shown in the last row of Table 9.7 are the sums of the number offunctions that transform into themselves less the number going into the negative ofthemselves. The decomposition into the IRs of C4v is �(�) = �1+�2+�5. Thisresult is also shown in Table 9.6 in the first column.

We can also determine the symmetry functions for the IRs of the decomposi-tion by employing the symmetry-function-generating machine and Table 9.7. The(un-normalized) symmetry function for �1 is f (�1) ∝ (A+ B+ C+ D). For the

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9.10 Energy bands in the plane-wave approximation 271

Table 9.7 The action table and characters of the four-dimensional representationof C4v based on the vectors A, B, C, and D: (κ, 1, 0) = A, (κ,−1, 0) = B,(κ, 0,+1) = C, and (κ, 0,−1) = D. The characters for the four-dimensionalrepresentation are listed in the last row.

�(�) E C (1)2 C (1)

4 (C (1)4 )3 iC (2)

2 iC (3)2 i C (1)

2 C (2)2

A A B D C B A D CB B A C D A B C DC C D A B C D B AD D C B A D C A Bχ 4 0 0 0 2 2 0 0

�2 symmetry function, f (�2) ∝ (A+ B− C− D). To obtain the symmetry func-tions for �5 we must use the matrix elements of the E IR. These matrix elementsare given in Appendix E, in Section E.4. Using the diagonal matrix elements of Egives the row-1 function as f1(�5) ∝ (2A+ 2B+ C+ D). The row-2 function isf2(�5) ∝ (2A+ 2B− C− D). These results should not be interpreted as a simplevector sum, since, for example, the vector sum of A+ B+ C+ D = (4κ, 0, 0). Touse the results, we note that the Bloch wavefunctions of the set exp[i(2π/a)A · r],exp[i(2π/a)B · r], exp[i(2π/a)C · r], and exp[i(2π/a)D · r] transform under theoperations of gk in exactly the same way as A, B, C, and D. The utility of aresult such as f (�1) ∝ (A+ B+ C+ D) is that ψ�1

κ (r) ∝ {exp[i(2π/a)A · r]+ exp[i(2π/a)B · r] + exp[i(2π/a)C · r] + exp[i(2π/a)D · r]} is a symmetryfunction belonging to the �1 IR of gk. The sum of Bloch waves, ψ�1

k (r), can betaken in a literal mathematical sense:

ψ�1k (r) ∝

[ei(2π/a)A·r + ei(2π/a)B·r + ei(2π/a)C·r + ei(2π/a)D·r

]= eikx x

[cos

(2πy

a

)+ cos

(2π z

a

)], (9.100)

where kx is the x-component of the k-vector, (2πκ/a). Similarly,

ψ�2k (r) ∝ [

ei(2π/a)A·r + ei(2π/a)B·r − ei(2π/a)C·r − ei(2π/a)D·r]= eikx x

[cos

(2πy

a

)− cos

(2π z

a

)], (9.101)

is a Bloch wave belonging to the �2 IR of gk. Since the Bloch waves are of the formexp(ik · r) unk(r), for the previous example unk(r) = [cos(2πy/a)− cos(2π z/a)].The degeneracies evident in Table 9.6 hold for −1/2 < κ < 1/2, but changeat the Brillouin-zone center and boundaries, κ = 0 and κ = ±1/2. At �

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272 Space groups and crystalline solids

(k = κ = 0), the degree of degeneracy is maximal, since all bands associ-ated with K-vectors of the same length are degenerate, ε(0,0,0)0 = 0, ε(±1,0,0)0 =ε(±1,0,0)0 = ε(0,±1,0)0 = 1, and ε(±1,±1,0)0 = ε(±1,0,±1)0 = ε(0,±1,±1)0 = 2. At �,gk = Oh and the reducible representation, �1, associated with the set of reciprocalvectors {(±1, 0, 0), (±1, 0, 0), (0,±1, 0)} is six-dimensional. The decompositionis �1 = �1g(A1g) + �3g(Eg) + �4u(T1u). As k moves away from κ = 0 alongthe [100] direction, the IRs of Oh evolve as determined by decomposing theminto the IRs of C4v: �1g(A1) → �1(A1), �3g(Eg) → �1(A1) + �2(B1), and�4u(T1u)→ �1(A1)+�5(E).

At k = X = (π/a, 0, 0), κ = (1/2, 0, 0). The group of the k-vector increasesin size because +κ and −κ are equivalent, κ = −κ + (1, 0, 0), and the inversionoperation is now a member of the group. As a result, the group of the k-vectorchanges from C4v (along �) to D4h (at X). To find the appropriate operations forgk use Table E.1 and identify any operation that takes x into x or −x . There areeight such operations, namely E , C (1)

2 , C (2)2 , C (3)

2 , C (1)4 , (C (1)

4 )3, C (1)2 , and C (2)

2 .Together with inversion, there are in total 16 operations in gk. The character tablefor D4h is shown in Table 9.8.

At X the energies are given by

ε(l,m,n)κ =(

l + 1

2

)2

+ m2 + n2, (9.102)

with κ = 1/2. The energies for the first 19 bands are

ε(0,0,0) = ε(−1,0,0) = 1

4, (9.103)

ε(−1,±1,0) = ε(−1,0,±1) = ε(0,0,±1) = ε(0,±1,0) = 5

4, (9.104)

ε(1,0,0) = ε(−1,±1,±1) = ε(0,±1,±1) = 9

4. (9.105)

The states at X associated with n = (0, 0, 0) and (−1, 0, 0) having respectivelyn+ κ = (1/2, 0, 0) and n + κ = (−1/2, 0, 0)) are degenerate and correspond toBloch waves exp(+iπx/a) and exp(−iπx/a). These two waves are basis func-tions for �a, a two-dimensional representation of gk. Using Table E.1, we find thecharacters of the representation as shown in Table 9.8 for �a.

Decomposition of �a into the IRs of D4h yields

�a = X1g(A1g)+ X4u(A2u). (9.106)

The symmetry functions can be obtained by using the symmetry-function-generating machine operating on ei(πx/a). The results are f (X1g) ∝ cos(πx/a)and f (X4u) ∝ sin(πx/a).

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Table 9.8 The character table for D4h. The operations of Oh corresponding to those of D4 are C (1)2 , C (1)

4 , (C (1)4 )3, C (2)

2 , C (3)2 ,

C (1)2 , and C (2)

2 . At X, �a is the two-dimensional representation based on n+ κ = (1/2, 0, 0) and (−1/2, 0, 0). At X, �b is theeight-dimensional representation based on n+ κ = (−1/2,±1, 0), (−1/2, 0,±1), (1/2, 0,±1), and (1/2,±1, 0). The lastthree rows show how x, y, and z transform under the operations of D4h. The matrices for the E representation of D4h aregiven at the bottom of the table.

D4h E C (1)2 C (1)

4 (C (1)4 )3 C (2)

2 C (3)2 C (1)

2 C (2)2 i iC (1)

2 iC (1)4 (iC (1)

4 )3 iC (2)2 iC (3)

2 i C (1)2 i C (2)

2

X1g(A1g) 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1X4g(A2g) 1 1 1 1 −1 −1 −1 −1 1 1 1 1 −1 −1 −1 −1X2g(B1g) 1 1 −1 −1 1 1 −1 −1 1 1 −1 −1 1 1 −1 −1X3g(B2g) 1 1 −1 −1 −1 −1 1 1 1 1 −1 −1 −1 −1 1 1X5g(Eg) 2 −2 0 0 0 0 0 0 2 −2 0 0 0 0 0 0X1u(A1u) 1 1 1 1 1 1 1 1 −1 −1 −1 −1 −1 −1 −1 −1X4u(A2u) 1 1 1 1 −1 −1 −1 −1 −1 −1 −1 −1 1 1 1 1X2u(B1u) 1 1 −1 −1 1 1 −1 −1 −1 −1 1 1 −1 −1 1 1X3u(B2u) 1 1 −1 −1 −1 −1 1 1 −1 −1 1 1 1 1 −1 −1X5u(Eu) 2 −2 0 0 0 0 0 0 −2 2 0 0 0 0 0 0�a 2 2 2 2 0 0 0 0 0 0 0 0 2 2 2 2�b 8 0 0 0 0 0 0 0 0 0 0 0 4 4 0 0

x x x x x −x −x −x −x −x −x −x −x x x x xy y −y −z z y −y z −z −y y z −z −y y −z zz z −z y −y −z z y −y −z z −y y z −z −y y

E matrices. Basis functions are y and z.

E =(

1 00 1

), C (1)

2 =( −1 0

0 −1

), C (1)

4 =(

0 −11 0

), (C(1)

2 )3 =(

0 1−1 0

),

C (2)2 =

(1 00 −1

), C (3)

2 =( −1 0

0 1

), C (1)

2 =(

0 11 0

), C (2)

2 =(

0 −1−1 0

).

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274 Space groups and crystalline solids

The states at X for n = (−1,±1, 0), (−1, 0,±1), (0, 0,±1), and (0,±1, 0)with κ = (1/2, 0, 0) are also degenerate and correspond to the eight Bloch waves

ei(

2πa

)[ x

2±y], (9.107)

ei(

2πa

)[ x

2±z], (9.108)

ei(

2πa

)[− x

2±y], (9.109)

ei(

2πa

)[− x

2±z]. (9.110)

These Bloch waves form an eight-dimensional reducible representation, �b, of gk.The characters of �b can be determined using the transformation properties of x ,y, and z as shown in Table 9.8.

For example, P(iC (2)2 ) (x, y, z)→ Pi (−x, y,−z) = (x,−y, z), therefore

P(iC (2)2 ) e

i(

2πa

)[ x

2±y]→ ei(

2πa

)[ x

2∓y], (9.111)

P(iC (2)2 ) e

i(

2πa

)[ x

2±z]→ ei(

2πa

)[ x

2±z], (9.112)

P(iC (2)2 ) e

i(

2πa

)[− x

2±y]→ ei(

2πa

)[− x

2∓y], (9.113)

P(iC (2)2 ) e

i(

2πa

)[− x

2±z]→ ei(

2πa

)[− x

2±z]. (9.114)

Only the functions in (9.112) and (9.114) are transformed into themselves, so thecharacter of P(iC (2)

2 ) = 4. Proceeding in this way, we find the characters listed for�b in Table 9.8. The decomposition of �b into the IRs of D4h is

�b = X1g(A1g)+ X4u(A2u)+ X2g(B1g)+ X3u(B2u)+ X5g(Eg)+ X5u(Eu).

(9.115)

Some texts use the notation “+” and “−” instead of “g” and “u”, for example X+1and X−5 instead of X1g and X5u . The symmetry functions for these representationscan be generated from the function exp{i(2π/a)[x/2+ y]}. The results are

f (X1g) ∝ cos

(πx

a

)[cos

(2πy

a

)+ cos

(2π z

a

)], (9.116)

f (X4u) ∝ sin

(πx

a

)[cos

(2πy

a

)+ cos

(2π z

a

)], (9.117)

f (X2g) ∝ cos

(πx

a

)[cos

(2πy

a

)− cos

(2π z

a

)], (9.118)

f (X3u) ∝ sin

(πx

a

)[cos

(2πy

a

)− cos

(2π z

a

)], (9.119)

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9.10 Energy bands in the plane-wave approximation 275

f (X5g) row-1 ∝ sin

(πx

a

)sin

(2π z

a

), (9.120)

f (X5g) row-2 ∝ sin

(πx

a

)sin

(2πy

a

), (9.121)

f (X5u) row-1 ∝ cos

(πx

a

)sin

(2πy

a

), (9.122)

f (X5u) row-2 ∝ cos

(πx

a

)sin

(2π z

a

). (9.123)

The symmetry functions for the E IR are found from the matrix elementsgiven in Table 9.8 using the functions exp{i(2π/a)[x/2 + y]} and exp{i(2π/a)[x/2 + z]} as initial generating functions. The symmetry functions are the zeroth-order (constant-potential) wavefunctions for the energy band states. Since they areconstructed from degenerate functions, the energies of the symmetry functions arealso degenerate. Degeneracy associated with functions belonging to different IRs is“accidental” degeneracy. If we now allow V (r) to be non-constant so that at leastsome matrix elements, VKK′ , other than K = K′ are non-zero then these matrixelements will alter the energies and lift the accidental degeneracies. We know thatthere can be no non-zero matrix elements between states that belong to differentIRs. Therefore bands of different IRs may cross. Two bands with the same IR(and same row) may, and usually will, have non-zero interactions (matrix elementsbetween them). As a result, we expect degenerate bands of the same IRs to repelone another near intersection. This leads to energy gaps, particularly at points ofhigh symmetry. For example, in Fig. 9.9(a), bands 1 and 2 belong to the �1 IR.These bands become degenerate at the X point. Hence one expects that, to firstorder, the secular equation would be of the form

det

∣∣∣∣ E0 − ε M12

M∗12 E0 − ε

∣∣∣∣(

f1

f2

)= 0, (9.124)

where f1 and f2 are the zeroth-order wavefunctions, M12 is the matrix elementbetween them, and E0 is the degenerate energy at X for M12 = 0. The wavefunc-tions are then ψ± ∝ f1 ± f2 and the energies are ε = E0 ± |M12|. Thus an energygap of 2|M12| develops at X. According to (9.106) the degenerate states split intoX1g(A1g) + X4u(A2u), where X1g ∝ f1 + f2 and X4u ∝ f1 − f2. Figure 9.11 is aschematic diagram of energy bands showing the gaps that develop when there arenon-zero matrix elements between bands with the same IR.

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276 Space groups and crystalline solids

0.0

0.5

1.0

1.5

2.0

1

2

3

4

Δ1

Δ1

Δ1 + Δ3 + Δ5

Δ1 + Δ3 + Δ5

Γ (κ,0,0) X

Figure 9.11 Gap formation. Two bands belonging to the same IR interact andrepel one another, forming an energy gap. Shown is a schematic diagram of �1bands forming gaps at � and X.

References

[9.1] M. S. Dresselhaus, G. Dresselhaus, and A Jorio, Group Theory Application to thePhysics of Condensed Matter (Berlin: Springer-Verlag, 2008).

[9.2] M. Lax, Symmetry Principles in Solid-State and Molecular Physics (New York:Wiley, 1974).

[9.3] P. Jacobs, Group Theory with Applications in Chemical Physics (Cambridge:Cambridge University Press, 2005).

Exercises

9.1 Consider a lattice vector, V = ra + sb + tc, and the reciprocal-latticevector, V∗ = ha∗ + kb∗ + lc∗, and show that V · V∗ = 2π(hr + ks + lt).

9.2 (a) For face-centered cubic crystals each primitive cube has atoms on eachof its eight corners and atoms at the center of each of the six faces. Howmany atoms are there per unit cell?

(b) Give a set of coordinates for basis atoms of the unit cell.(c) The primitive direct-lattice vectors for the body-centered cubic lattice are

given in Table 9.1. Using the equations in (9.27), calculate the primitivereciprocal-lattice vectors.

9.3 (a) The important energy bands for perovskite crystals such as SrTiO3 andKTaO3 are derived from the d-orbitals of the transition metal ions andthe p-orbitals of the oxygen ions. How many p–d energy bands areassociated with these orbitals?

(b) Consider an electron in a simple-cubic crystal having lattice constanta = 3.5 Å with V (r) = 0 (empty lattice). What energy bands

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Exercises 277

(reciprocal-lattice vectors) are included in the range 0 ≤ EKk ≤ 100 eVat � in the first Brillouin zone?

9.4 A plane may be defined by its intercepts with the axes of the primitive vec-tors a, b, and c. Denote these intercepts as ia, ib, and ic. The Miller indices,(hkl), for the set of parallel planes are obtained from the reciprocals of theintercepts: h = (αa/ ia), k = (αb/ ib), and l = (αc/ ic), where a, b, and care the lattice constants, and α is an integer chosen so that h, k, and l arethe smallest integers having the same ratios as the reciprocal intercepts. Forexample, a plane, P , with intercepts a, 2b, and 3c has reciprocal intercepts a,b/2, and c/3, so α = 6 and (hkl) = (632). The Miller indices are the samefor all planes parallel to P (not containing the origin), and therefore specifya set of parallel planes.(a) Find the Miller indices for the sets of planes whose intercepts ia, ib, and

ic are respectively 2a, 3b, 5c; 2a, 2b, 0c; and a, b, 2c.(b) Show that the reciprocal-lattice vector K(hkl) = ha∗ + kb∗ + lc∗ is

perpendicular to the set of parallel planes designated by the Miller indices(hkl).

(c) Using the general equation K(p1, p2, p3) ·R(m1,m2,m3) = 2πn, wherepi , mi , and n are integers, show that the distance from the origin to theclosest (hkl) plane is given by 2π/|K(hkl)|.

(d) Show that the spacing between consecutive (hkl) planes is uniform.(e) What is the spacing between (123) planes for a cubic lattice with lattice

constant a?9.5 X-rays incident on a crystal scatter from the electrons of the atoms causing

interference patterns. The electron charge density, ρ(r), must be a periodicfunction, and therefore can be represented as a Fourier series in reciprocal-lattice vectors,

ρ(r) =∑

K

ρ(K)eiK·r.

(a) Show that the probability, Pkk′ , that an initial X-ray with wavefunc-tion (1/V )1/2 exp(ik · r) is scattered by ρ(r) into the final state(1/V )1/2 exp(ik′ · r) is proportional to

Pkk′ =∑

K

ρ(K)δ(k′ − k−K),

where δ(k′ − k−K) = 1 for k′ − k = K and 0 otherwise.(b) From (a) we see that the maxima for Pkk′ occur for k′ − k = K (if ρ(K)

is not zero). If the energy of the X-ray is conserved (elastic scattering),

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278 Space groups and crystalline solids

show that the maxima occur according to Bragg’s interference equation,2d sin θ = λ. Here, θ is the angle between k and the lattice plane, P ,whose normal is K, and d is the spacing between planes parallel to P .

9.6 Prove that the reciprocal lattice of the body-centered cubic Bravais latticeis the face-centered cubic lattice using the primitive lattice vectors and thedefinitions of a∗, b∗, and c∗.

9.7 For a cubic Bravais lattice consider the plane-wave approximation (Table 9.6)for constant V for energy bands along the �-line.(a) Show that the (dimensionless) energy is

εnκ = 3κ2 + 2κ(nx + ny + nz)+ n2x + n2

y + n2z .

(b) Calculate εnκ for the eight reciprocal-lattice vectors K(a/(2π)) = n =(nx , ny, nz) = (±1,±1,±1), and construct a table similar to Table 9.6showing the energy and which bands are degenerate.

(c) Identify the IRs for each group of degenerate states.9.8 For the representation based on the functions A = exp[i(2π/a)(κ + 1, κ +

1, κ − 1) · r], B = exp[i(2π/a)(κ + 1, κ − 1, κ + 1) · r], and C =exp[i(2π/a)(κ − 1, κ + 1, κ + 1) · r], do the following.(a) Find the symmetry functions for the �1(A1) and �3(E) IRs.(b) Identify the unk(r) functions of the Bloch waves.(c) Plot a graph of the dimensionless energy εnκ versus κ for these energy

bands.9.9 (a) For a body-centered cubic lattice with constant lattice potential show that

the band energies are given by

εnκ =(

a

)2 (2m

�2

)[Enk − V0]

= (κx + nb + nc)2 + (κy + na + nc)

2 + (κz + na + nb)2,

where κ = (a/2π)(kx , ky, kz) and Kn(a/(2π)) = (a/(2π))[naa∗ +nbb∗ + ncc∗].

(b) What is the group of the wavevector for k along the �-line (excluding �

and N ), that is, for κ = (t, t, 0)?(c) Along the �-line, k = (2π/a)(t, t, 0) (κ = (t, t, 0), 0 < t < 1/2). Find

the energy, εnκ, for the smallest and second-smallest vectors of the formK+ k.

(d) Show that each of these vectors is a basis vector for the �1 IR, anddetermine the Bloch waves for the IRs of (c).

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Exercises 279

(e) Find the third-smallest vectors of the form K+ k and decompose the rep-resentation based on these vectors into the IRs of gk and determine εnκ.

(f) Determine the Bloch waves for each IR in (e).(g) Plot a graph of εnκ for the states in (c) and (e) for t = 0 to 1/2 (k from �

to N ).

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10

Application of space-group theory: Energy bandsfor the perovskite structure

In Chapter 9 nearly free-electron energy bands were discussed. In this chapter wediscuss the energy bands of a crystal for which the electronic states are derived fromatomic-like orbitals. For molecules this approach is the LCAO model. Applied tocrystalline solids, it is referred to as the tight-binding model. In either case thewavefunctions are linear combinations of atomic-like orbitals. We shall analyzethe energy bands of a class of perovskite oxides known as the d-band perovskites.1

10.1 The structure of the ABO3 perovskites

We consider the cubic ABO3 oxides having the perovskite structure with the tran-sition metal B ions. Examples include insulators, such as SrTiO3, BaTiO3, andKTaO3, and metals, such as NaWO3 and ReO3 (a perovskite without an A ion).The structure of these oxides is shown in Fig. 10.1. Many oxides (e.g., BaTiO3 andSrTiO3) undergo structural phase transitions at lower temperatures, leading to non-cubic lattices and ferroelectric and piezoelectric properties. Here we are concernedwith the cubic phase.

As illustrated in Fig. 10.1, the B ions are at the center of a cube, the oxygen ionsare on the faces of a cube, and the A ions are on the corners of a cube.

The five ions of the unit cell are labeled A, B, 2, 3, and 5 in Fig. 10.1. Theorigin is taken at the B site and the oxygen ions are located at a(1, 0, 0), a(0, 1, 0),and a(0, 0, 1). The A ion is located at a(1, 1, 1), where 2a is the lattice constant.For the crystal as a whole, the B ions are on a cubic lattice whose sites are ata(2nx , 2ny, 2nz), the oxygen ions are located at a(2nx+1, 2ny, 2nz), a(2nx , 2ny+1, 2nz), and a(2nx , 2ny, 2nz+1), and the A ions are at a(2nx+1, 2ny+1, 2nz+1),where nx , ny , and nz are positive or negative integers or zero.

1 Perovskites for which electron–electron correlation is strong can not be described by band theory. This includesmany of the insulating and magnetic perovskites.

280

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10.1 The structure of the ABO3 perovskites 281

4

t2t3

t5

1

2 3

A

B

5

6

Figure 10.1 The ABO3 perovskite structure. The B ions are at the centers of thecubes. The O ions are on the faces of the cubes. The A ions are at the corners ofthe cubes. The five ions comprising the unit cell are labeled B, A, and 2, 3, and5 (oxygen ions). The numbering of the oxygen ions is chosen so that the actiontable in Appendix E may be used. The vectors t2 = aex , t3 = aey , and t5 = aezjoin the B ion to the oxygen ions in a unit cell.

The electronic binding of the d-band perovskites includes both ionic and cova-lent contributions. The chemically important energy bands are derived from thed-orbitals of the transition metal ion (B ion) and the p-orbitals of the oxygen ions.The energy bands associated with the A ions, usually derived from their s-orbitals,are at higher energies. Typically, the A-ion energy bands are unoccupied by elec-trons and do not contribute strongly to the electronic or chemical properties. That isnot to say that the A ions are unimportant. The ionic charge of the A ion plays a cen-tral role in determining the relative energies of the d- and p-orbitals, and its ionicradius has a strong influence on the stability of the structure. For a more completediscussion of the d-band perovskites the reader is directed to the literature [10.1].

The Bravais lattice is simple cubic. The B ions are surrounded by 6 oxygenions, and the A ions are surrounded by 12 oxygens. Both the B ions and the Aions are at sites of cubic (Oh) symmetry. The oxygen ions, however, are at siteswith D4h symmetry. As a result, the p-orbitals divide into two groups. Thosewhose lobes are parallel to the B–O axis form σ -bonds with the d-orbitals, andthose whose lobes are perpendicular to the B–O axis form π -bonds with thed-orbitals. Since the environment (symmetry) of the sigma p-orbitals differs from

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282 Application of space-group theory: Energy bands for the perovskite structure

that of the pi p-orbitals the LCAO diagonal energies will be different. The d-ionorbitals separate into the t2g and eg types due to the octahedral arrangement of theoxygen ions.

10.2 Tight-binding wavefunctions

The LCAO method discussed in connection with molecular wavefunctions in pre-vious chapters can be applied to crystals. The method is most useful for compoundswhose properties strongly reflect the character of the atomic structure of its con-stituents, that is, when the energy bands can be associated with the atomic states ofthe ions of the crystal.

Neglecting the A ion, the perovskite unit cell includes the five d-orbitals of theB ion and nine p-orbitals (three for each of the three oxygen ions). This makes atotal of 14 orbitals, so there will be 14 energy bands for the crystal.

The one-electron Hamiltonian is of the form

H(r) =∑

m

{HB(r− Rm)+

∑i

HO(r− Rm − ti )

}+ Vint(r), (10.1)

where HB(r− Rm) is the atomic Hamiltonian for an isolated B ion located at Rm ,HO(r− Rm − ti ) is the atomic Hamiltonian for an isolated oxygen ion located atRm + ti , and Vint(r) is the effective potential due to interactions between the ions.The vectors ti , i = 2, 3, and 5, are vectors from the origin to the O ions labeled 2,3, and 5 in Fig. 10.1.

Because of translational invariance, the crystalline wavefunctions are charac-terized by a wavevector, k, just as in the case of the nearly free electrons in theprevious chapter. The amplitudes of the orbitals in different unit cells can differ atmost by a phase factor. From the space-group results of Chapter 9, we know thatthe phase factor for translation by a lattice vector R is exp(ik · R).

The LCAO or tight-binding wavefunction has the form

ψk(r) =∑Rm

eik·Rm

⎧⎨⎩∑α

Akα φdα(r− Rm)+∑

i

∑β

Bkβi φpβ[r− (Rm + ti )]⎫⎬⎭ .

(10.2)

In (10.2), Rm is a lattice vector from the origin to a B ion and ti is a vector fromthe B ion to an oxygen ion. For the unit cell at Rm , the three oxygen sites are atRm+ti with i = 2, 3, or 5. The terms Akα exp(ik·Rm) and Bkβi exp(ik·Rm) are theamplitudes of the dα and pβ orbitals located at Rm and Rm+ ti , respectively. In thiscase, α denotes one of the five d-orbitals, dxy , dxz , dyz , dz2 , or dx2 , β indicates oneof the three p-orbitals, px , py , or pz , and i specifies the i th oxygen-site vector, ti .

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10.3 The group of the wavevector, gk 283

Y

X

R

Σ

Λ

MZ

S

Z

TΔΓ

ky

kz

kx

Figure 10.2 The Brillouin zone for the simple-cubic lattice. The high-symmetrylines and points are labeled.

Equation (10.2) can also be written in the Bloch-wave form, uk(r) exp(ik · r),where uk(r) =∑

m exp[−ik · (r−Rm)]{∑α Akα φdα(r−Rm)+∑β Bkβ φdβ[r−(Rm + t j )]}. It is clear from its form that uk(r) has the periodicity of the lattice.

10.3 The group of the wavevector, gk

The space group for these perovskites is symmorphic, and therefore it may be con-sidered as the direct product of the group of the wavevector, gk, and the translationsubgroup. The matrix elements of the IRs are then �λ

k(A)rs = exp(ik · R)�λ(A)rs ,where �λ(A)rs is the rs matrix element of the λth IR of gk for the operation A, andR is the translation lattice vector.

The Brillouin zone for the simple cubic lattice is shown in Fig. 10.2.The group of the wavevector, gk, for any k in the zone can be found from

Table E.2 in Appendix E, by selecting the operations that leave the k-vectorunchanged or transformed to an equivalent vector, k+K. For example, for k =(0, 0, k) (along a line equivalent to the �-line shown in Fig. 10.2) with 0 < k <

π/a, the operations in Table E.2 under which k is invariant are just those thatleave z invariant. If an operation, A, listed in Table E.2 carries z into −z, then theoperation i A carries z into z.

The operations under which k = (0, 0, k) remains invariant are E , iC (1)2 , iC (2)

2 ,C (3)

2 , i C (5)2 , i C (6)

2 , C (3)4 , and (C (3)

4 )3. These eight operations form gk, a group thatis isomorphic to the group C4v. The correspondence to the usual C4v elements isshown in Table 10.1.

For the point X, k = (0, 0, π/a), and k and −k are equivalent, since they arerelated by a reciprocal-lattice vector: −k + (0, 0, 2π/a) = +k. Using Table E.2we find the operations that carry k into k or −k (z into z or −z) to be i , E ,C (1)

2 , C (2)2 , C (3)

2 , iC (1)2 , iC (2)

2 , iC (3)2 , C (5)

2 , C (6)2 , i C (5)

2 , i C (6)2 , C (3)

4 , (C (3)4 )3, iC (3)

4 , andi(C (3)

4 )3. This set of 16 operators forms a group isomorphic to D4h . By identifying

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284 Application of space-group theory: Energy bands for the perovskite structure

Table 10.1 Operations of Oh forming a C4v group

gk operator Usual element of C4v

E EC (3)

2 C2

C (3)4 and (C (3)

4 )3 2C4

iC (1)2 and iC (2)

2 2σv

iC (5)2 and iC (6)

2 2σd

Table 10.2 Group of the wavevector for lines and points of high symmetry in theBrillouin zone

Symmetrygk group of the

wavevectorOperation inTable E.2k Point Line

(0, 0, 0) � Oh All(0, 0, k) � C4v z→ z(0, 0, π/a) X D4h z→ z, z→−z(k, k, 0) � C2v x → x , y→ y and

x → y, y→ x(π/a, π/a, 0) M D4h x →±x , y →±y and

x →±y, y →±x(k, k, k) � C3v +x , +y, +z→ any

ordering of +x , +y,or +z (all positive)

(π/a, π/a, π/a) R Oh all

the operations under which k is invariant or carried into an equivalent k we canfind the group of the wavevector for the various points and lines of symmetry.Some selected results are summarized in Table 10.2. For Table 10.2, 0 < k < π/a.

10.4 Irreducible representations for the perovskite energy bands

We choose the origin at one of the B-ion sites, and refer to this unit cell as thecentral cell. Since no symmetry operation can transform a d-orbital into a p-orbital,each set of orbitals separately forms a basis for a representation of gk. The B siteof the central unit cell is invariant under the operations of gk, and therefore thed-orbitals of the central cell transform among themselves. The d-orbitals are at asite of cubic symmetry and therefore are basis functions for three-dimensional T2g

and two-dimensional Eg representations.

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10.4 Irreducible representations for the perovskite energy bands 285

(b)(a)

C2B

B

−px(r + aex)

px(r − aex)

Unit cell atR = (−2a, 0, 0)

Unit cell atR = (0, 0, 0)

t2t3

t5

5

6

2

4

1 3

Figure 10.3 (a) A P(C2) operation on px (r − aex ) transforms the orbital to thenegative of an equivalent orbital, px (r + aex ). The operation acting on the sitesis equivalent to a translation by the lattice vector R = −2a(1, 0, 0) that carriessite 2 into site 4. As a result, the contribution of px (r − aex ) to the character ofthe P(C2) operation is − exp(2ikx a), where 2a is the lattice constant. (b) Atomslabeled 2, 3, and 5 are in the central cells. Atoms 4, 1, and 6 are (respectively) theequivalent sites in neighboring cells. The numbering of the oxygen sites is chosento match the numbering used in the action table, Table E.2, of Appendix E.

The O sites behave differently. Under the operations of gk the O sites in thecentral unit cell are transformed among themselves, but also into O sites in adjacentunit cells. If an atomic site in the central unit cell is transformed into an equivalentsite in a neighboring cell, the operation is equivalent to a lattice translation. Indetermining the character of a point-group operation that transforms an atomicsite into an equivalent site, the equivalent site is regarded as being the same asthe original site. However, the character of the operation must also be multipliedby the character for the translation. If the original site and the transformed site areseparated by the lattice vector R, then the phase factor to be assigned is exp(ik·R).This situation is illustrated in Fig. 10.3, which shows a px orbital in the central cellrotated into a −px orbital in the neighboring cell. The contribution to the characterfor the operation is therefore (−1) exp(ik · R). For the case shown in Fig. 10.3,k · R = −2kxa.

The nine p-orbitals of the central cell are basis functions for a nine-dimensionalrepresentation. However, the representation can be immediately reduced. Thep-orbitals can be distinguished as σ and π orbitals with respect to the axis betweenthe B ion and the oxygen ion. Those p-orbitals in the unit cell with lobes paral-lel to the B–O axes are σ and those perpendicular to the B–O axis are π orbitals.Since no symmetry operation can transform a sigma orbital into a pi orbital (or vice

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286 Application of space-group theory: Energy bands for the perovskite structure

versa), the p-orbital representation can be divided into a three-dimensional (sigma)and a six-dimensional (pi) representation.

The IRs of the energy bands for k corresponding to the various points and lines ofhigh symmetry can be determined by decomposing the representations based on thep- and d-orbitals. As mentioned earlier, the d-orbital representation decomposesinto T2g + Eg. For the p-orbitals we have to consider which orbitals remain on thesame site or are moved to an equivalent site in a neighboring unit cell. Only orbitalsthat remain unmoved or are moved to an equivalent site contribute to the characterof an operation.

10.4.1 The σ and π p-orbitals

The sigma orbitals in the central cell are px(r−aex), py(r−aey), and pz(r−aez),and the equivalent orbitals in adjacent unit cells are px(r + aex), py(r + aey),and pz(r+ aez), respectively. In Fig. 10.3(b) the atomic sites for these orbitals arelabeled 1 through 6. The central cell oxygens are at sites numbered 2, 3, and 5, andthe equivalent sites in neighboring cells are labeled 4, 1, and 6, respectively. Thenumbering is chosen to agree with the action table, Table E.2, of Appendix E. Withthis assignment we can simplify the notation a bit. We write px(r− aex) = px(2),py(r− aey) = py(3), and pz(r− aez) = pz(5). The equivalent orbitals are px(4),py(1), and pz(6), respectively. To obtain the character for an operation of gk weuse the action table, Table E.2, paying attention to the equivalent orbitals and theirphase factors.

10.4.2 Irreducible representations at �(k = 0)

At �(k = 0) the equivalent-site phase factors are all unity and the group of thewavevector is Oh . The contribution of each sigma orbital to the character of anoperation can easily be found. For example, the contribution to the character ofP(C (1)

2 ) is obtained by applying P(C (1)2 ) to the three sigma orbitals. The p-orbitals

transform as x , y, and z. According to Table E.2, under the P(C (1)2 ) operation,

x → x and 2 → 2; therefore the contribution to the character of the operationapplied to px(2) is +1. For P(C (1)

2 ) applied to py(3), we have y →−y and 3→ 1.Since site 1 is equivalent to site 3, the contribution from this orbital to the characterof the operation is −1. For P(C (1)

2 ) applied to pz(5), z → −z and 5 → 6. Sincesite 6 is equivalent to site 5 the orbital makes a contribution of −1 to the characterof P(C (1)

2 ). The sum of the contributions gives the three-dimensional, p-orbitalrepresentation a character of −1 for P(C (1)

2 ). Since P(C (2)2 ) and P(C (3)

2 ) are in thesame class as P(C (1)

2 ), they have the same character. By proceeding in this mannerwe can find the characters of all of the operations. The character table for Oh andthe p-orbitals is given in Table 10.3.

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10.4 Irreducible representations for the perovskite energy bands 287

Table 10.3 Characters for the IRs of Oh and the characters of the representationsbased on the p- and d-orbitals at �. The names of the IRs are given for severalconventions. Those in parentheses are the labels commonly used in solid-statephysics.

IRs E 8C3 3C2 6C2 6C4 i 6iC3 3iC2 6i C2 6iC4

A1g(�1, �+1 , �1g) 1 1 1 1 1 1 1 1 1 1

A2g(�2, �+2 , �2g) 1 1 1 −1 −1 1 1 1 −1 −1

Eg(�12, �+12, �3g) 2 −1 2 0 0 2 −1 2 0 0

T1g(�′15, �

+15, �4g) 3 0 −1 −1 1 3 0 −1 −1 1

T2g(�′25, �

+25, �5g) 3 0 −1 1 −1 3 0 −1 1 −1

A1u(�′1, �−1 , �1u) 1 1 1 1 1 −1 −1 −1 −1 −1

A2u(�′2, �−2 , �2u) 1 1 1 −1 −1 −1 −1 −1 1 1

Eu(�′12, �

−12, �3u) 2 −1 2 0 0 −2 1 −2 0 0

T1u(�15, �−15, �4u) 3 0 −1 −1 1 −3 0 1 1 −1

T2u(�25, �−25, �5u) 3 0 −1 1 −1 −3 0 1 −1 1

IR decompositions� pσ = T1u(�15) 3 0 −1 −1 1 −3 0 1 1 −1� pπ = T1u(�15)+ T2u(�25) 6 0 −2 0 0 −6 0 2 0 0�d = T2g(�

′25)+ Eg(�12) 5 −1 1 1 −1 5 −1 1 1 −1

The same process can be applied to the oxygen π orbitals. There are six π

orbitals: py(r − aex), pz(r − aex), px(r − aey), pz(r − aey), px(r − aez), andpy(r−aez). We label these orbitals as py(2), pz(2), px(3), pz(3), px(5), and py(5),respectively. The equivalent sites are 4, 1, and 6, respectively, and the correspond-ing equivalent orbitals are py(4), pz(4), px(1), pz(1), px(6), and py(6). UsingTable E.2, we can find the characters for the operations. For example, accordingto Table E.2, P(C (1)

2 ) transforms x → x , y → −y, and z → −z, and 2 → 2,3 → 1, and 5 → 6. It follows that the contribution of py(2) and pz(2) orbitals tothe character of P(C (1)

2 ) is−2. For the pairs, px(2) and pz(2), and px(5) and py(5)the contributions sum to zero. On adding all of the contributions we find that thecharacter of P(C (1)

2 ) for the pi orbitals is −2. The characters of the operations ofthe Oh group are listed in Table 10.3.

Some important conclusions can be extracted from the decomposition of the p-and d-orbital representations shown at the bottom of Table 10.3. First, we note thatthe p- and d-orbitals have no IRs in common. As a result, the eigenstates are com-posed of either d-orbital or p-orbital functions. That is, there are no interactionsbetween, or mixing of, the p- and d-orbitals at �. Second, the p-sigma and p-piorbitals both contain the �15(T1u) IR. Therefore the �15 eigenstates may be admix-tures of both types of p-orbitals. Third, we see that the p-pi orbital representation

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288 Application of space-group theory: Energy bands for the perovskite structure

contains the �25(T2u) IR once and that no other representation contains �25. Thismeans that the three �25 symmetry functions are eigenstates and are composedexclusively of p-pi orbitals. Finally, we can characterize the 14 energy-band statesat � as follows.

(a) �(perovskite) = 2�15(T1u)+ �25(T2u)+ �′25(T2g)+ �12(Eg).(b) The block-diagonal secular equation consists of

a. Three 2× 2, �15 blocks that have the same eigenvalues.That is, two triply degenerate energies, Ea(15) and Eb(15).

b. Three 1× 1 blocks yielding a triply degenerate energy, E(25) = E(T2u).c. Three 1× 1 blocks yielding a triply degenerate energy, E(25′) = E(T2g).d. Two 1× 1 blocks yielding a doubly degenerate energy, E(12) = E(Eg).

(c) For the 14 band energies at �, there are at most five independent eigenvalues.(d) For �15 and �25 the wavefunctions are purely of p-orbital composition. For �′25

and �12 the wavefunctions are purely of d-orbital composition.

10.4.3 Symmetry functions at �

The symmetry functions for the central cell can be generated by means of thesymmetry-function-generating machine. As an example, let us find the eigenstatescorresponding to the �25 IR. We know from the decompositions that they are com-posed of p-pi orbitals, so we choose the function pz(2) as the “arbitrary” functionfrom which we will generate �25 symmetry functions. We shall need the diagonalmatrix elements of the �25(T2u) IR. These elements are given in Section E.4 ofAppendix E and are summarized in Table 10.4.

A function, f i (� j ), transforming according to the i th row of the � j IR isgiven by

f i (� j ) ∝∑

A

D� j

i i (A)PA f (r), (10.3)

where A denotes an element of gk, PA is the operator, D� j

i i (A) is a diagonal matrixelement of the � j IR, and f (r) is an arbitrary function of the orbitals.

Applying (10.3) to the function f (r) = pz(2) gives the following results:

f 1(�25) =∑

R

D�2511 (R)PR pz(2)

∝ pz(2)+ pz(2)+ pz(4)+ pz(4)+ py(4)− py(4)

− px(6)+ px(6)− pz(1)− pz(3)+ py(2)

− py(2)− px(6)+ px(6)− pz(1)− pz(3)

= 2pz(2)+ 2pz(4)− 2pz(1)− 2pz(3)

= 4pz(2)− 4pz(3). (10.4)

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10.4 Irreducible representations for the perovskite energy bands 289

Table 10.4 Diagonal matrix elements of the T2 IR for the O group. The diagonalelements for all eight of the C3 operations are zero and therefore omitted from thetable. The character for the operation i A is the same or the negative of that forthe operation A for a g or u representation, respectively.

E C12 C2

2 C32 C1

2 C22 C3

2 C42 C5

2 C62

Row 1 1 −1 −1 1 0 0 0 0 1 1Row 2 1 −1 1 −1 0 0 1 1 0 0Row 3 1 1 −1 −1 1 1 0 0 0 0

C14 (C1

4)3 C2

4 (C24)

3 C34 (C3

4)3

Row 1 0 0 0 0 −1 −1Row 2 0 0 −1 −1 0 0Row 3 −1 −1 0 0 0 0

The last line of (10.4) results since sites 2 and 4, and sites 1 and 3, are equivalent.The normalized function is

f 1(�25) = pz(2)− pz(3)√2(1− S23)

, (10.5)

where S23 is the overlap integral∫

pz(2)∗ pz(3)dr.Using the row-2 and row-3 diagonal matrix elements with py(2) and px(5) as

the generators gives the partner wavefunctions

f 2(�25) = py(2)− py(5)√2(1− S23)

, (10.6)

f 3(�25) = px(3)− px(5)√2(1− S23)

. (10.7)

These functions are repeated in each unit cell, so the total eigenfunctions are

ψ1(�25) = 1√N

∑Rn

pz(r− Rn − t2)− pz(r− Rn − t3)√2(1− S23)

, (10.8)

ψ2(�25) = 1√N

∑Rn

py(r− Rn − t2)− py(r− Rn − t5)√2(1− S23)

, (10.9)

ψ3(�25) = 1√N

∑Rn

px(r− Rn − t3)− px(r− Rn − t5)√2(1− S23)

, (10.10)

where N is the number of unit cells in the crystal, and the vectors t2, t3, and t5 areshown in Fig. 10.3(b).

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290 Application of space-group theory: Energy bands for the perovskite structure

We also know the cell eigenfunctions for the d-orbitals. They are just the �′25(t2g)

and �12(eg) d-electron states, namely the dxy , dxz , and dyz , and dx2 and dz2 orbitals,respectively. The total LCAO wavefunctions for the �′25 and �12 band states are

ψ1(�′25) =1√N

∑Rn

dxy(r− Rn), (10.11)

ψ2(�′25) =1√N

∑Rn

dxz(r− Rn), (10.12)

ψ3(�′25) =1√N

∑Rn

dyz(r− Rn), (10.13)

and

ψ1(�12) = 1√N

∑Rn

dz2(r− Rn), (10.14)

ψ2(�12) = 1√N

∑Rn

dx2(r− Rn). (10.15)

For the �15(T1u) wavefunction at k = 0 the wavefunction is a mixture of sigmaand pi p-orbitals. The T1u diagonal matrix elements are given in Section E.4 ofAppendix E. The symmetry functions can be generated using (10.3). The resultsare given in Table 10.5. The wavefunctions and energies can be calculated using theLCAO method. For simplicity we use the orthogonal Löwdin orbitals discussed inChapter 6. The �15 states are admixtures of π and σ orbitals belonging to the samerow of the T1u IR. For row 1, we have a secular equation involving the functionspx(r− Rn − t2) and [px(r− Rn − t3)+ px(r− Rn − t5)]/

√2. If we include only

nearest-neighbor (nn) p–d interactions, the secular matrix eigenvalue equation isdiagonal because the p–p interactions are second-nearest-neighbor interactions(which we are ignoring).

Figure 10.4(a) shows schematically the cell functions and the correspondingenergies. Figure 10.4(b) shows the �25 functions in the central cell and neighboringunit cells.

10.4.4 Irreducible representations along the �-line

Along the symmetry line � in the Brillouin zone the group of the wavevector, gk,is C4v. The phase factor for equivalent oxygen sites in neighboring cells along the�-line is exp(ik · R) = exp(ikz Rz), where Rz is the component of a lattice vector,R, along the z-axis. The phase factor for translation along the x- and y-directionsis +1, since kx = ky = 0.

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10.4 Irreducible representations for the perovskite energy bands 291

Table 10.5 Symmetry functions for the central cell for the p- and d-orbitals of aperovskite. The notation is pα( j) = pα(r− t j ). See Fig. 10.3 for definitions ofthe t j vectors. LCAO energies are given in the last column. Using onlynearest-neighbor (nn) interactions, the T1u(�15) π states and T2u(�15) energiesare accidentally degenerate. The Löwdin orbitals discussed in Chapter 6 areassumed, so that the overlap integrals between different orbitals vanish.

IR Row d-Orbital type

Unit-cellsymmetryfunction

LCAO energy (Löwdinorbitals, nn interactions)

T2g 1 t dxy(0) Et =∫

dxy(0)∗ H dxy(0)dr2 t dxz(0) Et3 t dyz(0) Et

Eg 1 e dx2(0) Ee =∫

dx2(0)∗ H dx2(0)dr2 e dz2(0) Ee

p-Orbital type

T2u 1 π [px (3)− px (5)]/√

2 Eπ =∫

px (3)∗ H px (3)dr2 π [py(2)− py(5)]/

√2 Eπ

3 π [pz(2)− pz(3)]/√

2 Eπ

T1u 1 σ px (2) Eσ =∫

px (2)∗ H px (2)dr1 π [px (3)+ px (5)]/

√2 Eπ

2 σ py(3) Eσ

2 π [py(2)+ py(5)]/√

2 Eπ

3 σ pz(5) Eσ

3 π [pz(2)+ pz(3)]/√

2 Eπ

The operations of gk are summarized in Table 10.6, and the characters for theseoperations are listed for the IRs for C4v. The characters based on the d- andp-orbital representations are given at the bottom of the table, together with thedecompositions into the IRs of C4v.

Using the symmetry functions, the secular equation for the �-band states willblock- diagonalize as follows.

1. One 3× 3 �1 block associated with p-sigma, p-pi, and d-Eg orbitals.2. One 2× 2 �3 block associated with p-pi and d-Eg orbitals.3. One 1× 1 �4 block associated with T2g orbitals.4. Two 4× 4 �5 blocks associated with p-sigma, p-pi, and d-T2g orbitals.

Some conclusions can be drawn from the decompositions. There is only one �4

symmetry function, and therefore it must be an eigenstate. The p- and d-orbitals

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292 Application of space-group theory: Energy bands for the perovskite structure

(a)

(Γ12) Ee

(Γ25) Et

(Γ15, Γ25) Eπ

(Γ12) Eσ

(b)

(Γ25)

Figure 10.4 (a) LCAO energies at � using Löwdin orbitals and nearest-neighborinteractions. The central cell symmetry functions are shown schematically foreach row of the IR. These functions are repeated in each unit-cell of the crystal.(b) The central unit-cell and neighboring unit-cell functions for the �25 band.

are mixed in forming the �1, �3, �4, and �5 band wavefunctions. However, wenote that the T2g and Eg d-orbitals are never mixed together in any wavefunction,since they belong to different IRs of Oh .

10.4.5 Symmetry functions for k on the �-line

The wavevector, k, is (0, 0, kz), with 0 < kz < π/(2a); that is, k is along a �-line.We use the symmetry-function-generating machine to find the symmetry functions.For example, let us find the �4 eigenstate. Since the IR is one-dimensional, we need

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10.4 Irreducible representations for the perovskite energy bands 293

Table 10.6 The character table for C4v . The last four rows give the characters forthe p- and d-orbital representations and their decompositions. �t is the 3× 3representation based on the T2g d-orbitals and �e is the 2× 2 representationbased on the Eg d-orbitals.

C4v E C (3)2 C (3)

4 (C (3)4 )3 iC (1)

2 iC (2)2 i C (5)

2 i C (6)2

�1(A1) 1 1 1 1 1 1 1 1�2(A2) 1 1 1 1 −1 −1 −1 −1�3(B1) 1 1 −1 −1 1 1 −1 −1�4(B2) 1 1 −1 −1 −1 −1 1 1�5(E) 2 −2 0 0 0 0 0 0

�π = �1(A1)+�3(B1)+ 2�5(E) 6 −2 0 0 2 2 0 0

�σ = �1(A1)+�5(E) 3 −1 1 1 1 1 1 1�t = �5(E)+�4(B2) 3 −1 −1 −1 −1 −1 1 1�e = �1(A1)+�3(B1) 2 2 0 0 2 2 0 0

only the characters. This state is derived from the T2g d-orbitals, and therefore wecan use dxy as the generating function.

Application of (10.3) to dxy gives

f (�4) ∝∑

A

χ�4(A) PA dxy

= +1(dxy)+ 1(dxy)− 1(−dxy)− 1(−dxy)

−1(−dxy)− 1(−dxy)+ 1(dxy)+ 1(dxy)

= 8dxy. (10.16)

Thus, for the model that uses only the p- and d-orbitals, the normalized eigen-function is just dxy . The LCAO energy of this eigenstate is Et (see Table 10.5).From this result we can conclude that the �4(dxy) band extending from � alongthe (0, 0, kz) direction (the � symmetry line) is “flat”. That is, it has no dispersion.

A more complex example is the symmetry functions for the �1 bands. Theseinvolve the p-sigma, p-pi, and Eg d-orbitals. Each orbital type can be used togenerate a �1 symmetry function. For the p-sigma functions we have

fz(�1) ∝∑

A

χ�1(A) PA pz(5) = 8pz(5),

fx(�1) ∝∑

A

χ�1(A) PA px(2) = 2px(2)− 2px(4)+ 2py(3)− 2py(1)

→ 0 (equivalent orbitals cancel out),

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294 Application of space-group theory: Energy bands for the perovskite structure

Table 10.7 Symmetry functions and energies for k along the �-line of symmetry,k = (0, 0, kz)

C4v IR Row Orbital type Symmetry functionLCAO energy (Löwdin orbitals,

nn interactions)

�1(A1) 1a pσ pz(5) Eme ±√(Ege/2)2 + 2(pdσ)2S2

z

1b dz2 dz2 E±ab(�1) = mixture of 1a and 1b1c pπ [pz(2)+ pz(3)]/

√2 Ec(�1) = Eπ

�3(B1) 1a pπ [pz(2)− pz(3)]/√

2 Ea(�3) = Eπ

1b dx2 dx2(0) Eb(�3) = Ee�4(B2) 1 dxy dxy E(�4) = Et�5(E) 1a, 2a pσ px (2), py(2) Ea(�5) = Eσ doubly degenerate

1b, 2b pπ px (3), py(3) Eb(�5) = Eπ doubly degenerateE±c (�5) = Em

1c, 2c pπ px (5), py(5) ±√(Eg/2)2 + 4(pdπ)2S2

z

1d, 2d dxz dxz(0), dyz(0) Mixture of 1c and 1d or2c and 2d, doubly degenerate

Eme = (Ee + Eσ )/2, Em = (Et + Eπ )/2, 2a is the lattice spacing,Ege = Ee − Eσ , Eg = Et − Eπ , Sz = sin(kza).

fy(�1) ∝∑

A

χ�1(A) PA py(3) = 2py(3)− 2py(1)− 2px(4)+ 2px(2)

→ 0 (equivalent orbitals cancel out).

(Note that the phase factors are exp(ikz �Rz) = 1 for all of the equivalent p-orbitals, since �Rz = 0 for the operations of C4v.) The symmetry function derivedfrom the p-sigma orbitals is therefore pz(5). By proceeding in the same way wefind that for the p-pi orbitals the normalized �1 symmetry function is [pz(2) +pz(3)]/

√2. Table 10.7 gives the symmetry functions and LCAO energies for all of

the band states for k along the �-line of symmetry.The calculation of the LCAO-model energy from the symmetry functions is dis-

cussed in Chapter 6. For example, for the �1 band energies, we have a 3 × 3matrix secular equation corresponding to the three symmetry functions. The matrixelements are

M11 =∫

pz(5)∗ H pz(5)dr = Eσ , (10.17)

M12 =∫

pz(5)∗ H

1√2[pz(2)+ pz(3)]dr

= 0 (nn approximation), (10.18)

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10.4 Irreducible representations for the perovskite energy bands 295

M13 =∫

pz(5)∗ H[dz2(r)+ dz2(r− 2aez)]dr

= 2i(pdσ)eikza sin(kza), (10.19)

M22 = 1

2

∫[pz(2)+ pz(3)]∗ H[pz(2)+ pz(3)]dr = Eπ , (10.20)

M23 =∫

1√2[pz(2)+ pz(3)]∗ H dz2 dr

= 0 (nn approximation), (10.21)

M33 =∫

d∗z2 H dz2 dr = Ee, (10.22)

Mi j = M∗j i . (10.23)

The eigenvalues are determined from det(M − E) = 0. The resulting three�1(A1) energies are shown in Table 10.7. The unit-cell eigenvectors are

�±(r) = [Eσ − E±kz(�1)]pz(5)− 2i(pdσ) exp(ikza)Szdz2(r)√[Eσ − E±kz(�1)]2 + 4(pdσ)2S2

z

, (10.24)

�π(r) = 1

2[pz(2)+ pz(3)]. (10.25)

The total �1 crystal wavefunctions are

�±(r) = 1√N

∑R

�±(r− R)eikz Rz , (10.26)

�π(r) = 1√N

∑R

�π(r− R)eikz Rz , (10.27)

where R is a lattice vector and N is the total number of unit cells.For the �5 bands we have two 4 × 4 matrix secular equations that yield the

same eigenvalues. To obtain the symmetry functions we need the diagonal matrixelements of the E IR for C4v. Note that this E representation is not the same as theE of the Oh group. Since x and y are bases for the E IR of C4v, we can generatethe matrices from Table E.2. The results for the diagonal elements are shown inTable 10.8.

Table 10.8 Diagonal elements for the E IR of C4v

E C (3)2 C (1)

4 (C (1)4 )3 iC (1)

2 iC (2)2 iC (5)

2 iC (6)2

Row 1 1 −1 0 0 −1 1 0 0Row 2 1 −1 1 −1 0 0 1 1

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296 Application of space-group theory: Energy bands for the perovskite structure

The symmetry functions generated and the LCAO energies are shown inTable 10.7. For our nn interaction model the 4 × 4 matrix is block-diagonal, con-sisting of two 1 × 1 blocks and a 2 × 2 block. px(2) and px(3) have zero forthe p–d matrix elements, and are therefore eigenstates with energies Eσ and Eπ ,respectively. The px(5) orbital interacts with the dxz orbitals and the 2 × 2 matrixelements are

M11 = Eπ , (10.28)

M12 = 2i(pdπ)eikza Sz = M∗21, (10.29)

M22 = Et . (10.30)

The eigenvalues of det(M− E) = 0 are

E±(�5) = Em ±√(Eg/2)2 + 4(pdπ)2S2

z , (10.31)

Em = 1

2(Et + Eπ), (10.32)

Eg = Et − Eπ . (10.33)

The unit-cell eigenfunctions are

�± = 1√N

{[Eπ − E±(�5)] px(5)− 2i(pdσ)eikza Szdxz(r)}, (10.34)

where

N = [Eπ − E±(�5)]2 + 4(pdσ)2S2z . (10.35)

The row-2 functions produce the same eigenvalues, and the cell wavefunctions canbe obtained from (10.34) by substituting py(5) for px(5) and dyz(r) for dxz(r). Fork along the �-line, the 14 energy bands are shown in Fig. 10.5.

10.4.6 Irreducible representations along the �-line

Along the �-line k = (t, t, 0) and the translation phase factor is exp(ik · R) =exp(i t Rx + i t Ry) = exp[2i ta(nx + ny)]. The group of the wavevector, gk, is C2v,as indicated in Table 10.1. The elements of Oh which leave (t, t, 0) invariant areshown in Table 10.9.

The p-sigma orbitals are bases for a three-dimensional representation of C2v.The p-pi are bases for a six-dimensional representation, and the d-orbitals arebases for a five-dimensional representation. The characters of these representationscan be found in the usual manner, namely by applying the operations of gk to theorbitals and determining the number of orbitals that are transformed into a multipleof themselves or a multiple of equivalent orbitals. The results are summarized in

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10.4 Irreducible representations for the perovskite energy bands 297

Table 10.9 The character table for C2v . The characters and decompositions of p-and d-orbital representations are given in the last three rows.

C2v usual symbols E C2 σv σ ′v k = (t, t, 0)

Oh elements in gk E C (6)2 i C (5)

2 iC (3)2 Basis functions

�1(A1) 1 1 1 1 z�2(A2) 1 1 −1 −1 Rotation about the

(1, 1, 0) symmetry axis,[pz(2)− pz(3)]/

√2

�3(B1) 1 −1 1 −1 x�4(B2) 1 −1 −1 1 y�σ = �1 +�3 +�4 3 −1 1 1 x(2), y(3), z(5)�π = 2�1 +�2 +�3 + 2�4 6 0 0 2 x(3), x(5), y(2), y(5),

z(2), z(3)�d = 2�1 +�2 +�3 +�4 5 1 1 1 dxy , dxz , dyz , dz2 , dx2

Ee

Et

Em Eg

ΔΓ X

Δ1

Δ3

Δ5

Δ4

Δ1, Δ3, Δ5

Δ5 Δ5

Δ1

Figure 10.5 Energy bands along the symmetry line, �.

Table 10.9. In the nearest-neighbor approximation there is additional block-diagonalization among the symmetry functions belonging to the same rowof the same IR. For example, for the �1 states, the 5 × 5 secular equa-tion block-diagonalizes into two 2 × 2 blocks (1a and 1b, and 1c and 1d,

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298 Application of space-group theory: Energy bands for the perovskite structure

Table 10.10 Symmetry functions and LCAO energies for k along �, k = (t, t, 0)

C2v IR Row Orbital typeUnit-cell symmetry

functionLCAO energy (Löwdin orbitals,

nn interactions)

�1(A1) 1a pσ [px (2)+ py(3)]/√

2 E(�1)± = mixture 1a and 1b

1b dz2 dz2(0) = Eme ±√(Ege/2)2 + 2(pdσ)2s2

1c pπ [px (3)+ py(2)]/√

2 E(�1)± = mixture of 1c and 1d

1d dxy dxy(0) = Em ±√(Eg/2)2 + 8(pdπ)2s2

1e pπ [px (5)+ py(5)]/√

2 E(�1) = Eπ

�2(A2) 1a pπ [pz(2)− pz(3)]/√

2 E(�2)± = mixture of 1a and 1b

1b dxz , dyz [dxz(0)− dyz(0)]/√

2 = Em ±√(Eg/2)2 + 4(pdπ)2s2

�3(B1) 1a pσ pz(5) E(�3) = Eσ

1b pπ [pz(2)+ pz(3)]/√

2 E(�3)± = mixture of 1b and 1c

1c dxz , dyz [dxz(0)+ dyz(0)]/√

2 = Em ±√(Eg/2)2 + 4(pdπ)2s2

�4(B2) 1a pσ [px (2)− py(3)]/√

2 E(�4)± = mixture of 1a and 1b

1b dx2 dx2(0) = Eme ±√(Ege/2)2 + 6(pdσ)2s2

1c pπ [px (3)− py(2)]/√

2 E(�4) = Eπ

1d pπ [px (5)− py(5)]/√

2 E(�4) = Eπ

Eme = (Ee + Eσ )/2, Em = (Et + Eπ )/2, 2a is the lattice spacing,Ege = Ee − Eσ , Eg = Et − Eπ , s = sin(ta), t = kx = ky , kz = 0.

respectively) and a 1 × 1 block for 1e. The additional block-diagonalizationis indicated in Table 10.10 by the mixtures of the orbitals shown in the lastcolumn.

The symmetry functions and energies for k along � are given in Table 10.10.

10.5 LCAO energies for arbitrary k

The nearest-neighbor LCAO model we have been using to determine the approx-imate energies for the perovskite bands can be solved analytically for an arbitraryk-vector [10.1]. For this model only the nearest-neighbor p–d interactions areretained, and orthogonal Löwdin orbitals are used. The resulting energies for the14 bands are

Eπ (triply degenerate), (10.36)

E±π (αβ) = Em ±√(Eg/2)2 + 4(pdπ)2(S2

α + S2β), αβ = xy, xz, yz

(six bands), (10.37)

Eσ (one band), (10.38)

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10.5 LCAO energies for arbitrary k 299

Eσ∗

Γ X M R

σ∗

σ0

σ

(a)

Em

Eπ∗

π∗

π0

π

Eg

–14

–12

–10

–8

–6

–4

–2

Γ X M R

Ene

rgy

E(e

V)

(b)

Figure 10.6 The 14 energy bands for d-band perovskite oxides. (a) The LCAOmodel with nearest-neighbor interactions, and (b) bands including next-nearest-neighbor oxygen–oxygen interactions. The π∗ and the σ ∗ bands are the “conduc-tion bands”, whereas the π , π0, σ , and σ0 are the “valence bands”. The valencebands are separated from the conduction bands by an energy gap, Eg.

E±σ (±) = Eme ±√(Ege/2)2 + 2(pdσ)2(S2

x + S2y + S2

z ± S2),

(four bands), (10.39)

where

Sα = sin(kαa), (10.40)

S2 ≡√

S4x + S4

y + S4z − S2

x S2z − S2

y S2z − S2

x S2y , (10.41)

Em = 1

2(Et + Eπ), Eme = 1

2(Ee + Eσ ), (10.42)

Eg = Et − Eπ , Ege = Ee − Eσ . (10.43)

For this model there are four flat bands (i.e., bands with no dispersion), includingone Eσ band and a triply degenerate Eπ band. The next-nearest-neighbor inter-actions are the oxygen–oxygen interactions. These interactions are much weakerbut serve to add dispersion to the flat bands. Figure 10.6 shows the energy bandswith and without the p–p interactions. It can be seen that the addition of the p–pinteractions makes significant changes only in the flat bands.

Figure 10.7 shows a comparison of the LCAO-model results with the bandscalculated numerically using the local-density approximation (LDA) [10.2]. The

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300 Application of space-group theory: Energy bands for the perovskite structure

–14

–12

–10

–8

–6

–4

–2

0SrTiO3

Γ X M R

Ene

rgy

(eV

)

Figure 10.7 A comparison of the LCAO model including p–p interactions(heavy lines) with calculations using the local-density approximation (LDA) forSrTiO3 [10.2] (thin lines). The LCAO parameters employed (in units of eV) areEe = −4.52, Et = −6.52, Eπ = −10.95, Eσ = −12.10, (pdσ) = 2.35,(pdπ) = 1.60, (ppπ) = 0.50, and (ppσ) = −0.05. Note that the calculatedLDA bands give an energy gap between the conduction and valence bands that istoo small compared with experiment. To correct this, the conduction bands havebeen shifted up about 1.5 eV at every k-point.

general features of the two models are in reasonable agreement. The π∗ (conduc-tion bands) are the most important bands in determining the electronic, chemical,and optical properties of a d-band perovskite. These bands are nearly identical forthe LCAO and LDA models, and therefore the LCAO model is quite useful forunderstanding the nature of the d-band perovskites.

10.6 Characteristics of the perovskite bands

The general properties of a d-band perovskite can be understood from the natureof the energy bands shown in Fig. 10.5. Consider, for example, SrTiO3. A simpleionic description of the crystal has one Ti4+, one Sr2+, and three O2− ions. The Ti4+

ion has a closed-shell [Ar] core, and Sr2+ has a closed-shell [Kr] core. Each O2−

ion has a [Ne] closed-shell configuration. The crystal is therefore an ionic insulatorheld together by Coulomb forces (Madelung potentials).

According to the band description, the oxygen p-orbitals and Ti d-orbitals over-lap and form the energy bands we have been discussing. The number of electrons

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References 301

to be distributed among the energy bands is 18; 4 from Ti, 2 from Sr, and 4 fromeach of the three oxygen atoms. Each energy band accommodates two electrons(per unit cell), one with spin up and one with spin down. The 18 electrons willoccupy the lowest-lying energy bands. Referring to Fig. 10.6(a), we see that theelectrons occupy the following bands:

2 electrons in each of the two σ bands = 42 electrons in each of the three π bands = 62 electrons in each of the three π0 bands = 62 electrons in the non-degenerate σ0 band = 2

Total number of electrons = 18

Therefore, all of the valence bands are occupied, and there are no occupied statesin the conduction bands. The occupied bands are separated from the unoccupiedbands by a sizable energy gap, Eg, which is a few electron volts for most per-ovskites. The crystal is an insulator because there are no available empty states inthe lower (valence) bands.

Next, consider the perovskite oxide NaWO3. For this material the metal ionscontribute 7 electrons and the oxygen ions 12, for a total of 19 electrons to beassigned to the energy bands. The additional electron (compared with the case ofSrTiO3) will occupy the π∗ bands. The π∗ bands can accommodate 3 × 2 = 6electrons, and therefore are only partially occupied. As a result NaWO3 is a metal.

Insulating perovskites can be made semiconducting or metallic by doping or bymaking the composition deficient in one of the ions. In addition, n-type SrTiO3 andNax WO3 (0.24 < x < 0.49) and several other perovskites are superconductors.

The π -band energies, E±π (xy), E±π (xz), and E±π (yz), each depend only on twocomponents of the wavevector, k. As a result each is mathematically equivalentto a two-dimensional band. This feature has important physical consequences.For example, the density of electronic states has a jump discontinuity at theconduction-band edges (at Et and Em + [(Eg/2)2 + 8(pdπ)2]1/2) and a logarith-mic infinity at E = Em ± [(Eg/2)2 + 4(pdπ)2]1/2. The effects of these “VanHove” singularities are clearly evident in the optical properties of the d-bandperovskites [10.1, 10.3].

References

[10.1] T. Wolfram and S. Ellialtıoglu, Electronic and Optical Properties of d-BandPerovskites (Cambridge: Cambridge University Press, 2006).

[10.2] E. Mete, R. Shaltaf, and S. Ellialtıoglu, “Electronic and structural properties of a4d-perovskite: Cubic phase of SrZrO3”, Phys. Rev. B 68, 035119 (2003) (4 pages).

[10.3] T. Wolfram, “Two-dimensional character of the conduction bands of d-bandperovskites”, Phys. Rev. Lett. 29, 1383–1387 (1972).

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302 Application of space-group theory: Energy bands for the perovskite structure

Exercises

(All problems refer to the perovskite structure.)10.1 At R (k = (π/(2a))(1, 1, 1)) in the Brillouin zone, answer the following.

(a) What group is gk?(b) What is the phase factor for translation by a lattice vector Rn =

2a(nx , ny, nz)?(c) What is the phase factor for translation from the origin to a nearest-

neighbor unit cell?10.2 (a) Find the characters of the p- and d-orbital representations at R.

(b) Decompose the p- and d-orbital representations into the IRs of gk at R.(Note that the p-orbitals are “g” under inversion because of the phasefactor of −1 between adjacent unit cells.)

10.3 (a) For Exercise 10.2, describe the form of the block-diagonalized secularequation and the degeneracies at R. Label the blocks as IRs of Oh .

(b) What symmetry functions will also be eigenfunctions?10.4 For Exercise 10.3 use the symmetry-function-generating machine to find

the 14 symmetry functions at R. Label them as IRs of Oh .10.5 Use the nearest-neighbor, tight-binding model to find the energies of the

14 bands at R found in Exercise 10.4.10.6 (a) According to symmetry-derived selection rules what electric-dipole

transitions are allowed for insulating SrTiO3 at � (k = 0) in theBrillouin zone?

(b) Find the transition energies of the allowed optical transitions using theresults in Table 10.5.

(c) Repeat (a) and (b) for k at R in the Brillouin zone using the resultsfrom Exercises 10.4 and 10.5.

10.7 Calculate the tight-binding energies of the T2u states at � (k = 0) inthe Brillouin zone including the second-nearest-neighbor interactions (i.e.,including the p–p interactions).

10.8 What energy bands are occupied in the cubic perovskite ReO3? Is thiscompound an insulator or a metal?

10.9 If a cubic ABO3 perovskite undergoes a transition to a tetragonal structurewith D4h symmetry, what additional energy splitting would be expected at� (k = 0) in the Brillouin zone?

10.10 Assuming the structure is elongated along the z-axis, which p–d interac-tions would be reduced? Make a sketch of how Eg and T2g states split andlabel the levels according to the IRs of D4h .(Hint: Make use of the basis functions for the IRs of D4.)

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Exercises 303

10.11 The density of electronic states of the j th energy band, ρ j (E), can becalculated from the expression

ρ j (E) = − 1

πIm

{1

!

∫dk

E − E jk + iε

},

where ! is the volume of the Brillouin zone, E jk is the j th band energyas a function of k, ε is a positive infinitesimal, and Im{. . .} means theimaginary part of the enclosed expression. Consider one of the π∗ bands(whose energy is given by (10.36) with the + sign), and show that ρπ(E)

has a jump discontinuity at E = Et of the form

ρ(E) ={

0 for E = Et − ε,

Eg/[4π(pdπ)2

]for E = Et + ε,

in the limit as ε → 0. Hint: Make use of the result that

− 1

πlimε→0

Im

{1

x − s + iε

}= δ(x − s).

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11

Applications of space-group theory: Lattice vibrations

The atoms of a crystal execute small, oscillatory motions about their equilib-rium positions called lattice vibrations. These vibrations are stimulated by thermalenergy or by external agents such as electromagnetic and mechanical forces. Aswith molecular vibrations, the atomic motions of the lattice can be expressed aslinear combinations of the normal modes of motion. Classically, the energy con-tained in a given normal mode is unrestricted. In quantum theory the energy in anormal mode is quantized in discrete units of �ω. A quantum (�ω) of energy ina normal mode of vibration is called a phonon. More loosely, the lattice vibrationwave in a crystal is also called a phonon.

Because of the translation symmetry of an (infinite) crystal the normal modesare characterized by a wavevector, k. In the case of lattice vibrations we associatea vector with the physical displacement of each atom from its equilibrium posi-tion. The Cartesian components of displacements transform in the same way asthe p-orbitals and therefore the application of space-group theory to lattice vibra-tions is analogous to finding the tight-binding energy bands of a crystal with onlyp-orbitals on each atom.1 The method of analysis of lattice vibrations is the sameas that employed in Chapter 10 for tight-binding energy bands. Instead of energybands we obtain “phonon branches”.

There are, however, a couple of major differences between the treatment oflattice vibration and that of tight-binding energy bands. In the case of lattice vibra-tions there is a requirement that three of the branches have a zero eigenvalue at �(k = 0) in the Brillouin zone (for the translation modes). In addition, in formingeigenvectors, a vibrational symmetry function does not mix with the translationsymmetry function even if these two functions belong to the same row of thesame IR.

1 If the unit cell contains more than one atom the analogous energy-band problem would have two or moreinequivalent p-orbitals at each atomic site.

304

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11.1 Eigenvalue equations for lattice vibrations 305

11.1 Eigenvalue equations for lattice vibrations

To describe the motions of the atoms of a crystal with more than one atom perunit cell we need several indices. The first index locates the unit cell, the secondindex specifies the various atoms in the unit cell, and the third index specifies theCartesian components of the vector from the origin (in the central cell) to the atom.For example, the displacement vector of ξ(Rλ

i ) specifies the displacement vectorof the λ-atom in the unit cell located at Ri . Its Cartesian components are ξα(Rλ

i )

For clarity, in the discussion we shall use the following notation:

Atoms within the unit cell Cartesian coordinate indicesλ, μ, or ν (superscript) α, β, or γ (subscript)Unit-cell locations Displacements from equilibriumRi , R j , Rk or Rn ξλ

i ≡ ξ(Rλi )

(i , j , k, or n subscript) bra 〈a1, a2, . . . , an|n-component row vectorket |a1, a2, . . . , an〉n-component column vector

Equilibrium atomic positions Cartesian components of vectorsRλ

i = Ri + dλ ξλαi ≡ ξα(Rλ

i ); Rλναi j ≡ Rλ

αi − Rνα j

(dλ locates the λ-atomwithin a unit cell)Rλν

i j ≡ Rλi − Rν

j

The instantaneous position of the ν-atom in the nth unit cell is Rνn+ξ(Rν

n, t), whereξ(Rν

n, t) is the vector displacement from equilibrium of the ν-atom in the cell atRn at the atomic position Rν

n = Rn + dν and t denotes time.For the harmonic approximation the potential energy due to the vibratory motion

of the atoms is assumed to be quadratic in the displacements,

U = 1

2

∑λαi

∑νβ j

fαβ(Rλνi j )ξ

λαiξ

νβ j , (11.1)

where fαβ(Rλνi j ) are force constants. Specifying the unit-cell vectors, Ri and R j , the

atoms, λ and ν, and the components α and β uniquely specifies which interactionforce constant is involved. The force constant for the interaction between λ and ν

atoms depends only on the difference Rλα − Rν

β and the components α and β. Theforce constants fαβ(Rλν

i j ) (in units of dynes/cm) give the restoring force on the atomat Rλ

i in the α-direction due to the motion of the atom at Rνj in the β-direction.

To simplify the notation we define f λναβi j ≡ fαβ(Rλν

i j ), keeping in mind that thedependence on i and j is only through the difference |Rλ

i − Rνj |.

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306 Applications of space-group theory: Lattice vibrations

The force on the λ-atom in the unit cell at Ri in the α-direction due to all of theatoms in motion is

Fα(Rλi ) = −

∂U

∂ξλαi

= −∑νγ j

f λναγ i jξ

νγ j . (11.2)

The equations of motion are

−∑νγ j

f λναγ i jξ

νγ j = mλ ∂2

∂t2ξλαi (t), (11.3)

where mλ is the mass of the λ-atom. We can simplify (11.3) by taking

ξλα j = ξλ

α (0)eik·R j eiωt , (11.4)

where ξλα (0) is the α-component of the displacement of the λ-atom in the central

unit cell. Using (11.4) in (11.2) gives∑νγ j

f λναγ i j [ξν

γ (0)eik·R j ] = mλω2ξλ

α (0)eik·Ri . (11.5)

Multiplying both sides of (11.5) by exp(−ik · Ri ) yields∑νγ j

f λναγ i jξ

νγ (0)e

ik·(R j−Ri ) = mλω2ξλα (0). (11.6)

Now define Rp = R j − Ri to obtain∑νγ p

f λναγ pξ

νγ (0)e

ik·Rp = mλω2ξλα (0), (11.7)

since the sum over j in (11.6) is the same as the sum over p in (11.7) exceptfor rearrangement of the terms. Equation (11.7) involves only the displacementswithin the central unit cell. It can written as a 3n × 3n matrix equation, where n isthe number of atoms per unit cell. In matrix form we have

[F(k)−Mω2]ξ = 0, (11.8)

where ξ is a 3n-vector whose components are ξλα (0), and the matrix elements of

F(k) are

F(k)λναγ =∑

p

f λναγ peik·Rp , (11.9)

Mλμαβ = δλμδαβmλ.

Equation (11.9) can be put into standard eigenvalue form,

[D(k)− ω2 I]η = 0, (11.10)

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11.1 Eigenvalue equations for lattice vibrations 307

with

D(k) =M−1/2 F(k) M−1/2

η =M1/2 ξ.

The D matrix of (11.10) is called the dynamic matrix. The eigenvectors are Blochlattice waves described by η(k, ζ ), where ζ denotes the various eigenvectors andη(k, ζ ) is a 3n-vector whose components are ηλ

α(k, ζ ) with α = x , y, or z (ζ =1, 2, . . . , 3n) and the eigenvalues are ω2(k, ζ ). The atomic displacements in theunit cell at Rp associated with a particular normal mode (ζ ) are given by

ξλα (Rp, ζ ) = [M−1/2η(k, ζ )]λαeik·Rp . (11.11)

The axially symmetric lattice-dynamics model

In principle it is possible to calculate the force constants, f λναγ p, numerically from

a model for the electron energy as a function of the atomic positions, but moreoften an empirical force-constant model that has a limited range of interactionsbetween the atoms is employed. The parameters are then determined using someexperimental data. Usually the force constants are neglected beyond the pth-nearest neighbors, where p is typically from 2 to 4. There are several differentlattice-dynamics models. Here, we shall discuss only one, the axially symmetricmodel [11.1].

The basis for the axially symmetric (AS) model is the assumption that the vibra-tional potential energy can be taken as a sum of pair-wise atomic interactions.Consider an atom of type λ at Rλ

i interacting with an atom of type ν at Rνj , and

write the vibrational potential energy of the pair as

U λνi j =

Cλνi j

(Rλνi j )

2

{[ξ(Rλi )− ξ(Rν

j )] · Rλνi j

}2

+ K λνi j

(Rλνi j )

2

{[ξ(Rλi )− ξ(Rν

j )] × Rλνi j

}2, (11.12)

where Cλνi j = Ci j (Rλν

i j ), K λνi j = Ki j (Rλν

i j ), and Rλνi j = |Rλ

i − Rνj |. The first term

of (11.12) is the energy associated with stretching of the distance between the twoatoms and Cλν

i j is a stretching force constant. The second term is the energy asso-ciated with relative motion normal to the line joining the two atoms and K λν

i j isa bending force constant. (For molecules, C and K are called the bond-stretchingand bond-bending force constants, respectively.) The bending force constant is thesame for any direction of motion normal to the line between the two atoms (at rest),and hence the interaction is axially symmetric for the pair of atoms.

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308 Applications of space-group theory: Lattice vibrations

The cross-product term in (11.12) can be simplified as follows. Consider vectorsv and w. We may write

v2 = [v · w]2|w|2 +

[v× w]2|w|2

or

[v× w]2|w|2 = v2 − [v · w]2

|w|2 .

Using this result in (11.12) gives

Uλνi j =

J λνi j

|Rλνi j |2[(ξλ

i − ξνj ) · Rλν

i j ]2 + K λνi j (ξ

λi − ξν

j )2

=∑αβ

{J λν

i j

|Rλνi j |2

Rλναi j Rλν

βi j + δαβ K λνi j

}(ξλ

αi − ξνα j )(ξ

λβi − ξν

β j ), (11.13)

where J λνi j = [Cλν

i j − K λνi j ]. The vibrational potential energy for the entire crystal is

then

U = 1

4

∑λνi j

Uλνi j . (11.14)

The force in the γ -direction acting on the μ-atom located at Rμn is

Fγ (Rμn ) = −

∂U

∂ξμγ n= −

∑νβ j

[Jμν

nj

Rμν

βnj Rμν

γ nj

|Rμν

nj |2+ δγβ Kμν

nj

](ξ

μβn − ξν

β j )

= −mμω2ξμγ n. (11.15)

We take ξλi = ξ(Rλ

i ) = ξλ(0) exp(ik · Rλi ), where ξλ(0) is the displacement vector

for the λ-atom in the central unit cell. (This procedure is slightly different thanthat in Section 11.1, where we took ξλ

i = ξλ(0) exp(ik · Ri ). Either assumptionis permissible. The eigenvectors in Section 11.1 will contain the intra-cell phasefactors. In the present case, the intra-cell phase is already assumed. For example,ξλ(0) exp(ik·Rλ

i ) = ξλ(0) exp(ik·[Ri+dλ]) = ξλ(0)′

exp(ik·Ri ), where ξλ(0)′ =

ξλ(0) exp(ik · dλ) and dλ is the vector from the origin of the unit cell to the λ-atom.) We may choose Rn as the origin of the central unit cell so that the equationof motion is

∑νβ j

[Jμν

j

Rμν

β j Rμν

γ j

|Rμν

j |2+ δγβ Kμν

j

][ξμ

β (0)− ξνβ (0) exp(ik · Rν

i )] = mμω2ξμγ (0).

(11.16)

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11.2 Acoustic-phonon branches 309

The 3n × 3n dynamic matrix may be expressed as [11.2]

Dλναβ(k) =

1√mλmν

{[∑μ

Aλμαβ (0

+)

]δλν − Aλν

αβ(k)

}, (11.17)

where 0+ indicates the limit as k→ 0 and A is a 3n × 3n matrix whose elementsare

Aλναβ =

∑s

{− J (Rλν

s )

(Rλνs )2

∂2

∂kα ∂kβ+ K (Rλν

s )δαβ

}Gλν

s , (11.18)

where

Gλνs =

∑n(s)

eik·Rλνn(s) . (11.19)

In (11.18) and (11.19), the index s denotes a shell of ν-neighbors that are equidis-tant (at the distance |Rλν

s | = Rλνs ) from the λ-atom in the central cell. The sum over

s is over all of the different shells. The sum over n(s) is over all of the atoms in thesth shell. The notation Rλν

n(s) means a vector from the λ-atom in the central cell tothe ν-atom in the sth shell (with both atoms at their equilibrium positions). Thus thesum over all the atomic sites is expressed as a sum over s and n(s). In applicationsof the model, the sum over s is truncated for s greater than some selected cut off(e.g., the third-nearest neighbors). The function G in (11.19) serves as a generatorfrom which the matrix elements of D can be generated by differentiation. For exam-ple, for a cubic crystal with one atom per unit cell the nearest neighbors are locatedat (±a, 0, 0), (0,±a, 0), and (0, 0,±a), so G11(1) = {exp(ikxa)+ exp(−ikxa)+ exp(ikya)+exp(−ikya)+exp(ikza)+exp(−ikza)} = 2{cos(kxa)+cos(kya)+cos(kza)}. For the second-nearest neighbors, G11(2) = 4{cos(kxa) cos(kya) +cos(kxa) cos(kza)+ cos(kya) cos(kza)}.

Some properties of the dynamic matrix are evident from (11.16) through (11.19).In particular, it can be seen that Dλν

αβ(k) = Dλνβα(k) = Dνλ

βα(k)∗ and that Dλν

αβ(0) =Dλν

βα(0) = Dνλβα(0), so the dynamic matrix is Hermitian and at k = 0 it is

symmetric.

11.2 Acoustic-phonon branches

As a simple, tutorial example consider a simple-cubic lattice with one atom perunit cell. Since there is only one type of atom, we may omit the upper super-scripts. The dynamic matrix is a 3 × 3 and the eigenvalue equation, (11.10), isa 3 × 3 with Mαβ = mδαβ . For this simple case the η-vectors are proportionalto the ξ-vectors, so the normalized eigenvectors are the same in either system.The three displacement components, ξα (or the three mass-weighted components,

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310 Applications of space-group theory: Lattice vibrations

ηα = √mξα), are basis functions for a three-dimensional representation of thegroup of the wavevector, gk.

11.2.1 k = 0 irreducible representations

At � in the Brillouin zone (k = 0) the group of the wavevector is Oh . The displace-ment components, ξα (α = x , y, and z), are basis functions for a three-dimensionalrepresentation of Oh . They transform under the operations of the group as x , y, andz. From the character table for O (Table E.1 in Appendix E) it is evident that x , y,and z are the basis functions for the three rows of the T1 representation. Since x , y,and z are “ungerade” (odd under inversion) functions, they are the basis functionsfor the T1u IR of the Oh group. Therefore there are three degenerate phonon nor-mal modes whose eigenvectors are ξα (α = x , y, and z) or, in ket form, |1, 0, 0〉,|0, 1, 0〉, and |0, 0, 1〉. At � the phase factor exp(ik · Rm) = 1, and the atoms havethe same displacement vector in every unit cell. Clearly these modes correspond totranslation in the x-, y-, and z-directions. For a crystal undergoing translation as aunit there are no restoring forces exerted on any atom, and therefore ω = 0 for allthree modes for k = 0. (This places restrictions on the force constants discussedbelow.)

11.2.2 k along the �-line

The group of the wavevector, gk, for k along various symmetry lines of a simplecubic lattice was discussed in Chapter 9. The same results apply to lattice vibrationsor any other excitations of the crystal characterized by a wavevector. In Chapter 9,we found that, along the �-line, k = (0, 0, t), gk is the group C4v. The charactertable for C4v (Table 1.6 in Chapter 1) shows that z is the basis function for the�1(A1) IR, and x and y are the basis functions for the two-dimensional �5(E)IR. Therefore the representation based on ξα decomposes into �1(A1) + �5(E).The secular equation (11.10) is diagonal. The diagonal elements are the mass-weighted squares of the frequencies corresponding to the �1(A1) normal modeand two (degenerate) �5(E) normal modes. The cell eigenvectors are |0, 0, 1〉 for�1(A1), and |1, 0, 0〉 and |0, 1, 0〉 for �5(E). The displacement ξz is parallel (orantiparallel) to k for the �1(A1) phonon branch. This branch is therefore called thelongitudinal acoustic branch. The �5(E) modes have displacements perpendicularto k. These two branches are called the transverse acoustic branches.

The translation phase factor is exp(ikz Rmz), so there is wave-like motion alongthe z-axis, but the displacements are uniform in any x–y plane in lattice space. AtX in the Brillouin zone, k = (π/c)(0, 0, 1). Consider any lattice vector, Rmz =c(0, 0,m), where c is the lattice constant and m is any integer. The phase factor

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11.2 Acoustic-phonon branches 311

(a)

xy

z

(b)

xy

z

Figure 11.1 Displacements for the acoustic modes at X in the Brillouin zone. Thegroup of the wavevector is D4h . The phase of the displacements varies by 180◦from unit cell to unit cell along the z-axis. The displacements in an x–y planeare uniform. (a) Longitudinal mode. Compression and dilation between adjacentx–y planes. (b) Transverse mode. Shear between adjacent x–y planes. The twotransverse modes are degenerate.

is exp(ikz Rzm) = exp(imπ) = (−1)m . Thus, at X, the displacements in alternatex–y planes are 180◦ out of phase with one another. The displacement vectors at Xare shown schematically in Fig. 11.1.

11.2.3 k along the �-line

For k along the �-line of symmetry (k = (t, t, 0)), the group of the wavevectoris C2v. The character table for the group for C2v (Table 10.8) assumes that thetwo-fold axis is in the z-direction. If we take the coordinate z-axis along the [110]direction and the x- and y-axes perpendicular to the �-line, the decompositionof the representation based on the components of ξ is evident from the charactertable. ξz is the basis function for the �1(A1) IR (longitudinal branch), ξy is thebasis for the �4(B2) IR (transverse branch), and ξx is the basis for the �3(B1) IR(transverse branch). With this choice of the coordinate axes the vibrational dis-placement eigenvectors in ket notation are |0, 0, 1〉, |0, 1, 0〉, and |1, 0, 0〉 for the�1(A1), �4(B2), and �3(B1) modes, respectively. If we use our original coordi-nate system, where the �-line is along the [110] direction (as shown in Fig. 10.2),then k = (t, t, 0). In this system the eigenvectors are 2−1/2|1, 1, 0〉 for the �1(A1)

branch, 2−1/2|−1, 1, 0〉 for the �3(B1) branch, and |0, 0, 1〉 for the �4(B2) branch.The translation phase factor is exp(ik·Rm) = exp[ict (mx+my)], where mx and my

are integers, and c is the lattice constant. At M the group of the wavevector changesto D4h . The longitudinal acoustic branch is M′2(A2u). The two transverse acous-tic branches are M′5(Eu). The two modes are shown in Fig. 11.2. Figure 11.2(a)

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312 Applications of space-group theory: Lattice vibrations

(a)

x

y (b)

x

y

Figure 11.2 Atomic displacements for the acoustic modes at M in the Brillouinzone. The group of the wavevector is D4h . The displacements are in phase alongthe line mx = my + 2n. However adjacent lines, mx = my + n and mx = my +n+1, have opposite phases. The phase is constant in the z-direction. (a) M′2(A2u)

(longitudinal acoustic) and (b) one of the M′5(Eu) modes (transverse acoustic).

shows the M′2(A2u) (longitudinal acoustic) mode, and Fig. 11.2(b) shows one ofthe M′5(Eu) modes (transverse acoustic). At M, k = (π/c)(1, 1, 0) and the phasefactor is exp[iπ(mx + my)] = (−1)mx+my . Along the line x = y in the x–y planein lattice space, mx = my , so the phase factor is (−1)2n = 1. For adjacent linesmx = my + n and mx = my + n + 1 the relative displacements are 180◦ out ofphase as shown in Fig. 11.2.

Phonon energies

Group theory provides the IRs of the phonon branches from which importantinformation such as the symmetry functions, degeneracies, and optical selectionrules can be obtained. In many cases it also provides the eigenvectors for thelattice modes. However, group theory alone can not give the energies of thephonons. A force-constant model in which the force constants are treated as param-eters chosen to fit experimental data may be employed. Data may be obtainedfrom neutron inelastic scattering, X-ray diffuse scattering, or optical measure-ments. The measured elastic constants can also be used to help determine theparameters [11.1].

The frequencies of the acoustic branches can be expressed in terms of the matrixelements of the dynamic matrix, D. For the simple-cubic lattice with one atom perunit cell,

ω2(ζ ) = 〈ηζ |D(k)|ηζ 〉,where ηζ is any eigenvector. For example, For k = (0, 0, t), the eigenvalues are

ω2(�1) = 〈0, 0, 1|D(k)|0, 0, 1〉 = Dzz(k),

ω2(�5) = 〈1, 0, 0|D(k)|1, 0, 0〉 = Dxx(k) = Dyy(k).

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11.2 Acoustic-phonon branches 313

For k = (t, t, 0) the eigenvalues are

ω2(�4) = 〈0, 0, 1|D(k)|0, 0, 1〉 = Dzz(k),

ω2(�1) = 1

2〈1, 1, 0|D(k)|1, 1, 0〉 = 1

2{Dxx(k)+ Dyy(k)+ Dxy(k)+ Dyz(k)}

= Dxx(k)+ Dxy(k),

ω2(�3) = 1

2〈−1, 1, 0|D(k)| − 1, 1, 0〉 = Dxx(k)− Dxy(k).

Figure 11.3 shows the results for the axially symmetric model compared withexperimental results for the acoustic modes of copper [11.3]. The upper curve isthe longitudinal acoustic branch, and the lower two curves are the two transverseacoustic branches. All of the acoustic branches have ω = 2πν proportional to |k|as k → 0. The proportionality constant depends on the direction of k and thebranch (as is evident in Fig. 11.3). The linearity of the acoustic dispersion curvesin the long-wavelength limit is a general property of phonons and elastic waves.The slope of the dispersion curve as k → 0 determines the velocity of sound forthe branch.

0

1

2

3

4

5

6

7

8

XKΓ

ν(T

Hz)

L

T2

T1

Cu

Figure 11.3 Dispersion curves for face-centered cubic copper along the [110]directions �K and KX in the Brillouin zone. The solid lines are for the axiallysymmetric lattice-dynamics model [11.1] with parameters k1(1) = 30.0, k1(2) =0.25, k1(3) = 1.50, CB(1) = −2.06, CB(2) = −0.38, and CB(3) = 0.03 in thenotation used in [11.1]. The circles are inelastic-neutron-scattering data [11.3].The upper curve is the longitudinal acoustic branch, and the lower two curves aretransverse acoustic branches. The frequency ν = ω/(2π), and kmax =

√2π/c =

K , where c is the lattice constant for copper.

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314 Applications of space-group theory: Lattice vibrations

11.3 Optical branches: Two atoms per unit cell

When a crystal’s unit cell contains more than one atom, additional vibrationbranches occur. If the unit cell contains n atoms, there will be 3n branches. Threeof the branches are always acoustic branches because there are three translationmodes in the limit as k → 0. The remaining 3n − 3 branches are called opticalbranches. The frequencies of the optical branches are non-zero at k = 0. In fact, inmany cases the frequencies lie in the infrared and can be excited optically, whichexplains the name “optical branches”.

To explore the nature of these optical branches we shall analyze the lattice vibra-tions of a simple-cubic structure with two atoms per cell. Atom “a” with mass mis at the center of the unit cell cube, and atom “b” with mass M is at the cornerof the cube, as shown in Fig. 11.4. With two atoms per unit cell, the secular equa-tion, (11.10), is a 6 × 6 matrix equation. The η-vector is a six-component vector,ηλα =√

mλ ξλα , where mλ = m for λ = a and mλ = M for λ = b; and α = x , y,

or z. The lattice constant is 2c. The vector joining atoms a and b in the unit cell ist = c(1, 1, 1), as shown in Fig. 11.4.

11.3.1 Vibration modes at � (k = 0)

At k = 0, gk is the Oh group. If we choose the a-atom site as the origin then thea-site is invariant under all the operations of the group. The b-atom site transformsinto itself or one of the seven other equivalent sites in adjacent unit cells. Thetranslation phase factor is 1 at k = 0. The a and b atoms are never transformedinto one another, so each of the sets of displacements {ξ a

x , ξ ay , ξ a

z } and {ξ bx , ξ b

y , ξ bz }

87

6 5

43

2 1

a

b

t

2c

x

y

z

Figure 11.4 A unit cell with two atoms. Atom a is at c(0, 0, 0) and atom b is atc(1, 1, 1), where 2c is the lattice constant for the simple-cubic Bravais lattice. Theopen circles labeled 1 through 8 indicate equivalent atom-b sites in adjacent unitcells. The vector t = c(1, 1, 1).

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11.3 Optical branches: Two atoms per unit cell 315

is a basis function for a three-dimensional representation of Oh . Since the dis-placements transform as x , y, and z, each set is the basis for the T1u irreduciblerepresentation. That is, the decomposition of the six-dimensional representationbased on displacements is 2T1u . The set {ξ a

x , ξ ay , ξ a

z } is the basis for one T1u , andthe set {ξ b

x , ξ by , ξ b

z } is the basis for the other T1u . These two sets of symmetry func-tions belong to the same IR. Therefore the eigenvectors for the normal vibrationmodes are of the form Aξ a

α + Bξ bα . At k = 0, three of the modes correspond to

translation along the x-, y-, and z-axes with normalized ket eigenvectors,

displacement eigenvectors mass-weighted eigenvectorsξ (trans)

x = (1/√

2)|1, 0, 0, 1, 0, 0〉, η(trans)x = (1/

√m + M)|√m, 0, 0,

√M, 0, 0〉,

ξ (trans)y = (1/

√2)|0, 1, 0, 0, 1, 0〉, η(trans)

y = (1/√

m + M)|0,√m, 0, 0,√

M, 0〉,ξ (trans)

z = (1/√

2)|0, 0, 1, 0, 0, 1〉, η(trans)z = (1/

√m + M)|0, 0,

√m, 0, 0,

√M〉.

The remaining three (vibrational) eigenvectors must be orthogonal to one anotherand to the translation eigenvectors. The orthogonality requirements are sufficientto determine the remaining three eigenvectors,

vibrational eigenvectorsη

optx = (1/

√m + M)|√M, 0, 0,−√m, 0, 0〉,

ηopty = (1/

√m + M)|0,√M, 0, 0,−√m, 0〉,

ηoptz = (1/

√m + M)|0, 0,

√M, 0, 0,−√m〉.

At k = 0 the frequencies of the translation modes are zero, but the vibration eigen-states are optical modes and they have non-zero ω. At k = 0 the optical modes arecharacterized by the motions of the two atoms being 180 degrees out of phase. Aschematic representation of the symmetry functions is shown in Fig. 11.5.

α

β

γ

M

ξa

ξb

m

Figure 11.5 A schematic representation of the unit-cell optical-mode displace-ments at k = 0 for M > m. The vectors ξa and ξb are the physical displacementvectors, ξb = −(M/m)ξa . For the vibration modes the center of mass is at rest.

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316 Applications of space-group theory: Lattice vibrations

11.3.2 Matrix elements of D (k = 0)

Since we were able to find the eigenvectors at k = 0, the dynamic matrix D(0) iscompletely specified in terms of the three optical frequencies. For the simple-cubicstructure we are considering, these optical frequencies are degenerate. Thereforethe dynamic matrix is completely specified in terms of a single frequency that weshall call ω0.

The method of finding the elements of D(0) is the same as that employed inChapter 1. The matrix of eigenvectors expressed as kets (column vectors) can beused to diagonalize D(0). Let S be the matrix of eigenvectors,

S = (ηoptx ,ηopt

y ,ηoptz ,ηT

x ,ηTy ,η

Tz )

= 1√M + m

⎛⎜⎜⎜⎜⎜⎜⎜⎝

√M 0 0

√m 0 0

0√

M 0 0√

m 00 0

√M 0 0

√m

−√m 0 0√

M 0 00 −√m 0 0

√M 0

0 0 −√m 0 0√

M

⎞⎟⎟⎟⎟⎟⎟⎟⎠. (11.20)

The inverse matrix, S−1, is the transpose of S. The similarity transformationS−1 D(0)S diagonalizes D(0),

S−1 D(0)S =

⎛⎜⎜⎜⎜⎜⎜⎜⎝

(ω0)2 0 0 0 0 00 (ω0)2 0 0 0 00 0 (ω0)2 0 0 00 0 0 0 0 00 0 0 0 0 00 0 0 0 0 0

⎞⎟⎟⎟⎟⎟⎟⎟⎠

. (11.21)

It then follows that

D(0) = S[S−1 D(0)S]S−1 = 1

M + m×⎛

⎜⎜⎜⎜⎜⎜⎜⎝

M(ω0)2 0 0 −√Mm(ω0)2 0 00 M(ω0)2 0 0 −√Mm(ω0)2 00 0 M(ω0)2 0 0 −√Mm(ω0)2

−√Mm(ω0)2 0 0 m(ω0)2 0 00 −√Mm(ω0)2 0 0 m(ω0)2 00 0 −√Mm(ω0)2 0 0 m(ω0)2

⎞⎟⎟⎟⎟⎟⎟⎟⎠

.

(11.22)

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11.3 Optical branches: Two atoms per unit cell 317

If we rearrange the rows and columns of D(0) it takes the block-diagonal form

1

M + m

∣∣∣∣∣∣∣∣

[M(ω0)2 −√Mm(ω0)2

−√Mm(ω0)2 m(ω0)2

][

M(ω0)2 −√Mm(ω0)2

−√Mm(ω0)2 m(ω0)2

][

M(ω0)2 −√Mm(ω0)2

−√Mm(ω0)2 m(ω0)2

]

∣∣∣∣∣∣∣∣.

(11.23)

Therefore the dynamic matrix D(0) is completely determined if the frequencyof the optical mode is known. Often, ω0 can be measured by optical or neutron-scattering experiments. Conversely, the optical frequency can be expressed in termsof the dynamic matrix.

From the matrix elements of the dynamic matrix in (11.22) it is clear that

(ω0)2 = m + M

MDaa

αα. (11.24)

Other relationships that can be deduced from (11.22) include

Daaαα(0) =

M

mDbb

αα(0),

Dabαα(0) = Dba

αα(0),

Dbaαα(0) = −

√M

mDaa

αα(0),

Dλναβ(0) = 0 for α �= β. (11.25)

As a result of these relationships the dynamic matrix has only one independentmatrix element.

As mentioned before, the “longitudinal” and “transverse” designations are fuzzyfor k = 0, since the direction of the wavevector is indeterminate. However, wecan define the k = 0 branches as the limit as k → 0. For this to make sense, thelimiting frequency must be independent of the direction of the wavevector. Thispresents no problem here because the three optical frequencies are degenerate atk = 0 (see the discussion in the box below).

The dynamic matrix presented in Section 11.1 accounts only for the short-rangeforces assuming the ions are “rigid”. This model works well for metals and non-ioniccrystals. For ionic crystals such as CsCl the long-range forces due to the charges onthe ions and the polarizability of the ions cause the longitudinal and transverseoptical modes to have different frequencies even for very small values of thewavevector. The lattice vibrations for these ionic crystals are usually analyzed usingwhat is called a lattice “shell model”. The shell model allows the ion cores to moverelative to their outer shell of electrons and includes the subsequent polarization of

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318 Applications of space-group theory: Lattice vibrations

the ions caused by such movements [11.4]. For a cubic crystal with the CsCl structurethe three optical modes must be degenerate if the wavevector is identically zero.However, an applied electromagnetic field will establish a preferred direction so thatthe symmetry is no longer cubic. In the case of optical absorption the electric field ofthe light wave establishes a preferred direction and conservation of momentummeans that the stimulated lattice wave has k unequal to zero. The same is true forinelastic scattering of neutrons. In order to distinguish a transverse mode from alongitudinal mode, there must be a preferred direction. Generally any probe of thelattice modes will establish a preferred direction, and therefore the k = 0 point is notaccessible to experiment. Mathematically, k = 0 is a singular point. The results of theshell model should be interpreted as the limit as k→ 0.

11.3.3 k along the �-line

Along the �-line, k = (0, 0, t), the group of the wavevector is C4v. Again, thesets of displacements {ξ a

x , ξ ay , ξ a

z } and {ξ bx , ξ b

y , ξ bz } provide bases for two reducible

representations of C4v. According to the character table z is the basis function forthe �1(A1), while x and y are bases for the �5(E) IR. Therefore the representationbased on {ξ a

x , ξ ay , ξ a

z } decomposes into �1(A1)+�5(E).Under the operations of gk the b-atoms transform into themselves, but also

into b-atoms on equivalent sites. We must consider the translation phase fac-tors, exp(ikz δRz), where δRz is the z-component of the change in the positionof an atom due to the symmetry operation. However, these factors are alwaysunity for k along the �-line because an operation that leaves kz unchanged hasδRz = 0. By definition, the operations of gk for k = (0, 0, t) do not changethe z-components of the position vectors. For example, consider the C2 rota-tion about the z-axis. This operation carries the b-atom at site 1 into site 3. Thetranslation of site 1 into site 3 is a translation by 2c(1, 1, 0), so the phase fac-tor exp(ik · R) = exp[i(0, 0, t) · 2c(1, 1, 0)] = exp(0) = 1. Since all the phasefactors are unity, the representation based on {ξ b

x , ξ by , ξ b

z } also decomposes into�(A1) + �5(E). At the X point in the Brillouin zone, where new operations thattransform k into k+K are included, gk changes from C4v to D4h .

As a result of the above consideration, we see that the dynamic matrix has theblock-diagonal form: a 2× 2 �1(A1) block whose symmetry functions are ξ a

z andξ b

z ; a 2 × 2, row-1, �5(E) block whose symmetry functions are ξ ax and ξ b

x ; and a2×2, row-2, �5(E) block whose symmetry functions are ξ a

y and ξ by . The normalized

η-symmetry functions and the normalized ξ -symmetry functions are the same inthis case,

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11.3 Optical branches: Two atoms per unit cell 319

Table 11.1 Phonon eigenvalues for k = (0, 0, t), for the simple-cubic structure,with two atoms per unit cell

Mode Eigenvalue (ω2) Degeneracy

LA �112 (Daa

zz + Dbbzz )−

√[ 12 (Daa

zz − Dbbzz )]2 + |Dab

zz |2 1

LO �112 (Daa

zz + Dbbzz )+

√[ 12 (Daa

zz − Dbbzz )]2 + |Dab

zz |2 1

TA �512 (Daa

xx + Dbbxx )−

√[ 12 (Daa

xx − Dbbxx )]2 + |Dab

xx |2 2

TO �512 (Daa

xx + Dbbxx )+

√[ 12 (Daa

xx − Dbbxx )]2 + |Dab

xx |2 2

Dλναβ = Dλν

αβ(k), k = (0, 0, t).

LA (TA), longitudinal (transverse) acoustic; LO (TO), longitudinal (transverse) optical.

IR Symmetry functions�1(A1) η1 = |0, 0, 1, 0, 0, 0〉

η2 = |0, 0, 0, 0, 0, 1〉�5(E) row 1 η3 = |1, 0, 0, 0, 0, 0〉

η4 = |0, 0, 0, 1, 0, 0〉�5(E) row 2 η5 = |0, 1, 0, 0, 0, 0〉

η6 = |0, 0, 0, 0, 1, 0〉

The eigenfunctions are linear combinations of the symmetry functions that trans-form according to the same row of the same IR. The two �1(A1) symmetryfunctions have displacements parallel to the wavevector, and therefore will mix toform the longitudinal modes, one acoustic and one optical. The four �5(E) modesare transverse modes, two acoustic and two optical. The secular equation for thenormal modes is of the form∣∣∣∣ 〈ηi |D|ηi 〉 − ω2 〈ηi |D|η j 〉

〈η j |D|ηi 〉 〈η j |D|η j 〉 − ω2

∣∣∣∣ = 0, (11.26)

where i = 1 and j = 2 for the �1(A1) modes, and i = 3 and j = 4 (or i = 5and j = 6) for the �5(E) modes. Table 11.1 lists the eigenvalues in terms of thedynamic matrix elements.

11.3.4 k along the �-line

Along the �-line of symmetry, the group of the wavevector is C2v, and k =(t, t, 0). The translation phase factor, exp[ik(δRx + δRy)], is unity for all of the

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320 Applications of space-group theory: Lattice vibrations

operations of C2v. A couple of examples show why this is so. Consider the phasefactor for the b-atom under the operation P(C (6)

2 ). (The equivalence of the “barred”symbols to the usual C2v symbols is given in Table 10.9.) Site 1 is translated intosite 5. The translation vector is δR = 2c(0, 0,−1) and kx δRx = ky δRy = 0. Nextconsider the effect of P(i C (5)

2 ) on site 1. P(C (5)2 ) translates site 1 into site 7, and

P(i) carries site 7 back to site 1. Therefore site 1 is invariant, and the phase factoris again unity.

Since the phase factors are all unity, we can identify the IRs for the vibrationalmodes directly from the point-group character table for C2v. It should be noted thatthe character table uses the z-axis as the principal axis. So in the C2v charactertable the z-axis is along the [110] direction. Consider coordinates in which thez-axis is parallel to the �-line and the x- and y-axes are perpendicular to the�-line. Table 10.9 shows that z is the basis for the �1(A1) IR, x is the basis for the�3(B1) IR, and y is the basis for the �4(B2). The normalized symmetry functionsfor z along the �-line and for the original x-, y-, and z-axes are shown below. Thenormalized symmetry functions for η-vectors and ξ-vectors are the same.

C2v Symmetry functions Symmetry functionsIR z along [110] z along [001]�1(A1) |0, 0, 1, 0, 0, 0〉 η1 = (1/

√2)|1, 1, 0, 0, 0, 0〉

|0, 0, 0, 0, 0, 1〉 η2 = (1/√

2)|0, 0, 0, 1, 1, 0〉�3(B1) |1, 0, 0, 0, 0, 0〉 η3 = (1/

√2)| − 1, 1, 0, 0, 0, 0〉

|0, 0, 0, 1, 0, 0〉 η4 = (1/√

2)|0, 0, 0,−1, 1, 0〉�4(B2) |0, 1, 0, 0, 0, 0〉 η5 = |0, 0, 1, 0, 0, 0〉

|0, 0, 0, 0, 1, 0〉 η6 = |0, 0, 0, 0, 0, 1〉

Since there are two symmetry functions for each IR, the block-diagonal form ofthe secular equation is three 2× 2 blocks. Each block of the secular equation takesthe form of (11.26). For the �1(B1) IR modes, i = 1 and j = 2. For the �3(B1)

modes, i = 3 and j = 4. For the �4(B2) modes, i = 5 and j = 6. The eigenvaluesare shown in Table 11.2.

11.4 Lattice vibrations for the perovskite structure

The tight-binding or LCAO energy bands of the simple-cubic, perovskite structurewere discussed in Chapter 10. Here we are concerned with the lattice vibrations.The perovskite unit cell contains one A, one B, and three O atoms (here O doesnot necessarily indicate oxygen) as shown in Fig. 11.6(a). A fixed x, y, z coordi-nate system is erected at each of the sites in the central unit cell (see Fig.11.6(b)).Various atomic displacements are denoted by

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11.4 Lattice vibrations for the perovskite structure 321

Table 11.2 Phonon eigenvalues for k = (0, t, t), for thesimple-cubic structure, with two atoms per unit cell. All havedegeneracies equal to 1.

Mode Eigenvalue (ω2) z along [001]

LA �112 (Daa

xx + Daaxy + Dbb

xx + Dbbxy)

−√[

12 (Daa

xx + Daaxy − Dbb

xx − Dbbxy)]2 + |Dab

xx + Dabxy |2

LO �112 (Daa

xx + Daaxy + Dbb

xx + Dbbxy)

+√[

12 (Daa

xx + Daaxy − Dbb

xx − Dbbxy)]2 + |Dab

xx + Dabxy |2

TA �312 (Daa

xx − Daaxy + Dbb

xx − Dbbxy)

−√[

12 (Daa

xx − Daaxy − Dbb

xx + Dbbxy)]2 + |Dab

xx − Dabxy |2

TO �312 (Daa

xx − Daaxy + Dbb

xx − Dbbxy)

+√[

12 (Daa

xx − Daaxy − Dbb

xx + Dbbxy)]2 + |Dab

xx − Dabxy |2

TA �412 (Daa

zz + Dbbzz )−

√[12 (Daa

zz − Dbbzz )

]2 − |Dabzz |2

TO �412 (Daa

zz + Dbbzz )+

√[12 (Daa

zz − Dbbzz )

]2 − |Dabzz |2

Dλναβ = Dλν

αβ(k), k = (t, t, 0); Dλλxx = Dλλ

yy , Dλνxy = Dλν

yx = Dνλxy .

LA, longitudinal acoustic, LO, longitudinal optical,

TA, transverse acoustic, TO, transverse optical.

ξB for the B atom,ξA for the A atom,ξ(2), ξ(3), ξ(5), for the three O atoms.

11.4.1 k = 0 irreducible representations

At � (k = 0) the group of the wavevector is Oh . The sets of displacement com-ponents for the A and B atoms, {ξ A

x , ξ Ay , ξ

Az } and {ξ B

x , ξ By , ξ B

z }, are each bases forthe �15(T1u) IR of Oh . For the O-atom displacements we note that the set ξ (2)

x ,ξ (3)

y , ξ (5)z may be classified as σ -type displacements since they are along the B–O

internuclear axis of the unit cell. By contrast, the set ξ (2)y , ξ (2)

z , ξ (3)x , ξ (3)

z , ξ (5)x , ξ (5)

y

are π -type displacements since they are displacements perpendicular to theB–O internuclear axis of the unit cell. No operation of Oh can transform aσ -displacement into a π -displacement and vice versa. Therefore, the sets of σ -and π -displacements form separate representations. The displacements transform

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322 Applications of space-group theory: Lattice vibrations

(a)

A

B

O

O

O

c

c

c

(b)

x

y

z

O2

O5

O3

A

B

(c)

(0,0,0)BO1

O2

O3

O4

O5

O6

c

(d)

A8A7

A6 A5

A4A3

A2 A1

B

Figure 11.6 (a) The unit cell containing one B, one A, and three O atoms. TheB–O distance is c. (b) The coordinate system for the atoms of the unit cell. (c) Thesix equivalent O atoms surrounding the B atom. The pairs of equivalent O-atomsites are 2 and 4, 1 and 3, and 5 and 6. (d) The eight equivalent sites for the Aatom. The lattice spacing is 2c and the origin is at the B site.

in the same way as p-orbitals. We may use the results of Table 10.3 in Chapter 10to obtain the decompositions based on the displacements. They are

�(A) = �15(T1u),�(B) = �15(T1u),�σ (O) = �15(T1u),�π(O) = �15(T1u)+ �25(T2u),�(ABO3) = 4�15(T1u)+ �25(T2u).

The above decomposition indicates that the secular equation will block-diagonalizeinto three degenerate 1 × 1 blocks for the �25(T2u) modes and three 4 × 4 blocksfor the �15(T1u) modes.

11.4.2 Symmetry functions for the perovskites at �

The symmetry functions for the O-atom displacements are analogous to those forthe p-orbitals. Therefore, the π and σ symmetry functions may be taken directly

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11.4 Lattice vibrations for the perovskite structure 323

from Table 10.5 by substituting ξ ( j)α for pα(r−R j ). The symmetry functions for the

A and B atoms are just the displacement components, ξ Ax , ξ A

y , ξ Az and ξ B

x , ξ By , ξ B

z .The IR �25(T2u) occurs only once in the decomposition, and hence the symmetry

functions are eigenstates (normal modes). The frequency is triply degenerate.For the remaining vibrational modes it is necessary to combine four symmetry

functions belonging to a specific row of �15(T1u) to form an eigenvector. The fourfrequencies of the �15 4 × 4 block are triply degenerate. That is, each of the three4× 4 blocks yields the same four frequencies. Before transformation by the matrixof symmetry functions, the dynamic matrix is a 15× 15 matrix. After transforma-tion by the symmetry functions, it consists of three uncoupled 5×5 blocks, each ofwhich consists of a 1× 1 �25(T2u) block and a 4× 4 �15(T1u) block. Furthermore,each 4 × 4 block must yield a translation mode with zero frequency. Therefore ifwe use the translation mode as one of the symmetry functions, each 4× 4 block isfurther block-diagonalized into a 1× 1 block and a 3× 3 block.

We shall represent a five-column vector as the ket whose entries correspond tothe amplitudes of displacement of the A, B, O(2), O(3), and O(5) atoms in theα-direction. With this choice, a ket such as vα = |α; 1, 0, 0, 0, 1〉 represents a vec-tor whose displacement amplitudes are unity for ξ A

α , zero for ξ Bα , ξ (2)

α , and ξ (3)α , and

unity for ξ (5)α . Expanded into a 15-component vector such as vy = |y; 1, 0, 0, 0, 1〉,

this is |0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0〉. In the mass-weighted η-system thenormalized ket vy = (m A + m)−1/2|y;√m A, 0, 0, 0,

√m〉. Note that a normalized

ket involving only one type of atom is the same in the ξ- and η-systems, but whentwo or more different atom types are involved the kets in the ξ- and η-systems arenot the same. Table 11.3 summarizes the results.

Table 11.3 IRs and vibration symmetry functions at k = 0 for the cubicperovskites. m A is the mass of the A atom, m B is the mass of the B atom, and mis the mass of each O atom. The kets’ entries are ordered as|α; ξ A

α , ξ Bα , ξ

(2)α , ξ

(3)α , ξ

(5)α 〉.

Atom IR Normalized symmetry functions α = x, y, z/row 1, 2, 3

A �15(T1u) ηAα (�25) = |α; 1, 0, 0, 0, 0〉

B �15(T1u) ηBα (�15) = |α; 0, 1, 0, 0, 0〉

O-σ �15(T1u) ησα (�15) = |α; 0, 0, 1, 0, 0〉

O-π �15(T1u) ηπx (�15) = 2−1/2|x; 0, 0, 0, 1, 1〉 row 1

ηπy (�15) = 2−1/2|y; 0, 0, 1, 0, 1〉 row 2

ηπz (�15) = 2−1/2|z; 0, 0, 1, 1, 0〉 row 3

O-π �25(T2u) ηπx (�25) = 2−1/2|x; 0, 0, 0, 1,−1〉 row 1

ηπy (�25) = 2−1/2|y; 0, 0, 1, 0,−1〉 row 2

ηπz (�25) = 2−1/2|z; 0, 0, 1,−1, 0〉 row 3

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324 Applications of space-group theory: Lattice vibrations

11.4.3 Eigenfrequencies of the optical modes at k = 0

Without any further analysis, we already have 6 of the 15 normal modes at �.The three translation eigenvectors are known and, since �25(T2u) occurs only oncein the decomposition, we know that the three symmetry functions, ηα(�25) =2−1/2|α; 0, 0, 0, 1,−1〉 with α = x , y, and z, are also eigenvectors. The �25

eigenvalue expressed in terms of these dynamic matrix elements is

ω2(�25) = 〈ηπα (�25)|D(0)|ηπ

α (�25)〉 (α = x, y, z)

= 1

2〈α; 0, 0, 0, 1,−1|D(0)|α; 0, 0, 0, 1,−1〉

= 1

2[D33

αα(0)+ D55αα(0)− D35

αα(0)− D53αα(0)]. (11.27)

The eigenvalue is the same for each α, so ω2(�25) is triply degenerate.

Constraints on the dynamic matrixThe requirement that the translation frequency vanish (ω2(tα) = 0) places constraintson the general dynamic matrix defined by (11.9) and (11.10). In particular, if ηt is atranslation in any direction, it requires that Dηt = 0. For a translation in an arbitrarydirection, (tx , ty, tz), one requires∑

νβ

Dλναβ(0)tβ

√mν = 0 (ηνβ = tβ

√mν).

Since tβ is arbitrary it follows that each term (β = x, y, z) must vanish separately, so∑ν

Dλναβ(0)

√mν = 0.

For the axially symmetric model these constraints are automatically satisfied. (In(11.16) set k = 0 and note that ξμβ = ξνβ for any translation.)

The eigenvectors for the translation modes are

ξtα = 1√5|α; 1, 1, 1, 1, 1〉 (α = x, y, z), (11.28)

ηtα = 1√mT|α;√m A,

√m B,√

m,√

m,√

m〉, (11.29)

where m A, m B , and m are the masses of the A, B, and O atoms, respectively,and mT is the total mass of the unit cell, mT = m A + m B + 3m. The translationeigenvalue is ω2(tα) = 0.

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11.4 Lattice vibrations for the perovskite structure 325

Symmetry functions orthogonal to translations

It is useful to construct symmetry functions that are orthogonal to the translationeigenfunctions. Since the symmetry functions do not mix different rows of the rep-resentations, we now limit our discussion to the row-1 functions. We already havetwo of the five row-1 eigenvectors. The remaining three eigenvectors must be con-structed as linear combinations of the �15 row-1 symmetry functions. They mustbe mutually orthogonal and orthogonal to ηx(�25) and ηt x . The vectors ηx(�25)

and ηt x are already orthogonal to each other. We can not determine the remain-ing eigenvectors from group theory alone. Knowledge of the dynamic matrix isrequired. However, we can form new symmetry functions that are mutually orthog-onal and orthogonal to ηx(�25) and ηt x . To accomplish this, we may employ theGram–Schmidt process.

The Gram–Schmidt orthogonalization processSuppose we have n vectors, v1, v2, . . . , vn , and wish to form n mutually orthogonalvectors, u1,u2, . . . ,un . The Gram–Schmidt process gives a recipe for the u-vectorsin terms of the original v-vectors:

u1 = v1,

u2 = v2 − (u1 · v2)

(u1 · u1)u1,

u3 = v3 − (u1 · v3)

(u1 · u1)u1 − (u2 · v3)

(u2 · u2)u2,

un = vn −n−1∑i=1

(ui · vn)

(ui · ui )ui .

The ui vectors are mutually orthogonal but not normalized.

For our analysis we take the original vectors to be

ηtα =1√mT|x;√m A,

√m B,√

m,√

m,√

m〉, (11.30)

ηBx (�15) = |x; 0, 1, 0, 0, 0〉, (11.31)

ησx (�15) = |x; 0, 0, 1, 0, 0〉, (11.32)

ηπx (�15) = 1√

2|x; 0, 0, 0, 1, 1〉, (11.33)

and apply the Gram–Schmidt process to obtain four mutually orthogonal, normal-ized symmetry functions:

ηtx =1√mT

M1/2|x; 1, 1, 1, 1, 1〉, (11.34)

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326 Applications of space-group theory: Lattice vibrations

Figure 11.7 Symmetry functions for the k = 0 vibration modes of the ABO3perovskites. The center of mass remains stationary for all of the vibration sym-metry functions. (a) Relative physical displacements for the �25(T2u) symmetryfunctions. The B atom and one of the three O atoms are stationary. (b) Relativephysical displacements for the �15(T1u) symmetry functions, M−1/2 η2x . (c) Rel-ative physical displacements for the �15(T1u) symmetry functions, M−1/2 η3x . (d)Relative physical displacements for the �15(T1u) symmetry function, M−1/2 η4x .The B atom and one of the three O atoms are stationary. The numerical ratios ofthe displacements shown are for SrTiO3 with m A = 88, m B = 48, and m = 16 inatomic units.

η2x = 1√N2

M1/2|x; 1,−c2, 1, 1, 1〉, (11.35)

η3x = 1√N3

M1/2|x; 1, 0,−c3, 1, 1〉, (11.36)

η4x = 1√N4

M1/2|x;−c1, 0, 0, 1, 1〉, (11.37)

where

mT = m A + m B + 3m, c1 = 2m

m A, c2 = mT − m B

m B, c3 = m A + 2m

m,

N2 = m A + c22m B + 3m, N3 = m A + (c2

3 + 2)m, N4 = c21m A + 2m.

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11.5 Localized vibrations 327

The symmetry functions in (11.34)–(11.37) are the mass-weighted displace-ments of the atoms of the unit cell. The physical displacement vectors are obtainedfrom these equations by omitting multiplication by the square root of the massmatrix, M1/2. The relative displacements of the atoms are shown in Fig. 11.7 forthe masses of SrTiO3. The center of mass is stationary for all of the vibrationsymmetry functions. That is,

∑ν mνdν = 0, where dν is the displacement vec-

tor for the ν-atom of the unit cell. The symmetry functions ηtx (translation) and η2x

(�25) are eigenfunctions. The remaining three symmetry functions are not eigen-vectors. The linear combination of η2x , η3x , and η4x required in order to constructthe eigenvectors is determined by the dynamic matrix, D. The secular equation is

det

⎡⎣〈η2x |D(0)|η2x〉 − ω2 〈η2x |D(0)|η3x〉 〈η2x |D(0)|η4x〉〈η3x |D(0)|η2x〉 〈η3x |D(0)|η3x〉 − ω2 〈η3x |D(0)|η4x〉〈η4x |D(0)|η2x〉 〈η4x |D(0)|η3x〉 〈η4x |D(0)|η4x〉 − ω2

⎤⎦ = 0.

(11.38)

The solutions of (11.38) are k = 0 optical modes. The eigenvalues and eigenvec-tors can be obtained in terms of the dynamic-matrix elements by employing theformulas for the roots of a cubic equation. We shall not reproduce the results here.

11.5 Localized vibrations

Impurities in a lattice can produce vibrational modes that are highly localized inthe neighborhood of the impurity. These modes can have frequencies that lie abovethe top of the acoustic branches, in gaps between branches, or above the opti-cal branches of the pure crystal. In this section we illustrate the nature of theselocalized modes, starting with the simplest model possible. The extension to morecomplex models is then outlined.

We consider a simple-cubic crystal with one atom per unit cell, with restoringforces due to nearest-neighbor atoms only. The host-crystal atoms have mass m andthe impurity atoms are isotopes with mass m ′. If the concentration of impurities issmall, we can approximate the system by a single impurity in an otherwise perfectcrystal. For the isotope of the host crystal the force constants would be expectedto be essentially the same as the force constants for the regular atoms. Thereforethe entire perturbation is the difference in mass, �M = m ′ − m. The problem ofdetermining the localized vibrational modes is analogous to the case of isotopicallysubstituted molecules discussed in Chapter 2. However, in the molecular case thenumber of vibrational modes of the host molecule is small. For a crystal, the num-ber of vibrational modes is essentially infinite. Nevertheless, the same approach isapplicable.

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328 Applications of space-group theory: Lattice vibrations

x

y

z

aex aey

aez

Figure 11.8 A model for the simple-cubic crystal with nearest-neighbor“springs”. The nearest-neighbor vectors are aex , aey , and aez . The white circlesrepresent normal atoms, and the gray circle indicates the isotopic atom. The springconstant is f and the lattice constant is a. The masses of the atoms throughout thecrystal are m, except at Ri = 0, where the mass is m′.

Assume the impurity is located at R0. The equations of motion for the atomicdisplacement from equilibrium at Ri , d(Ri ), are

− f∑α

{[dα (Ri , t)− dα (Ri + aeα, t)]+ [dα (Ri , t)− dα (Ri − aeα, t)]}

= mω2[1+ εδRi ,R0]∂2

∂t2dα(Ri , t) (α = x, y, and z), (11.39)

where f is the effective “spring constant” (force/cm) between nearest-neighboratoms (see Fig. 11.8). For our simple-cubic crystal the nearest neighbors in theα-direction are located at±aeα, where a is the lattice spacing and ε = (m ′−m)/m.

For the simple spring model we are considering, the motion of an ion in theα-direction does not produce a force on the neighboring ion in a perpendiculardirection. As a consequence the displacements in the x-, y-, and z-directions areuncoupled, but each obeys equivalent equations of motion. Consequently we maydrop the subscript α.

Therefore the model is essentially three, uncoupled, infinite, one-dimensionallines of atoms with an isotopic mass at the origin. Each line has the same set ofnormal modes. Because of the isotope, the crystal no longer possesses translationsymmetry, and the eigenstates are not characterized by a wavevector.

For simplicity, let us consider a line of 2N + 1 atoms extending from −Na to+Na, where a is the lattice constant, and with the isotopic mass located at the

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11.5 Localized vibrations 329

origin. Later we shall take the limit as N tends to infinity. The notation d(n, t)represents the displacement of the atom located at na, where n is a positive ornegative integer, or zero, and t is time. For harmonic motion we can take d(n, t) =d(n) exp(−iωt), so Eq. (11.39) reduces to

{[d(n)− d(n + 1)] + [d(n)− d(n − 1)]} − ω2

γ[1+ εδn0]d(n) = 0, (11.40)

where γ = f/m.Let the (2N + 1)-column vector d represent the 2N + 1 displacements,

d =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

d(N )

d(N − 1)...

d(0)...

d(−N + 1)d(−N )

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

. (11.41)

Each entry is a displacement at a lattice site, so the secular equation is a (2N + 1)× (2N + 1) matrix equation,[

K− ω2

γI− ω2

γ�

]d = 0, (11.42)

where K is a matrix with elements

Knn′ = [2δnn′ − δn+1,n′ − δn−1,n′ ], (11.43)

I is the unit matrix, and the elements of � are

�nn′ = εδnn′δn0. (11.44)

In Eq. (11.42) we can factor out the matrix representing the perfect crystal andobtain [

K− ω2

γI

] [I− ω2

γG(ω2) �

]d = 0, (11.45)

where G(ω2) = [K− (ω2/γ ) I]−1. The eigenvalues are determined by

det

{[K− ω2

γI

] [I− ω2

γG(ω2)�

]}= det

[K− ω2

γI

]

× det

[I− ω2

γG(ω2)�

]= 0. (11.46)

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330 Applications of space-group theory: Lattice vibrations

The eigenvalues of the first determinant are those of the perfect crystal. The eigen-values of the second factor are those associated with the isotopic mass. Thereforethe solutions we seek are given by the condition

det

[I− ω2

γG(ω2)�

]= 1− ε

ω2

γG00(ω

2) = 0, (11.47)

and the displacements satisfy the equation[I− ω2

γG(ω2

iso)�

]d = 0, (11.48)

where ω is evaluated at the eigenvalue of (11.47), ωiso. Equation (11.48) yields

d(n) = ω2iso

γεGn0(ω

2iso)d(0). (11.49)

To find ω2iso we need to obtain expressions for the lattice Green’s function, G00(ω

2).The eigenvectors of the matrix [K − (ω2/γ ) I] are the column vectors, d0(kp),where

d0(kp) = 1√2N + 1

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

eNkpa

e(N−1)kpa

e(N−2)kpa

...

ekpa

e0

e−kpa

...

e(−N+1)kpa

e−Nkpa

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

, (11.50)

and [K− ω2

γI

]d0(kp) =

[[2− 2 cos(kpa)] − ω2

γ

]d0(kp), (11.51)

where kpa = pπ/(2N + 1), p = −N , −N + 1, . . ., 0, . . ., N − 1, N . We form the(2N + 1) × (2N + 1) unitary matrix U and its transpose complex conjugate, U†,from the eigenvectors d0(kp),

Upm = d0(kp)m = 1√2N + 1

eikpma, (11.52)

U †ns = U ∗sn =

1√2N + 1

e−iks na, (11.53)

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11.5 Localized vibrations 331

and U† U = I. The transformed matrix U†[K − ω2 I]U = �(ω2) is a diagonalmatrix of the eigenvalues,

�pp′ = {[2− 2 cos(kpa)] − ω2/γ }δpp′ . (11.54)

The inverse of �(ω2) is

�−1pp′ =

δpp′

[2− 2 cos(kpa)] − ω2/γ, (11.55)

and, since U�−1 U† = [K− (ω2/γ ) I]−1 = G(ω2), we have a prescription for thelattice Green’s-function matrix. In detail, we have

G(ω2)mn = 1√2N + 1

∑p

∑s

eikpmaδpse−iks na

[2− 2 cos(kpa)] − ω2/γ

= 1√2N + 1

∑p

eikp(m−n)a

[2− 2 cos(kpa)] − ω2/γ. (11.56)

In the limit of large N , the sum in (11.56) may be converted into an integral,

G(ω2)mn = a

∫ π/a

−π/adk

eikp(m−n)a

[2− 2 cos(kpa)] − ω2/γ. (11.57)

To determine the isotope eigenvalue we need only G(ω2)00, which is given by

G(ω2)00 = a

∫ π/a

−π/a

dk

[2− 2 cos(ka)] − ω2/γ

= a

∫ π/a

−π/a

dk

4 sin2(ka/2)− ω2/γ

= 1

∫ π/2

0

dx

sin2x − ω2/(4γ )

= −1/4√(ω2/(4γ ))[ω2/(4γ )− 1] . (11.58)

Substituting (11.58) into (11.47) gives the condition for an isotopic, local-mode,eigenvalue,

1+ ε λ√λ(λ− 1)

= 0, (11.59)

where λ = ω2/(4γ ).

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332 Applications of space-group theory: Lattice vibrations

11.5.1 Eigenvalues and eigenvectors for isotopic local modes

From (11.59) we see that there is a singularity at λ = 1 or at ω2 = 4γ . The perfectcrystal has an acoustic branch described by ω2 = [2− 2 cos(ka)] = 4 sin2(ka/2).As ka/2 varies from 0 to π/2, ω2 varies from 0 to 4γ , and therefore ω2 = 4γoccurs at the maximum frequency of the acoustic branch. It is clear from (11.59)that there are no real solutions to the local-mode equation for λ less than 1 (ω2 lessthan 4γ ) because the square-root denominator becomes purely imaginary. Thus, alocal mode can exist only above the top of the acoustic branch. Second, we alsonote that there are no solutions for positive values of ε = (m ′ − m)/m. Negativevalues of ε occur when the isotopic mass is less than the mass of the normal atom.However, physically ε must be greater than −1, since the isotope mass must bepositive. Equation (11.59) can be solved for λ:

λ = 1

1− ε2(−1 < ε < 0). (11.60)

The frequency, ωiso, satisfying the above condition for λ must be greater than 4γ ,and therefore lies above the top of the pure-crystal vibration band. As ε tends to−1+, the frequency tends to infinity, because the isotopic mass is approaching zero.As ε tends to 0−, the frequency approaches that of the top of the acoustic band,since the mass is approaching the normal mass.

To examine the eigenvector for the isotope mode, we may employ (11.49), toobtain

d(n)

d(0)= ω2

iso

γεGn0(ω

2iso).

Since Gn0(ω2) is real for λ = ω2/(4γ ) > 1, it follows that Gn0 = G∗0n = G0n .

Therefore, the displacements for the isotopic mode are symmetrical about theorigin: d(n) = d(−n). From (11.57) we have

Gn0(ω2) = − 1

∫ π/2

0

dx cos(2nx)

λ− sin2(x)(λ > 1). (11.61)

In particular,

G10(ω2) = (1− 2λ)G00(ω

2)− 1

2= (−1/4)(1− 2λ)√

λ(λ− 1)− 1

2. (11.62)

Using (11.60) gives

d(±1)

d(0)= ω2

iso

γεG10(ω

2iso) =

−(1+ ε)

1− ε, −1 < ε < 0. (11.63)

Since −1 < ε < 0, (11.63) shows that the displacements for the neighbors areopposite to that of the isotopic atom. The displacements of adjacent atoms are 180◦

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11.5 Localized vibrations 333

out of phase. The mode is similar to an optical mode except that the amplitudes ofthe displacements decrease with distance from the isotope site. For example, forε = −0.5, d(n)/d(0) = (−1/3)|n|. This corresponds to exponential decay of theamplitude, |d(n)/d(0)| = exp(−|nθ |) with θ = 1.098 612 3 (see Exercise 11.9).For very small negative values of ε the amplitudes of the displacements are signifi-cant over a long distance, but, as ε approaches−1, significant displacements occuronly at the isotope site. That is, the mode becomes highly localized.

The simple model discussed above serves as an introduction to the generallocal-mode problem. For more realistic force-constant models the determina-tion of the eigenvalues and eigenvectors usually requires numerical calculations.Nevertheless, the analytic approach to finding solutions is essentially the same.

In the most general case we are faced with solving the matrix equation

[F+�F− ω2 M− ω2 �M]d = 0, (11.64)

where now F includes all of the forces that one atom can exert on another when theyare in motion. In addition, there may be several atoms per unit cell. The impuritymay be a foreign atom rather than an isotope, and therefore changes in the forceconstants, �F, also occur. The vector d will be a column vector with 3nN entries,where n is the number of atoms per unit cell and N is the total number of unitcells in the crystal. The eigenvalue equation is a 3nN × 3nN equation. However,assuming that the perturbations in force constants and mass are localized at andnear the impurity ion, and that the concentration of impurities is small, the Green’s-function approach is still useful in reducing the problem to a more manageable size.

Writing the matrix equation as

{(D− ω2)−M−1/2[ω2 �M−�F]M−1/2}q = 0 (11.65)

leads to the eigenvalue equation

det{I−G(ω2) [ω2ε−�f]} = 0, (11.66)

where ε ≡ M−1/2 �M M−1/2, �f ≡ M−1/2 �F M−1/2 and G(ω2) = (D − ω2)−1.If the range of the perturbations affects m neighboring cells, then (11.66) will bea 3nm × 3nm determinant. However, it should be kept in mind that the number ofsolutions with real values of ω2 may be less than 3nm, or there may be none at all.In general the Green’s function, G(ω2), is complex when ω2 lies within a band offrequencies of the host crystal. When a solution of (11.66) is a complex number,ω = ω(real)+ i�, the local mode is not a stationary state. Rather, it is interpretedas a mode with a finite lifetime that is proportional to 1/�. Such finite-lived modesare called continuum resonances.

The lattice Green’s function has the full space-group symmetry of the host crys-tal, but the introduction of an impurity removes the translation operations, so that

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334 Applications of space-group theory: Lattice vibrations

the solutions of (11.66) will be characterized by the point-group symmetry of theperturbations, �F or �M, depending on which has the lower symmetry.

References

[11.1] G. W. Lehman, T. Wolfram, and R. E. De Wames, “Axially symmetric model forlattice dynamics of metals with application to Cu, Al, and ZrH2”, Phys. Rev. 128,1593–1599 (1962).

[11.2] C. B. Walker, “X-ray study of lattice vibrations in aluminum”, Phys. Rev. 103, 547–557 (1956).

[11.3] E. C. Svensson, B. N. Brockhouse, and J. M. Rowe, “Crystal dynamics of copper”,Phys. Rev. 155, 619–632 (1967).

[11.4] B. G. Dick, Jr. and A. W. Overhauser, “Theory of the dielectric constants of alkalihalide crystals”, Phys. Rev. 112, 90–103 (1958).A. D. B. Woods, W. Cochran, and B. N. Brockhouse, “Lattice dynamics of alkalihalide crystals”, Phys. Rev. 119, 980–999 (1960).

Exercises

11.1 Consider the perovskite structure with k = (0, 0, t), 0 ≤ t ≤ π/c.(a) What is the group of the wavevector, gk?(b) Identify the operations of gk in terms of the operations listed in Table E.2

of Appendix E.(c) Construct a table showing how the 2, 3, and 5 O sites of the central unit

cell transform under the operations of gk. Mark the sites that transforminto themselves or an equivalent site. Include in the table rows showinghow x , y, and z transform among themselves.

(d) Use the table to find the character of each of the operations of gk for the9 × 9 representation based on the nine displacement vectors of the threeO atoms.

(e) What are the translation phase factors?(f) Decompose the 9× 9 representation into the IRs of gk.(g) Decompose the two 3 × 3 representations based on the displacement

vectors for the A and B atoms.(h) Discuss the form of the block-diagonal dynamic matrix.

11.2 Consider the perovskite structure with k = (0, 0, t), 0 ≤ t ≤ π/c.(a) Use the information in Exercise 11.1 to find 15 symmetry functions.(b) Make a sketch of the displacements associated with the symmetry

functions.(c) Identify the transverse and longitudinal symmetry functions.(d) Which symmetry functions are eigenfunctions?

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Exercises 335

(e) Show that

ω2(B1) = 1

2[D22

zz + D33zz − D23

zz − D32zz ],

where, for Dnn′zz , n and n′ indicate the O-atom site numbers.

11.3 For the perovskite structure (in the phonon ground state) answer the follow-ing.(a) What vibration modes at k = 0 are infrared active?(b) Which modes are Raman active?(c) Absorption of a photon requires conservation of wavevector as well as

conservation of energy. Therefore absorption of a photon resulting in thecreation of a phonon is not a vertical transition. That is, if the initialstate has k = 0, the final phonon state can not have k = 0. Calcu-late the ratio of the wavevector of a 1,000-cm−1 infrared photon to thesmallest reciprocal-lattice vector for a cubic crystal with lattice constanta = 4 Å. What can be concluded about the neglect of the conservation ofmomentum?

11.4 For the CsCl structure with k along the � line of symmetry (k = (t, t, t),with 0 ≤ t ≤ π/c):(a) Find gk and its elements in terms of the operators in Table E.1.(b) Find the IRs of the vibrations.(c) Find the six symmetry functions, and identify the transverse and longitu-

dinal functions.(d) Find expressions for the longitudinal acoustic and optical modes in terms

of the matrix elements of the dynamic matrix, D.11.5 For the axially symmetric model, translations automatically give zero fre-

quencies. However, for the general model, Eqs. (11.2)–(11.10), the require-ment that a translation have zero frequency places constraints on its dynamicmatrix. For the CsCl structure, |1, 1, 1, 1, 1, 1〉 represents a translation inξ -space. Show that this requires

∑β

Daaαβ = −

√m B

m A

∑β

Dabαβ,

∑β

Dbbαβ = −

√m A

m B

∑β

Dbaαβ,

m A

∑β

Daaαβ = m B

∑β

Dbbαβ.

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336 Applications of space-group theory: Lattice vibrations

11.6 According to the theory of elastic media, the velocity of sound, v, for a cubiccrystal is given by the solutions of

det{E− ρv2 I} = 0,

where E is the 3 × 3 elastic-constant matrix, ρ is the mass density, v is thevelocity of sound for a given direction of the wavevector k, and I is the unitmatrix. The matrix elements of E are given by

E11 = c11 l2 + c44(m2 + n2),

E22 = c11 m2 + c44(l2 + n2),

E33 = c11 n2 + c44(l2 + m2),

E12 = E21 = (c12 + c44)lm,

E13 = E31 = (c12 + c44)ln,

E23 = E32 = (c12 + c44)mn,

where l, m, and n are the direction cosines of the wavevector k, and ci j arethe elastic constants for the crystal.

Find the sound velocities for propagation along the [100], [110], and [111]directions in terms of the elastic constants. (Hint: For the [111] direction, usesymmetry functions to diagonalize the E matrix.)

11.7 For the perovskite SrTiO3, the elastic constants are c11 = 3.162, c12 = 1.035,and c44 = 1.212, all in units of 1012 dynes/cm2. Calculate the sound velocitiesin cm/s for k in the three directions [100], [110], and [111]. In calculating thedensity ρ, assume the lattice constant is 4 Å.

11.8 (a) Using the axially symmetric model (Eqs. (11.17)–(11.19)) for a cubiccrystal with one atom per unit cell, find the dynamic-matrix elements,Dαβ(k) including nearest- and next-nearest-neighbor interactions.

(b) Find the longitudinal and transverse frequencies in terms of the elementsof the dynamic matrix for k = (t, 0, 0).

(c) Repeat part (b) for k = (t, t, 0).11.9 For the localized vibration modes discussed in the chapter, do the following.

(a) Show that, for |n| > 1, ∣∣∣∣d(|n|)d(0)

∣∣∣∣ = (−1)|n|e−|n|θ ,

where e−θ = (1+ ε)/(1− ε) and θ is a real positive number.(b) Show that, for ε = −0.5, θ = 1.098 612 3.

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12

Time reversal and magnetic groups

12.1 Time reversal in quantum mechanics

In this chapter we investigate the effects of time-reversal symmetry on quantumoperators and systems. For Newton’s classical equations the force on an object isrelated to the second time derivative, ∂2r/∂t2, of the spatial variable, r. Substitu-tion of −t for t does not change the equations of motion. Therefore, if r(t) is asolution of the equations of motion, so is r(−t). On the other hand, the velocity ormomentum is reversed (flows backward), meaning that p(t) = −p(−t) under timereversal. The oddness of the momentum under time inversion applies not only tolinear momentum but also to angular momentum.

In quantum mechanics we are concerned with Schrödinger’s equation and theeffect of time reversal on the quantum operators and wavefunctions. The timeevolution of the wavefunction, �(r, t), is determined by the unitary operator,exp(−i Ht), where H is the system Hamiltonian. The evolution of � from 0 tot is given by �(t) = exp[−i Ht]�(0). If H itself is time-independent and tis reversed, the wavefunction evolves backward in time according to �(−t) =exp[+i Ht]�(0).

12.1.1 Properties of the time-reversal operator, T

An appropriate time-reversal operator, T, when applied to a system must accom-plish a number of things. If we denote space coordinates by r, velocity by v, linearmomentum by p, angular momentum by J, current by I , charge density by ρ,electric field by E, magnetic field by Hf, and the spin by S, then

T f (r)→ f (r),

T E→ E,

T ρ → ρ,

337

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338 Time reversal and magnetic groups

T(v, J, I )→−(v, J, I ),

T(Hf, S)→−(Hf, S).

The appropriate time-reversal operator depends on the Hamiltonian. If H includesno spin operators or external magnetic field, T may be chosen as the complex-conjugation operator, K0, since K0[exp(−i Ht)] = [exp(−i Ht)]∗ = [exp(i Ht)] =[exp(−i H(−t)]. When spin-dependent terms are present in H, T must also reversethe spins, since they are angular-momentum operators. Finally, if an externalmagnetic field is also present, T must also reverse the direction of the field.For stationary states of Schrödinger’s equation, �(r, t) = exp(−i Et) ψ(r) andT�(r, t) = exp(i Et)Tψ(r, 0). We can define the time-reversal operator by therequirement that it commutes with coordinates and functions of coordinates, butanticommutes with momentum operators,

T r = r T, (12.1)

T V (r) = V (r)T, (12.2)

T p = T (−i�∇) = i�∇T = −p T, (12.3)

T J = T (r× p) = −(r× p)T = −J T, (12.4)

T S = −S T, (12.5)

where J is any angular-momentum operator, and S is the total spin vector operator.If a one-electron Hamiltonian is of the form

H = p2

2m+ V (r), (12.6)

without spin operators, then (12.1)–(12.5) are satisfied by T = K0, the complex-conjugate operator.1 Since the Hamiltonian of (12.6) is real, K0 commutes withH. However, if H contains spin operators, as it does in the case when spin–orbit interactions are included, K0 is not adequate, because it does not reversethe spin angular momentum as required by (12.5). A Hamiltonian with spin–orbitinteractions is of the form

H = p2

2m+ V (r)+ �

4m2c2

[∇V (r)× p · σ]. (12.7)

In this instance the time-reversal operator can be taken as a product of operators,2

T = σy K0, (12.8)

1 Representing the time-reversal operator by complex conjugation is valid only in a spatial-coordinate represen-tation such as Schrödinger’s equation.

2 T could also be chosen as σy K0 or = eiθ σy K0, where θ is an arbitrary, but fixed, real number. In particular,T could be taken as iσy K0.

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12.1 Time reversal in quantum mechanics 339

where σy is the Pauli spin matrix,

σy =(

0 −ii 0

). (12.9)

We can demonstrate that T = σy K0 satisfies (12.1)–(12.5). Since σy operatesonly on the spin, we need only show that σy K0 anticommutes with the spin vectoroperator σ and commutes with the spin–orbit term of the Hamiltonian of (12.7).First consider a one-electron Hamiltonian with spin–orbit interaction, with σ =exσx + eyσy + ezσz , where ex , ey , and ez are unit vectors along the x-, y-, andz-directions, respectively. We have

σy K0(exσx) = exσyσ∗x K0 = ex

(0 −ii 0

)(0 11 0

)K0

= ex

( −i 00 i

)K0 = −(exσx)σy K0, (12.10)

σy K0(eyσy) = eyσyσ∗y K0 = ey

(0 −ii 0

)(0 i−i 0

)K0

= −ey

(1 00 1

)K0 = −(eyσy)σy K0, (12.11)

σy K0(ezσz) = ezσyσ∗z K0 = ez

(0 −ii 0

)(1 00 −1

)K0

= ez

(0 ii 0

)K0 = −(ezσz)σy K0. (12.12)

On collecting the components together, we have

T σ = −σ T. (12.13)

T also commutes with the spin–orbit term for real V (r),

σy K0[∇V (r)× p·σ] = σy[∇V (r)× (−p) · σ]K0

= [∇V (r)× (−p) · (−σ)]σy K0

= [∇V (r)× p · σ]σy K0, (12.14)

and therefore commutes with the Hamiltonian.The above proof establishes a valid time-reversal operator for a single electron.

The form of the many-electron Hamiltonian is

H =∑

i

{p2

i

2m+ �

4m2c2[∇i V (r1, r2, . . . , rN )× pi · σi ]

}+ V (r1, r2, . . . , rN ).

(12.15)

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340 Time reversal and magnetic groups

For the N -electron system we may take the time-reversal operator as

T = �K0, � =N∏

n=1

σny, (12.16)

where σny is the y-component of the spin operator which operates only on the nthelectron’s spinor.

12.2 The effect of T on an electron wavefunction

Let �(r1, r2, . . . , rN ; σ1z, σ2z, . . . , σN z) = �(ri , σi z) be an antisymmetrized sta-tionary state of an N -electron system, where ri is a space coordinate and σi z is thespin state of the i th electron. The spin, σi z , may be either α (spin up) or β (spindown). For the Hamiltonian of (12.15), T = �K0 = �iσiy K0 is a symmetryoperation. Since

T α = iβ and T β = −iα, (12.17)

it follows that

T�(ri , σi z) = i n(α)(−i)n(β) �(ri ,−σi z)∗, (12.18)

where n(α) is the number of σi z that are equal to α, n(β) is the number of σi z thatare equal to β in �(ri , σi z), and −σi z is the opposite spin state of σi z (that is, ifσi z = α then −σi z = β and vice versa). Note that, for �(ri ,−σi z), n(α) is thenumber of β spins and n(β) is the number of α spins. As a result,

T2� = [i n(α)(−i)n(β)]∗[i n(β)(−i)n(α)]� = i2N� = (−1)N �, (12.19)

where N = n(α)+ n(β). The final result is

T2 � = ±� (+ for N even,− for N odd). (12.20)

Since (12.20) holds for every N -electron wavefunction, we can infer that the T

operator satisfies the condition

T2 = ±1. (12.21)

If � is an eigenstate with eigenvalue λ, and T commutes with the Hamiltonian, thenT� is also an eigenstate with the same eigenvalue. The question is whether T� isa linearly independent state or simply � multiplied by a phase factor, eiθ . If T� =eiθ � then � and T� are equivalent, and time reversal does not require additionaldegeneracy. It turns out (as we show below) that if the number of electrons is oddthen T� is orthogonal to � and therefore is an independent state. In this case, atleast two different states have the same eigenvalue, λ. For atoms and molecules thismeans that the energy level is at least doubly degenerate (Kramers’ theorem [12.1]).

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12.3 Time reversal with an external field 341

Kramers’ theorem [12.1]If the Hamiltonian commutes with the time-reversal operator, the eigenvalues of asystem with an odd number of electrons are at least doubly degenerate. In general thedegeneracy is even; that is, the degeneracy of the eigenvalue is 2n, where n is apositive integer greater than zero. The degeneracy is not lifted by an electric field or acrystal field, but may be lifted by application of an external magnetic field.

To show that T� is orthogonal to �, we show that the (Hermitian) scalarproduct 〈�|T�〉 = 0 if the number of electrons is odd. Consider two arbitraryfunctions, f and g. We have that

〈T g|T f 〉 = 〈�K0 g|�K0 f 〉 = 〈K0 g|K0 f 〉 = 〈g| f 〉∗ = 〈 f |g〉. (12.22)

The second equality results because � is a unitary operator, and applying a unitaryoperator to both functions does not change the scalar product. If we choose f = �

and g = T�, then (12.22) yields

〈T g|T f 〉 = 〈T(T�)|T�〉 = 〈T2 �|T�〉 = −〈�|T�〉 = −〈 f |g〉, (12.23)

since T2 = −1 when N is odd. Both (12.22) and (12.23) can be true if and only if

〈�|T�〉 = 0. (12.24)

Equation (12.24) shows that � and T� are orthogonal states when N is odd. If Nis even, the same proof yields only the tautology 〈�|T�〉 = 〈�|T�〉. It should benoted that (12.24) depends only on having time reversal as a symmetry operator.

12.3 Time reversal with an external field

In the presence of an external electric field the Hamiltonian will contain a potential-energy term, eE, where E is the electric field. The time-reversal operator commuteswith this term. In general, however, electromagnetic fields are accommodated inquantum systems by the replacement of p with (p − eA/c), where A is a vectorpotential, c is the velocity of light, and e is the magnitude of the electron charge. Ifthe fields are time-varying, the requirements of the time-reversal operator are

T E(r, t) = E(r,−t) (E is the electric field), (12.25)

T A(r, t) = −A(r,−t) (A is the magnetic vector potential), (12.26)

T Hf(r, t) = −Hf(r,−t) (Hf is the magnetic field strength). (12.27)

If T satisfies Eqs. (12.25)–(12.27) and commutes with the Hamiltonian,and �(r,Hf, t) is an eigenstate of the Hamiltonian, then T�(r,Hf, t) =�(r,−Hf,−t) is also an eigenstate with the same eigenvalue. If a 180◦ rotation, δ,

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342 Time reversal and magnetic groups

about an axis perpendicular to the direction of Hf commutes with the Hamiltonianin the absence of the magnetic field, then the time-reversal operator may be takenas δ �K0.

Unitary and antiunitary operatorsAn operator U is unitary if U−1 is the transpose, complex conjugate of U. U is alinear operator that satisfies U(c ψ(r)+ d φ(r)) = c Uψ(r)+ d Uφ(r) (c and dconstants) and 〈Uψ(r)|Uφ(r)〉 = 〈ψ |φ〉.

An operator A is an antiunitary operator that satisfiesA[c ψ(r)+ d φ(r)] = c∗ Aψ(r)+ d∗ Aφ(r) and 〈Aψ(r)|Aφ(r)〉 = 〈ψ(r)|φ(r)〉∗.

The time-reversal operator, T, is an antiunitary operator. T = � K0 is a product ofa unitary operator, �, and an antiunitary operator, K0.

The product of two unitary or two antiunitary operators is unitary. The product of aunitary operator with an antiunitary operator is antiunitary.

12.4 Time-reversal degeneracy and energy bands

For crystalline solids the symmetry elements belong to a space group. The wave-functions for the energy bands are one-electron Bloch waves characterized by awavevector k, as discussed in Chapter 9. Let ψk,α and ψk,β represent the Blochwaves for spin-up and spin-down states. The two states are degenerate in theabsence of spin-dependent terms in the Hamiltonian and an external magnetic field.

Applying T = σy K0 to the Bloch wavefunctions gives

σyK0 ψk,α(r) = σy[e−ik·r u∗k,α(r)]α = e−ik·r u∗k,α(r)(iβ), (12.28)

σyK0 ψk,β(r) = σy[e−ik·r u∗k,β(r)]β = e−ik·r u∗k,β(r)(−iα). (12.29)

We define

i u∗k,α(r) = u−k,β(r), (12.30)

−i u∗k,β(r) = u−k,α(r), (12.31)

so that

Tψk,α = ψ−k,β, (12.32)

Tψk,β = ψ−k,α. (12.33)

If ψk,s is an eigenfunction of a Hamiltonian, Tψk,s = ψ−k,−s is also an eigenstatewith the same eigenvalue. It follows that every one-electron eigenvalue is at leastdoubly degenerate (Kramers’ theorem). To distinguish Bloch waves of differentenergy bands we can add the index n so that ψn,k,s is a wavefunction for the nthband with wavevector k and spin state s.

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12.4 Time-reversal degeneracy and energy bands 343

If T is a symmetry operator of the space group, the states ψn,k,s and ψn,−k,−s

have the same energy, so

E(n, k, s) = E(n,−k,−s). (12.34)

If the space group also contains spatial inversion, then

E(n, k, s) = E(n,−k, s). (12.35)

In this case the states ψn,k,s and ψn,−k,s and their time-reversed partners ψn,−k,−s

and ψn,k,−s all have the same energy. The energy as a function of k is shownin Figs. 12.1(a) and (b). If the group of the Hamiltonian does not have spatialinversion, then only (12.34) holds. In this case only the time-reversed pairs haveequal energy, as illustrated in Figs. 12.1(c) and (d). For a “spinless” Hamiltonian,the energy will display inversion in k-space even if inversion is not a symme-try operator, since in this case �n,k and �n,−k are time-reversed partners. Ingeneral, if k is not a point of high symmetry, and there are no spin-dependentterms in the Hamiltonian, �n,k,s and �n,−k,−s have the same energy. In addition,E(n,−k, s) = E(n, k, s), so the energy eigenvalue is four-fold degenerate withE(n,±k, s) = E(n, k,±s) (Figs. 12.1(a) and (b)).

If the Hamiltonian does not contain spin-dependent terms but does contain thetime reversal and spatial inversion, then the gradient of the energy ∇k E(n, k, s)must vanish at k = K/2, where K is any reciprocal-lattice vector (including K =0). To see this, note that the i th Cartesian component of the gradient is

∂E(n, 12 K, s)

∂ki= lim+�ki→0

{E(n, ( 1

2 K+�ki ), s)− E(n, ( 12 K−�ki ), s)

+2�ki

}.

(12.36)

Under spatial inversion,

E(n, k, s) = E(n,−k, s). (12.37)

This yields

E

(n,

(1

2K−�ki

), s)= E

(n,

(−1

2K+�ki

), s). (12.38)

We also have

E(n, k, s) = E(n, (k+K), s), (12.39)

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344 Time reversal and magnetic groups

K2

K2

iΔk

iΔk

(a)

(b)

K2

K2

iΔk

iΔk

(c)

(d)

E(n, −k, ±s) E(n, +k, ±s)

E(n, −k, ±s) E(n, +k, ±s)

E(n, −k, −s)

E(n, −k, +s)

E(n, +k, +s)

E(n, +k, −s)

E(n, +k, +s)

E(n, +k, −s)E(n, −k, +s)

E(n, −k, −s)

Figure 12.1 A schematic representation of energy bands near k = K/2:(a) and (b), time reversal and inversion symmetry for convex and concave bands,∇k E(n,k, s) = 0 at k = K/2; (c) and (d), time reversal but no inversionsymmetry for convex and concave bands, ∇k E(n,k, s) �= 0 at k = K/2.

so (12.38) becomes

E

(n,

(K2−�ki

), s)= E

(n,

(−K

2+K+�ki

), s)

= E

(n,

(K2+�ki

), s), (12.40)

and therefore the numerator of (12.36) vanishes. This proves that in the absence ofspin-dependent terms in the Hamiltonian the gradient vanishes at k = K/2, whereK is any reciprocal-lattice vector.

If the Hamiltonian has spin-dependent terms, ∇k E is not required to van-ish because of symmetry. The time-reversed pairs still have the same energy,(E(n, k, s) = E(n,−k,−s) and E(n, k,−s) = E(n,−k, s)), but E(n,−k,−s) �=E(n, k,−s). This situation is illustrated in Figs. 12.1(c) and (d).

A test for “additional” degeneracy (resulting from time reversal) that employsonly the characters of the point group gk (group of the wavevector) of the spacegroup has been developed by Herring [12.2]. Denote an operator that carries thewavevector, k, into −k or its equivalent by the symbol Qi , and denote the set of

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12.4 Time-reversal degeneracy and energy bands 345

such operators as {Q}. It follows that Q2i carries k into itself or an equivalent vector

and therefore must be an operator in gk. The Herring test involves the sum of thecharacters of the squares of the operators,

∑i χ(Q

2i ). The sum has three possible

values, namely hQ, 0, and −hQ, where hQ is the number of operators in {Q}. Theadditional degeneracy of the energy levels belonging to �α due to the addition ofthe time-reversal symmetry depends upon the value of

∑i χ

α(Q2i ).

Theorem 12.1 (Herring rules for additional degeneracy of energy bands) Let �α bean IR of a point group, gk, and Qi an operator that transforms k into−k or−k+K(where K is a reciprocal-lattice vector). The sum of the characters,

∑i χ

α(Q2i ),

has three possible values, namely hQ, 0, and −hQ, where hQ is the number ofQi operators. The additional degeneracy of the levels belonging to �α due to theaddition of the time-reversal operator is determined by the value of

∑i χ

α(Q2i ).

(a)∑

i χα(Q2

i ) = hQ.Time reversal leads to no additional degeneracy for integral spin, but doublesthe degeneracy for half-integral spin.

(b)∑

i χα(Q2

i ) = 0.Time reversal leads to a doubling of the degeneracy both for integral and forhalf-integral spin.

(c)∑

i χα(Q2

i ) = −hQ.Time reversal leads to a doubling of the degeneracy for integral spin but noadditional degeneracy for half-integral spin.

To employ the Herring test, we need to find the Q operators. The group gk con-tains all of the point-group operations that transform k into k or k+K, where K isa reciprocal-lattice vector. Denote the set of operators in gk by {O j }, j = 1, 2, . . . ,h(gk). We consider two cases.

Case (1) The group of the wavevector, gk, contains the inversion operator, Pinv.We note that Pinvk = −k, and therefore Pinv can be a member of gk only ifk = 0 or K/2, where K is a reciprocal-lattice vector. If k = 0 or K/2, thenany O j in gk satisfies the condition that O j k = −k or its equivalent. In thiscase the set {Qi } = {O j } = gk, so

∑i χ(Q

2i ) =

∑j χ(O2

j ).Case (2) The group of the wavevector, gk, does not contain the inversion oper-

ator. For this case Pinv is not a member of gk; however, if k �= 0 or K/2,then (Pinv O j )k = Pinv(k+K) = −k+K′. Therefore the set of operators{Pinv O j } = Pinv gk = {Qi }. When an operator in Pinv gk is squared, usually(but not always) the inversion operator drops out. If Pinv commutes with theoperators of gk, then (Pinv O j )

2 = P2inv O2

j = O2j . In this instance we have

again that∑

i χ(Q2i ) =

∑j χ(O2

j ).

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346 Time reversal and magnetic groups

Table 12.1 The character table for the double groupC3. The last column lists the type of IR whentime-reversal symmetry is added.

C3 E C23 C3 IR type

A 1 1 1 aE 1 ω ω2 bE 1 ω2 ω b

ω = e2π i/3

In summary, if Pinv commutes with the operators of gk, then∑

i χ(Q2i ) =∑

j χ(O2j ) regardless of the k-vector.

As an example of the Herring rules we consider gk = C3. The character table isgiven in Table 12.1.

The group C3 does not contain the inversion operator, hence {Q} = {Pinv(gk)},where Pinv is the inversion operator. The set {Pinv(gk)}= {i , iC3, iC2

3 }. The squareof the operators is the set {E , C2

3 , C3}, which is just the operators of the group C3.For the squares of the various operators of C3 we have χ(E2) = χ(E),

χ([C3]2) = χ(C3), and χ([C23 ]2) = χ(C3).∑

i Q2i = 3 for �A, and therefore this IR is of type (a) according to the Herring

rules. For this IR time reversal leads to no additional degeneracy for integral spin,but doubles the degeneracy for half-integral spin. For the two E IRs,

∑i Q2

i =1+ eiπ/3 + e2iπ/3 = 0, so they are IRs of type (b). For these two IRs, time reversalleads to a doubling of the degeneracy both for integral and for half-integral spin.

12.5 Magnetic crystal groups

Crystals with magnetic (ferromagnetic, ferrimagnetic, or antiferromagnetic) orderare not invariant under the time-reversal operator, T, since this operation reversesthe direction of the spins. However, in some cases (e.g., antiferromagnets) the crys-tal may be invariant under a space-group operation {T|τ}, that is, time reversal plusa translation. Also, even when T itself is not in the point symmetry group, productoperators such as T U, where U represents an ordinary (unitary) operation, may be.The aim of the discussion here is to express the magnetic point groups in terms ofthe 32 ordinary point groups (type I groups) and certain invariant subgroups.

Invariant subgroups

Before beginning our discussion of magnetic crystal groups, we need to introducethe concept of an invariant subgroup or normal divisor. Theorem 12.2 provides adefinition.

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12.5 Magnetic crystal groups 347

Theorem 12.2 (Invariant subgroups)1. A collection of elements of G0 that themselves form a smaller group is a sub-

group of G0. There may be several distinct subgroups. A subgroup Gu is aninvariant subgroup if it consists of complete classes of G0.

2. Let Gu be an invariant subgroup of G0. Suppose G0 has h0 elements and Gu hashu elements, then h0/hu is the index of Gu relative to G0. The index is a posi-tive integer. That is, hu is a divisor of h0; hence Gu is called a normal divisorof G.

3. An invariant subgroup has the property that {X Gu} = {Gu X}, where X is anyelement of G0. If we consider the ordering of the elements in the sets to beirrelevant, we may write X Gu = Gu X. This property follows from the fact thatGu consists of complete classes.

4. If Gu is an invariant subgroup of G0, and X is a member of G0 but not of Gu,then X Gu = {G0 − Gu} (the order of elements is irrelevant).

5. Any subgroup of index 2 is an invariant subgroup.

As an example of an invariant subgroup, consider the group C4v. The multipli-cation table is shown in Table 12.2. The classes for this group are {E}, {C4, C3

4 },{C2}, {σv1, σv2}, and {σd1, σd2}. (For a detailed description of the operations, seeTable 1.1 in Chapter 1.) Inspection of Table 12.2 shows that {E , C2} is an invariantsubgroup (index 4) since it consists of entire classes. Another invariant subgroupis {E , C4, C3

4 , C2} (index 2).For the purpose here we shall be concerned only with invariant subgroups of

index 2. For the C4v group there are several index-2, invariant subgroups: {E ,C2, C4, C3

4 }, {E , C2, σv1, σv2}, and {E , C2, σd1, σd2}. The second subgroup isisomorphic (equivalent) to the third.

12.5.1 Types of magnetic crystal groups

Type II magnetic groups

For diamagnetic or paramagnetic crystals the time-averaged magnetic moment van-ishes. The time-reversal operator, T, is an element of the magnetic point group,G. For these systems, G is the combination of the unitary operations {U} of G0

and antiunitary operations {T U}, where “{. . .}” denotes a set of operators. Thesegroups, called type II magnetic groups, have

G = {U} + {T U} = G0 + T G0 (type II). (12.41)

For type II groups, G0 is the invariant subgroup of G of index 2. For each of the32 ordinary point groups we can form a type II magnetic group, G0 + T G0, andtherefore there are 32 such magnetic groups.

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348 Time reversal and magnetic groups

Table 12.2 The multiplication table for the group C4v

C4 E C4 C2 C34 σv1 σv2 σd1 σd2

E E C4 C2 C34 σv1 σv2 σd1 σd2

C4 C4 C2 C34 E σd1 σd2 σv2 σv1

C2 C2 C34 E C4 σv2 σv1 σd2 σd1

C34 C3

4 E C4 C2 σd2 σd1 σv1 σv2

σv1 σv1 σd2 σv2 σd1 E C2 C34 C4

σv2 σv2 σd1 σv1 σd2 C2 E C4 C34

σd1 σd1 σv1 σd2 σv2 C4 C34 E C2

σd2 σd2 σv2 σd1 σv1 C34 C4 C2 E

Type III magnetic groups

For magnetic crystals with a net magnetic moment the time-reversal operator T

is not in the point group G, but operators such as T U can be. For example, if aferromagnet has a C2 symmetry axis perpendicular to the magnetic direction, thenT C2 leaves the magnetic moments unchanged. In this case T is not a symmetryoperation, but T C2 is.

Let G0 be one of the 32 ordinary point groups. We may write G0 = Gu+{G0−Gu}, where Gu is an invariant subgroup of G0 of index 2 and {G0−Gu} is the set ofoperators not in Gu. (Note that {G0 − Gu} is not a group, since it does not containthe identity operator.) A type III magnetic point group, G, can be formed from thegroup of unitary operators, Gu, and the set of antiunitary operators T {G0 − Gu}:

G = G0 + T {G0 − Gu} (type III). (12.42)

Magnetic, type II and type III, point groups can also be constructed as

G = Gu + A Gu, (12.43)

where A is any antiunitary operator in G. If G contains T then we may chooseA = T. In this instance G = Gu + T Gu, which is a type II magnetic point groupwith Gu an invariant subgroup of G of index 2. If T itself is not in G we maychoose A = T Ui , where Ui is any member of {G0 − Gu}. In this case G =Gu+T Ui Gu. However, for an invariant subgroup of index 2, Ui Gu = {G0−Gu}(see Theorem 12.2). Therefore we arrive at the same construction as (12.42).

In summary, the three types of groups are

type I, G = G0 (32 ordinary point groups), (12.44)

type II, G = G0 + T G0 (paramagnetic, diamagnetic), (12.45)

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12.5 Magnetic crystal groups 349

type III, G = Gu + T (G0 − Gu) (ferromagnetic, ferrimagnetic,

antiferromagnetic), (12.46)

type III, G = Gu + A Gu (A any member of T {G0 − Gu}). (12.47)

There are 32 type I (ordinary point groups), 32 type II magnetic point groups,and 58 (distinct) type III magnetic point groups. The type II and type III groups areshown in Table 12.3 for different crystal systems. Also indicated are the magneticorderings corresponding to the magnetic point groups.

The magnetic point groups are called Shubnikov [12.3] point groups or colorgroups. For type II and type III, the magnetic point groups contain antiunitary oper-ators, and the representations do not obey the usual matrix-multiplication rules.This aspect of magnetic groups is discussed in Section 12.6.

As an example of the construction of a type III magnetic group, consider thesquares shown in Fig. 12.2. In Fig. 12.2(a) we have a square with identical atoms(with no magnetic moments) on each of the four corners. In Fig. 12.2(b) we haveatoms with their magnetic vectors alternating in direction. For Fig. 12.2(a), thesimple square of non-magnetic atoms, the two-dimensional covering group is G0 =C4v. The operations of the group for Fig. 12.2(a) shown in Fig. 12.2(c) include theeight operators E , C2, C4, C3

4 , σd1, σd2, σv1, and σv2. The set of symmetry elements{E , C2, σv1, σv2} = Gu forms an invariant subgroup of C4v of index 2, and thesquare in Fig. 12.2(b) remains invariant under these operations. For the square inFig. 12.2(b) time reversal is not a symmetry operation since it reverses the magneticvectors. In addition, the operators of the set {C4, C3

4 , σd1, σd2} = {G0−Gu} are alsonot symmetry operations for the square in Fig. 12.2(b). However, the operators forthe set T {C4, C3

4 , σd1, σd2} = T {G0−Gu} are symmetry operations for the square

12

34

(a)

12

34

(b)

σd1

σd2

σv1

σv2

C2, C4, C43

(c)

Figure 12.2 (a) A square with C4v symmetry. (b) Symmetry reduced by magneticvectors. (c) Rotation and reflection operations of C4v .

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350 Time reversal and magnetic groups

in Fig. 12.2(b). In this case G0 = Gu+ (G0−Gu) is a point group for Fig. 12.2(a),and G = Gu+T (G0−Gu) is a magnetic point group for Fig. 12.2(b). The alternateconstruction of G for C4v (Eq. (12.43)) gives the same results. Note that we couldhave formed a type III magnetic point group with Gu = {E , C2, σd1, σd2} and{G0−Gu} = {C4, C3

4 , σv1, σv2}. This magnetic group is isomorphic to the previousone, and corresponds to simply rotating the square in Fig. 12.2(b) by 90◦.

In general, a magnetic point group can also be specified in Schönflies notationas G0(Gu). For example, D6h(C6v) is the type III group for which G0 = D6h ,Gu = C6v, and G = C6v + T (D6h − C6v). The type II and type III magnetic pointgroups for different crystal systems are shown in Table 12.3 using the Hermann–Mauguin/international notation.

Hermann–Mauguin/international notationInternational notation specifies some of the symmetry elements of a group, but givesonly the information necessary to deduce the entire group of symmetry elements.

The first integer, n, specifies the principal n-fold symmetry axis. Subsequentintegers indicate lesser rotation axes (in descending order).

Inversion through a point followed by a rotation is indicated as “n”. Mirror planesthat contain a rotation axis or that are perpendicular to an n-fold axis are indicated bythe symbol “m”. If the mirror plane is perpendicular to a rotation axis, then a slash isplaced between the axis number and the “m”. For example, “m/n” indicates an n-foldrotation axis with a mirror plane perpendicular to it.

For magnetic groups, underlined characters indicate complementary operations,that is, operations of the type T O, where O is a rotation or mirror operation.

Indications of screw axes and glide planes can be added. These are shown byadding an integer subscript to the notation. The integer indicates the fraction of thetranslation along the axis. For example, 21 is a 180◦ rotation about the two-fold axisand translation by half of the lattice constant along the axis, whereas 62 indicates a60◦ rotation about a six-fold axis and translation by 2/6 = 1/3 of a lattice constantalong the axis.

The possible screw axes are 21, 31, 31, 42, 61, 62, and 63. The symbols a, b, and cindicate glide-plane axes along the a, b, and c crystal axes, respectively.

12.6 Co-representations for groups with time-reversal operators

For a group that contains antiunitary operators the usual representation matricesare inadequate. Instead, a different type of representation called a co-representationmust be employed. The algebra of co-representations was initially worked out byWigner [12.4]. For a different discussion see Lax [12.5]. Here we shall give briefsummary of co-representations and a few examples.

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12.6 Co-representations for groups with time-reversal operators 351

Table 12.3 Type II and type III (32+ 58 = 90) magnetic point groups. Eachgroup is designated by specifying the ordinary point group, G0, and the unitary,invariant subgroup (index 2), Gu. The international notation for each group andthe type of magnetic ordering are given in the last two columns. The 32 type Igroups (ordinary groups not involving the time-reversal operator) are notincluded. F, ferromagnetic; AF, antiferromagnetic.

Crystalsystem

GroupG0

SubgroupGu

Number ofelements

Internationalnotation

Magneticorder

Triclinic C1 C1 1 1 FS2(Ci ) S2 2 1 F

Ci 1 AFMonoclinic C1h C1h 2 m F

C1 m FC2 C2 2 2 F

C1 2 FC2h C2h 4 2/m F

Ci 2/m FC2 2/m AFC1h 2/m AF

Rhombic C2v C2v 4 2mm AFC1h 2mm FC2 2mm F

D2 D2 4 222 AFC2 222 F

D2h D2h 8 mmm AFC2h mmm FC2v mmm AFD2 mmm AF

Tetragonal C4 C4 4 4 FC2 4 AF

S4 S4 4 4 FC2 4 AF

C4h C4h 8 4/m FC2h 4/m AFC4 4/m AFS4 4/m AF

D2d D2d 8 42m AFC2v 42m AFD2 42m AFS4 42m F

C4v C4v 8 4mm AFC2v 4mm AFC4 4mm F

D4 D4 8 42(422) AFD2 42 AFC4 42 F

D4h D4h 16 4/mmm AFD2h 4/mmm AFC4h 4/mmm F

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352 Time reversal and magnetic groups

Table 12.3 (cont.)

D2d 4/mmm AFC4v 4/mmm AFD4 4/mmm AF

Rhombohedral C3 C3 3 3 FS6 S6 6 3 F

C3 3 AFC3v C3v 6 3m AF

C3 3m FD3 D3 6 32 AF

C3 32 FD3d D3d 12 3m AF

S6 3m FC3v 3m AFD3 3m AF

Hexagonal C3h C3h 6 6 FC3 6 AF

C6 C6 6 6 FC3 6 AF

C6h C6h 12 6/m FS6 6/m AF

C3h 6/m AFC6 6/m AF

D3h D3h 12 6m2 AFC3v 6m2 AFD3 6m2 AFC3h 6m2 F

C6v C6v 12 6mm AFC3v 6mm AFC6 6mm F

D6 D6 12 62(622) AFD3 62 AFC6 62 F

D6h D6h 24 6/mmm AFD3d 6/mmm AFC6h 6/mmm FD3h 6/mmm AFC6v 6/mmm AFD6 6/mmm AF

Cubic T T 12 23 AFTh Th 24 m3 AF

T m3 AFTd Td 24 43m AF

T 43m AFO O 24 43(432) AF

T 43 AFOh Oh 48 m3m AF

Th m3m AFTd m3m AFO m3m AF

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12.6 Co-representations for groups with time-reversal operators 353

In the discussion here we will, for simplicity, use Ui for PUi , a unitary operator,and Ai for PAi , an antiunitary operator. We shall consider magnetic point groups oftype II and type III that have the form G = Gu+A0 Gu, where A0 is any antiunitaryoperator belonging to T (G0 − Gu). The notation to be used is summarized below.

Gs A space group.G0 A unitary point group obtained from Gs by setting all translations to zero,

or any type I, ordinary point group.G A magnetic point group.Gu An invariant subgroup of G0 (or G for type II) of index 2.Ai Antiunitary operators of G (or of the set {A0 Gu}).Ui Unitary operators of G (or Gu).� An IR of an ordinary point group or subgroup.� A representation of a magnetic point group.D An IR of a magnetic group.

12.6.1 Basis functions for co-representations

We consider a magnetic group G = Gu + A0 Gu. The rules for the operators areas follows. (1) The product of two unitary operators or two antiunitary operators isunitary and is a member of Gu and G. (2) The product of a unitary operator andan antiunitary operator is an antiunitary operator that is a member of A0 Gu andG. (3) The product of any two operators in G is a member of G. (4) Since thenumber of operators in Gu (index 2) is equal to the number in A0 Gu, we can labelthe operators as Ui and Ai = A0 Ui , i = 1, 2, . . . , hu, where hu is the number ofoperators belonging to Gu (half the number belonging to G).

Let fλ (λ = 1, 2, . . . , l) be a set of basis functions for �, an l-dimensional IRof the unitary group, Gu. Define a second set of the functions as {A0 fλ} = {gλ}.For the discussion in this section, assume that the functions of fλ are orthogonal tothose of A0 fλ.

Since � is an l-dimensional IR of Gu based on the functions of fλ (λ = 1, 2,. . . , l),

Ui fλ =∑μ

fμ �(Ui )μλ. (12.48)

It follows that

Ui gλ = Ui A0 fλ = A0 A−10 Ui A0 fλ = A0

∑μ

�(A−10 Ui A0)μλ fμ

=∑μ

�A0(Ui )μλ A0 fμ =∑μ

�A0(Ui )μλ gμ, (12.49)

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354 Time reversal and magnetic groups

where

�A0(Ui ) ≡ �(A−10 Ui A0)

∗. (12.50)

Since A−10 UiA0 is a unitary operator, it is a member of Gu. If � is irreducible then

�A0 is also irreducible. To prove this, assume that � is irreducible, but �A0 canbe reduced by the unitary transformation matrix V. In that case, V�A0(Ui )V−1 =[V∗ �(A−1

0 UiA0)μλ(V∗)−1]∗, which implies that �(A−1

0 UiA0) is reduced by theunitary transformation V∗ for all Ui in Gu. But (A−1

0 UiA0) is a unitary operatorbelonging to Gu and the set, {A−1

0 UiA0} = Gu. Therefore, � is reducible by V∗.That conclusion contradicts the assumption that � is irreducible. Therefore, if � isirreducible, so is �A0 .

For the antiunitary operators,

Ai fλ = A0Ui fλ = A0

∑μ

�(Ui )μλ fμ =∑μ

�(Ui )∗μλA0 fμ

=∑μ

�(Ui )∗μλ gμ, (12.51)

Ai gλ = (A0Ui )(A0 fλ) = (A0UiA0) fλ

=∑μ

�(A0UiA0)μλ fμ. (12.52)

Define the row vector (bra) by

〈 f, g| = 〈 f1, f2, . . . , fl, g1, g2, . . . , gl |. (12.53)

Then we can express (12.48) to (12.52) by the 2× 2 supermatrix form,

〈 f, g|Ui = 〈 f, g|D(Ui ),

〈 f, g|Ai = 〈 f, g|∗D(Ai ), (12.54)

where D(Ui ) is a 2l × 2l supermatrix (l is the dimensionality of �) given by

D(Ui ) =(

�(Ui ) 00 �A0(Ui )

)(type b) (12.55)

and

D(Ai ) =(

0 �(A0 Ui A0)

�(Ui )∗ 0

)(type b). (12.56)

Noting that A0 Ui = Ai and that Ui = A−10 Ai , the matrix in (12.56) can also be

written as

D(Ai ) =(

0 �(Ai A0)

�(A−10 Ai )

∗ 0

)(type b). (12.57)

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12.6 Co-representations for groups with time-reversal operators 355

The notation “(type b)” is explained below. All of the operator products in (12.56)and (12.57) are unitary elements of Gu. The co-representation matrices are unitary,since they consist of unitary matrices, all of which belong to an IR of Gu. For a typeII group where T itself is in G we may choose A0 = T, and, since T commuteswith Ui , (12.55) and (12.56) reduce to

D(Ui ) =(

�(Ui ) 00 �(Ui )

)(type b)

and

D(Ai ) =(

0 �(Ui )

�(Ui )∗ 0

)(type b).

The matrices D(Ui ) and D(Ai ) above do not form an ordinary matrix representa-tion because they do not obey the group product rules. For example,

D(A) D(U) =(

0 �(A A0) �(A−10 U A0)

�(A−10 A)∗ �(U) 0

)

=(

0 �(A U A0)∗

�(A−10 A)∗ �(U) 0

), (12.58)

but

D(A U) =(

0 �(A U A0)

�(A−10 A U)∗ 0

)�= D(A) D(U). (12.59)

In order to have a consistent multiplication system, the co-representations obey therules

D(U1) D(U2) = D(U1 U2), (12.60a)

D(U) D(A) = D(U A), (12.60b)

D(A) D(U)∗ = D(A U), (12.60c)

D(A1) D(A2)∗ = D(A1 A2). (12.60d)

For example,

D(A) D(U)∗ =(

0 �(A A0) �(A−10 U A0)

�(A−10 A)∗�(U)∗ 0

)

=(

0 �(A U A0)

�(A−10 A U)∗ 0

)= D(A U). (12.61)

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356 Time reversal and magnetic groups

The multiplication rule for the product of two co-representation matricesIf the first matrix represents an antiunitary operator, the complex conjugate of thesecond representation matrix is used in the matrix product; otherwise, ordinary matrixmultiplication is used.

Given the product rules of (12.60a)–(12.60d), the problem of constructing irre-ducible corepresentations revolves around the question of whether or not � isequivalent to �A0 . � and �A0 are equivalent if there is a fixed unitary matrix Zsuch that �(U) = Z �(A−1

0 U A0)∗ Z−1 for all U in Gu. If � is an IR of Gu and is

inequivalent to �A0 , then the IRs, namely D(U) and D(A), are given by (12.55)and (12.57).

In general, the forms of the IRs of the magnetic point groups can be expressedin terms of the IRs of the invariant subgroups of Gu. There are three possibilities:(a) � and �A0 are equivalent to the same real representation; (b) � and �A0 areinequivalent; and (c) � and �A0 are equivalent, but both can not be expressed asreal representations. For details see [12.4, 12.6].

Equations (12.55) and (12.57) refer to a type (b) IR of the magnetic point group.The IRs for type (a) and type (c), for which � and �A0 are equivalent, are

D(U) = �(U) (type a), (12.62)

D(A) = ±�(A A−10 ) Z (type a), (12.63)

�(U) = Z �(A−10 U A0)

∗ Z−1, (12.64)

Z Z∗ = �(A20). (12.65)

Note that in this case D consists of single, l × l matrices, rather than 2l × 2lsupermatrices. The “+” sign holds for integer-spin systems and the “−” for half-integer-spin systems.

For type (c),

D(U) =(

�(U) 00 �(U)

)(type c), (12.66)

D(A) =(

0 −�(A A0) Z�(A A−1

0 ) Z 0

)(type c), (12.67)

�(U) = Z �(A−10 U A0)

∗ Z−1 and Z Z∗ = −�(A20). (12.68)

The group properties of these co-representations can be verified by multiplyingmatrices and using the rules of (12.60a)–(12.60d). However, one must employ theequivalence of � and �A0 and the equation Z Z∗ = �(A2

0) for type (a) or Z Z∗ =−�(A2

0) for type (c). For example, for the type (a) matrices,

D(A) D(U)∗ = ±�(A A−10 ) Z �(U)∗. (12.69)

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12.7 Degeneracies due to time-reversal symmetry 357

From (12.64) we have �(U)∗ = Z∗ �(A A20) Z)Z−1∗, and, from (12.65), Z−1∗ =

�(A−20 )Z and Z = �(A2

0)Z∗−1. Substitution of these relations into (12.69) gives

D(A) D(U)∗ = ±�(A A−10 A2

0 A−10 U A0) Z

= ±�(A U A−10 ) = D(A U) (type a). (12.70)

As a second example, consider the type (b) matrices. We have that

D(A1) D(A2)∗=

(0 �(U1)

�(U1)∗ 0

)(0 �(U2)

�(U2) 0

)

=(

�(U1) �(U2) 00 �(U1)

∗ �(U2)∗

)

=(

�(U1 U2) 00 �(U1 U2)

)= D(A1 A2) (type b). (12.71)

12.7 Degeneracies due to time-reversal symmetry

A method to determine the relationship of � and �A0 that uses only the charactersof the squares of operations of {A0 Gu} is the Frobenius–Schur test described inTheorem 12.3 below.

Theorem 12.3 (Frobenius–Schur test for magnetic point groups) Let �α be anIR of the invariant subgroup Gu, and let Ai = A0 Ui be an operator belongingto the set {A0 Gu}, where A0 is an arbitrary, but fixed, antiunitary operator ofthe magnetic point group, G. The sum of the characters,

∑i χ

α(A2i ), has three

possible values, namely hu, 0, and−hu, where hu is the number of operators in Gu.The additional degeneracies of the levels belonging to �α due to the addition of thetime-reversal operator is determined by the value of

∑i χ

α(A2i ).

(a)∑

i χα(A2

i ) = hu.Time reversal leads to no additional degeneracy for integral spin, but doublesthe degeneracy for half-integral spin.

(b)∑

i χα(A2

i ) = 0.Time reversal leads to a doubling of the degeneracy both for integral and forhalf-integral spin.

(c)∑

i χα(A2

i ) = −hu.Time reversal leads to a doubling of degeneracy for integral spin, but noadditional degeneracy for half-integral spin.

Tinkham [12.7] has given an informative discussion of the magnetic group forantiferromagnets such as MnF2, FeF2, and CoF2 (rutile structure). The crystal

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358 Time reversal and magnetic groups

c

aa

x

y

zη-axis

ξ-axis

Figure 12.3 The magnetic unit cell for MnF2. The dark gray circles represent theMn ions and the light gray circles represent the F ions. The unit cell containstwo magnetic metal ions; the central shaded ion and the corner metal ions (eachshared by eight other cells). The crystal is tetragonal, with lattice constants a = bunequal to c. The F ions lie in an x–y plane along diagonals in the planes. TheMn–F nearest-neighbor spacing in all x–y planes is the same. (Note that the widthof the figure has been elongated for improved presentation. The width and depthare both a.)

structure is illustrated in Fig. 12.3. In the non-magnetic state (above the Néeltemperature) there are symmetry operations involving a non-primitive translation,τ0 = (1/2)(a, a, c). The space-group operations include

1. {E |0} 9. {i |0}2. {C2z|0} 10. {C2z|0}{i |0}3. {C2ξ |0} 11. {C2ξ |0}{i |0}4. {C2η|0} 12. {C2η|0}{i |0}5. {C4z|τ0} 13. {C4z|τ0}{i |0}6. {C−1

4z |τ0} 14. {C−14z |τ0}{i |0}

7. {C2x |τ0} 15. {C2x |τ0}{i |0}8. {C2y|τ0} 16. {C2y|τ0}{i |0}.

The subscripts x , y, z, ξ , and η indicate the axis of rotation, with ξ and η beingrotations about axes parallel to the (110) and (110) directions, respectively. If wenow set τ0 = 0, the first eight operations are those of the group D4. Operations9–16 are the products of the first eight operations times the inversion operator.Therefore the point group is G0 = D4 × i = D4h . The time-reversal operator itselfis not a symmetry operator, so the magnetic group G is of type III. The appropriateinvariant subgroup is D2h = Gu, and the magnetic group is G = D4h(D2h) =D2h + T(D4h − D2h).

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12.7 Degeneracies due to time-reversal symmetry 359

x

y

z

C2 character table

IR E C2z

A 1 1B 1 −1

Figure 12.4 A ferromagnet with G = D2(C2), and the character table for C2.

In detail,

G = (E,C2z,C2x ,C2y, i, iC2z, iC2x , iC2y)

+T(C2ξ ,C2η,C4z,C−14z , iC2ξ , iC2η, iC4z, iC−1

4z ),

where T = �K0. We could also use the form G = Gu + A0 Gu. In this case A0 isany of the operators in T(G0−Gu). For the discussion here we choose A0 = T C4z .

The Frobenius–Schur test of Theorem 12.3 gives∑

i χ�(A2

i ) = ±[4χ(E) +4χ(C2)] = ±8 for the A1g, A1u , A2g, and A2u IRs of Gu(D2h), and therefore theseIRs are of type (a) for integer spin and type (c) for half-integer spin. Eg and Eu IRshave

∑i χ�(A2

i ) = 0 and are IRs of type (b).As a simple example of a co-representation consider the magnetic group D2(C2).

The group G = C2+T(D2−C2) is applicable to a ferromagnet of the type shownin Fig. 12.4. This group has been discussed in connection with the ferromagneticsuperconductor UGe2 [12.8]. The operators of G are E , C2z , T C2y , and T C2x . Forintegral spin we take T2 = +1. The operator rules are C2i C2 j = C2k , where i ,j , and k can take the values x , y, or z but no two may be the same. For exampleC2xC2y = C2yC2x = C2z . The type of co-representations may be determined usingthe Frobenius–Schur test. We have that (T C2y)

2 = T2 C22y = E , and (T C2x)

2 =E , so that

∑i χ(A

2i ) = 2χ(E) = hu. Therefore the co-representations are of type

(a). The co-representation D(U) = �(U), and D(A) = �(A A−10 )Z , where A is

T C2y or T C2x and A0 is an arbitrary operator from the set T(D2 − C2) (T C2y orT C2x in this case). We arbitrarily choose A0 = T C2x .

We must also have Z Z∗ = �(A20) and �(U) = Z�(A−1

0 U A)∗Z−1, where Uis E or C2z . The quantity �(A2

0) = �(KC2x KC2x)∗ = �(E) = 1. It follows

that |Z2| = 1, so Z = exp(iφ), where φ is a real, positive number. To test this

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360 Time reversal and magnetic groups

choice we ask whether �(U) = Z�(A−10 UA)∗Z−1 for the U operators of C2. Since

the IRs of C2 are one-dimensional, real numbers, Z �(A−10 U A0)

∗ Z−1 reduces to�(C2x T−1UTC2x). For U = E , we obtain �(C2x T−1 ET C2x) = �(E). For U =C2z , we obtain �(C2x T−1 C2z T C2x) = �(C2xC2zC2x) = �(C2xC2y) = �(C2z).Therefore (12.64) is satisfied.

For the A irreducible representation of the invariant subgroup C2:

D(E) = �(E) = 1,D(C2z) = �(C2z) = 1,D(T C2y) = �(T C2yA−1)Z = �(T C2yC2x T)Z = �(C2z)Z = eiφ ,D(T C2x) = �(T C2x A−1)Z = �(T C2xC2x T)Z = �(E)Z = eiφ .

For the B irreducible representation of the invariant subgroup C2:

D(E) = �(E) = 1,D(C2z) = �(C2z)Z = −1,D(T C2y) = �(T C2yA−1)Z = �(T C2yC2x T) = �(C2z)Z = −eiφ ,D(T C2x) = �(T C2x A−1)Z = �(T C2xC2x T)Z = �(E)Z = eiφ .

The character table for G = D2(C2) with T2 = +1 is then

IR E C2z T C2x T C2y

A 1 1 w w w = eiφ

B 1 −1 w −w

Since the IRs are one-dimensional, the D(U) and D(A) matrices are the same asthe characters in the table. These two representations obey the multiplication rulesgiven by Eqs. (12.60a)–(12.60d).

For the case T2 = −1,∑

χ(A2i ) = −2, so the IRs are of type (c). The matrix

elements are

D(U) =(

�(U) 00 �(U)

), D(A) =

(0 −�(A A−1

0 )Z�(A A−1

0 )Z 0

)

and Z Z∗ = −�(A20) = −�(T C2x T C2x) = +�(E) = 1, so again Z = eiφ .

We have that

�(T C2x A−10 )Z = �(T C2xC2x T−1)eiφ = �(E)eiφ = eiφ, (12.72)

�(T C2y A−10 )Z = �(T C2yC2x T−1)eiφ = �(C2z)e

iφ = −eiφ. (12.73)

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Exercises 361

The co-representation matrices are

D(E) =(

1 00 1

), D(T C2x) =

(0 −eiφ

eiφ 0

),

D(C2) =(

1 00 1

), D(T C2y) =

(0 eiφ

−eiφ 0

)

for the A IR, and

D(E) =(

1 00 1

), D(T C2x) =

(0 −eiφ

eiφ 0

),

D(C2) =( −1 0

0 −1

), D(T C2y) =

(0 eiφ

−eiφ 0

)

for the B IR.

References

[12.1] H. A. Kramers, “Théorie générale de la rotation paramagnétique dans les cristaux”,Koninkl. Ned. Akad. Wetenschap. 33, 959–972 (1930).

[12.2] C. Herring, “Effect of time-reversal symmetry on energy bands of crystals”, Phys.Rev. 52, 361–365 (1937).

[12.3] S. V. Shubnikov and N. V. Belov, Colored Symmetry (Oxford: Pergamon Press,1964).

[12.4] E. P. Wigner, Group Theory and Its Application to the Quantum Mechanics ofAtomic Spectra (New York: Academic Press, 1959), Chapter 26.

[12.5] M. Lax, Symmetry Principles in Solid State and Molecular Physics (New York:Wiley, 1974).

[12.6] P. Jacobs, Group Theory with Applications in Chemical Physics (Cambridge:Cambridge University Press, 2005).

[12.7] M. Tinkham, Group Theory and Quantum Mechanics (New York: McGraw-Hill,1964).

[12.8] I. A. Fomin and P. L. Kapitza, “Gap nodes in the superconducting phase of theitinerant ferromagnet UGe2”, arXiv:cond-mat/0207152v1 (2002) (7 pages).

Exercises

12.1 If the set of operators {R1, R2, . . . , Rm} forms a class of the group G, provethat χ(R2

i ) is the same for all operators of the class.12.2 Consider the type II magnetic group G = O+T O , where O is the octahedral

group. Using the Frobenius–Schur test, classify the IRs of G as types (a), (b),or (c). Show that there are no additional degeneracies due to time reversal foreither integer or half-integer spin.

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362 Time reversal and magnetic groups

12.3 Consider the type III magnetic group Oh(O) for which G = O+T(Oh−O),where O is the octahedral group and Oh = O×i . Using the Frobenius–Schurtest, classify the IRs of G as types (a), (b), or (c). Show that there are noadditional degeneracies due to time reversal for either integer or half-integerspin.

12.4 For G = C4v(C2v), what is the magnetic point-group type (a, b, or c)? Findthe B1 irreducible representation for the magnetic point group for T2 = 1(integral or no spin).

12.5 For an antiferromagnet with the rutile structure (e.g., MnF2) what is the dif-ference between the group G0 and the pure point-group operations of thespace group?

12.6 Consider the energy bands of a simple cubic crystal along the � symmetryline.(a) What is the group of the wavevector, gk?(b) What are Q operators?(c) Classify the IRs using the Herring rules (Theorem 12.1).(d) What additional degeneracies result from time reversal?

12.7 Given a group, G, with an invariant subgroup, Gu, of index 2, prove that theset {XGu} = {(G − Gu)}, where X is any member of G not in Gu.

12.8 The orthogonality theorem of group theory states that, if �α and �β are IRs ofthe same group,

∑j �

α(U j )∗pq �

β(U j )rs = δαβδprδqs(h/ lα), where the sumover j is over all the operators, U j , of the group, h is the order of the group,and lα is the dimensionality of �α. For the magnetic group G = Gu+A0 Gu,use the orthogonality theorem to show that

∑i χ(A

2i ) = 0 if � is inequivalent

to �A0 .12.9 For G = Gu + A0 Gu, assume that � is equivalent to �A0 , then there

exists a unitary matrix Z , such that Z �A0 Z−1 = �. Show that in this case∑i χ(A

2i ) = ±hu for Z Z∗ = ±(�A0)2.

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13

Graphene

13.1 Graphene structure and energy bands

Graphene is a two-dimensional monolayer of carbon atoms. It is the basic build-ing block for several carbon allotropes such as graphite, carbon nanotubes, andfullerenes. The atoms are arranged in a hexagonal array as shown schematically inFig. 13.1(a). With the origin of coordinates chosen at the center of the hexagon, theprimitive translation vectors of the Bravais lattice and the reciprocal lattice are

a = a

2(√

3ex + ey), b =a

2(−√3ex + ey),

a∗ = 2π

a

(1√3

ex + ey

), b∗=2π

a

(− 1√

3ex + ey

),

where a is the spacing between the adjacent hexagon centers, and ex and ey areunit vectors along the x- and y-axes.

There are two carbon atoms per unit cell (A and B carbons). The A and B car-bons are located at RA(m, n) = ma + nb + τA and RB(m, n) = ma + nb + τB,respectively, where m and n are positive or negative integers including 0, τA =(a/2)(ex/

√3+ ey), and τB = (a/

√3)ex .

Each A carbon has three B carbons as nearest neighbors, and each B carbonhas three A carbons as nearest neighbors, as shown in Fig. 13.1(a) by the heavylines. The vectors from an A carbon to its neighboring B carbons are −(a/√3)ex ,(a/2)(ex/

√3+ ey), and (a/2)(ex/

√3− ey). The vectors from B to A carbons are

the negatives of the vectors from A to B carbons.The outer configuration of carbon is 2s22p2. In the honeycomb lattice the s2 p2

configuration becomes sp2 + pz configuration for the interacting carbon atoms,i.e., the s- and p-sigma orbitals (px and py) hybridize to form low-lying, bondingbands and high-lying, antibonding bands. The carbon pz orbitals form the π and π∗

bands situated in energy above the three lowest hybrid bands and below the threehighest hybrid bands. There are two atoms per unit cell and eight outer electrons

363

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364 Graphene

b aτA

τB

(a)

AB

B

B

x

y

(b)a∗b∗

K1

K6

K5

K4

K3

K2

MΓ Σ

Λ

kx

ky

Figure 13.1 (a) The two-dimensional graphene lattice showing the primitive lat-tice vectors a and b. The two inequivalent atoms of the unit cell are shown as (A)solid and (B) open circles. The heavy lines indicate the three nearest neighbors ofan A carbon. (b) The reciprocal lattice and the Brillouin zone, showing the vec-tors a∗ and b∗. The corners of the zone are labeled Ki (i = 1 to 6). There are twoinequivalent corners. Even-numbered Ki are equivalent and odd-numbered Ki areequivalent. Odd–odd and even–even pairs of Ki are related by a reciprocal-latticevector.

including spin. Six of the electrons occupy the 2s–2p hybrid valence bands. Theremaining two occupy the pz π valence bands. Consequently the π valence bandsare filled and the π∗ antibonding, conduction bands are empty. The core states, 1s,lie at much lower energies, and have only a small effect on the pi bands.

The simplest model for the pi bands is an LCAO-like “tight-binding” modelinvolving only the pz orbitals and only interactions between the nearest-neighbororbitals. The Bloch wavefunction is of the form

�k(r) = �Ak(r)+�Bk(r), (13.1)

�Ak(r) = CAk√N

∑m,n

eik·[R(m,n)+τA] pz(r− [R(m, n)+ τA]), (13.2)

�Bk(r) = CBk√N

∑m,n

eik·[R(m,n)+τB] pz(r− [R(m, n)+ τB]), (13.3)

where N is the number of carbon atoms, k is the wavevector, and pz(r− R) isa p-orbital oriented along the z-axis at R. The coefficients CAk and CBk are theamplitudes of the pz orbitals at the A and B sites in the unit cell.

Using orthogonalized Löwdin orbitals, the LCAO secular equation is(εp − Ek HAB(k)H ∗AB(k) εp − Ek

)(CAk(r)CBk(r)

)= 0, (13.4)

where

εp =∫

pz(RA)∗ H pz(RA) dr, (13.5)

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13.1 Graphene structure and energy bands 365

HAB(k) =∫

�Ak(r)∗ H �Bk(r)dr

= (ppπ){

e−ikx a/√

3 + ei[kx/√

3+ky ]a/2 + ei[kx/√

3−ky ]a/2}, (13.6)

(ppπ) ≡∫

pz(RA)∗ H pz(RB) dr, (13.7)

and in (13.6) and (13.7) only nearest-neighbor interactions are retained.We may write

HAB(k) = h(k)eiθ , (13.8)

with h(k) = |HAB(k)| and θ = tan−1(Im HAB/Re HAB), where Im and Re denotethe imaginary and real parts, respectively. The secular equation then takes the form(

εp − Ek h(k)eiθ

h(k)e−iθ εp − Ek

)(CAk(r)CBk(r)

)= 0. (13.9)

Solution of the eigenvalue equation (13.9) gives

!±k =εp − Ek

|(ppπ)| = ±h(k)|(ppπ)|

= ±[

1+ 4 cos2

(kya

2

)+ 4 cos

(√3kxa

2

)cos

(kya

2

)]1/2

, (13.10)

and the eigenvectors are

V+(k) = 1√2

(eiθ/2

−e−iθ/2

), (13.11)

V−(k) = 1√2

(eiθ/2

e−iθ/2

). (13.12)

The energy surface and dispersion curves for !±k versus k, and the correspondingdensity of states, are shown in Figs. 13.2(a), (b), and (c), respectively.

The density-of-states expression for graphene from [13.1] is

ρ(!) ={ [√|!|/(2π2)]K (m), 1 ≤ |!| ≤ 3,[√|!|/(2π2)](1/m)K (1/m), |!| ≤ 1,

where

m2 = (1+ |!|2)2 − (!2 − 1)2/4

4|!|and K (m) is the complete elliptic function of the first kind.

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366 Graphene

KK KKK K

M M

Γ

Γ –3

–1

0

1

3

K Λ Γ Σ M K DOS

(b) (c)

Ωk

Ω+

Ω−

π∗ band

π band

Figure 13.2 (a) The energy surface generated by ! for k ranging over theentire first Brillouin zone. (b) !±k versus k along symmetry lines. (c) Thecorresponding density of states.

As can be seen from Fig. 13.2, !±k vanishes at the K-points in the first Brillouinzone. This is not immediately obvious from the dispersion formula (13.10), but canbe seen from the expression for HAB in (13.6) as follows. The K-points are

K1 = −K4 = 4π

3aey, (13.13a)

K2 = −K5 = 2π

3a

(√3ex + ey

), (13.13b)

K3 = −K6 = 2π

3a

(√3ex − ey

). (13.13c)

At k = Ki the three exponential factors of HAB become the three roots of unity(1, exp(i2π/3), and exp(−i2π/3)), and the sum of the three roots is zero. WhenHAB vanishes, !±k vanishes. The lower band (labeled !−k ) is the valence band, andthe upper band (labeled !+k ) is the conduction band. For the model consideredhere the two bands are mirror reflections of one another.1 For an infinite sheetof graphene the valence band is completely occupied and the conduction band isempty. The Fermi level bisects the two bands at !±K = 0. At absolute zero thereare no electrons in the conduction (π∗) band, but also there is no gap between thefilled and empty states. At finite temperatures electrons will be thermally excitedinto the bottom of the conduction band and there will be holes in the top of thevalence band. As a result graphene is a semi-metal.

It is of some interest to explore the behavior of the energy bands near a K-pointbecause the energy surfaces consist of six cones, each centered at one of the

1 If atomic-like orbitals are employed, the overlap causes the valence band to be narrower than the conductionband.

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13.1 Graphene structure and energy bands 367

Ω+

Ω−

ky

kx

Figure 13.3 A schematic representation of !± near the K-symmetry points. Thesurfaces are cones near each K-point. !± is proportional to ±|δk| as |δk| → 0.

K-points in graphene. Let k = K1 + δk = (4π/(3a))ey + δkx ex + δky ey;then

HAB(k)→−√

3a

2(ppπ)(i δkx − δkx)

=√

3a

2(ppπ)|δk|e−i(π/2+η), (13.14)

where η is tan−1(δky/δkx). The dimensionless energy,

!±K1+δk = ±h(K1 + δk)

(ppπ)→±

√3a

2|δk|, (13.15)

depends linearly on the magnitude of δk. Therefore, near a K-point the !± versus ksurfaces are cones, as illustrated schematically in Fig. 13.3. The cell wavefunctionsnear K1 are

V+(K1 + δk) = 1√2

(e−iη/2

−ieiη/2

), (13.16)

V−(K1 + δk) = 1√2

(e−iη/2

ieiη/2

). (13.17)

The above results obtained for k near K1 hold for each of the six K-points in thefirst Brillouin zone. Because of the phase factor, exp(±iη/2), the cell wavefunctionchanges sign when η changes by 2π . Therefore, if k adiabatically follows a pathin k-space that circles a K-point, the wavefunction goes into the negative of itself(acquires a Berry phase of −1) [13.2, 13.3]. The k-vector must circle a K-pointtwice for the wavefunction to return to its original state. In a magnetic field thisfeature leads to half-integer quantization of the Landau levels. If atomic orbitals

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368 Graphene

are used instead of Löwdin orbitals to calculate the energy bands, the overlap ofthe pz orbitals causes the valence band to be narrower than the conduction band.However, the essential features of the bands near the K-points, including lineardispersion, are preserved.

Near the K-points, E(K + δk) ∝ ξr , where r = (δk2x + δk2

y)1/2 and ξ =

(√

3a/2)(ppπ). The radial effective mass, m∗r , is given by

�2

m∗r= ∂2 E

∂r2+ 1

r

∂E

∂r∝ 1

r, (13.18)

m∗r ∝ r. (13.19)

The radial effective mass is independent of the angle on the energy cone. An elec-tron (or hole) near K behaves as a massless particle as r → 0 (k → K, a Diracpoint). The radial group velocity of an electron or a hole near a K-point is

vg = 1

∂Ek

∂(|δk|) = ±√

3a

2

1

�|(ppπ)|. (13.20)

For graphene |(ppπ)| is approximately 3 eV and a = 2.46 Å, so vg = 9.74× 107

cm/s. This velocity far exceeds the group velocity in typical semiconductors.

13.2 The analogy with the Dirac relativistic theory for massless particles

For electrons or holes with a k-vector near any of the K-points, the energy disper-sion is linear, the group velocity is constant, and the effective mass is zero. Thedescription of the electrons (holes) is similar to the massless particles described bythe relativistic Dirac equation [13.4]. Near a K-point the secular equation can bewritten as

− ivg�

(0 ∂/∂x − i∂/∂y∂/∂x + i∂/∂y 0

)(�Ak(r)�Bk(r)

)=Ek

(�Ak(r)�Bk(r)

), (13.21)

where �Ak(r) and �Bk(r) are the sublattice Bloch waves (Eqs. (13.1)–(13.3)). Thisis the same equation as that satisfied by massless Dirac particles, except that vg = c,the velocity of light, for the Dirac particles.

The two-dimensional states near a K-point are often referred to as Dirac statesand the K-points are called Dirac points.

There are two inequivalent K-points, and the others are related to these two byreciprocal-lattice vectors. K1 is equivalent to K3 and K5, while K2 is equivalent toK4 and K6. For example, K1 = a∗ − K3. The Dirac states at K1 and −K1 = K4

are inequivalent, but their energies are degenerate. The Dirac form can be extendedto a four-component matrix equation describing the energy bands of electrons nearK and −K. In this case the equations are analogous to those of the Dirac–Weylequations for massless neutrinos [13.5].

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13.3 Graphene lattice vibrations 369

13.3 Graphene lattice vibrations

The group of the wavevector, gk, for an infinite monolayer of graphene at varioushigh-symmetry points is given in Table 13.1.

At �, gk is D6h . The D6h group is the direct product of D6 with inversion. Theaction table for D6 is shown in Table 13.2.

The action tables for various symmetry points in the Brillouin zone can beobtained by selecting the operations in Table 13.2 that leave k invariant or carryit into an equivalent k-vector. The operators for the group of the wavevector at theM- and K-points and on the �-line are given in Table 13.3.

Table 13.1 The group of the wavevector, gk, for an infinite monolayer ofgraphene. For a k-vector at a general point in the Brillouin zone gk = C1h.

Space group � K M � �

P6/mmm D6h D3h D2h C2v C2v

Table 13.2 The action table for D6. D6h = D6 × i. The coordinate system andnumbering of the sites are shown below the table, where the operations aredefined. The left-hand system of rotation is used, so the result is that of PR ratherthan rotation of the axes.

E C2 C3 C−13 C6 C−1

6 C (1)′2 C (2)′

2 C (3)′2 C (1)′′

2 C (2)′′2 C (3)′′

2

x x −x ξ−− ξ−+ ξ+− ξ++ −x ξ+− ξ++ ξ−+ x ξ−−y y −y ζ+− ζ−− ζ++ ζ−+ y ζ−− ζ+− ζ++ −y ζ−+z z z z z z z −z −z −z −z −z −z1 1 4 3 5 2 6 6 4 2 1 3 52 2 5 4 6 3 1 5 3 1 6 2 43 3 6 5 1 4 2 4 2 6 5 1 34 4 1 6 2 5 3 3 1 5 4 6 25 5 2 1 3 6 4 2 6 4 3 5 16 6 3 2 4 1 5 1 5 3 2 4 6

ξss′ ≡ s 12 x + s′

√3

2 y and ζss′ ≡ s√

32 x + s′ 12 y with s, s′ = {+,−}

x

y

1

2

34

5

6

C2, C3, C6

C(1)2

C(2)2

C(3)2

C(1)2

C(2)2

C(3)2

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370 Graphene

Table 13.3 gk operations for M, K, and �. D2h = D2 × i and D3h = D3 × i.Additional operations (not shown) for D2h and D3h include the operations listedtimes the inversion operator. (See Appendix E for Dn operator symbols.)

M(D2h) E C2 C (1)′2 C (2)′′

2K(D3h) E C3 C−1

3 C (1)′′2 C (2)′′

2 C (3)′′2

�(C2v) E C (2)′2 iC2 iC (1)′′

2

Table 13.4 The character table for D6. D6h = D6 × i

D6(622) E C2 2C3 2C6 3C ′2 3C ′′2x2 + y2, z2 A1 1 1 1 1 1 1

Rz , z A2 1 1 1 1 −1 −1B1 1 −1 1 −1 1 −1B2 1 −1 1 −1 −1 1

(xz, yz) (x, y), (Rx , Ry) E1 2 −2 −1 1 0 0(x2 − y2, xy) E2 2 2 −1 −1 0 0

Even though graphene is a two-dimensional sheet, we must include thez-coordinate in order to account for the “out-of-plane” normal modes ofvibration.

13.3.1 Out-of-plane normal modes at �

We begin by examining the out-of-plane vibrations of graphene. At k = 0 thegroup of the wavevector, gk, is D6h . We make use of the action table (Table 13.2)and the equivalences of sites 2, 4, and 6 to find the symmetry functions based onthe transformation properties of the displacements. We see that the x1, y1, x2, andy2 transform among themselves and are therefore basis functions for a 4× 4 repre-sentation. At the �-point all the phase factors are unity. For the 2×2 representationbased on z1 and z2 we find that

χ(E) = 2, χ(C2) = 0, χ(C3) = 2,

χ(C6) = 0, χ(C ′2)= 0, χ(C ′′2 ) = −2.(13.22)

Using the character table, Table 13.4, we find that the decomposition into IRs ofD6 is A2 + B1. If we include the inversion operations of D6h , the decompositionis A2u + B1g. We can use the symmetry-function-generating machine to find theout-of-plane modes, starting with z1 as the trial function.

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13.3 Graphene lattice vibrations 371

Table 13.5 The characters for the �(x, y) representation

�(x, y) E C2 2C3 2C6 3C ′2 C (1)′′2 C (2)′′

2 C (3)′′2

x1 1 0 − 12 0 0 − 1

2 1 − 12

x2 1 0 − 12 0 0 − 1

2 1 − 12

y1 1 0 − 12 0 0 1

2 −1 12

y2 1 0 − 12 0 0 1

2 −1 12

Trace 4 0 −2 0 0 0 0 0

We have that fz(A2) ∝ ∑R χ A2(PR)PRz1 = 2z1 + 2z2 + 2z3 + 2z4 + 2z5 +

2z6. Using the equivalence of z1, z3, and z5, and the equivalence of z2, z4, and z6,gives fz(A2) ∝ (z1 + z2). If we include the inversion operations, the normalizedsymmetry function for the unit cell is fz(A2u) = (z1+z2)/

√2. Similar calculations

give fz(B1g) = (z1 − z2)/√

2. Clearly fz(A2u) corresponds to translation in thez-direction. It is an acoustic mode at �. The function fz(B1g) is an optical vibrationat �. Both symmetry functions must also be normal modes, since A2u and B1g occuronly once in the decomposition.

13.3.2 In-plane normal modes at �

Consider the matrix representation based on the functions x1, x2, y1, and y2. Thetraces of the 4×4 representation matrices can be found from the action table. Usingthe equivalence of carbons 1, 3, and 5, and that of carbons 2, 4, and 6, we find theresults in Table 13.5.

The decomposition of �(x, y) into the IRs of D6 yields

�(x, y) = E1 + E2. (13.23)

The coordinates x and y are basis functions for the E1 IR. The matrix elements canbe read directly from the action table,

E =(

1 00 1

), C2 =

( −1 00 −1

), C3 =

( − 12 −√3/2√3/2 − 1

2

),

C−13 =

( − 12

√3/2

−√3/2 − 12

), C6 =

( 12 −√3/2√

3/2 12

),

C−16 =

( 12

√3/2

−√3/2 12

), C (1)′

2 =( −1 0

0 1

),

C (2)′2 =

( 12 −√3/2−√3/2 − 1

2

), C (3)′

2 =( 1

2

√3/2√

3/2 − 12

),

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372 Graphene

C (1)′′2 =

( − 12

√3/2

−√3/2 12

), C (2)′′

2 =(

1 00 −1

),

C (3)′′2 =

( − 12 −√3/2−√3/2 1

2

). (13.24)

Note that x1 and y1 do not form bases for a representation of E1 because someoperations of the group transform carbon 1 into carbon 2 or its equivalent and viceversa. Instead, the basis functions for one representation of E1 are (x1+x2)/

√2 and

(y1 + y2)/√

2. The functions (x1 − x2)/√

2 and (y1 − y2)/√

2 are basis functionsfor the representation of E2. The first set clearly corresponds to translation in thex- and y-directions (acoustic branches at �). The second set of basis functionscorresponds to vibration modes (optical branches at �). If we include the inversionoperations, the normal modes are

E1u(�−5 ) =

x1 + x2√2

(row 1), (13.25a)

E1u(�−5 ) =

y1 + y2√2

(row 2), (13.25b)

E2g(�+6 ) =

x1 − x2√2

(row 1), (13.25c)

E2g(�+6 ) =

y1 − y2√2

(row 2), (13.25d)

A2u(�−2 ) =

z1 + z2√2

, (13.25e)

B1g(�+4 ) =

z1 − z2√2

. (13.25f)

The commonly used �-labels are shown in parentheses. That these functions aresymmetry functions can be verified by using the symmetry-function-generatingmachine and the atomic-site equivalences.

At � all of the symmetry functions are also eigenvectors, since no two symmetryfunctions transform according to the same row of the same IR. The normal modesare shown schematically in Fig. 13.4.

13.3.3 Lattice vibrations of graphene at M

From the action table we can find the operations that leave k(M) =(2π/a)(3−1/2, 0, 0) invariant or carry k(M) into an equivalent wavevector (−k(M)

is equivalent to k(M)). The group consists of E , C2, C (2)′′2 , C (1)′

2 , and i times theseoperators. Noting that C2 = Cz

2, C (2)′′2 = C x

2 , and C (1)′2 is equivalent to C y

2 , wecan identify the group of the wavevector as D2h . The phase factors between carbonatoms are given in Table 13.6.

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13.3 Graphene lattice vibrations 373

Table 13.6 Phase angles and phase factors betweencarbon atoms for k(M)

R j − Ri k(M) · (R j − Ri ) exp[ik(M) · (R j − Ri )]R2 − R1 φ eiφ

R3 − R1 0 1R4 − R1 −2φ −eiφ

R5 − R1 −3φ −1R6 − R1 −2φ −eiφ

R3 − R2 −φ e−iφ

R4 − R2 −3φ −1R5 − R2 −4φ −e−iφ

R6 − R2 −3φ −1

φ = π/3, n − m phase angle is negative of m − n.

(a) A2u

x

y

(b) B1g

(c) E1u (d) E1u

(e) E2g(f) E2g

Figure 13.4 Normal modes of graphene at � (k = 0). Black circles with whitecenters represent displacements along the +z-direction (perpendicular to theplane of the paper); the white circles with black centers represent displacementsalong the−z-direction: (a) the acoustic mode, A2u , in the z-direction; (b) the out-of-plane, optical mode, B1g; (c) and (d) in-plane, acoustic normal modes, E1u ;and (e) and (f) in-plane, optical normal modes, E2g . The E2g modes produce the“G-band” observed in spectroscopic experiments.

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374 Graphene

Table 13.7 The character table for D2. D2h = i× D2

D2(222) E Cz2 C y

2 Cx2

x2, y2, z2 A1 1 1 1 1xy Rz , z B1 1 1 −1 −1xz Ry , y B2 1 −1 1 −1yz Rx , x B3 1 −1 −1 1

A representation of D2, �(M), with bases x1, x2, y1, y2, z1, and z2, has thefollowing characters (note that all of the phase factors involved in calculating thetraces are +1):

χ(E) = 6, χ(i) = 0,

χ(C2) = χ(C Z2 ) = 0 χ(iC Z

2 ) = 2,

χ(C (1)′2 ) = χ(CY

2 ) = 0, χ(iCY2 ) = 2,

χ(C (2)′′2 ) = χ(C X

2 ) = −2, χ(iC X2 ) = 0.

From Table 13.7 we find that the decomposition into the IRs of D2h gives

�(M) = [B1u + B2g] + A1g + B1g + B2u + B3u, (13.26)

where the two IRs in brackets correspond to the out-of-plane vibrations involvingonly z1 and z2. Since no IR occurs more than once, all of the symmetry functionswill also be normal modes (eigenvectors). Using the symmetry-function-generatingmachine and the appropriate phase factors, we find

B1u(M−2 ) =1√2(z1 − z2eiφ) (TA), (13.27a)

B2g(M+4 ) =1√2(z1 + z2eiφ) (TO), (13.27b)

A1g(M+1 ) =1√2(x1 + x2eiφ) (LO), (13.27c)

B2g(M+3 ) =1√2(x1 − x2eiφ) (LA), (13.27d)

B3u(M−4 ) =1√2(y1 − y2eiφ) (TA), (13.27e)

B1g(M+2 ) =1√2(y1 + y2eiφ) (TO), (13.27f)

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13.3 Graphene lattice vibrations 375

where TA indicates a transverse acoustic mode, TO indicates a transverse opticalmode, LA indicates a longitudinal acoustic mode, and LO indicates a longitudinaloptical mode. (The M labels commonly used for the IRs are shown in parentheses.)

Notes on the symmetry functionsThe symmetry functions displayed in Eqs. (13.27a)–(13.27f) specify the amplitudesof displacements for carbon atoms labeled 1 and 2 in the central unit cell (shown inFig. 13.1). The symmetry function can also be considered as a two-component vector,Vi . For example, for the A1g mode the displacements in the unit cell centered atR(m, n) are

Vk(A1g)mn = 1√2

[ex

−ex eiφ

]eik(M)·R(m,n).

The total normalized A1g normal mode is

1√N

∑m,n

Vk(A1g)mn,

where N is the total number of carbon atoms in the graphene sheet.

Had we considered the equivalent M point, −k(M), the symmetry functions’phase factor e−iφ would instead be eiφ . Since k(M) and −k(M) are equiva-lent, the modes are degenerate. In fact, the Bloch waves for these two states aretime-reversed partners since the time-reversal operator is the complex-conjugateoperator. Therefore, one can take combinations of the two Bloch waves to formatomic displacements that are real. For example, the Bloch waves for k(M) and−k(M) are N−1/2ζ(k) exp(ik · R) and N−1/2ζ(−k) exp(−ik · R), where ζ(k) isthe unit-cell eigenvector and R is a vector from the origin to any of the hexagoncenters. The sum of the two waves gives the stationary state (standing wave),

ζ(k)eik·R + ζ(−k)e−ik·R = ζ(k)eik·R + ζ(k)∗e−ik·R

= 2 Re{ζ(k)} cos(k · R)− 2 Im{ζ(k)} sin(k · R),

(13.28)

where Re{ζ(k)} and Im{ζ(k)} are the real and imaginary parts of ζ(k), respec-tively. The difference of the two waves gives the vector (not normalized),

−2 Re{ζ(k)} sin(k · R)+ 2 Im{ζ(k)} cos(k · R). (13.29)

For k = k(M) the phase factor sin(k · R) = 0 for all R, so the sum of the Blochwaves is proportional to Re{ζ(k)} cos(k ·R). Consider a line of hexagon centers inthe y-direction with one of the centers as the origin. The phase factor k(M) ·R = 0for this line since Rx = 0 and k(M) is parallel to the x-axis. The adjacent centerson a line to the right, also oriented along the y-axis, will have Rx = +

√3a/2,

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376 Graphene

(a)n = −1

−π

origin

n = 0

0

n = 1

π = k(M) R

(b)

x

y

k(M)

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

Figure 13.5 (a) Stacks of hexagons along the y-axis. The factor k(M) · R is con-stant for cells in a given stack, where R is a lattice vector of the cell (vectorfrom the origin to the center of a hexagon) and k(M) is oriented along the x-axis.(b) Displacements for the B1u stationary state constructed from the symmetryfunctions for k(M) and −k(M). The large circles indicate a displacement twiceas large as that for the small circles. The circles with the dot in the center representpositive displacement perpendicular to the plane of the paper and the circles with“×” in the center represent negative displacement perpendicular to the surface.

giving k(M) · R = π . The line to the left has k(M) · R = −π . In general, asindicated in Fig. 13.5(a), k(M) · R = nπ , where n is the number of lines awayfrom the line containing the origin.

Consider the B1u mode with unit-cell symmetry function

1√2

(z1 − z2eiφ

) = 1√2

[z1 − z2 cos

3

)− i z2 sin

3

)]. (13.30)

The amplitudes of the displacements at the atomic sites for the sum of the Blochwaves from k(M) and k(−M) are proportional to(

z1 − z2

2

)cos[k(M) · R] =

(z1 − z2

2

)(−1)n, (13.31)

where n is the line number as illustrated in Fig. 13.5(a). The displacement patternis shown in Fig. 13.5(b).

13.3.4 Lattice vibrations of graphene at K

The group of the wavevector is D3h . The operations can be selected from thosein Table 13.2 that leave k(K1) = (4π/(3a))ey unchanged or that convert it to anequivalent k-vector. The equivalent k-vectors are k(K3) = (2π/(3a))(3ex−ey) andk(K5) = (2π/(3a))(−3ex − ey). They are shown in Fig. 13.6. The operations are

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13.3 Graphene lattice vibrations 377

Table 13.8 The character table for D3h. The upper labels for the operations arethose from Table 13.2. The lower labels are the conventional operations used forthe D3h character table.

E iC2 2C3 2iC6 3C ′2 3iC ′′2

D3h = D3 × σh(6m2) E σh 2C3 2S3 3C ′2 3σv

x2 + y2, z2 A′1 1 1 1 1 1 1Rz A′2 1 1 1 1 −1 −1

A′′1 1 −1 1 −1 1 −1z A′′2 1 −1 1 −1 −1 1

(x2 − y2, xy) (x , y) E ′ 2 2 −1 −1 0 0(xz, yz) (Rx , Ry) E ′′ 2 −2 −1 1 0 0

x

y

a∗ b∗

K1

K3K5

K3 = K1 − b∗

K5 = K1 − a∗

Figure 13.6 Equivalent K-points in the Brillouin zone. K1, K3, and K5 are relatedby reciprocal-lattice vectors. The wavevector k = K1.

E , C3, C−13 , C (1)′

2 , C (2)′2 , C (3)′

2 and the inversion operations iC2, iC6, iC−16 , iC (1)′′

2 ,iC (2)′′

2 , and iC (3)′′2 . This collection of operations is isomorphic to D3h . Typically,

D3h is presented in terms of other operators. The correspondence is shown at thetop of the character table in Table 13.8. The action table for the group applied tographene is shown in Tables 13.9 and 13.10.

A representation of D3h , �(K1), with bases x1, x2, y1, y2, z1, and z2 has thefollowing characters: χ(E) = 6, χ(iC2) = 2, χ(2C3) = 0, χ(iC6) = 2, and 0for all other operators. For example, let us calculate the character for iC6. UsingTable 13.9 we have

P(iC6)x1 →(−1

2x5 +

√3

2y5

)→

(−1

2x1 +

√3

2y1

)e−2π i/3,

P(iC6)y1 →(−1

2y5 −√

3

2x5

)→

(−1

2x1 −

√3

2y1

)e−2π i/3,

P(iC6)z1 →−z5 →−z1e−2π i/3,

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378 Graphene

Table 13.9 The action table for D3h

E iC2 C3 C−13 iC6 iC−1

6 C (1)′2 C (2)′

2 C (3)′2 iC (1)′′

2 iC (2)′′2 iC (3)′′

2

x x x ξ−− ξ−+ ξ−+ ξ−− −x ξ+− ξ++ ξ+− −x ξ++y y y ζ+− ζ−− ζ−− ζ+− y ζ−− ζ+− ζ−− y ζ+−z z −z z z −z −z −z −z −z z z z1 1 1 3 5 5 3 6 4 2 4 6 22 2 2 4 6 6 4 5 3 1 3 5 13 3 3 5 1 1 5 4 2 6 2 4 64 4 4 6 2 2 6 3 1 5 1 3 55 5 5 1 3 3 1 2 6 4 6 2 46 6 6 2 4 4 2 1 5 3 5 1 3

ξss′ ≡ s 12 x + s′

√3

2 y and ζss′ ≡ s√

32 x + s′ 12 y with s, s′ = ±

x

y

1

2

34

5

6

C2, C3, C6

C(1)2

C(2)2

C(3)2

C(1)2

C(2)2

C(3)2

P(iC6)x2 →(−1

2x6 +

√3

2y6

)→

(−1

2x2 +

√3

2y2

)e2π i/3,

P(iC6)y2 →(−1

2y6 −√

3

2x6

)→

(−1

2y2 −√

3

2x2

)e2π i/3,

P(iC6)z2 →−z6 →−z2e2π i/3.

The character for the operation is then −2(e−2π i/3 + e2π i/3) = +2. The decompo-sition of �(K1) into the IRs of D3h gives

�(K1) = A′1 + A′2 + E ′ + E ′′. (13.32)

From the decomposition we see that no two modes belong to the same row ofthe same IR and therefore the symmetry functions will also be the normal modes(eigenvectors). The symmetry functions can be obtained by use of the symmetry-function-generating machine, taking care to include the phase factors for equivalentatoms.

The symmetry functions for the lattice vibrations at k(K1) are

A′1(K+1 ) =

1

2(x1 − iy1 − x2 − iy2), (13.33a)

A′2(K−2 ) =

1

2(x1 − iy1 + x2 + iy2), (13.33b)

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13.3 Graphene lattice vibrations 379

Table 13.10 Phase angles and phase factors betweencarbon atoms for k(K1)

Translation Phase angle Phase factorR j − Ri k · (R j − Ri ) exp[ik · (R j − Ri )]R2 − R1 −θ − 1

2 − (√

3/2)i

R3 − R1 −2θ − 12 + (√

3/2)i

R4 − R1 −2θ − 12 + (√

3/2)i

R5 − R1 −θ − 12 − (√

3/2)i

R6 − R1 0 1

R3 − R2 −θ − 12 − (√

3/2)i

R4 − R2 −θ − 12 − (√

3/2)i

R5 − R2 0 1

R6 − R2 θ − 12 + (√

3/2)i

θ = 2π/3, exp(±3iθ) = 1, n − m phase angle is the negative ofm − n.

E ′(K+3 ) =1

2(x1 + iy1 + x2 − iy2) (row 1),

(13.33c)E ′(K+3 ) =

1

2(x1 + iy1 − x2 + iy2) (row 2),

E ′′(K−3 ) =1√2(z1 + z2) (row 1),

(13.33d)E ′′(K−3 ) =

1√2(z1 − z2) (row 2).

The time-reversed state has k = −K1, so the cell symmetry functions are thecomplex conjugates of those in (13.33a)–(13.33d). Symmetry functions with realamplitudes of displacement can be formed by taking combinations of the stateswith K1 and −K1. For example, if we take the sum of the two Bloch waves forE ′′(K−3 ) we have

E ′′(K−3 )(row 1) ∝ (z1 + z2) cos(K1 · R). (13.34)

If we take the difference of the two Bloch waves we have for the A′2(K−2 ) stationary

state, we obtain

A′2(K−2 ) ∝ (x1 + x2) sin(K1 · R)+ (y1 − y2) cos(K1 · R). (13.35)

Sketches of the displacements are shown in Fig. 13.7.Since K1 is oriented along the y-axis, the hexagon centers parallel to the x-axis

have constant phase. The line containing the origin has phase K1 · R = 0, the line

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380 Graphene

(a)

origin0

4π/3

2π/3

−2π/3

−4π/3

n = −3

n = −2

n = −1

n = 0

n = 1

n = 2

n = 3

(b)

x

y

K1

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

×

(c)

Figure 13.7 (a) The graphene lattice showing the constant-phase lines and loca-tions of the carbon atoms. (b) The E ′′ (row 1) stationary state. The large circlesindicate a displacement twice as large as that for the small circles. The circleswith the dot in the center represent positive displacement perpendicular to theplane of the paper and the circles with “×” in the center represent negativedisplacement perpendicular to the surface of the paper. (c) The A′2 stationarystate. The displacement vector makes a 30◦ angle with the adjacent side of thehexagon.

above has K1 · R = θ = 2π/3, and for the line below K1 · R = −θ = −2π/3. Ingeneral, K1 · R = n(2π/3), where n labels the lines as shown in Fig. 13.7(a).

Matrix elements of the E ′′ IRThe basis functions for the E ′′ IR are xz and yz. Using Table 13.2 together withTable 13.8, we find

E =(

1 00 1

), iC2 =

(1 00 −1

),

C3 =( − 1

2 −√3/2√3/2 − 1

2

), C−1

3 =( − 1

2

√3/2

−√3/2 − 12

),

iC6 =( 1

2 −√3/2√3/2 1

2

), C−1

6 =( 1

2

√3/2

−√3/2 12

),

C (1)′2 =

(1 00 −1

), C (2)′

2 =( − 1

2

√3/2√

3/2 12

),

C (3)′2 =

( − 12 −√3/2−√3/2 1

2

), C (1)′′

2 =( 1

2 −√3/2−√3/2 − 1

2

),

C (2)′′2 =

( −1 00 1

), C (3)′′

2 =( 1

2

√3/2√

3/2 − 12

).

These matrices can be used to generate the E ′′ symmetry functions by means of thesymmetry-function-generating machine,

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Exercises 381

f (E ′′, row i) =∑

R

Dii (PR)∗ PR z1.

The results are given in (13.33d).In concluding this section we note that in the case of graphene all of the vibra-

tional eigenvectors at high-symmetry points in the Brillouin zone are determinedby symmetry alone.

13.3.5 Finite sheets of graphene

The analysis of the electronic states and lattice vibrations given above is for an infi-nite sheet of two-dimensional graphene. Under experimental conditions graphenesamples are often quite small. Furthermore, the manner in which the sample issecured to the substrate becomes relevant. For example, if two edges of the sheetare pinned by a massive holder, the displacements of the edge atoms will be atten-uated. To deal with this circumstance, boundary conditions may be imposed on thepropagating-wave solutions obtained for the infinite sheet. For example, the waveand its time-reversed partner can be combined with certain amplitudes to satisfyappropriate boundary conditions. This procedure will lead to stationary states witha finite number of “allowed” wavevectors and hence discrete frequencies (or ener-gies). Even for a “free”, finite sheet there are many different possible boundaryconditions, depending on how the edges terminate. In addition, unsaturated bondsat the edges may form bonds with foreign atoms (hydrogen or oxygen, for exam-ple). On the other hand, if the sheet is larger than about 20 atoms on a side, theeigenvalues will be reasonably represented by the energies of the infinite sheetevaluated at the “allowed” (quantized) wavevectors.

References

[13.1] J. P. Hobson and W. A. Nierenberg, “The statistics of a two-dimensional, hexagonalnet”, Phys. Rev. 89, 662 (1953).

[13.2] S. Pancharatnam “Generalized theory of interference, and its applications. Part I.Coherent pencils”, Proc. Indian Acad. Sci. 44, 247–262 (1956).

[13.3] M. V. Berry, “Quantal phase factors accompanying adiabatic changes”, Proc. Roy.Soc. London A 392, 45–57 (1984).

[13.4] J. W. McClure, “Diamagnetism of graphite”, Phys. Rev. 104, 666–671 (1956).[13.5] C. W. J. Beenakker, “Colloquium: Andreev reflection and Klein tunneling in

graphene”, Rev. Mod. Phys. 80, 1337–1354 (2008).

Exercises

13.1 Consider the Bloch wave with a Dirac wavevector, exp(iKi · r), and showthat the time-reversed state is of the form exp(iK j · r), where Ki and K j areinequivalent Dirac points.

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382 Graphene

13.2 Use Eq. (13.21) to show that the graphene effective Hamiltonian for k near aDirac point can be written as

H = �vF σ · k

where σ is the Pauli spin operator, k = kx ex + kyey is the two-dimensionalwavevector near a Dirac point, and vF is the group velocity at the Fermi level(i.e., at ! = 0). The operator σ appearing in the Hamiltonian above is calledthe pseudo-spin.

13.3 (a) Find an expression for the energy of the graphene energy bands includingthe overlap integral, Sπ , between nearest-neighbor pz orbitals. AssumeSπ > 0 and (ppπ) < 0.

(b) Find the energy for k near a Dirac point.(c) Plot a graph of (E − εp) versus ka along the lines � → M → K in the

Brillouin zone. Use Sπ = 0.13, (ppπ) = −3 eV, and εp = 0.13.4 For Exercise 13.3, show that the radial effective mass, m∗r , and the group

velocity near a Dirac point do not depend on the overlap integral. r = (δk2x +

δk2y)

1/2 and �2/m∗r = ∂2 E/∂r2 + (∂E/∂r)/r .13.5 Consider the Z -modes (out-of-plane lattice vibrations) of graphene. Assume

the restoring force is of the form fi j = − f (δzi − δz j ), where δzi is thedisplacement of the i th carbon atom in the z-direction from its equilibriumposition.(a) Considering only nearest-neighbor interactions show that the lattice

vibration eigenvalue equation is

det

∣∣∣∣ 3 f − mω2(k) − f L(k)f L(k)∗ 3 f − mω2(k)

∣∣∣∣ = 0,

where L(k) = [exp(iKi ·Ra)+exp(iKi ·Rb)+exp(iKi ·Rc)] and Ra,Rb,and Rc are the nearest-neighbor vectors.

(b) Find dispersion formulas for the two branches and the eigenvectorsfor the modes. Identify the acoustic and optical branches and anydegeneracies. Give the symmetry labels of the branches at �.

(c) Plot a graph of mω2(k)/ f versus the wavevector, k, for the path � →M→ K→ �.

13.6 (a) For Exercise 13.5 find the lattice vibration eigenvectors for k along the �

symmetry line.(b) Give the frequencies in terms of the dynamic matrix.

13.7 For Exercise 13.5 sketch the relative displacements for the row-2 E ′ latticevibration mode at K1 using the sum of the lattice wave and its time-reversedstate.

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14

Carbon nanotubes

The structure of a “single-walled” carbon nanotube (SWCNT) can be formed byrolling up a graphene sheet into a seamless tube as shown in Fig. 14.1. Multi-walledcarbon nanotubes consist of two or more concentric SWCNTs or scroll-wrappedgraphene sheets. Typical SWCNTs have diameters around 2 nm, but the range isfrom 0.7 to 50 nm. Some tubes are open at the ends and some are capped with abucky hemisphere.

Nanostructures are produced by many different methods, including arc dis-charge, chemical vapor deposition, laser ablation, and flame burning of carbon-containing gases and solids. Some of these methods are becoming commerciallypractical for producing large quantities of carbon nanotubes.

Carbon nanotubes are of great scientific and technological interest because oftheir unique properties [14.1]. The sp2 carbon–carbon bond is stronger than the sp3

carbon–carbon bond of diamond. The tensile strength of an SWCNT is 10 to 100times stronger than that of steel. Perhaps the most important potential applicationis in electronics, since SWCNTs can be grown as metallic threads that can carry anelectrical current density that is 1,000 times greater than that of copper. The energygaps between the valence and conduction bands in SWCNTs range from 0 to 2 eV.Semiconducting nanotubes have been fabricated as transistors that operate at roomtemperature and are capable of digital switching using just a single electron. Light-emitting diodes and photodetectors using single SWCNTs have been demonstratedunder laboratory conditions. Multi-walled carbon nanotubes with connected innertubes are superconducting, with Tc as high as 12 K. Carbon nanotubes have highthermal conductivity, about a factor of 10 greater than that of copper, along the tubeaxis, but have low thermal conductivity across the diameter. They are stable up to3,800 ◦C.

383

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384 Carbon nanotubes

Figure 14.1 A single-walled carbon nanotube.

14.1 A description of carbon nanotubes

There are many different ways in which a graphene sheet can be rolled into ananotube. The various nanotubes so produced have different properties. Thereforeit is important to develop a systematic method for characterizing any nanotubestructure in a relatively simple manner [14.2].

Figure 14.2 illustrates a portion of a long, unrolled nanotube superimposed on agraphene lattice. When the area is rolled up to form a tube the “chiral” vector Ch

lies on the tube circumference and τ is parallel to the tube axis. The angle labeledϕ is the chiral angle. This is the angle between Ch and a1. At the top of the diagramthe two inequivalent carbons in the unit cell, 1 and 2, are shown, together withthe primitive-lattice vectors, a and b. The chiral vector initiates and terminateson lattice centers, and therefore can be expressed as a lattice vector. Followingconvention, we use primitive vectors a1 = a and a2 = −b. Then we have

Ch = na1 + ma2 ≡ (n,m), (14.1)

where n and m are integers. The vector τ is perpendicular to Ch. It can beexpressed as

τ = ra1 + sa2 ≡ (r, s). (14.2)

If dt is the diameter of a single wall nanotube, then we have

πdt = |Ch| = a√

n2 + nm + m2, (14.3)

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14.1 A description of carbon nanotubes 385

2−5a

C h

a

4a

ϕ

b a

a

y

x

τ

R1

1

2

2a2

1

2

Figure 14.2 A sketch of a portion of an unrolled (4, 2) carbon nanotube in the x–yplane. The full nanotube is enclosed by τ, Ch, and the dashed lines. The nanotubeis superimposed on a planar graphene array. The chiral angle is the angle betweenCh and a = a1. When the area (cell) is rolled up to form a tube the “chiral”vector Ch lies on the tube circumference and τ is parallel to the tube axis. τ is theprimitive translation vector for the cell. The angle labeled ϕ is the chiral angle. Atthe top of the diagram two inequivalent carbons, 1 and 2, are shown, together withthe primitive-lattice vectors for graphene, a1 = a and a2 = −b. Near the bottomR is the graphene lattice vector with the smallest (non-zero) projection onto Ch.

where a is the lattice spacing (|a1| = |a2| = a). The tangent of the angle the chiralvector, Ch, makes with the x-axis is

tan θx = n − m√3(n + m)

. (14.4)

For the chiral angle, ϕ,

tanϕ = tan(30◦ − θx) =√

3 m

2n + m. (14.5)

The SWCNTs having n = m (ϕ = 30◦) are called “armchair” nanotubes, whereasthose with m = 0 (ϕ = 0) are referred to as “zigzag” nanotubes. The names referto the bond patterns seen on circling the circumference of the tube as illustrated inFig. 14.3. These two structures are achiral because these nanotubes possess mir-ror symmetry elements. All other (n,m) nanotubes have chirality. Because of thesymmetry, we need only consider 0 ≤ m ≤ n to describe all chiral nanotubes.

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386 Carbon nanotubes

(a)

(b)

Figure 14.3 Single-walled carbon nanotubes. (a) A single-walled carbon nan-otube with the armchair structure. The chiral angle is 30◦. (b) A single-walledcarbon nanotube with the zigzag structure. The chiral angle is zero.

The indices m and n determine the properties of a nanotube. As will be shownlater, if n−m = 3p, where p is an integer, then the SWCNT is metallic. If n−m =3p ± 1, the nanotube is semiconducting.

14.2 Group theory of nanotubes

A single-walled carbon nanotube can be described group-theoretically as a one-dimensional system. In essence we treat the area in Fig. 14.2 determined by Ch

and τ as the unit cell. The vector τ is is the primitive-lattice vector for the one-dimensional unit cell. Given the chiral vector, (n,m), we can determine τ. Wehave Ch = (n + m)(

√3a/2)ex + (n − m)(a/2)ey . A vector perpendicular to Ch

extending in the positive y-direction is

V = −(n − m)a ex + (n + m)(√

3a) ey

∝ (2n + m) a1 − (2m + n) a2. (14.6)

The vector τ must be parallel to V, and |τ| must be equal to the smallest distancebetween lattice points along the axis defined by V. With these requirements, wemay write

τ = (2m + n) a1 − (2n + m) a2

D(n,m), (14.7)

with

D(n,m) = gcd of (2n + m) and (2m + n), (14.8)

where “gcd” stands for “greatest common denominator”.

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14.2 Group theory of nanotubes 387

Ch

τ

ϕ

Figure 14.4 Ch = (4, 1), D(4, 1) = 3, and τ = (2,−3).

In Fig. 14.2, Ch = (4, 2), d = 2, and D(4, 2) = 2, so we have that τ =[(4 + 4)a1 − (8 + 2)a2]/2 = (4,−5). This result may be verified visually onFig. 14.2 (recall that a1 = a and a2 = −b). Figure 14.4 shows another examplewith Ch = (4, 1). In this case d = 1, n − m = 3, so D(4, 1) = 3 and we have thatτ = (2,−3).

Some useful results for nanotubes are the following (a is the lattice spacing, anda/√

3 is the C–C distance):

a1 = a

2(√

3 ex + ey) = a2

4π(2b1 + b2),

a2 = a

2(√

3 ex − ey) = a2

4π(b1 + 2b2),

b1 = 2π

a

(1√3

ex + ey

)= 4π

3a2(2a1 − a2),

b2 = 2π

a

(1√3

ex − ey

)= 4π

3a2(−a1 + 2a2),

ex = a1 + a2√3a=√

3a

4π(b1 + b2),

ey = a1 − a2

a= a

4π(b1 − b2),

a1 · a1 = a2 · a2 = a2,

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388 Carbon nanotubes

b1 · b1 = b2 · b2 = 16π2

3a2,

a1 · a2 = a2

2,

b1 · b2 = −b2

2,

Nh = 2(n2 + m2 + nm)

D(n,m),

Ch = na1 + ma2 = (n,m),

|Ch| = a√

n2 + m2 + nm = a

√Nh D(n,m)

2,

τ = ra1 + sa2 = (r, s) = a2

4π[(2r + s)b1 + (2s + r)b2],

r = 2m + n

D(n,m),

s = − 2n + m

D(n,m),

|τ| =√

3|Ch|D(n,m)

=√

3a√

n2 + m2 + nm

D(n,m)=√

3a2 Nh

2D(n,m),

κ1 = 2π

|Ch|2 Ch = 4π

a2 Nh D(n,m)(na1 + ma2)

= 1

Nh(−sb1 + rb2),

|κ1|2 = 8π2

a2 Nh D(n,m),

κ2 = 2π

|τ|2 τ = 4πD(n,m)

3a2 Nh(ra1 + sa2)

= 1

Nh(−mb1 + nb2),

|κ2|2 = 8π2

3a2 Nh,

R = ca1 + da2 = (c, d), rc − sd = 1, 0 < mc − nd ≤ Nh,

where the greatest common denominator of (c, d) must be 1, and Nh is the numberof hexagons.

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14.2 Group theory of nanotubes 389

The Dirac points are given by

K1 = 4π

3aey = 4π

3a2(a1 − a2) = 1

3(b1 − b2),

K2 = 2π

3a(√

3 ex + ey) = 4π

3a2a1 = 1

3(2b1 + b2).

The number of atoms in an SWCNT can be calculated in terms of n and m. Theunit cell for the SWCNT is the rectangle whose sides are Ch and τ. The area of thecell is

|Ch||τ| =√

3|Ch|2D(n,m)

=√

3a2(n2 + nm + m2)

D(n,m). (14.9)

The area of one of the hexagons is√

3a2/2, and the number of carbon atoms istwice the number of hexagons, so the number of atoms in the SWCNT is

N (n,m) = 2Nh = 4(n2 + nm + m2)

D(n,m). (14.10)

For Fig. 14.2, with Ch = (4, 2), and Fig. 14.4, with Ch = (4, 1), the unit cellsdefined by Ch and τ contain 4(16 + 4 + 8)/2 = 56 and 4(16 + 4 + 1)/3 = 28carbon atoms, respectively.

If we regard the nanotube as a one-dimensional system then the primitive-latticevector is τ. Any translation by an integral multiple of τ is a translation symmetryoperation if the tube is of infinite length. For a tube of finite length we may imposeperiodic boundary conditions.

When rolled up into a tube, translations along the chiral vector, Ch, becomerotations about the tube axis. In the unrolled state we can define reciprocal-latticevectors κ1 and κ2 analogous to those of graphene by the requirements that κ1 ·Ch =2π , κ2 · τ = 2π , κ1 · τ = 0, and κ2 · Ch = 0. Since Ch and τ are perpendicular, κ1

is parallel to Ch and κ2 is parallel to τ. A wave propagating along κ1 must return toits original phase and amplitude after having made one or multiple trips around thetube circumference; therefore, an allowed wavevector will satisfy k1 · Ch = 2πη,where η is an integer.

We can write

κ1 = 2πCh

|Ch|2 =2π

|Ch|e1, (14.11)

where e1 is a unit vector parallel to Ch. For κ2 we find

κ2 = 2πτ

|τ|2 =2π

|τ| e2, (14.12)

where e2 is a unit vector parallel to τ.

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390 Carbon nanotubes

kx

y26κ1

26κ2

κ1κ2

k

κ2

κ1

b1

b2

Figure 14.5 An unrolled (3, 1) nanotube with κ1 = (7b1 + 5b2)/26 andκ2 = (b1 − 3b2)/26. The Brillouin zone of the nanotube is the segmentκ2. For the rolled-up nanotube, κ1 becomes an angular quantum number.(4π/a2)|Ch|/|κ1| = Nh = 26 for the example shown. The inset shows the vectorsin the first Brillouin zone of the graphene sheet.

It is useful to express the reciprocal-lattice vectors for the SWCNT in termsof the reciprocal-lattice vectors for the graphene sheet. To be consistent with theliterature, we use the graphene reciprocal-lattice vectors b1 and b2, where b1 =a∗ = (2π/a)(ex/

√3 + ey) and b2 = −b∗ = (2π/a)(ex/

√3 − ey). The result

is that

κ1 = 1

Nh(mb1 − nb2), (14.13)

κ2 = 1

Nh(−sb1 + rb2), (14.14)

where r and s are the coefficients of τ = (r, s), and n and m are the coefficientsof Ch = (n,m). Figure 14.5 shows schematically the reciprocal-lattice vectors forCh = (3, 1) on the reciprocal lattice for graphene.

In developing a more complete group theory of the nanotube we need to considerall of the possible symmetry operations. For an infinitely long tube, translationby any graphene lattice vector is a symmetry operation that corresponds to atranslation-and-rotation operation. Such an operation can be written in the space-group notation as {R(θ)|t}. We wish to express {R(θ)|t} in terms of Ch and τ.We can solve for the primitive-lattice vectors, a1 and a2, in terms of Ch and τ.This gives

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14.2 Group theory of nanotubes 391

a1 = 1

Nh

[βCh + m D(n,m)τ

], (14.15)

a2 = 1

Nh

[αCh − nD(n,m)τ

], (14.16)

where Nh = 2(n2 + m2 + mn)/D(n,m), α = 2m + n, and β = 2n + m.Let t be a lattice vector for the unrolled tube, t(c, d) = ca1+da2. Then, in terms

of Ch and τ,

t(c, d) = cβ + dα

NhCh + mc − nd

Nhτ ≡ u

NhCh + v

Nhτ, (14.17)

where u = (cβ + dα)/D(n,m) and v = mc − nd.In the rolled-up tube configuration, (v/Nh)τ represents a translation along the

tube axis, and (u/Nh)Ch represents a rotation about the tube axis. Clearly, ifu/Nh = 1 the rotation is one trip around the diameter or a rotation of 2π . Thereforeu/Nh is a measure of the rotation angle in units of 2π . In space-group notation, wedefine a rotation by the notation Cu

N ,

CuN = 2π

u

Nh. (14.18)

In space-group notation, t(c, d) is a screw operation,

t(c, d)→{

CuN

∣∣∣ v

Nhτ

}. (14.19)

Given Ch = (n,m), the smallest angle of rotation for a symmetry operation cor-responds to the values of c and d that result in a vector R = ca1 + da2 that hasthe smallest projection onto Ch. For the example in Fig. 14.2, R = aey . Sinceaey = a1 − a2, it follows that the smallest projection onto Ch occurs for c = 1 andd = −1. With (n,m) = (4, 2) in Fig. 14.2, we find that the projection of R ontoCh, C(R), is given by

C(R) = R · Ch

|Ch| =a√28

. (14.20)

If we divide this result by |Ch|, we obtain C(R)/|Ch| = 1/28, meaning that therotation about the tube axis is 2π/28. We can also calculate this rotation angleusing 2π(u/Nh). For the example we are considering, n = 4, m = 2, c = 1,d = −1, and D(4, 2) = 2, we have

u = cβ + dα

D(n,m)= 1, (14.21)

Nh = 28, (14.22)

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392 Carbon nanotubes

u

Nh2π = 2π

28. (14.23)

In general, the minimum C(R) is such that the corresponding rotation is 2π/Nh.This requires that R = ca1 + da2 satisfies the equation

c(2n + m)+ d(2m + n) = D(n,m). (14.24)

For example, for Ch = (4, 2), we have 10c + 8d = 2, which is satisfied by c = 1and d = −1. For Ch = (8, 3) we have 19c + 14d = 1, which is satisfied by c = 3and d = −4.

Since {CuNh|(v/Nh)τ} is a symmetry operation, it leaves the (infinitely long)

nanotube unchanged. Therefore, {CuN |(v/Nh)τ}q (with q an integer) is also a sym-

metry operation. Nh such screw operations correspond to a rotation of 2π , so{Cu

N |(v/Nh)τ}Nh = {E |vτ}, where E is the identity operation.In addition to these screw operations, for achiral tubes (m = 0 or m = n),

rotations about axes perpendicular to the tube axis are also symmetry operations

Mh

Mv

Cn

6

3

5

2

4

1

2′

3′

4′

5′

6′

1′

Ci

Figure 14.6 An achiral (3, 3) nanotube unfolded. The center of inversion of theunit cell is indicated by Ci . The axis of the rolled tube is a Cn axis of symmetry.A vertical plane containing the Cn–Mv line is a mirror plane. A horizontal planecontaining the Ci –Mh line is another mirror plane. When rolled up the arrowslabeled 1 through 6 match up with the arrows on the opposite side of the unrolledtube. The axes through the tube connecting the primed and unprimed numbers aretwo-fold rotation axes perpendicular to the tube axis. Since n is 3, there are threeaxes passing through the centers of hexagons (1, 3, and 5) and three axes bisectingC–C bonds (2, 4, and 6).

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14.3 One-dimensional nanotube energy bands 393

(Fig. 14.6). If Ch = (n, 0) or (n, n), there are n axes through the center of a hexagonand carbon atoms on either side of the tube and n axes bisecting a C–C bond oneither side of the tube. In the case of the achiral tubes, inversion is also a symmetryoperation. The point group for the nanotube is obtained from the space group bysetting the translation part of each operation to zero. Therefore, the point group ofthe achiral nanotubes is Dn × i .

For an achiral tube the nature of the group depends on whether n is even or odd.For n even, Dn × i = Dnh , whereas when n is odd Dn × i = Dnd . Some charactertables for the space groups and point groups are given in Appendix C. A morecomplete discussion of the space groups of nanotubes can be found in references[14.2] and [14.4].

14.3 One-dimensional nanotube energy bands

The one-dimensional energy bands for a nanotube can be deduced from those ofgraphene. For an infinitely long, single-walled nanotube the wavevector k2 in thedirection of κ2 is a continuous wavevector. The vector k1 along the κ1 directionis quantized, taking on the values |k1| = η|κ1|, where η = 1, 2, . . . , Nh. If weexpress the graphene energy bands !±(kx , ky) as !±(k1, k2) and impose the dis-crete values of |k1|, we obtain a series of one-dimensional energy bands, one (±)pair for each value of η or a total of 2Nh bands. That is, there is one energy band foreach carbon atom in the nanotube unit cell. The process is illustrated in Fig. 14.7,which shows the graphene energy surface intersected by a plane passing thoughone of the quantized k1 values and parallel to the k2 vector. The intersection of theplane with the graphene energy surface produces a pair of one-dimensional energybands.

The slices through the energy surface are along the lines k1(η)+ k2, where

k1(η) = η κ1 = k1 e1 (η = 1, 2, . . . , Nh) (14.25)

and

k2 ≡ k2τ

|τ| = k2 e2 (−κ2/2 ≤ k2 ≤ κ2/2), (14.26)

where e1 and e2 are unit vectors in the κ1 and κ2 directions, respectively.The energy bands for graphene (13.10) are given by

!±k = ±|HAB|/(ppπ) = ±√

eik·R1 + eik·R2 + eik·R3, (14.27)

where R1 = −(a/√

3) ex , R2 = (a/2)[(ex/√

3) + ey], and R3 = (a/2)×[(ex/√

3)− ey].

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394 Carbon nanotubes

κ2κ1

η = 12

Nh

Slice sheet

Grapheneenergy surfaces

k2

k1

Ω

One-dimensionalenergy bands

Figure 14.7 Intersection of the plane k1 = κ1η = constant (η = 1, 2, . . . ,Nh) with the graphene energy surface, !(k1,k2). The nanotube energy bandsare shown by the heavy curves between 0 and +κ2/2.

To find the nanotube’s one-dimensional bands, we express the scalar productsk · Ri in the system of orthogonal coordinates in which the unit vectors are e1 =Ch/|Ch| and e2 = τ/|τ|. We have

e1 = a

2

√3(n + m) ex + (n − m) ey

|Ch| , (14.28)

e2 = a

2

(m − n) ex +√

3(n + m) ey

|Ch| , (14.29)

|Ch| =√

m2 + n2 + mn, (14.30)

Ri = (e1 · Ri ) e1 + (e2 · Ri ) e2 (i = 1, 2, or 3), (14.31)

k = k1 e1 + k2 e2. (14.32)

The scalar products are then

k · Ri = k1(Ri1)+ k2(Ri2), (14.33)

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14.3 One-dimensional nanotube energy bands 395

where Ri j = Ri · e j . We find that

k · R1 = k1(R11)+ k2(R12) = a2

2

−k1(m + n)+ k2(n − m)/√

3

|Ch| , (14.34)

k · R2 = k1(R21)+ k2(R22) = a2

2

k1n + k2(2m + n)/√

3

|Ch| , (14.35)

k · R3 = k1(R31)+ k2(R32) = a2

2

k1m − k2(2n + m)/√

3

|Ch| . (14.36)

The reciprocal-lattice vectors for a nanotube are

κ1 = 2π

|Ch| e1, κ2 = 2πD(n,m)√3|Ch|

e2, (14.37)

and the quantized values of k1 = k1(η) are

k1a = κ1aη (η = 1, 2, . . . , Nh). (14.38)

If the results of (14.34) through (14.38) are substituted into (14.27), one obtainsthe one-dimensional energy bands for an arbitrary nanotube,

!±(n,m, η, k2) = ±∣∣∣∣exp

{i

a2

2|Ch|[−k1(n − m)+ k2(n − m)√

3

]}

+ exp

{i

a2

2|Ch|[k1n + k2(2n + m)/

√3]}

+ exp

{i

a2

2|Ch|[k1m + k2(2n + m)/

√3]}∣∣∣∣ . (14.39)

In (14.39) η = 1, 2, . . . , Nh, and k2 is confined to the range ±|κ2/2| so that k2

always lies within the first Brillouin zone. On the other hand, the quantized valuesof k1 may span several Brillouin zones as illustrated in Figs. 14.5 and 14.8.

As simple examples consider the armchair and zigzag nanotubes. A schematicdiagram of the armchair, (n, n), nanotube on the extended graphene Brillouin zoneis shown in Fig. 14.8. The parameters for the (n, n) armchair nanotube are asfollows:

Ch =√

3na ex ,

D(n, n) = 3n,

R = (1, 0),

τ = a ey,

κ1 = 2π√3 na

ex ,

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396 Carbon nanotubes

κ2 = 2π

aey,

Nh = 2n,

b = 4π√3 a

,

|� −K1| = b√3= 4π

3a,

|� −M| = b

2= 2π√

3a,

k1(η) = κ1η = 2πη√3na= bη

4n(η = 1, 2, . . . , 2n),

Nhk1(2n) = 4π√3 a= b,

n 1+n 2n

12 3

Nh = 2n quantizedvalues of k1

b

M2M1 Γ2Γ1

.... ....

Ran

ge o

f k2

Figure 14.8 The armchair (n, n) nanotube on the graphene extended Brillouinzone, showing the range of the wavevector k2 and the Nh = 2n quantized k1values (indicated by the short vertical lines), with k1(η) = ηκ1, η = 1, 2, . . . , 2n,and −π/a ≤ k2 ≤ π/a. Since cos(pπ/n) = cos[(2n − p)π/n], η = p andη = 2n − p (excluding the cases p = n and p = 2n) are degenerate energybands. For η = n and also η = 2n the bands are non-degenerate. All of the one-dimensional energy bands can be obtained by taking η = 1, 2, . . . , n. The bandscorresponding to η = 0 (equivalent to η = 2n) and η = n give non-degeneratebands. The points η = n and k2 = ±b/(2

√3) = ±(2/3)π/a are Dirac K-points,

and therefore the armchair nanotube is always a conductor (has no gap betweenthe valence and conduction bands).

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14.3 One-dimensional nanotube energy bands 397

and

−π

a≤ k2 ≤ π

a,

where b is the magnitude of b1 or b2.Using the armchair parameters in (14.39) yields the one-dimensional energy

bands,

!±(n, n, η, k2) = ±√

1+ 4 cos2

(k2a

2

)+ 4 cos

(πηn

)cos

(k2a

2

)(η = 1, 2, . . . , 2n and −π/a ≤ k2 ≤ π/a). (14.40)

Since cos(πp/n) = cos[(2n − p)π/n] it follows that the bands corresponding toη = p and η = 2n−p (p = 1, 2, . . ., n−1) are degenerate. Note that η = n and η =2n are non-degenerate. All of the bands can be obtained by taking η = 1, 2, . . . , n,with the understanding that η = n and η = 2n are non-degenerate and all otherbands are doubly degenerate. The slice line for η = n passes through the Dirac K1-and K2-points for k2a = ±2π/3 and therefore the valence and conduction bandstouch at ! = 0. The armchair nanotubes are therefore metallic. Figure 14.9 showsthe energy bands for the (3, 3) case.

The parameters for the (n, 0) achiral, zigzag, tube (Fig. 14.10) are as follows:

Ch = na e1,

D(n, 0) = n,

R = (1,−1),

τ = √3 a e2,

κ1 = 2π

nae1,

κ2 = 2π√3a

ey,

Nh = 2n,

b = 4π√3a

,

� −K1 = b√3= 4π

3a,

� −M = b

2= 2π√

3a,

k1(η) = κ1η = 2πη

na=√

3bη

2n(η = 1, 2, . . . , 2n),

|κ2| = b

2,

|κ1|Nh =√

3b,

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398 Carbon nanotubes

–3

–2

–1

0

1

2

3

Γ K2 X

Ω±(k

2,η)

6

1, 5

2, 4

3

3

2, 4

1, 5

η = 6

0 k2a 2π/3 π

Figure 14.9 One-dimensional, armchair, energy bands along the �–X directionfor Ch = (3, 3). Curves with η = 5 and η = 1 are degenerate and also thosewith η = 4 and η = 2 are degenerate. The slice line for η = 3 passes through theDirac K2-point, and therefore the valence and conduction bands touch at ! = 0.The (3, 3) nanotube is metallic. The one-dimensional Brillouin zone extends from−κ2/2 to κ2/2, which corresponds to −π ≤ k2a ≤ π . The energy bands aresymmetric in k2a. (Note that X denotes the X-point of the nanotube Brillouinzone, not the X-point of the two-dimensional graphene zone.)

and

− π√3a≤ k2 ≤ π√

3a.

Using (14.39) gives the energy dispersion curves for this case,

!±(n, 0, η, k2) = ±√√√√1+ 4 cos2

(πη

n

)+ 4 cos

(√3k2a

2

)cos

(πη

n

)(η = 1, 2, . . . , 2n and − π√

3 a≤ k2 ≤ π√

3a

). (14.41)

The zigzag energy bands are also symmetric in k2 and, again, cos(πp/n) =cos[(2n− p)π/n], so η = p and η = 2n− p (η = 1, 2, . . . , n− 1) are degenerate.The bands for η = n and 2n are non-degenerate.

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14.3 One-dimensional nanotube energy bands 399

Γ1

Γ

Γ2

K

K2

1

2

Γ1

1K

K2

k values

Range of k

2n

n divisible by 3(n,0)

(4,0)8 quantized

n

b

b

b 2

1

2

1

2

123

(a)

(b)

/

}

}

Figure 14.10 The Brillouin zone of a zigzag (n, 0) nanotube on extendedgraphene zones. (a) A (4, 0) zigzag tube with eight quantized values for k1, noneof which passes through the Dirac points, K1 and K2. (b) When n is divisible by3 the slice lines pass through K1 for η = 2n/3 and through K2 for η = 4n/3.In both (a) and (b), |κ1| =

√3b/(2n) and |κ2| = b/2, where b is the length of a

graphene reciprocal-lattice vector nanotube.

As a last example we consider the chiral nanotube (4, 2). The parameters for thisSWCNT are as follows:

|Ch| =√

28a,

D(4, 2) = 2,

|τ| = √21a,

κ1 = 2π√28a

e1,

κ2 = 2π√21a

e2,

Nh = 28,

k1 = η|κ1| (η = 1, 2, . . . , 28),

and

− π√21≤ k2a ≤ π√

21.

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400 Carbon nanotubes

0

1

2

3

Ω+

−π√

21k2a → π

(c)

−π

√21k2a → π

(b)

15

16

17

18

19

20

21, 2322

24

25

26

27

28

0

1

2

+

−π√

21k2a → π

(a)

1

2

3

4

5, 7 6

8

9

10

11

12

13

14

Figure 14.11 One-dimensional conduction energy bands for the (4, 2) chiralnanotube. (a) The first 14 bands (η = 1 to 14) and (b) the bands for η = 15to 28. (c) All 28 conduction bands. For Löwdin orbitals the valence bands aremirror reflections through ! = 0.

The dimensionless energy, !±(n,m, η, k2), determined by use of (14.39) for theone-dimensional energy bands (the conduction bands are shown in Fig. 14.11), isgiven by

!±(4, 2, η, k2) = ±∣∣∣∣exp

(−i3πη

14+ ik2a

2√

21

)+ exp

(iπη

7+ i2k2a√

21

)

+ exp

(iπη

14− i5k2a

2√

21

)∣∣∣∣ . (14.42)

To convert from ! to (E − εp) we need only multiply by |(ppπ)|. Variouscalculations show that |(ppπ)| is in the range from 2.4 to 3.2 eV.

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14.4 Metallic and semiconducting nanotubes 401

14.4 Metallic and semiconducting nanotubes

If the slice line determined by ηκ1e1+ k2e2 cuts through any of the K-points in thegraphene Brillouin zone the one-dimensional valence and conduction energy bandswill touch at !± = 0. That is, there will be no energy gap between the lower andupper bands, and the nanotube will be metallic. If none of the slices through theenergy surface cut through a K-point, then all the one-dimensional energy bandswill have an energy gap between the valence and conduction bands. In this case thenanotube will be semiconducting. Nanotube parameters are given in Table 14.1.

The Dirac points for the graphene Brillouin zone have the vectors V1 = K1+Kr,and V2 = K2 +Kr, where K1 and K2 are Dirac vectors in the central unit cell, andKr is any reciprocal-lattice vector, Kr = pb1 + qb2, where p and q are arbitraryintegers. We need only consider the Dirac points K1 and K2, where

K1 = 4π

3aey = b1 − b2

3,

K2 = 2π

3a

(√3 ex + ey

)= 2b1 + b2.

If we express Vi (i = 1 or 2) as the vector Aκ1 + Bκ2, then A = (Ki · κ1)/κ21 . A

necessary condition for a slice line to pass through a Dirac point is that A must bea positive integer. For V1 we find that

A = (K1 +Kr) · κ1

κ21

= n − m

3+ (pn + qm) = integer. (14.43)

The condition is satisfied if n − m is divisible by 3. For V2 we find that

A = (K2 +Kr) · κ1

κ21

= 2n + m

3+ (pn + qm) = integer. (14.44)

Again the condition is satisfied if n − m is divisible by 3.It is expedient to use the abbreviation NBZ (nanotube Brillouin zone) to refer to

the rectangle in the reciprocal-lattice space of graphene. This area is bounded by0 ≤ k1 ≤ Nhκ1 and −κ2/2 ≤ k2 ≤ κ2/2. The requirement that n − m be divisibleby 3 is necessary but not sufficient because the vector V1 or V2 must also lie in oron the boundary of the NBZ. It turns out that in all cases at least one Dirac pointlies within the NBZ if n − m is divisible by 3. As a consequence this conditionis necessary and sufficient to insure that !± = 0 within the NBZ and that thenanotube is metallic. In the case of (n, n) armchair nanotubes, n − m = 0, so theyare all metallic. For the (n, 0) zigzag nanotubes, n − m = n, and hence they aremetallic only if n is a multiple of 3. Figure 14.12 shows the bands for (2, 0) and(3, 0). (The X points in Fig. 14.12 refer to the one-dimensional Brillouin zone ofthe carbon nanotube.) For the armchair κ1 is parallel to the x-axis, whereas for the

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402 Carbon nanotubes

Table 14.1 Nanotube parameters

Ch = (n,m)na1 + ma2 |Ch| D(n,m) τ κ1 κ2units of a Nh |τ| |κ1| |κ2|(n, n)

√3na 3n a1 − a2 (b1 + b2)/(2n) (b1 − b2)/2

armchair 2n a b/(2n)√

3 b/2

(n, 0) na n a1 − 2a2 (2b1 + b2)/(2n) −b2/2

zigzag 2n√

3a√

3b/(2n) b/2

(2, 1)√

7a 1 4a1 − 5a2 (5b1 + 4b2)/14 (b1 − 2b2)/14

14√

21a√

21b/14√

7b/14

(3, 1)√

13a 1 5a1 − 7a2 (7b1 + 5b2)/26 (b1 − 3b2)/26

26√

39a√

39b/26√

13b/26

(3, 2)√

19a 1 7a1 − 8a2 (8b1 + 7b2)/38 (2b1 − 3b2)/38

38√

57a√

57b/38√

19b/38

(4, 1)√

21a 3 2a1 − 3a2 (3b1 + 2b2)/14 (b1 − 4b2)/14

14√

7a√

7b/14√

21b/14

(4, 2)√

28a 2 4a1 − 5a2 (5b1 + 4b2)/28 (2b1 − 4b2)/28

28√

21a√

21b/28 b/√

28

(4, 3)√

37a 1 10a1 − 11a2 (11b1 + 10b2)/74 (3b1 − 4b2)/74

74√

111a√

111b/74√

37b/74

(5, 1)√

31a 1 7a1 − 11a2 (7b1 + 11b2)/62 (b1 − 5b2)/62

62√

93a√

93b/62√

31b/62

(5, 2)√

39a 3 3a1 − 4a2 (4b1 + 3b2)/26 (2b1 − 5b2)/26

26√

13a√

13b/26√

39b/26

(5, 3) 7a 1 11a1 − 13a2 (13b1 + 11b2)/98 (3b1 − 5b2)/98

98√

147a√

147b/98 7b/98

(5, 4)√

61a 1 13a1 − 14a2 (14b1 + 13b2)/122 (4b1 − 5b2)/122

122√

183a√

183b/122√

61b/122

a = |a1| = |a2|, b = |b1| = |b2|, b2 = 16π2/(3a2), a1 · a2 = a/2, b1 · b2 = −b/2,D(n,m) = gcd of (2n + m) and (2m + n), r = (2m + n)/D(n,m),s = −(2n + m)/D(n,m), τ = ra1 + sa2, κ1 = (−sb1 + rb2)/Nh ,κ2 = (mb1 − nb2)/Nh .

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14.5 The nanotube density of states 403

–3

–2

–1

0

1

2

3

Ω±(k

2,η)

4

2

1, 3

1, 3

2

η = 4

0√

3k2a π

–3

–2

–1

0

1

2

3

Ω±(k

2,η)

6

1, 5

3

2, 4

2, 4

3

1, 5

η = 6

0√

3k2a π

(a) (b)

Figure 14.12 (a) One-dimensional, zigzag, energy bands for (2, 0). None of theslice lines cut through a Dirac point (n is not a multiple of 3). Two degenerate,flat bands (η = 1 and η = 3) are produced. (b) One-dimensional, zigzag, energybands for (3, 0). One of the slice lines cuts through the K2 Dirac point in theBrillouin zone, and therefore the valence and conduction bands touch at ! = 0for k2 = 0. Here κ2 = 2π/(

√3a) and −π/√3 ≤ k2a ≤ π/

√3.

zigzag κ1 is parallel to the a1 axis. All other nanotubes have κ1 oriented betweenthe x-axis and the a1 axis.

14.5 The nanotube density of states

The density of states (DOS), ρ(E), is defined so that ρ(E) d E is the number ofstates in the range between E and E + d E . The function ρ(E) has dimensions of(energy)−1. If the energy, E , is a function of k2 only, as it is for the one-dimensionalnanotube energy bands, then

ρ(E(k2)) d E = ρ(k2) dk2. (14.45)

The density of k2-points, ρ(k2), is a constant,

ρ(k2) = 1

|κ2| . (14.46)

It follows that

ρ(E(k2)) = 1

|κ2||d E(k2)/dk2| . (14.47)

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404 Carbon nanotubes

The absolute value of d E(k2)/dk2 is introduced into (14.47) because the densityof states is positive irrespective of the sign of the slope of the energy band. Thelength of the unit cell along the κ2-direction is |τ| and if we consider the density ofstates per unit length (along the axis of the nanotube), ρl, we obtain

ρl(E(k2)) = 1

|κ2||τ||d E(k2)/dk2| =1

2π |d E(k2)/dk2| , (14.48)

where we have used the fact that |κ2||τ| = 2π .In the previous sections on graphene we showed that near a Dirac K-point the

dimensionless energy, !±(K+ δk), behaved as

!±(K+ δk) = ±√

3a

2|δk|, (14.49)

describing a conical energy surface centered at the Dirac K-points. For a metallicnanotube one of the slice lines passes through two inequivalent K-points. The one-dimensional energy band near a K-point is given by

!±(K+ δk2) = ±E±(K+ δk2)

(ppπ)→±

√3a

2|δk2|. (14.50)

The situation for the conduction band is illustrated in Fig. 14.13.We can restrict our attention to the right-hand side of the energy band where

δk2 ≥ 0, remembering that there will be a factor of 2 multiplying ρl to account for

k2

k2δ

k2δΔ

constant

Slice lin

e for

η

E and E+ ΔEStates betweenOne-dimensional

energy band

E

kx

k1ky

E

0

Dirac K-point

Figure 14.13 (a) A slice through a Dirac cone. (b) Linear dispersion near a DiracK-point; �E/�δk2 = constant is the slope of the one-dimensional energy band.

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14.5 The nanotube density of states 405

the left-hand side of the dispersion curve in Fig. 14.13. Using (14.48) and (14.49),we obtain

d E(k2)

dk2= d E(K+ δk2)

d(δk2)= d{(√3a/2)(ppπ)δk2}

d(δk2)=√

3a

2(ppπ), (14.51)

for ! near zero, and

ρl(E) = 8

2π(√

3a/2)(ppπ)= 8

π√

3a(ppπ). (14.52)

The factor of 8 arises from three considerations: (1) a factor of 2 because boththe right- and the left-hand side of the energy curve in Fig. 14.13 contribute, (2)a factor of 2 because the slice line for a metallic nanotube always passes throughtwo inequivalent Dirac K-points (see Figs. 14.9 and 14.10), and (3) a factor of 2to account for the two spin states. The density of states (DOS) near the top of thevalence band at a Dirac K-point will also be described by (14.52) for small negativevalues of E .

The DOS for graphene (the background curves in Fig. 14.14) vanishes at theDirac K-points. In contrast, Eq. (14.52) shows that the DOS for a metallic nano-tube is a non-zero constant over the range of E for which the energy band can beconsidered as linear in δk2. The DOS for graphene has a single peak (logarithmicsingularity) but the nanotube DOS has many spikes (singularities). These spikesresult whenever a one-dimensional energy band has a minimum or maximum, sinceat such points d E/dk2 vanishes. To see the spiky nature of the nanotube DOSconsider the achiral, (n, 0) nanotube whose dimensionless energy is

!±(k2, η) = ±√√√√1+ 4 cos2

(πη

n

)+ 4 cos

(√3 k2 a

2

)cos

(πη

n

). (14.53)

We may limit discussion to the upper branch since ρ(!) = ρ(|!±|). For simplicitywe drop the superscript, with the understanding that ! = !+. We have∣∣∣∣d!(k2, η)

dk2

∣∣∣∣ =∣∣∣∣∣ 1

2!(2√

3a) sin

(√3k2a

2

)cos

(πη

n

)∣∣∣∣∣ , (14.54)

and

cos

(√3k2a

2

)≡ f (!) = !2 − 1− 4A2

η

4Aη

, (14.55)

where

Aη ≡ cos(πη/n). (14.56)

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406 Carbon nanotubes

The DOS is then described by

ρl(!) = 8(1/(2π))!

|(√3a)Aη

√1− f (!)2| , !min(η) ≤ ! ≤ !max(η), (14.57)

where !min and !max are, respectively, the minimum and maximum values of theenergy band described by !(η). The factor of 8 was explained above.

For numerical calculations of the DOS for (n, 0) nanotubes a more efficientmethod is to use the expression

ρl(!) = 16√3πa

Re

{!√

(α2 −!2)(!2 − β2)

}, (14.58)

where α = 2Aη + 1 and β = 2Aη − 1. (a) Choose k2 in the range 0 ≤ k2a ≤π/√

3, (b) compute !(k2) from (14.53), and (c) compute ρl(!) using (14.58). Thisprocedure gives ρl(!) versus ! and converges well at the Dirac points. If n is even,the band for η = n/2 and (3/2)n will have a delta-function DOS at ! = 1. Thisband can not be numerically computed from (14.57) or (14.58). A disadvantage ofthis method is that each band has a different range of !, so calculating the total ofall bands is more difficult.

Several features are apparent. First, (14.52) shows that if n is even the band forη = n/2 and (3/2)n has !+ = 1 irrespective of k2. That is, the band is flat, andtherefore the DOS tends to infinity (a delta-function DOS). Second, (14.56) showsthat the DOS will have a spike (square-root singularity) at the value of !+ forwhich 1 − f (η)2 or sin(

√3k2a/2) = 0. From (14.54) we see that this occurs for

every band (every η) at �(k2 = 0) except for bands that include Dirac points. Inthe case for which !+ = 0, the DOS must be determined by a limiting processsince both the numerator and the denominator in (14.55) tend to zero. The limit-ing process yields 8/(

√3πa). Since ρl(E) = ρl(!)|(ppπ)|, we regain the results

of (14.52).Figure 14.14 shows the total DOS for the metallic zigzag nanotube (6, 0), and

the semiconducting nanotube (7, 0).

14.6 Curvature and energy gaps

When a graphene sheet is rolled into a nanotube, the angles between the C–Cbonds are altered. Two effects result from this. First, the pz orbitals on nearest-neighbor carbon atoms are slightly tilted with respect to one another. This leads toa small sigma interaction between the pz orbitals. Second, and more importantly,the sp2 bonds are twisted, leading to an interaction between the 2s and pz orbitals.Except for the armchair nanotubes [14.2], both these effects lead to the formationof a small energy gap (of the order of meVs) between the conduction and valence

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Exercises 407

0

1

2

3

4

5(a) (b)

–3 –2 –1 0 1 2 3Ω

DO

S

0

1

2

3

4

5

–3 –2 –1 0 1 2 3Ω

DO

S

Figure 14.14 The total DOS for (a) the metallic zigzag nanotube (6, 0), and(b) the semiconducting zigzag nanotube (7, 0). The background curve is the DOSfor graphene.

bands near the Dirac points. Clearly, the hybridization is more pronounced forsmall-diameter tubes than for large-diameter tubes. The gap induced by the tubecurvature is proportional to 1/d2

t , where dt is the diameter of the nanotube [14.3].The gap is important for nanotubes that would otherwise be metallic, but not forsemiconducting nanotubes, whose energy gap is already much larger than that dueto curvature.

References

[14.1] Carbon nanotube: en.wikipedia.org/wiki/Carbon_nanotube.[14.2] R. Saito, G. Dresselhaus, and M. S. Dresselhaus, Physical Properties of Carbon

Nanotubes (London: Imperial College Press, 1998).[14.3] A. Kleiner and S. Eggert, “Curvature, hybridization, and STM images of carbon

nanotubes”, Phys. Rev. B 64, 113402 (2001) (4 pages).[14.4] M. S. Dresselhaus, G. Dresselhaus, and A. Jario, Group Theory and Applications

to the Physics of Condensed Matter (Berlin: Springer-Verlag, 2007), pp. 538–542.

Exercises

14.1 (a) For armchair nanotube energy bands show that ! = ±1 at the one-dimensional zone boundary irrespective of η (2n-fold degeneracy).

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408 Carbon nanotubes

(b) Show that the valence band touches the conduction band for η = n andk2a/2 = π/3.

(c) Show that the touching point is at the Dirac point, K2 = (2π/a) ×(ex/√

3+ ey/3).14.2 (a) Derive a formula for the one-dimensional energy bands for Ch = (2, 1).

(b) Find κ1 and κ2 and specify the first Brillouin zone for the nanotube.(c) Make a graph of the energy bands for −κ2/2 ≤ k2 ≤ κ2/2 for η =

1, 2, . . . , Nh.(d) Are the bands symmetric in the variable k2?(e) Characterize the degeneracies at k2 = 0 and k2 = ±κ2/2.

14.3 (a) For the (2, 1) nanotube express Ch, κ1, and κ2 in terms of the reciprocal-lattice vectors b1 and b2.

(b) Construct a graph in reciprocal-lattice space that shows the quantized k1

vectors and the k2 vector.14.4 (a) For an (n,m) nanotube show that the reciprocal-lattice vector pb1 +

qb2 = (np+mq)κ1+ (pr +qs)κ2, where p and q are arbitrary integers.(b) Show that an arbitrary vector from the origin to a K2 Dirac point any-

where in the reciprocal-lattice space (b1, b2 space) is given by V2 =[((2n + m)/3)(np + mq)κ1 + (1/D)[m + p(2m + n)− q(2n + m)]κ2.

(c) Show that D(n,m) is a multiple of 3 if n − m is divisible by 3.14.5 (a) For the (7, 4) nanotube, use the results of Exercise 14.4 to show that the

slice line with η = 17 passes through a K2 Dirac point.(b) What is the value of k2 when the slice line intersects K2?

14.6 Consider an (n, n) armchair nanotube and two nearest-neighbor carbon atomson the rim as shown in Fig. 14.15.

(a) Unrolled nanotube1 2

rc

(b)

1 2

z

y

p2z

p2y

θ

(c)Top view nanotube

1 2

rc

rc

rr

Figure 14.15

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Exercises 409

Take the origin at the tube center in the y–z plane. Carbon atoms 1 and 2 liein the y–z plane, and the x-direction is perpendicular to the plane of the paper(along the tube axis). Figure (a) is a portion of the (n, n) tube unrolled in they–z plane. Figures (b) and (c) are top views of the rolled nanotube showingthe locations of carbon atoms 1 and 2. Assume that the distance betweencarbon atoms 1 and 2, rc, is fixed.(a) Show that the direction cosines for the vector from carbon atom 2 to

carbon atom 1 are α = 0, β = (1− r2c /(4r2))1/2 and γ = −rc/(2r).

(b) The 2s–2p interaction vanishes by symmetry for graphene. Show that the2s1–2p2 interaction for carbon atoms 1 and 2 of the nanotube is given by

(spσ)[β(sin θ)+ γ cos θ)] = (spσ)

[−√

1− r2c

4r2sin θ + rc

2rcos θ

],

where θ = 2π/n.(c) Show that, for large n, the interaction is inversely proportional to the tube

diameter.

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Appendix A

Vectors and matrices

A.1 Vectors

A vector in n-dimensional space is specified by n components that give the projec-tions of the vector onto the n unit vectors of the space: V = (v1, v2, v3, . . . , vn).Two vectors are equal if all of the components are equal. The components may bereal or complex numbers. Vectors obey the following rules:

U+ (V+W) = (U+ V)+W,

U+ V = V+ U,

rU+ sU = (r + s)U,

rU+ rV = r(U+ V),

r(sU) = rsU

(r and s are any real or complex numbers). The inner product or (Hermitian) scalarproduct of two n-dimensional vectors U and V is

n∑i=1

u∗i vi = (U,V) = (V,U)∗.

The magnitude, or length, of V is |V| = (V,V)1/2. It is a positive number. It is zeroif and only if V = 0, and V = 0 if and only if vi = 0 for all i . A “normalized”vector has |V| = 1.

The scalar product, (U,V)/|V|, is the projection of U onto V, and the cosine ofthe angle, φ, between two vectors is

−1 ≤ cosφ = (U,V)

|U||V| ≤ 1.

If cosφ = 1, U and V are parallel vectors. If cosφ = 0, U and V are orthogonal.The scalar product of two vectors is unchanged if both vectors are subjected to the

410

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A.2 Properties of matrices 411

same symmetry operation. For example, if U and V are subjected to a rotation Ror operator PR ,

(PRU, PRV) = (RU, RV) = (U,V).

A set of m normalized, mutually orthogonal, n-dimensional vectors,U1,U2, . . . ,Um , is an orthonormal set. If m = n then the set “spans” the n-dimensional vector space, and each Ui (i = 1, 2, . . . , n) is a basis vector forthe space. Any vector W in this n-dimensional vector space can be constructedfrom a linear combination of basis vectors: W =∑n

i=1 ri Ui , where ri is a constantcoefficient.

If a set of m vectors is such that no vector in the set can be expressed as a linearcombination of the other vectors, then the vectors of the set are said to be linearlyindependent. An n-dimensional vector space has exactly n linearly independentvectors.

A.2 Properties of matrices

A.2.1 Rectangular matrices

A rectangular matrix is an array of elements with lr rows and lc columns. Takingthe product of two rectangular matrices A and B requires that lc(A) = lr(B), andthe resultant matrix, AB, is an lr(A)× lc(B) matrix. The product is

(AB)ik =lc(A)∑j=1

Ai j B jk (lc(A) = lr(B)).

Rectangular matrices and allowed products obey the associative law, but the trace,determinant, and inverse are not defined for non-square matrices.

The scalar product of two vectors is equivalent to matrix multiplication of a(1× l) matrix by an (l × 1) matrix to produce a 1× 1 matrix (a scalar).

Rectangular matrices can be partitioned into rectangular submatrices. For exam-ple, let U and V be matrices that can be multiplied in the order UV. Partition thesematrices as shown below:

U =(

u11 u12 u13

u21 u22 u23

), V =

⎛⎝ v11

v12

v13

⎞⎠ .

The product is

UV =(

u11v11 + u12v12 + u13v13

u21v11 + u22v12 + u23v13

),

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412 Appendix A. Vectors and matrices

where the us and vs are themselves matrices of appropriate dimensions. That is,the partitioned matrices follow the same multiplication law as regular rectangularmatrices. The partitioning is rather arbitrary, provided that all of the products of thesubmatrices meet the requirement that the number of columns of the left submatrixis equal to the number of rows of the right submatrix. For example, the number ofcolumns of u11 and u21 must equal the number of rows of v11. Partitioned matricesare sometimes called supermatrices.

A.2.2 Definitions and properties of square matrices

Some of the properties of square matrices are listed below.

1. Let the elements of A be Ai j . The transpose of A, denoted by AT, is obtainedby interchanging the rows and columns of A: [AT]i j = A ji .

2. The Hermitian conjugate of A is A† = [AT]∗.3. A matrix is unitary if A−1 = A† = [AT]∗.4. The rows (or the columns) of a unitary matrix form a set of n orthogonal

vectors.5. The product of two unitary matrices is a unitary matrix.6. A matrix for which A = A† is called self-adjoint or Hermitian. The

eigenvalues of a Hermitian matrix are real numbers.7. The trace, spur, or character of a matrix is the sum of its diagonal elements:

Tr{A} = ∑i Aii . Also, Tr{AB} = Tr{BA}, where “Tr{ }” denotes the trace

of the enclosed matrix.8. The trace of A is invariant under a similarity transformation: Tr{A} =

Tr{S−1AS}.9. The determinant of the product of two square matrices is the product of the

individual determinants: det{AB} = det{A} det{B}. Therefore the determinantis invariant under a similarity transformation: det{S−1AS} = det{A}. (Thedeterminant of A is often denoted by |A|.)

10. The identity matrix, also called E, 1, I, or the unit matrix is a diagonal matrixwith elements Ei j = δi j .

11. If a square matrix has a non-zero determinant, then a unique inverse matrix,A−1, can be constructed that has the property that AA−1 = A−1A = E.The inverse of a product of two matrices is the product of the inverses in theopposite order: (AB)−1 = B−1A−1.

12. If the elements of A are Ai j and those of B are Bi j , then the elements of A+ B

are (A+ B)i j = Ai j + Bi j .13. If c is a number, the elements of cA are cAi j . Tr{cA} = c Tr{A}, but

det{cA} = cn det{A}, where n is the dimensionality of A.

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A.2 Properties of matrices 413

14. The form of a matrix equation remains unchanged by a similarity transforma-tion of all of the matrices. If A = f (B), then A′ = f (B′) with A′ = S−1AS

and B′ = S−1BS, where f is a function. For example, if A = sin(B), thenA′ = sin(B′).

15. The secular equation for an n-dimensional matrix A is det{A − λ} = 0. Thisequation has n eigenvalues, λk (k = 1, 2, . . ., n), which are the only values of λfor which the secular equation can be satisfied. For λ = λk , the matrix equationA ξk = λk ξk has a non-trivial solution for the n-dimensional eigenvector ξk .When normalized, the eigenvectors form an orthonormal set of vectors thatspan an n-dimensional space.

16. Diagonalizing a matrix. Transforming a matrix to a diagonal form. This isusually accomplished by a similarity transformation: d = V−1AV, where d is adiagonal matrix whose elements are the eigenvalues of A. That is, di j = λi δi j .The normalized eigenvectors, ξk , arranged as columns of an n × n matrixproduce a unitary transformation, V, that diagonalizes A: V−1AV = d, whereVi j = (ξ j )i .

17. Since the determinant of a matrix is invariant under similarity transformation,it follows from item 13 in this list that det{A} =∏

k λk .18. If f (A) is a function of the matrix A, then f (A) = V f (λ)V−1, where f (λ)i j =

f (λi )δi j and V is the transformation that diagonalizes A (see item 16 in thislist). As examples,

[cos(A)]i j =∑

k

V−1ik cos(λk)Vkj ,

[A− ω]−1i j =

∑k

V−1ik (λk − ω)−1Vkj ,

ln[det{A}] = ln

(∏k

λk

)=∑

k

ln(λk).

19. The direct product of two square matrices A and B is denoted by A ⊗ B orA× B. The matrix elements of A× B use two sets of indices,

(A× B)ik, jl = Ai j Bkl .

If A and B are square matrices then A×B is a square matrix. However, A andB need not have the same dimensions.1 If A has lA rows and columns and B

has lB rows and columns then A× B has lAlB rows and columns.

1 The concept of the direct product of two matrices can be extended to rectangular matrices with appropriatedimensionalities.

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414 Appendix A. Vectors and matrices

20. Direct-product matrices multiply in the same manner as ordinary matrices ifsubscripts i and k are considered the “row” indices, with j and l the “column”indices,

[(A× B)(C× D)]ik, jl =∑mn

[A× B]ik,mn [C× D]mn, jl .

Here the number of columns of (A × B) should equal the number of rows of(C× D).

21. From the rule of multiplication, it follows that

(A× B)(C× D) = (AC)× (BD).

22. The direct product of two diagonal matrices is a diagonal matrix.

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Appendix B

Basics of point-group theory

B.1 Definitions

Group theory is a very broad field of study. We shall look only at a narrow part ofthe field. We are concerned here with the application of group theory to the analysisof physical and chemical systems. For our discussions a group consists of elements(or operators) that mathematically represent operations that leave a system in anequivalent state. For our purposes group multiplication is the sequential applica-tion of symmetry operations or the multiplication of square matrices representingtwo symmetry operations. The point group of interest in the analysis of atoms,molecules, and solids is the covering group, which consists of the elements (oroperators) of rotation, reflection, and inversion under which the atom, molecule, orsolid remains invariant. For crystalline solids the group (space group) is enlargedto include rotations, reflections, and inversion combined with translations underwhich the crystalline solid remains invariant.

A group is a collection of distinct elements that possess the following fourcharacteristics.

(1) Closure. The product of any two elements is an element of the group. If A andB ∈ G and AB = C , then C ∈ G (the symbol ∈ means “belongs to” or “is amember of the set that follows”).

(2) Every group must contain the identity element, E , which commutes with allelements of G: E A = AE = A for all A ∈ G.

(3) Elements of the group obey the associative law: A(BC) = (AB)C .(4) Each element has an inverse. If A ∈ G, then A−1 ∈ G, where AA−1 =

A−1 A = E .

The order of a group is the number of elements in the group, usually denotedby h.

415

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416 Appendix B. Basics of point-group theory

A subgroup, G, is a collection of elements of G (less than hG in number) thatsatisfy the four requirements listed above. The subgroup G is an invariant or normalsubgroup if XiGX−1

i = G, for every Xi in G. If this is the case, G consists ofentire classes of elements. (The ordering of the elements appearing in XiGX−1

i isirrelevant as long as all of the elements of G occur once and only once.)

Rotations. A rotation of a system or object about an axis by 2π/n that leaves thesystem or object in a configuration indistinguishable from its original configurationis a symmetry operation usually denoted by Cn . Here n denotes the number of suchrotations resulting in the identity: (Cn)

n = E . The axis of rotation is said to be ann-fold axis of symmetry.

Center of symmetry and inversion. If a system or object possesses a center ofsymmetry, then it is invariant under the transformation of coordinates (x, y, z)→(−x,−y,−z). That is, the system or object is invariant under the operation, i , ofinversion.

Reflection. If a system reflected through a line or plane of symmetry results in anequivalent system, the system is symmetric under the reflection. Such a reflectionoperation is an element of the covering group of the system. Elements of reflectionare usually denoted by the symbol, σ . σh denotes reflection in a plane that bisectsa principal rotation axis. σv denotes a reflection in a plane that includes a principalaxis of rotation. σd denotes a reflection in a diagonal plane that includes the prin-cipal axis and bisects a pair of two-fold axes perpendicular to the principal axis. Areflection operation is its own inverse; that is, σ 2 = E .

Improper rotation. A rotation by 2π/n followed by a reflection in a planeperpendicular to the rotation axis is an improper rotation denoted by Sn .

Group multiplication table. The rows and columns of a group multiplicationtable are labeled by the group elements. The table entries list the products of therow element times a column element in the order row × column, as illustrated:

E A1 A2 . . . Ah

E ↓ . . .

A1 ↓ . . .

A2 →→ A2 A1 . . . A2 Ah...

......

Ah Ah A1 . . .

Rearrangement theorem. If all of the elements of G are multiplied by any ele-ment of G the result is G (ignoring the order of the elements so produced). That is,AG = G if A ∈ G.

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B.1 Definitions 417

Proof. Show that every element in G is also in AG. Show that no element of Goccurs more than once in AG. Let C be an arbitrary element of G, then A−1C ∈ G.Therefore A(A−1C) = C is in AG. Suppose D occurs twice in AG, that is, thereexist two distinct elements of G, say B1 and B2, for which AB1 = AB2 = D.Multiplying by A−1 on the left gives B1 = B2. This contradicts the assumptionthat B1 and B2 are distinct. As a result of the rearrangement theorem, it followsthat every row of the multiplication table contains each element of G once andonly once. Also, every column of the multiplication table contains each element ofG once and only once.

Conjugate elements and classes

Let A, B, C ∈ G. If C = B−1 AB, then A is conjugate to C . If C is conjugate to Aand D is conjugate to A, then C is conjugate to D.

Proof. If C = B−1 AB and D = X−1 AX , then A = (BC B−1) and D =(X−1 B)C(B−1 X) = T−1CT , where T ∈ G. Therefore C is conjugate to D.

A complete set of distinct elements that are mutually conjugate forms a class.The identity is in a class by itself. An element can not be in more than one class.

Proof. Let K1 and K2 be two different classes of G. Assume that A ∈ K1 andalso A ∈ K2. For some element X ∈ G, we have X−1 AX = B ∈ K1, and forsome element Y ∈ G, Y−1 AY = D ∈ K2. Then, B and D are conjugates, andtherefore belong to the same class. This contradicts the assumption that K1 and K2

are different classes. It follows that in the collection of all of the distinct classes ofG every element of G occurs once and only once.

If G is an invariant subgroup of G then XiGX−1i = G for every Xi in G. To

prove this we show that every element of G occurs once and only once in theproduct XiGX−1

i . First we show that every element of G is contained in the productXiGX−1

i . Let C be an arbitrary element of G, then B = X−1i C Xi must also be an

element in G, since C and B are conjugates and belong to the same class. ThusC = Xi B X−1

i ∈ XiGX−1i . Therefore every element of G occurs at least once in

the product XiGX−1i . Since XiGX−1

i has the same number of elements as G, noelement of XiGX−1

i can occur more than once. Consequently, every element of Goccurs once and only once in XiGX−1

i .

Cosets

Let G be a normal subgroup (invariant subgroup) of a group, G: G =(S1, S2, S3, . . . , Sn), where the ordering of the elements is irrelevant. MultiplyingG by the elements Xi (i = 1, 2, . . . , hG) of the group G produces hG left cosets(and hG right cosets), where hG is the number of elements in G:

Li = XiG = {Xi S1, Xi S2, Xi S3, . . . , Xi Sn} (left coset),

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418 Appendix B. Basics of point-group theory

Ri = GXi = {S1 Xi , S2 Xi , S3 Xi , . . . , Sn Xi } (right coset).

If Xi is a member of G, then Li = Ri = G by virtue of the rearrangement theorem.Therefore we now restrict our attention to Xi in G but not in G. (Note that for apoint group G this restriction means that no coset can be a subgroup, since nonecan contain the identity element.) In this case neither Li nor Ri contains elements incommon with those of G. To show this for Li , assume that some Xi S j is a memberof G, say Sk . Then we have Xi = Sk S−1

j , which belongs to G. This contradicts ourassumption that Xi is not in G. A similar proof holds for a right coset.

It also follows that two cosets of G are either identical or have no elementsin common. Assume that two cosets Li = XiG and L j = X jG have a com-mon element. Then, Xi Sk = X j Sl for some elements Sk and Sl of G. As a resultX−1

j Xi = Sl S−1i is then a member of G, so X−1

j XiG = G by virtue of the rearrange-ment theorem. Multiplying on the left by X j gives XiG = X jG, so the two cosetsare identical if they have a common member. Among the hG cosets generated byXiG (i = 1, 2, . . . , hG), hs of them are identical to G (except for the ordering ofthe elements), where hs is the number of elements in G. The remaining hG−hs leftcosets have no elements of G. However, some of these cosets may be identical. Ifwe consider the collection of distinct cosets together with G, every element of Gcan be found once and only once in the collection, Suppose there are n− 1 distinctcosets, then we have

G = {G, L1, L2, L3, . . . , Ln−1}.Note that each term on the right-hand side of the equation contains hs elements.Therefore there are nhs elements on the right-hand side. There are hG terms on theleft-hand side of the equation, so hG = nhs or hG/hs = n (an integer). (A similarproof can be constructed for right-hand cosets.) The above considerations showthat the order, hs, of an invariant (normal) subgroup is a divisor of the order of thefull group.

The factor group

Consider a group G with an invariant subgroup G and n − 1 distinct cosets so that

G = {G, L1, L2, L3, . . . , Ln−1}.We may regard each G and Li as elements of a more complex group called the fac-tor group of G with respect to the normal divisor G. Li = {Xi S1, Xi S2, . . . , Xi Shs }is regarded as a single complex element. Two complex elements are the same ifthey contain the same elements irrespective of the order of the elements. In addi-tion, for the complex group multiplication Li L j , GLi , or LiG only the distinctterms are retained. For this group of complexes G plays the role of the identity,

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B.1 Definitions 419

since G = G−1, GG = G, and LiG = (XiG)G = XiGG = XiG = Li . We define theinverse, L−1

i , as G−1 X−1i so that Li L−1

i = XiGG−1 X−1i = XiGX−1

i = Xi X−1i G =

G (the identity). In a similar fashion L−1i Li = G. Finally we show that, for the

elements of G, Xi X j = Xk , then Li L j = Lk , and Lk is a member of the factorgroup. That is, the factor group is closed under group multiplication. We have

Li L j = XiGL j = Xi X jG = XkG = Lk .

If there are n − 1 distinct cosets, then the factor group has n (complex) elements.The number of elements in a class is a divisor of hG , the number of elements

in G.Given two groups, G1 and G2, whose elements commute, the direct product is

denoted by G1 ⊗ G2 or G1 × G2. If Ai (i = 1, 2, . . . , h1) denotes the elements ofG1 and B j ( j = 1, 2, . . . , h2) the elements of G2, then the h1 times h2 elements ofG1×G2 are all of the products, Ai B j . The order of G1×G2 is h1h2. See AppendixA, items 19–21 in the list in Section A.2.2.

Point groups

Abelian group. All of the elements of an Abelian group commute with eachother. That is, the order of the product of two elements is irrelevant: AB =B A for all A and B ∈ G.

Cyclic group. A cyclic group of order n, denoted as Cn , consists of the elementsA, A2, A3, . . . , An−1, An , where An = E . All cyclic groups are Abelian.

Ci and Cs . These are groups that have two elements: Ci = {E, i} and Cs ={E, σd}

Cn groups. A cyclic group of this type is the covering group of an n-sided, regu-lar polygon. The elements include n rotations by 2π/n about a perpendicularaxis through the center of the polygon.

Cnv groups. A group of this type is the covering group for a regular polygonprism. It contains the n rotations of Cn plus reflections in n vertical planesthat contain the rotational axis and bisect the prism. For n odd, there are nσv reflections For n even there are n/2 σv and n/2 σd reflections. The planesmake an angle of π/n with one another.

Dn groups. A group of this type is also a covering group of an n-sided prismwhose cross-section is a regular polygon. It includes the elements of Cn , andit has n polygon bisector lines through the center cross-section. If n is even,the elements include n/2 rotations by π about bisector lines through verticesand n/2 rotations by π about bisector lines through parallel sides. If n is odd,the elements include n rotations by π about the bisector lines through thevertices.

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420 Appendix B. Basics of point-group theory

Cnh groups. The elements of a group of this type are those of Cn and also nproduct elements, σhCn .

Sn groups. The group S2n+1 (odd number of elements) is the same as Cnh; S2n

contains 2n improper rotations.Dnd groups. The elements of these groups include the 2n rotations of Dn and

the 2n product elements σd times the elements of Dn .Dnh groups. The elements of these groups includes the 2n rotations of Dn and

the 2n product elements σh times the elements of Dn .Groups associated with polyhedra. The five polyhedra of importance are

depicted in Fig. B.1. The covering groups of the cube, octahedron, andtetrahedron are most often encountered.

O and Oh groups. There are 24 elements in the octahedral, O, group (seeAppendix E): E , 8C3, 6C4, 3C2 (=C2

4 ), and 6C2. The Oh group has 48 ele-ments, namely the 24 elements of O and i times each of the O elements:Oh = O × i . These two groups apply to the symmetries of the cube and theoctahedron. An alternate definition for Oh is the elements of O plus i , 6S4,8S6, 3σh , and 6σd .

T , Th, and Td groups. The T group has 12 elements: E , 4C3, 4C23 , and 3C2. The

elements of Th are the elements of T plus i times the elements of T : Th =T × i . The elements of Td are the elements of T plus σd times the elementsof T : Td = T × Cs . These groups apply to the symmetries of tetrahedra (seeAppendix F).

Tetrahedron Cube Octahedron

Icosahedron Dodecahedron

Figure B.1 The five regular convex polyhedra.

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B.2 Matrix representations of groups 421

B.1.1 Symmetry operators

Rotation, reflection, translation, or inversion of coordinates can be accomplishedby a real, orthogonal transformation R such that r′ = Rr or, in terms of compo-nents, r ′i =

∑i Ri jr j . The inverse transformation is R−1r′ = r, where R−1 = RT

with superscript T denoting the transpose. With a group G = {Aα} consisting ofrotations, reflections, and inversion we may associate a transformation matrix Rα

that induces the operation corresponding to Aα. The set of transformation matri-ces {Rα} form a group isomorphic to G. These matrices transform (rotate, reflect,translate, or invert) the coordinates of a system.

We may also define operators that transform a function. These operators, denotedby P(Rα), transform a function, f (r), in a manner defined by

P(Rα) f (r) = f (R−1α r). (B.1)

The set of operators {P(Rα)} form a group that is isomorphic to G. Conceptually,we may think of P(Rα) as an operator that rotates (reflects, translates, or inverts)the function but leaves the coordinate system unchanged.

B.2 Matrix representations of groups

A matrix representation of a group G is a set of matrices {D(Ai )} that obey thegroup multiplication table of G. That is, if Ai A j = Ak , then D(Ai )D(A j ) =D(Ak). It follows that D(A−1

i ) = [D(Ai )]−1. The representation matrices maybe assumed to be square, unitary matrices with non-vanishing determinants. Thenumber of distinct representation matrices is less than or equal to the number ofelements, h, of G. If there is a one-to-one correspondence between Ai and D(Ai )

the representation is said to be faithful and the set of representation matrices isisomorphic to G. If the representation matrices obey the multiplication table of G,but their number is less than h, the set is homomorphic to G.

The character, χ , of a representation matrix D(Ai ) is the sum of the diagonalelements, namely the trace of the matrix: χ(Ai ) =∑

α Dαα(Ai ). The character ofa representation matrix is unchanged by a similarity transformation, S.

Let D′(Ai ) = S−1 D(Ai )S (similarity transformation). Then

χ ′(Ai ) =∑αβγ

S−1αβ Dβγ (Ai ) Sγα =

∑βγ

[∑α

Sγα S−1αβ

]Dβγ (Ai )

=∑βγ

δγβ Dβγ (Ai ) =∑β

Dββ(Ai ) = χ(Ai ).

Hence, χ [D(Ai )] = χ [D′(Ai )]. It follows that the characters of elements oroperators belonging to the same class are equal.

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422 Appendix B. Basics of point-group theory

A matrix representation that can be block-diagonalized (cast in a form consistingof smaller unconnected blocks along the diagonal) by a similarity transformationis “reducible”. A matrix representation that can not be reduced to smaller blocksby a similarity transformation is “irreducible”. The same block-diagonalizationmust be achieved for all of the representation matrices by the same similaritytransformation.

For any group, G, it is always possible to find a similarity transformation, S,that block-diagonalizes the representation matrices of G into blocks that are allirreducible. For physical systems whose states are characterized by eigenvectorsand eigenvalues, the irreducible representations (IRs) are associated with quantitiesthat are conserved in the dynamics of the system. The IRs of a quantum system areassociated with the “good” quantum numbers.

Consider a set of functions { f (r)k}, k = 1, 2, . . . , n, that transform amongthemselves under the operators P(Rα). That is,

P(Rα) f (r)k =n∑

m=1

f (r)m Dmk(Rα), (B.2)

where the set of Dmk(Rα) are real constants. If we define an n-column vector, F ,whose columns are the functions f (r)1, f (r)2, . . . , f (r)n , then the Dmk(Rα)s arethe matrix elements of an n × n matrix transformation,

P(Rα)F = FD(Rα). (B.3)

In bra–ket notation the transformation that rotates the functions for fixedcoordinate axes is P(Rα)〈 f1, f2, f3, . . . , fn| = 〈 f1, f2, f3, . . . , fn|D(Rα).

Rotating the coordinate axes would give the transformation Rα| f1, f2, f3, . . . , fn〉= D(Rα)| f1, f2, f3, . . . , fn〉. The transformation R−1

α | f1, f2, f3, . . . , fn〉 =D(R−1

α )| f1, f2, f3, . . . , fn〉 is equivalent to the transformation given in Eq. (B.2).We have R−1

α fk = ∑m D(R−1

α )km fm = ∑m fm D(Rα)mk or P(Rα)F =

F D(Rα), which is the same as Eq. (B.3). A clockwise rotation is the inverse of acounter-clockwise rotation, so we have that P(RR.H.

α ) f = RL.H. f , where the super-scripts R.H. and L.H. refer to right-hand and left-hand screw-rule conventions,respectively.

The set of matrices D(Rα) forms an n × n matrix representation of G that maybe reducible or irreducible, depending on the set of functions f (r)k . The proof thatD(Rα) forms a group is as follows. Let R and S be elements of G, then

P(S) P(R) fk = P(S)∑

m

fm Dmk(R) =∑m,n

Dmk(R) fn Dnm(S)

=∑

n

fn

[∑m

Dnm(S) Dmk(R)]=∑

n

fn [D(S)D(R)]nk .

(B.4)

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B.3 Properties of representations 423

On the other hand, since T = S R ∈ G according to (B.2),

P(T ) fk =∑

n

fn Dnk(T ) =∑

n

fn Dnk(S R). (B.5)

On comparing (B.4) and (B.5) we can conclude that D(S R) = D(S)D(R), andtherefore the matrices form a representation of G.

The functions { f (r)k} are called the basis functions for the D representation,and f (r)m is called the basis for the mth row of the D representation. The p-fold, degenerate wavefunctions of a quantum-mechanical system (solutions toSchrödinger’s equation, for example) are basis functions for a p-dimensionalrepresentation of the group of the Hamiltonian.

Note that Rα and P(Rα) are defined differently. Rα rotates the coordinate sys-tem with the function fixed, while P(Rα) rotates the function with the coordinatesystem fixed.

B.3 Properties of representations

1. Any two representations related by a similarity transformation are equivalent. IfD and D′ are matrix representations of G, they are equivalent if D ′ = S−1 D S.Equivalent here means that the matrices of D and D′ satisfy the same groupmultiplication table, and have the same classes and characters.

2. For non-magnetic groups, any matrix representation with non-vanishingdeterminants is equivalent to a representation for which the matrices are all uni-tary and D(R)−1 = D(R)∗T. Therefore we may assume that all of the representationmatrices are square, unitary matrices with non-zero determinants.

3. The characters of the elements (operators) belonging to the same class areequal.

B.3.1 Irreducible representations

1. The number of inequivalent IRs of a group is equal to the number of classes ofthe group.

2. The sum of the squares of the dimensions of all the inequivalent IRs of agroup is equal to the order of the group:

∑i l2

i = h, where li is the dimension ofthe i th IR.

3. The characters of the elements (operators) of the IRs of a group areorthogonal:

1

h

∑R

χ i (R)∗ χ j (R) = δi j , (B.6)

where i and j label the IRs of the group.

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424 Appendix B. Basics of point-group theory

4. Schur’s lemma. Any matrix that commutes with all the matrices of an IR is aconstant matrix, that is, a multiple of the unit matrix. Conversely, if a non-constantmatrix commutes with all of the matrices of a representation, the representation isreducible.

5. Let D(i) and D( j) be two inequivalent IRs of the same group with dimensionsl1 and l2, respectively. Suppose that there exists a matrix H with l1 rows and l2

columns that satisfies the condition

D( j)(R)H = H D(i) (R) (for all R in the group). (B.7)

Then (a) if l1 �= l2, H = 0 or (b) if l1 = l2, either H = 0 or det(H) �= 0. If det(H) �=0, H has an inverse, and therefore D(i) and D( j) are equivalent representations,contradicting the initial assumption. Therefore a matrix H satisfying (B.7) is a nullmatrix.

6. The great orthogonality theorem. Let D(i), i = 1, 2, . . . , Nc, represent the i thIR of a group, where Nc is the number of classes of the group. The matrix elementsof different IRs are orthogonal,∑

R

D(i)mn(R)∗ D( j)

pq (R) = h

liδi j δmp δnq . (B.8)

We sketch the proof of this theorem here. Let D(i) and D( j) be two inequivalentIRs of G. The matrix

H =∑

R

D( j)(R)K D(i)(R−1), (B.9)

where K is an arbitrary, rectangular matrix of dimensions l1 rows × l2 columns,satisfying item 5 in the list above, namely that D( j)(P)H = H D(i)(P) for all P inthe group. The proof is as follows:

D( j)(P)H =∑

R

D( j)(P)D( j)(R) K D(i)(R−1)

=∑

R

D( j)(P)D( j)(R) K D(i)(R−1)D( j)(P−1)D(i)(P)

={∑

R

D( j)(P R) K D(i)(R−1 P−1)

}D(i)(P) = H D(i)(P).

According to item 5 in the list above, H = 0. Since the matrix K is arbitrary, wecan select its elements. Suppose K has only one non-zero element, Krs = δrα δsβ .Then we have

Hγ ν =∑

R

∑m,n

D( j)γm(R) δmα δnβ D(i)

nν (R−1) =∑

R

D( j)γ α (R) D(i)

βν(R−1)

=∑

R

D(i)νβ (R)∗ D( j)

γ α (R) = 0, (B.10)

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B.3 Properties of representations 425

where the unitary feature D(i)(R−1) = D(i)(R)∗T is used to arrive at the finalresult in (B.10). This establishes the orthogonality of the matrix elements ofinequivalent IRs.

For equivalent IRs we use the matrix

H =∑

R

D(i)(R) K D(i)(R−1), (B.11)

which commutes with all of the matrices of D(i). Employing Schur’s lemma (item4 in the list above), we have H = c I, where c is a constant and I is an li × li unitmatrix. Writing (B.11) in subscript form gives

Hγ ν =∑

R

∑m,n

D(i)γm(R) Kmn D(i)

nν (R−1) = c δγ ν. (B.12)

Now choose the arbitrary matrix K to have only a single non-zero element, Kmn =δm,α δn,β , so that

Hγ ν =∑

R

∑m

D(i)γm(R) δmα δnβ D(i)

nν (R−1)

=∑

R

D(i)γ α(R) D(i)

βν(R−1) = c δγ ν. (B.13)

For the diagonal element we have

Hγ γ =∑

R

D(i)γ α(R) D(i)

βγ (R−1) = c. (B.14)

Choose α = β and sum over the index γ . The right-hand side yields∑R

∑γ

D(i)γ α(R) D(i)

αγ (R−1) =∑

R

D(i)αα(E) =

∑R

I = h. (B.15)

The left-hand side of (B.14) gives ∑γ

c = li c,

where li is the dimensionality of the D(i) IR. Using (B.15) and (B.16) gives c =h/ li . Using this result in (B.14) with α = β yields∑

R

D(i)αγ (R) D(i)

γ α(R−1) =∑

R

[D(i)αγ (R−1)]∗ D(i)

αγ (R) = h

li. (B.16)

Equations (B.10) and (B.16) can be combined to give the general result∑R

D(i)νβ (R)∗ D( j)

γ α (R) = h

liδi j δβα δγ ν. (B.17)

Renaming the indices in (B.17) as ν = m, β = n, γ = p, and α = q gives (B.8).

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426 Appendix B. Basics of point-group theory

B.4 The projection operator (symmetry-function-generating machine)

Suppose we have a set of normalized basis functions { f (α)(r)k} for the αth IR ofa group, G, where k = 1, 2, . . . , lα. The function f (α)(r)m , hereafter denoted assimply f (α)

m , is a basis function for the mth row of the αth IR. The functions in theset { f (α)

k } are called partners or partner basis functions. Under the operations ofG, f (α)

m transforms into linear combinations of the set of partner functions, and thecoefficients are the matrix elements of the αth IR of G:

P(R) f (α)m =

∑n

f (α)n D(α)

nm (R). (B.18)

If we multiply both sides of (B.18) by [D(β)

n′m′(R)]∗ and sum over all the operationsof the group, we arrive at∑

R

P(R) f (α)m D(β)

n′m′(R)∗ =∑

n

f (α)n

[∑R

D(α)nm (R) D(β)

n′m′(R)∗]. (B.19)

Applying the orthogonality theorem, (B.8), to the right-hand side of (B.19) yields∑n

f (α)n

h

lαδαβ δnn′ δmm′ = h

lαf (α)

n′ δαβ δmm′ . (B.20)

Equating the result of (B.20) to the left-hand side of (B.19) gives∑R

D(β)

n′m′(R)∗ P(R) f (α)m = h

lαf (α)

n′ δαβ δmm′ . (B.21)

We define the projection operator, Pn′m′ , as

P (β)

n′m′ =∑

R

D(β)

n′m′(R)∗ P(R), (B.22)

and have that

P (β)

n′m′ f (α)m = h

lαf (α)

n′ δαβ δmm′ . (B.23)

Consider a function that is an arbitrary linear combination of the basis functionsof all of the IRs of a group,

F =∑γ

∑q

c(γ )q f (γ )q , (B.24)

where c(γ )n are constant coefficients. If we operate on F with P (β)

n′m′ , we obtain

P (β)

n′m′ F =∑γ

∑q

c(γ )q

[P (β)

n′m′ f (γ )q

]

=∑γ

∑q

c(γ )q

h

lγf (γ )

n′ δγβ δqm′ = c(β)m′h

lβf (β)

n′ . (B.25)

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B.5 Character and action tables 427

Equation (B.25) shows that when the operator P (β)

n′m′ operates on an arbitrary linearcombination of the basis functions it projects out a function that is proportionalto a basis function that belongs to the n′th row of the βth IR of G. The projectionoperator can therefore be used to generate the basis functions (symmetry functions)starting with an arbitrary function that P (β)

n′m′ can operate on. To employ this tool,one assumes an arbitrary function, applies P (β)

n′m′ to the arbitrary function, and thennormalizes the result. The function so produced will be either a null function or afunction that can serve as the basis function for the n′th row of the βth IR of G.The operator P (β)

n′m′ is sometimes referred to as the symmetry-function-generatingmachine. The diagonal projection operator

P (β)nn =

∑R

D(β)nn (R)∗ P(R) (B.26)

may also be used to generate symmetry functions. In the case that D(β) is a one-dimensional IR, D(β)

nn (R)∗ = χ(β)(R)∗, and (B.26) becomes

P (β)nn =

∑R

χ(β)(R)∗ P(R), (B.27)

where χ(β)(R) is the character of the operation for the βth, one-dimensionalIR of G.

B.5 Character and action tables

The character table for a group presents many useful pieces of information regard-ing the properties of the group. A typical character table is shown in Fig. B.2. Thecharacter table contains six essential pieces of information, as indicated in Fig. B.2.

(1) The name of the group (C4v in this case).(2) Names of the IRs of the group.(3) Classes of the group as headings for the columns.(4) The number of elements in each class.(5) The character of the class for a particular IR shown as a row.(6) Some of the basis functions for the IRs. Included are x , y, z, and infinitesimal

rotations about the major axis, Rx , Ry , and Rz . Other bilinear bases may alsobe listed.

All one-dimensional IRs are named Ai or Bi , where the index i is used todistinguish various IRs when needed. Two-dimensional IRs are labeled Ei . Three-dimensional IRs are most often labeled Ti . If the group has a center of inversion,the labels of the IRs carry an additional subscript, either “g” for symmetric underinversion or “u” for antisymmetric under inversion. For example, for the Oh group,

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428 Appendix B. Basics of point-group theory

x2 + y2, z2 z

Rz

x2 − y2

xy

xz, yzx, y

Rx, Ry

}

C4v E C2 2C4 2σv 2 σd

A1 1 1 1 1 1A2 1 1 1 −1 −1B1 1 1 −1 1 −1B2 1 1 −1 −1 1

E 2 −2 0 0 0

(6) Basis functions

� �

(1) Name of the group

(3) Elements of the group by class

(4) Number ofelementsin the class�

(5) Characters ofthe classes fordifferent IRs

(2) Names of the IRs

Figure B.2 The character table for C4v , illustrating the types of informationcontained therein.

Basis functions

x2 + y2, z2 z,Rz

xz, yzx, y

Rx, Ry

}

x2 − y2, xy

C5(5) E C5 C25 C3

5 C45

A 1 1 1 1 1

E ′{1 ω ω2 ω3 ω4

1 ω4 ω3 ω2 ω

E ′′{1 ω2 ω4 ω ω3

1 ω3 ω ω4 ω2

ω = e2πi/5

Pairs of one-dimensional IRswhose bases are degeneratedue to time-reversal symmetry

Figure B.3 The character table for C5, showing pairs of one-dimensional IRsgrouped together to form E representations.

one of the three-dimensional IRs is T2g, indicating a three-dimensional IR for basisfunctions that are symmetric under inversion.

For cyclic groups and direct products with cyclic groups, pairs of one-dimen-sional basis functions may be degenerate due to time-reversal symmetry. In thesecases the two IRs are grouped together and listed as a two-dimensional, E-type,representation. This feature is illustrated for C5 in Fig. B.3.

Some groups are specified as a direct product. An example is Oh = O × i . Let{Rα} denote the elements of O . Then the elements of Oh are {Rα, i Rα}. For suchgroups, the character table can be divided into four sectors. Denote the table ofcharacters for the O group as X , then the table of characters of Oh is

gu

(X XX −X

),

where g indicates that the basis functions for the upper rows of IRs are symmetricunder inversion and u indicates that the basis functions for the lower rows of IRsare antisymmetric under inversion. For more detail see the character table for Oh

in Appendix C.

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B.5 Character and action tables 429

B.5.1 The action table

For molecular vibrations the covering-group operations transform both the dis-placements and the atomic position numbers relative to a fixed coordinate system.A representation, �, of the group that is based on atomic displacements and atomicposition numbers will consist of 3Na × 3Na matrices, where Na is the number ofatoms in the molecule. The action of an operator on displacements can be deter-mined by considering separately the way in which the vector r transforms under theoperators of the group and the way in which the atomic position numbers changerelative to the fixed coordinate system. A representation based on atomic displace-ments and atomic position numbers is the direct product of a representation �r

based on rx , ry , and rz , and �(ap) based on the atomic position numbers. That is,�(molecule) = �r × �(ap) (“ap” stands for “atomic positions”).

A similar result holds for molecular electronic states, in which case �r is a rep-resentation based on how the orbitals alone transform. For vibrations of crystallinesolids, Na is the number of atoms per unit cell. For the LCAO electronic statesof a crystalline solid, �r is a representation that is based on the way in which theorbitals in the central unit cell transform under the group of the wavevector.

An “action table” is a convenient, compact method for obtaining the informationneeded to construct �r×�(ap). Use of an action table is first discussed in Chapter 1.Action tables for the octahedral and tetrahedral groups are given in Appendices Eand F, respectively.

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Appendix C

Character tables for point groups

C1(1) E

A 1h = 1

C2(2) E C2

x2, y2, z2, xy z, Rz A 1 1

yz, zxx, yRx , Ry

}B 1 −1

h = 2

C3(3) E C3 C23

x2 + y2, z2 z, Rz A 1 1 1yz, zxx2 − y2, xy

}x, yRx , Ry

}E

{11

ω ω2

ω2 ω

ω = e2π i/3

h = 3

C4(4) E C2 C4 C34

x2 + y2, z2 z, Rz A 1 1 1 1x2 − y2, xy B 1 1 −1 −1

yz, zxx, yRx , Ry

}E

{11

−1 i −i−1 −i i

h = 4

430

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Character tables for point groups 431

C5(5) E C5 C25 C3

5 C45

x2 + y2, z2 z, Rz A 1 1 1 1 1

xz, yzx, yRx , Ry

}E ′

{11

ω ω2 ω3 ω4

ω4 ω3 ω2 ω

x2 − y2, xy E ′′{

11

ω2 ω4 ω ω3

ω3 ω ω4 ω2

ω = e2π i/5

h = 5

C6(6) E C6 C3 C2 C23 C5

6

x2 + y2, z2 z, Rz A 1 1 1 1 1 1B 1 −1 1 −1 1 −1

xz, yzx, yRx , Ry

}E ′

{11

ω ω2 ω3 ω4 ω5

ω5 ω4 ω3 ω2 ω

x2 − y2, xy E ′′{

11

ω2 ω4 1 ω2 ω4

ω4 ω2 1 ω4 ω2

ω = e2π i/6

h = 6

C2v(2mm) E C2 σv σ ′v

x2, y2, z2 z A1 1 1 1 1xy Rz A2 1 1 −1 −1xz Ry, x B1 1 −1 1 −1yz Rx , y B2 1 −1 −1 1

h = 4

C3v(3m) E 2C3 3σv

x2 + y2, z2 z A1 1 1 1Rz A2 1 1 −1

yz, zxx2 − y2, xy

}x, yRx , Ry

}E 2 −1 1

h = 6

C4v(4mm) E C2 2C4 2σv 2σd

x2 + y2, z2 z A1 1 1 1 1 1Rz A2 1 1 1 −1 −1

x2 − y2 B1 1 1 −1 1 −1xy B2 1 1 −1 −1 1

xz, yzx, yRx , Ry

}E 2 −2 0 0 0

h = 8

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432 Appendix C. Character tables for point groups

C5v(5m) E 2C5 2C25 5σv

x2 + y2, z2 z A1 1 1 1 1Rz A2 1 1 1 -1

yz, zxx, yRx , Ry

}E1 2 2 cos x 2 cos(2x) 0

x2 − y2, xy E2 2 2 cos(2x) 2 cos(4x) 0

x = 2π/5h = 10

C6v(6mm) E C2 2C3 2C6 3σd 3σv

x2 + y2, z2 z A1 1 1 1 1 1 1Rz A2 1 1 1 1 −1 −1

B1 1 −1 1 −1 −1 1B2 1 −1 1 −1 1 −1

yz, zxx, yRx , Ry

}E1 2 −2 −1 1 0 0

x2 − y2, xy E2 2 2 −1 −1 0 0

h = 12

C1h(m) E σh

x2, y2, z2, xy x, y, Rz A′ 1 1yz, zx Rx , Ry, z A′′ 1 −1

h = 2

C2h(2/m) E C2 σh i

x2, y2, z2, xy Rz Ag 1 1 1 1z Au 1 1 −1 −1

xz, yz Rx , Ry Bg 1 −1 −1 1x, y Bu 1 −1 1 −1

h = 4

C3h = C3 × σh(6) E C3 C23 σh S3 (σhC2

3)

x2 + y2, z2 Rz A′ 1 1 1 1 1 1z A′′ 1 1 1 −1 −1 −1

x2 − y2, xy x, y E ′{

11

ω ω2 1 ω ω2

ω2 ω 1 ω2 ω

xz, yz Rx , Ry E ′′{

11

ω ω2 −1 −ω −ω2

ω2 ω −1 −ω2 −ω

ω = e2π i/3

h = 6

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Character tables for point groups 433

C4h = C4 × i(4/m) h = 8C5h = C5 × σh(10) h = 10C6h = C6 × i(6/m) h = 12

S2(1) E i

x2, y2, z2, xy, yz, zx Rx , Ry, Rz Ag 1 1x, y, z Au 1 −1

h = 2

S4(4) E C2 S4 S34

x2 + y2, z2 Rz A 1 1 1 1z B 1 1 −1 −1

xz, yzx2 − y2, xy

}x, yRx , Ry

}E

{11

−1 i −i−1 −i i

h = 4

S6 = C3 × i(3) h = 6

D2(222) E Cz2 C y

2 Cx2

x2, y2, z2 A1 1 1 1 1xy Rz , z B1 1 1 −1 −1xz Ry , y B2 1 −1 1 −1yz Rx , x B3 1 −1 −1 1

h = 4

D3(32) E 2C3 3C ′2x2 + y2, z2 A1 1 1 1

z, Rz A2 1 1 −1yz, zxx2 − y2, xy

}x, yRx , Ry

}E 2 −1 0

h = 6

D4(422) EC2 = C24 2C4 2C ′2 2C ′′2

x2 + y2, z2 A1 1 1 1 1 1Rz , z A2 1 1 1 −1 −1

x2 − y2 B1 1 1 −1 1 −1xy B2 1 1 −1 −1 1

xz, yzx, yRx , Ry

}E 2 −2 0 0 0

h = 8

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434 Appendix C. Character tables for point groups

D5(52) E 2C5 2C25 5C ′2

x2 + y2, z2 A1 1 1 1 1z, Rz A2 1 1 1 −1

yz, zxx, yRx , Ry

}E1 2 2 cos x 2 cos(2x) 0

x2 − y2, xy E2 2 2 cos(2x) 2 cos(4x) 0

x = 2π/3h = 10

D6(622) E C2 2C3 2C6 3C ′2 3C ′′2x2 + y2, z2 A1 1 1 1 1 1 1

Rz , z A2 1 1 1 1 −1 −1B1 1 −1 1 −1 1 −1B2 1 −1 1 −1 −1 1

xz, yzx, yRx , Ry

}E1 2 −2 −1 1 0 0

x2 − y2, xy E2 2 2 −1 −1 0 0

h = 12

D2d(42m) E C2 2S4 2C ′2 2σd

x2 + y2, z2 A1 1 1 1 1 1Rz A2 1 1 1 −1 −1

x2 − y2 B1 1 1 −1 1 −1xy z B2 1 1 −1 −1 1

xz, yzx, yRx , Ry

}E 2 −2 0 0 0

h = 8

D3d = D3 × i (3m) h = 12D2h = D2 × i (mmm) h = 8

D3h = D3 × σh (6m2) E σh 2C3 2S3 3C ′2 3σv

x2 + y2, z2 A′1 1 1 1 1 1 1Rz A′2 1 1 1 1 −1 −1

A′′1 1 −1 1 −1 1 −1z A′′2 1 −1 1 −1 −1 1

x2 − y2, xy x, y E ′ 2 2 −1 −1 0 0xz, yz Rx , Ry E ′′ 2 −2 −1 1 0 0

h = 12

D4h = D4 × i (4/mmm) h = 16D5h = D5 × σh (10m2) h = 20D6h = D6 × i (6/mmm) h = 24

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Character tables for point groups 435

T (23) E C2 4C3 4C S′3

r2 A 1 1 1 1

x2 − y2, 3z2 − r2 E{

11

1 ω ω2

1 ω2 ω

xy, yz, zxx, y, zRx , Ry, Rz

}T 3 −1 0 0

h = 4

Th = T × i(m3) E 4C3 4C23 3C ′2 i 4S6 4S5

6 3σh

x2 + y2 + z2 Ag 1 1 1 1 1 1 1 1

x2 − y2, 3z2 − r2 Eg

{11

ω ω2 1 1 ω ω2 1ω2 ω 1 1 ω2 ω 1

xy, xz, yz Rx , Ry, Rz Tg 3 0 0 −1 3 0 0 −1Au 1 1 1 1 −1 −1 −1 −1

Eu

{11

ω ω2 1 −1 −ω −ω2 −1ω2 ω 1 −1 −ω2 −ω −1

x, y, z Tu 3 0 0 1 −3 0 0 1

h = 48

O(432) E 8C3 3C2 6C2 6C4

A1 1 1 1 1 1A2 1 1 1 −1 −1

x2 − y2, 3z2 − r2 E 2 -1 2 0 0Rx , Ry, Rzx, y, z

}T1 3 0 −1 −1 1

xy, yz, xz T2 3 0 −1 1 −1

h = 24

Oh = O × i (m3m) E 8C3 3C2 6C ′2 6C4 i 8iC3 3iC2 6iC ′2 6iC4

x2 + y2 + z2 A1g 1 1 1 1 1 1 1 1 1 1A2g 1 1 1 −1 −1 1 1 1 −1 −1

x2 − y2, 3z2 − r2 Eg 2 −1 2 0 0 2 −1 2 0 0Rx , Ry , Rz T1g 3 0 −1 −1 1 3 0 −1 −1 1xy, xz, yz T2g 3 0 −1 1 −1 3 0 −1 1 −1

A1u 1 1 1 1 1 −1 −1 −1 −1 −1A2u 1 1 1 −1 −1 −1 −1 −1 1 1Eu 2 −1 2 0 0 −2 1 −2 0 0

x, y, z T1u 3 0 −1 −1 1 −3 0 1 1 −1T2u 3 0 −1 1 −1 −3 0 1 −1 1

h = 48

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436 Appendix C. Character tables for point groups

Td(43m) E 8C3 3C2 6σd 6S4

r2 A1 1 1 1 1 1A2 1 1 1 −1 −1

x2 − y2, 3z2 − r2 E 2 −1 2 0 0Rx , Ry, Rz T1 3 0 −1 −1 1

xy, yz, xz x, y, z T2 3 0 −1 1 −1

h = 24

C∞v(∞m) E 2Cφ σv

x2 + y2, z2 z A1(�+) 1 1 1

Rz A2(�−) 1 1 −1

xz, yzx, yRx , Ry

}E1(�) 2 2 cosφ 0

x2 − y2, xy E2(�) 2 2 cos(2φ) 0. . . . . . . . . . . .

h = 4

D∞h(∞/mm) E 2Cφ C ′2 i 2iCφ iC ′2x2 + y2, z2 A1g(�

+g ) 1 1 1 1 1 1

A1u(�−u ) 1 1 1 −1 −1 −1

Rz A2g(�−g ) 1 1 −1 1 1 −1

z A2u(�+u ) 1 1 −1 −1 −1 1

xz, yz Rx , Ry E1g(�g) 2 2 cosφ 0 2 2 cosφ 0x , y E1u(�u) 2 2 cosφ 0 −2 −2 cosφ 0

x2 − y2, xy E2g(�g) 2 2 cos(2φ) 0 2 2 cos(2φ) 0E2u(�u) 2 2 cos(2φ) 0 −2 −2 cos(2φ) 0

. . . . . . . . . . . . . . . . . . . . .

h = 8

O ′ E R 8C3 8RC3 6C2 12σd 6S4 6RS4

�1(A1) 1 1 1 1 1 1 1 1�2(A2) 1 1 1 1 1 −1 −1 −1�3(E) 2 2 −1 −1 2 0 0 0�4(T1) 3 3 0 0 −1 −1 1 1�5(T2) 3 3 0 0 −1 1 −1 −1�6 2 −2 1 −1 0 0

√2 −√2

�7 2 −2 1 −1 0 0 −√2√

2�8 4 −4 −1 1 0 0 0 0

Double group

h = 48

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C.1 Some character tables for nanotubes 437

C.1 Some character tables for nanotubes

(See Group Theory Applications to the Physics of Condensed Matter, M. S.Dresselhaus, G. Dresselhaus, and A. Jorio, Springer 2008.)

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Character table for the group of the wavevectors k = 0 and k = π/τ for achiral carbon nanotubes

D2nh {E |0} . . . 2Sp . . . 2Sn n{C ′2|0} n{C ′′2 |0} {E |0} . . . 2S′p . . . {σh |0} n{σ ′v|0} n{σ ′′v |τ/2}A1g 1 . . . 1 . . . 1 1 1 1 . . . 1 . . . 1 1 1A2g 1 . . . 1 . . . 1 −1 −1 1 . . . 1 . . . 1 −1 −1B1g 1 . . . (−1)p . . . (−1)n −1 1 1 . . . (−1)p . . . (−1)n −1 1B2g 1 . . . (−1)p . . . (−1)n 1 −1 1 . . . (−1)p . . . (−1)n 1 −1...

......

...... . . .

......

...... . . .

......

......

Eμg 2 . . . 2 cos(μpπ/n) . . . 2(−1)μ 0 0 2 . . . 2 cos(μpπ/n) . . . 2(−1)μ 0 0...

......

...... . . .

......

......

... . . ....

......

A1u 1 . . . 1 . . . 1 1 1 −1 . . . −1 . . . -1 −1 −1A2u 1 . . . 1 . . . 1 −1 −1 −1 . . . −1 . . . -1 1 1B1u 1 . . . (−1)p . . . (−1)n −1 1 −1 . . . −(−1)p . . . −(−1)n 1 −1B2u 1 . . . (−1)p . . . (−1)n 1 −1 −1 . . . −(−1)p . . . −(−1)n −1 1...

......

...... . . .

......

...... . . .

......

......

Eμu 2 . . . 2 cos(μpπ/n) . . . 2(−1)μ 0 0 −2 . . . −2 cos(μpπ/n) . . . −2(−1)μ 0 0...

......

...... . . .

......

......

... . . ....

......

The values of p and μ span the integer values between 1 and n − 1. Sm = {Cu2n|vτ/(2n)}m , S′m = {iCu

2n|vτ/(2n)}m .

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Character table for the group of the wavevectors k = 0 and k = π/τ for chiral carbon nanotubes

DN {E |0} 2S1 2S2 . . . 2SN/2−1 2SN/2 (N/2){C ′2|0} (N/2){C ′′2 |0}A1 1 1 1 . . . 1 1 1 1A2 1 1 1 . . . 1 1 −1 −1B1 1 −1 1 . . . (−1)(N/2−1) (−1)N/2 1 −1B2 1 −1 1 . . . (−1)(N/2−1) (−1)N/2 −1 1E1 2 2 cos(2π/N ) 2 cos(4π/N ) . . . 2 cos[2(N/2− 1)π/N ] −2 0 0E2 2 2 cos(4π/N ) 2 cos(8π/N ) . . . 2 cos[4(N/2− 1)π/N ] 2 0 0...

......

...... . . .

......

...

EN/2−1 2 2 cos[2(N/2− 1)π/N ] 2 cos[4(N/2− 1)π/N ] . . . 2 cos[2(N/2− 1)2π/N ] 2 cos[(N/2− 1)π ] 0 0

Sm = {Cu2n|vτ/(2n)}m .

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Character table for the group of the wavevectors 0 < k < π/τ for achiral carbon nanotubes

C2nv {E |0} 2S1 2S2 . . . 2Sn−1 2Sn n{σ ′v|r ′} n{σ ′′v |r ′′}A′ 1 1 1 . . . 1 1 1 1A′′ 1 1 1 . . . 1 1 −1 −1B ′ 1 −1 1 . . . (−1)(n−1) (−1)n 1 −1B ′′ 1 −1 1 . . . (−1)(n−1) (−1)n −1 1E1 2 2 cos(π/n) 2 cos(2π/n) . . . 2 cos[2(n − 1)π/n] −2 0 0E2 2 2 cos(2π/n) 2 cos(4π/n) . . . 2 cos[4(n − 1)π/n] 2 0 0...

......

...... . . .

......

...

En−1 2 2 cos[(n − 1)π/n] 2 cos[2(n − 1)π/n] . . . 2 cos[(n − 1)2π/n] 2 cos[(n − 1)π] 0 0

For zigzag nanotubes with n odd, r ′ = r ′′ = τ/2. Sm = {Cu2n|vτ/(2n)}m .

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Character table for the group of the wavevectors 0 < k < π/τ for chiral carbon nanotubes

CN {E |0} S1 S2 S3 . . . St . . . SN−1

A 1 1 1 1 . . . 1 . . . 1B 1 −1 1 −1 . . . (−1)t . . . −1E+1 1 ω ω2 ω3 . . . ωt . . . ωN−1

E−1 1 ω∗ ω∗2 ω∗3 . . . ω∗t . . . ω∗(N−1)

E+2 1 ω2 ω4 ω6 . . . ω2t . . . ω2(N−1)

E−2 1 ω∗2 ω∗4 ω∗6 . . . ω∗2t . . . ω∗2(N−1)

......

......

... . . .... . . .

...

E+N/2−1 1 ω(N/2−1) ω2(N/2−1) ω3(N/2−1) . . . ωt (N/2−1) . . . ω(N/2−1)(N−1)

E−N/2−1 1 ω∗(N/2−1) ω∗2(N/2−1) ω∗3(N/2−1) . . . ω∗t (N/2−1) . . . ω∗(N/2−1)(N−1)

Sm = {CuN |vτ/N }m , ω = e2π i/N .

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Appendix D

Tensors, vectors, and equivalent electrons

A tensor may be described as a multidimensional array of components (numbers).The rank of a tensor is the number of indices required to specify a particularcomponent. A zeroth-rank tensor is a scalar. A first-rank tensor is a vector withcomponents Vi . A second-rank tensor is a matrix with elements Mi j . An nth-ranktensor has components labeled by n indices: Ti(1)i(2)i(3)...i(n).

In science tensors are used to express relationships between vectors and combi-nations of vectors. For example, in electromagnetic theory the relationship betweenthe dielectric displacement, D, and the electric intensity, E, is the dielectric tensor,Di =∑

j εi j E j , where εi j is a second-rank tensor.The coupling of two quantum systems of angular momenta J1 and J2 can be

characterized by eigenfunctions whose quantum numbers are J2 = (J1 + J2)2 =

j ( j + 1) and Jz = j1z + j2z . These quantum numbers label the representationD( j1) × D( j2) based on the product of the eigenfunctions of the two isolatedsystems. The decomposition of the direct-product representation is

D( j1) × D( j2) =| j1+ j2|∑

k=| j1− j2|D(k). (D.1)

The vector r transforms according to D(1) so that the product of the componentsof two vectors r1 and r2 transforms as a second-rank tensor. The component Ti j

transforms as the product r1i r2 j , which in turn transforms according to D(1)×D(1).The direct product D(1) × D(1) is a reducible representation that decomposes intothe IRs D(0)+D(1)+D(2) of the full rotation group. Similarly, a second-rank tensorcan be decomposed into a sum of irreducible tensors,

T = T (1) + T (2) + T (3), (D.2)

which transform as S, P , and D functions, respectively. The parts of the decompo-sition T (i) (i = 1, 2, and 3) are called irreducible tensors because they transform

442

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Appendix D. Tensors, vectors, and equivalent electrons 443

according to the IRs of the full rotation group. The total second-rank tensordescribed above (Cartesian coordinates) is a sum of parts that transform as a scalarand as an axial vector, and a set of five degenerate d-orbitals. This type of decompo-sition allows the determination of selection rules. For example, the matrix elementof T between two atomic-like states, ψ(α) and ψ(β),

〈α|T |β〉 =∫

ψ(α)∗ T ψ(β) dτ, (D.3)

vanishes unless D(α) × D(β) contains D(0), D(1), or D(2). If α = β, then the matrixelement does not vanish because of symmetry considerations since D(α) × D(α)

contains D(0).Combinations of the Cartesian components can be used to express the irreducible

forms of the tensor. For example, the scalar product r1 ·r2 =∑k r1k r2k is invariant

under all of the operations of the full rotation group, and T (1) may be taken asTr{T} = T11 + T22 + T33.

The cross product, r1×r2, is an axial vector. Its components are (r1y r2z−r1z r2y),(r1x r2z − r1z r2x ), and (r1x r2y − r1y r2x ), which are symmetric under inversion, butantisymmetric in the interchange of r1 and r2. The elements (T13−T31), (T12−T21),and (T12 − T21) constitute the part of T that transforms as D(1). The remainingelements that transform as D(2) may be patterned after the five angular functionsof the d-orbitals (the spherical harmonics for l = 2),

(T12 + T21) (analogous to xy)(T13 + T31) (analogous to xz)(T23 + T32) (analogous to yz)(T11 − T22) (analogous to x2 − y2)(2T33 − T11 − T22) (analogous to 2z2 − x2 − y2).

It is useful to define T (2) in such a way that it is traceless. The components are thentaken as

T (2)i j ≡ Si j = 1

2(Ti j + Tji )− 1

3Tr(T ),

where the last term ensures that Tr{S} = 0. Only five of the six components areindependent, since we have required the trace to vanish.

Higher-rank tensors can be constructed in a similar fashion by consideringthe transformation properties of higher-order products of the components such asr1i r2 j r3k . A third-rank tensor, Ti jk , with 27 components transforms as D(1)×D(1)×D(1), so it decomposes into D(0) + 3D(1) + 2D(2) + D(3). Thus it decomposes intoseven distinct tensors, namely one scalar, three first-rank, two second-rank, and one

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444 Appendix D. Tensors, vectors, and equivalent electrons

third-rank irreducible tensor. A fourth-rank tensor has 81 components and decom-poses into 3D(0)+6D(1)+6D(2)+3D(3)+D(4). From the decomposition rule it isclear that an nth-rank tensor in Cartesian coordinates will always contain exactlyone nth-rank, irreducible tensor.

D.1 Spherical component vectors

It is often convenient to use vectors specified by their spherical-like componentsrather than their Cartesian components. To this end we define a vector, V, andorthogonal unit vectors, e0, e−1, and e1, that form the basis vectors for the D(1) IR:

V = −V−1 e1 + V0 e0 − V1 e−1, (D.4)

e0 = ez, (D.5)

e1 = − 1√2(ex + iey), (D.6)

e−1 = 1√2(ex − iey), (D.7)

eμ · eν = (−1)μδμ,−ν. (D.8)

Equation (D.4) can be written as

V =∑ν

(−1)νV−ν eν, (D.9)

so that the scalar product between two vectors V and X is

V · X =∑μ

∑ν

(−1)μ+νV−μX−ν(eμ · eν) =∑μ

(−1)μV−μXμ. (D.10)

Vectors defined in this way are basis functions for the D(1) IRs of the full rotationgroup and transform as

PR Vμ(r) =∑ν

D(1)νμ(R−1) Vν(r) (ν = 0,±1).

The spherical vectors can be used to construct irreducible tensors of any rank.In particular, irreducible tensor operators can be defined by their transformationproperties. If j denotes the angular quantum number then a tensor operator, T ( j)

m ,is of rank (2 j + 1) (the magnetic quantum number, m, takes on 2 j + 1 values).These tensors transform among themselves as

PR T ( j)m =

∑m′

T ( j)m′ D( j)

m′m(R)

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D.2 Representations based on products of wavefunctions 445

when subjected to a rotation, R. Analogous to case of the Cartesian tensors, thetransformation properties correspond to those of products of components of thespherical vectors.

D.2 Representations based on products of wavefunctions

Consider the product of two equivalent, one-electron wavefunctions for a spher-ically symmetric system, �

j× jmm′ = φ

jm(r) φ

jm′(r

′), where j and m denote theangular-momentum and magnetic quantum numbers. The �

j× jmm′ functions are bases

for a direct-product representation,

PR�j× j

mm′ = (PR φ jm(r))(PR ψ

jm′(r

′))

=(∑

λ

D( j)λm(R−1) φ

) (∑ω

D( j)ωm′(R−1) ψ j

ω

)

=∑λ

∑ω

(D( j)(R−1)× D( j)(R−1)

)λm,ωm′

�j× jλω , (D.11)

where (D( j)(R−1)× D( j)(R−1)

)λm,ωm′

= D( j)λm(R−1) D( j)

ωm′(R−1) (D.12)

and D( j)(R) is the IR matrix for the rotation R and angular momentum J 2 =j ( j + 1).

We can define basis functions that are symmetric or antisymmetric in theinterchange of indices m and m ′ (equivalent to interchanging r and r′):

�j× j±

mm′ =1

2

[�

j× jmm′ ±�

j× jm′m

]. (D.13)

Under the operations of the full rotation group these functions transform as

PR �j× j±

mm′ =1

2

∑λ

∑ω

[D( j)

λm D( j)ωm′ ± D( j)

λm′ D( j)ωm

]�

j× jλω (D.14)

(for simplicity in notation we omit the operation symbol R−1 in D(R−1).For the upper (+) sign the quantity in brackets is symmetric in the interchange

of λ and ω. For the lower (−) sign the quantity in brackets is antisymmetric in theinterchange of λ and ω. Therefore we may write

PR �j× j+

mm′ =1

2

∑λ

∑ω

[D( j)

λm D( j)ωm′ + D( j)

λm′ D( j)ωm

]1

2

[�

j× jmm′ +�

j× jm′m

]

= 1

2

∑λ

∑ω

[D( j)

λm D( j)ωm′ + D( j)

λm′ D( j)ωm

]�

j× j+mm′ . (D.15)

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446 Appendix D. Tensors, vectors, and equivalent electrons

Similarly,

PR �j× j−

mm′ =1

2

∑λ

∑ω

[D( j)

λm D( j)ωm′ − D( j)

λm′ D( j)ωm

]1

2

[�

j× jmm′ −�

j× jm′m

]

= 1

2

∑λ

∑ω

[D( j)

λm D( j)ωm′ − D( j)

λm′ D( j)ωm

]�

j× j−mm′ . (D.16)

We can combine (D.15) and (D.16) in the form

PR �j× j±

mm′ =∑λ

∑ω

D j× j±λω,mm′ �

j× j±λω , (D.17)

where

D j× j±λω,mm′ =

1

2

[D( j)

λm D( j)ωm′ ± D( j)

λm′ D( j)ωm

].

Equation (D.15) shows that the functions �j× j+

mm′ transform among themselvesunder the operations of the group and do not mix with the antisymmetric func-tions. Equation (D.16) confirms that the functions �

j× j−mm′ do not mix with the

symmetric functions. Therefore, the �j× j±

mm′ functions are bases for a symmetric(antisymmetric) representation of the full rotation group, and the IR matrices areD j× j±

λω,mm′ .Let S = R−1, then the characters of these representations are

χ j (S)± =∑λ

∑ω

D j× j±λω,λω(S)

= 1

2

∑λ

∑ω

[D( j)

λλ (S) D( j)ωω(S)± D( j)

λω (S) D( j)ωλ (S)

]. (D.18)

For the left-hand side of (D.18) we have

1

2

∑λ

∑ω

[D( j)

λλ (S) D( j)ωω(S)

]= 1

2

{∑λ

D( j)λλ (S)

}{∑λ

D( j)λλ (S)

}

= 1

2[χ j (S)]2, (D.19)

and

1

2

∑λ

∑ω

[D( j)

λω (S) D( j)ωλ (S)

]= 1

2

∑λ

D( j)λλ (S

2) = 1

2[χ j (S2)]. (D.20)

Substituting (D.19) and (D.20) into (D.18) gives

χ j (S)± = 1

2[χ j (S)]2 ± 1

2χ j (S2). (D.21)

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D.2 Representations based on products of wavefunctions 447

This may be written as

χ j (θ)± = 1

2[χ j (θ)]2 ± 1

2χ j (2θ), (D.22)

χ j (θ) = sin[( j + 12)θ]

sin(θ/2), (D.23)

where θ is the angle of rotation of the operation S.Equation (D.23) can be rewritten as

χ0(θ) = 1, (D.24)

χ j (θ) = 1+j∑

n=1

2 cos(nθ), for j ≥ 1. (D.25)

Equation (D.21) leads to restrictions on the direct-product decompositions. Forexample, for j = 1,

χ1(θ)± = 1

2[χ1(θ)]2± 1

2χ1(2θ) = 1

2[1+ 2 cos(θ)]2± 1

2[1+ 2 cos(2θ)]. (D.26)

With the plus sign we find

χ1(θ)+ = 1+ [1+ 2 cos(θ)+ 2 cos(2θ)] = χ0(θ)+ χ2(θ), (D.27)

whereas with the minus sign we obtain

χ1(θ)− = 1+ 2 cos(θ) = χ1(θ). (D.28)

Equations (D.27) and (D.28) indicate the decompositions of the IRs D j× j± as

D(1×1)+ = D(0) + D(2), (D.29)

D(1×1)− = D(1). (D.30)

These results express the vector coupling model for a (spinless) two-electron wave-function. Equation (D.29) shows that two p-states can be combined to form asymmetric S and D state, but not a symmetric P state. Equation (D.30) shows thatthe only antisymmetric state that can be formed is a P state. The direct-productrule gives

D(1×1) = D(0) + D(1) + D(2), (D.31)

but gives no information about the parity of the two-electron states that can beformed. The notation D(i×i) or D(i×i) is often used to indicate a symmetric rep-resentation and D[i×i] or D[i×i] to indicate an antisymmetric representation. Also,the decomposition may be written as, for example,

D(1×1) = D(0) + [D(1)] + D(2), (D.32)

where the square brackets indicate that D(1) has odd parity.

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448 Appendix D. Tensors, vectors, and equivalent electrons

Next we consider the case where j = 2 (a pair of electrons in d-states).Equations (D.21) and (D.25) give

χ(2×2)(θ)± = 1

2[χ(2)(θ)]2 ± 1

2χ2(2θ)

= 1

2[1+ 2 cos(θ)+ 2 cos(2θ)]2 ± 1

2[1+ 2 cos(2θ)+ 2 cos(4θ)].

(D.33)

Using the minus sign in (D.33) gives

χ(2×2)(θ)− = 2+ 4 cos(θ)+ 2 cos(2χ(2×2)(θ)+)+ 2 cos(3θ)

= [1+ 2 cos(θ)] + [1+ 2 cos(θ)+ 2 cos(3θ)]= χ(1) + χ(3). (D.34)

The plus sign yields

χ(2×2)(θ)+ = 1+ [1+ 2 cos(θ)+ 2 cos(2θ)]+ [1+ 2 cos(θ)+ 2 cos(2θ)+ 2 cos(3θ)+ 2 cos(4θ)]= χ(0) + χ(2) + χ(4). (D.35)

Equations (D.34) and (D.35) indicate the decomposition as

D(2×2)+ = D(0) + D(2) + D(4), (D.36)

D(2×2)− = D(1) + D(3), (D.37)

or

D(2×2) = D(0) + [D(1)] + D(2) + [D(3)] + D(4), (D.38)

where the square brackets indicate antisymmetric representations. In general, in thedecomposition formulas, an odd j does not have a symmetric representation.

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Appendix E

The octahedral group, O and Oh

E.1 Elements of the O group

The octahedral or O group consists of the 24 rotations about various axes that leavea cube or octahedron unchanged. The symmetry axes are shown in Fig. E.1. Thesymmetry elements are listed below.

(1) Four C3 axes: 120- and 240-degree rotations about cube body diagonals.These symmetry axes are shown in Fig. E.1(a). We label the elements Ck

3

and (C23)

k , k = 1, 2, 3, and 4, where the superscript k identifies the particu-lar axis. There are in total eight such rotations, and these eight elements form aclass.

(2) Three C4 axes: 90-, 180-, and 270-degree rotations about the x-, y-, and z-axes in Fig. E.1. We label these elements as Ck

4 , Ck2 , and (C3

4)k , k = 1, 2, and

3. The six elements Ck4 and (C3

4)k , k = 1, 2, and 3, form a class. The three

elements Ck2 , k = 1, 2, and 3, form a separate class. The axes are shown in

Fig. E.1(b).(3) Six C2 axes: 180-degree rotations about axes through the origin bisecting face

edges. We label these elements as Ck2 , k =1, 2, . . . , 6. The six elements form a

class. The axes are shown in Fig. E.1(c).

The Oh group is the direct-product group O × i . All rotations are clockwise(left-hand screw rule) so that they correspond to operations rather than elementsof the group. The rotations and matrices representing them are appropriate for theoperators PR applied to a function.

E.2 The character table for O

The character table for the group O is given in Table E.1.

449

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450 Appendix E. The octahedral group, O and Oh

Table E.1 The character table for O. Oh = O × i (m3m).

O(432) E 8C3 3C2 6C2 6C4

A1 1 1 1 1 1A2 1 1 1 −1 −1

(x2 − y2, 3z2 − r2) E 2 −1 2 0 0(Rx , Ry , Rz), (x , y, z) T1 3 0 −1 −1 1(xy, yz, xz) T2 3 0 −1 1 −1

12

3 4

(a) (b) (c)

C3 axes1

2

3

C2, C4 axes

12 3

4 5

6C2 axes

Figure E.1 Symmetry axes and operations of the O group.

�x

� y

�z

1

2

3

4

5

6

B

BL6

B ≡ metal ionL ≡ ligand

Figure E.2 An octahedrally coordinated ion. The ligands are labeled 1 through 6.

E.3 The action table for octahedral coordination

The geometry for a central ion surrounded by an octahedron of ligands is shown inFig. E.2. The ligands are labeled 1 through 6. The action table for the octahedralcomplex is shown in Table E.2.

E.4 The irreducible-representation matrices of the O group

The individual matrices for the IRs of the O group are given in Tables E.3, E.4,and E.5. Denote the operators of the O group as the set {R}, then the operators ofOh are {R, i R}, where i is the inversion operator. If the matrices {Mα(R)} formthe �α IR of the O group, then for the Oh group �α

g has Mα(i R) = Mα(R) and �αu

has Mα(i R) = −Mα(R).

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Table E.2 The action table for the octahedral complex in Fig. E.2 for the O group. The first section shows how the coordinates, x,y, and z, transform under the group operations. The entries in boldface roman indicate coordinates that transform into a multipleof themselves. The second sections shows how the ligand positions are transformed under the group operations. The entries inboldface roman indicate the ligands that are unchanged by the operation. The third and fourth sections show how the d-orbitalstransform under the group operations.

E C13 C1

32

C23 C2

32

C33 C3

32

C43 C4

32

C12 C2

2 C32 C1

2 C22 C3

2 C42 C5

2 C62 C1

4 C14

3C2

4 C24

3C3

4 C34

3

x −y −z −z y −y z z y x −x −x −x −x −z z −y y x x z −z −y yy z −x x −z −z −x x z −y y −y z −z −y −y −x x −z z y y x −xz −x y −y −x x −y y x −z −z z y −y −x x −z −z y −y −x x z z

1 6 2 4 5 5 2 4 6 3 1 3 6 5 3 3 2 4 5 6 1 1 4 22 1 6 6 3 1 5 5 3 2 4 4 4 4 6 5 1 3 2 2 5 6 1 33 5 4 2 6 6 4 2 5 1 3 1 5 6 1 1 4 2 6 5 3 3 2 44 3 5 5 1 3 6 6 1 4 2 2 2 2 5 6 3 1 4 4 6 5 3 15 4 3 1 4 2 1 3 2 6 6 5 3 1 4 2 6 6 3 1 4 2 5 56 2 1 3 2 4 3 1 4 5 5 6 1 3 2 4 5 5 1 3 2 4 6 6

xy −yz xz −xz −yz yz −xz xz yz −xy −xy xy −xz xz yz −yz xy xy −xz xz yz −yz −xy −xyxz xy −yz yz −xy −xy −yz yz xy −xz xz −xz −xy xy xz xz yz −yz xy −xy −xz −xz −yz yzyz −xz −xy −xy xz −xz xy xy xz yz −yz −yz yz yz xy −xy xz −xz −yz −yz −xy xy xz −xz

3z2−r

2

2√3

=b

x2−y

2

2=

a

−1 2b+√ 3 2

a−

1 2a−√ 3 2

b

−1 2b−√ 3 2

a−

1 2a+√ 3 2

b

−1 2b−√ 3 2

a−

1 2a+√ 3 2

b

−1 2b+√ 3 2

a−

1 2a−√ 3 2

b

−1 2b+√ 3 2

a−

1 2a−√ 3 2

b

−1 2b−√ 3 2

a−

1 2a+√ 3 2

b

−1 2b−√ 3 2

a−

1 2a+√ 3 2

b

−1 2b+√ 3 2

a−

1 2a−√ 3 2

b

ba

ba

ba

−1 2b−√ 3 2

a1 2

a−√ 3 2

b

−1 2b−√ 3 2

a1 2

a−√ 3 2

b

−1 2b+√ 3 2

a1 2

a+√ 3 2

b

−1 2b+√ 3 2

a1 2

a+√ 3 2

b

b−a

b−a

−1 2b−√ 3 2

a1 2

a−√ 3 2

b

−1 2b−√ 3 2

a1 2

a−√ 3 2

b

−1 2b+√ 3 2

a1 2

a+√ 3 2

b

−1 2b+√ 3 2

a1 2

a+√ 3 2

b

b−a

b−a

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452 Appendix E. The octahedral group, O and Oh

Table E.3 The E representation

E =⎛⎝ 1 0

0 1

⎞⎠ C1

3 =⎛⎝ − 1

2 −√3/2

√3/2 − 1

2

⎞⎠ C1

32 =

⎛⎝ − 1

2

√3/2

−√3/2 − 12

⎞⎠

C23 =

⎛⎝ − 1

2

√3/2

−√3/2 − 12

⎞⎠ C2

32 =

⎛⎝ − 1

2 −√3/2

√3/2 − 1

2

⎞⎠ C3

3 =⎛⎝ − 1

2 −√3/2

√3/2 − 1

2

⎞⎠

C33

2 =⎛⎝ − 1

2

√3/2

−√3/2 − 12

⎞⎠ C4

3 =⎛⎝ − 1

2

√3/2

−√3/2 − 12

⎞⎠ C4

32 =

⎛⎝ − 1

2 −√3/2

√3/2 − 1

2

⎞⎠

C12 =

⎛⎝ 1 0

0 1

⎞⎠ C2

2 =⎛⎝ 1 0

0 1

⎞⎠ C3

2 =⎛⎝ 1 0

0 1

⎞⎠

C12 =

⎛⎝ 1

2 −√3/2

−√3/2 − 12

⎞⎠ C2

2 =⎛⎝ 1

2 −√3/2

−√3/2 − 12

⎞⎠ C3

2 =⎛⎝ 1

2

√3/2

√3/2 − 1

2

⎞⎠

C42 =

⎛⎝ 1

2

√3/2

√3/2 − 1

2

⎞⎠ C5

2 =⎛⎝ −1 0

0 1

⎞⎠ C6

2 =⎛⎝ −1 0

0 1

⎞⎠

C14 =

⎛⎝ 1

2 −√3/2

−√3/2 − 12

⎞⎠ C1

43 =

⎛⎝ 1

2 −√3/2

−√3/2 − 12

⎞⎠ C2

4 =⎛⎝ 1

2

√3/2

√3/2 − 1

2

⎞⎠

C24

3 =⎛⎝ 1

2

√3/2

√3/2 − 1

2

⎞⎠ C3

4 =⎛⎝ −1 0

0 1

⎞⎠ C3

43 =

⎛⎝ −1 0

0 1

⎞⎠

For Eu , M(i R) = −M(R). For Eg , M(i R) = M(R). The basis functions are x2 − y2 and(3z2 − r2)/

√3.

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E.4 The irreducible-representation matrices of the O group 453

Table E.4 The T1 representation

E =(

1 0 00 1 00 0 1

)C1

3 =(

0 −1 00 0 1−1 0 0

)C1

32 =

(0 0 −1−1 0 0

0 1 0

)

C23 =

(0 0 −11 0 00 −1 0

)C2

32 =

(0 1 00 0 −1−1 0 0

)C3

3 =(

0 −1 00 0 −11 0 0

)

C33

2 =(

0 0 1−1 0 0

0 −1 0

)C4

3 =(

0 0 11 0 00 1 0

)C4

32 =

(0 1 00 0 11 0 0

)

C12 =

(1 0 00 −1 00 0 −1

)C2

2 =( −1 0 0

0 1 00 0 −1

)C3

2 =( −1 0 0

0 −1 00 0 1

)

C12 =

( −1 0 00 0 10 1 0

)C2

2 =( −1 0 0

0 0 −10 −1 0

)C3

2 =(

0 0 −10 −1 0−1 0 0

)

C42 =

(0 0 10 −1 01 0 0

)C5

2 =(

0 −1 0−1 0 0

0 0 −1

)C6

2 =(

0 1 01 0 00 0 −1

)

C14 =

(1 0 00 0 −10 1 0

)C1

43 =

(1 0 00 0 10 −1 0

)C2

4 =(

0 0 10 1 0−1 0 0

)

C24

3 =(

0 0 −10 1 01 0 0

)C3

4 =(

0 −1 01 0 00 0 1

)C3

43 =

(0 1 0−1 0 0

0 0 1

)

For T1u , M(i R) = −M(R). For T1g , M(i R) = M(R). The basis functions are x(z2 − y2)

(row 1), y(z2 − x2) (row 2), and z(x2 − y2).

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454 Appendix E. The octahedral group, O and Oh

Table E.5 The T2 representation

E =(

1 0 00 1 00 0 1

)C1

3 =(

0 0 −11 0 00 −1 0

)C1

32 =

(0 1 00 0 −1−1 0 0

)

C23 =

(0 −1 00 0 1−1 0 0

)C2

32 =

(0 0 1−1 0 0

0 1 0

)C3

3 =(

0 0 1−1 0 0

0 −1 0

)

C33

2 =(

0 −1 00 0 −11 0 0

)C4

3 =(

0 1 00 0 11 0 0

)C4

32 =

(0 0 11 0 00 1 0

)

C12 =

( −1 0 00 −1 00 0 1

)C2

2 =( −1 0 0

0 1 00 0 −1

)C3

2 =(

1 0 00 −1 00 0 −1

)

C12 =

(0 −1 0−1 0 0

0 0 1

)C2

2 =(

0 1 01 0 00 0 1

)C3

2 =(

0 0 10 1 01 0 0

)

C42 =

(0 0 −10 1 0−1 0 0

)C5

2 =(

1 0 00 0 10 1 0

)C6

2 =(

1 0 00 0 −10 −1 0

)

C14 =

(0 −1 01 0 00 0 −1

)C1

43 =

(0 1 0−1 0 0

0 0 −1

)C2

4 =(

0 0 10 −1 0−1 0 0

)

C24

3 =(

0 0 −10 −1 01 0 0

)C3

4 =( −1 0 0

0 0 −10 1 0

)C3

43 =

( −1 0 00 0 10 −1 0

)

For T2u , M(i R) = −M(R). For T2g , M(i R) = M(R). The basis functions are xy, yz,and zx .

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Appendix F

The tetrahedral group, Td

F.1 Elements of the Td group

The octahedral or Td group consists of the 24 rotations, rotations, and reflectionsthat leave a tetrahedron unchanged. For visual purposes it is useful to show thetetrahedron inscribed within a cube as illustrated in Fig. F.1. The symmetry axesare also shown. The symmetry elements in addition to the identity are listed below.

(1) C3. Four three-fold axes: 120- and 240-degree clockwise rotations about cubebody diagonals. We label the elements C (k)

3 and (C (k)3 )2, k = 1, 2, 3, and 4,

where the superscript k identifies the particular axis. There are in total eightsuch rotations, and these eight elements form the C3 class. The C3 axes areshown in Fig. F.1(a).

(2) σd . Six reflections. The reflection planes are defined by a pair of face diagonalson opposite sides of the circumscribed cube. We label these elements as σ

(k)d ,

k = 1, . . . , 6. The six elements form the σd class. The σd reflection planes areshown in Fig. F.1(b).

(3) C2. Three two-fold axes: 180-degree clockwise rotations about the x-, y-, andz-axes. We label these elements as C (k)

2 , k = 1, 2, and 3. The three elementsCk

2 , k = 1, 2, and 3, form the C2 class. The C2 axes are shown in Fig. F.1(c).(4) S4. Six improper rotations. Clockwise rotations by 90 and 270 degrees about a

C2 axis followed by a reflection in the plane perpendicular to the axis of rota-tion. These elements are labeled as S(k)

4 and (S(k)4 )3, k = 1, 2, and 3. These six

elements form the S4 class. The rotation axes and reflection planes are shownin Fig. F.1(d).

F.2 Character table for Td

The character table for the Td group is given in Table F.1.

455

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456 Appendix F. The tetrahedral group, Td

(a) C(1)3

C(2)3

C(3)3

C(4)3

0

1

2

3

4

(b)

σ(1)d σ

(2)d σ

(3)d

σ(4)d σ

(5)d σ

(6)d

(c)

C(1)2 C

(2)2 C

(3)2

(d)

C(1)4 + σyz = S

(1)4 C

(2)4 + σxz = S

(2)4 C

(3)4 + σxy = S

(3)4

Figure F.1 Symmetry elements for the Td group. (a) C3 rotation axes and (in thelast column on the right) a tetrahedron inscribed within a cube. (b) σd reflectionplanes. (c) C2 rotation axes. (d) Axes and reflection planes for S4 elements.

Table F.1 The character table for Td

Basis functions Td E 8C3 3C2 6S4 6σd

A1 1 1 1 1 1A2 1 1 1 −1 −1

(3z2 − r2, x2 − y2) E 2 −1 2 0 0[x(z2 − y2), y(z2 − x2), z(x2 − y2)] T1 3 0 −1 1 −1(xy, xz, yz) T2 3 0 −1 −1 1

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F.4 The irreducible-representation matrices of the Td group 457

F.3 The action table for tetrahedral coordination

The geometry for a central ion surrounded by a tetrahedron of ligands is shownin Fig. F.1(a). The ligands are labeled 1 through 4. The action table is shown inTable F.2.

F.4 The irreducible-representation matrices of the Td group

The individual matrices for the E , T1, and T2 IRs are given in Tables F.3, F.4,and F.5.

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Table F.2 The action table for ligands. The symmetry operations are listed across the top row. The first column lists the functions.Subsequent columns list the function produced when a symmetry operation is applied to the function.

E C13 C1

32

C23 C2

32

C33 C3

32

C43 C4

32

C12 C2

2 C32 σ 1

d σ 2d σ 3

d σ 4d σ 5

d σ 6d S1

4 S14

3S2

4 S24

3S3

4 S34

3

x z y −z y z −y −x −y x −x −x x x −z z y −y −x −x z −z −y yy x z x −z −x −z −z z −y y −y z −z y y x −x −z z −y −y x −xz y x −y −x −y x y −x −z −z z y −y −x x z z y −y −x x −z −z1 1 1 4 3 4 2 3 2 4 3 2 1 4 3 1 1 2 3 2 2 4 4 32 1 6 6 3 1 5 5 3 2 4 4 4 4 6 5 1 3 2 2 5 6 1 33 4 2 1 4 3 3 2 1 2 1 4 2 3 1 3 4 3 4 1 4 2 1 24 2 3 3 1 2 1 4 4 1 2 3 4 1 4 2 3 4 2 3 1 3 2 1

xy xz yz −xz −yz yz −xz −yz xz −xy −xy xy xz −xz −yz yz xy xy xz −xz −yz yz −xy −xyxz yz xy yz −xy −xy −yz xy −yz −xz xz −xz xy −xy xz xz yz −yz −xy xy xz −yz yz −yzyz xy xz −xy xz xz xy −xz −xy yz −yz −yz yz yz −xy xy xz −xz −yz −yz −xy −xy −xz xz

3Z

2−r

2

2√3=

bX

2−Y

2

2=

a

−1 2b−√ 3 2

a−

1 2a+√ 3 2

b

−1 2b+√ 3 2

a−

1 2a−√ 3 2

b

−1 2b−√ 3 2

a−

1 2a−√ 3 2

b

−1 2b+√ 3 2

a−

1 2a+√ 3 2

b

−1 2b−√ 3 2

a−

1 2a+√ 3 2

b

−1 2b+√ 3 2

a−

1 2a−√ 3 2

b

−1 2b−√ 3 2

a−

1 2a+√ 3 2

b

−1 2b+√ 3 2

a−

1 2a−√ 3 2

b

ba

ba

ba

−1 2b−√ 3 2

a1 2

a−√ 3 2

b

−1 2b−√ 3 2

a1 2

a−√ 3 2

b

−1 2b+√ 3 2

a1 2

a+√ 3 2

b

−1 2b+√ 3 2

a1 2

a+√ 3 2

b

b−a

b−a

−1 2b−√ 3 2

a1 2

a−√ 3 2

b

−1 2b−√ 3 2

a1 2

a−√ 3 2

b

−1 2b+√ 3 2

a1 2

a+√ 3 2

b

−1 2b+√ 3 2

a1 2

a+√ 3 2

b

b−a

b−a

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F.4 The irreducible-representation matrices of the Td group 459

Table F.3 The E representation

E =⎛⎝ 1 0

0 1

⎞⎠ C (1)

3 =⎛⎝ − 1

2

√3/2

−√3/2 − 12

⎞⎠ C (1)

3

2 =⎛⎝ − 1

2 −√

3/2

√3/2 − 1

2

⎞⎠

C (2)3 =

⎛⎝ − 1

2 −√

3/2

√3/2 − 1

2

⎞⎠ C (2)

3

2 =⎛⎝ − 1

2

√3/2

−√3/2 − 12

⎞⎠ C (3)

3 =⎛⎝ − 1

2 −√

3/2

√3/2 − 1

2

⎞⎠

C (3)3

2 =⎛⎝ − 1

2

√3/2

−√3/2 − 12

⎞⎠ C (4)

3 =⎛⎝ − 1

2 −√

3/2

√3/2 − 1

2

⎞⎠ C (4)

3

2 =⎛⎝ − 1

2

√3/2

−√3/2 − 12

⎞⎠

C (1)2 =

⎛⎝ 1 0

0 1

⎞⎠ C (2)

2 =⎛⎝ 1 0

0 1

⎞⎠ C (3)

2 =⎛⎝ 1 0

0 1

⎞⎠

σ(1)d =

⎛⎝ 1

2 −√

3/2

−√3/2 − 12

⎞⎠ σ

(2)d =

⎛⎝ 1

2 −√

3/2

−√3/2 − 12

⎞⎠ σ

(3)d =

⎛⎝ 1

2

√3/2

√3/2 − 1

2

⎞⎠

σ(4)d =

⎛⎝ 1

2

√3/2

√3/2 − 1

2

⎞⎠ σ

(5)d =

⎛⎝ −1 0

0 1

⎞⎠ σ

(6)d =

⎛⎝ −1 0

0 1

⎞⎠

S(1)4 =

⎛⎝ 1

2 −√

3/2

−√3/2 − 12

⎞⎠ (S(1)

4 )3 =⎛⎝ 1

2 −√

3/2

−√3/2 − 12

⎞⎠ S(2)

4 =⎛⎝ 1

2

√3/2

√3/2 − 1

2

⎞⎠

(S(2)4 )3 =

⎛⎝ 1

2

√3/2

√3/2 − 1

2

⎞⎠ S(3)

4 =⎛⎝ −1 0

0 1

⎞⎠ (S(3)

4 )3 =⎛⎝ −1 0

0 1

⎞⎠

The basis functions are (x2 − y2) (row 1), and (3z2 − r2) (row 2).

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460 Appendix F. The tetrahedral group, Td

Table F.4 The T1 representation

E =(

1 0 00 1 00 0 1

)C (1)

3 =(

0 0 −1−1 0 0

0 1 0

)(C (1)

3 )2 =(

0 −1 00 0 1−1 0 0

)

C (2)3 =

(0 −1 00 0 −11 0 0

)(C (2)

3 )2 =(

0 0 1−1 0 0

0 −1 0

)C (3)

3 =(

0 1 00 0 −1−1 0 0

)

(C (3)3 )2 =

(0 0 −11 0 00 −1 0

)C (4)

3 =(

0 1 00 0 11 0 0

)(C (4)

3 )2 =(

0 0 11 0 00 1 0

)

C (1)2 =

(1 0 00 −1 00 0 −1

)C (2)

2 =( −1 0 0

0 1 00 0 −1

)C (3)

2 =( −1 0 0

0 −1 00 0 1

)

σ(1)d =

( −1 0 00 0 −10 −1 0

)σ(2)d =

( −1 0 00 0 −10 1 0

)σ(3)d =

(0 0 −10 −1 0−1 0 0

)

σ(4)d =

(0 0 10 −1 01 0 0

)σ(5)d =

(0 1 01 0 00 0 −1

)σ(6)d =

(0 −1 0−1 0 0

0 0 −1

)

S(1)4 =

(1 0 00 0 10 −1 0

)(S(1)

4 )3 =(

1 0 00 0 −10 1 0

)S(2)

4 =(

0 0 10 1 0−1 0 0

)

(S(2)4 )3 =

(0 0 −10 1 01 0 0

)S(3)

4 =(

0 −1 01 0 00 0 1

)(S(3)

4 )3 =(

0 1 0−1 0 0

0 0 1

)

The basis functions are x(z2 − y2) (row 1), y(z2 − x2) (row 2), and z(x2 − y2).

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F.4 The irreducible-representation matrices of the Td group 461

Table F.5 The T2 representation

E =(

1 0 00 1 00 0 1

)C (1)

3 =(

0 0 11 0 00 1 0

)(C (1)

3 )2 =(

0 1 00 0 11 0 0

)

C (2)3 =

(0 0 −11 0 00 −1 0

)(C (2)

3 )2 =(

0 1 00 0 −1−1 0 0

)C (3)

3 =(

0 0 1−1 0 0

0 −1 0

)

(C (3)3 )2 =

(0 −1 00 0 −11 0 0

)C (4)

3 =(

0 0 −1−1 0 0

0 1 0

)(C (4)

3 )2 =(

0 −1 00 0 1−1 0 0

)

C (1)2 =

(1 0 00 −1 00 0 −1

)C (2)

2 =( −1 0 0

0 1 00 0 −1

)C (3)

2 =( −1 0 0

0 −1 00 0 1

)

σ(1)d =

(1 0 00 0 10 1 0

)σ(2)d =

(1 0 00 0 −10 −1 0

)σ(3)d =

(0 0 −10 1 0−1 0 0

)

σ(4)d =

(0 0 10 1 01 0 0

)σ(5)d =

(0 1 01 0 00 0 1

)σ(6)d =

(0 −1 0−1 0 0

0 0 1

)

S(1)4 =

( −1 0 00 0 −10 1 0

)(S(1)

4 )3 =( −1 0 0

0 0 10 −1 0

)S(2)

4 =(

0 0 10 −1 0−1 0 0

)

(S(2)4 )3 =

(0 0 −10 −1 01 0 0

)S(3)

4 =(

0 −1 01 0 00 0 −1

)(S(3)

4 )3 =(

0 1 0−1 0 0

0 0 −1

)

The basis functions are xy (row 1), xz (row 2), and yz (row 3).

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Appendix G

Identifying point groups

Applying group theory requires identifying the point group appropriate to the sym-metry of the system or molecule. There are many schemes using flow charts toidentifying the point group of a molecule (see, for example, reference [G.1]).

G.1 Definitions of symmetry elements

E : The identity.C p

n : Rotation by 2πp/n (p = 1, 2, . . ., n), Cnn = E .

Sn: Improper rotation. A rotation of 2π/n followed by a reflection in a planeperpendicular to the rotation axis. S1 = S is a reflection through a plane.S2 = i (inversion).

σ : Reflection through a plane.σh: Reflection through a horizontal plane that is perpendicular to the principal

axis of rotation. The principal axis is Cn with the largest value of n.σv or σd : Reflection through a plane that contains the principal axis of rotation

and is perpendicular to a σh plane if one exists. For molecules, if both σv andσd planes exist, the σv planes are assigned to the planes containing the mostatoms and the σd contain the bisectors of the bond angles.

i : The inversion operation. A center of symmetry (of inversion). If all Cn andSn axes pass through a common point, the point is a center of symmetry orcenter of inversion.

G.2 Special point groups

Cubic: Tetrahedral, octahedral/cubic, isohedral, and dodecahedral. A moleculewith two or more Cn axes with n > 2 is a member of the cubic group.

462

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G.2 Special point groups 463

Tetrahedral: Four three-fold axes and no four-fold axes.

T : Three two-fold axes (no reflection planes, no center of inversion).Th: Three two-fold axes and a center of inversion. Th = T × i .Td : Three two-fold axes and six reflection planes.

Cubic/octahedral: Four three-fold axes and three four-fold axes.

O: No center of inversion.Oh: A center of inversion. Oh = O × i .

Icosahedral: Six five-fold axes.

I : No center of inversion.Ih: A center of inversion. Ih = I × i .

Linear molecules have an axis about which any rotation is a symmetry element.Any plane containing this axis is a reflection plane. The groups are

C4∞: No center of inversion.D4∞: A center of inversion.

Yes

NoOne axis ofrotation Cn

Sn

Yes

Sn

No

σ plane

Yes

Cs

No

Yes

C1

No

Ci

YesNo

Inversioncenter

n C2 axesperpendicular toprincipal axis Cn

Yes

Cnh

No

YesNo

Cn Cnv

σ planeperpendicular tothe principal axis

n σ planesparallel to theprincipal axis

Yes

Dnh

No

YesNo

Dn Dnv

σ planeperpendicular tothe principal axis

n σ planesparallel to theprincipal axis

Figure G.1 A flow chart for identifying point groups with no more than one Cnwith n > 2. The thick arrows show the identification of a Cnv group.

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464 Appendix G. Identifying point groups

G.3 Other point groups

If the group has two or more Cn axes (n > 2) it is a member of one of the specialgroups. A flow chart for identifying other point groups not listed in Section G.2above is shown in Fig. G.1.

As an example of the use of the flow chart, consider a molecule (e.g., NH3).It has only a C3 axis and three σ reflection (mirror) planes. The thick arrows inFig. G.1 indicate the flow ending at Cnv. Since n = 3, in this case the group is C3v.

The identification of the elements of a group is not always unique. There can beequivalences among rotations, inversion, and improper rotations, depending on thegeometry of the system. For example, D6h = D6 × i has the operations (a) E , C2,2C3, 2C6, 3C ′2, 3C ′′2 , and i times these operators. The operations can also be takenas (b) E , C2, 2C3, 2C6, 3C ′2, 3C ′′2 , i , 2S3, 2S6, σh , 3σd , and 3σv. In the analysisof a system the choice of the group elements for D6h as (a) or (b) yields the sameresults.

References

[G.1] D. C. Harris and M. D. Bertolucci, Symmetry and Spectroscopy: An Introductionto Vibrational and Electronic Spectroscopy (New York: Dover Publications, Inc.,1978).

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Index

Abelian groups, 251, 419accidental degeneracy, 255action table

AB2 molecule, 41, 63, 190diatomic molecules electronic states, 195–197, 207graphene energy bands, 369–372, 377, 378, 427,

429H2O electronic states, 183, 184octahedral complex, 213, 214, 222, 223, 450perovskite energy bands, 281, 285, 286plane wave energy bands, 270, 271squarene electronic states, 170, 172squarene vibrations, 7–9, 23, 169tetrahedral complex, 228, 230, 238, 457, 458

angular momentumcrystal field, 93, 112diatomic molecules, 198, 202, 203, 205double group, 150, 198–199, 202, 203, 205multi-electron systems, 95, 98–100nuclear-shell model, 150operators, 100, 112, 129, 131, 133, 155, 338quantum numbers, 66–67, 93, 100, 112, 135, 140,

150, 198–199, 442, 444rotation group, 67, 85spin, 123, 127, 129–132, 135, 138, 140tensors and vectors (Appendix D), 442, 444time reversal, 337, 338

antiferromagnetic structures, 346, 351–352anti-Stokes transitions, 33antisymmetric, 79, 99–100, 181, 427, 428, 443

representations, 447, 448functions, 96, 98, 99, 101, 109, 131, 132, 145, 156,

172–174, 195, 445, 447vibration, 32–33, 94

antiunitary operators, 342, 348–350, 354, 356axial vector, 89, 443

band, see energy bandsbasis functions

AB2 molecules, 39–42Bloch-wave states, 254, 259, 261

character table, 428co-representations, 353crystal field, 96, 97, 101–103D4h splitting, 273, 302D∞h , 195definition, 423diatomic molecules, 195, 197, 210direct product, 82–85, 98, 99, 428double group, 1513 F ground state, 99, 101, 102, 108generation of (symmetry-function-generating

machine), 22, 23, 36, 47, 48, 63, 83, 84, 89, 108,180, 213, 223, 264, 269, 270, 272, 288, 291, 302,370, 372, 374, 378, 380, 426

graphene, 371, 372, 380half-integer angular momentum, 104, 127, 140H2O, 182–184hybrid, 167irreducible representation for O group, 451–454,

456irreducible representation for Td group, 459–461Jahn–Teller, 117, 118Löwdin orbitals, 165multi-electron, 97, 99, 101octahedral complex, 212, 213, 217, 222perovskite (vibrations), 284, 285, 296phonon, 310, 315, 318plane-wave states, 268, 270, 273, 274product wavefunction, 444projection operator, 426rotation group, 66, 71, 74–76, 78, 79, 85, 88, 89,

99, 108space groups, 250, 254, 259squarene (electronic states), 168–171squarene (vibrational states), 5, 7–10, 12, 14, 15,

17, 20, 21, 23–26, 32, 33tetrahedral complex, 229, 233, 238vector model, 97–99

basis vectors, see basis functionsBloch theorem, 249–250

465

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466 Index

Bloch waves (functions), 249–252, 255, 259, 260,263, 381

Brillouin zone, 250character, 338–339compatibility, 260, 264graphene, 368, 375, 379lattice vibrations, 283plane waves, 264, 268, 270–272point-group symmetry, 252–254representation by, 255, 259, 263space group, 283time reversal, 307, 308, 364

block-diagonal matrices, 15, 21, 22, 26–28, 37, 43,50, 175, 229, 295, 317, 323, 334, 422

block-diagonalized secular equation, 175, 184, 198,213, 238, 288, 290, 297, 302, 320, 322

bonds, 31, 117, 168, 184, 199, 204, 208, 209, 213,307, 383, 385, 393, 462

Bragg condition, reflection, 247, 278Bravais lattice, 239–247, 249, 253, 259, 260, 267,

278, 281, 314, 363Brillouin zone, 249, 250, 252, 253, 260, 267, 270,

290, 304cubic, 248, 260graphene, 364, 366, 367, 369, 377, 381, 382, 395,

401hexagonal, 247, 257–258, 260, 266, 271, 276,

283–284, 310–313, 318nanotube, 390, 395, 396, 398, 399, 401–403, 408perovskite, 290, 302squarene, 243

Cartesian, 5, 9, 31, 305cell, see unit cellcenter of mass, 29, 45, 64, 315, 326character

C4v group, 16, 19, 22, 23, 32, 248, 262crystal field, 96, 107, 108crystal double group, 150, 151D(l), 75, 76, 85, 121definition, 16, 18, 412, 428diatomic molecule, 196, 197, 209direct product, 11, 79, 82, 96, 97, 446graphene, 373, 377, 393H2O, 62, 183, 188Herring rules, 345inversion, 78, 213irreducible representations, 18magnetic groups, 350, 357non-symmorphic group, 264Oh , 78, 79, 88, 108, 211, 261, 273, 287octahedral complex, 213, 222, 223, 228orthogonality, 16, 18, 20perovskite (electronic), 284–286, 289, 291, 293,

296plane wave, 270, 272, 273projection operator, 427rotations, 73, 74

same class, 423similarity transformation, 421, 422squarene (electronic), 183time reversal, 342, 344, 345

character tablesdescription of, 427–428for all point groups (Appendix C), 430–437for nanotubes, 437C2, 42, 360C2v , 187, 297, 310, 320C3, 346C3v , 171, 261, 263C4v , 16, 26, 33, 190, 261, 262, 293, 310, 428C5, 374C∞h , 193, 194D(1/2)–D(2), 87D2, 374D4, 89, 117, 118D3h , 376, 377D4h , 272, 273D6, 369, 370D∞h , 193–196, 205O , Oh groups, 76, 107, 108, 211, 212, 261, 287,

310, 449, 450O ′ crystal double group, 152, 153Td , 229, 456

chiral nanotubes, 385, 386class, 15–18, 73, 75, 81, 85, 151, 170, 215, 246, 251,

257, 280, 286, 361, 417–418, 421, 423, 427, 428,455

character of, 17, 20definition of, 15elements of, 17, 418, 449

co-representations, 350, 356, 357conduction band, 301, 366, 368, 404, 408configuration

bond order, 204crystal field, 98, 111d1, 91, 92, 94, 117, 226d2, 95, 96, 99, 103, 107, 114–116, 226d3, 121, 153, 155, 226, 236d4, 92, 95, 226, 227, 236d5, d6, 94, 227, 236d7, 236d8, 107, 227, 236d9, 94, 95, 117d10, 94diatomic molecule, 204double group, 152, 153graphene, 363H2O electronic states, 186heteronuclear molecule, 208high/low-spin states, 226, 227, 236, 238molecular, 199, 201–202, 204, 209, 220nanotubes, 391octahedral complex, 220perovskite energy bands, 300spin–orbit interaction, 144, 149

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Index 467

squarene electronic states, 179Zeeman effect, 155

conjugate-group elements, 338, 375correlation diagram, 179, 192, 201covering group, 62, 64, 77, 83, 96, 166, 190, 193, 198,

200, 206, 211, 244, 349, 416, 419, 422, 429covering operations, 3, 75, 77, 168, 227crystal double group, 150–153crystal field, 90, 93–96, 99, 152, 153, 211, 222

potential, 92, 103–105, 109splitting, 91, 107, 109, 120, 153, 219, 260

cyclic groups, 419, 428

decompositionof a representation, 14, 15, 17, 20–22, 27, 28, 35,

36, 42, 43, 75, 79, 83–85, 96, 108, 117, 118, 121,125, 132, 140, 144, 145, 151–153, 170, 172, 183,184, 213, 231, 260, 270–273, 287, 288, 291, 293,297, 311, 315, 321–323, 370, 371, 374, 378

of a tensor, 442–444, 447–448degeneracy, 27, 90, 116, 148, 154, 198, 221, 263, 268,

271, 275, 340, 342, 344–346, 357, 408accidental, 254energy bands, 396time reversal, 99

diagonal matrix, 21, 23, 26, 40, 47, 50, 52, 109, 171,176, 224, 244, 266, 271, 288–290, 295, 331, 413,414

diagonalization of a matrix, 2, 43, 198diamagnetic, 226, 238, 347, 348dimensionality of a representation, 19, 72, 354, 362,

412, 425Dirac

points, 368, 381, 401, 408relativistic quantum theory, 139

direct product, 76–83, 89, 96–98, 108, 119, 120, 140,144, 145, 157, 193, 202, 205, 252, 259, 283, 369,413, 419, 428, 442, 445, 447, 449

double groups, 150–153, 346double-valued representation, 87

effective mass, 368, 382eigenvalues, 1–2, 21, 26–31, 39, 43, 52, 99–102, 105,

107, 113, 114, 124–125, 129, 130, 136, 139–143,145, 156, 165, 167, 175, 176, 179, 185, 189, 191,194, 206, 213, 219, 232, 235, 244, 266, 288, 295,296, 306, 312, 319–321, 327, 329, 330, 333, 340,381, 412, 413, 422

eigenvectors/eigenfunctions, 23, 27, 32, 37, 46, 48,64, 111, 126, 184, 198, 293, 312, 323, 332, 342,375, 413

electric-dipole transition, 187, 205, 237electron, 86, 94, 107, 113, 117, 123, 126, 144, 187,

190, 199, 201, 203, 215, 220, 225–227, 231, 236,241, 265, 276, 277, 301, 341, 368, 383

crystal-field splitting of, 90, 91, 93, 96, 98, 109electron–electron interactions, 103, 107–109, 159,

280

lone pair, 184multi-electron systems, 66, 74, 82, 95, 96, 98,

129–131, 138–140, 143, 153, 158–160, 171, 181,195, 339, 340

one-electron approximation, 66, 74, 82, 282, 338,339

spin, 123, 125, 129, 131, 138–139, 143, 144two-electron wavefunctions, 82, 96–101, 103–105,

132, 447empty-lattice approximation, 266, 276energy bands, 243, 244, 254, 260, 262–264, 266–270,

272–282, 284, 286, 294, 296–303bcc, 278cubic, 362degeneracy, 268, 272, 275, 278, 345, 396–398graphene, 363–368, 382, 393LCAO or tight-binding, 291, 293, 304, 320, 363LDA, 299–300nanotube, 393–400, 403–409theory, 261, 280time reversal, 342, 344wavefunctions/states, 275, 288, 290–291, 294

equivalentatomic sites, 77, 285–287, 289, 314, 318, 334, 372,

378electrons, 98, 136, 145–149, 156, 294, 296,

442–447groups, 4, 347irreducible co-representations, 356irreducible representation, 15, 252, 372, 423oxygen sites, 228, 290, 322representations, 12, 121, 356, 362, 423, 424wavevectors, 253–254, 256, 258, 261, 272, 283,

290, 345, 364, 368, 369, 372, 375Euler angles, 72even modes, 34exclusion principle, 96, 98, 99, 145, 150, 221

face-centered cubic, 247–249, 276, 278, 313factor group, 259, 418, 419faithful representation, 421Fermi, 150, 366, 382ferromagnetic, 346, 359force constant, 40, 50, 54, 59–60, 305–307, 327Frobenius–Schur test, 357, 359, 361, 362full rotation group, 66, 68, 71, 75, 85, 442–444

representations of, 71, 85, 87, 444

glide plane, 246, 350group, 11, 32, 73, 117, 168, 169, 182, 263, 270, 289,

290, 295, 296, 302, 310–312, 314, 318–321, 325,345, 350, 353, 355, 356, 362, 368–370, 372–374,377, 382, 389, 417, 420–428, 438–441, 445,450–452, 456, 457

Abelian, 251, 419classes of a, 15, 18covering, 62, 63, 77, 83, 96, 166, 190, 193, 198,

200, 206, 211, 227, 244

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468 Index

cyclic, 251, 252, 419, 428definition/properties of, 4, 5, 15direct-product, 76, 79, 81, 82, 96double, 150–153elements of a, 4, 8, 10, 16, 17, 33, 38, 41, 68factor, 259, 418full rotation, 66, 68, 71, 75, 85, 87inversion, 77, 78magnetic, 337, 346–351, 357–359, 361multiplication table, 8, 10, 16, 37octahedral, 74–77, 79, 83–85, 88, 99, 108, 117, 121,

151, 152, 211, 212, 221, 224, 259, 260, 361, 449operators, 7, 10, 15, 19, 21, 38order of, 10, 171, 362, 415, 418, 423point, 3, 12, 17, 39, 245, 252, 259, 346, 348, 349,

351, 353, 356, 357, 393, 415, 430, 462–464representations of a, 12, 443rotation, 70, 95, 97, 127, 150, 151space, 239, 241, 244–246, 252–260, 264, 268, 269,

280, 282, 304, 333, 342–344, 346, 353, 357, 362,393, 415

symmorphic, 249, 250, 252, 255, 258, 259theory, 1, 2, 4, 10, 14, 17, 18, 20, 29, 30, 34, 39, 79,

88, 95, 98, 156, 165, 200, 386tetrahedral, 455of the wavevector (gk), 253–255, 258–261, 264,

269, 271, 279, 283–384

Hamiltonian, 15, 75, 76, 79, 82, 83, 91, 95, 158,337–344, 423

atomic, 282crystal-field, 91, 109effective, 382eigenfunctions of, 249, 254electronic, 245group of, 82, 96, 98Hartree–Fock, 161, 162hydrogen-like, 93, 98, 129many-electron, 339matrix elements of, 163–165, 174–177, 189, 194,

198, 206, 244molecular, 198one-electron, 66, 282, 338, 339perturbation, 209spin–orbit, 338spinless, 343two-electron, 96, 103, 108

Hartree–Fock equations, 159–160Hermitian, 21, 27, 29, 35, 37, 83, 309, 341, 410, 412

adjoint, 412matrix, 309

hole, 92, 95, 107, 117, 368homomorphic, 421Hund’s rules, 98, 99, 107, 120, 204, 210, 222, 223,

225, 226, 237hybrid orbitals, 167, 168, 217, 232hydrogen, 39, 54, 123, 140, 157, 183–185, 208, 381

angular symmetry of orbitals, 161

hydrogen-like orbitals/hydrogenic orbitals, 130,161, 167, 193

improper rotation, 416inequivalent

carbons, 384, 385electrons, 145, 304irreducible representations, 72, 356, 362, 423, 424wavevectors, 256–257, 364, 368, 382, 404, 405

integralGreen’s function, 331interaction, 162, 178, 215, 217, 233overlap, 162–164, 167, 179, 189, 191, 215, 233,

237, 238, 289, 291, 382internal coordinates, 5, 31invariant, 96, 111, 238

under full rotation group, 66, 95, 443under gk, 270, 283–284, 369, 372under inversion, 172, 203under Oh , 296, 314under point group, 253, 415under reflection, 203, 243under similarity transformation, 412, 413under space group, 245, 320under subgroups, 346–349, 351, 353, 356–360,

362, 416–418under time reversal, 346under translation, 241, 249, 263, 346

inverse, 5, 10, 16, 26, 70, 167, 245, 254, 316, 331,409, 411, 412, 415, 416, 419, 421, 422, 424

inversion, 4, 33, 34, 66, 77–79, 89, 94, 99, 100, 114,117, 168, 172–174, 180, 195, 196, 199, 203–206,213, 227, 245, 272, 302, 310, 337, 343–346, 350,358, 369–372, 377, 392, 393, 415, 416, 421,427–428, 462–464

irreduciblerepresentations, 14, 17, 22, 25, 42, 72, 197, 198,

268, 310, 360, 362, 450, 457tensors, 442, 444

isotope, 3, 327, 328, 331–333isotopic substitution, 54, 56, 58

Jahn–Teller effects, 116

Koopman’s theorem, 160

lattice, 241, 243, 249–255, 259, 262–264, 278,280–285, 290, 294, 295, 299, 302, 304, 305, 307,309–313, 317, 320, 322, 327–331, 333, 336, 350,358, 378, 380–382, 384–387, 390

Bravais, 239, 241, 242, 259, 260, 312reciprocal, 243–244, 246–247, 252–254, 257, 258,

265–268, 276, 277, 283, 335, 343–345, 363, 364,368, 376, 389, 395, 399, 401, 402, 408

LCAO method, 161, 162, 282, 290longitudinal, 311–313, 317, 319, 334–336, 375

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Index 469

magnetic, 67, 72, 74, 75, 90, 100, 112, 138, 150,153–155, 226, 280, 337, 338, 341–344, 346–350,353, 356–359, 361, 367, 444

matrix elements, 2, 10, 21–23, 27, 28, 34, 43, 50–55,58, 60, 79, 83, 84, 88, 89, 99, 103–104, 109, 111,113, 114, 120, 127, 129, 139, 161–165, 171, 174,176, 179, 185, 189, 206, 215, 224, 237, 238, 255,259, 270, 271, 275, 283, 289, 294, 296, 306, 309,312, 317, 319, 327, 335, 336, 361, 371, 413, 422,425, 426

matrix representations, 8, 10, 41, 80, 81, 85, 170, 421Miller indices, 277mirror plane, 257, 350, 392, 464molecular, 9, 59, 158, 182, 183, 211, 223, 327, 361

bond, 163, 168, 179, 185, 222complex, 90, 116, 227, 228configurations, 192, 199–201, 204, 209, 210, 227Green’s function, 52–54orbitals, 192, 193, 198, 199, 202, 209, 210,

219–222, 225, 236, 237state/level, 32, 159, 161, 171, 179–181, 185–188,

190, 192, 193, 195, 198–203, 205, 209–210, 212,213, 219–221, 224, 225, 233, 236–238, 282, 429

vibrations, 1–3, 5, 15, 23, 29, 32, 35, 40, 52, 53,165, 219, 304, 429

nanotube, 363, 383–407curvature and energy gap, 406, 407density of states, 403–406energy bands, 393–401metallic and semiconducting, 401

Néel temperature, 358neutron scattering, 313, 317normal divisor (factor subgroup), 347, 418normal modes of vibration

AB2 molecules, 39, 44–51, 53AB3 molecules, 64crystalline solids, 304–328graphene, 370–381infrared-active, 31, 34, 42parity, 32Raman-active, 33squarene, 1–30

octahedral complexcorrelation diagram, 115d1 crystal-field theory, 90d2 crystal-field theory, 94d4 splitting of energy levels, 117eg–t2g splitting, 76, 94high- and low-spin states, 226Jahn–Teller effects, 116ligand-field theory, 211

octahedral group (O and Oh ), 74–77, 83, 361,449–454

action table, 450character table, 449elements and operators, 449

opticalinfrared absorption, 31–34, 42, 314modes, 316–317, 324, 326, 333properties, 31, 158, 301Raman scattering, 33, 34, 43, 84

optical modes of vibrationoptical branches, CsCl structure, 314, 327optical branches, graphene, 372optical branches, perovskite structure, 320

orderbond order, 204, 205group, of, 10, 170, 415, 418levels of a transition-metal complex, 235magnetic, 346–350molecular levels, 222

orthogonal/orthogonality ofatomic/atomic-like orbitals, 93characters of different irreducible representations,

16, 423Löwdin orbitals, 166matrix elements between different irreducible

representations, 21, 111, 275matrix elements of different irreducible

representations, 423–425orthonormal vectors, 411, 413overlap integral/matrix, 162–167, 174, 176, 179, 189,

191, 194, 198, 215, 233, 237, 238, 289, 291, 382oxygen

ABO3 phase factor, 285, 286atoms, 183, 185bands in perovskites, 299–301H2O molecules, 184LCAO interactions, 294π and ζ orbitals, 286, 287sites in perovskites, 190, 191

paramagnetic, 226type II magnetic group, 347type III magnetic group, 348

parity, 32, 94, 150, 199, 261, 447partner basis functions, 173, 343, 375, 426Pauli, 155

exclusion principle, 96, 98, 99, 145, 150, 159, 221spin matrices, 123–128, 130, 339, 382

periodic boundary conditions, 239, 245, 251, 264,381, 389

point groupsbasic theory (Appendix B), 415character tables (Appendix C), 430covering group, 3, 62, 64, 415, 416, 419, 420, 422,

429irreducible representation, 12–20, 42, 72, 127irreducible representations for O , Oh (Appendix

E), 449irreducible representations for T , Td (Appendix F),

455polarizability, 33, 317primitive

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470 Index

lattice vector, 239–243, 246–248, 251, 276, 277,363, 364, 384–386, 390, 391

reciprocal-lattice vector, 246–248, 257, 276, 277,363

projection operator, 22, 426–427proper rotation, 66

quantumcoupling of two systems, 442phonon, 304probability, 32, 33theory, 126, 139, 304, 341tunneling, 118

quantum numbers, 66, 72, 82, 93, 100, 112, 122, 422,442, 444

diatomic molecules, 198j–m j coupling, 140, 141L–S coupling, 130, 135

rearrangement theorem, 38, 345, 417, 418reciprocal lattice, 243–244, 246–249, 343, 344

body-centered cubic, 247, 276, 278Brillouin zone, 247energy bands, 265–268face-centered cubic, 248, 278graphene, 363, 364, 366, 372, 389Herring rules, 345hexagonal, 247, 248, 257, 259, 276nanotube, 395, 401, 402, 408simple cubic, 248, 276, 277, 284, 335

representationsC4v , 9–11, 126, 318, 428, 431character of, 16, 17, 170compatibility, 260constructing, 8–12co-representation, 350–357D4h , 116, 272, 363decomposition of, 13–15, 27, 28, 42, 75, 76, 83–84,

173, 183, 191, 213, 229, 269, 286–287dimensionality, 423–424direct product, 76, 79–83, 96, 97, 121equivalent, 4, 9, 12, 72, 121, 211, 252, 356, 423,

425faithful, 421full rotation group, 68–72, 85half-integer, for, 86, 127, 130, 150–152, 359irreducible, 12, 17, 22, 25, 47, 75, 197, 198, 268,

315, 360, 362, 450, 453matrix, 8–12, 19, 41, 85, 170, 195, 250, 421non-symmorphic, 264number and type of irreducible representations,

15–19O and Oh , 310, 450–454orthogonality, 17–20, 423–425perovskite, 284plane-wave, 265–267regular, 37–38reducible, 12–15, 72

space group, 258, 268symmorphic group, 255Td , 459–461unitary D(l) matrices, 72–86

rotation group, 66–72, 75characters of a rotation, 76decomposition, 97representations of, 70, 71, 85, 87, 127, 446–448

Russell–Saunders coupling (L–S coupling), 95, 96,99, 129–132, 135

scalarconstant, 2, 35, 41operator, 125, 253product, 18, 125, 341, 394, 410, 411, 444tensor, 442–444

Schrödinger equation, 123, 150Schrönflies notation, 350Schur’s lemma, 424, 425screw

axis, 246, 350operation, 391, 392rule, 422, 449

selection rules, 33–34, 82–85, 94, 113, 205, 302, 312,443

similarity transformation, 12–16, 19, 21, 26, 29, 30,50, 165, 227, 316, 421–423

simple cubic lattice, 262, 266, 269, 283, 309, 310, 312singularity, 332, 405, 406Slater determinant, 74, 98, 101, 106, 107, 121, 131,

159, 161, 182space groups, 239, 242, 244–246, 248, 249, 268, 282,

304, 333, 342, 343, 346, 353, 358, 369, 391, 393,415

definition of, 244non-symmorphic, 246, 264symmorphic, 246, 249, 252, 255, 258, 283

spherical harmonics, 67, 76, 85, 88, 104, 119, 121,150, 158, 161, 443

spin–orbitHund’s rules, 98, 99interaction, 67, 138–140j– j states, 95, 143–149j–m j states, 140L–S states, 129molecular orbitals, 199nuclear-shell model, 149splitting, 140time reversal, 338–340Zeeman effect, 153

spinors, 124, 125, 127, 129squarene, 1–3, 6, 7, 11, 15, 16, 20, 22–23, 26, 28–30,

32, 33, 36, 38, 68, 168–171, 175, 179, 180star of the wavevector (of k), 255–258, 261Stokes transition, 33subgroup, 246, 250–256, 258–260, 267, 283,

346–349, 362, 416, 418factor group, 259, 418

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Index 471

invariant, 346–349, 351, 353, 356–358, 360, 362,417–418

symmetry-function-generating machinecrystal field, 108direct product, 84graphene, 370, 372, 375, 378, 380H2O, 46, 49LCAO model, 181octahedral complex, 215, 223perovskite energy bands, 288plane wave, 269, 270projection operator, 426, 427space group, 264, 272, 288squarene, 22, 23tetrahedral complex, 231

tensor, 33, 83, 442–448tetrahedral

group, 455–461splitting in crystal field, 235

time-reversal symmetryantiunitary operators, 342band theory, 342–345C5, 428co-representation, 350external field, 341Frobenius–Schur test, 357Herring test, 344–346Kramers’ theorem, 427–428magnetic group, 346–352MnF2, 357–358operator, 338–342partners, 375quantum systems, 337–340

trace of a matrix, see charactertransformation

AB2 molecule, 50, 51, 60ABO3, 323Bloch waves, 272graphene, 370inversion, 240LCAO model, 165, 176, 180matrices, 421point group, 421–423rotation group, 66, 69similarity, 121, 412, 413spin, 128, 145squarene, 12–16, 19, 21, 26, 30tensor, 443, 444time reversal, 354two atoms per unit cell, 316

translationAB2 vibration modes, 43, 44, 49perovskite vibration modes, 323, 325

space-group operations, 241, 244–246squarene electronic states, 170squarene vibration modes, 2, 23–33

unit cell, 239, 241, 243, 247, 262, 266, 276, 280–282,284–286, 289–292, 295, 296, 298, 301, 302,304–312, 314–315, 320–322, 324, 327, 333, 334,336, 358, 363, 364, 371, 375, 376, 384, 386, 389,392–393, 401, 404, 429

unitaryco-representations, 353–357Hamiltonian, 166magnetic group, 346–349matrices, 12, 59, 330, 356, 362, 412, 413, 421, 423,

425rotation group, 85time reversal, 337, 341–342

valencegraphene bands, 364, 366, 408H2O states, 186M L4 complex, 228nanotube bands, 383, 396–398, 400, 401, 405, 407perovskite bands, 300–301

variational principle, 159vector

angular momentum, 86, 95basis, 21displacement, 2, 5–9, 21–23, 26, 40, 41, 44–45electric dipole, 32inversion, 77normalized, 24orthogonal, 17–20, 25, 31rotation group, 68, 70–73

vibrations (lattice vibrations)axially symmetric model, 307–308eigenvalue equation, 305graphene, 369–381localized modes, 327–337perovskites, 320–327simple cubic crystal, 314two atoms per unit cell, 316–320

vibrations (molecular vibrations)AB2 molecules, 39–52isotopically substituted AB2 molecules, 52–60isotopically substituted H2O molecules, 60, 61squarene, 1–34

water (H2O molecule)force constants, 61isotopically substituted vibrational frequencies, 60vibrational frequencies, 61

Zeeman effect, 153–155