applications of symmetry and group theory (artigo)

Acta Appl Math (2012) 118:3–24DOI 10.1007/s10440-012-9675-5

Applications of Symmetry and Group Theoryfor the Investigation of Molecular Vibrations

Jaan Laane · Esther J. Ocola

Received: 25 April 2011 / Accepted: 16 November 2011 / Published online: 8 February 2012© Springer Science+Business Media B.V. 2012

Abstract The application of symmetry and mathematical group theory is a powerful tool forinvestigating the vibrations of molecules. In this paper, we present an overview of the meth-ods utilized. First we briefly discuss the quantum mechanical nature of vibrations and theexperimental methods used. We then present the principal concepts for applying group the-ory to molecules. The symmetry operations which are used to comprise groups are describedand then used to determine the point groups of molecules. The properties of character tablesare presented and the method for obtaining a reducible representation for all the motions ofa molecule is detailed. This can then be broken down to obtain the irreducible representa-tion which contains the symmetry species of the individual vibrations. The determinationof symmetry adapted linear combinations is outlined and the basis for spectroscopic selec-tion rules is presented. The paper concludes by examining how matrix algebra along withsymmetry concepts simplifies calculations with molecular force constants.

Keywords Symmetry · Group theory · Molecular vibrations · Potential energy · Symmetryadapted linear combinations · Selection rules

Mathematics Subject Classification (2000) 20-01 · 58D19

1 Introduction

One of the notable developments in chemistry during the second half of the twentieth centurywas the emergence of symmetry and group theory as a means to better understand molecularspectroscopy, molecular bonding, and molecular reactions and dynamics. The well knownbook series by Nobel laureate Gerhard Herzberg [1–3] published in 1945 to 1966 alreadyused these concepts to present the principles of molecular spectroscopy, but this was donein a fairly complicated way. A major advance came when our late colleague F.A. Cottonat Texas A&M University first published his book Chemical Applications of Group Theory

J. Laane (�) · E.J. OcolaDepartment of Chemistry, Texas A&M University, College Station, TX 77843-3255, USAe-mail: [email protected]

mailto:[email protected]

4 J. Laane, E.J. Ocola

Fig. 1 Natural symmetry of asnowflake

in 1963. The third edition [4] appeared in 1993. This text presented symmetry and grouptheory in a way that the average chemist could understand. Another very good and earlybook, although somewhat more advanced, was the text by Tinkham [5]. Following these,numerous other books of varying depth have been published on this topic [6–20].

In the present paper we will attempt to present a condensed version of how the use ofsymmetry and group theory can be applied to better understand molecular vibrations andto simplify the calculations that are utilized for these purposes. Everyone has some kind ofconception of symmetry. Thus, when looking at the snowflake in Fig. 1, it is evident that itcan be cut in half many ways leaving two identical sides. Similarly it can be rotated by 60◦after which it looks just the same. These same ideas can be applied to molecules as we shallsee.

This paper is organized as follows. In Sect. 2, we present some basic concepts for thestudy of molecular vibrations. In Sect. 3, we examine the applications of symmetry andgroup theory to molecules. In Sect. 4, we describe the implementation of symmetry adaptedlinear combinations for the representation of vibrations. Section 5 describes the origin ofspectroscopic selection rules based on symmetry, and Sect. 6 examines the effect of symme-try on the vibrational potential energy. A brief conclusion is given in Sect. 7.

2 Molecular Vibrations

2.1 Vibrations of a Diatomic Molecule

A two-atom or diatomic molecule can have six different individual motions. It can translatein three directions, it can rotate about two axes, or it can have a vibration where the distancebetween the atoms rapidly gets larger or smaller (perhaps 1014 times per second). There areonly two molecular rotations since one of the Cartesian axes passes through the atoms andhence there is no moment of inertia about this axis (I = ∑

mir2i where mi = mass of atom

and ri = distance from axis). Our interest is in the molecular vibration, which is generallyapproximated by the harmonic oscillator model. Here, the bond is pictured to be similar toa spring so that the potential energy function for the vibration can be written

V = 1

2kx2 (1)

Applications of Symmetry and Group Theory for the Investigation 5

where x = R − Re is the bond displacement from its equilibrium position Re and k is aforce constant (or spring constant). Use of this potential function in the Schrödinger waveequation

− �2

2μ

d2ψ

dx2+ V ψ = Eψ (2)

allows the quantum energy levels E to be calculated along with their corresponding wavefunctions. The constant � = h

2πwhere h is the well known Planck’s constant, and the ψ are

the wave functions. The reduced mass μ is given by

μ = m1m2

m1 + m2(3)

where m1 and m2 are the masses of the two atoms in the molecule. Equation (2) can besolved exactly to yield

E =(

v + 1

2

)

hν and ψv = NvHv(α12 x)e− αx2

2 (4)

where the frequency ν is given by

ν = 1

2π

(k

μ

) 12

(5)

and the quantum number v = 0,1,2, . . . ,Nv is a normalization constant, Hv is the Hermite

polynomial of degree v, and α = (μk)12

�. The infrared absorption spectrum of a diatomic

molecule is generally presented on a wavenumber scale in cm−1 (ν) where

ν = ν

c= 1

2πc

(k

μ

) 12

(6)

and typically the v = 0 → 1 quantum transition is observed at this value since all of theenergy levels are calculated to have a spacing of hν.

2.2 Vibrations of Water

For larger molecules the vibrational problem becomes multi-dimensional, requiring 3N − 6coordinates for an N atom non-linear molecule. Of the 3N degrees of freedom, six representthe translations and rotations. Vibrational motions are best described by internal coordinatessuch as molecular bonds and bonding angles. As an example, Fig. 2 shows the three internalcoordinates for the two-bond stretching and one-angle bending motions of water. For thiscase the quadratic potential energy function is

2V = fR(ΔR1)2 + fR(ΔR2)

2 + fα(Δα)2 + 2fRR(ΔR1)(ΔR2)

+ 2fRα(Δα)(ΔR1) + 2fRα(Δα)(ΔR2), (7)

where ΔR1 and ΔR2 represent the stretching of the two bonds and where Δα is the bend-ing of the HOH angle. The force constants fR and fα are the primary constants for thestretching and bending motions, respectively, and fRR and fRα are interactive constants.The detailed methods for calculating the vibrational frequencies of polyatomic moleculeswere nicely presented by Wilson, Decius, and Cross [21].


Fig. 2 Internal coordinates ofwater

Fig. 3 Infrared and Ramantransitions for a molecularvibration

2.3 Experimental Methods

There are two primary methods for studying molecular vibrations which generally have theirquantum state separations in the 100 to 4000 cm−1 range. Figure 3 shows the possible tran-sitions for a diatomic molecule. Infrared absorption experiments typically involve a directtransition from the vibrational ground state (v = 0) to the first excited quantum state (v = 1).Many commercial instruments are available and these utilize the concept of Fourier trans-form infrared (FT-IR) spectrometry where two light beams are separated by a beam splitterand then combined to give an intensity versus path difference data set. Fourier transform isthen used to convert this to an intensity versus wavenumber spectrum.

Raman spectroscopy utilizes a monochromatic laser such as an argon-ion laser at514.5 nm wavelength or a frequency doubled Nd:YAG laser at 532 nm. The laser induces thesample to produce a quasi-excited or virtual state which then scatters the radiation isotrop-ically as either Rayleigh light or Raman radiation. The Rayleigh scattering, which has thesame frequency as the laser exciting line, is about a million times more intense than theRaman scattering which is shifted in frequency. Raman spectra are generally collected asStokes scattering when molecules are excited from the lowest ground state vibrational leveland these have frequencies of

νRAMAN = νLASER − νVIBRATION (8)

so that the vibrational frequency can be determined. As will be discussed later, the selectionrules for which transitions can be observed are different for infrared and Raman spectra. Asan example of how the two types of spectra differ, Fig. 4 presents the observed spectra foran organic molecule, 3-cyclopenten-1-one. This molecule has 30 vibrations and the bandscorresponding to many of them are shown in the figure. Note that both types of spectra areneeded in order to observe most of the vibrations.


Fig. 4 Infrared and Ramanspectra for the3-cyclopenten-1-one molecule

Table 1 Symmetry elements andoperations for molecular systems Element Symbol Operation

Identity E Nothing

Proper rotation axis Cn Rotate by 360°/n n = 2,3,4, . . .

Plane of symmetry σ Reflection through plane(σv = vertical; σh = horizontal;σd = dihedral)

Center of inversion i Inversion though center

Improper rotation axis Sn Cn followed by σh

3 Applications of Symmetry and Group Theory

The principal concepts for applying mathematical group theory to chemical systems are thefollowing:

1. Symmetry operations performed on molecules make up the members of a mathematicalgroup.

2. All the properties and requirements for a mathematical group apply. This includes thefact that the product of any two operations must also be a member of the group andthat any operation X must have an inverse such that XX−1 = E where E is the identityoperation.

3. A valid symmetry operation performed on a molecule produces a result where themolecule appears to be identical to its original position.

4. Symmetry operations are carried out with respect to symmetry elements.

Table 1 presents a list of the symmetry elements and the corresponding symmetry opera-tions. Examples of these will be shown for the benzene molecule (C6H6) which is a hexagonof six carbon atoms with one hydrogen atom attached to each. Figure 5 first shows the op-erations about the C6 proper rotation axis. A C6 operation is a rotation of 60◦, a C2

6 = C3

operation is a rotation by 120◦, etc. Note that C66 gets you back to the original arrangement

of atoms, so this is equivalent to the identity E. Figure 6 shows the two types of C2 rotationaxes which benzene also possesses. Note that there are a total of six C2 axes perpendicularto the C6 axis and this will be the basis for the molecule having D6h point group symmetry.


Fig. 5 Symmetry operations about the C6 axis for benzene

Fig. 6 The two types of C2 axesfor benzene which areperpendicular to the C6 axis

Fig. 7 The three types ofreflection planes for benzene

Fig. 8 The inversion (top) andimproper rotation (bottom)operations of benzene

Figure 7 shows three different planes of symmetry, σh (horizontal), σv (vertical), and σd

(dihedral). Figure 8 shows the effect of the inversion and improper rotation operations. Notethat whether the atoms wind up right side up or inverted is important. In total, benzene has


Tabl

e2

Cha

ract

erta

ble

and

redu

cibl

ere

pres

enta

tion

Γfo

rth

ebe

nzen

em

olec

ule

ofD

6hsy

mm

etry

D6h

E2C

62C

3C

23C

′ 23C

′′ 2i

2S3

2S6

σh

3σd

3σv

A1g

11

11

11

11

11

11

x2

+y

2,z

2

A2g

11

11

−1−1

11

11

−1−1

Rz

B1g

1−1

1−1

1−1

1−1

1−1

1−1

B2g

1−1

1−1

−11

1−1

1−1

−11

E1g

21

−1−2

00

21

−1−2

00

(Rx

,Ry

)(x

z,y

z)

E2g

2−1

−12

00

2−1

−12

00

(x2

−y

2,x

y)

A1u

11

11

11

−1−1

−1−1

−1−1

A2u

11

11

−1−1

−1−1

−1−1

11

z

B1u

1−1

1−1

1−1

−11

−11

−11

B2u

1−1

1−1

−11

−11

−11

1−1

E1u

21

−1−2

00

−2−1

12

00

(x,y

)

E2u

2−1

−12

00

−21

1−2

00

Γ36

00

0−4

00

00

120

4


Fig. 9 Symmetry operations forthe ammonia moleculedemonstrating the product rule

Table 3 Product table for C3v

C3v E C3 C23 σv(A) σv(B) σv(C)

E E C3 C23 σv(A) σv(B) σv(C)

C3 C3 C23 E σv(B) σv(C) σv(A)

C23 C2

3 E C3 σv(C) σv(A) σv(B)

σv(A) σv(A) σv(B) σv(C) E C3 C23

σv(B) σv(B) σv(C) σv(A) C23 E C3

σv(C) σv(C) σv(A) σv(B) C3 C23 E

24 symmetry operations that can be performed, and these are shown as the heading for theD6h character table in Table 2. Note that the coefficients for the operations either indicatehow many there are of the same type (3C ′

2, 3C ′′2 , 3σd , 3σv) or how many can be performed

about the same axis (2C6 = C6, C56 ; 2C3 = C3, C2

3 ; 2S6 = S6, S56 ; 2S3 = S3, S2

3 ). Utilizationof character tables will be discussed later.

One of the rules for mathematical groups is that the product of any two operations is alsoa member of the group. Figure 9 shows the ammonia molecule (NH3) which belongs to theC3v point group which has two C3 operations and three reflection planes. The effect of aC3 operation followed by a reflection through plane A is shown. It is also shown that thisproduct is equivalent to a reflection through a different plane B . In other words C3σv(A) =σv(B). Table 3 presents the product table for C3v symmetry. Note that each product is amember of the group and that each has an inverse.

As can be seen, it is important to identify the symmetry point group of a molecule. Table 4summarizes the possible point groups for molecules. Scheme 1 shows in a concise way howto determine the appropriate point group. This scheme precludes having a molecule of Sn

symmetry which is extremely rare.

Scheme 1

1. Decide if molecule is octahedral (Oh), tetrahedral (Td), or linear (D∞h with center ofsymmetry; C∞v with no inversion center).

2. If not, look for highest order proper axis (Cn). If present, go to 4.3. If no Cn, look for σ which gives a Cs point group. If no σ , the molecule has C1 or the

minimal symmetry.4. Look for nC2 axes perpendicular to the highest order Cn axis. If present, go to 7.5. Look for σh reflection plane perpendicular to Cn. If present, the symmetry is Cnh.


Table 4 Summary of point groups

Point group Important symmetry elements Order of the group

C1 E 1Ci i 2Cs σ 2Cn Cn n

S†n Sn (very rare) n

Cnv Cn, σv 2n

Cnh Cn, σh 2n

Dn Cn, ⊥C2 (very rare) 2n

Dnd Cn, ⊥C2, σd (also S2n) 4n

Dnh Cn, ⊥C2, σ12 4n

C∞v Linear molecules without center of inversion ∞D∞h Linear molecules with center of inversion ∞Td Terahedral symmetry 24Oh Octahedral symmetry 48

†n must be even, or else Sn = Cnh

Fig. 10 Examples of moleculeswith octahedral, tetrahedral, andlinear symmetries

6. If no σh, look for nσv planes. If present, the symmetry is Cnv . If not, the symmetry pointgroup is Cn.

7. Look for σh. If present, the symmetry is Dnh.8. If no σh, look for n dihedral planes bisecting C2 axes. If present, the symmetry is Dnd . If

not, it is Dn.

Figure 10 shows Oh, Td , D∞h, and C∞v molecules and lists some of their operations.Figures 11 and 12 show C2v , C3v , D3h, and D2h molecules along with the steps which leadto their identification.

We now turn to the use of character tables. Based on Schur’s lemma in the representationtheory of finite groups [22, 23], matrix elements of irreducible representations of a finitegroup satisfy certain orthogonality relations. Here the great orthogonality theorem applies.As given by Nowick [24] it can be stated as follows:

Theorem 1 (The Great Orthogonality Theorem [24]) Express the αth irrep of a group bythe set of matrices [D(α)(R)]ij where R is one of the operations of the group G , and take the


Fig. 11 Examples of C2v and C3v molecules

Fig. 12 Examples of D3h and D2h molecules

dimensionality of this irrep to be lα . Then the components of the various irreps of group Gobey the equation

∑

R

[D(α)(R)]∗ij [D(β)(R)]pq = (h/lα)δαβδipδjq

where ∗ denotes the complex conjugate, h is the order of the group G and the δ’s are allKronecker δ’s.

This rather complicated equation may be interpreted as follows. The fact that the orderof the group is h means that each irrep has h matrices. Suppose we treat the correspondingelements of these matrices, for example [D(α)(R)]ij for given i and j , as we vary the groupoperation R, as a vector in h-dimensional space. There are l2

α such vectors for irrep α and∑α l2

α in total. But group theory also tells us that∑

α l2α = h, so that there are h such vectors

in all. All members of this collection of h vectors are orthogonal to each other (defined sothat complex conjugates are used in the dot products, where necessary). Further it tells usthat the magnitude of each other vector (the dot product with itself) equals h/lα .


Fig. 13 Effect of a C2 operationon the translational motions ofwater

Table 5 The character table for the C2v symmetry point group. The reducible representation Γ for water isalso given

C2v E C2 σv(xz) σ ′v(yz)

A1 1 1 1 1 z x2, y2, z2

A2 1 1 −1 −1 Rz xy

B1 1 −1 1 −1 x, Ry xz

B2 1 −1 −1 1 y, Rx yz

Γ 9 −1 3 1

Next we turn to the characters χ(α)(R) of the various irrep matrices. From the definitionof the character, it readily follows that the characters obey their own orthogonality relation

∑

R

χ(α)(R)∗χ(β)(R) = hδαβ

Again, we may think of the characters of each irrep α as forming an h-dimensional vector.These vectors are all orthogonal to each other and that the magnitude of each is equal to h,the order of the group.

The sum of the diagonal elements of a matrix representation is called a character. Thenumber of irreducible representations of a finite group is equal to the number of conjugacyclasses of that group. Each character takes a constant value on a conjugacy class. We willuse the water molecule (Figs. 2 and 13) of C2v symmetry as an example. Table 5 showsthe character table for any C2v molecule which has C2, σv , and σ ′

v operations in additionto the identity. The character table lists four symmetry species and each has either a +1 or−1 value for the operations shown. A positive value means that the motion (or orbital) ofinterest is symmetric with respect to that operation while a −1 means it is antisymmetric.This is readily understood if we examine the translational and rotational motions of water.Figure 13 shows the effect of a C2 operation on the three translational motions. The C2

axis is taken to be the z direction and the molecule is in the xz plane. It is evident that theC2 operation will leave the direction of Tz unchanged, but the vectors of Tx and Ty changedirection. Hence, we can place a +1 character under C2 for the Tz operation while Tx and


Table 6 Characters for molecular motions of water


Tz 1 1 1 1 A1

Tx 1 −1 1 −1 B1

Ty 1 −1 −1 1 B2

Rz 1 1 −1 −1 A2

Vibrations:

Symmetric stretch 1 1 1 1 A1

Antisymmetric stretch 1 −1 1 −1 B1

Bend 1 1 1 1 A1

Fig. 14 Effect of C2 and σv

operations on a rotation of thewater molecule

Fig. 15 The three vibrations ofwater: symmetric stretching(A1), antisymmetric stretching(B1), and angle bending (A1),respectively

Ty have −1 characters. The σv(xz) reflection leaves Tz and Tx in the same direction (+1characters) while Ty switches direction. Similarly σ ′

v(yz) leaves Tz and Ty in their originaldirection but the direction of Tx is changed. Table 6 summarizes these results and shows thatthe Tz, Tx , and Ty characters match symmetry species A1, B1, and B2, respectively. Notethat in Table 5 this fact is already identified by the location of the x, y, and z to the right ofthe character table. Figure 14 shows the effect of the C2 and σv operations on the rotationabout the z axis and Table 6 presents the characters for each of the operations. These showRz to be of symmetry species A2, and Table 5 can be seen to identify each of the rotationalsymmetries.

When a vibrational description is known, we can use this “+ or − method” to also deter-mine the symmetry species for vibrations. Figure 15 shows the three vibrations of water. Thebehavior of each motion (symmetric or antisymmetric) upon each operation is also summa-


Fig. 16 The effect of a C4operation on the Cartesiancoordinates of PtCl2−

4

rized in Table 6. For example, the antisymmetric stretch is antisymmetric with respect to theC2 and σ ′

v(yz) operations, but symmetric to reflection in the σv(xz) plane and thus has B1

symmetry. The other two vibrations have A1 symmetry species.We now address how symmetry operations are represented mathematically. Figure 16

shows the P tCl2−4 molecular ion of D4h symmetry. The fifteen Cartesian coordinate vectors

needed to represent all the motions for the five atoms are also shown, and the effect of a C4

operation on the positions of all these vectors is presented. A matrix for the C4 operationcan be used to transform a column matrix of the original coordinates to the new coordinatepositions, and this is shown below.

⎡

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

0 0 0 0 1 0 0 0 0 0 0 0 0 0 00 0 0 −1 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 1 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 1 0 0 0 0 0 0 00 0 0 0 0 0 −1 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 1 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 1 0 0 0 00 0 0 0 0 0 0 0 0 −1 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 1 0 0 00 1 0 0 0 0 0 0 0 0 0 0 0 0 0

−1 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 1 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 1 00 0 0 0 0 0 0 0 0 0 0 0 −1 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 1

⎤

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

⎡

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

x1

y1

z1

x2

y2

z2

x3

y3

z3

x4

y4

z4

x5

y5

z5

⎤

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

=

⎡

⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

y2

−x2

z2

y3

−x3

z3

y4

−x4

z4

y1

−x1

z1

y5

−x5

z5

⎤

⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

(9)This shows that x1 is moved to where y2 was previously, y1 goes to −x2, and so on as

a result of the C4 operation. Fortunately, it is only necessary to have the character or trace(sum of the diagonal elements) of the C4 matrix in order to analyze molecular motions. Notethat for PtCl2−

4 the character for the C4 operation is +1 and this arises from the fact that thez5 vector on the Pt atom has not moved. In all cases, contributions to the character onlycome from unshifted atoms when an operation is performed. Although vectors x5 and y5 arealso on the P t atom, these do not contribute to the character since they are turned 90◦ fromtheir original direction. What simplifies matters a great deal is that each type of operationproduces the same contribution to the character per unshifted atom. For a C4 operation thisis always +1 since one vector goes into itself and the other two do not contribute. Figure 17shows the effect of E, σ , and C2 operations on any unshifted atom and why they contribute+3, +1, and −1, respectively, to the character. Table 7 shows the contribution per unshiftedatom for all types of the operations.

We will now apply this approach to the motions of water using Table 5. For the C2

and σ ′(yz) operations only the oxygen atom is unshifted while all three atoms are unshifted


Fig. 17 Effect of E, σ , and C2operations on the Cartesiancoordinates of an unshifted atom

Table 7 Contribution per unshifted atom (for atom vectors)

Operation Contribution Operation Contribution

E 3 σ (any) 1

C2 −1 i −3

C3 0 S3 −2

C4 1 S4 −1

C5 1 + 2 cos(72◦) S5 −1 + 2 cos(72◦)

C6 2 S6 0

CL 1 + 2 cos(360◦/L) SL −1 + 2 cos(360◦/L)

for the σ(xz) reflection. Utilizing the contribution per unshifted atom information in Table 7then results in the reducible representation Γ in Table 5. This can then be broken down to anirreducible representation consisting of the symmetry species as components. The followingformula is used:

nγ =(

1

g

)∑

j

gj xγ

j xjg , (10)

where

γ = particular symmetry species such as A1 or A2,

nγ = number of species γ represented in reducible representation Γ ,

g =∑

gj = total number of symmetry operations (add coefficients in heading),

gj = number of operations in class j (coefficient for particular column);

this is often one (implied),

xγ

j = number in character table for species γ for operation class j ;

this is +1 or −1 for non-degenerate species,

xjg = character in reducible representation for operation j .


As an example, to calculate how many of the A2 symmetry species are in Γ , we see thateach gj is one and g = 4. Then

nA2 = 1

4[1 · 1 · 9 + 1 · 1 · −1 + 1 · −1 · 3 + 1 · −1 · 1] = 1. (11)

The total result is that the 3N = 9 motions of water can be described as

Γ = 3A1 + A2 + 3B1 + 2B2. (12)

Table 5 shows the symmetry species for the translations and rotations are

ΓTRANS = A1 + B1 + B2. (13)

and

ΓROT = A2 + B1 + B2. (14)

When these are subtracted from Γ above, we are left with the symmetry species for thevibrations as

ΓVIB = Γ − ΓTRANS − ΓROT = 2A1 + B1. (15)

These vibrations were shown in Fig. 15.We will now apply this approach to the 12 atom benzene molecule using Table 2 where

the reducible representation Γ is also shown. Γ results from examining the number of un-shifted atoms for each operation and then using Table 7 to get the total contribution. Theresult for the 36 motions of benzene is

Γ = 2A1g + 2A2g + 2B2g + 2E1g + 4E2g + 2A2u + 2B1u + 2B2u + 4E1u + 2E2u. (16)

Note that E type symmetry species represent two motions. Table 2 shows the translationsto be of symmetry A2u + E1u and rotations to be A2g + E1g . Thus the symmetry species forthe 30 vibrations of benzene are

ΓVIB = Γ − ΓTRANS − ΓROT = 2A1g + A2g + 2B2g + E1g + 4E2g + A2u

+ 2B1u + 2B2u + 3E1u + 2E2u. (17)

Previously in Fig. 4 we presented the spectra of an organic molecule of C2v symmetry.When symmetry considerations are applied here, we find the vibrational symmetry speciesare

ΓVIB = 10A1 + 5A2 + 9B1 + 6B2. (18)

When the vibrations are numbered, as they are in Fig. 4, we start with the first symmetryspecies and label the vibrations from highest frequency down. The first vibration (ν1) isthen the highest frequency A1 mode which happens to be a C–H bond stretching vibrationwith frequency 3079 cm−1. Vibration ν10 is a bending motion of the five-membered ringof A1 symmetry at 623 cm−1. The next vibration numbered is ν11 which is the highestfrequency vibration with A2 symmetry and so on. The last numbered vibration ν30 is thelowest frequency B2 motion at 83 cm−1.


4 Symmetry Adapted Linear Combinations (SALCs)

In our discussion of water we pointed out that vibrations are best represented by internalcoordinates as shown in Fig. 2. We now show how the application of symmetry enables us toprovide better descriptive pictures of the vibrations. During a molecular vibration equivalentbonds or angles stretch or bend simultaneously, but not necessarily in the same direction.Thus, for water, the two oxygen-hydrogen molecular bonds R1 and R2 will either stretchat the same time or one will stretch while the other contracts. This is shown in Fig. 15.The method of symmetry adapted linear combinations (SALCs) allows us to calculate howequivalent internal coordinates combine. The formula is

Sγ = η∑

Op

χγ

OpOp(In1), (19)

where Sγ is the SALC with symmetry species γ , χγ

Op is the value in the character tablefor species γ and operation Op, and Op(In1) is the resulting internal coordinate when theoperation Op is performed on the initial internal coordinate of choice In1, and η is thenormalization constant. This process is shown below when the bond stretching coordinateR1 of water is selected as In1:


Op(R1) R1 R2 R1 R2(20)

Thus R1 stays in its own position upon E and σv(xz) operations but moves to where R2 wasupon C2 and σ ′

v(yz) operations. The result above in combination with (19) yields

SA1 = η(2R1 + 2R2), (21)

SB1 = η(2R1 − 2R2), (22)

and

SA2 = SB2 = 0. (23)

The normalization condition requires that

∑c2i = 1, (24)

where the ci are the coefficients for the internal coordinates. Thus, (21) and (22) become

SA11 = 2− 1

2 (ΔR1 + ΔR2) (25)

and

SB13 = 2− 1

2 (ΔR1 − ΔR2) (26)

The Δ have been inserted to indicate the vibrations actually involve the change in the bonddistances R1 and R2. The B1 SALC is labeled as S3 since

SA12 = Δα (27)

is numbered first since it belongs to A1 symmetry. In this case, there is only one angle tobend and α has nothing to combine with so the SALC is the same as the internal coordinate.


Fig. 18 The effect of symmetryoperations on the internalcoordinates of ethylene

Fig. 19 SALCs representing the C–H bond stretching and angle bending vibrations of ethylene

To further demonstrate SALCs we examine the C–H bond stretching and HCH anglebending internal coordinates of ethylene shown in Fig. 18. The effect of each operation isshown. When used together with a D2h character table, the SALCs shown in Fig. 19 result.For this molecule and in many cases, the SALCs shown are actually very good representa-tions for the so-called normal coordinates which are the precise descriptions of the vibra-tional motions. Mixing or “vibrational coupling” of motions of the same symmetry speciesis allowed, but no interactions between different symmetry species may occur.

5 Spectroscopic Selection Rules

The vibrational spectra of molecules involve transitions between quantum states and thesewere shown for infrared absorption and Raman scattering transitions in Fig. 3. For an in-frared transition to occur the molecule must have a change in dipole moment (a change


in charge distribution) during the vibration, and this can be determined by symmetry con-siderations. Namely, the following transition moment integral Rif must not be zero for an“allowed transition”:

Rif =∫

ψ∗f Mσ ψidτ = 〈f |σ |i〉. (28)

The expression on the right is in the more concise Dirac notation. Here, ψi and ψf arethe wave functions for the initial and final states of the transition, and ψ∗

f is the complexconjugate of ψf . Also here Mσ is the dipole moment expression for the σ direction (σ =x, y, or z) in terms of the vibrational coordinate Q:

Mσ = M0σ +

(∂Mσ

∂Q

)

0

Q +(

∂2Mσ

∂Q2

)

0

Q2 + · · · . (29)

Then

〈f |σ |i〉 = M0σ 〈f |i〉 +

(∂2Mσ

∂Q2

)

0

〈f |Q|i〉 + · · · . (30)

The first term is zero due to the fact that the wave functions are orthogonal. It can readilybe shown that Rif is non-zero only if the direct product of the symmetries of ψi , Q, andψf is totally symmetric where all the characters for the symmetry species are +1 (A1 forC2v symmetry, for example). For vibrations the initial state is totally symmetric and thesymmetry species for Q is the same as Tx , Ty , or Tz. Thus for Rif to be non-zero, ψf

must have the same symmetry species as Tx , Ty , or Tz. This is because the direct productof a symmetry species times itself is totally symmetric (e.g., B1 ⊗ B1 = A1 for C2v). Thisprovides a simple way to identify infrared active vibrations since they are those that have anx, y, z at the right hand side of a character table. For C2v these are A1, B1, and B2. For D6h

they are A2u and E1u.For Raman transitions the polarizability of a molecule must change during a vibration.

This essentially means that the shape of the molecule must change. This is generally truefor more symmetric vibrations but often not true for antisymmetric motions such as theantisymmetric C–O bond stretching of carbon dioxide. Mathematically, the product of twointegrals must be non-zero:

Pif = 〈f |ρ|j〉〈j |σ |i〉, (31)

where i and f are the initial and final states and j is the virtual state shown in Fig. 3.For the second integral to be non-zero j must have the same symmetry as σ since i istotally symmetric. Then in the first integral the direct product of ρ and j must match thatof the vibrational level f . Hence, the vibration must have the same symmetry as the tensorelement σ ⊗ ρ to be Raman active. Each character table lists these at the right side of thetable. For C2v all of the symmetry species correspond to Raman active vibrations. For D2h

symmetry only the A1g , E1g , and E2g vibrations are Raman active. In Raman experimentsdifferent vibrations can also be determined to be either Raman polarized or depolarizeddepending on how the intensities vary with laser beam polarization. As it turns out, only thetotally symmetric vibrations have polarized Raman bands. There are also other spectroscopicprocesses such as two-photon absorption and hyper Raman for which the selection rules canbe determined in a similar way.

To demonstrate the application of selection rules we will examine the predicted spectralfeatures of two organic molecules with the same formula but belonging to different symme-try point groups. The cis- and trans-dichloroethylene molecules of C2v and C2h symmetry,


Fig. 20 Cis- andtrans-dichloroethylene molecules

Table 8 Infrared and Raman bands expected for isomers of 1,2-dichloroethylene

Isomer Symmetry Raman bands Raman polarized Infrared bands

cis C2v 12 5 10

trans C2h 6 5 6

respectively, are shown in Fig. 20. The cis molecule has vibrations

Γcis(C2ν) = 5A1 + 2A2 + 4B1 + B2, (32)

IR,R(P) R IR,R IR,R

while the trans has

Γtrans(C2h) = 5Ag + Bg + 2Au + 4Bu. (33)

R(P) R IR IR

For each symmetry species, the infrared (IR) and/or Raman (R) activity are shown andthe polarized Raman bands [R(P)] are also indicated. Table 8 summarizes the number ofbands expected for the infrared and Raman spectra. As can be seen, the two molecules areeasy to distinguish based on symmetry selection rules. The experimental spectra for thesemolecules can be found elsewhere [25].

6 Effect of Symmetry on Potential Energy

We now wish to show how symmetry helps to simplify the computation of vibrational fre-quencies and force constants. We will use the water molecule with its internal coordinatesshown in Fig. 2 as an example. First we note that the internal coordinates can be representedby a column matrix R which is defined by a transformation from Cartesian coordinates X:

R = BX. (34)

For water X is a column matrix of the nine Cartesian coordinates required to define thepositions of the three atoms. B is the 3 × 9 transformation matrix which will not be shownhere. For water,

R =⎡

⎣ΔR1

ΔR2

Δα

⎤

⎦ . (35)


These internal coordinates can be transformed to symmetry coordinates using

S = UR, (36)

where for water

S =⎡

⎣S1

S2

S3

⎤

⎦ (37)

is based on the definitions in (25)–(27) so that

U =⎡

⎢⎣

2− 12 2− 1

2 00 0 1

2− 12 −2− 1

2 0

⎤

⎥⎦ . (38)

The potential energy is given by

2V = R†FR. (39)

For water the potential function in terms of internal coordinates is given in (7) so that

F =⎡

⎣fR fRR fRα

fRR fR fRα

fRα fRα fα

⎤

⎦ . (40)

This can be converted to a symmetric set of force constants using the similarity transforma-tion

Fsym = UFU†. (41)

For water this becomes

Fsym =⎡

⎣F11 F12 0F21 F22 00 0 F33

⎤

⎦ =⎡

⎢⎣

(fR + fRR) 212 fRα 0

212 fRα fα 00 0 (fR − fRR)

⎤

⎥⎦ . (42)

Then in terms of symmetry coordinates we have

2V = S†FsymS. (43)

The solution of the vibrational problem also requires the consideration of the kinetic energywhich can be written as

2T = R†G−1R. (44)

where G−1 is the reciprocal of the G matrix which is determined from the geometry of themolecule and the masses of the individual atoms. Details for its calculation can be foundelsewhere [21]. The same symmetry considerations apply to the G matrix as they do for F.Namely

Gsym = UGU†. (45)


Once the F and G matrices are available, the vibrational eigenvalues λ can be calculated andthen related to the vibrational frequencies ν:

λ = 4π2ν2. (46)

In internal coordinates,

|GF − Eλ| = 0, (47)

where E is the identity matrix. For water this would be a 3 ⊗ 3 determinant that needsto be solved (or matrix diagonalization methods can be used). However, use of symmetrycoordinates results in

|GsymFsym − Eλ| = 0, (48)

and the resulting determinant can be broken up in terms of the individual symmetry species.For water a 2×2 determinant for A1 results and a simple 1⊗1 determinant for the B1 block.For larger molecules this symmetry factoring is even more significant. For example, forbenzene and its vibrations given in (17), instead of having a 30 × 30 matrix or determinantto deal with, symmetry factoring results in ten separate symmetry blocks with the largestbeing a 4 × 4 matrix.

7 Conclusions

In this paper we have tried to convey how powerful the application of symmetry and grouptheory is to the analysis of molecular vibrations. However, we have only touched the surfaceas the understanding of electronic and rotational spectroscopies is also greatly facilitated bysimilar methods. We have also not delved deeply into demonstrating how the computationof vibrational frequencies and force constants is simplified to a great extent by symmetryconsiderations. Symmetry and group theory are also invaluable for understanding molecularbonding based on molecular orbital theory where molecular orbitals are generated from thelinear combination of atomic orbitals (MO-LCAO theory).

Acknowledgements The authors wish to thank the Robert A. Welch Foundation (Grant A-0396) for finan-cial support. Yi-Ching Wang assisted with adapting the manuscript to LaTeX format.

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