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QUANTUM MECHANICS WITH APPLICATIONS by Iraj R. Afnan School of Chemical and Physical Sciences Flinders University Energy bands in a one dimensional solid

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Page 1: pdfs.semanticscholar.org€¦ · Contents Preface v 1 Introduction1 1.1 Classical Physics. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .2 1.1.1 Newtonian Mechanics

QUANTUM MECHANICS

WITH

APPLICATIONS

by

Iraj R. Afnan

School of Chemical and Physical Sciences

Flinders University

Energy bands in a one dimensional solid

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Contents

Preface v

1 Introduction 1

1.1 Classical Physics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.1.1 Newtonian Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.1.2 Maxwell’s Electrodynamics . . . . . . . . . . . . . . . . . . . . . . . 3

1.2 The Particle Nature of Radiation . . . . . . . . . . . . . . . . . . . . . . . 4

1.2.1 Blackbody radiation . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.2.2 Photoelectric effect . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.2.3 Compton scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.3 The Wave Nature of Particles . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.4 Atomic Spectra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2 Classical Mechanics - A Review 11

2.1 Hamilton’s Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2.2 The Euler-Lagrange Equation . . . . . . . . . . . . . . . . . . . . . . . . . 14

2.3 Hamiltonian Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

2.4 Symmetry and Conservation Laws . . . . . . . . . . . . . . . . . . . . . . . 18

2.4.1 Homogeneity of time . . . . . . . . . . . . . . . . . . . . . . . . . . 18

2.4.2 Homogeneity of space . . . . . . . . . . . . . . . . . . . . . . . . . . 19

2.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

3 Wave Packet and the Uncertainty Principle 23

3.1 Superposition of Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

3.2 Fourier Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

3.3 The Uncertainty Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

3.4 The Schrodinger Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

3.5 Physical Interpretation of Ψ(~r, t) . . . . . . . . . . . . . . . . . . . . . . . 40

3.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

i

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ii CONTENTS

4 The Schrodinger Equation 53

4.1 Method of Separation of Variables . . . . . . . . . . . . . . . . . . . . . . . 54

4.2 Method of Fourier Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . 55

4.3 Particle in a Box . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

4.4 General Properties of the Wave Function ψn(x) . . . . . . . . . . . . . . . 61

4.5 Symmetry Under Inversion - Parity . . . . . . . . . . . . . . . . . . . . . . 65

4.6 Eigenstates of the Momentum . . . . . . . . . . . . . . . . . . . . . . . . . 67

4.7 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

5 Simple One Dimensional Problems 73

5.1 Free-Particle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

5.2 Potential Step . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

5.3 Potential Well . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80

5.3.1 Bound state problem . . . . . . . . . . . . . . . . . . . . . . . . . . 81

5.3.2 Symmetry of the Hamiltonian . . . . . . . . . . . . . . . . . . . . . 82

5.3.3 Eigenstates of the Hamiltonian . . . . . . . . . . . . . . . . . . . . 83

5.3.4 The scattering problem . . . . . . . . . . . . . . . . . . . . . . . . . 86

5.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

6 Application of Quantum Mechanics 91

6.1 Barrier Penetration and α-decay . . . . . . . . . . . . . . . . . . . . . . . . 91

6.1.1 Potential Barrier . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93

6.1.2 α-decay . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

6.2 The Deuteron . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98

6.2.1 The binding energy of the deuteron . . . . . . . . . . . . . . . . . . 99

6.2.2 The deuteron wave function . . . . . . . . . . . . . . . . . . . . . . 102

6.3 The δ-Function Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

6.4 The Diatomic Molecule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

6.5 The Atomic Clock . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110

6.6 One Dimensional Solids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

6.6.1 Translational symmetry – Bloch’s Theorem . . . . . . . . . . . . . . 112

6.6.2 The Kronig-Penney Model . . . . . . . . . . . . . . . . . . . . . . . 115

6.7 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118

7 Molecular Vibration – The Harmonic Oscillator 121

7.1 The One Dimensional Harmonic Oscillator . . . . . . . . . . . . . . . . . . 122

7.2 Vibrational Spectrum of Diatomic Molecule . . . . . . . . . . . . . . . . . 129

7.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131

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CONTENTS iii

8 Central Force Problem I 1338.1 Cartesian Tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1368.2 Angular Momentum in Quantum Mechanics . . . . . . . . . . . . . . . . . 1398.3 Rotational Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1468.4 The Radial Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1488.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150

9 Central Force Problem II 1559.1 The Three-Dimensional Square Well . . . . . . . . . . . . . . . . . . . . . . 1559.2 The Three-Dimensional Harmonic Oscillator . . . . . . . . . . . . . . . . . 1599.3 The Coulomb Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1639.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171

10 Scattering by a Central Potential 17310.1 The Cross Section . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17410.2 Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17710.3 The Square Well Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . 17810.4 The Scattering Amplitude . . . . . . . . . . . . . . . . . . . . . . . . . . . 18010.5 The Optical Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18410.6 The Phase Shifts for Two-Body Scattering . . . . . . . . . . . . . . . . . . 18510.7 Coulomb Scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18710.8 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190

11 Matrix Formulation of Quantum Mechanics 19311.1 Operators and Observables . . . . . . . . . . . . . . . . . . . . . . . . . . . 19311.2 Dirac Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20211.3 Representation of Operators . . . . . . . . . . . . . . . . . . . . . . . . . . 206

11.3.1 The Coordinate Representation . . . . . . . . . . . . . . . . . . . . 20911.3.2 The Momentum Representation . . . . . . . . . . . . . . . . . . . . 21211.3.3 Angular Momentum Representation . . . . . . . . . . . . . . . . . . 215

11.4 Spin 12

Particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21911.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224

12 Symmetry and Conservation 22912.1 Translation in Space and Time . . . . . . . . . . . . . . . . . . . . . . . . . 23012.2 Rotation in Three-Dimensions . . . . . . . . . . . . . . . . . . . . . . . . . 23512.3 Addition of Angular Momentum . . . . . . . . . . . . . . . . . . . . . . . . 24312.4 Space Inversion and Parity . . . . . . . . . . . . . . . . . . . . . . . . . . . 24712.5 Time Reversal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24912.6 Isospin and the Pauli Principle . . . . . . . . . . . . . . . . . . . . . . . . . 25212.7 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254

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iv CONTENTS

13 Approximation Methods for Bound States 25713.1 Perturbation Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25713.2 The Helium Atom in Perturbation Theory . . . . . . . . . . . . . . . . . . 26113.3 Anharmonic Linear Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . 263

13.3.1 Occupation Number Representation . . . . . . . . . . . . . . . . . . 26513.3.2 Lowest Order Contribution . . . . . . . . . . . . . . . . . . . . . . . 268

13.4 Van der Waal’s Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27013.5 The Variational Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27313.6 The Helium Atom - Variational Method . . . . . . . . . . . . . . . . . . . 27513.7 Degenerate Perturbation Theory . . . . . . . . . . . . . . . . . . . . . . . . 27813.8 The Spin-Orbit Interaction . . . . . . . . . . . . . . . . . . . . . . . . . . . 28013.9 The Zeeman Effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28413.10Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 289

14 Scattering Theory; Revisited 29314.1 Formal Theory of Scattering . . . . . . . . . . . . . . . . . . . . . . . . . . 29314.2 The Born Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30214.3 Electron Atom scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30514.4 Unitarity of the T-matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . 30814.5 The T -matrix for Separable Potentials . . . . . . . . . . . . . . . . . . . . 31014.6 Spin Dependent Scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . 31914.7 Effective Range Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . 32014.8 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 326

A Complex Analysis in a Nutshell 329A.1 Functions of a Complex Variable . . . . . . . . . . . . . . . . . . . . . . . 329A.2 Analytic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 332A.3 Cauchy’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 333A.4 Taylor and Laurent Series . . . . . . . . . . . . . . . . . . . . . . . . . . . 335A.5 Residue Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 336A.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 336

B Atomic Units 339

C Numerical Solution of the Schrodinger Equation 341C.1 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 345

References 346

Index 349

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Preface

The present set of lecture notes are a byproduct of lectures given by the author to un-dergraduate students at Flinders University over the period 1970 to 2004. The materialin the lecture notes (eBook) was to cover a two semester course. The first semester,at second year level, consisted of two lectures plus one tutorial per week for a periodof thirteen weeks, while the second semester, at third year level, was given at the rateof three lectures plus one tutorial per week. The aim of the lectures was to bring thestudent to a level that would allow them to read the literature in atomic, molecular andnuclear physics and be able to perform some simple calculations based on non-relativisticquantum mechanics for one- and two-body systems.

With the limited knowledge of calculus, the first semester concentrates on systems inone dimension with extensive applications to atomics, molecules, solids and nuclei. InChapter 2 we review classical mechanics and the role of space symmetry in conservationlaws. Although this chapter is not a prerequisite to later chapters, it gives the studentthe insight into the relation between classical and quantum physics. The emphasis inthe following chapters is to illustrate the mathematical structure of quantum mechanicswith examples rather than mathematical proofs. The first semester brings the studentsto the level of writing the Schrodinger equations for central potentials, e.g. the Coulombpotential.

The second semester of the course commences by considering the solution of theSchrodinger equation as a second order differential equation for simple potentials thatadmit analytic solutions. This allows us to study the properties of many of the specialfunctions encountered in problems in atomic and nuclear physics. These functions arethen used in the application of quantum mechanics to problems in atomic and nuclearphysics. The Dirac notation is introduced more as a shorthand notation than as a trea-tise on Hilbert spaces to understand the importance of the different representations of theSchrodinger equation. Since symmetry plays a fundamental role in the simplification ofproblems encountered in quantum physics, a chapter is devoted to space symmetry and thecorresponding conservation laws with emphasis on rotational symmetry. As most prob-lems in physics require the introduction of approximation methods, a chapter is devotedto time independent perturbation and variational methods. The last chapter concentrateson scattering theory based on the Lippmann-Schwinger equation. Here, the solution ofthe Lippmann-Schwinger equation is illustrated by considering a rank one separable po-

v

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vi CONTENTS

tential for which one can derive an analytic solution for both the scattering amplitudeand bound state and scattering wave function.

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Chapter 1

Introduction

Abstract : In this chapter we briefly review the failure of classical physics andthe experiments and ideas that led to the formulation of quantum mechanicsin 1925.

The first quarter of the 20th century saw the introduction of two new developmentsin physics that have revolutionized our understanding of the subject. These two develop-ments are:

1. Relativity

2. Quantum Mechanics

Although the first of these two developments has had minor influence on our daily activityuntil recently, the second has dominated much of the technological development of the20th century. It is quantum mechanics that has set the framework for our theoreticalunderstanding of the basic properties of atoms, molecules and solids. Having this theo-retical understanding, it is hoped that we can predict the nature of chemical reactions,the behavior of semiconductors, and the properties of superconductors at low and hightemperatures. Thus quantum mechanics forms the cornerstone of both the chemical andelectronics industry.

However, the influence of this theory goes beyond science and technology. In factits influence extends to all aspects of human thinking from Art to Philosophy. WhenHeisenberg, in 1927[1], came out with the idea of quantum theory based on the uncertaintyprinciple, he found that his greatest opposition came from the German philosophers of thetime. This was basically because the uncertainty principle violated not only the micro-causality of Newton and Lagrange, but also the basic philosophical ideas of Kant[2].

In the present eBook, I want to concentrate on the scientific influence of quantummechanics, and in particular the tools it gives us to understand nature at the microscopiclevel. In fact, over the next chapters in this eBook, we will develop the ideas and the

1

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2 CHAPTER 1. INTRODUCTION

mathematical tools required to understand the properties of atoms, molecules, nuclei,solids and finally the subatomic world of particle physics. We will see how the symmetriesof nature play a vital role in unfolding the underlying dynamics of each of the systems wewill consider.

However, before we set out on our exploration into the world of quanta, I would liketo give you a glimpse of the experiments and the models that led to this most excitingtheory. To appreciate its novel ideas, we should go back and very briefly summarize thetheories that existed some one hundred years ago.

1.1 Classical Physics

At the turn of the 20th century, most physical phenomena were considered to be governedby one of two theories;

1. Mechanics as originally formulated by Newton and developed by Euler, Lagrange,Hamilton and others.

2. Electrodynamics which is a unification of electricity and magnetism as observed byFaraday and formulated by Maxwell.

1.1.1 Newtonian Mechanics

In any introductory course in physics the student is introduced to the famous equation~F = m~a. This in fact is a statement of Newton’s second law and should be written, inthe case of a single particle, as

d~p

dt= ~F , (1.1)

where ~p is the momentum of the particle, ~F is the force on that particle and t stands fortime. Thus the motion of the particle is governed by Eq.(1.1), which states that the rateof change in momentum of a body is equal to the force exerted on it. A more practicalform of this equation is

md2~r

dt2= ~F , (1.2)

where we have taken the momentum ~p to be given in terms of the velocity ~v or the positionof the particle ~r by the relation

~p = m~v = md~r

dt. (1.3)

In Eq.(1.2), we have a second order differential equation which for certain simple forcescan be integrated to give the position of the particle as a function of time. In general, theintegrated equation has two arbitrary constants, which as we will see, are determined bythe initial condition, e.g. the position and velocity of the particle at some initial time t0.

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1.1. CLASSICAL PHYSICS 3

Example: For a particle with mass m in a gravitational force ~F = −mgy, the motionin the vertical direction (i.e along the y-axis) is governed by the second order differentialequation

md2y

dt2= −mg .

Upon integration, this equation gives the familiar result

y = y0 + v0t−1

2gt2

where y0 and v0 are the position and velocity of the particle at some initial time t0 = 0.Here we observe that given y0 and v0, we can determine the position of the particle atany time t ≥ t0. This result is in general true and its implications are enormous. Theresult basically implies that if we know the position and velocity of all the particles in theuniverse at a given time, we could predict the behavior of the universe from that instantof time onward, i.e. we can tell what the future has in store for us. In practice that is notquite true, for we can never state the initial condition with sufficient precision to uniquelydetermine the behavior of the system for all times in the future.1

1.1.2 Maxwell’s Electrodynamics

In a brilliant first unification, Maxwell[4], with the help of the experimental results ofFaraday, was able to write four equations for the electric and magnetic field of any dis-tribution of charges and currents. These equations, known as Maxwell’s equations, werenot only able to describe electricity and magnetism, but also predicted that an oscillatingelectric and magnetic field will generate radiation, presently known as electromagnetic ra-diation, light being a special example. Hertz in 1887 put Maxwell’s prediction to the testby observing the radiation from an oscillating electric and magnetic field, and showingthat this radiation has the same behavior as light. The electric and magnetic fields forthis radiation are governed by the equations(

1

c2

∂2

∂t2−∇2

)~E (~r, t) = 0 (1.4)(

1

c2

∂2

∂t2−∇2

)~B (~r, t) = 0 . (1.5)

These equations are a special form of Maxwell’s equation known as the wave equations.With this result it was finally established that light is a wave as Huygens predicted, andnot a particle as Newton thought.

1This phenomena is often referred to as The Butterfly Effect. For a popular description of this andother related phenomena see [3].

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4 CHAPTER 1. INTRODUCTION

1.2 The Particle Nature of Radiation

At the turn of the 20th century, many physicists thought it was merely a question of timebefore the minor problems still unresolved would be overcome, to render physics to thestatus of a dead subject. One of these minor problems was the spectrum of blackbodyradiation.

1.2.1 Blackbody radiation

A blackbody is a closed box for which there is thermodynamic equilibrium between theradiation in the cavity and the walls of the box. If we now make a hole in the wall, withthe hole being small enough not to disturb this equilibrium, then the spectrum of theradiation emitted through the hole is known as blackbody radiation. The name ‘blackbody’is a result of the fact that the radiation from the box is emitted and not reflected radiation,a property common with any black surface that does not reflect light. The spectrum ofthis radiation is shown in Figure 1.1 on the left, where we have the energy density de inthe frequency range between ω and ω+dω plotted versus the frequency ω. The solid curvein Figure 1.1 on the right represents the experimental result, while the dashed curve wasthe best that the classical theory of Rayleigh-Jeans could do in predicting the spectrumof the radiation to be

de =kTω2

π2c3dω , (1.6)

where T is the temperature, k = 8.617 × 105 eV K−1 is Boltzmann’s constant, c is thevelocity of light, and ω is the frequency of the radiation emitted by the blackbody.

Ene

rgy

dens

ity

Ene

rgy

dens

ity

ω ω1 2 3 1

4000

3000

3500

Rayleigh-Jeans

Planck

Figure 1.1: The Planck distribution for the energy density. The figure on the left givesthe distribution for the energy at a temperature of 3000, 3500, and 4000K, while thefigure on the right is a comparison of the Planck distribution and the Rayleigh-Jeansresult as defined in Eq. 1.6.

In this classical theory it was assumed that the energy in the cavity was the electro-magnetic waves of Maxwell, in equilibrium with the walls. For this classical result, the

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1.2. THE PARTICLE NATURE OF RADIATION 5

total energy,∫de, was infinite, and therefore unacceptable. To overcome this discrepancy

between theory and experiment, Max Planck, in 1901[5], suggested that the energy of theradiation emitted or absorbed by the walls of the cavity is proportional to its frequencyi.e.

E = hν = hw h =h

2π, (1.7)

where h is Planck’s constant,indexPlanck!constant h = 0.6582 × 10−15 eV sec.2 Planckthen derived the energy density to be

de =h

π2c3

ω3

ehω/kT − 1dω , (1.8)

which gives a perfect fit to the experimental data, and is known as the Planck’s distributionlaw.

This quantization of the energy that is emitted and/or absorbed by the walls of thecavity, removed the discrepancy between theory and experiment at the cost of introducing,for the first time, the concept that energy is transferred only in discreet quantities (∆E =hω). Because of the small value of h, the amount of energy transferred is too small to beobserved in our day to day activities, but is important at the microscopic level of atomsand molecules.

1.2.2 Photoelectric effect

A second minor problem with classical physics involved the properties of the electronsemitted when radiation was incident on a metallic surface, see Figure 1.2. Here theelectromagnetic radiation incident on a metallic surface results in the emission of electrons.In particular, it was observed that:

1. The energy of the individual electrons emitted was independent of the intensity ofthe radiation, but was linearly proportional to the frequency of the radiation, i.e.,

Ee ∝ hω ,

where Ee is the energy of the electron and ω the frequency of the radiation.

2. The number of electrons emitted was independent of the frequency of the radiationω, provided the frequency was greater than some critical frequency. However, thenumber of electrons emitted was dependent on the intensity of the radiation, i.e.

Ne ∝ Nγ ,

2In units of energy-length, we have

hc = 197.327 MeV-fm = 197.327 eV-nm .

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6 CHAPTER 1. INTRODUCTION

where Ne is the number of electrons emitted while Nγ is the intensity of the radiationor the number of photons.

Solid

W

e

γ

Figure 1.2: The photoelectric effect, schematic diagram on the right, while on the leftwe illustrate the atomic potential in a solid that gives rise to the work function W .

In 1905, Einstein[6] took Planck’s postulate one step further by assuming that theelectrons emitted are the result of the fact that the radiation of frequency ω, consisted ofphotons of energy E = hω, and that the electron absorbs the full energy of the photon.In other words, the energy of the electron Ee is given by

Ee =1

2mev

2 = hω −W , (1.9)

where v is the velocity of the electron emitted. In Eq. (1.9), W is the work function, anddepends on the type of surface being radiated. W was introduced in order to explain thepresence of a critical frequency below which no electrons are emitted from the surface.

Here again, we have an experimental observation whose explanation relies on the factthat radiation, which is electromagnetic in nature, comes in quanta with fixed energyE = hω, i.e., electromagnetic radiation consists of bundles of energy like particles, andthis energy in total is transferred to a single electron. This result was in contradiction withMaxwell’s equations which was, and is still, considered one of the major achievements ofnineteenth century science.

1.2.3 Compton scattering

Some twenty years later Compton[7] was scattering radiation from the electrons in anatom. Here, we find that the energy of the scattered radiation as a function of thescattering angle, can only be explained if we take the collision between the radiation andthe electron to satisfy energy and momentum conservation. For the radiation, Comptontook Planck’s postulate, i.e. E = hω, while for the momentum of the radiation hepostulated

p =hω

c=h

λ, (1.10)

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1.3. THE WAVE NATURE OF PARTICLES 7

where λ is the wave length of this radiation. This postulate is equivalent to assumingthat the radiation can be represented by particles called photons. These photons travelwith the velocity of light and according to relativity, are therefore massless. Their energycan then be written as3

E =√p2c2 +m2c4

= pc for m = 0 . (1.11)

which is consistent with the definition of the momentum as given in Eq. (1.10). Thekinematics for this reaction can be treated as if the photons in the initial and final stateare a particle with energy hω and momentum hω/c, i.e., the conservation of energy andmomentum are given by

hω +mc2 = hω′ +√p2c2 +m2c4

c=

hω′

ccos θ + p cosφ (1.12)

0 =hω′

csin θ − p sinφ .

Thus Compton scattering established the fact that radiation can transfer momentum toan electron just like a particle does.

The above three experiments established the fact that electromagnetic radiation insome instances behaves like massless particles with energy E = hω = pc, while in otherinstances it is known to behaves like a wave. This result, which violates classical physicsi.e. Maxwell’s equations, is an illustration of particle-wave duality of radiation.

1.3 The Wave Nature of Particles

Having established that electromagnetic radiation satisfies particle-wave duality, we canask the question: Can particles that are governed by Newton’s Laws sometime behave likewaves? i.e., can particles satisfy the same duality that radiation seems to have? In 1923,this hypothesis was made by the young French nobleman Louis de Broglie. In fact, in aPh.D. thesis submitted in 1924, de Broglie worked out the consequence of his hypothesis.The basis of this hypothesis[8] is that a particle with momentum p has a wave length λgiven by

λ =h

por p =

h

λ=hν

c=hω

c. (1.13)

Note, this is identical to the relation between the momentum and the wave length ofthe radiation as given in Eq. (1.10). The main consequence of this hypothesis is that abeam of particles incident on a diffraction grating will produce an interference pattern.

3Here, we take the relativistic relation between the energy E and the momentum p.

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8 CHAPTER 1. INTRODUCTION

The problem is that the wave length λ is very short and thus to observe the interferencepattern requires a very fine grating.

While de Broglie was working out the consequences of his postulate in France, on theother side of the Atlantic Ocean, two American scientists, Davisson and Germer, whilecarrying out an experiment on electron scattering from a metal target, discovered thediffraction pattern resulting from the scattering of electrons from a crystal[9].

We thus have established that matter, e.g. electrons and neutrons can behave likewaves under the right conditions. This implies that their motion cannot be governed byNewton’s Second Law ~F = m~a, since this equation does not admit solutions that describethe behavior of a wave.

We now have a complete complementarity, to the extent that both matter and radia-tion can behave like particles or waves depending on the circumstance. In Chapter 3 wewill make use of this particle-wave duality to develop quantum mechanics as formulatedby Schrodinger.

1.4 Atomic Spectra

Another of the minor problems to be resolved at the turn of the 20th century was thequestion of the radiation emitted by matter. Here it was found that the frequenciesof this form of radiation suggested that the energy emitted comes in certain discreetquantities. In fact, for hydrogen, the energy of the radiation emitted could be given by

Enm = R(

1

n2− 1

m2

)n < m , (1.14)

where n and m are integers and R = 13.6058 eV. For n = 1 and m = 2, 3, . . . we havea series of possible frequencies or energies for the radiation. This series is known as theLyman series. On the other hand the series resulting from taking n = 2 and m = 3, 4, . . .is known as the Balmer[10] series. These results were based on detailed experimentalobservations, and there was no theoretical justification for them at the time.

Since radiation is described by Maxwell’s equations, one may be tempted to use theseequations to understand the radiation. However, Maxwell’s equations state that radiationresults from the acceleration of charged particles have a continuous spectrum, i.e. all fre-quencies are possible. Clearly, one needs to go beyond the classical theories to understandthe fundamental structure of the atom that is emitting the radiation.

In 1911 Rutherford[11] suggested a model for explaining the results of the scatteringof α-particles from a thin gold foil. The experimental observation was that the number ofparticles scattered at large angles was consistent with the hypothesis that most of the massof the atom is concentrated at its centre and that the centre is positively charged. At thesuggestion of Rutherford, Bohr[12] constructed a planetary model of the atom in whichthe electrons are in orbits around the nucleus, and he allowed only certain orbits. Theradiation then was the result of the electron jumping from one orbit to the next. In this

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1.5. PROBLEMS 9

way only certain frequencies were allowed. This model, which was based on Newtonianmechanics with an additional quantization rule, gave the observed spectrum of hydrogen.However, the extension of Bohr’s ideas to more than one electron atom faced problemsthat could not be overcome, and by the mid 1920’s it was getting clear that a new theorywas required for the description of microscopic systems which seem to exhibit the effectsof quantization of energy and particle-wave duality.

1.5 Problems

1. Given the Planck energy density distribution

de

dω=

h

π2c3

ω3

ehω/kT − 1,

where k = 8.617× 10−5 eV K−1 is the Boltzmann constant and T the temperaturein degrees Kelvin. Use MAPLE or Mathematica to plot this energy distribution forthe following temperatures: 5, 300, 1000, and 5000 degrees Kelvin. From the plots,find the wave length at which the distribution is maximum.

2. Use energy and momentum conservation to show that in Compton scattering, thefinal energy of the photon E2, is expressed as a function of the initial energy of thephoton E1 and the angle of deflection θ by

E2 =E1

1 + (E1/mc2)(1− cos θ),

where c is the velocity of light and m is the mass of the electron.

Use either MAPLE or Mathematica to plot the energy of the final photon as afunction of the angle θ, given the incident photon has a wave number of 0.07 A.Note 1 A=10−8 cm.

Hint: Use relativistic kinematics for the electron, i.e. the energy of the electronis given by E =

√p2c2 +m2c4, where p is the momentum and m the mass of the

electron.

3. We have a mono-energetic source of X-rays (photons) of energy 50 KeV. However,for a specific experiment we need a beam of X-rays of energy 48 KeV.

(a) Can we use Compton scattering to generate the beam of X-rays with therequired energy? How?

(b) At what angle to the original beam do we extract the final beam to get theenergy of 48 KeV?

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10 CHAPTER 1. INTRODUCTION

(c) Plot, using MAPLE or Mathematica the energy of the final beam as a functionof the extraction angle.

4. The energy released by the hydrogen atom in 2p→ 1s transition is given by

∆E = R(

1

n2− 1

m2

)where n = 1, m = 2 and R = 13.6 eV.

(a) What is the wave length of the light emitted as a result of this transition?Assume the atom does not recoil.

(b) To what accuracy do you have to measure the wave length of the emittedradiation to observe the effect of the recoil of the atom? Take the mass ofthe atom to be m, with mc2 = 940 MeV. Here c is the velocity of light. (Useenergy momentum conservation.)

(c) In a nuclear transition the typical energy release is a factor of 106 larger thanthe atomic transition. To what accuracy do you have to measure the energyof the photon in a nuclear transition, to detect the effect of the recoil of theatom?

Hint: We have that hc = 197.3 eV-nm.

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Chapter 2

Classical Mechanics - A Review

Abstract : This chapter is devoted to a short review of classical mechanics withthe aim of defining the classical Hamiltonian, and establishing the relationbetween space symmetry and conservation laws.

In the present chapter, I would like to review classical mechanics with the aim ofintroducing the Euler-Lagrange formulation of mechanics based on Hamilton’s principleof least action, and the Hamiltonian formulation of mechanics. The motivation is twofold:

1. The Euler-Lagrange formulation of mechanics allows us to examine the relationshipbetween the symmetry of space-time and the conservation laws in mechanics. Thesame correspondence between symmetry and conservation can be utilized exten-sively in quantum mechanics to reduce the complexity of the problems at hand.

2. In the Hamiltonian formulation of mechanics, the independent variables are themomentum and the position or coordinate of the particles, rather than the velocityand coordinate. In quantizing a classical system we find that the momentum andthe coordinate are the canonical variables needed for quantization. This allows usto introduce quantization rules that are general enough to be applied not only toclassical mechanics, but also to electromagnetic theory.

2.1 Hamilton’s Principle

In the Newtonian formulation of mechanics, the equations of motion or the Laws ofmechanics are presented in terms of two vectors, the momentum ~p, and the force ~F .On the other hand, the Euler-Lagrange formulation of mechanics is based on two scalarquantities – the kinetic energy, T , and the potential energy, V . The fact that we havereplaced two vector quantities by two scalar quantities is balanced by the introduction of aprinciple called Hamilton’s principle or the principle of least action. The basic idea of this

11

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12 CHAPTER 2. CLASSICAL MECHANICS - A REVIEW

principle is as follows. Consider two points in space-time, P1 and P2, which correspondto the position of the particle at times t1 and t2 with t1 < t2, see Figure 2.1. The motionof the particle from P1 to P2 can be along any one of an infinity of possible paths. InFigure 2.1 we have three such paths. The integral of the difference between the kineticenergy, T , and potential energy, V , along any one of these paths is called the action, S,i.e.

P1

P2 t2

t1

t

x

1

32

Figure 2.1: Possible trajectories for the motion of a particle starting at the point P1 attime t1 and ending at the point P2 at time t2 with t1 < t2.

S[q, q] ≡t2∫t1

(T − V ) dt ≡t2∫t1

L (q, q; t) dt , (2.1)

where L (q, q; t) is called the Lagrangian for the system. Here q is the position of theparticle while the velocity of the particle is given by

q ≡ dq

dt.

Note that both q and q are functions of time, t, i.e. q(t) and q(t). Since the value of theaction S depends on the path, we have indicated this by making the action a functionalof q and q.

Hamilton’s principle states that the path which the particle actually takes is the onefor which the action S, defined in Eq. (2.1), is a minimum. The problem now is how todetermine this path for which S is a minimum.

In calculus, if we want to calculate the minimum of a function y(t), we first take atime t1 and calculate y(t1) and y(t1 + δt), where δt is a small change in the time at t1.If y(t1) < y(t1 + δt), then we calculate y(t1 − δt). If y(t1 − δt) > y(t1), then we haveestablished that y(t) is a minimum at t = t1. Otherwise we calculate y(t1 − 2δt) and

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2.1. HAMILTON’S PRINCIPLE 13

compare it with y(t1 − δt), and so on. Alternatively, we can calculate

δy = y(t1 + δt)− y(t1) =

(∂y

∂t

)t=t1

δt ,

where the final result is obtained by a Taylor series1 expansion of y(t1 + δt) about t = t1,

keeping the first non-zero term in the final result. If δy = 0 and δt 6= 0,(∂y∂t

)t=t1

= 0,

then we have a minimum of the function y(t).In our case, the action S is a function of the path along which we are carrying out the

integral, i.e. [q(t), q(t)]. To change the path by an infinitesimal amount is equivalent totaking

q(t)→ q(t) + δq(t) (2.2)

for every time t, with t1 < t < t2. However, we need to keep the initial and final points P1

and P2 unchanged, i.e. q(t1)→ q(t1) and q(t2)→ q(t2) or δq(t1) = 0 = δq(t2). A changein the path along which the particle moves, changes not only the position of the particlebut also its velocity, i.e.,

q(t)→ q(t) + δq(t) . (2.3)

We now can calculate the change in the action S, i.e.

δS = S [q + δq, q + δq]− S [q, q]

=

t2∫t1

L (q + δq, q + δq) dt−t1∫t1

L (q, q) dt

≡t2∫t1

δL (q, q) dt . (2.4)

To calculate δL(q, q), we need to expand L(q + δq, q + δq) about L(q, q), and keep thelowest order terms in δq and δq since we are taking an arbitrarily small variation in thepath, i.e.,

L(q + δq, q + δq) = L(q, q) +∂L

∂qδq +

∂L

∂qδq .

Here, we have dropped terms of the order (δq)2 or higher, since δq is small. Therefore,we can write the variation in the action S, as

δS =

t2∫t1

(∂L

∂qδq +

∂L

∂qδq

)dt .

Making use of the fact that

δq = δ

(dq

dt

)=

d

dt(δq) ,

1See Appendix A for the definition of a Taylor series.

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14 CHAPTER 2. CLASSICAL MECHANICS - A REVIEW

we can rewrite the change in the action δS as

δS =

t2∫t1

(∂L

∂qδq +

∂L

∂q

d

dt(δq)

)dt .

Integrating the second term on the right hand side by parts, we get

δS =∂L

∂qδq

]t2t1

+

t2∫t1

[∂L

∂q− d

dt

(∂L

∂q

)]δq dt .

Since we have taken the points P1 and P2 to be fixed, then δq(t1) = 0 and δq(t2) = 0.This renders the first term on the right hand side to be zero, and we have by Hamilton’sPrinciple that

δS =

t2∫t1

[∂L

∂q− d

dt

(∂L

∂q

)]δq dt = 0 . (2.5)

This result is valid for any variation δq in the path taken between P1 and P2, and thereforethe quantity in the square bracket should be zero if the variation in the action is to bezero, i.e.,

d

dt

(∂L

∂q

)− ∂L

∂q= 0 . (2.6)

This equation is known as the Euler-Lagrange equation or the equation of motion for thisone particle system.

2.2 The Euler-Lagrange Equation

In the above section, we found that Hamilton’s principle or the principle of least action,when applied to the motion of a single particle in one dimension with kinetic energy T andpotential energy V , gives a differential equation for the Lagrangian L, where L ≡ T − V .This result can be generalized to a system of n-particles in more than one dimension.The Lagrangian in this case is a function of all the coordinates and velocities, and is thedifference between the kinetic energy of the system, T , and its potential energy V , i.e.,

L = L (q1, . . . , qN , q1, . . . , qN ; t) ≡ T − V , (2.7)

where N = 3n for n particles in three dimensional space. The action S for this system isnow given by

S[q1, . . . , qN , q1, . . . , qN ] =

t2∫t1

L (q1, . . . , qN , q1, . . . , qN ; t) dt , (2.8)

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2.2. THE EULER-LAGRANGE EQUATION 15

and the Euler-Lagrange equations become

d

dt

(∂L

∂qi

)− ∂L

∂qi= 0 i = 1, . . . , N . (2.9)

Here, we observe that the number of equations is the same as was the case for Newton’sequation, e.g., for one particle in three dimension, Newton’s equation is a vector equationthat can be written in terms of components of the vectors. This would give three equationsfor the three components. In the case of the Euler-Lagrange equation, we would haveN = 3, and therefore three equations as well.

m

q

q=0

k

Figure 2.2: Massm connected to a spring with spring constant k. The equilibrium positionof the mass is q = 0.

To illustrate the form of these equations for a familiar system, consider the problemof a mass at the end of a spring. This is a one dimensional system with the kinetic energygiven by

T =1

2mq2 ,

while the potential energy is given by

V =1

2mw2q2 =

1

2kq2 ,

with the frequency w =√

km

. Here k is the spring constant. The Lagrangian for thissystem can now be written as

L = T − V =1

2mq2 − 1

2kq2.

We now have that∂L

∂q= mq ,

andd

dt

(∂L

∂q

)= m

dq

dt≡ mq ,

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16 CHAPTER 2. CLASSICAL MECHANICS - A REVIEW

while∂L

∂q= −kq.

Thus the Euler-Lagrange equation, Eq. (2.6), is given by

mq + kq = 0 .

This is the equation of motion for a simple harmonic motion, and is identical to the resultwe get from Newton’s equation in one dimension, i.e., F = ma.

Having established the fact that the Euler-Lagrange approach, based on Hamilton’sprinciple of minimum action, is identical to Newton’s second Law, the question is; whatare the advantages and disadvantages of the two methods? The main advantage of theEuler-Lagrange method is that we can add any number of constraints to the motionwithout changing the formalism. All we need to do is minimize the action subject tothe constraint. In addition, the approach can be extended to electrodynamics providingus with a classical theory of fields. In particular, Maxwell’s equations turn out to bethe Euler-Lagrange equations for the electromagnetic field. Finally, one can quantize thetheory based on the action using the Feynman path integral approach[13]. The maindisadvantage of the Euler-Lagrange method is it’s failure, in it’s present form, when thereare dissipative forces present, e.g. friction.

2.3 Hamiltonian Mechanics

In the above formulation of mechanics we have concentrated on the coordinate and velocityof the particle. In quantum mechanics it is advantageous to work with the coordinateand momentum. The Lagrangian, as we have seen above, is a function of the position,q, and velocity, q. To construct a function that will replace the Lagrangian and is afunction of position and momentum, we need to carry out a Legendre transformation onthe Lagrangian L(q, q).We first define the canonical momentum p as

p ≡ ∂L

∂q. (2.10)

Then the Hamiltonian H(p, q), which is a function of the position q and the canonicalmomentum p, is defined in terms of the Lagrangian by the Legendre transformation

H(p, q) = q∂L

∂q− L(q, q)

= qp− L(q, q) . (2.11)

Assuming the velocity q is a function of the position q and the momentum p, we can write

dH(p, q) = qdp+ p

(∂q

∂qdq +

∂q

∂pdp

)− ∂L

∂qdq − ∂L

∂q

(∂q

∂qdq +

∂q

∂pdp

)

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2.3. HAMILTONIAN MECHANICS 17

= qdp− ∂L

∂qdq +

(p− ∂L

∂q

) (∂q

∂qdq +

∂q

∂pdp

)

Making use of the definition of the canonical momentum, Eq. (2.10), we can write

dH(p, q) = qdp− ∂L

∂qdq

= qdp− pdq , (2.12)

since

p =d

dt

(∂L

∂q

)=∂L

∂q,

where we have made use of the Euler-Lagrange equation, Eq. (2.6). Since the HamiltonianH is a function of the position q and momentum p, we can write

dH(p, q) =∂H

∂pdp+

∂H

∂qdq . (2.13)

Comparing Eqs. (2.12) and Eq. (2.13), we can write the equation of motion in terms ofthe Hamiltonian H as,

q =∂H

∂pand p = −∂H

∂q. (2.14)

These two equations are known as Hamilton’s equations of motion.

For microscopic systems, the underlying forces are non-dissipative, and the potentialenergy is a function of the position only. In this case the Lagrangian is given by

L =1

2mq2 − V (q) , (2.15)

and the canonical momentum p is given as

p =∂L

∂q= mq , (2.16)

and the Hamiltonian is given by

H =p2

2m+ V (q) . (2.17)

In this case the Hamiltonian is the sum of the kinetic energy p2/2m and the potential en-ergy V (q). We will use this form for the Hamiltonian in conjunction with the quantizationrules to discuss the corresponding quantum mechanical systems.

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18 CHAPTER 2. CLASSICAL MECHANICS - A REVIEW

2.4 Symmetry and Conservation Laws

Newton, in his formulation of mechanics, introduced three laws. Only the second law givesthe equation of motion for the system. The first and third laws are basically statementsabout the symmetry of space in which the system is placed. Having derived the equationof motion, i.e. the Euler-Lagrange equations, from Hamilton’s principle, we will turnour attention to the conservation laws. In particular we will show that the standardconservation laws are a direct result of the symmetries of the system and the space inwhich it is placed. Here, we will consider only the simplest of these space-time symmetries,and the corresponding conservation laws.

2.4.1 Homogeneity of time

What do we mean by homogeneity in time? If we consider an experiment being performedtoday, and then again in a week’s time, then for the results of the experiment to bemeaningful, we should get the same results in both cases. This means that the resultsof the experiments do not depend on the time they were performed. If they did, itwould make very little sense to perform the experiments. In making this statement, weshould also make sure the conditions under which the experiment were performed havenot changed - at least those conditions that can have influence on the final result of theexperiments. These conditions can be stated mathematically, by making the statementthat the equations of motion do not change under translation in time.

If our Lagrangian has no explicit time dependence, i.e. the potential energy or theforces in the system are time independent, then ∂L

∂t= 0, and the time derivative of the

Lagrangian is given by

dL

dt=

∑i

∂L

∂qi

dqidt

+∑i

∂L

∂qi

dqidt

=∑i

∂L

∂qiqi +

∑i

∂L

∂qiqi .

Making use of the Euler-Lagrange equations Eq. (2.9), we get

dL

dt=

∑i

d

dt

(∂L

∂qi

)qi +

∑i

∂L

∂qiqi

=d

dt

[∑i

(∂L

∂qi

)qi

],

ord

dt

[∑i

(∂L

∂qi

)qi − L

]= 0 . (2.18)

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2.4. SYMMETRY AND CONSERVATION LAWS 19

The quantity in the square bracket is nothing but the Hamiltonian which is the sum ofthe kinetic and potential energy of the system. Making use of Eq. (2.11) we have

dH(p, q)

dt= 0 ,

i.e. the total energy of the particles does not change with time. Thus, we have found thatfor systems where the Lagrangian has no explicit time dependence, the total energy ofthe system does not change with time. We thus can conclude that: Homogeneity of time,which is a symmetry of space-time, leads to the conservation of energy.

2.4.2 Homogeneity of space

The homogeneity of space means that if we move our experiment from one corner of aroom to the next corner, the properties of the space in which we are doing the experimentdo not change. Mathematically this means that if we displace a system at ~q by an amount~ε, i.e.,

~q → ~q + ~ε ,

then the Lagrangian does not change, i.e.

δL = 0 .

For a system consisting of one particle in three-dimensions, we have in rectangular coor-dinates

δL =3∑i=1

∂L

∂qiδqi =

3∑i=1

∂L

∂qiεi ,

where εi, (i = 1, 2, 3) are the components of the vector ~ε. Making use of the Euler-Lagrange equations and the fact that δqi = εi, we can write the above change in theLagrangian as,

δL =3∑i=1

d

dt

(∂L

∂qi

)δqi

=d

dt

(3∑i=1

pi εi

)= 0 .

In writing the second line we have made use of the fact that the εi’s i = 1, 2, 3 are not afunction of time t. Since the εi are arbitrary, the above result means that the componentsof the particles momenta are constant, i.e.

~p = constant .

For space to be homogeneous, we should have no external forces acting on the system.For a system consisting of one particle, this homogeneity of space gives us the result that

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20 CHAPTER 2. CLASSICAL MECHANICS - A REVIEW

the momentum of the particle is constant, i.e. does not change with time. This is nothingbut Newton’s first law. Thus Newton’s first law is a statement that space is homogeneousand has nothing to do with the properties of the system itself.

Let us now consider a system of two particles. For simplicity, let us first consider theproblem in one dimension. Then qi , i = 1, 2, labels the position of the two particles. Thetransformation of translation of the total system by a distance ε is now given by

qi → qi + ε (i = 1, 2)

where both particles are displaced by the same amount ε, i.e., the whole apparatus forthe experiment has been moved by ε. The change in the Lagrangian is then,

δL = ε2∑i=1

∂L

∂qi.

Making use of the equation of motion, i.e. the Euler-Lagrange equation, we get

δL = ε2∑i=1

d

dt

(∂L

∂qi

)

= εd

dt

(2∑i=1

∂L

∂qi

)= 0 . (2.19)

Making use of the definition of momentum Eq. (2.10), and the fact that ε 6= 0 and is aconstant, we can write this equation as,

d

dt(p1 + p2) = 0

and therefore,p1 + p2 = constant .

In general, for n-particles in three dimension, we have

n∑i=1

~pi = constant , (2.20)

i.e. the total momentum of the system is a constant of the motion if the space in whichthe system is placed is homogeneous. An alternative way of examining this result is tomake use of the fact that the kinetic energy is a function of the velocity qi only. In thiscase

∂L

∂qi= −∂V

∂qi= Fi ,

where Fi is the force on the ithe particle. Using this result, we can write Eq. (2.19) forthe case when δL = 0 as

~F1 + ~F2 = 0 .

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2.5. PROBLEMS 21

This is a statement of Newton’s third Law, i.e., for every action there is an equal andopposite reaction.

In the above analysis we have examined two properties of space-time, and the corre-sponding conservation laws. In fact within the framework of the special theory of rela-tivity, time and space combine to form a four dimensional space where the homogeneityof space-time gives the conservation of the total four-momentum of the system. This isidentical to the conservation of energy and momentum.

In addition to the homogeneity of space, we can also accept the isotropy of space,i.e. the orientation of the apparatus in the room does not effect the final result of theexperiment. This isotropy of space gives rise to the conservation of the total angularmomentum of the system.

We will come back to a more detailed discussion of the relation between symmetriesand the corresponding conservation laws in quantum systems in a later chapter.

2.5 Problems

1. Consider a particle of mass m moving along the x-axis under the influence of anexternal force

F = F0 sinωt .

(a) Integrate the equation of motion to determine the position of the particle atany time t, given the initial position is given to be x = 0 and the initial velocityto be v0.

(b) Use MAPLE or Mathematica to plot the position of the particle as a functionof time for 0 < t < 50 sec. given m = 0.05 Kg. and v0 = 3 m/sec. Takeω = π/2 and F0 = 2 N.

2. Consider a particle of mass m in an external potential that is proportional to theposition, i.e., V = Aq, where A is a constant.

(a) Write the Lagrangian for this system.

(b) What is the equation of motion, i.e. the Euler-Lagrange equation for thissystem?

(c) Integrate the equation of motion to determine the position of the particle as afunction of time.

(d) Write the Hamiltonian for this system.

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22 CHAPTER 2. CLASSICAL MECHANICS - A REVIEW

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Chapter 3

Wave Packet and the UncertaintyPrinciple

Abstract : We devote this chapter to the concept of the superposition ofwaves to construct wave packets that satisfy an uncertainty principle andthe Schrodinger equation.

In Chapter 1, we discussed a number of experimental observations that raised ques-tions regarding the validity of classical Newtonian mechanics in the realm of microscopicphysics. The general conclusions we drew from these observations were that particlessometimes behave like waves, while radiation, that commonly exhibits a wave behavior,can transfer energy and momentum like a particle. In an attempt to understand this dualbehavior of matter and radiation, let us attempt to form a localized object like a particle,out of a superposition of waves.

3.1 Superposition of Waves

To examine the wave nature of matter, let us consider a particle of momentum p andenergy E. In accordance with de Broglie’s postulate, the de Broglie wave length λ forthis particle is related to its momentum p by the postulate

p =h

λ≡ hk where h ≡ h

2π, (3.1)

and the energy of the particle is given in terms of the frequency by

E = hω with ω = 2πν ,

23

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24 CHAPTER 3. WAVE PACKET AND THE UNCERTAINTY PRINCIPLE

where ν is the frequency1. We now can write a wave in one dimension in one of threeforms:

sin(kx− ωt) , cos(kx− ωt) or ei(kx−ωt) .

In all three cases the wave has a unit amplitude and is spread over all space. This canhardly be considered a particle, since we conceive a particle as something localized inspace. Let us now take the sum of two waves, i.e.,

Ψ(x, t) = Ψ1(x, t) + Ψ2(x, t) ,

where Ψ1 and Ψ2 are two waves with different wave numbers and frequencies. Using thesin(kx− ωt) representation for the waves in one dimension, we get,2

Ψ(x, t) = sin(k1x− ω2t) + sin(k2x− ω2t)

= 2 cos

(k1 − k2

2x− ω1 − ω2

2t

)sin

(k1 + k2

2x− ω1 + ω2

2t

).

We now define the average wave number and average frequency as

k =1

2(k1 + k2) and ω =

1

2(ω1 + ω2) , (3.2)

while the relative wave number and relative frequency are defined as

∆k =1

2(k1 − k2) and ∆ω =

1

2(ω1 − ω2) . (3.3)

This allows us to write the superposition of two waves as

Ψ(x, t) = 2 cos(∆k x−∆ω t) sin(kx− ωt) . (3.4)

Since in general, ∆k k and ∆ω ω, we have in the above expression for the sum oftwo waves, a product of a wave with a short wave length λS = 2π/k, and one with a longwave length λL = 2π/∆k. Furthermore, the wave with the longer wave length modulates(or envelopes) the wave with the shorter wave length. The velocity of the two waves are

vp =ω

k(phase velocity) , (3.5)

and

vg =∆ω

∆k(group velocity) . (3.6)

1In these lectures we always work with ω and refer to it as frequency.2Note that

sinA+ sinB = 2 cosA−B

2sin

A+B

2.

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3.1. SUPERPOSITION OF WAVES 25

20 40 60 80 100 120x

-1

-0.5

0.5

1

Ψ( )x

20 40 60 80 100 120x

-2

-1

1

2Ψ( )x

0

Figure 3.1: The plot on the left is of the wave sin(kx− ωt) as a function of x for fixed t.On the right, we have the sum of two waves of equal amplitude but of a different wavenumber and frequency.

This superposition of the two waves allows us to construct a function with an amplitudethat is suppressed at some points in space in comparison to other points (see plot on theright in Figure 3.1). This suggests that we might be able to localize our wave by theprocess of a superposition of a number of waves with different wave numbers. In otherwords, we can write a wave that is localized in space as

Ψ(x, t) =∫dk g(k) ei(kx−ωt) . (3.7)

In writing Eq. (3.7), we have made use of the ei(kx−ωt) representation of a wave ratherthan the sin(kx − ωt) representation. This, we will see, allows us to make use of toolsdeveloped over a century ago by the French mathematician Fourier (1768-1830)[14] forthe analysis of functions of the form presented in Eq. (3.7).

Before we proceed further with our discussion of how we are to choose the functiong(k) to get a description of a particle as a localized wave in space, let us re-examinethe physical meaning of the phase and group velocity as defined in Eqs. (3.5) and (3.6)respectively. We know that for a wave, the velocity is given by the ratio of the frequencydivided by the wave number, i.e.

v =ω

k.

Making use of this definition of velocity of a wave we can write the phase velocity as

vp =ω

k=hω

hk=E

p.

On the other hand, for a particle of mass m, the energy and momentum of the particle aregiven by E = 1

2mv2 and p = mv. If we now assume that the energy and the momentum

of the particle are E and and p, then we have

vp =E

p=

12mv2

mv=v

2.

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26 CHAPTER 3. WAVE PACKET AND THE UNCERTAINTY PRINCIPLE

In a similar manner we can write the group velocity as

vg =∆ω

∆k=h∆ω

h∆k≈ dE

dp= v .

The above results suggest, very strongly, that the group velocity of the wave is in fact thevelocity of the particle.

3.2 Fourier Integrals

To get a better understanding of the mathematics required to construct an object which,on the one hand is the sum of waves, and on the other hand a quantity that is localizedin space, we need to study integrals of the form given in Eq. (3.7). This integral is oftenreferred to as the integral representation of the function Ψ(x, t). This subject comes underthe heading of Fourier series and Fourier integrals in mathematics. Here I will introducesome of the basic ideas and results we need in order to examine the integral in Eq. (3.7),and in this way determine what functions we can take for g(k) in order to make Ψ(x, t)localized in space.

Consider the periodic function f(x) with a period 2a, i.e.

f(x+ 2a) = f(x) . (3.8)

Because of the periodic nature of this function, we can write it as a superposition of sin kxand cos kx since these functions are also periodic, i.e.

f(x) =∑n

An cos knx+∑n

Bn sin knx . (3.9)

To guarantee the aperiodicity of f(x), i.e. f(x) = f(x+ 2a) we should have

cos knx = cos kn(x+ 2a) and sin knx = sin kn(x+ 2a) .

To satisfy these two conditions, we need to take 2akn = 2πn or kn = nπa

, with (n =0, 1, . . .). As a result when x → x + 2a, then knx → knx + 2kna = knx + 2nπ, andtherefore the value of the function f(x) remains unchanged.

Making use of the fact that

cos knx =1

2

(eiknx + e−iknx

)( 3.10a)

and

sin knx =1

2i

(eiknx − e−iknx

)( 3.10b)

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3.2. FOURIER INTEGRALS 27

we can rewrite the function f(x) as a sum over exponentials of the form

f(x) =∞∑n=0

1

2(An − iBn) eiknx +

∞∑n=0

1

2(An + iBn) e−iknx

=+∞∑

n=−∞Cne

inπx/a . (3.11)

In this way we can write any periodic function of period 2a in terms of the sum overeinπx/a or cos(nπx/a) and sin(nπx/a) with n taking all integer values between −∞ and+∞. Making use of the orthonormality relation3

1

2a

+a∫−a

dx eiπxa

(n−m) = δnm =

1 n = m0 n 6= m

, (3.12)

we can determine the coefficients of the expansion Cn, in Eq. (3.11). To achieve this wemultiply both sides of Eq. (3.11) by e−imπx/a, with m an integer, and integrate over xbetween −a and +a. This gives, using the orthogonality in Eq. (3.12), the coefficient Cmto be

Cm =1

2a

+a∫−a

dx f(x) e−imπx/a . (3.13)

This result allows us to write any periodic function, f(x), of period 2a, as an infinitesum of exponentials of the form eiknx, with kn = nπ

awhere n is an integer. The series in

Eq. (3.11) is known as the Fourier series for the function f(x), and the coefficients Cnare the Fourier coefficients.

In the problem we are considering, kn will play the role of the wave number, or hknthe momentum. Since the momentum of a particle is a continuous variable, we need totake the limit of the above results for the Fourier series, as the sum over n is replaced byan integral over k. Since

kn =nπ

aand ∆k = kn+1 − kn =

π

a

kn can be a continuous variable in the limit as a→∞. This in turn will make the resultvalid for any function f(x), which is exactly what we want. This can be achieved byintroducing ∆n = 1 in Eq. (3.11), see Figure 3.2, so that

f(x) =∑n

Cn ∆n einπx/a

=a

π

∑n

Cnπ∆n

aeinπx/a .

3We have made use of the fact that∫dx eiax =

∫dx cos ax+ i

∫dx sin ax

to write Eq. (3.12).

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28 CHAPTER 3. WAVE PACKET AND THE UNCERTAINTY PRINCIPLE

n0 1 2-1-2

∆n

Figure 3.2: The n-axis which we convert, in the limit a→∞ to the k-axis.

Since kn = πna

, we can define ∆k = π∆na

and the limit a → ∞ gives ∆k → dk. Wenow rewrite f(x) as:

f(x) =∑k

g(k)√2π

eikx ∆k

where the function g(k) is defined by the relation,

g(k)√2π

=aCnπ

.

We should note that in writing the above expression we have converted∑n to

∑k, and

this is the first step in converting the sum into an integral when a→∞. We are now ina position to take the limit as a→∞, in which case ∆k → dk in the above sum. In thisway we have converted the series expansion for the periodic function f(x) to an integralfor a non-periodic function in the form

f(x) =1√2π

+∞∫−∞

dk g(k) eikx . (3.14)

This is known as the Fourier decomposition of the non-periodic function f(x). The func-tion g(k) can now be written using Eq. (3.13) as

g(k) =√

2π(aCnπ

)=

1√2π

+a∫−a

dx f(x) e−i(nπ/a)x

→ 1√2π

+∞∫−∞

dx f(x) e−ikx for a→∞ . (3.15)

In Eq. (3.15) we have the Fourier decomposition of g(k) in terms of the function f(x). Inother words, if we consider Eq. (3.14) as the Fourier transform, then Eq. (3.15) may beconsidered the inverse transform. By substituting the result of Eq. (3.15) into Eq. (3.14),we get

f(x) =1

+∞∫−∞

dk eikx+∞∫−∞

dx′e−ikx′f(x′)

=

+∞∫−∞

dx′f(x′)

1

+∞∫−∞

dk eik(x−x′)

. (3.16)

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3.2. FOURIER INTEGRALS 29

This result, if it is to be valid, suggests that the quantity in the curly bracket is zeroexcept when x = x′ in which case it is one. We now define the Dirac δ-function, in onedimension, as

δ(x− x′) ≡ 1

+∞∫−∞

dk eik(x−x′) . (3.17)

This function, often called a distribution rather than a function, plays a central role inQuantum Mechanics. It was first introduced by Dirac for this specific purpose, and hasthe property that

+∞∫−∞

dx′ δ(x− x′)f(x′) = f(x) , (3.18)

In other words δ(x − x′) 6= 0 for x = x′, otherwise it is zero. We can write our Diracδ-function as a limit of the form

δ(x) = limL→∞

1

+L∫−L

dk eikx

= limL→∞

1

eixL − e−ixL

ix

= limL→∞

sinLx

πx. (3.19)

In fact, if we take any function that has a peaked and normalize it so that the area underthe peak equal to one, then the Dirac δ-function can be defined as the limit when thewidth of the peak tends to zero while the area under the peak stays fixed at one, e.g.

δ(x) = limα→0

1

π

α

x2 + α2, ( 3.20a)

andδ(x) = lim

α→∞

α√πe−α

2x2

. ( 3.20b)

A property of the Dirac δ-function that is very useful is

+∞∫−∞

dx f(x) δ(g(x)) =∑i

f(xi)

|g′(xi)|, ( 3.20c)

where xi are the positions of the zeros of the function g(x) and

g′(xi) =dg(x)

dx

∣∣∣x=xi

. (3.21)

Other properties of the Dirac δ-function will be introduced as we need them.

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30 CHAPTER 3. WAVE PACKET AND THE UNCERTAINTY PRINCIPLE

We thus have established the procedure for determining the function g(k) given the factthat we want a function f(x) that is localized in space. In particular we have introducedthe Fourier integral representation for the function f(x) given by

f(x) =1√2π

+∞∫−∞

dk g(k) eikx ( 3.22a)

with the inverse transform giving the function g(k) in terms of f(x) as

g(k) =1√2π

+∞∫−∞

dx f(x) e−ikx . ( 3.22b)

We note here, the symmetry between the Fourier transform and its inverse. At this stagethis symmetry was introduced by construction. However, when we proceed to use itin Quantum Mechanics, we will find that it is natural to have this symmetry betweenthe Fourier transform and its inverse if we are going to give a physical meaning to thistransform.

3.3 The Uncertainty Principle

With the above result in hand, let us consider the following integral

ψ(x) =1√2π

+∞∫−∞

dk φ(k) eikx

as a possible function that is localized in space, which could represent the wave natureof a particle. To get the localization in space, we need to carefully select our weightingfunction φ(k). We know from the results of Sec. 3.2 on Fourier transforms, that thefunction φ(k) is the inverse Fourier transform and is given by

φ(k) =1√2π

+∞∫−∞

dx ψ(x) e−ikx .

If we want to consider ψ(x) to be the amplitude of a ‘wave’ that represents a localized par-ticle at x = x0, then one possible choice for ψ(x) is the Gaussian function (see Figure 3.3),i.e.,

ψ(x) = N e−(x−x0)2/4σ2

, (3.23)

where N is a normalization factor. As we will see, the width of the peak in the functionis related to σ. Since we want ψ(x) to be the amplitude of the wave, then the intensity of

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3.3. THE UNCERTAINTY PRINCIPLE 31

the wave at point x will be given by |ψ(x)|2. Since the wave is going to represent a singleparticle, we need to choose our normalization N such that the integral of the intensityover all space is one, i.e.,

+∞∫−∞

dx |ψ(x)|2 = 1 .

1 2 3 4

0.2

0.4

0.6

0.8

1

x

Figure 3.3: This is a plot of the function e−(x−x0)2for x0 = 2, and corresponds to a

function peaked at x = 2 with a maximum height of one.

This gives us a value for the normalization constant N of 4

N2 =1

σ√

2π. (3.24)

The intensity of the wave that is localized at x = x0 is then given by

|ψ(x)|2 =1

σ√

2πe−(x−x0)2/2σ2

. (3.25)

We now define the width of the peak in |ψ(x)|2 by first determining the value of x1 forwhich

|ψ(x1)|2 = |ψ(x0)|2 e−1 ,

where x0 is the point at which |ψ(x)|2 is maximum. In this case, this corresponds to

(x1 − x0)2 = 2σ2 or |x1 − x0| =√

2 σ .

4We have made use of the Gaussian integral

+∞∫−∞

du e−u2

=√π .

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32 CHAPTER 3. WAVE PACKET AND THE UNCERTAINTY PRINCIPLE

We now define the width to be5

∆x = 2|x1 − x0| = 2√

2 σ . (3.26)

This can be considered as the uncertainty in the position of the particle about the valuex0.

We now make use of the inverse Fourier transform to determine φ(k) in terms of ψ(x)to be

φ(k) =1√2π

+∞∫−∞

dx ψ(x) e−ikx

=N√2π

+∞∫−∞

dx e−(x−x0)2/4σ2

e−ikx

To simplify this integral, we change the variable of integration to ξ = x−x0 or x = ξ+x0.We then have dξ = dx so that φ(k) can be written as

φ(k) =N√2π

e−ikx0

+∞∫−∞

dξ e−(ξ2

4σ2 +ikξ

).

To evaluate this integral, we need to recast it into an integral over a Gaussian function.This can be achieved by completing the square in the exponent under the integral sign.This gives

φ(k) =N√2π

e−ikx0 e−σ2k2

+∞∫−∞

dξ e−( ξ2σ

+iσk)2

=N√2π

e−ikx0 e−σ2k2

+∞∫−∞

du e−u2

=√

2N σe−ikx0 e−σ2k2

, (3.27)

where in the second line we have again changed the variable of integration such thatu = ξ

2σ+ iσk and then dξ = 2σdu. We now can write the magnitude of φ(k) as

|φ(k)|2 = 2N2 σ2 e−2σ2k2

=

√2

πσ e−2σ2k2

. (3.28)

5This definition of the width is chosen such that:√

2σ∫−√

dx e−x2/2σ2

= Erf(1) = 0.842701 ,

and independent of σ.

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3.3. THE UNCERTAINTY PRINCIPLE 33

In the next chapter we will come back to a discussion of the function φ(k) and we willbe able to give it a physical meaning similar to that of ψ(x). We now can determine thewidth of the Gaussian given by |φ(k)|2 using the same definition of width used to extractthe width of |ψ(x)|2 above. This is given by

∆k =2√2 σ

. (3.29)

This width can be considered as an uncertainty in the wave number k. If we combine theresults of Eqs. (3.26) and (3.29) for ∆x and ∆k, we get

∆k ∆x = 4 > 1

or after multiplication by h we have

∆p ∆x > h , (3.30)

which is a statement of the Heisenberg uncertainty relation. We have thus found that ifwe want to localize the position of a particle to an uncertainty ∆x, we have to take thesum of waves with different wave numbers ( i.e. momenta), to get this localization inspace. As a result of this superposition, we have an uncertainty in the momentum of theparticle of magnitude ∆p with ∆x∆p > h.

In Figure 3.4 we have the plot of |ψ(x)|2 and |φ(k)|2 as given in Eqs. (3.25) and (3.28).We observe that both functions have a Gaussian form, and as we reduce the width of thepeak in ψ(x) we increase the width of the peak in φ(k). (For a more detailed examinationof how ψ(x) and φ(k) depend on σ, see the problems at the end of this chapter).

From the above analysis we come to the conclusion that if we treat the particle asthe superposition of simple waves, then the momentum of the particle is uncertain. Infact, the harder we work to localize the position of the particle, the more uncertain itsmomentum becomes. This is not consistent with classical physics, where a particle bydefinition is localized in space and has a definite momentum at any instant of time. Thusthe concept of associating a wave property with a particle has put us at odds with theclassical concept of a particle being localized in space and having a definite momentum.This suggests that the experimental observation of the wave nature of matter is going toput us in contradiction with classical Newtonian mechanics.

So far we have considered the superposition of stationary (time independent) wavesto form our localized wave packet. To see how such a wave packet propagates with time,we need to consider the superposition of traveling waves, e.g., for waves traveling alongthe positive x-axis, we use the basic wave

ei(kx−ωt) .

To study the propagation of such a wave we need to know the relationship between theangular frequency ω and the wave number k, e.g., for electromagnetic waves, we have

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34 CHAPTER 3. WAVE PACKET AND THE UNCERTAINTY PRINCIPLE

-2 0 2 4 6x

0.4

0.8

| (x)|ψ 2

-2 -1 0 1 2k

0.4

0.8

| (k)|φ 2

σ = 0.8

-2 0 2 4 6x

0.4

0.8

| (x)|ψ 2

-2 -1 0 1 2k

0.4

0.8

| (k)|φ 2

σ = 0.6

-2 0 2 4 6x

0.4

0.8

| (x)|ψ 2

-2 -1 0 1 2k

0.4

0.8

| (k)|φ 2

σ = 1.0

Figure 3.4: Here we have a plot of |ψ(x)|2 as a function of x, and the corresponding|φ(k)|2 as a function of k. We have taken x0 = 2 and σ = 1.0, 0.8, 0.6.

E = hω = pc = hkc and thus ω = kc. This allows us to describe an electromagnetic wave(e.g. light) as

eik(x−ct)

where c is the velocity of the electromagnetic wave, i.e., the velocity of light. In this caseour wave packet takes the form

Ψ(x, t) =1√2π

+∞∫−∞

dk φ(k)eik(x−ct)

= Ψ(x− ct) . (3.31)

Because the wave packet is a function of (x − ct), it travels along the x-axis with time

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3.3. THE UNCERTAINTY PRINCIPLE 35

without changing its form. To illustrate this, consider the case when the wave Ψ(x, t)has its maximum, f0, at x = 0 when t = 0, i.e. Ψ(0, 0) = f0. At a later time t > 0,the peak of the wave packet will have moved to x. Since Ψ(x, t) is a function of (x− ct),then Ψ(x, t) = Ψ(0, 0) = f0, if x = ct. This means that the peak of the wave travels withvelocity of light c, and the shape of the wave packet has not changed with time.

For particles of mass m, the relation between ω and k is more complicated. Forexample,

E = hω =

√p2c2 +m2c4 for relativistic particles

p2

2m = h2k2

2m for non-relativistic particles

,

where c is the velocity of light. In general we can assume the frequency ω, to be a functionof the wave number k, and we can write our wave packet as

Ψ(x, t) =1√2π

+∞∫−∞

dk φ(k) ei(kx−ω(k)t) . (3.32)

Supposing φ(k) is peaked sharply about k = k0, then the major contribution to the integralcomes from values of k close to k0. This allows us to expand ω(k) in the integrand in aTaylor series about k = k0. If we retain all terms up to quadratic terms in (k − k0), wecan write the frequency ω(k) as a quadratic in (k − k0) of the form

ω(k) ≈ ω(k0) + (k − k0)

(dω

dk

)k=k0

+1

2(k − k0)2

(d2ω

dk2

)k=k0

≡ ω0 + vg(k − k0) + β(k − k0)2 , (3.33)

where the group velocity, vg, is given by

vg =

(dω

dk

).

If we now assume that φ(k) in Eq. (3.32) is a Gaussian, i.e.

φ(k) = Nke−σ2(k−k0)2

,

and change the variables of integration in Eq. (3.32) to η = k − k0, we get

Ψ(x, t) =Nk√2π

ei(k0x−ω0t)

+∞∫−∞

dη eiηx e−σ2η2

e−ivgηte−iβη2t

=Nk√2π

ei(k0x−ω0t)

+∞∫−∞

dη e−(σ2+iβt)η2

ei(x−vgt)η .

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36 CHAPTER 3. WAVE PACKET AND THE UNCERTAINTY PRINCIPLE

Completing the squares in the exponent of the integrand, we get

Ψ(x, t) =Nk√2π

ei(k0x−ω0t)

√σ2 + iβt

exp

− (x− vgt)2

4(σ2 + iβt)2

+∞∫−∞

du e−u2

,

where

u =√σ2 + iβt η − i(x− vgt)

2√σ2 + iβt

.

Since the above integral over u is√π, we can write the intensity of our wave packet as

|Ψ(x, t)|2 =N2k

2

(1

σ4 + β2t2

) 12

exp

−σ

2(x− vgt)2

2(σ4 + β2t2)

. (3.34)

This is a wave packet whose peak travels with velocity vg which is the velocity of theparticle. However, the width of this wave increases with time. In fact at t = 0 the widthis 2√

2σ, whereas at a time t later, it has increased to

2√

(1 +

β2t2

σ4

) 12

.

Furthermore, if the wave packet is wide initially, the rate at which it spreads is smallerbecause the rate of increase in width is inversely proportional to σ, the width at timet = 0.

3.4 The Schrodinger Equation

In the last section, we showed how we can construct a function, Ψ(x, t), which can describethe time evolution of the system (particle, e.g. an electron) that is consistent with wave-particle duality. This wave-particle duality was exhibited by the observation that theobject or particle described by the function Ψ(x, t) was localized in space like a particleand its development with time was also consistent with a particle having velocity vg.On the other hand Ψ(x, t) was nothing more than the superposition of waves. The twoconsequences of satisfying the above wave-particle duality are;

1. There is a limitation on the accuracy of our knowledge of the position and momen-tum of the particle described by Ψ(x, t). This limitation is stated in the form of theuncertainty principle, Eq.(3.30), and illustrated in Figure 3.4.

2. The uncertainty in the position of the particle, i.e. the width of the wave packet,changes with time. The rate of change depends on the shape of the wave packetand the relation between the frequency ω and the wave number k.

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3.4. THE SCHRODINGER EQUATION 37

In this section, we will show that this wave packet Ψ(x, t) which describes the motionof a free particle in space, satisfies a differential equation first proposed by Schrodingerin 1926[15]. The particular form of the differential equation is a consequence of the non-relativistic nature of the relation between the energy and the momentum of the particle.Thus for a free particle, i.e., for a particle that is not under the influence of any externalforce, we have

E = hω(k) =p2

2m=h2k2

2m.

Since our wave packet is given by

Ψ(x, t) =1√2π

+∞∫−∞

dk φ(k)ei(kx−ωt) ,

we can differentiate it with respect to time to get

ih∂Ψ

∂t=

1√2π

+∞∫−∞

dk φ(k) hω(k) ei(kx−ωt) . (3.35)

On the other hand, differentiating the wave packet with respect to position x, twice, weget

− h2

2m

∂2Ψ

∂x2=

1√2π

+∞∫−∞

dk φ(k)h2k2

2mei(kx−ωt)

=1√2π

+∞∫−∞

dk φ(k) hω(k) ei(kx−ωt) . (3.36)

We therefore have, on comparing the results of Eqs. (3.35) and (3.36), that the wavepacket amplitude Ψ(x, t), satisfies the differential equation

ih∂Ψ

∂t= − h2

2m

∂2Ψ

∂x2, (3.37)

which describes the evolution of the wave packet that represents a free particle in space-time.

For a particle in an external potential V (x), the total energy E = hω is the sum of

the kinetic energy p2

2mand the potential energy V (x), i.e.,

E = hω =p2

2m+ V (x)

=h2k2

2m+ V (x) . (3.38)

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38 CHAPTER 3. WAVE PACKET AND THE UNCERTAINTY PRINCIPLE

In this case we can follow the same procedure implemented above, for a free particle, tofind that the function Ψ(x, t) satisfies the relation(

− h2

2m

∂2

∂x2+ V (x)

)Ψ(x, t) =

1√2π

+∞∫−∞

dk φ(k)

(h2k2

2m+ V (x)

)ei(kx−ωt)

=1√2π

+∞∫−∞

dk φ(k)hω ei(kx−ωt)

=1√2π

+∞∫−∞

dk φ(k)ih∂

∂tei(kx−ωt)

= ih∂

∂tΨ(x, t) .

In writing the second line in the above equation we have made use of the relation betweenthe energy and wave number as given in Eq. (3.38). We now can write the partial differ-ential equation that determines the evolution of a wave packet describing a particle in anexternal potential V (x). This equation is of the form

ih∂Ψ

∂t=

(− h2

2m

∂2

∂x2+ V (x)

)Ψ(x, t) , (3.39)

and is known as the Schrodinger equation for the wave amplitude or wave function Ψ(x, t).In deriving this equation, our starting point was de Broglie’s postulate which states

that a particle behaves like a wave with momentum p = hλ

= hk, with k being the wavenumber corresponding to the energy E = hω. In an attempt to localize the particlein coordinate space, we found it necessary to take a linear superposition of waves withdifferent wave numbers, each wave having a weighting determined by φ(k). This procedureled us to the uncertainty relationship between the position and the momentum of theparticle, i.e.,

∆p ∆x > h .

We then made use of the relationship between the energy E = hω and the momentump = hk (Eq. (3.38)) to derive an equation (a partial differential equation) that describesthe properties of the wave amplitude Ψ(x, t) in space-time. This equation is known as theSchrodinger equation.

From classical mechanics we know that the Hamiltonian for the system is given by(see Eq. (2.17))

H =p2

2m+ V (x)

and is equal to the total energy E, i.e.

E = H(p, x) .

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3.4. THE SCHRODINGER EQUATION 39

One procedure to quantize a system described by the above classical Hamiltonian is tocarry out the following substitution for the energy and momentum

E → ih∂

∂t, ( 3.40a)

p → −ih ∂∂x

. ( 3.40b)

In this way, the classical Hamiltonian is converted to a differential operator which wedefine as H, i.e.,

H(p, x)→ − h2

2m

∂2

∂x2+ V (x) ≡ H ( 3.40c)

The above substitutions may be considered the general rule for deriving the quantummechanical equivalent to a classical system described by the Hamiltonian H(p, x). TheSchrodinger equation can now be written as

ih∂Ψ

∂t= H Ψ , (3.41)

where H is referred to as the quantum mechanical Hamiltonian for the system.So far we have considered one space dimension. In three dimensions, the same substi-

tution is used to go from classical to quantum mechanics. However, now the momentumis a vector, and therefore

~p→ −ih~∇ ,

where ~∇, the gradient operator, is defined in rectangular coordinates as

~∇ = ı∂

∂x+

∂y+ k

∂z.

Here ı, , and k are unit vectors along the x, y and z axis. The quantum Hamiltonian fora particle in a potential V (~r) is then given by

H = − h2

2m∇2 + V (~r) , (3.42)

where ∇2, the Laplacian, is given in rectangular coordinates as

∇2 = ~∇ · ~∇ =∂2

∂x2+

∂2

∂y2+

∂2

∂z2.

The Schrodinger equation in this case is still given by Eq. (3.41) with the Hamiltoniandefined by Eq. (3.42).

In the above discussion, we have shown that the amplitude of the wave packet, resultingfrom the need to describe a particle’s motion in terms of a wave, satisfies the Schrodinger

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40 CHAPTER 3. WAVE PACKET AND THE UNCERTAINTY PRINCIPLE

equation, and is consistent with Heisenberg’s uncertainty relation. However, it should bepointed out that the procedure followed is by no means a derivation of the Schrodingerequation. On the other hand, if we postulate the quantization rules given in Eqs. (3.40a)- (3.40c), we can write the Schrodinger equation, Eq. (3.41), for any system, given theclassical Hamiltonian for that system. To demonstrate the generality of the quantizationrules we have introduced, let us consider the case of a free relativistic particle, so that therelation between the energy, i.e. the Hamiltonian, and the momentum is given by

E =√p2c2 +m2c4 .

Using the quantization rules given in Eqs. (3.40), the corresponding quantum mechanicalequation could be

ih∂Ψ

∂t=√−h2c2∇2 +m2c4 Ψ(~r, t) .

This equation is highly non-local because the differential operator ∇2 is under the squareroot sign. An alternative starting point would be to consider

E2 = p2c2 +m2c4 . (3.43)

Upon quantization, the relation between the energy and momentum gives the equation

−h2 ∂2Ψ(~r, t)

∂t2=(−h2c2∇2 +m2c4

)Ψ(~r, t) ,

or (1

c2

∂2

∂t2−∇2 +

m2c2

h2

)Ψ(~r, t) = 0 . (3.44)

This equation, known as the Klein-Gordon equation[16, 17], reduces to the wave equationfor a zero mass particle, and to the Schrodinger equation in the non-relativistic limit, i.e.pm 1. The main problem with this equation is the fact that it is second order in the

time derivative which gives rise to possible problems in the interpretation of Ψ(~r, t). Wewill come back to this point when we consider relativistic quantum mechanics.

3.5 Physical Interpretation of Ψ(~r, t)

In the above, we have referred to Ψ(x, t) as the amplitude of the wave packet. This wasthe result of the fact that we had constructed this wave packet in terms of a superpositionof waves with a wave number k and an amplitude φ(k). However, since our wave packetis to describe a particle, we need to interpret the function Ψ(x, t) in terms of properties ofparticles. In electromagnetic waves, and light in particular, the intensity is considered tobe the magnitude squared of the amplitude, and is a measure of how much light is presentat a given point in space and time. Since our particle, as a superposition of waves, is

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3.5. PHYSICAL INTERPRETATION OF Ψ(~R, T ) 41

not localized in space any more, we could give |Ψ(x, t)|2 an interpretation similar to theintensity of light. However, rather than using the word intensity, we make use of the wordprobability density to specify the probability of finding the particle at the point x at timet. This probability density, ρ(x, t), is given in analogy with intensity as

ρ(x, t) = |Ψ(x, t)|2 . (3.45)

In this way, we can reconcile the fact that a particle is in fact a point object, while at thesame time we can get interference behaviour by using a beam of particles, e.g. electronsin an electron microscope. With the above interpretation for Ψ(x, t), we require that

+∞∫−∞

dx |Ψ(x, t)|2 = 1 , (3.46)

i.e., the probability of finding the particle anywhere between x = −∞ and x = +∞, ata given time t, be equal to one. For this integral to be finite, we require that in onedimension

limx→±∞

Ψ(x, t)→ x−12−ε ,

where ε > 0 and infinitesimal.Although we have given a meaning to the magnitude of the function Ψ(x, t), the phase

of this function is important. In fact, to get any form of interference we need to add twofunctions, such as

Ψ(x, t) = a1Ψ1(x, t) + a2Ψ2(x, t) ,

where a1 and a2 are in general complex constants. It is clear from this result that therelative phase of Ψ1 and Ψ2, and the value of the constants a1 and a2 determine thedegree of interference. Finally, given the fact that Ψ(x, t) satisfies the partial differentialequation that is second order in x, the function Ψ(x, t) has to be a continuous functionof x.

To examine the properties of the probability function ρ(x, t), let us consider theSchrodinger equation and its complex conjugate, i.e.,

ih∂Ψ

∂t=

[− h2

2m

∂2

∂x2+ V (x)

]Ψ(x, t) , ( 3.47a)

and

− ih ∂Ψ∗

∂t=

[− h2

2m

∂2

∂x2+ V (x)

]Ψ∗(x, t) . ( 3.47b)

In writing Eq. (3.47b) we have assumed the potential V (x) to be a real function. If wenow multiply Eq. (3.47a) from the left by Ψ∗(x, t) and Eq. (3.47b) from the right byΨ(x, t) and subtract the second equation from the first equation, we get

ih∂

∂t(Ψ∗Ψ) = ih

∂ρ

∂t= − ∂

∂x

h2

2m

(Ψ∗∂Ψ

∂x− ∂Ψ∗

∂xΨ

). (3.48)

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42 CHAPTER 3. WAVE PACKET AND THE UNCERTAINTY PRINCIPLE

The result in Eq. (3.48) can be written as an equation for the conservation of probabilityby recasting it into the form of a continuity equation, i.e.,

∂ρ

∂t+∂j

∂x= 0 , (3.49)

where the probability current j(x, t) is given by

j(x, t) =h

2im

(Ψ∗∂Ψ

∂x− ∂Ψ∗

∂xΨ

). (3.50)

To see how the continuity equation, Eq. (3.49), is a statement of conservation of proba-bility, let us integrate Eq. (3.49) over all space, to get

+∞∫−∞

dx∂ρ

∂t= −

+∞∫−∞

dx∂j

∂x

or

d

dt

+∞∫−∞

dx ρ(x, t) = −j(x, t)∣∣∣x=+∞

x=−∞. (3.51)

Since Ψ(x, t)→ 0 for x→ ±∞, then j(x, t)→ 0 for x→ ±∞, and the right hand side ofEq. (3.51) is zero. We therefore have

d

dt

+∞∫−∞

dx ρ(x, t) = 0 . (3.52)

Since ρ(x, t) is the probability density, then∫ρ(x, t) dx is the probability of having a

particle at time t. This integrated probability being time independent is a statementof the fact that the number of particles in the system does not change with time, i.e.,particles are not created or destroyed, and we have a conservation of particle number orconservation of probability.

Having established that we can take Ψ(x, t) to be the probability amplitude of findingthe particle at position x at time t, we can now calculate the average position of theparticle as

〈x〉 =

+∞∫−∞

dx x ρ(x, t)

=

+∞∫−∞

dxΨ∗(x, t)xΨ(x, t) . (3.53)

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3.5. PHYSICAL INTERPRETATION OF Ψ(~R, T ) 43

In fact, making use of the fact that the average of xn is given by

〈xn〉 =

+∞∫−∞

dxΨ∗(x, t)xn Ψ(x, t) , (3.54)

we can calculate the average of any function f(x) that can be written as a power seriesin x as

〈f(x)〉 =

+∞∫−∞

dxΨ∗(x, t) f(x) Ψ(x, t) . (3.55)

What if we want to calculate the average momentum of the particle? Since we do not knowthe momentum as a function of position we can not use the above procedure. However,we could try the classical definition, i.e.,

〈p〉 = md

dt〈x〉 = m

d

dt

∞∫∞

dx ρ(x, t)x = m

∞∫∞

dx x∂ρ

∂t. (3.56)

Making use of the continuity equation, Eq.(3.49), we can write

〈p〉 = m

∞∫∞

dx x∂ρ

∂t= −m

∞∫∞

dx x∂

∂xj(x, t) . (3.57)

Integrating by parts, we get

〈p〉 = −mxj(x, t)∣∣∣+∞−∞

+m

+∞∫−∞

dx j(x, t) .

Since the amplitude Ψ(x, t) goes to zero faster than x−1 as x → ±∞, then x j(x, t) → 0as x→ ±∞ and we have

〈p〉 = m

+∞∫−∞

dx j(x, t) =h

2i

+∞∫−∞

dx

(Ψ∗∂Ψ

∂x− ∂Ψ∗

∂xΨ

)

=

+∞∫−∞

dx Ψ∗(x, t)

(−ih ∂

∂x

)Ψ(x, t) . (3.58)

To get the last line of Eq. (3.58), we have integrated by parts and used the fact thatΨ(x, t) → 0 as x → ±∞. Thus, to calculate the average momentum using the wavefunction Ψ(x, t) we have to write the momentum p in terms of the coordinate x as

p = −ih ∂∂x

, (3.59)

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44 CHAPTER 3. WAVE PACKET AND THE UNCERTAINTY PRINCIPLE

theˆon top of the p indicates that it represents the momentum as a differential operator.In a similar manner, we can show that the average of pn is given by

〈pn〉 =

+∞∫−∞

dx Ψ∗(x, t)

(−ih ∂

∂x

)nΨ(x, t) , (3.60)

and the average of any function of momentum, f(p), that can be written as a power seriesin p is given by

〈f(p)〉 =

+∞∫−∞

dx Ψ∗(x, t) f

(−ih ∂

∂x

)Ψ(x, t) . (3.61)

Here, we can raise the question of what is meant by the function f(−ih ∂

∂x

). In general,

any analytic function f(p) can be written as

f(p) = a0 + a1p+ a2p2 + · · ·

with the ai, i = 0, 1, · · · as complex constants. This allows us to define any function ofan operator in terms of a power series in the operator. e.g., a function of the momentumoperator is given by the power series

f

(−ih ∂

∂x

)= a0 + a1

(−ih ∂

∂x

)+ a2

(−ih ∂

∂x

)2

+ · · · . (3.62)

This is a well defined operator for which we can calculate the above integral with theexpectation that the resultant series for 〈f(p)〉 can be summed to give a finite result.

An alternative way of writing the average momentum makes use of the fact that wecan write a Fourier expansion for Ψ(x, t), i.e

Ψ(x, t) =1√2π

+∞∫−∞

dk Φ(k, t) eikx ,

where

Φ(k, t) = φ(k) e−iωt .

We now can write the average momentum in terms of the function Φ(k, t) as

〈p〉 =1

+∞∫−∞

dx dk dk′ e−ikxΦ∗(k, t)

(−ih ∂

∂x

)Φ(k′, t) eik

′x

=1

+∞∫−∞

dk dk′ Φ∗(k, t) hk′ Φ(k′, t)

+∞∫−∞

dx ei(k′−k)x

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3.5. PHYSICAL INTERPRETATION OF Ψ(~R, T ) 45

=

+∞∫−∞

dk dk′ Φ∗(k, t) hk′ Φ(k′, t) δ(k − k′)

=

+∞∫−∞

dk Φ∗(k, t) hk Φ(k, t)

=

+∞∫−∞

dk |Φ(k, t)|2 hk

≡+∞∫−∞

dk ρ(k, t) hk . (3.63)

This suggests that we can consider Φ(k, t) to be the probability amplitude of finding theparticle with momentum p = hk at time t, and ρ(k, t) = |Φ(k, t)|2 as the correspondingprobability of finding the particle with momentum p = hk at time t. In other words,while Ψ(x, t) describes the particle in coordinate space, Φ(k, t) describes the same particlein momentum space. We will see later, when we consider problems in Atomic and SolidState Physics, that experiments often give a direct measurement of Φ(k, t) and not Ψ(x, t),and to get a physical image of what the experimental measurement gives, we will have toFourier transform the experimental results ( or data ) from momentum space to coordinatespace. From this point on we will refer to Ψ(x, t) as the wave function in coordinate space,and Φ(k, t) as the wave function in momentum space.

So far we have established that in coordinate space, the average position and theaverage momentum of the particle are given by integrals involving the function Ψ(x, t),and by the operators corresponding to the position and momentum of the particle, i.e.

x = x and p = −ih ∂∂x

, (3.64)

On the other hand, in momentum space, the average momentum of the particle is givenby an integral involving the wave function Φ(k, t) and the momentum operator p =p = hk. To complete the symmetry between the two spaces, we need to determine theaverage position of the particle in terms of the momentum space wave function Φ(k, t).To establish this relation, we commence with the definition of the average position asgiven in Eq. (3.53), i.e.,

〈x〉 =

+∞∫−∞

dx Ψ∗(x, t)xΨ(x, t) ,

and write the coordinate space wave function Ψ(x, t) in terms of the momentum spacewave function Φ(k, t). This allows us to write the average position as

〈x〉 =1

+∞∫−∞

dx dk dk′ Φ∗(k, t) e−ikx xΦ(k′, t) eik′x

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46 CHAPTER 3. WAVE PACKET AND THE UNCERTAINTY PRINCIPLE

=1

+∞∫−∞

dx dk dk′ Φ∗(k, t) Φ(k′, t)

(−i ∂∂k′

)ei(k

′−k)x

=

+∞∫−∞

dk dk′Φ∗(k, t) Φ(k′, t)

(−i ∂∂k′

)δ(k′ − k) .

We now integrate by parts over k′ with the result that

〈x〉 =

+∞∫−∞

dk dk′ Φ∗(k, t) δ(k′ − k)

(i∂

∂k′

)Φ(k′, t)

=

+∞∫−∞

dk Φ∗(k, t)

(ih∂

∂p

)Φ(k, t) . (3.65)

Thus in momentum space we have, for the position and momentum operators,

x = ih∂

∂pand p = p . (3.66)

The results of these quantization rules for the energy, position, and the momentum of aparticle are summarized in Table 3.1. These quantization rules will allow us to quantizeany system for which we know the classical Hamiltonian. Furthermore, the resultantequations can be written in coordinate space or momentum space. To get a better feelingfor the behavior of quantum systems, we will initially consider all examples in coordinatespace.

If we now consider the momentum and position operators as p and x, then it is clearfrom the above definitions of these operators that

p x Ψ(x, t) 6= x p Ψ(x, t) .

In fact, in coordinate space we have that

p x Ψ(x, t) = −ih ∂∂x

(xΨ(x, t))

= −ihΨ(x, t)− ihx ∂Ψ

∂x= −ihΨ(x, t) + x p Ψ(x, t) . (3.67)

or(xp− px) Ψ(x, t) = ihΨ(x, t) .

Since this result is valid for any function Ψ(x, t), we have that

xp− px = ih . (3.68)

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3.5. PHYSICAL INTERPRETATION OF Ψ(~R, T ) 47

Table 3.1: The quantization rules for the energy, position and momentum in coordinateand momentum space.

Classical variable Quantum Mechanical operatorCoordinate space Momentum space

E ih ∂∂t ih ∂∂t

x x = x x = ih ∂∂p

p p = −ih ∂∂x p = p

If we now define for any two operators A and B, the commutator[A, B

]≡ AB − BA ,

we can write[x, p] = ih . (3.69)

This is known as the commutation relation between the coordinate x and the canonicalmomentum p.

We are now in a position to outline the procedure for quantizing any classical theory.To illustrate this, let us consider a Lagrangian L(q, q) for a single particle in one-dimension.For the position q we can define the canonical momentum by the standard procedure, i.e.,

p =∂L

∂q. (3.70)

The corresponding classical Hamiltonian is now given by

H(p, q) = qp− L(q, q) . (3.71)

The quantization procedure involves replacing the position and momentum variable bythe corresponding position and momentum operators, i.e.,

q → q and p→ p . (3.72)

These operators then satisfy the commutation relation

[q, p] = ih . (3.73)

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48 CHAPTER 3. WAVE PACKET AND THE UNCERTAINTY PRINCIPLE

If we now make this substitution for the coordinate and corresponding momentum in theHamiltonian, i.e.,

H(p, q)→ H (p, q) , (3.74)

then the states of the system are described by the Schrodinger equation

ih∂Ψ

∂t= H (p, q) Ψ , (3.75)

where Ψ can be the wave function in either coordinate or momentum space. Although theabove procedure for quantization has been carried out for one particle in one-dimension,the generalization to more than one particle in more than one dimension is simply achievedby giving the position and momentum operators as vector operators with each set ofposition and momentum operators having a particle label, i.e.

p→ ~pi and x→ ~ri with i = 1, · · · , n , (3.76)

where n is the number of particles.In the event that we have a Lagrangian that is Lorentz invariant, the corresponding

equation to Eq. (3.75) will be the Klein-Gordon equation given in Eq. (3.44). Finally, wenote that this procedure for quantizing a classical theory can be applied to fields, and thuswe are able to derive the quantum theory of radiation by quantizing Maxwell’s equationfor the electromagnetic fields from a Lagrangian for the electromagnetic fields. In thisway, we have a unified theory for both particles and waves, and a procedure for quantizingthe corresponding classical theory.

The above quantization program is known as canonical quantization since the com-mutation relation is written between canonical variables, in this case, the position andmomentum. There is an alternative quantization program based on path integrals. Here,the quantization is achieved by summing over all possible paths each weighted by a factorof

exp− ihS

where S is the classical action defined in the last chapter.

3.6 Problems

1. A free electron bounces elastically back and forth in one dimension between twowalls that are L = 0.50 nm apart.

(a) Assuming that the electron is represented by a de Broglie standing wave witha node at each wall, show that the permitted de Broglie wave lengths areλ = 2L/n, (n = 1, 2, . . .).

(b) Find the values of the kinetic energy of the electron for n = 1, 2, and 3.

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3.6. PROBLEMS 49

(c) Use MAPLE or Mathematica to plot the wave function for n = 1, 2, 3, 4. Canyou see any difference between the odd n and even n wave functions if the wallsare at x = ±L

2?

2. Use MAPLE or MATHEMATICA to plot two waves with wave numbers 0.9 and 1.1,and frequencies 0.85 and 1.0. Show that the sum of these waves has an amplitudethat is not uniform in space. What happens if you add a third wave with wavenumber 1.0 and frequency 1.1?

3. Given the function f(x) that is localized in space between −1,+1, and definedas:

f(x) =

1− |x| for |x| < 10 for |x| ≥ 1

.

(a) Plot f(x) as function of x.

(b) Find the Fourier transform g(k) using Eq. (3.22b) of the lecture notes.

(c) Plot g(k) as function of k for −6 < k < +6.

4. The momentum distribution for a wave packet is given by

g(k) =N

k2 + β2.

where β is a constant and N is the normalization of the wave packet.

(a) Calculate the normalization of the packet.

(b) Calculate the width of the packet in coordinate space by calculating f(x).

(c) Taking the width of the wave packet to be the width of the distribution whenthe height is e−1 of the maximum, show that

∆k∆x > 1

is independent of β.

Hint: You may use the following integrals. See Appendix A for the method ofevaluation of such integrals.

+∞∫−∞

dk1

(k2 + β2)2=

π

2β3and

+∞∫−∞

dkeikx

k2 + β2=π

βe−β|x|

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50 CHAPTER 3. WAVE PACKET AND THE UNCERTAINTY PRINCIPLE

5. A beam of electrons is to be fired over a distance of 104 km. If the size of the initialpacket is 1 mm, what will be its size upon arrival, if its kinetic energy is:

(a) 13.6 eV,

(b) 100 MeV.

(c) Write a MAPLE or Mathematica program that calculates the spreading inthe width of a wave packet with energy over the distance of 104 km. (Thisprogram should cover both of the energies considered in the previous parts ofthis question.)

Note, the relation between kinetic energy and momentum is not always K.E. =p2/2m. When the momentum of the particle becomes comparable to its mass, weneed to use the relativistic relation between the energy and momentum.

6. Suppose that V (x) is complex, i.e. V (x) = VR(x) + iVI(x).

(a) Derive an expression for the rate of change of the density ∂ρ∂t

, for the Hamilto-nian

H = − h2

2m

∂2

∂x2+ V (x)

(b) Calculated

dt

∫dx ρ(x, t) .

(c) Show that for absorption,d

dt

∫dx ρ(x, t)

must be negative. What does this tell us about V (x)?

7. The coordinate space wave function for a particle is given by

ψ(x) =N

x2 + a2.

(a) Calculate N , the normalization of the wave function ψ(x).

(b) For what values of n is the integral 〈xn〉 defined?

(c) Use the above coordinate space wave function to calculate 〈xn〉.(d) Calculate the momentum space wave function φ(k).

(e) Calculate 〈p2〉 directly in coordinate and momentum space.

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3.6. PROBLEMS 51

(f) Use the definitions

∆x =√〈x2〉 − 〈x〉2

∆p =√〈p2〉 − 〈p〉2

to calculate ∆x∆p. Is this consistent with the Heisenberg uncertainty relation?

8. Show that the operator relation

eipa/h x e−ipa/h = x+ a

is valid. The operator eA is defined to be

eA =∞∑n=0

An

n!.

[Hint: Calculate eipa/h x e−ipa/h f(p) where f(p) is any function of p, and use therepresentation x = ih d

dp.]

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52 CHAPTER 3. WAVE PACKET AND THE UNCERTAINTY PRINCIPLE

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Chapter 4

The Schrodinger Equation

Abstract : In this chapter we turn to a solution of the Schrodinger equation fora simple one dimensional potential as a tool for understanding the meaning ofthe wave function with an introduction to symmetry and conservation laws.

In Chapter 2 we showed that a particle of mass m moving in one dimension in anexternal potential V (x), is described by the classical Hamiltonian

H(p, x) =p2

2m+ V (x) . (4.1)

If this system is then quantized, the wave function for the particle Ψ(x, t) satisfies theSchrodinger equation

ih∂Ψ

∂t= H(p, x)Ψ(x, t)

=

(− h2

2m

∂2

∂x2+ V (x)

)Ψ(x, t) . (4.2)

In writing the second line of Eq. (4.2), we made use of the fact that in quantizing thesystem, the momentum operator p is replaced by −ih ∂

∂x, while the coordinate operator x

is replaced by x.In the present chapter we would like to discuss some of the general properties of

the wave function Ψ(x, t). To extract these general properties we will make use of thesimple example of a particle in a box. As a first step in that direction, we will showhow Eq. (4.2), a partial differential equation in two variables, can be reduced to a setof two ordinary differential equations. We then proceed to show that the solution ofthe Schrodinger equation can form a basis for the expansion of any function. This is ageneralization of the results of Fourier where the sin and cos functions formed the basisfor the expansion of a periodic function. We also demonstrate that the symmetries of the

53

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54 CHAPTER 4. THE SCHRODINGER EQUATION

system are reflected in the wave function Ψ(x, t) that describes the quantum mechanicalbehavior of the system. Finally, we consider the condition under which two or morequantities can be measured at the same time to any desired accuracy.

Although all the results demonstrated in this chapter are based on the simple exampleof a particle in a box, the results hold true in general for any potential V (x), and formthe basis for the mathematical structure of quantum mechanics. In Chapter11 we willreturn to this mathematical structure and prove these results in their more general form.

4.1 Method of Separation of Variables

Before we can proceed to a general discussion of the solutions of the Schrodinger equation,we need to solve the above partial differential equation. This can be achieved in one oftwo ways such that the time and space dependence of the solution are separated. The firstmethod is that of separation of variables, while the second approach involves the Fourierdecomposition of the time dependence.

To implement the method of separation of variables, we rewrite the Schrodinger equa-tion as (

ih∂

∂t− H

)Ψ(x, t) = 0 (4.3)

orLΨ(x, t) = 0 . (4.4)

The operator L is the sum of two parts. The first part depends on time, while the secondpart depends on the coordinate x. Under these circumstances, the solution to Eq. (4.4),Ψ(x, t), is the product of two functions - one depending on time only, the other dependingon the coordinate x, i.e.

Ψ(x, t) = F (t)ψ(x) . (4.5)

With this form for the wave function we can write Eq. (4.3) as

ih∂F

∂tψ(x)− F (t)Hψ(x) = 0 .

If we now divide this equation by F (t)ψ(x), we can rewrite it as

ih

F (t)

∂F

∂t=

1

ψ(x)Hψ(x) . (4.6)

The left hand side of this equation depends on time only, while the right hand side dependson the position x only. Since time and position are independent variables, each side ofEq. (4.6) must be a constant for this equality to be valid, i.e.

1

ψ(x)Hψ(x) = E and

ih

F (t)

dF

dt= E .

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4.2. METHOD OF FOURIER ANALYSIS 55

Note, we have replaced the partial derivative by the total derivative since the functionF depends on time only. At this stage, E is a constant with the same dimension as theHamiltonian, i.e., it has the dimension of energy. The above procedure allows us to reducethe one partial differential equation in two variables, Eq. (4.2), to two ordinary differentialequations in one variable. These two equations are

H ψ(x) = E ψ(x) ( 4.7a)

and

ihdF

dt= E F (t) . ( 4.7b)

In Eq. (4.7b) we have a first order linear differential equation with the solution given by1

F (t) = F0 e−iEt/h . (4.8)

This means that our total wave function is of the form

Ψ(x, t) = e−iEt/h ψ(x) . (4.9)

In writing Eq. (4.9), we have incorporated, without any loss of generality, the constant F0

into the function ψ(x). This is justified on the grounds that the total wave function willbe normalized. Here we observe that the solution given in Eq. (4.9) is general enough,provided that the potential V (x), and therefore the Hamiltonian H is time independent.To complete our solution of the Schrodinger equation we need to solve Eq. (4.7a) and forthat we have to specify Hamiltonian H, and therefore the potential V (x).

4.2 Method of Fourier Analysis

An alternative procedure for reducing partial differential equations to ordinary differentialequations, involves the application of Fourier transforms. In this case we can reduce thedimensionality of the partial differential equation by the Fourier decomposition of thesolution in one of the variables. For the Schrodinger equation, with a Hamiltonian thathas no explicit time dependence, it is most convenient to Fourier decompose the timevariable in the wave function, i.e.,

Ψ(x, t) =1√2π

+∞∫−∞

dwΨ(x,w) e−iwt . (4.10)

1It is straight forward to show, by substitution, that the function F (t), given in Eq. (4.8), is a solutionto the first order differential equation given in Eq. (4.7b).

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56 CHAPTER 4. THE SCHRODINGER EQUATION

Then by straight forward differentiation we have that:

ih∂Ψ

∂t=

1√2π

+∞∫−∞

dw hwΨ(x,w) e−iwt , (4.11)

and

H Ψ(x, t) =1√2π

+∞∫−∞

dw H Ψ(x,w) e−iwt . (4.12)

Therefore, the Schrodinger equation can be written as

1√2π

+∞∫−∞

dw[hw − H

]Ψ(x,w) e−iwt = 0 . (4.13)

This in general is only valid if [hw − H

]Ψ(x,w) = 0 , (4.14)

where w is a parameter of the differential equation since it only appears in hw. Equation(4.14), which now is an ordinary differential equation in x, is identical to Eq. (4.7a), if wetake

E = hw and ψE(x) = Ψ(x,w) . (4.15)

In writing Eq. (4.15) we had to label the function ψ(x) with the subscript E since thefunction Ψ depends on the variable ω and therefore E = hω. Our solution for the timedependent Schrodinger equation is now of the form

Ψ(x, t) =1√2πh

+∞∫−∞

dE ψE(x) e−iEt/h . (4.16)

In fact, this way of writing the time dependent wave function is a special case of a moregeneral form which is given by

Ψ(x, t) =1√2πh

∫dE C(E)ψE(x) e−iEt/h (4.17)

which satisfies the Schrodinger equation for any function C(E). This result is a conse-quence of the fact that the operator L = ih ∂

∂t− H is linear. The question then is, what

is a linear operator? An operator L is said to be linear if

L (Ψ1 + Ψ2) = LΨ1 + LΨ2 , (4.18)

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4.3. PARTICLE IN A BOX 57

where in our case, Ψ1 and Ψ2 are functions of (x, t). A more general definition of a linearoperator is 2

L (aΨ1 + bΨ2) = aLΨ1 + bLΨ2 , (4.19)

where a and b are any two complex numbers.At this stage, we should point out that the integral over E (4.17) consists of a sum

over a set of discrete energies, and an integral with a finite lower limit, i.e.,∫dE · · · =

∑n

· · ·+∞∫E0

dE · · · . (4.20)

The fact that we have a sum over discrete energies is a result of the fact that Eq. (4.7a)will have solutions only for certain values of the energy En < E0.

x

V(x)

∞ ∞

-a +a0

Figure 4.1: The potential for a particle in a one dimensional box with sides at x = ±a.

4.3 Particle in a Box

To illustrate some of the general properties of the wave function ψ(x),3 we need to solveEq. (4.7a) for a specific potential V (x). The simplest such potential that also exhibits theproperties of the wave function ψ(x), is the potential that corresponds to a particle in aone dimensional box of length 2a, see Figure 4.1 i.e.,

V (x) =

+∞ x < −a

0 −a < x < a+∞ x > a

. (4.21)

2In Chapter 12 we will come across an anti-linear operator for which

LaΨ = a∗LΨ

where a is a complex number, and a∗ is the complex conjugate of a.3To simplify the notation, we have for the moment dropped the E subscript on ψ(x).

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58 CHAPTER 4. THE SCHRODINGER EQUATION

Since the potential is infinite for |x| > a, the particle is constrained to move in theregion |x| < a. In other words,

ψ(x) = 0 for |x| ≥ a . (4.22)

We therefore have to solve the time independent Schrodinger equation,

Hψ(x) = Eψ(x)

in the domain |x| < a. In this region since the potential is zero, the time independentSchrodinger equation takes the form

− h2

2m

d2ψ

dx2= Eψ(x) |x| < a

or, upon multiplication by −2m/h2, we have

d2ψ

dx2= −k2ψ(x) for |x| < a , (4.23)

where k2 = 2mE/h2. This equation is a second order linear differential equation, andtherefore has two independent solutions. These could be sin kx and cos kx. The generalsolution to this second ordered differential equation can be written as a linear combinationof the two solutions, i.e.,

ψ(x) = B sin kx+ C cos kx . (4.24)

The constants B and C, and the parameter k can now be determined by the boundaryconditions on the general solution given in Eq. (4.24).

The first set of boundary conditions are that the wave function ψ(x) is zero at x = ±a.This follows from the fact that the particle is constrained to the region −a < x < +a,i.e., ψ(x) = 0 for |x| > a. The application of these two boundary conditions gives us twoequations for the constants B, C and k, which are

B sin ka+ C cos ka = 0 for x = a ( 4.25a)

−B sin ka+ C cos ka = 0 for x = −a . ( 4.25b)

The second set of boundary conditions for a second order differential equation are that thederivative of the wave function should be zero for x = ±a. These boundary conditionsresult in Eqs. (4.25a) and (4.25b) with B and C interchanged. Since the final wavefunction is to be normalized (see Eq. (4.36)), this second set of boundary conditions giveno additional constraints on the wave function and are not considered any further for thisproblem.

If we now add the two Eq.s (4.25a) and (4.25b), we get the condition

2C cos ka = 0 . (4.26)

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4.3. PARTICLE IN A BOX 59

There are two possible solutions to this equation. These are: (i) C = 0 which gives a zerowave function and therefore is of no interest to us. (ii) C 6= 0, in which case cos ka = 0.For this condition to be satisfied, ka can take only certain values, i.e.

ka =nπ

2or k =

2awith n = 1, 3, 5, · · · . (4.27)

On the other hand, if we subtract Eq. (4.25b) from Eq. (4.25a), we get

2B sin ka = 0 . (4.28)

Here again, there are two possible solutions with the non-trivial solution correspondingto B 6= 0, in which case sin ka = 0 and ka can take on only certain values, i.e.

ka =nπ

2or k =

2awith n = 2, 4, . . . . (4.29)

We can combine the results in Eqs. (4.27) and (4.29) by noting that when C 6= 0, k isgiven by Eq. (4.27) with n odd, while for B 6= 0, k is given by the same equation with neven. Furthermore, the parameter k is related to E by

k =

√2mE

h2 .

Thus, in general, we can write k as4

kn =

√2mEn

h2 =π

2a(n+ 1) for n = 0, 1, 2, . . . , (4.30)

where we have introduced a subscript for both k and E to indicate that these quantitiesdepend on n. This result proves that the energy En can have only certain discrete values,and these are given by

En =h2k2

n

2m=

h2π2

8ma2(n+ 1)2 with n = 0, 1, 2, . . . , (4.31)

while the corresponding wave functions are given by

ψn(x) =

Bn cos knx for n = 0, 2, . . .

Bn sin knx for n = 1, 3, . . .. (4.32)

In writing Eq. (4.32), we have: (i) Introduced a subscript for the wave function n. Thissubscript replaces the energy, given the fact that the energy is a function of n. (ii) Replacedthe constants B and C by the constant Bn, which corresponds to taking B → Bn for n aneven integer, or C → Bn for n an odd integer. In this way we have emphasized the fact

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60 CHAPTER 4. THE SCHRODINGER EQUATION

-1.0 -0.5 0.5 1.0x

-1.0

-0.5

0.5

1.0

ψ ( )x 0

-1.0 -0.5 0.5 1.0x

-1.0

-0.5

0.5

1.0

ψ ( )x1

-1.0 -0.5 0.5 1.0x

-1.0

-0.5

0.5

1.0

ψ ( )x2

-1.0 -0.5 0.5 1.0x

-1.0

-0.5

0.5

1.0

ψ ( )x3

-1.0 -0.5 0.5 1.0x

-1.0

-0.5

0.5

1.0

ψ ( )x5

-1.0 -0.5 0.5 1.0x

-1.0

-0.5

0.5

1.0

ψ ( )x4

Figure 4.2: Here we have a plot of the wave function for a particle in a box with a = 1,h = m = 1, and n = 0, 1, . . . , 5..

that we have one arbitrary constant, with the overall magnitude of the wave function, Bn,to be determined.

As stated earlier, the third constant in the wave function Bn, is determined by thenormalization of the wave function, i.e. the overall magnitude of the wave function Bn,is determined by the condition that,

+∞∫−∞

dx ψ∗n(x)ψn(x) =

+a∫−a

dx ψ∗n(x)ψn(x) = 1 , (4.33)

or

B2n

+a∫−a

dx sin2 knx = 1 for n = odd , ( 4.34a)

4We have replaced n→ (n+ 1) in order to have n = 0, 1, 2, . . .. In this way the lowest energy, i.e., theground state, corresponds to n = 0. This is a convention used in most books on the subject.

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4.4. GENERAL PROPERTIES OF THE WAVE FUNCTION ψN(X) 61

and

B2n

+a∫−a

dx cos2 knx = 1 for n = even . ( 4.34b)

Evaluating these integrals gives us 5

Bn =1√a. (4.35)

Therefore the normalized wave function for a particle in a one dimensional box is givenfor |x| < a by

ψn(x) =1√a

cos knx for n = even

sin knx for n = odd, (4.36)

while ψn(x) = 0 for |x| ≥ a. The wave functions ψn(x) for a particle in a box areillustrated in Figure 4.2 for n = 0, · · · , 5 for the case when a = 1 and h = m = 1.

4.4 General Properties of the Wave Function ψn(x)

Having determined the general solution of the Schrodinger equation for the simple caseof a particle in a box, we proceed in this section to examine some of the mathematicalproperties of this solution that hold true in general for all solutions of the Schrodingerequation for any potential. In this way we can illustrate the mathematical structureof Quantum Mechanics without getting involved in the mathematical foundation of thetheory.

With the solution of the Schrodinger equation for a particle in a box, we can nowcalculate the following integral

+∞∫−∞

dx ψ∗n(x)ψm(x) =1

a

+a∫−a

dx cos knx cos kmx for n and m even

=1

a

+a∫−a

dx sin knx sin kmx for n and m odd

=1

a

+a∫−a

dx sin knx cos kmx for n odd and m even .

5We make use of the following integrals,

+π/2∫−π/2

dx sin2 nx =

+π/2∫−π/2

dx cos2 nx =π

2for n integer .

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62 CHAPTER 4. THE SCHRODINGER EQUATION

Here we find that for n 6= m, the integral is zero,6 i.e.,

+∞∫−∞

dx ψ∗n(x)ψm(x) =

0 n 6= m

1 n = m

≡ δnm . (4.37)

In writing the last line in the above equation, we have defined a new function δmn thattakes the value one if n = m, and zero otherwise. The motivation for introducing thisnew function is that we will encounter integrals of the above form repeatedly in QuantumMechanics.

To give a physical meaning to the result in Eq. (4.37), let us assume we want to evaluatethe integral on the left-hand-side of Eq. (4.37) numerically. This involves replacing theintegral by a sum, i.e.

+∞∫−∞

dx ψ∗n(x)(x)ψm(x) =N∑i=1

ψ∗n(xi)ψm(xi) .

This sum looks like the product of a row matrix of length N and a column matrix of lengthN with the result that for n 6= m the product is zero, while for n = m the product is one.In linear algebra this property is known as orthogonality and normalization. In otherwords, the functions ψn(x) are normalized and orthogonal, i.e., they are orthonormal.Note, the process of replacing the integral in Eq. (4.37) by a sum, replaces the functionψn(x) by a column matrix ψn(xi) in which the elements of the matrix are labeled by xi,and these cover the domain over which the function ψn(x) is defined.

A second property the functions ψn(x) have is that any function f(x) that satisfiesthe boundary condition f(a) = 0 = f(−a), can be written as

f(x) =∞∑n=0

An ψn(x) . (4.38)

In this case, because the wave function ψn(x) is sin knx or cos knx, the above expansionreduces to the Fourier series for the function f(x) since Eq. (4.38) can be written as

f(x) =∑

n= even

An√a

cos knx+∑

n= odd

An√a

sin knx .

6We have made use of the fact that

+π/2∫−π/2

dx sinnx sinmx =

+π/2∫−π/2

dx cosnx cosmx = 0 for n 6= m,n and m integer .

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4.4. GENERAL PROPERTIES OF THE WAVE FUNCTION ψN(X) 63

However, this result is true for any set of functions ψn(x) that are the solution of anequation (e.g. the time independent Schrodinger equation) of the form

Hψn = Enψn (4.39)

provided H is a linear Hermitian operator.7 Here again we can recast Eq. (4.39) inthe language of linear algebra. In particular, we consider the problem of eigenstates andeigenvalues of a matrix. If we think of the Hamiltonian H as a matrix of dimension N×N ,then Eq. (4.39) suggests that we refer to the wave function ψn(x) as the eigenstate of thematrix H, with En as the corresponding eigenvalue, i.e. the eigenstate ψn(x) is taken tobe a column matrix of length N.

This equivalence between Quantum Mechanics (i.e. the solution of the Schrodingerequation) and linear algebra allows us to make use of many of the results derived in linearalgebra when examining quantum mechanical systems. To illustrate this analogy, considerthe eigenstates of a real symmetric matrix where the eigenstates belonging to differenteigenvalues are orthogonal. In the case of matrices, this orthogonality takes the form ofa multiplication of a row matrix by a column matrix, while in quantum mechanics thisorthogonality takes the form given in Eq. (4.37). These two forms become identical if wereplace the integral in Eq. (4.37) by a sum over quadratures, which is how we performintegration on a computer.

To illustrate the matrix nature of the Hamiltonian H, we first consider the integral∫dx ψ∗m(x) H ψn(x) .

Making use of the fact that ψn(x) is an eigenstate of H with eigenvalue En, and theeigenstates form an orthonormal basis, we can write

+a∫−a

dxψ∗m(x)Hψn(x) =

+a∫−a

dxψ∗m(x)Enψn(x)

= En

+a∫−a

dxψ∗m(x)ψn(x) = En δmn . (4.40)

7The matrix H is Hermitian if (H∗)T ≡ H† = H, where (H∗)T in the complex conjugate transpose,i.e., if we define the (n,m) elements of the matrix H as hnm, i.e.

[H]nm = hnm then [(H∗)T ]nm = h∗mn .

Therefore, for a Hermitian matrix we have

hnm = h∗mn .

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64 CHAPTER 4. THE SCHRODINGER EQUATION

The quantity we have calculated in Eq. (4.40) we refer to as the matrix element of theHamiltonian H between wave functions which, in this case, are the eigenstates of thissame Hamiltonian. This gives us the energy of the system in a given state n. In otherwords, the Hamiltonian matrix is diagonal and the elements of this diagonal matrix arethe energies of the system. In Quantum Mechanics terminology this can be stated as: TheHamiltonian H is an operator corresponding to the energy observable En. On the otherhand in the language of linear algebra we say that ψn is an eigenstate of the Hermitianmatrix H with eigenvalue En.

Let us now consider a system described by the wave function f(x). This wave functioncan be written as in Eq. (4.38) in terms of the functions ψn(x), which in this case aresolutions of the Schrodinger equation for a given potential, e.g. the potential given inEq. (4.21). To give a physical meaning to the coefficients An in Eq. (4.38), we firstmultiply Eq. (4.38) from the left by ψ∗m(x) and integrate over the coordinate x. Becauseof the orthonormality ( see Eq. (4.37) ) of the wave function ψn(x), we can write thecoefficients An, as

An =

+∞∫−∞

dxψ∗n(x)f(x) =

+a∫−a

dx ψ∗n(x) f(x) . (4.41)

We now consider the integral ∫dx f ∗(x) H f(x) .

Using first the expansion for the function f(x) in Eq. (4.38), we can then write the matrixelement of the Hamiltonian H with respect to the state f(x) as∫

dx f ∗(x)Hf(x) =∑n

∑m

A∗nAm

∫dxψ∗n(x)Hψm(x) .

Making use of the orthonormality of the eigenstates of the Hamiltonian H, we can writethis integral as ∫

dx f ∗(x)Hf(x) =∑nm

A∗nAmEm

∫dxψ∗n(x)ψm(x)

=∑n

|An|2En . (4.42)

On the other hand, the normalization of the wave function f(x) implies that∫dx f ∗(x)f(x) =

∑N

|An|2 = 1 . (4.43)

From the above results given in Eqs. (3.42) and (3.43), we may conclude that:

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4.5. SYMMETRY UNDER INVERSION - PARITY 65

1. From the structure of Eq. (4.43) we may consider |An|2 as a probability. In partic-ular, this |An|2 is the probability that the particle described by the wave functionf(x) is in a state with energy En. In this case the normalization is another way ofstating that the probability of finding the particle in any state ψn is one.

2. If |An|2 is a probability of finding the particle with energy En, then∑n|An|2En is

the average energy of the particle described by the function f(x).

Since |An|2 is a probability, then An, defined in Eq. (4.40), is the probability amplitude.In particular, for a particle described by the wave function f(x), An is the probabilityamplitude of finding the particle in a state with energy En.

4.5 Symmetry Under Inversion - Parity

Before we can commence our discussion on symmetry we should make sure we all agreeas to what is meant by symmetry.

Consider a chain with beads spaced at intervals of a cm along the chain. If you movethe chain by a distance a along its length while we blink our eyes, then we will not be ableto tell if you have done anything to the chain, provided of course, that we can’t see theends of the chain. Under these conditions we would say that the chain has a symmetrycorresponding to translation of the chain by a distance a cm.

Our problem of a particle in a box, discussed in Sec. 4.3, has a symmetry underinversion, i.e., when x → −x, the Hamiltonian for the system does not change. This isclear from Figure 4.1. This symmetry is known as parity, and the operator for inversion,P , has the property that

Pψ(x) = ψ(−x) . (4.44)

The wave functions for the particle in a box as given in Eq. (4.36) can be divided into twoclasses - those that change sign under the parity operation, and those that do not changesign, i.e.,

ψn(−x) =

+ψn(x) for n even

−ψn(x) for n odd. (4.45)

In other words

Pψn(x) = ψn(−x) = (−1)nψn(x) . (4.46)

Thus, the wave function or eigenstate ψn, which is an eigenstate of the Hamiltonian, isalso an eigenstate of the parity operator P . This is established by the fact that when theoperator P acts on the eigenstate ψn(x) we get the corresponding eigenvalue (−1)n timesthe same eigenstate. This means we can label the states of the particle ψn(x), by boththe energy and the parity. In this case it turns out that n can label both the energy and

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66 CHAPTER 4. THE SCHRODINGER EQUATION

the parity of the state, but this is a special case. We could have written the eigenstate ofH and P as ψE,π where E labels the energy, and π = ±1 labels the parity of the state.

The reason we can construct one eigenstate which is both an eigenstate of the Hamil-tonian H and an eigenstate of parity P is because of the symmetry of the HamiltonianH under the parity operation. In this case the Hamiltonian does not change under thetransformation x→ −x in H. This leads us to the concept of invariance of the theory un-der that symmetry. To illustrate this, consider the time dependent Schrodinger equationfor the Hamiltonian H, i.e.

ih∂Ψ

∂t= HΨ(x, t) . (4.47)

If we operate on this equation with the parity operator, P , we get

ihP∂Ψ

∂t= ih

∂(PΨ)

∂t= P HΨ , (4.48)

where we have assumed that the parity operator does not depend on time and can betaken inside the differentiation. But in general the action of the operator P on a state Ψgives a new state, i.e.,

PΨ = Ψ′ and Ψ = P−1Ψ′ , (4.49)

where in writing the second statement we have assumed that the operator P has aninverse, which in the case of the parity operator is true. This is also true for the operationof moving the chain with beads. In this case if the operator corresponds to moving thechain to the right by a cm, then the inverse would correspond to moving the chain to theleft by a distance a cm. With the definition of Ψ′ and the inverse operator for parity, wenow can write Eq. (4.48) as

ih∂Ψ′

∂t= P HΨ = P HP−1Ψ′ (4.50)

But for the theory to be invariant under a given symmetry, the equations of motion haveto keep their form ( i.e. not change ) under the transformation of the symmetry. For ourequations to be the same under parity, we require that

P HP−1 = H . (4.51)

Then both Ψ and Ψ′ satisfy the same identical equation, and therefore the solution isthe same and the physics has not changed as a result of the transformation. Thus thecondition for the theory to be invariant under the symmetry transformation is

P H − HP ≡ [P , H] = 0 . (4.52)

The fact that P commutes with the Hamiltonian implies that H does not change underthe transformation x → −x. The fact that our equation does not change under paritymeans that the parity of a given state is a constant of the motion, and this is emphasizedby the fact that the wave function is labeled by both the energy and the parity.

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4.6. EIGENSTATES OF THE MOMENTUM 67

4.6 Eigenstates of the Momentum

The time independent Schrodinger equation, Hψn = Enψn, is an eigenvalue problem inwhich the eigenvalues, En are the energies of the system described by the HamiltonianH. In actual fact, to every observable, e.g., position and momentum, we have a Hermi-tian operator, and given that this operator is both linear and Hermitian, we can set upan eigenvalue problem with the eigenvalues being the observables corresponding to thatoperator. As an example let us consider the momentum. The corresponding operator incoordinate space is

p = −ih ddx

(4.53)

The eigenvalue problem we can now set up is

p χp = p χp ,

where χp is the eigenstate, and p is the corresponding eigenvalue of the operator p. Sincethe momentum operator in coordinate space is the differential operator given in Eq. (4.53),we can write this eigenvalue problem as a first order linear differential equation of theform

−ihdχpdx

= p χp .

The solution to this first order differential equation is obtained by multiplying both sidesof the equation by dx, dividing by χp and integrating both sides, i.e.,

−ih∫ dχp

χp=∫p dx .

Since p is a constant, the result of the integration is

logχp =i

hxp+ C,

where C is a constant of integration of the differential equation. This result can berewritten as

χp(x) = N eipx/h = N eikx ,

where the constant of integration is given by N.8 To normalize this wave function we haveto calculate the integral

+∞∫−∞

dxχ∗p′(x) χp(x) = N2

+∞∫−∞

dx ei(p−p′)x/h

= N2 (2πh) δ(p− p′) ,8The normalization constant can be written in terms of the constant C as

C = logN .

Note: We have used log for the natural logarithm.

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68 CHAPTER 4. THE SCHRODINGER EQUATION

where we have made use of the definition of the Dirac δ-function, Eq. (3.17), to write thesecond line in the above expression. The normalization of the wave function N , is nowgiven as,

N =1√2πh

.

We are now in a position to write the normalized eigenstates of the momentum operatoras

χp(x) =1√2πh

eipx/h . (4.54)

We should note that this function is, up to normalization, the space part of the waveei(kx−wt) if we observe that p = hk.

To understand the relation between the eigenstate of the momentum operator anda wave with wave number k and frequency w, we next consider the time independentSchrodinger equation for a free particle, i.e. V = 0. This Schrodinger equation in onedimension takes the form

− h2

2m

d2ψ

dx2= Eψ(x) (4.55)

ord2ψ

dx2= −k2ψ(x) with k2 =

2mE

h2 . (4.56)

This second order differential equation is the space part of the wave equation and has asolution of the form

ψk(x) = N eikx . (4.57)

Taking into consideration the fact that p = hk, we have established that the eigenstatesof the momentum operator are identical to the eigenstates of the Hamiltonian for a freeparticle. This result is a consequence of the fact that the Hamiltonian for a free particleH = p2

2mcommutes with the momentum operator p, i.e.,

[H, p] = 0 . (4.58)

When this condition is satisfied, we can find one eigenstate that is an eigenstate of bothoperators.

The physical implication of this result is that a measurement of the momentum ofthe particle does not in any way effect the energy of the particle, and both the energyand momentum can be measured to any desired accuracy. Thus when two operatorscommute, there is no uncertainty principle between the corresponding observables andwe can construct one wave function that is an eigenstate of both operators. Note that ifthe particle is in an external force, i.e., V 6= 0, then the Hamiltonian will not commutewith the momentum operator and in that case we can not measure both the energy andmomentum to any desired accuracy.

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4.7. PROBLEMS 69

At this stage let us go back to our description of a wave packet in which we wrote

ψ(x) =1√2π

+∞∫−∞

dk φ(k) eikx . (4.59)

If we now change variables from k to p = hk we have

ψ(x) =1√2πh

+∞∫−∞

dp φ(p) eipx/h

=

+∞∫−∞

dp φ(p)χp(x) . (4.60)

We now can understand this result as an expansion of the wave packet in terms of thecomplete set of eigenstates of the momentum operator. In this case, φ(p) is the probabilityamplitude of finding the particle in a state with momentum p, and χp(x) is the wavefunction for this momentum eigenstate. Thus, what we considered previously as a Fourierdecomposition of the wave function ψ(x), we now can interpret as nothing more thanan expansion of the wave function ψ(x) in terms of the eigenstates of the momentumoperator, χp(x).9

4.7 Problems

1. You are given the following operators

(a) O1ψ(x) = x2ψ(x) (b) O2ψ(x) = xd

dxψ(x)

(c) O3ψ(x) = λψ∗(x) (d) O4ψ(x) = eψ(x)

(e) O5ψ(x) = dψ(x)dx

+ a (f) O6ψ(x) =

x∫−∞

dx′ (ψ(x′)x′) .

Which of these are linear operators?

2. Calculate the following commutators

(a)[O2, O6

](b)

[O1, O2

],

9The difference between φ(p) and φ(k) is the normalization. Thus φ(k) has a normalization of 1/√

2π,while φ(p) has a normalization of 1/

√2πh.

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70 CHAPTER 4. THE SCHRODINGER EQUATION

where the operators Oi, i = 1, 2, 6 are defined in the preceding problem. Theprocedure is to calculate [A, B] by expressing A(Bψ)− B(Aψ) in the form Cψ.

3. Consider the two operators a and a† defined such that

a ψ(x) =1

2

(x+ i

d

dx

)ψ(x)

a† ψ(x) =1

2

(x− i d

dx

)ψ(x) .

(a) Show that the operators a and a† are linear operators.

(b) Calculate the commutation relation[a, a†

].

(c) Find the eigenvalues and eigenstates of the operator a − a†. How are theseeigenvalues and eigenstates related to the eigenvalues and eigenstates of themomentum operator?

4. Solve the Schrodinger equation for a particle in a box with sides at x = 0 and x = awith the boundary condition that

ψ(0) = ψ(a) = 0 .

(a) What are the eigenvalues and the normalized eigenstates?

(b) Use MAPLE or Mathematica to plot the wave function for the ground stateand first excited state.

(c) Use MAPLE or Mathematica to calculate the probability of finding the particlein the first excited state between x = 0 and x = a/2.

(d) Does the wave function have a definite symmetry under the transformationx↔ −x? Why?

5. Given: The one dimensional potential that represents a particle in a box is

V (x) =

+∞ x < −a0 −a < x < 0+∞ x > 0

.

(a) Write the time independent Schrodinger equation for this potential in the dif-ferent regions.

(b) What are the boundary conditions on the wave function?

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4.7. PROBLEMS 71

(c) Solve the Schrodinger equation in the different regions and normalize the solu-tion.

(d) What are the energies (i.e., eigenvalues) the particle can have in this potential?

(e) Does the wave function for the particle have a definite symmetry under thetransformation x→ −x? Why?

(f) Plot the wave normalized wave function for the lowest three eigenstates, giventhe mass of the particle is one, and a = 2.

6. A particle is in the ground state of a box with sides at x = ±a. Very suddenly thesides of the box are moved to x = ±b (b > a).

(a) Write the ground state normalized wave function for the particle in the initialbox with sides at x = ±a.

(b) Write the ground state normalized wave function for a particle in the final boxwith sides at x = ±b.

(c) What is the probability that the particle which is initially in the ground stateof the box with sides at ±a, will be found in the ground state for the finalpotential with sides at ±b?

(d) What is the probability that the particle in the ground state of the box withsides at x = ±a, will be found in the first excited state of the final box withsides at x = ±b?

In the last case, the simple answer has a simple explanation. What is it?

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72 CHAPTER 4. THE SCHRODINGER EQUATION

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Chapter 5

Simple One Dimensional Problems

Abstract : This chapter is devoted to the application of the Schrodinger equa-tion to one dimensional models of interesting phenomena. In this way we canget an understanding of quantum effects, and the role of symmetry, withoutthe complexity of the mathematics encountered in three dimensional problems.

In this chapter we will consider the solution of the Schrodinger equation for simple onedimensional problems. The main motivation for considering systems in one-dimension isthat the complexity of the mathematical formulation is reduced, yet the main ideas ofa quantum system are retained. In particular, we will find that for a one-dimensionalproblem, with a time independent potential, the Schrodinger equation reduces to anordinary second order differential equation. The formulation of quantum mechanics inone-dimension will allow us to consider the reflection and transmission of a wave by apotential well as a simple example of a scattering problem. In particular, we will test theconcept of current conservation within the framework of a scattering experiment, and therelation between the scattering matrix and bound states in the potential.

5.1 Free-Particle

All scattering experiments involve a beam of particles, e.g., electrons, protons, and pho-tons, incident on a target with the scattered particles being collected and analyzed bya detector. In Figure 5.1 we illustrate a typical experimental set-up where the incidentbeam is generated by an accelerator. This incident beam is then scattered from a targetT , and the scattered particles are detected by the detector D.

Before the incident beam reaches the target it is traveling in a potential free zone(i.e., no external force on the particle), and therefore is described by the solutions of the

73

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74 CHAPTER 5. SIMPLE ONE DIMENSIONAL PROBLEMS

Accelerator T

D

θ

Figure 5.1: An illustration of a typical scattering experiment in which the incident beamis generated by an accelerator, is scattered by a target T, and is detected by the detectorD. The scattering angle is θ.

Schrodinger equation for a free particle. The Hamiltonian for a free particle is given by1

H = − h2

2m

d2

dx2, (5.1)

and the Schrodinger equation takes the form

d2ψ

dr2+ k2ψ = 0 with k2 =

2mE

h2 , (5.2)

where E is the energy of the particles in the incident beam. The solution of Eq. (5.2) isgiven by

ψRk (x) = eikx ψLk (x) = e−ikx . (5.3)

Both of the above solutions satisfy the Schrodinger equation, Eq. (5.2). The correspondingtime dependent solutions of the Schrodinger equation are

Ψ(x, t) = ψ(x) e−iEt/h .

If we now write the energy in terms of the the frequency, i.e., E = hω, we can write thetwo energy dependent solutions as

ΨRk (x, t) = ei(kx−ωt) and ΨL

k (x, t) = e−i(kx+ωt) . (5.4)

The most general solution could be a linear combination of the above two solutions.However, before we can consider such general solutions, let us examine the properties ofthe individual solutions. These solutions represent waves, and we would like to determinethe direction of propagation of one of these waves, e.g. ΨR

k .

To determine the direction of propagation of the wave, we can think of a championsurfer that can stay at a specific point in front of the crest of a perfect wave for an

1From this point on we will drop the ˆ from all operators and leave it to the reader to differentiatebetween operators and functions.

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5.2. POTENTIAL STEP 75

indefinite period of time. If at time t0 he is at a position x0, then his height above thesurface of the sea is given by2

ΨRk (x0, t0) = ei(kx0−ωt0) .

At a time t > t0, the surfer still maintains the same height above the surface of the sea,and his position x, will be determined by the condition:

Ψ(x, t) = Ψ(x0, t0) .

This is equivalent to the condition

kx0 − ωt0 = kx− ωt ,

orx− x0 =

ω

k(t− t0) .

Thus for t > t0 we have that x > x0, and we can say the champion surfer, and thereforethe wave, is traveling along the positive x-axis, i.e. to the right. This establishes the factthat the function ΨR

k (x, t) represents a wave that is propagating to the right. In a similarmanner we can show that ΨL

k (x, t) represents a wave moving to the left.This means that the time independent wave function for a wave propagating to the

right is given by ψRk (x) while a wave proceeding to the left is denoted by ψLk (x) with

ψRk (x) = eikx and ψLk (x) = e−ikx . (5.5)

The general solution to the Schrodinger equation, Eq. (5.2), is now taken as a linearcombination of a wave traveling to the right and one traveling to the left, i.e.,

ψk(x) = Aeikx +Be−ikx . (5.6)

The constants A and B are to be determined by the boundary conditions imposed by theexperimental set-up. These will be specified as we examine each individual problem insubsequent sections.

5.2 Potential Step

We would like now to make use of the general solution to the time independent Schrodingerequation as given in Eq. (5.6) to illustrate the difference between the classical problemand corresponding quantum mechanical problem for the simplest system of a potentialbarrier and potential well. In this way we hope to also illustrate the problem of scattering

2The surface of the sea is taken to be the level of water when there are no waves.

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76 CHAPTER 5. SIMPLE ONE DIMENSIONAL PROBLEMS

from a barrier or a potential well, and how current conservation plays an important rolein all scattering experiments.

As the simplest scattering problem, let us consider a plane wave incident from the lefton a potential step, see Figure 5.2. The potential is taken to be of the form

V (x) =

0 x < 0

V0 x > 0. (5.7)

x

V(x)

V0

Figure 5.2: Plot for a potential barrier of height V0.

Before we consider this problem within the framework of quantum mechanics, let usconsider the classical problem. Here we recall that the force is given by

F = −dVdx

,

which in this case is an impulse at x = 0 directed towards the negative x-axis. Let us firstconsider the case when E < V0 and the particle is traveling to the right. Its momentumwould be p =

√2mE for x < 0. When it gets to x = 0, it is reflected by the barrier, i.e. it

bounces back with momentum p′ = −√

2mE. On the other hand, for E > V0, the particle

keeps going to the right but for x > 0, its momentum is reduced to p′ =√

2m(E − V0).The change in the momentum of the particle as it goes from x < 0 to x > 0 is a result ofthe impulse at x = 0.3

To get the corresponding quantum mechanical solution, we need to solve the timeindependent Schrodinger equation for the potential in Eq. (5.7). The solution to theSchrodinger equation for this potential is given, considering the boundary condition thatwe have a beam of particles incident from the left, as

ψ(x) =

1 eikx +Re−ikx for x < 0

T eik′x for x > 0

, (5.8)

3Recall that Newton’s second law is given by F = dp/dt.

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5.2. POTENTIAL STEP 77

where the wave number for x < 0 is k, while the wave number for x > 0 is k′. These wavenumbers are given in terms of the incident particle’s energy E, and barrier height V0 as

k =

√2mE

h2 and k′ =

√2m

h2 (E − V0) . (5.9)

In writing the solution of the Schrodinger equation in Eq. (5.8), we have taken the incidentwave to have unit amplitude while the reflected wave has an amplitude R. For thetransmitted wave, the amplitude is taken to be T .

For E > V0, we expect a reflected and a transmitted wave, while in the classical casefor E > V0, we only had a particle traveling along the positive x-axis with a change inmomentum at x = 0. To calculate the incident, reflected and transmitted currents, andthus examine current conservation at x = 0, we will make use of the definition of thecurrent as given in Eq. (3.50), i.e.,

j(x) =h

2im

ψ∗dψ

dx− dψ∗

dxψ

.

In this case the current for x < 0 is given, using the wave function in Eq. (5.8), by

j(x) =h

2mi[(e−ikx +R∗eikx)(ikeikx − ikR e−ikx)− complex conjugate]

=h

2mi[2ik(1− |R|2)]

=hk

m(1− |R|2) . (5.10)

This current consists of two parts, one corresponds to the incident beam and is given by

hk

m=

p

m= v ,

which is the velocity of the incident particle. The second component of the current forx < 0 is the reflected current which is proportional to |R|2, where R is the amplitude ofthe reflected wave. Here we note that unlike the classical case, R 6= 0 for E > V0, and wehave a reflected current.

On the other hand the current in the region x > 0 is given in terms of the amplitudefor the transmitted wave, i.e. T , and can be written using the definition of the current as

j(x) =hk′

m|T |2 . (5.11)

Now the conservation of current at x = 0 requires that the current for x < 0 be equal tothe current for x > 0, and since these currents, as given in Eqs. (5.10) and (5.11) do notdepend on x, we have

hk

m(1− |R|2) =

hk′

m|T |2 or

hk

m=hk

m|R|2 +

hk′

m|T |2 . (5.12)

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78 CHAPTER 5. SIMPLE ONE DIMENSIONAL PROBLEMS

In other words, the sum of the reflected and transmitted current is equal to the incidentcurrent. At this stage we note that current conservation does not determine the amplitudeof the reflected and transmitted wave, it only puts a constraint on R and T in the formof Eq. (5.12).

We therefore turn to the boundary condition and the properties of our differentialequation to determine the constants R and T . Since the Schrodinger equation is a secondorder linear differential equation, we require that the wave function and its derivative becontinuous at all points, and in particular at x = 0. Thus from the continuity of the wavefunction we have that

1 +R = T . (5.13)

On the other hand the continuity of the derivative of the wave function requires that

k(1−R) = k′T . (5.14)

We now can solve these two linear algebraic equations (i.e., Eqs. (5.13) and (5.14)) re-sulting from the requirement of continuity of the wave function and its derivative, todetermine the amplitude of the reflected and transmitted waves R and T , to be

R =k − k′

k + k′, (5.15)

and

T =2k

k + k′. (5.16)

If we now compare these results with the classical result we find that:

1. For E > V0 we have a reflected as well as a transmitted wave, while classically wehad no reflected particle.

2. The transmitted current jT = hk′

m|T |2 and the reflected current jR = hk

m|R|2 add up

to give the incident current jI = hkm

, i.e.,

hk

m=hk

m|R|2 +

hk′

m|T |2 .

This result, which is a statement of current conservation, is consistent with theamplitudes for the reflected and transmitted current as given in Eqs. (5.15) and(5.16).

3. In the limit, when the energy of the incident particles is much greater than theheight of the barrier, i.e. E V0, the change in the momentum of the particle atx = 0 is expected to be small and k′ ≈ k. In this case the amplitude of the reflectedwave is much smaller than the amplitude of the transmitted wave, R T , and thereflected wave is negligible, which is close to the classical limit.

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5.2. POTENTIAL STEP 79

4. For the case when the energy of the incident beam is less than the barrier height,i.e. E < V0, the wave number for x > 0 is given by

k′ =

√2m

h2 (E − V0)

= iκ , (5.17)

where κ is real and given by the expression

κ =

√2m

h2 (V0 − E)

i.e., the momentum p′ = hk′ = ihκ becomes imaginary. This means that for x > 0,the wave function for the transmitted wave is given by,

ψ(x) = T e−κx for x > 0 . (5.18)

This wave function decays exponentially, and the corresponding current is zero ata detector place at a point x 0. In other words we have a decaying solution ofthe Schrodinger equation similar to the one we encountered when we had a particlebound in a well. The amplitude of the reflected wave in this case is given by,

R =k − iκk + iκ

, (5.19)

and the corresponding reflected current is given by

jR =hk

m|R|2 =

hk

m= jI . (5.20)

In other words all of the incident current is reflected by the barrier despite the factthat the wave function is not zero for x > 0. The fact that the wave function is notzero for x > 0 is an indication that the particle penetrates this region. In fact, theamplitude of the transmitted wave T is not zero and is given by

T =2k

k + iκ. (5.21)

To understand this apparent lack of current conservation, we should first recall thatthe wave in the region x > 0 is decaying exponentially, and the wave number k′ inthis region is given by

k′ = iκ where κ =

√2m

h2 (V0 − E) .

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80 CHAPTER 5. SIMPLE ONE DIMENSIONAL PROBLEMS

However, despite the fact that the wave function is not zero for x > 0, a calculationof the current in this region using Eq. (3.50) gives a zero current, proving theconservation of current. This is a result of the fact that we have a decaying wavefunction which goes to zero. The rate at which it goes to zero is governed by thevalue of κ. The larger the height of the barrier, the faster the decay of the wavefunction, and therefore the smaller the penetration depth. This non-zero penetrationof the particle is a purely quantum mechanical effect. We will come back to thisproblem of barrier penetration and current conservation when the barrier is of finitelength.

5.3 Potential Well

In the last section we considered scattering from a potential step, and when we comparedthe classical and quantum mechanical results we found two new features:

1. For E > V0 there was a reflected wave.

2. For E < V0 the particle penetrated the classically forbidden region x > 0. Inparticular, we found that in this region, the wave function decays exponentially,and in fact does not look like a wave any more.

x

x = -a x = a

V(x)

-V0

Figure 5.3: Plot for an attractive potential well of depth V0 and width 2a.

To see if these effects are present in other systems, and in particular if a particle whenplaced in a potential well does in fact penetrate the classically forbidden region, let usconsider a potential well of the form illustrated in Figure 5.3, i.e.,

V (x) =

0 for |x| > a

−V0 for |x| < a, (5.22)

where V0 > 0 is the depth of the potential well.

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5.3. POTENTIAL WELL 81

5.3.1 Bound state problem

Let us first consider the case when E < 0, in which case we expect the particle to bebound inside the well. The Schrodinger equation now takes the form

− h2

2m

d2ψ

dx2= E ψ for |x| > a , ( 5.23a)

while

− h2

2m

d2ψ

dx2− V0 ψ = E ψ for |x| < a . ( 5.23b)

These equations can be written, after multiplication by 2m/h2, as

d2ψ

dx2− α2ψ = 0 where α2 =

2m|E|h2 for |x| > a

. (5.24)

d2ψ

dx2+ k2ψ = 0 where k2 =

2m

h2 (V0 − |E|) for |x| < a

The general solution to these two equations is of the form

ψ(x) =

A e−αx +B eαx for |x| > a

C sin kx+D cos kx for |x| < a. (5.25)

To determine the constants A,B,C and D we need to make use of the boundary conditionsfor this problem. These boundary condition are: (i) The wave function and its derivativeshould be continuous for all values of x, and in particular at x = ±a. (ii) For the wavefunction of a bound state to be normalizable, we require that the wave function go tozero at x → ±∞. Let us first consider the region outside the well, i.e. |x| > a. Theboundary condition that the wave function be normalizable requires that for x < −a wehave A = 0, while for x > a we have B = 0, i.e.,

ψ(x) =

B eαx for x < −a

C sin kx+D cos kx for − a < x < a

A e−αx for x > a

. (5.26)

To determine the other constant, we make use of the continuity of the wave function andits first derivative at x = ±a. Thus at x = −a we have

B e−αa = −C sin ka+D cos ka

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82 CHAPTER 5. SIMPLE ONE DIMENSIONAL PROBLEMS

and

αB e−αa = kC cos ka+ kD sin ka ,

while at x = +a we have

A e−αa = C sin ka+D cos ka

and

−αA e−αa = kC cos ka− kD sin ka .

From these four equations we get

α = kC cos ka+D sin ka

−C sin ka+D cos ka

= kD sin ka− C cos ka

D cos ka+ C sin ka

which, after cross multiplication, can be written as

(D cos ka− C sin ka)(D sin ka− C cos ka) = (D cos ka+ C sin ka)(D sin ka+ C cos ka) .

This can be simplified to give us the condition that

CD = −CD or CD = 0 . (5.27)

This means that either D = 0 or C = 0, i.e. the solution for |x| < a is either sin kx orcos kx. These two solutions differ by the fact that sin(−kx) = − sin(kx), while cos(−kx) =cos(kx), i.e., one solution is an odd solution while the second is an even solution.

5.3.2 Symmetry of the Hamiltonian

At this stage we should realize that the Hamiltonian for this potential is invariant underthe transformation x → −x. This means that the parity operator P commutes with theHamiltonian, i.e.,

[H,P ] = 0 , (5.28)

and the eigenstates of H, i.e. the wave function ψ(x), should also be an eigenstate of theparity operator P . This implies that

Pψ(x) = πψ(x) , (5.29)

where π = ±1. For the case when π = +1, the wave function does not change under thetransformation x→ −x. This solution, known as the even solution, is given by,

ψπ=+1(x) = D cos kx for − a < x < a , (5.30)

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5.3. POTENTIAL WELL 83

where we have labeled the wave function ψ(x) by its parity of π = +1. For π = −1, thewave function ψ(x) should change sign under the transformation x→ −x. This solution,known as the odd solution, is given by

ψπ=−1(x) = C sin kx for − a < x < a . (5.31)

The even solution corresponding to π = +1 must now satisfy the boundary conditionsat x = ±a, i.e., the wave function and its derivative must be continuous, so that atx = −a,

B e−αa = D cos ka

αB e−αa = kD sin ka ,

while for x = +a the boundary condition gives us

A e−αa = D cos ka

−αA e−αa = −kD sin ka .

In both cases, if we divide the second equation by the first equation, we get

α = k tan ka . (5.32)

On the other hand, for the odd solution, i.e. π = −1, we have for the boundarycondition at x = −a

−B e−αa = C sin ka

αB e−αa = kC cos ka .

In a similar way we get a set of boundary conditions for x = a. Both of the boundaryconditions give the same expression, as was the case for the even solution, to be

− α = k cot ka . (5.33)

Both of these transcendental equations, Eqs. (5.32) and (5.33), when solved will giveus values for the energy E for which we can have a solution to the Schrodinger equation.In the case of bound state, i.e., E < 0, only certain energies will be allowed and these willbe the eigenstates of the Hamiltonian.

5.3.3 Eigenstates of the Hamiltonian

To see how we can solve these transcendental equations, let us consider the case of theeven parity solution i.e. π = +1. In this case the wave number inside the well is given by

k2 =2m

h2 (V0 − |E|) , (5.34)

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84 CHAPTER 5. SIMPLE ONE DIMENSIONAL PROBLEMS

while the corresponding quantity outside the well is

α2 =2m

h2 |E| . (5.35)

This allows us to write a relation between k and α of the form

k2a2 = η2 − α2a2 , (5.36)

where η is given in terms of the potential depth V0, and the potential width 2a, by

η2 =2mV0a

2

h2 . (5.37)

With this relation between the wave number k and α, we can rewrite the transcendentalEq. (5.32) in a form that will allow us to get a graphical solution, i.e.√

η2 − y2

y2= tan y , (5.38)

where y = ka. To solve this equation graphically we need to plot the right and left handside of Eq. (5.38), i.e.

F (y) = tan y and F (y) =

√η2 − y2

y2. (5.39)

The points at which the two curves intersect give those values of y for which we have asolution that gives the allowed energies for the positive parity solutions. (See left plot inFigure 5.4). In a similar way we can write Eq. (5.33) as

F (y) = − cot y and F (y) =

√η2 − y2

y2. (5.40)

These are plotted on the right hand side in Figure 5.4, and the intercept gives the eigen-values for the negative parity states.

Since |E| < V0, we have that

F (y) =

√η2 − y2

y2=

0 for y = η

→∞ for y → 0, (5.41)

and thus an increase in η will lead to intercepts at greater values of y. We may thenconclude that as η increases, the number of solutions increase and therefore the numberof bound states increases. From Eq. (5.41) and Figure 5.4, we can draw the followinggeneral conclusions:

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5.3. POTENTIAL WELL 85

2 4 6

-8

-4

4

8

0

positi parity

F(y)

y

2 4 6

-8

-4

4

8

0

F(y)

y

e parity

Figure 5.4: A plot of F (y) for the positive parity (on the left), and for the negative paritysolutions (on the right).

1. For the even solutions we have that

F (y) =

√η2 − y2

y2= tan y

and from the graph on the left in Figure 5.4 we see that we always have at least onesolution, and therefore one bound state.

2. For η < π/2, we have

V0a2 <

h2

2m

π2

4.

In this case Eq. (5.38) has a solution, see graph on the left-hand side of Figure 5.4.This bound state corresponds to the even solution, π = +1. For the odd solution,π = −1, the equation

F (y) =

√η2 − y2

y2= − cot y ,

has no solution, see graph on the right-hand side of Figure 5.4. As a result, forη < π/2 there is only one solution which is even under the parity transformation.

3. The ground state of the system corresponds to an even solution, while the firstexcited state corresponds to the odd parity solution.

If we compare the bound states problem for this potential with that of a particle in abox as discussed in the last section, we find that:

1. The wave function for a potential well extends into the region |x| > a which was notthe case for a particle in a box. On the other hand the shape of the wave functionsare similar. This is mainly due to the fact that the potential in both examples havethe same symmetry.

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86 CHAPTER 5. SIMPLE ONE DIMENSIONAL PROBLEMS

2. For a potential well, we have only a finite number of bound states in contrast to theparticle in a box where we had an infinite number of bound states. However as wewill see, for a potential well we have scattering states which were not present in thecase of a particle in a box.

5.3.4 The scattering problem

We now turn to the case when the energy E > 0, and consider a beam of particles ofunit amplitude incident from the left. Making use of the general results of the potentialstep discussed at the beginning of this chapter, we observe that for x < −a we have anincident wave from the left and a possible reflected wave. On the other hand, for x > a wehave only a transmitted wave. In the region in between, i.e −a < x < a, we expect wavesto travel both to the right and to the left. We therefore can write our wave function as:

ψ(x) =

eikx +R e−ikx for x < −a

A eik′x +B e−ik

′x for − a < x < a

T eikx for x > a

, (5.42)

where the wave number outside and inside the well are given by k2 = 2mE/h2 andk

′2 = 2m(E + V0)/h2, respectively. Here again we have taken the amplitude of theincident wave to be one, while that of the reflected and transmitted wave are taken to beR and T , respectively. We now can calculate the current in the three regions to be

j(x) =

hkm (1− |R|2) for x < −a

hk′m (|A|2 − |B|2) for − a < x < a

hkm |T |

2 for x > a

, (5.43)

and since the current in the three regions do not depend on the position x, the conservationof current will now require that

hk

m(1− |R|2) =

hk′

m(|A|2 − |B|2) =

hk

m|T |2 . (5.44)

Here we note that |R|2 + |T |2 = 1, i.e., the transmitted current plus the reflected current isequal to the incident current, i.e. no particles were created or destroyed in the scatteringoff the attractive square well. This is a statement of conservation of particle number orunitarity. We will come back to this conservation of particle number in a more generaldiscussion of scattering in Chapter14.

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5.3. POTENTIAL WELL 87

We note here that although the potential has reflection symmetry, the boundary con-dition that the incident beam is from the left has broken this symmetry and as a resultthe solution does not have the reflection symmetry we had in the bound state problem.Therefore, the constants R, A, B and T have to be determined by the boundary condi-tions that the wave function, and its first derivative be continuous at x = ±a. In factthese boundary conditions give

e−ika +R eika = A e−ik′a +B eik

′a ,

and

ik(e−ika −R eika) = ik′(A e−ik′a −B eik

′a)

for x = −a, while for x = +a, we have that

T eika = A eik′a +B e−ik

′a ,

and

ikT eika = ik′(A eik′a −B e−ik

′a) .

The boundary conditions on the wave function have given us exactly four linear equationsthat can be solved for the four unknown constants R, A, B and T . In particular, we havefor R and T

R = i e−2ika (k′2 − k2) sin 2k′a

2kk′ cos 2k′a− i(k′2 + k2) sin 2k′a, (5.45)

and

T = e−2ika 2kk′

2kk′ cos 2k′a− i(k′2 + k2) sin 2k′a. (5.46)

It is now a simple exercise to show that the above expression for the reflected and trans-mitted amplitudes satisfy the current conservation condition

|R|2 + |T |2 = 1 .

If we now define the cross section for reflection and transmission as

σR =jRjI

= |R|2 and σT =jTjI

= |T |2 (5.47)

then the sum of the two cross section is one, i.e.

σR + σT = 1 . (5.48)

Note that if E V0, then k′2 ≈ k2 and the amplitude for the reflected wave is

considerably reduced. On the other hand, as the energy E → 0, the wave number

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88 CHAPTER 5. SIMPLE ONE DIMENSIONAL PROBLEMS

k → 0, and the amplitude of the transmitted wave T → 0. However if sin 2k′a = 0, thencos 2k′a = 1, and T does not go to zero as k → 0. In this case

2k′a = nπ and λ =2π

k′=

4a

n. (5.49)

Or, if the width of the well

2a =λ

2n , (5.50)

then we get 100% transmission when the width of the well is an integer multiple of half awave length. In this limit, the reflected wave has an amplitude of zero, i.e., the wave goesthrough the potential well as if it is not there. In other words, the system represented bythe potential is transparent.

This effect, which corresponds to an increase in the transmission cross section with acorresponding reduction in the reflection cross section, is observed in noble gases and isknown as the the Ramsauer-Townsend effect. This is illustrated in Fig. 5.5

0.0 0.1 0.2 0.3 0.4 0.5E

0.2

0.4

0.6

0.8

1.0

|T| 2

0.0 0.1 0.2 0.3 0.4 0.5E

0.2

0.4

0.6

0.8

1.0

|R|2

Figure 5.5: The transmission and reflection probability for an attractive square well. Notethat at a specific energy the reflection probability goes to zero indicating zero scattering.

This effect of being able to change the transmission of a current by changing the widthcan have practical applications in switching devices at a microscopic level. Although in theabove discussion we concentrated on changing the width of the potential well to changethe transmission, we could have instead changed the potential depth and as a result thewave length of the particle in the well, to change the transmission amplitude. Note thatthe fact that we are considering a one dimensional system is not an approximation for aswitching device since most macroscopic switches are wires which are one dimensional. Infact we expect the next generation of computer chips to rely on solid state devices thatare based on this principle. The problem is how do we change the width or depth of thewell in a controlled manner.

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5.4. PROBLEMS 89

5.4 Problems

1. The wave function for a particle is given by

ψ(x) = Aeikx +B e−ikx .

What is the corresponding current?

2. Show that Eq. (5.12) is valid if we take the amplitude for the reflected and trans-mitted wave, R and T , from Eqs. (5.15) and (5.16).

3. Consider the one dimensional potential

V (x) =

+V0 for x < 00 for x > 0

.

(a) Write the most general solution of the Schrodinger equation for x > 0 and forx < 0, assuming E > V0.

(b) Does this solution have reflection symmetry?, i.e. symmetry under x → −x.Why?

(c) Consider the experimental situation when there is an incident beam of particlesfrom the left (i.e., from x = −∞). Calculate the amplitude for the reflectedand transmitted waves, given the amplitude for the incident wave is equal toone.

(d) Show that the incident current is equal to the sum of the reflected current andthe transmitted current.

4. Consider the square well potential

V (x) =

0 for x < −a−V0 for −a < x < +a0 for x > a

(a) Given the well size a = 5 × 10−15 m ≡ 5 fm, find the well depth V0, in MeV,that will support five bound states.

(b) Use the value of V0, determined in (a), to determine the energy and parity ofeach of the bound states.

(c) Plot using MAPLE or Mathematica the normalized wave function for theground state (i.e. the lowest energy state), and the first excited state.

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90 CHAPTER 5. SIMPLE ONE DIMENSIONAL PROBLEMS

We are given that hc = 197.3 MeV-fm, and mc2 = 940 MeV. Here, c is the velocityof light.

Hint: First use the graphical solution of the transcendental equation (5.32) and(5.33) to determine the well depth. Then use MAPLE or Mathematica to solve thetranscendental equation for the energy of each of the bound states.

5. Consider an arbitrary potential localized on a finite part of the x-axis. The solutionof the Schrodinger equation to the left and to the right of the potential region aregiven in the figure below.

A e + B e ikx -ikx C e + D e ikx -ikx

Show that if we write

C = S11A+ S12D

B = S21A+ S22D

i.e., if we relate the “outgoing” wave to the “incoming” wave by(CB

)=

(S11 S21

S12 S22

) (AD

),

that the following relations are valid:

|S11|2 + |S21|2 = 1

|S12|2 + |S22|2 = 1

S∗11S12 + S∗21S22 = 0 .

This is equivalent to the statement that the scattering matrix

S =

(S11 S12

S21 S22

)is unitary. (Hint, Use current conservation and the possibility that A and D arearbitrary complex numbers.)

6. Calculate the elements of the scattering matrix, S11, S12, S21, and S22 for the po-tential

V (x) =

0 for x < −aV0 for −a < x < a0 for x > a

and show that the scattering matrix is unitary.

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Chapter 6

Application of Quantum Mechanics

Abstract : Here we turn to some applications of quantum mechanics to systemsin atoms, nuclei and solids that can be modeled in terms of one dimensionalquantum systems.

In the present chapter we will develop simple one dimensional models of physicalsystems in atomic, nuclear and solid state physics. In all cases we try to simplify thephysical problem by constructing a one dimensional quantum mechanical model. In eachcase, the solution of the Schrodinger equation in one dimension will illustrate the majorfeatures of the system under consideration. Although the results of such models maynot be quantitative to the extent that they reproduce detailed experimental results, themain features are exhibited to get a qualitative understanding of the behavior of thesystem. In most cases further improvement will require an order of magnitude increasein computational power, which often renders such calculations beyond the scope of a firstcourse in quantum mechanics at the undergraduate level.

6.1 Barrier Penetration and α-decay

In 1896, Becquerel [18] accidentally discovered that uranium, when placed near unexposedphotographic plates, invariably results in the fogging of the plates when developed. Fur-ther examination of the phenomena led to the idea that some naturally occurring elementsare radioactive, i.e., they emit radiation. This radiation, which we now know emanatesfrom the nucleus of an atom, was found to be of three kinds:

1. α-rays – These are nuclei with two protons and two neutrons, i.e. they are oneof the isotopes of He. The nucleus that emits α particles loses two protons andtwo neutrons and as a result its atomic number is reduced by four, while its chargedecreases by two units of positive charge.

91

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92 CHAPTER 6. APPLICATION OF QUANTUM MECHANICS

2. β-rays – These are electrons or positrons that are emitted by the nucleus. Theseelectrons or positrons are not present in the nucleus (see problems at end of thischapter), but are created at the instant the nucleus decays. In this case the chargeon the nucleus changes by one unit depending on the charge of the the β-ray. If thenucleus emits an electron, the charge on the nucleus increases by one unit. However,if the β-ray is a positron, then the charge on the nucleus decreases by one unit.1

In the case of β decay, the number of neutrons protons changes by one, while theatomic number of the nucleus does not change.

3. γ-rays – These are electromagnetic radiation, or photons, that are emitted by thenucleus. In this case the number of neutrons and protons in the nucleus does notchange.

In all these decay modes, a nucleus in an excited state emits radiation and as a resultends in a state of lower energy. This process continues until the system, in this case anucleus, is in its ground state, or lowest energy state and cannot lose any further energy.We then say that the nucleus is stable. In Figure 6.1 we illustrate the decay mode of 211

83 Biwhich has an atomic number of 211 with 83 protons, and therefore 128 neutrons. Thisnucleus has two possible decay modes (i.e. ways of losing energy). The first is by α decayto 207

81 Tl, the second by β−-decay to 21184 Po. In turn, both 207

81 Tl and 21184 Po decay by β−- or

α-decay to 20782 Pb. This is a simple example of how a system in an excited energy state

loses its energy by any means it can to attain the lowest energy state, the ground state.

8

6

4

2

0

Ene

rgy

in M

eV αα

Bi

Pb207

82

Ti20781

Po21184

21183

β −

β−

Figure 6.1: A bimodal decay of 21183 Bi in which the daughter nuclei decay to the same final

nucleus 20782 Pb.

In the present section we would like to examine how a nucleus loses energy by α-decay. It was another 30 years after the discovery of radioactivity before Gamow [19],

1The magnitude of the charge on the proton is identical to the magnitude of the charge on the electron.

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6.1. BARRIER PENETRATION AND α-DECAY 93

Condon and Gurney [20] in 1928 gave a quantitative explanation of α-decay in terms of aquantum mechanical barrier penetration. This model for alpha-decay has since been usedto study the decay of nuclei, and in particular, the large variation in the lifetime (τ), orthe probability for decay, of some of these nuclei, e.g.,

23290 Th → 228

88 Ra + α τ = 2.03× 1010

21284 Po → 208

82 Pb + α τ = 4.3× 10−7

To understand this large variation in lifetime we will first consider the problem of barrierpenetration in one dimension. We will then modify this simple model to give a represen-tation of the decay of a nucleus by α-emission.

6.1.1 Potential Barrier

Consider the barrier of height V0 and width 2a centered about the origin, see Figure 6.2.The corresponding potential is given by

V (x) =

0 for |x| > a

+V0 for |x| < a(6.1)

x

V(x)

x = -a x = a

Figure 6.2: A plot of a potential barrier of height V0 and width 2a .

Since this potential is repulsive, we have no bound state, but as in the case of awell considered in the last chapter, we still have two possible domains for the energy E.These two energy domains correspond to: (i) Energies greater than the barrier height.(ii) Energies less than the barrier height. Classically, we expect the particle to be reflectedoff the barrier when the energy is less than the barrier height, and to go through thebarrier when the energy E is greater than the barrier height. In the quantum case we willfind that there is a transmitted and reflected wave in both cases. To consider these twopossibilities, we first consider the Schrodinger equation for |x| < a, i.e.,

d2ψ

dx2+

2m

h2 (E − V0)ψ(x) = 0 , (6.2)

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94 CHAPTER 6. APPLICATION OF QUANTUM MECHANICS

The solution in this region is given by

ψ(x) =

A eik

′x +B e−ik′x for E > V0

A e−κx +B eκx for E < V0

, (6.3)

where

κ2 =2m

h2 (V0 − E) while k′ =2m

h2 (E − V0) . (6.4)

The solution of the Schrodinger equation for |x| > a is identical to that of the potential wellconsidered in the last chapter (see e.g. Eq. (5.42)). Thus for E > V0, the wave functionfor the potential barrier is the same as that of the potential well, the only difference beingthe value of k′. For the potential barrier, k′ < k, while for the potential well, k′ > k.

The interesting aspect of this potential occurs in the case when 0 < E < V0. Theclassical solution for this problem of a beam of particles incident from the left is thatall the particles are reflected, because none of the particles have enough kinetic energyto overcome the potential energy barriers in the region |x| < a. However, the quantummechanical problem leads to a solution of the Schrodinger equation of the form

ψ(x) =

eikx +R e−ikx for x < −a

A e−κx +B eκx for −a < x < a

T eikx for x > a

(6.5)

To get the four constants (R, A, B and T ) fixed, we need to make use of the facts thatψ(x) and its derivative are continuous at x = ±a. The result in this case is the same asthe potential well if we make the substitution

k′ → iκ (6.6)

in Eq. (5.46), i.e., the transmission amplitude T , in this case is given by

T = e−2ika 2ikκ

2ikκ cos 2iκa− i(k2 − κ2) sin 2iκa

but we have that

cos 2iκa =1

2(e−2κa + e2κa) = cosh 2κa

and

sin 2iκa =1

2i(e−2κa − e2κa) = i sinh 2κa .

This allows us to write the amplitude for the transmitted wave as

T (k, iκ) = e−2ika 2kκ

2kκ cosh 2κa− i(k2 − κ2) sinh 2κa. (6.7)

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6.1. BARRIER PENETRATION AND α-DECAY 95

This means the transmitted current, which is defined as jT = hkm|T |2, is given by

jT =

(hk

m

)(2kκ)2

(k2 + κ2)2 sinh2 2κa+ (2kκ)26= 0 , (6.8)

i.e. the particles in the incident beam, or at least some of them, get through the barriereven though the energy of each particle is less than the barrier height, i.e. E < V0. Thiseffect is known as tunneling, and is purely a quantum effect and has no classical analog.It is this quantum effect that was required to understand α-decay as we will see in thenext section.

Before we proceed with the application of this tunneling effect to α-decay, we shouldhave a better understanding of how the tunneled or transmitted current depends on thebarrier height. To illustrate the relation between the transmitted current and the barrierheight, we consider the limit of κa 1, i.e. a 1, or V0 E. In this case we have

sinh 2κa→ 1

2e2κa .

In this limit the transmitted current becomes

jT = 4

(hk

m

) (2kκ

k2 + κ2

)2

e−4κa , (6.9)

which is what we expected, i.e., that as the height (V0) or thickness (a) of the barrierincreases, the transmitted current drops to zero exponentially. This in turn means thatthe transmitted current is a very sensitive function of the a and V0.

6.1.2 α-decay

There are several applications of tunneling in nature. The oldest and most famous isα-particle decay which was considered as the tunneling of α-particle2 from the inside ofthe nucleus to the outside. From inside the nucleus, the α-particle sees a potential thatcan be approximated by a square well, while from the outside, the α-particle sees theCoulomb potential due to a positively charged nucleus and the fact that the charge onthe α-particle is also positive. This allows us to write an approximation to the potentialthe α-particle sees that is of the form:

V (r) =

−V0 for r < R

+ZZ′e2

r for r > R

, (6.10)

where R is approximately the radius of the nucleus, Z and Z ′ are the charge of the finalnucleus and α-particle, and e is the charge on the proton.

2This was first considered by Gamow [19], Condon and Gurney[20] in 1928.

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96 CHAPTER 6. APPLICATION OF QUANTUM MECHANICS

1

-30

-10

10

30

50

r

V(r)

-V0

R

2 3 4 5 6

Figure 6.3: The nuclear potential as seen by an α-particle. In the figure we have illus-trated how we can replace the Coulomb barrier by a set of rectangular barriers whichwould correspond to replacing the integral in Eq. (6.11) by a sum, and therefore the totaltransmission |T | by a product of transmission for the different strips.

For the potential barrier we considered in Sec. 6.1.1, the κ in Eq. (6.9) is given by

κ =√

2mh2 (V0 − E), and is a constant since the potential was of constant height. However,

for the more realistic nuclear situation in α-decay, the potential is not a constant butdepends on the radial distance from the center of the nucleus r. As a result, the κ forthe potential in Eq. (6.10) is now dependent on r. This means the factor of 2κa inthe amplitude for the transmitted wave through the barrier should be replaced by theintegral of κ(x) over the barrier, i.e. we can write the exponential in the amplitude of thetransmitted wave as3

|T | ∼ exp

−+a∫−a

dx κ(x)

For α-decay the potential barrier is not constant in height, and the exponential factor is

|T | ∼ exp

−b∫

R

dr

√2m

h2 (V (r)− E)

. (6.11)

3We can replace the barrier in α-decay as given in Figure 6.3 by a series of barriers strips of thickness∆x. In this case the total transmission amplitude is, i.e.

|T | = |T1| |T2| |T3| . . . |TN |2

where |Ti| is the transmission amplitude for the ith strip in Figure 6.3, i.e.,

|Ti| = e−κi∆x .

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6.1. BARRIER PENETRATION AND α-DECAY 97

Here, b is the point at which V (b) = E outside the potential, i.e.,

E =ZZ ′e2

bor b =

ZZ ′e2

E.

This allows us to write the transmitted amplitude T as

|T | ∼ exp

−√

2m

h2 ZZ′e2

b∫R

dr

√1

r− 1

b

= exp

−√

2m

h2 ZZ′e2√b

cos−1(R

b

)1/2

−(R

b− R2

b2

)1/2 .

For R b this result simplifies to

|T | ∼ exp

−π2√

2m

h2 ZZ ′e2b

= exp

−πZZ ′e2

2h

√2m

E

. (6.12)

The probability for transmission through the barrier is equal to the rate the α-particlepresents itself at the barrier, v

2Rtimes the probability for transmission through the barrier

|T |2. This gives us, for the lifetime τ , the result4

τ−1 =v

2R|T |2

≈ v

2Rexp

− 4Z√E(MeV)

,

where we have taken the charge on the α-particle to be two, i.e. Z ′ = 2. This allows usto write the lifetime as

log10 τ−1 ≈ C1 − C2

Z

E1/2(MeV). (6.13)

where C1 and C2 are constants. This result is in remarkable agreement with the experi-mental lifetime of α emitting nuclei.

4In writing this result we have made use of the fact that

hc = 197.3 MeV fm e2 = 1.44 MeV fm

and the mass of the α particle is approximately four times the mass of the proton, which is

mpc2 = 938 MeV .

Here c is the velocity of light.

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98 CHAPTER 6. APPLICATION OF QUANTUM MECHANICS

At this stage we observe that as the energy of the emitted α-particle increases, theprobability for transmission through the barrier increases. This is due to the fact thatthe barrier width has decreased with increasing energy. Since the lifetime is inverselyproportional to the probability for transmission |T |2, the lifetime of the α-emitting nucleusdecreases. This result based on this simple one-dimensional model is in good agreementwith experiment.

6.2 The Deuteron

The deuteron is the simplest nuclear system we can consider. It consists of a bound stateof a proton and a neutron with a binding energy of 2.2246 MeV. The properties of thissystem can be used to examine the properties of the force between the neutron and proton.In the present section, we would like to determine how the binding energy of the deuteroncan be used to determine the depth of the potential that binds the proton-neutron system.

Before we examine the solution of the Schrodinger equation for the deuteron, let usassume that the force between the proton and the neutron is due to the exchange of a πmeson (pion) of mass mπ ≈ 140 MeV. From the energy-time uncertainty relation we have

∆E∆t ' h

and therefore

∆t ' h

∆E.

If we take the velocity of the pion to be approximately the velocity of light c, thenthe distance the pion can travel in the time ∆t will be approximately the range of thepotential, i.e.

a ' c∆t =hc

∆E∼ hc

mπc2' 200

140∼ 1.5 fm ,

where we have taken hc ≈ 200 MeV fm. Given the range of the force and the bindingenergy of the deuteron we need to estimate the depth of the potential. To determinethe kinetic energy of the proton-neutron system we make use of the position momentumuncertainty, i.e.,

∆p∆x = h .

Then taking ∆x to be the range of the potential, we can determine ∆p as

∆p ≈ h

a.

Assuming the relative momentum of the proton-neutron system is ∆p, the kinetic energycan now be written in terms of the uncertainty in the momentum as

(∆p)2

2m=

h2

2ma2=

h2c2

2(mc2) a2=

(200)2

103 × (1.5)2=

4× 104

2× 103' 20 MeV .

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6.2. THE DEUTERON 99

Here we have taken mc2 to be the reduced mass of the neutron-proton system. Since themass of the neutron is approximately equal to the mass of the proton, this reduced massis half the the mass of the proton which is approximately 1000 MeV, i.e. mc2 ≈ 500 MeV.The total energy of a system is the sum of the kinetic energy (a positive number) andthe potential energy ( usually a negative number). Given the energy of the system to be≈ −2.2 MeV,5 and the kinetic energy is approximately 20 MeV, we expect the potentialenergy to be about -22.2 MeV.

6.2.1 The binding energy of the deuteron

We now have to construct a one dimensional model for the neutron-proton system. Thefirst approximation is to take the potential between a neutron and a proton to be a squarewell of depth V0 and width a. This is illustrated in Figure 6.4. Since the distance betweenthe neutron and proton is r > 0, we need only consider the values of x > 0 when weconvert the problem from three- to one-dimension. As a result our model for the neutronproton system is the one-dimensional potential given by

- V

V(r)

rr = a

0

Figure 6.4: The radial potential between the neutron and proton in the deuteron.

V (x) =

+∞ for x < 0

−V0 for 0 < x < a

0 for x > a

, (6.14)

For this potential, the solution of the Schrodinger equation for x ≤ 0 is zero i.e. ψ(x) = 0for x ≤ 0. This leaves the domain x > 0 to play the role of the variable r which is the

5The binding energy of a system is a positive number, while the energy of a bound state is taken to bea negative number. Therefore, the binding energy of the system is the magnitude of the energy of thatsystem, i.e., for a bound state we have:

Binding energy of a system = −Energy of the system

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100 CHAPTER 6. APPLICATION OF QUANTUM MECHANICS

radial distance from the proton to the neutron. Thus for x > 0, the Schrodinger equationtakes the form

d2ψ

dx2= −

2mh2 (E + V0) ψ(x) for 0 < x < a

2mh2 E ψ(x) for x > a

, (6.15)

where m is the reduced mass of the proton-neutron system, i.e. m = mpmn/(mp + mn)where mp is the mass of the proton and mn is the mass of the neutron. Since the deuteronis a bound state of a proton and a neutron, we will consider the solution of the abovedifferential equation for E < 0. As the original problem is a three dimensional problemin which we have replaced the radial variable in spherical polar coordinates by the onedimension we are considering, we will replace x by r from this point on. The solution ofour Schrodinger equation is now of the form

ψ(r) =

A sin kr +B cos kr for 0 < r < a

C e−αr +D eαr for r > a. (6.16)

Here the parameters α and k are given in terms of the energy E and the well depth V0

by the relations

α =

√2m

h2 |E| and k =

√2m

h2 (V0 − |E|) . (6.17)

The constants A, B, C and D are to be determined by the boundary conditions onthe wave function and its derivative, and the normalization of this wave function. Theboundary condition at the origin requires that the wave be zero, i.e. ψ(0) = 0, Thiscondition is satisfied by requiring that B = 0. For a bound state, the wave function ψ(r)should go to zero as r →∞. This second boundary condition is satisfied if we take D = 0.The solution of the Schrodinger equation with these two boundary conditions satisfied cannow be written as

ψ(x) =

A sin kr for 0 < r < a

C e−αr for r > a. (6.18)

To determine the remaining constants A,C and the allowed energies or eigenvalues E, wemake use of the requirement that the wave function and its derivative to be continuousat r = a, i.e.,

A sin ka = C e−αa

kA cos ka = −αC e−αa . (6.19)

Dividing the second equation by the first equation we eliminate the constants A and C.The resulting transcendental equation is similar to that encountered in the last chapter

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6.2. THE DEUTERON 101

for an attractive potential well ( see Eq. (5.33) ), i.e.,6

k cot ka = −α . (6.20)

We can be solve this equation graphically by plotting the functions

F (y) = cot√y2 − α2a2 and F (y) = −

√α2a2

y2 − α2a2, (6.21)

where

y2 =2ma2V0

h2 , (6.22)

with y, α and k related by the equation

k2a2 = y2 − α2a2 . (6.23)

The graphical solution of Eq. (6.21) is given in Figure 6.5.

0.6 0.8 1.0 1.2 1.4

2

4

6

8

F(y)

y

Figure 6.5: The graphic solution of the transcendental equation. The intercept of the twocurves gives the solution.

The deuteron is a neutron-proton system with a binding energy of 2.2246 MeV. Thequestion is: can we find a well depth V0 that will give us a bound state with such a bindingenergy? Since there is only one such bound state, there is only one solution to the abovetranscendental equation. Let us call this solution y0. This value of y0, when used inEq. (6.22), gives the potential depth V0 provided we know the range of the potential a.We have estimated above that for a range a ≈ 1.5 fm, the well depth V0 ≈ 22.2 MeV.Since the binding energy is 2.2246 MeV, we have that V0 |E|. From the definition of yand α, we can establish that

y α2a2

6Note this equation is identical to the case of the odd solution encountered in the last chapter. This isa result of the fact that we imposed the boundary condition that the wave function be zero at the origin.

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102 CHAPTER 6. APPLICATION OF QUANTUM MECHANICS

so that

cot√y2 − α2a2 ≈ cot y and

√α2a2

y2 − α2a2 1 .

In this case F (y), as defined in Eq. (6.21) is approximately zero, i.e.,

F (y) = cot y ≈ 0 and y ≈ π

2.

In Figure 6.5 we have a plot of the Eq. (6.21) for a deuteron binding energy of 2.224 MeV,and find the graphical solution is at y = 1.38 radians. From this we can extract a valueof the well depth given a value for the range of the potential a. For example, we observehere that the binding energy of the deuteron does not determine both the depth andrange of our square well potential. It only determines the product V0a

2. To determinethe range and strength or depth of the potential we will need further information aboutthe scattering of neutrons from the proton. This problem will be considered when weconsider scattering in three-dimensions.

6.2.2 The deuteron wave function

We are now in a position to determine the relative wave function of the neutron andproton in the deuteron. Having determined the binding energy |E|, we now know thewave number k, and from the continuity of the wave function at r = a, we have that

A sin ka = Ce−αa .

This allows us to determine the constant C in terms of A, i.e.,

C = A eαa sin ka . (6.24)

We now have the wave function up to an overall constant A. This wave function is givenby

ψ(x) = A

sin kr for 0 < r < a

sin ka e−α(r−a) for r > a. (6.25)

Finally, to determine the overall amplitude of our wave function A, we have to make useof the fact that the wave function for a bound state has to be normalized if its square isto be interpreted as a probability, i.e.,7∫ +∞

0dr ψ∗(r)ψ(r) = 1 . (6.26)

7Note in three dimensions the radial integral in Eq. (6.26) should have an angle integration whichwould give a factor of 4π. We have chosen to ignore this for simplicity.

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6.3. THE δ-FUNCTION POTENTIAL 103

Making use of the wave function given in Eq. (6.25), we get

A2∫ a

0dr sin2 kx+ sin2 ka e2αa

∫ ∞a

dr e−2αr

= 1 .

Using the tabulated values of the integrals8 we find that A is given by

A2

a

2− 1

4ksin 2ka+

sin2 ka

= 1

or

A =

a

2− 1

4ksin 2ka+

sin2 ka

−1/2

. (6.27)

Given the parameters of the potential, we can determine the binding energy E, and thenormalized wave function for the deuteron.

6.3 The δ-Function Potential

Having considered two examples of the application of the one-dimensional Schrodingerequation to nuclear physics problems, we turn our attention to atoms and solids. Hereagain we find cases where some of the basic properties of molecules and solids can bemodeled in terms of the one dimensional solution of the Schrodinger equation. Since bothsolids and molecules are made up of atoms, we need to first consider a model for theatom that is simple enough to lend itself to analytic solutions. This model for an atom isthen used to construct first a model of a diatomic molecule, and then a one dimensionalsolid. In both cases we find that our model has some of the basic symmetries of the morecomplicated system. As a result, those features of the physical system that are due tosymmetry, are preserved in the simple one dimensional models we consider.

In a first step towards developing a one dimensional model of a molecule or a solid,we turn our attention to the case of a potential of the form

2m

h2 V (x) = −λaδ(x) . (6.28)

8 We have made use of the integrals∫ a

0

dx sin2 kx =1k

ka

2− 1

4sin 2ka

and ∫ ∞

a

dx e−2αa = +e−2αa

2α.

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104 CHAPTER 6. APPLICATION OF QUANTUM MECHANICS

This may be considered as the potential that determines the behavior of the valenceelectron in an atom. The Schrodinger equation for the bound state (E < 0) in thispotential takes the form

d2ψ

dx2− κ2ψ(x) = −λ

aδ(x)ψ(x) , (6.29)

where κ2 = 2mh2 |E|. The interest in this potential stems from the fact that one can combine

two such potentials to get a simple model of a molecule with one electron being shared bythe two nuclei in the molecule. Also, we can consider an infinite number of such potentialsthat are equally spaced along the x-axis. This is a simple model of a one dimensionallattice, and has the feature of being the potential the electron sees in a one dimensionalsolid or a one dimensional polymer.

The new feature of the Schrodinger equation for this potential is that it is singular atx = 0, and as a result the first derivative of the wave function is not continuous at x = 0.In fact if we integrate the above equation between −ε and +ε and take the limit as ε→ 0,we get

dx

∣∣∣∣∣+ε

−ε=dψ

dx

∣∣∣∣∣+ε

− dψ

dx

∣∣∣∣∣−ε

= −λaψ(0) . (6.30)

We will need to use this boundary condition to determine the constants in the solutionof the Schrodinger equation for this potential.

The solution of the Schrodinger equation for x 6= 0 is that of an equation with nopotential, i.e. for E < 0 it is given by

ψ(x) =

A eκx for x < 0

B e−κx for x > 0. (6.31)

The continuity of the wave function at x = 0 requires that A = B, while the discontinuityin the first derivative of the wave function should satisfy Eq. (6.30). This gives us

limε→0

[−Aκ e−κε − Aκ eκε

]= −λ

aA . (6.32)

Taking into consideration the fact that the exponentials inside the brackets are one in thelimit as ε→ 0, we get

2κ =λ

a. (6.33)

From this result we see that this δ-function potential supports one, and only one boundstate with the binding energy given by

|E| = h2

2mκ2 =

h2λ2

8ma2. (6.34)

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6.4. THE DIATOMIC MOLECULE 105

The wave function for this bound state is given in Eq. (6.31) with A = B and κ given byEq. (6.33). The overall normalization constant A is now determined by the fact that thewave function has to be normalized, i.e.

+∞∫−∞

dx |ψ(x)|2 = 2A2

∞∫0

dxe−2κx =A2

κ= 1 .

This gives, for the normalization constant, A =√κ, and the normalized wave function as

ψ(x) =√κ

eκx for x < 0

e−κx for x > 0. (6.35)

This wave function is illustrated in Figure 6.6, where the wave function is scaled so thatit has a unit amplitude at x = 0, which would correspond to κ = 1. This wave function issimilar to the wave function of a valence electron for large separation between the electronand the nucleus of the atom. This is the most important feature of the wave function ofan electron in an atom when it comes to constructing either molecules or solids. In thisway we have preserved that feature of the atomic wave function needed for molecular andsolid state physics. In the atomic case, the binding energy for this potential will be theionization energy of the valence electron.

-3 -2 -1 1 2 3

0.2

0.4

0.6

0.8

1.0

x

ψ (x)

Figure 6.6: The wave function for a δ-function potential.

6.4 The Diatomic Molecule

In a diatomic molecule we have two nuclei a distance R apart with one or more electronsbeing shared by the two atoms in the molecule. These electrons are in a potential with

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106 CHAPTER 6. APPLICATION OF QUANTUM MECHANICS

two centers. The simplest such molecule is the hydrogen molecular ion. In this case wehave one electron and two protons. Since the protons are much heavier than the electron,to first approximation we can assume the two protons are stationary a distance R apart,while the electron is shared by the two centers. In this way we can replace the three-bodyproblem (the electron and two protons) by an electron moving in an average field of thetwo protons. This problem can be approximated by a one dimensional problem in whichwe have one proton stationary at x = a while the second proton is at x = −a. This meansthat R, the distance between the two protons, is 2a. The potential energy the electronsees is the Coulomb potential due to the two protons, which is attractive. This we canapproximate by the potentials used in the last section for the valence electron in an atom,i.e., the potential is the sum of two δ-functions.

Let us now consider the problem of two δ-function potentials placed symmetricallywith respect to the origin, e.g., at x = ±a. This potential can be written as

2m

h2 V (x) = −λaδ(x+ a) + δ(x− a) , (6.36)

and is invariant under inversion i.e. x→ −x. As a result of this symmetry, the solutionsare eigenstates of the parity operator which means they are either symmetric (even) orantisymmetric (odd). We first consider the symmetric solution which is given by

ψe(x) =

C eκx for x < −a

A coshκx for −a < x < a

C e−κx for x > a

. (6.37)

Note, coshκx = 12(eκx + e−κx) and is a symmetric function, i.e., it is an even function of

x. The continuity of the wave function at x = a gives

A coshκa = C e−κa . (6.38)

On the other hand, the discontinuity in the derivative of the wave function at x = a gives( see Eq. (6.30) )

−κC e−κa − κA sinhκa = −C λ

ae−κa

or

A sinhκa = C

κa− 1

)e−κa . (6.39)

Combining the results of Eqs. (6.38) and (6.39), we get

tanhκa =

κa− 1

). (6.40)

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6.4. THE DIATOMIC MOLECULE 107

Since tanhκa is positive for κa positive, we expect the solution of this transcendentalequation to be restricted to those values of κ that satisfy the relation ( λ

κa− 1) > 0 or

λκa> 1, and therefore κ < λ

a. On the other hand since tanhκa < 1, the solution of the

transcendental equation is in the domain where ( λκa− 1) < 1 or λ

κa< 2 and therefore

κ > λ2a

. In other words, the solution of Eq. (6.40) is restricted to values of κ in thedomain

λ

2a< κ <

λ

a.

In Figure 6.7 we have a graphical solution of Eq. (6.40). Here again we have only onebound state with positive parity. At this point we note that κ > λ

2a, i.e., the even solution

has a binding energy greater than that of one δ-function potential as given in Eq. (6.33).

0.4 0.6 0.8 1.0 1.2

0.5

1.0

1.5

2.0

κa

tanh κa

-1+1/κa

Figure 6.7: This is a graphical determination of the solution to the equation tanhκa =1κa− 1. The numerical solution is κ = 0.639.

We now turn to the antisymmetric (odd) solution. This solution is given by

ψo(x) =

−D eκx for x < −a

B sinhκx for −a < x < +a

D e−κx for x > a

. (6.41)

In this case the boundary condition that the wave function is continuous at x = a givesus

B sinhκa = D e−κa , (6.42)

while the boundary condition on the derivative of the wave function at x = a gives us

−κD e−κa − κB coshκa = −D λ

ae−κa ,

which can be written as

B coshκa = D

κa− 1

)e−κa . (6.43)

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108 CHAPTER 6. APPLICATION OF QUANTUM MECHANICS

Combining the results of Eqs. (6.42) and (6.43), we get

cothκa =

κa− 1

)

or

tanhκa =

κa− 1

)−1

=κa

λ− κa. (6.44)

0.2 0.4 0.6 0.8 1.0

0.2

0.4

0.6

0.8

1.0

0.2 0.4 0.6 0.8 1.0

0.2

0.4

0.6

0.8

1.0

κa

tanh κa

κa/(λ−κa)

κα/(λ−κα)

tanh κa

κa

Figure 6.8: The graphical determination of the solution to the equation tanhκa = κaλ−κa

for λ = 2 (left graph), and λ = 1.1 (right graph).

In this case we are not always guaranteed a solution. If the slope of κa/(λ − κa)is greater than the slope of tanhκa at κa = 0 there is no solution (see Figure 6.8 onthe right), and therefore there is no bound state for the antisymmetric case. If thereis a solution to the transcendental equation it should be for values of κ that satisfy therelation

κa

λ− κa< 1 ,

or

2κa < λ and therefore κ <λ

2a. (6.45)

This means the binding energy of this state is less than that of the symmetric solution.The wave function for the two solutions is given below in Figure 6.9. Here we note thatthe behavior of the wave function for |x| > a is completely determined by the bindingenergy. This behavior is preserved if we go to a more realistic model for the molecule. Onthe other hand, the behavior of the wave function for |x| < a is determined for the twosolutions by the symmetry of the system under reflection, i.e., x→ −x. In the case of thereal molecule this symmetry corresponds to exchanging the two protons in the molecule.Since the two protons are identical, the exchange of the two protons is a good symmetryof the system and therefore the Hamiltonian. As a result, the qualitative behavior of thetwo solutions for |x| < a, which is governed by the symmetry of the system, will holdtrue for the real molecule as well. In particular, we expect the even solution (blue) tocorrespond to a state which is more bound than that of the odd solution (red). We note

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6.4. THE DIATOMIC MOLECULE 109

-4 -2 2 4

-0.4

-0.2

0.2

0.4

x

x = -a x = a

ψ (x)

Figure 6.9: This is a graph of the wave functions of the ground state (blue) and firstexcited state (red) for the potential with two attractive δ-functions at x = ±a.

here that we applied our boundary conditions at x = +a and symmetry of the systemtook care of the boundary condition at x = −a.

To get a better understanding as to why the symmetric solution is more bound thanthe antisymmetric solution, we write the total energy of the molecule as the sum of thekinetic energy and the potential energy i.e.,

E = K.E.+ P.E. . (6.46)

We know that the potential energy is negative for attractive potentials while the kineticenergy is always positive. To get a bound state we need to have the energy E < 0. Thepotential energy is given by the matrix element of the potential, i.e.,

P.E =∫dxψ∗(x)V (x)ψ(x) =

∫dx |ψ(x)|2 V (x) . (6.47)

Since V (x) < 0 for attractive potentials, and |ψS(x)|2 > |ψA(x)|2 because ψA(x) goesthrough zero, the P.E for the symmetric wave function is larger in magnitude than theP.E. for the antisymmetric wave function, i.e.,

(P.E)S =∫dx |ψS(x)|2 V (x) <

∫dx |ψA(x)|2 V (x) = (P.E)A . (6.48)

On the other hand, the K.E is given by

K.E =∫dxψ∗(x)

(− h2

2m

d2

dx2

)ψ(x)

=h2

2m

∫dx

∣∣∣∣∣dψdx∣∣∣∣∣2

> 0 . (6.49)

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110 CHAPTER 6. APPLICATION OF QUANTUM MECHANICS

But from Figure 6.9 it is clear that ∣∣∣∣∣dψSdx∣∣∣∣∣ <

∣∣∣∣∣dψAdx∣∣∣∣∣ ,

for −a < x < +a, and therefore

(K.E)S < (K.E)A . (6.50)

This allows us to compare the energy of the symmetric and the antisymmetric states,with the result that

ES = (K.E)S + (P.E)S < (K.E)A + (P.E)S

< (K.E)A + (P.E)A = EA . (6.51)

Thus our simple model provides us with a simple explanation for why the H+2 molecule

has a symmetric wave function for the ground state. We observe that this simple model,that has the basic symmetry of the real H+

2 , has given us a good qualitative understandingof this diatomic molecule. This is just an example of the powerful role symmetry plays inquantum systems.

6.5 The Atomic Clock

We now consider the above diatomic molecule in which we have two states with differentenergies, one corresponding to an even solution the other due to an odd solution. We nowdemonstrate how this system can be used to construct a frictionless oscillator that can beused as a clock.

Consider the above two-center potential with an initial condition in which the particleis localized to one of the two centers. Clearly this state is going to be a linear combinationof the symmetric (even) and antisymmetric (odd) wave functions, i.e.

ψi(x) = ψS(x) + αψA(x) . (6.52)

The constant α is determined by the requirement that the probability of finding theparticle along the negative x-axes (i.e. the nucleus on the left) should be zero, i.e.,∫ 0

−∞dx|ψi(x)|2 = 0 . (6.53)

The time dependence of this initial wave function is given by

Ψi(x, t) = ψS(x) e−iESt/h + α e−iEAt/hψA(x)

= e−ESt/hψS(x) + α ei(ES−EA)t/hψA(x)

. (6.54)

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6.6. ONE DIMENSIONAL SOLIDS 111

Since the symmetric and antisymmetric states have different energies, the time evolutionof the two eigenstates will be different. As a result, the relative phase of ψS(x) and ψA(x)changes with time. Thus if we take α at t = 0 so the wave function is that of an electronlocalized to the right hand nucleus (|Ψ(x, 0)|2 = 0 for x < 0 ), then at a later time t1 > 0,we have

ei(ES−EA)t1/h = −1 ,

and this corresponds to the electron being localized to the left hand side nucleus (|Ψ(x, t1)|2= 0 for x > 0). This oscillation of the electron between the left and right nucleus keepsgoing on with time, i.e., the electron hops from one nucleus to the other and back with afrequency

ω = 2(ES − EA)/h . (6.55)

This frequency depends on the energy difference between the symmetric state and theantisymmetric state. The period of this oscillation is directly related to the time it takesfor the state bound on one nucleus in the molecule to penetrate the region between thetwo nuclei. Such a system can be used as a very accurate clock and since there is nofriction in the system, the frequency does not change with time and the clock is stable.

6.6 One Dimensional Solids

Having demonstrated the success of predicting some of the features of a diatomic moleculein terms of a potential with two δ-function in one dimension, we turn our attention to thepossibility of determining some of the features of a solid by extending the above model tosolids. In a solid, the atoms are placed at regular sites with fixed separation between theatoms. In a diatomic molecule the valence electrons are shared by the two atoms, whilein a solid the valence electrons are shared by more than one atom. In fact, we will findthat the electrons move through the solid almost as if they were a gas of free electrons.The atoms, or ions9, in the solid are fixed in space. The most the ions can do is vibrateabout their equilibrium position. For the present analysis we will assume that the atomsor ions are fixed, and concentrate on the motion of the electron. Since the ions are fixedin space we can model the solid in terms of potential wells placed at regular intervals.In one dimension this would correspond to atoms placed at a regular spacing of lengtha. Thus if we neglect the ends of our one dimensional solid, the solid has the symmetrythat a translation by an amount a along the x-axis leaves the solid unchanged. Thissymmetry under translation will be present in the potential the electron feels, and shouldbe included in the analysis of the problem of the motion of the electron in the solid. Thistranslational symmetry is also present in two and three dimensional crystals. Thus ina three dimensional crystal, we have translational symmetry along three axes. We willfirst consider the constraint this translational symmetry places on the wave function. We

9Since each atom has lost its valence electron, what is left behind is a positively charged ion.

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112 CHAPTER 6. APPLICATION OF QUANTUM MECHANICS

will see that with the help of this symmetry, a problem that looks very difficult becomesalmost as simple as that of a diatomic molecule.

6.6.1 Translational symmetry – Bloch’s Theorem

In solids, and metals in particular, the atoms are fixed to certain positions in space, whilethe electrons are free to move. In fact, the valence electrons in metals behave like afree gas of electrons with a background potential that is periodic. This periodicity orsymmetry can be used to derive the general form of the wave function that insures the ofthe system symmetry is satisfied. This was first established by Felix Bloch and is knownas the Bloch’s Theorem [21].

In one dimension, this periodicity can be stated as a symmetry in the potential of theform

V (x) = V (x+ a) , (6.56)

and the Hamiltonian has the property that

H(x) = H(x+ a) . (6.57)

This means the wave function should also have this symmetry.10

1 2 3 4

0.4

0.8

x

ψ(x)

1 2 3 4

0.4

0.8

x

ψ'(x)

Figure 6.10: The wave function ψ(x), peaked at x = x0 = 1, and the translated wavefunction ψ′(x), peaked x = x0 + λ = 3.

Let us examine this symmetry in more detail to determine the operator for this sym-metry. Consider a system described by the wave function ψ(x). If we displace this systemby the transformation

x→ x+ λ , (6.58)

then the new system will be described by a new wave function ψ′(x). This new wavefunction is related to the old wave function ψ(x) by the action of a displacement operator,i.e.,

ψ′(x) = D(λ) ψ(x) , (6.59)

10Although solids are three dimensional systems, it is possible to form polymers that are chains ofmolecules that form one dimensional periodic systems.

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6.6. ONE DIMENSIONAL SOLIDS 113

where D(λ) is the operator that moves the system by the distance λ. From Figure 6.10it is clear that

ψ′(x) = ψ(x− λ)

= ψ(x)− λ ddxψ(x) +

λ2

2!

d2

dx2ψ(x)− · · ·

=

1− λ d

dx+

1

2!λ2 d

2

dx2− · · ·

ψ(x) . (6.60)

In writing the second line of Eq. (6.60), we have made use of the Taylor series expansionfor ψ(x − λ) about ψ(x). Since the momentum operator in coordinate space is given byp = −ih d

dx, we can write the last line of Eq. (6.60) as

ψ′(x) =

1− iλp

h− 1

2!

(iλp

h

)2

− · · ·

ψ(x)

= e−iλp/h ψ(x) . (6.61)

From this result we can read off the operator D(λ) to be11

D(λ) = e−iλp/h . (6.62)

If the Hamiltonian does not change under the transformation x→ x+λ, then the Hamil-tonian should commute with the operator D(λ)12 , i.e.,

[H,D(λ)] = 0 . (6.63)

11Note the operator D(λ) moves the system by a distance λ to the right. This is equivalent to movingthe coordinates system a distance λ to the left, see Figure 6.10.

12Given the time dependent Schrodinger equation

ih∂Ψ∂t

= H Ψ ,

we can operate on this equation with the translation operator D(λ) to get

ih∂

∂tD(λ)Ψ ≡ ih ∂Ψ′

∂t= D(λ)H Ψ

= D(λ)HD−1(λ)D(λ) Ψ= D(λ)HD−1(λ) Ψ′ .

Here, we have defined Ψ′ ≡ D(λ)ψ. Invariance of the equation under the transformation D(λ) (i.e., theshape of the wave function does not change) requires that both Ψ and Ψ′ satisfy the same equation. Thiscan be achieved only if

D(λ)HD−1(λ) = H or D(λ)H = HD(λ) .

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114 CHAPTER 6. APPLICATION OF QUANTUM MECHANICS

Because D(λ) is a function to the momentum operator p, D(λ) is defined in terms of apower series in the momentum p (see Eq. (3.60)). Therefore for Eq. (6.63) to be valid,the momentum operator has to commute with the Hamiltonian, i.e.,

[H, p] = 0 , (6.64)

This implies that the momentum is a constant of the motion, and that we can constructsimultaneous eigenstates of H and p, i.e.,

H ψ = E ψ (6.65)

p ψ = hk ψ . (6.66)

For solids, λ = a is the spacing between the atoms, while in free space, λ can take onany value. We now assume that the wave function of a one dimensional periodic systemis of the form

ψk(x) = eikxUk(x) , (6.67)

where Uk(x) is known as the Bloch function. Then the wave function at x+ a is given as

D−1(a)ψk(x) = ψk(x+ a)

= eik(x+a)Uk(x+ a)

= eikaeikxUk(x+ a)

.

But since pψk = hkψk and D−1(a) = eiap/h, we have that

D−1(a)ψk(x) = eika ψk(x) = eika eikx Uk(x) .

We therefore have that

Uk(x) = Uk(x+ a) , (6.68)

and

D(a)ψk(x) = e−ika ψk(x) , (6.69)

i.e., ψk(x) is also an eigenstate of the displacement operator D(a). At the same time wecan calculate the wave function at x+ a, given the wave function at x, by the applicationof D−1(a), i.e.

ψk(x+ a) = D−1(a)ψk(x) .

We emphasize here that the operator D(λ) was constructed to move the physical systemrepresented by a wave function and not the coordinate system defined by the observer.

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6.6. ONE DIMENSIONAL SOLIDS 115

6.6.2 The Kronig-Penney Model

Let us now turn to a specific example of a one dimensional solid first suggested by Kronigand Penney and referred to in the literature as the Kronig-Penney Model [22]. In thismodel we can explicitly determine the Bloch wave function Uk(x) and the correspondingeigenstates or spectrum. Consider the potential that is the sum of δ-functions potentialswith the same strength and at regular spacing of a. Each of these δ-functions wouldcorrespond to the potential an electron feels due to one of the ions in our one dimensionalcrystal. This potential is of the form

V (x) = − h2

2m

λ

a

+∞∑n=−∞

δ(x− na) . (6.70)

This potential consists of a series of repulsive δ-functions (spikes) with spacing a betweenthe spikes. The Schrodinger equation for this system is

− h2

2m

d2ψ

dx2+ V (x)ψ(x) = Eψ(x) . (6.71)

For E > 0, this equation can be written as

d2ψ

dx2+ k2

0ψ(x) = −λa

+∞∑n=−∞

δ(x− na)ψ(x) , (6.72)

where k20 = 2m

h2 E.From the symmetry of the problem, we know that the wave function can be written

asψ(x) = eikxUk(x) , (6.73)

with Uk(x+ a) = Uk(x). This implies that

ψ(x+ a) = eikaψ(x) . (6.74)

For −a < x < 0, the potential is zero and we can write the solution of the Schrodingerequation as

ψ1(x) = A eik0x +B e−ik0x for − a < x < 0 , (6.75)

while for 0 < x < a, we can write the wave function, making use of Eq. (6.74), as

ψ2(x) = eikaψ1(x− a)

= eikaA eik0(x−a) +B e−ik0(x−a)

for 0 < x < a . (6.76)

As illustrated in Figure 6.11, the wave function ψ1 is defined over the interval −a < x < 0,while the wave function ψ2 is defined over the interval 0 < x < +a.

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116 CHAPTER 6. APPLICATION OF QUANTUM MECHANICS

x = 0x = -a x = a

2

Figure 6.11: The domain for the wave functions ψ1(x) and ψ2(x).

The boundary condition at x = 0 for the wave function is now given as

ψ1(−ε) = ψ2(ε) as ε→ 0 , (6.77)

while that on the derivative of the wave function is

ψ′2(ε)− ψ′1(−ε) = −λaψ1(−ε) as ε→ 0 . (6.78)

Using our wave functions, as given in Eqs. (6.75) and (6.76), in these boundary conditions,we get

A+B = eikaA e−ik0a +B eik0a

, (6.79)

and

ik0 eika

A e−ik0a −B eik0a

− ik0(A−B) = −λ

a(A+B) . (6.80)

These two equations resulting from the application of the boundary conditions atx = 0, can be rewritten as

A

1− ei(k−k0)a

+B

1− ei(k+k0)a

= 0 , (6.81)

and

A

λ

ik0a− 1 + ei(k−k0)a

+B

λ

ik0a+ 1− ei(k+k0)a

= 0 . (6.82)

For these two homogeneous equations to have a solution, the determinant of coefficientsmust be zero, i.e.,

λ

ik0a+ 1 − ei(k+k0)a − λ

ik0aei(k−k0)a − ei(k−k0)a + e2ika

λ

ik0a− 1 + ei(k−k0)a − λ

ik0aei(k+k0)a + ei(k+k0)a − e2ika

= 0

or

2eika(e−ika + eika

)− 2eika

(eikoa + e−ik0a

)+λeika

ik0a

(eik0a − e−ik0a

)= 0 .

This result can be written as a transcendental equation of the form

cos ka = cos k0a−λ

2k0asin k0a . (6.83)

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6.6. ONE DIMENSIONAL SOLIDS 117

-30 -20 -10 10 20 30-1

1

3

5

k a0

(-λ / 2k a) sin k a + cos k a0 0 0

Figure 6.12: This is a graphical solution of the equation cos ka = λ2k0a

sin k0a + cos k0a.Since | cos ka| ≤ 1, only certain ranges of values of k0a are allowed. These correspond tothe energy bands in our one dimensional solid.

This is similar in form to Eq. (6.20) for the bound state of a particle in a square well (e.g.the deuteron), and can be solved graphically as illustrated in Figure 6.12.

The left hand side of Eq. (6.83) can take values between −1 and +1 for all valuesof the momentum k. On the other hand, the right hand side of this equation can havevalues greater than +1 and less than −1 as demonstrated in Figure 6.12. Thus the energyE = h2

2mk2

0 can take only those values for which the right hand side of Eq. (6.83) is between−1 and +1. This suggests that there are bands of energies allowed, and within each bandwe have continuous values for the energy E.

-20 -10 10 20-20

20

40

60

80

energy E

k

Figure 6.13: The energy-wave vector plot for a one dimensional Kronig-Penney model.

From the wave function of the electron as given in Eqs. (6.75) and (6.76) for a one di-mensional solid, we expect the electron to behave as a free particle. This can be confirmedby the relation between the energy E and the momentum k. In Figure 6.13 we present aplot of the energy E verses the momentum k. It is clear from this figure that for thoseenergies allowed the relation between the energy and momentum is almost that of a free

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118 CHAPTER 6. APPLICATION OF QUANTUM MECHANICS

particle. This suggests that the electrons within a given band behave like a gas of freenon-interacting electrons, and the mobility of these electrons determines the conductivityof the solid. The detailed properties of the band, i.e., the relation between the energyand momentum, are determined by the crystal structure. Thus a central problem in solidstate physics is to find crystal structures for which the band, and therefore the propertiesof the solid, are those useful for certain devices. Here again we find that a simple modelthat has the correct symmetry gives the general properties of the solid in the form ofenergy bands and the fact that the electrons within a band behave like a free gas. A moredetailed and therefore more realistic model for a solid will give the Bloch wave functionUk(x), and the relation between the energy E and momentum k. To achieve this willrequire more complex computational methods than those considered here.

6.7 Problems

1. Consider the potential

V (r) =

−V0 for r < R0V1

R0(R1 − r) for R0 < r < R1

0 for r > R1

as a model for α-decay.

(a) Calculate the lifetime when R0 = 5 fm., R1 = 20 fm., V0 = 50 MeV, andV1 = 10 MeV. The energy of the final α particle is 1 MeV.

(b) Use MAPLE or Mathematica to calculate the lifetime of the nucleus as a func-tion of the energy of the α particle.

2. Given: the binding energy of the deuteron is 2.2246 MeV and the range of theneutron proton potential is 1.5 fm:

(a) Assuming the potential is a square well, find the well depth V0.

(b) Use MAPLE or Mathematica to plot the wave function of the deuteron.

3. Consider the δ-function potential

V (x) =h2

2m

λ

aδ(x)

(a) Show that the scattering matrix (S-matrix as defined in the last chapter) isgiven in this case by

S(E) =1

c− λ

(c λλ c

),

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6.7. PROBLEMS 119

where

c = 2ika and k2 =2mE

h2 .

(b) Prove that this matrix is unitary.

(c) Show that for λ < 0 the S-matrix becomes infinite when E is the bound stateenergy for the potential.

4. Consider the Kronig-Penney potential

2m

h2 V (x) =λ

a

n=+∞∑n=−∞

δ(x− na)

with λ = 3π.

(a) Make a detailed plot of

cosx+λ

2

sinx

x

as a function of x = k0a.

(b) Show that forbidden energy bands start just above k0a = nπ.

(c) Show that the allowed energy bands get narrower as λ increases.

(d) Plot the energy h2k20/2m as a function of k.

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120 CHAPTER 6. APPLICATION OF QUANTUM MECHANICS

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Chapter 7

Molecular Vibration –The Harmonic Oscillator

Abstract : In the present chapter we turn to the one-dimensional harmonicoscillator potentials for which the Schrodinger equation admits analytic so-lutions. This will allow us to examine the vibrational spectrum of quantumsystems, and in particular, diatomic molecules.

Vibrational modes of excitation are a common feature in many quantum, as well asclassical systems. Invariably these vibrational modes are the result of the displacement ofthe system from equilibrium, with these deviation being small in magnitude. In quantumsystems, these vibrational excitations are present in molecules, nuclei, solids and molecularclusters. Usually, such vibrational excitations have the common feature that the energyspectrum has the unique signature whereby the spacing between the levels is constant andcan be represented by hω. In the present chapter, we will consider vibration in diatomicmolecules, but the formulation can be extended to more complicated system such as thevibrational spectrum of nuclei or the vibration of an atom in a solid about it’s equilibriumposition. The choice of diatomic molecules is dictated by the fact that we can model thisvibration in terms of a one dimensional harmonic oscillator.

In the last two chapters we restricted our analysis to systems for which the corre-sponding Schrodinger equation has the simple form(

d2

dx2± k2

)ψ(x) = 0 .

This equation had simple solutions, and to that extent, we avoided the question of howto solve the Schrodinger equation as a second order linear differential equation. We arenow at a stage to consider more complicated potentials and as a result need to considerthe problem of solving the Schrodinger equation for these more complicated potentials.

121

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122 CHAPTER 7. MOLECULAR VIBRATION – THE HARMONIC OSCILLATOR

In practical terms there are only a handful of potentials for which we have analyticsolutions. The more common practice is to solve the differential equation numericallyon a computer, or use one of the standard packages such as MAPLE or Mathematicato extract the numerical or analytic solution. One of the motivations for consideringthe solution of the Schrodinger equation for the harmonic oscillator is to determine themechanism by which we end up with a spectrum that is quantized.

7.1 The One Dimensional Harmonic Oscillator

The harmonic oscillator problem in Physics is a very good approximation to all phenomenawhich result from the system deviating by a small amount from equilibrium. A simpleexample of such a situation is the vibrational excitation of a diatomic molecule. Here thepotential between the two atoms in a diatomic molecule is attractive at large distancesand is the result of each atom polarization the other atom, i.e. the interaction is a dipole-dipole interaction. On the other hand, as the atoms get closer and their electronic cloudsoverlap, this interaction is highly repulsive. In other words the potential is of the form ofthe Lennard Jones potential [23] (see Figure 7.1)

V (r) = 4ε

[ (σ

r

)12

−(σ

r

)6]

(7.1)

where ε and σ are parameters of the potential that are adjusted to fit either scatteringdata, or the thermodynamic properties of the system. The lowest energy state of themolecule corresponds to an average separation of r0, where r0 is the distance at which thepotential is a minimum. If we now consider the lowest excitation of the molecule, thenthey are of two kinds: (i) the vibration of the two atoms about the equilibrium distancer0. (ii) the rotation of the molecule as a rigid object. For the vibrational motion, providedthe amplitude of vibration is small, the potential energy V (r) can be expanded about theequilibrium separation of the two atoms, i.e.,

2 4 6

-10

10

0 rr=R

30

V(r)

Figure 7.1: The Lennard-Jones potential for the He-He system.

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7.1. THE ONE DIMENSIONAL HARMONIC OSCILLATOR 123

V (r) = V (r0) + (r − r0)

(dV

dr

)r0

+1

2(r − r0)2

(d2V

dr2

)r0

+ . . . . (7.2)

Since V (r) is minimum at r = r0, then(dVdr

)r0

= 0, and if we take our zero of energy to

be such that V (r0) = 0 then the potential that describes the oscillation about equilibriumis given to lowest order by

V (x) =1

2mω2x2 , (7.3)

where x = (r − r0) and(d2Vdr2

)r0

= mω2. Thus to examine the low energy vibrational

spectrum of molecules we need to examine the potential V (x) given above. The othermotivation for considering this potential is that the Schrodinger equation for this potentialis slightly more complicated than the examples we have considered so far. In fact theSchrodinger equation takes the form

− h2

2m

d2ψ

dx2+

1

2mω2x2ψ = Eψ(x) , (7.4)

or after multiplication by 2hω

we have

− h

d2ψ

dx2+mω

hx2ψ =

2E

hωψ(x) . (7.5)

To simplify the differential equation, we define new variables η and a rescale energy ε as

η =

√mω

hx and ε =

2E

hω. (7.6)

Our Schrodinger equation now takes the simple form

d2ψ

dη2+ (ε− η2)ψ = 0 . (7.7)

To solve this second order differential equation, we first consider the asymptotic solutions,i.e. in the limit as |x| → ∞. Considering the fact that V (x) → ∞ as |x| → ∞ we mayconclude that the wave function should go to zero as |x| → ∞. To get the form of thewave function in this region, we will consider first the differential equation for η ε. Inthis limit, the Schrodinger equation reduces to

d2ψ

dη2− η2ψ = 0 . (7.8)

We now change the variable in this differential equation to z = η2 so that

dη=dz

dz= 2η

dz

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124 CHAPTER 7. MOLECULAR VIBRATION – THE HARMONIC OSCILLATOR

and

d2ψ

dη2= 2

dz+ 2η

d

(dψ

dz

)

= 2dψ

dz+ 4η2d

dz2

= 2dψ

dz+ 4z

d2ψ

dz2.

Our differential equation is now given by

d2ψ

dz2+

1

2z

dz− 1

4ψ = 0 (7.9)

or for z →∞ we haved2ψ

dz2− 1

4ψ = 0 . (7.10)

In this way we have reduced the Schrodinger equation to the form we are familiar with,and the solution for large z is given by

ψ = e−z/2 = e−η2/2 . (7.11)

We now make use of our knowledge of the asymptotic solution for large η to write thegeneral solution to the Schrodinger equation for the harmonic oscillator potential as

ψ(η) = u(η)e−η2/2 , (7.12)

with the hope that u(η) is some simple polynomial in η. To determine the equation u(η)satisfies, we substitute the above expression for ψ(η) in terms of u(η) in the Schrodingerequation for this potential, i.e. Eq. (7.7). Making use of the fact that

d2ψ

dη2= e−η

2/2

d2u

dη2− 2η

du

dη2+ (η2 − 1)u(η)

we can write the equation for u(η) as

d2u

dη2− 2η

du

dη+ (ε− 1)u(η) = 0 . (7.13)

At this stage it appears that we have replaced the Schrodinger equation by a more com-plicated differential equation. However, we will find that the above equation will givepolynomial solutions. In fact we will find that the requirement that the wave functiongoes to zero as |x| → ∞ will admit us not only a polynomial solution, but also that onlycertain energies are allowed, i.e. we will get a discreet spectrum of energy levels. To solve

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7.1. THE ONE DIMENSIONAL HARMONIC OSCILLATOR 125

the second order differential equation in Eq. (7.13), we use the method of series to writea series solution for the unknown function u(η) of the form

u(η) =∞∑n=0

anηn , (7.14)

with the constants an to be determined from the differential equation (7.13). To achievethis we need to determine the series for the first and second derivative of the functionu(η). For the first derivative we have

du

dη=

∞∑n=0

n an ηn−1

=∞∑n=0

(n+ 1) an+1 ηn .

In writing the second line above, we have changed the sum over n to run from 1 → ∞since the first term in the series in the first line is zero. We then replace n by (n+ 1) andrewrite the sum to run over 0→∞. We now can write the second term in our differentialequation as

2ηdu

dη=

∞∑n=0

2(n+ 1) an+1 ηn+1

=∞∑n=1

2n an ηn .

In a similar manner we can write the series for the second derivative of u(η) as

d2u

dη2=

∞∑n=0

(n+ 1)n an+1 ηn−1

=∞∑n=0

(n+ 2) (n+ 1) an+2 ηn .

We now can substitute the series for the function u(η) it’s first and second derivative inthe differential equation, Eq. (7.13) to get

∞∑n=0

(n+ 2) (n+ 1) an+2 ηn −

∞∑n=0

2n an ηn + (ε− 1)

∞∑n=0

an ηn = 0 . (7.15)

Since the right hand side of this equation is zero for any η, the coefficients of η`, for any`, must also be zero i.e. for

η0 : 2a2 + (ε− 1)a0 = 0 =⇒ a2 =1

2(1− ε)a0

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126 CHAPTER 7. MOLECULAR VIBRATION – THE HARMONIC OSCILLATOR

η1 : 6a3 − 2a1 + (ε− 1)a1 = 0 =⇒ a3 =1

6(3− ε)a1

η2 : 12a4 − 4a2 + (ε− 1)a2 = 0 =⇒ a4 =1

12(5− ε)a2

...

η` : (`+ 2)(`+ 1)a`+2 − 2`a` + (ε− 1)a` = 0 .

In this way we have made use of the differential equation to set up relations betweenthe coefficients an. These relations are unique to this differential equation and will allowus to write the solution with two undetermined constants, a0 and a1. This is expectedconsidering the fact that the solution of a second order linear differential equation has twoarbitrary constants. These constants are to be determined by the boundary condition ofthe physical system under consideration.

From the above relation between the constants an, we can deduce a general relationbetween a` and a`+2 that is of the form

a`+2 =1

(`+ 2)(`+ 1)(2`+ 1)− ε a` . (7.16)

Thus given a0 and a1, we can generate all the other coefficients of the power series. Thefact that we have two series, one for n = odd, the other for n = even, is the result of thefact that the odd n series corresponds to a final solution that is odd under parity, whilethe even n solution gives a wave function that is even under parity. Furthermore thesetwo solutions do not mix. This is the result of the fact that V (x) does not change underparity, i.e., x→ −x.

For arbitrary values of ε, the series is not finite, and for large N we have that

aN+2 =2

NaN for N 1

For the even solution, we can write the series as

u(η) = polynomial in η of order (N − 2)

+aN

ηN +

2

NηN+2 +

2

N

2

N + 2ηN+4 + · · ·

= polynomial in η of order (N − 2)

+aNη2

(η2)N2−1 +

(η2)N2

N2

+(η2)

N2

+1

N2

(N2

+ 1)+ · · ·

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7.1. THE ONE DIMENSIONAL HARMONIC OSCILLATOR 127

= polynomial in η of order (N − 2)

+aNη2(N

2− 1

)!

(η2)N2−1

(N2− 1)!

+(η2)

N2

(N2

)!+

(η2)N2

+1

(N2

+ 1)!+ · · ·

.

But we have that1 (η2)N2−1

(N2− 1)!

+(η2)

N2

(N2

)!+

(η2)N2

+1

(N2

+ 1)!+ · · ·

= eη2 − polynomial in η .

If we now make use of this solution in Eq. (7.12), we find that our solution for ψ(η) is ofthe form

ψ(η) = polynomial in η e−η2/2 + aN η2(N

2− 1

)! eη

2/2 . (7.17)

This solution goes to ∞ as η →∞ and does not satisfy the boundary condition that thewave function should go to zero as η → ∞. Thus to satisfy the boundary condition andhave a wave function that goes to zero as η → ∞, we need to guarantee that u(η) is afinite order polynomial in η. Since the coefficients in the power series are related by therecursion relation

a`+2 =(2`+ 1− ε)

(`+ 2)(`+ 1)a` ,

to turn this infinite series to a polynomial in η, we need to have a`+2 = 0 when a` 6= 0.This condition can be satisfied only if

ε = 2`+ 1 , (7.18)

where ` is an integer. This allows us to write the energy as

E` =hω

= hω(`+1

2) where ` = 0, 1, 2, · · · . (7.19)

Here we observe that the boundary condition that the wave function go to zero as η →∞has given a condition on the energy. This condition essentially states that only certainenergies are allowed, i.e., we have generated a quantized spectrum.

1We have made use of the fact that

ex = 1 + x+x2

2!+x3

3!+ · · ·

wheren! = n(n− 1)(n− 2) · · · 1 .

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128 CHAPTER 7. MOLECULAR VIBRATION – THE HARMONIC OSCILLATOR

We are now in a position write the allowed energies and corresponding wave functionfor our harmonic oscillator potential. The lowest energy state corresponds to taking thecoefficient a2 = 0, and this corresponds to ` = 0. In this case the energy, which is theground state energy, is given by

E0 =1

2hω (7.20)

with the corresponding ground state wave function being

ψ0 = a0e−η2/2 . (7.21)

The next eigenstate corresponds to a1 6= 0 while a3 = 0. Here ` = 1 and the energy isgiven by

E1 = hω(1 +1

2) =

3

2hω , (7.22)

while the wave function for this state is given by

ψ1 = a1ηe−η2/2 . (7.23)

We now observe that the ground state wave function with energy E0 = 12hω is even under

parity, while the wave function of the first excited state with energy E1 = 32hω is odd

under parity. For the second excited state we have that a2 6= 0 while a4 = 0. Here, a2 isgiven in terms of a0 by the relation

a2 =1− ε

2a0 = −`a0 .

In this case, ` = 2 and the energy is given by

E2 =5

2hω , (7.24)

while the wave function is given by

ψ2 = a0(1− 2η2) e−η2/2 . (7.25)

In this way we can continue to generate all the eigenvalues and eigenstates of the harmonicoscillator Hamiltonian.

An alternative way of generating the eigenstates is to examine the form of the differ-ential equation for u(η) when we take ε = 2`+ 1. This gives us a differential equation ofthe form

d2H`

dη2− 2η

dH`

dη+ 2`H`(η) = 0 . (7.26)

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7.2. VIBRATIONAL SPECTRUM OF DIATOMIC MOLECULE 129

This equation is known as the Hermite equation, and H`(η) is the Hermite polynomial.Special cases of the Hermite polynomial are

H0(η) = 1H1(η) = 2ηH2(η) = 4η2 − 2H3(η) = 8η3 − 12ηH4(η) = 16η4 − 48η2 + 12

These polynomials have the recursion relation

2ηH` = H`+1 + 2`H`−1 . (7.27)

The general solution for the one dimensional harmonic oscillator Hamiltonian can now bewritten as

ψ`(η) = (√π 2` `!)−

12 H`(η) e−η

2/2 , (7.28)

where η =√

mωhx. The corresponding energies are

E` = hω(`+ 1/2) . (7.29)

These wave functions given in Eq. (7.28) are orthonormal, i.e.,∫ +∞

−∞dη ψ`(η)ψ`′(η) = δ``′ , (7.30)

and form a complete set of states.The above solution of the Schrodinger equation for the harmonic oscillator potential

can be used in many applications where the system is not far from equilibrium as is thecase of the vibration of a diatomic molecule or the vibration of the lattice in a solid. Thethree dimensional harmonic oscillator has been used extensively in Nuclear Physics as ashell model of nuclei in analogy with the atomic shell model. We will find that the methodwe used to solve the Schrodinger equation in this case is very similar to the solution of thecorresponding radial equation in three dimensional problems. This will be the subject ofthe next chapter in which we consider the problem of an electron in the Coulomb field ofa nucleus.

7.2 Vibrational Spectrum of Diatomic Molecule

Having established the spectrum of the one dimensional harmonic oscillator, we turnour attention to the application of these results to the case of a diatomic molecule. Inparticular we will consider the spectrum of H2, HD and DD, where D is the isotope ofhydrogen in which the nucleus consists of a proton and a neutron, i.e., the deuteron.

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130 CHAPTER 7. MOLECULAR VIBRATION – THE HARMONIC OSCILLATOR

We showed in the last section that the energy levels of the harmonic oscillator aregiven by

E` = hω (`+ 1/2)

where the frequency ω is given by

ω =1√µ

√√√√(d2V

dr2

)r0

where µ =m1m2

m1 +m2

. (7.31)

Here µ is the reduced mass of the molecule, with m1 and m2 being the masses of the twoatoms in the molecule. Since the potential between the atoms in the molecule depends onthe wave function of the electron, and that does not depend on the mass of the nucleus,we can assume that (

d2V

dr2

)r0

= C

and is the same for all the three molecules under consideration. The reduce masses forthe three diatomic molecules are given by,

µH2 =mp

2, µHD =

2mp

3=

4

3mH2 , and µDD = mp = 2mH2 . (7.32)

where we have assumed that the mass of D is twice the mass of the proton (mp). Thisallows us to calculate the ratio of the frequency of vibration for these three molecules.Making use of the relation between the frequency ω and the reduced mass µ, we cancalculate in this model the ratio of the frequencies to be

ωH2

ωHD=

√4

3≈ 1.1547 and

ωH2

ωDD=√

2 ≈ 1.4142 . (7.33)

These numbers can be compared with the corresponding experimental values of

ωH2

ωHD= 1.1514 and

ωH2

ωDD= 1.4096 . (7.34)

This agreement is impressive considering the fact that we have a very simple modelfor a diatomic molecule. These results clearly illustrate that this simple approximationworks very well for simple diatomic molecules. The model also suggests a limitation onthe application of the simple one dimensional harmonic oscillator. From Figure 7.1, weexpect that for high excitation, the difference between the harmonic oscillator and the realspectrum should deviate. This is a result of the fact that the potential in Figure 7.1 is notthe same as a harmonic oscillator. This prediction is also in agreement with experimentalobservation. This deviation from a simple harmonic oscillator will be addressed when weexamine perturbation theory.

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7.3. PROBLEMS 131

This simple one dimensional model for a vibrational spectrum can also be used tounderstand the specific heat of solids due to the vibration of atoms in the solid abouttheir equilibrium position. This model of the solid is commonly referred to as the Einsteinsolid, and predicts the fact that the specific heat goes to zero as the temperature goes tozero. For nuclei, the vibrational motion is a more complicated, and one needs to extendthis one dimensional model to five dimensions to describe the vibration on the surface ofthe nucleus. Here again, the agreement between the model and experiment is very good.

7.3 Problems

1. Use either MAPLE of Mathematica to solve Eq. (7.13) for ε = 3, 5 and compare theresults with the analytic solution.

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132 CHAPTER 7. MOLECULAR VIBRATION – THE HARMONIC OSCILLATOR

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Chapter 8

Central Force Problem I

Abstract : In this chapter we commence with the solution of the Schrodingerequation in three dimension and consider the case when the potential hasspherical symmetry. This requires the introduction of spherical polar coordi-nates and the reduction of the three dimensional partial differential equationto a set of ordinary differential equation. The solution of the equations forthe angular dependence we allow us to examine the rotational spectrum ofdiatomic molecules.

So far we have concentrated our effort on one dimensional systems with the hopeof gaining some understanding of the fundamental ideas of quantum mechanics whilediscussing application to nuclei, atoms, molecules and solids. Before we proceed to aformal development of quantum theory it may be useful to examine some simple systemsin three dimensions. The hydrogen atom is, in a sense, the simplest example of a realatom, and it was the hydrogen atom that was first considered by the early developers ofquantum theory. However, we will find through further studies, that a full understandingof the hydrogen atom will strain our theoretical understanding of physics to the limit.

The simplest model of the hydrogen atom involves the system consisting of a protonand an electron with the Coulomb potential being the force of attraction that binds thetwo particles together to form the hydrogen atom. The classical Hamiltonian for this twoparticle system consists of the kinetic energy of the electron and proton and the potentialenergy, the Coulomb attraction, i.e.,

H =p2e

2me

+p2p

2mp

+ V (~re − ~rp) , (8.1)

where ~pe (~re) and ~pp (~rp) are the momenta (position) of the electron and proton respec-tively, while me and mp are the corresponding masses. The Coulomb potential is given

133

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134 CHAPTER 8. CENTRAL FORCE PROBLEM I

by1

V (~re − ~rp) = − e2

| ~re − ~rp |, (8.2)

with e being the magnitude of the charge of the proton or electron. The two are identicalin magnitude.

At this stage we could quantize our system by the substitution

~p→ −ih~∇ , (8.3)

where the gradient, ∇, in rectangular coordinates is given by

~∇ = ı∂

∂x+

∂y+ k

∂z. (8.4)

Here ı, and k are unit vectors in the x-, y- and z-directions. The Schrodinger equationfor this system now depends on the coordinates of both the electron and proton, i.e.,

HΨ(~r1, ~r2) = EΨ(~r1, ~r2) , (8.5)

and is a partial differential equation in six variables. This problem can be considerablysimplified if we introduce relative and center of mass coordinates. Since we are interestedin the relative motion of the electron and proton, we hope, and rightly so, to reduce theproblem to three dimensions, or variables. The relative and center of mass momenta ~pand ~P are defined as

~p =mp~pe −me~ppmp +me

~P = ~pe + ~pp , (8.6)

while the relative and center of mass coordinates ~r and ~R are defined as

~r = ~re − ~rp ~R =me~re +mp~rpme +mp

. (8.7)

From the above definition of relative and center of mass variables we have

(mp +me)~p+me~P = (me +mp)~pe

and−(mp +me)~p+mp

~P = (me +mp)~pp .

Therefore, we can write the momentum of the proton and electron in terms of the relativeand center of mass momenta as

~pe = ~p+me

me +mp

~P and ~pp = −~p+mp

me +mp

~P . (8.8)

1We will use units in which 14πε0

= 1. This corresponds to using Gaussian units which are moreconvenient for microscopic systems.

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135

We now can write the kinetic energy of the electron and proton in terms of the relativeand center of mass momenta as

p2e

2me

=1

2me

(p2 +

2me

me +mp

~p · ~P +m2e

(me +mp)2P 2

),

p2p

2mp

=1

2mp

(p2 − 2mp

me +mp

~p · ~P +m2p

(me +mp)2P 2

).

With these results in hand we can write the kinetic energy of the system as

p2e

2me

+p2p

2mp

=p2

2µ+P 2

2M, (8.9)

where the reduced mass µ and total mass M are given by

µ =memp

me +mp

M = me +mp . (8.10)

Here we observe that the kinetic energy of the proton and electron can be written as thesum of the kinetic energy of the center of mass plus the relative kinetic energy. Since thepotential energy depends on the relative position of the electron and proton in the atom,we can write the Hamiltonian in terms of the new coordinates as

H =P 2

2M+

(p2

2µ+ V (r)

), (8.11)

where the first term describes the motion of the center of mass of the atom as a whole,while the second term describes the relative motion of the electron and proton whichgives us the properties of the hydrogen atom. Since we are only interested in the internalstructure of the atom, we can take the center of mass momentum ~P to be zero. Theresultant coordinate system is known as the center of mass system.

For the hydrogen atom, mpme≈ 2000 and as a result

µ =me

1 + memp

≈ me , (8.12)

i.e., the reduced mass is basically the mass of the electron. This is equivalent to assum-ing that the inertial frame in which the proton is stationary is also the center of masscoordinate system.

From this point on, we will discuss the properties of the hydrogen atom and othertwo-body systems in the center of mass coordinate system where ~P = 0. These proper-ties are determined by the Schrodinger equation resulting from the quantization of theHamiltonian in Eq. (8.11) when ~P = 0, i.e.

p2

2µ+ V (r)

ψ(~r) = E ψ(~r) , (8.13)

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136 CHAPTER 8. CENTRAL FORCE PROBLEM I

where the momentum operator is given by

~p = −ih~∇ . (8.14)

Although the above derivation of the Schrodinger equation was for the Coulomb potential,the derivation is valid for any potential between two particles, provided the potential is afunction of the magnitude of the relative distance between the two particles that form thesystem. Such potentials are known as central potentials. The Schrodinger equation (8.13)for the potential V (r) is now a partial differential equation in three variables.

At this stage we observe that in rectangular coordinates the kinetic energy operatorin coordinate space is given as

p2

2µ= − h

2

2µ∇2 = − h

2

∂2

∂x2+

∂2

∂y2+

∂2

∂z2

, (8.15)

where∇2 is the Laplacian, while the potential energy operator for the case of the Coulombpotential takes the form

V (r) = − e2

[x2 + y2 + z2]12

. (8.16)

The form of this potential suggests that the Schrodinger equation, Eq. (8.13), in rectan-gular coordinates is very difficult if not impossible to solve because of the square rootin the potential energy. However, we note that the potential between the two particlesdepends on the magnitude of the relative distance between the two particles, and rotatingour coordinate system does not change this potential. At the same time this rotation ofthe coordinate system does not change the relative kinetic energy. In other words, theHamiltonian has a symmetry under rotation in three dimensional space. This suggeststhat the spherical polar coordinate system is the best choice for solving the Schrodingerequation. Before we write our differential equation in spherical polar coordinates, let uswrite p2 in a form that will allow us to separate the radial from the angular dependence.

8.1 Cartesian Tensors

In this section we will introduce a tool for simplifying identities in vector calculus thatis very powerful. Although all the results reported in this chapter can be derived usingstandard methods developed in several variable calculus, the power of the approach de-veloped in this section can be applied in diverse fields from electromagnetic theory togeneral relativity to hydrodynamics, and to that extent is a very useful tool. For thepresent chapter, the Cartesian tensor approach is no more than a shorthand notation forproducts of a combination of vectors and gradients.

The angular momentum in classical mechanics is given by

~L = ~r × ~p . (8.17)

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8.1. CARTESIAN TENSORS 137

To define the angular momentum in quantum mechanics, we make use of the quantizationrules in Table 3.1 to write the angular momentum in coordinate space as

~Lop = ~r × ~pop = −ih ~r × ~∇ . (8.18)

From this point on we will drop the ‘op’ subscript on ~L and ~p with the result

L2 = ~L · ~L = (~r × ~p ) · (~r × ~p ) . (8.19)

Note the momentum operator ~p does not commute with the coordinate operator ~r. Infact we have a generalization of Eq. (3.65) in the form

[ri, pj] = ihδij , (8.20)

with i, j = 1, 2, 3 labeling the components of the position and momentum vectors, e.g.~r = (r1, r2, r3), with r1 being the component of the position vector along the x-axis.

To calculate L2, we introduce tensor notation that is very useful in calculating vectoridentities involving ~∇ in vector products.2 In this tensor notation, the ith component ofthe angular momentum

~L = ~r × ~p , (8.21)

is written asLi = εijkrjpk , (8.22)

where the indices i, j, k = 1, 2, 3 label the Cartesian components of a vector and εijk isthe totally antisymmetric tensor that has the property

εijk =

0 if any two indices are equal+1 if (ijk) is an even permutation of (123)−1 if (ijk) is an odd permutation of (123)

. (8.23)

We have also made use of the notation whereby repeated indices in a product are summedover, e.g., the product of two vectors

~A = A1ı+ A2+ A3k and ~B = B1ı+B2+B3k

can be written as

~A · ~B =3∑i=1

AiBi ≡ AiBi .

In a similar manner, we can write the component of the angular momentum in Eq. (8.22)as3,

Li =∑jk

εijk rjpk ≡ εijk rjpk . (8.24)

2This notation is very useful in continuum mechanics and relativity. It is also used in advancedelectromagnetic theory.

3As an exercise write the components of the angular momentum L1, L2 and L3. Compare your answerwith you favourite method of calculating vector product. Do they agree?

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138 CHAPTER 8. CENTRAL FORCE PROBLEM I

Finally, to simplify complicated expressions we need to know that the totally antisym-metric tensor εijk has the property that∑

i

εijk εi`m ≡ εijk εi`m = δj`δkm − δjmδk` . (8.25)

With these results, we can turn to the calculation of ~L · ~L. In our new notation we have

L2 = Li Li = εijk rjpk εi`mr`pm

= εijk εi`m rj pk r` pm . (8.26)

Since the position r` and momentum pk operators don’t commute, we will keep the orderof these operators as they are in the original scalar product. We then have, making useof Eq. (8.25), that

L2 = (δj`δkm − δjmδk`) rj pk r` pm= rj pk rj pk − rj pk rk pj . (8.27)

Note, in writing the right hand side of this equation, we need to remember the the orderin which we write the components of the position and momentum vectors is important asthey don’t commute. We also need to make sure that we use different indices for writingthe two components of the angular momentum as there is an implicit sum assumed. Usingthe commutation relation of ri and pj as given in Eq. (8.20), we get

L2 = rj (−ihδjk + rjpk) pk − rj (−ihδkk + rkpk) pj

= −ih(~r · ~p ) + r2p2 + 3ih(~r · ~p )− rjrkpkpj= 2ih(~r · ~p ) + r2p2 − rk(ihδjk + pkrj)pj

= r2p2 + ih(~r · ~p )− (~r · ~p ) (~r · ~p ) . (8.28)

This result allows us to write the kinetic energy for relative motion as

p2

2µ=

1

2µr2

(~r · ~p ) (~r · ~p )− ih(~r · ~p ) + L2

. (8.29)

Furthermore, since ~p = −ih~∇, we have

~r · ~p = −ihr ∂

∂r, (8.30)

and we can be write the kinetic energy as

p2

2µ= − h

2

2µ∇2 = − h2

2µr2

r∂

∂r

(r∂

∂r

)+ r

∂r− L2

h2

= − h2

1

r2

∂r

(r2 ∂

∂r

)− L2

h2r2

. (8.31)

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8.2. ANGULAR MOMENTUM IN QUANTUM MECHANICS 139

We will show in the next section that L2 depends on the angle variable (θ, φ) in sphericalpolar coordinates. As a result, the above expression for the kinetic energy has effectivelyseparated the radial variable from the angle variable. This we will see will allow us tosolve the Schrodinger equation using separation of variables. We could have derived theabove result by writing ∇2 in spherical polar coordinates. The Hamiltonian for a centralpotential can now be written in spherical polar coordinates as:

H = − h2

1

r2

∂r

(r2 ∂

∂r

)− L2

h2 r2

+ V (r) . (8.32)

The corresponding Schrodinger equation takes the form[− h

2

1

r2

∂r

(r2 ∂

∂r

)− L2

h2 r2

+ V (r)

]ψ(~r ) = E ψ(~r ) . (8.33)

The Hamiltonian in Eq. (8.32) is the sum of two parts – one depends on the radial variabler, and the second which includes L2, depends on the angles θ and φ. As a result, we canuse the method of separation of variables to write

ψ(~r ) = R(r) Y (θ, φ) . (8.34)

Furthermore, since the angular momentum square, L2, is a linear Hermitian operator, wecan take Y (θ, φ) to be a solution of the eigenvalue problem

L2Y (θ, φ) = h2λ Y (θ, φ) , (8.35)

where λ is a parameter to be determined by the boundary condition on the Schrodingerequation. We now can write the radial Schrodinger equation as:[

− h2

1

r2

d

dr

(r2 d

dr

)− λ

r2

+ V (r)

]R(r) = E R(r) . (8.36)

In this way we have reduced the Schrodinger equation in three dimensions, to a radialequation in one dimension for R(r), and a partial differential equation in the two anglevariables that involves the eigenstates of L2. It is important to note at this stage thatthe angular part of the solution Y (θ, φ) does not depend on the choice of potential V (r),and will be a valid angular solution to the Schrodinger equation for any potential thatdepends on the radial variable r. In other words, the angular solution is valid for anypotential with spherical symmetry.

8.2 Angular Momentum in Quantum Mechanics

Since space is isotropic, rotational symmetry is present in all problems where externalforces do not break this symmetry of space. As a result, all atoms, molecules, and nuclei

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140 CHAPTER 8. CENTRAL FORCE PROBLEM I

in free space have spherical symmetry. To that extent, angular momentum in quantummechanics plays a central role in most of the above problems.

In classical mechanics angular momentum is defined as

~L = ~r × ~p , (8.37)

and the corresponding quantity in quantum mechanics results from the substitution ~p→−ih~∇, i.e.

~L = −ih~r × ~∇ = ~r × ~pop . (8.38)

The components of this angular momentum operator are given in rectangular coordinatesas

Lx = ypz − zpy = −ih(y∂

∂z− z ∂

∂y

)

Ly = zpx − xpz = −ih(z∂

∂x− x ∂

∂z

)(8.39)

Lz = xpy − ypx = −ih(x∂

∂y− y ∂

∂x

).

We have found that in quantum mechanics we cannot measure the position and momentumof a particle to any desired accuracy. This results in the position-momentum uncertaintyrelation, which can be represented mathematically by the commutation relation

[ri, pj] = ihδij . (8.40)

Using these commutation relations we can calculate the commutation relation betweenthe components of the angular momentum, ~L, e.g.

[Lx, Ly] = LxLy − LyLx= (ypz − zpy)(zpx − xpz)− (zpx − xpz)(ypz − zpy)= (ypzzpx − ypzxpz − zpyzpx + zpyxpz)

−(zpxypz − zpxzpy − xpzypz + xpzzpy)

= ypx(pzz − zpz) + xpy(zpz − pzz)

= ih(xpy − ypx) = ihLz . (8.41)

From this result we may conclude that one cannot measure simultaneously the x- andy-components of the angular momentum vector. Alternatively we can say that we can-not construct a state that is an eigenstate of two different components of the angularmomentum vector ~L. The above result can be written, in general, as

[Li, Lj] = ihεijk Lk , (8.42)

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8.2. ANGULAR MOMENTUM IN QUANTUM MECHANICS 141

where εijk is the totally antisymmetric tensor used in the previous section. We have usedan implicit sum over repeated indices on the right hand side of Eq. (8.42). We now definethe angular momentum square as

L2 = ~L · ~L = L2x + L2

y + L2z ≡

3∑i=1

L2i . (8.43)

This allows us to calculate the commutation relation between the angular momentumsquare, L2, and the component of the angular momentum, Lj, i.e.,

[L2, Lj] =∑i

[L2i , Lj] =

∑i

(L2iLj − LjL2

i

)=

∑i

(L2iLj − LiLjLi + LiLjLi − LjL2

i

)=

∑ik

(LiihεijkLk + ihεijkLkLi)

= ih∑ik

(εijkLiLk + εijkLkLi)

= ih∑ik

(εijkLiLk + εkjiLiLk) = 0 ,

where we have made use of the fact that εkji = −εijk. Thus we have established that

[L2, Lj] = 0 . (8.44)

In this case we may conclude that we can construct states that are eigenstates of bothL2 and one of the components of the angular momentum Lk. It is common practice totake Lk to be the z-component of the angular momentum. In other words we can write

L2 Yλm(θ, φ) = h2λYλm(θ, φ) ( 8.45a)

Lz Yλm(θ, φ) = hmYλm(θ, φ) . ( 8.45b)

We now have to find what the eigenvalues λ and m are, and what the eigenstateYλm(θ, φ) that determines the angular part of the wave function is. In other words, weneed to find the solution of the differential equations (8.45). This is best achieved bychanging variables to spherical polar coordinates, (see Figure 8.1) i.e.

x = r sin θ cosφ

y = r sin θ sinφ (8.46)

z = r cos θ .

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142 CHAPTER 8. CENTRAL FORCE PROBLEM I

x

y

z

θ

φ

r

Figure 8.1: Spherical polar coordinates for central potentials.

Taking the differential of these equations, we get

dx = sin θ cosφ dr + r cos θ cosφ dθ − r sin θ sinφ dφ

dy = sin θ sinφ dr + r cos θ sinφ dθ + r sin θ cosφ dφ (8.47)

dz = cos θ dr − r sin θ dθ .

Solving for dr, dθ, and dφ, we get

dr = sin θ cosφ dx+ sin θ sinφ dy + cos θ dz

dθ =1

r(cos θ cosφ dx+ cos θ sinφ dy − sin θ dz) (8.48)

dφ =1

r sin θ(− sinφ dx+ cosφ dy) .

With these results we can write Lx, Ly and Lz in spherical polar coordinates making useof the fact that

∂x=∂r

∂x

∂r+∂θ

∂x

∂θ+∂φ

∂x

∂φ,

with similar expressions for the partial derivatives with respect to y and z. This gives,after some algebra,

Lx = ih

(sinφ

∂θ+ cot θ cosφ

∂φ

)

Ly = ih

(− cosφ

∂θ+ cot θ sinφ

∂φ

)(8.49)

Lz = −ih ∂∂φ

,

and

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8.2. ANGULAR MOMENTUM IN QUANTUM MECHANICS 143

L2 = L2x + L2

y + L2z

= −h2

1

sin θ

∂θ

(sin θ

∂θ

)+

1

sin2 θ

∂2

∂φ2

. (8.50)

With Lz and L2 written in terms of θ and φ, we can write Eq. (8.45) as partial differentialequations. The second of these equations, Eq. (8.45b), is just a first order differentialequation of the form

− ih∂Yλm∂φ

= hmYλm . (8.51)

This equation can be integrated to give us4

Yλm(θ, φ) = eimφPλm(θ) . (8.52)

In writing this solution we have taken the constant to be a function of the other anglein the problem, i.e., θ. The boundary condition that Yλm(θ, φ + 2π) = Yλm(θ, φ) impliesthat m is an integer. We can now write the equation

L2 Yλm(θ, φ) = h2λYλm(θ, φ) , (8.53)

as 1

sin θ

∂θ

(sin θ

∂θ

)+

1

sin2 θ

∂2

∂φ2

Yλm(θ, φ) = −λYλm(θ, φ) . (8.54)

But we have that∂2

∂φ2Yλm(θ, φ) = −m2 Yλm(θ, φ) , (8.55)

and therefore we have1

sin θ

∂θ

(sin θ

∂θ

)− m2

sin2 θ

Pλm(θ) = −λPλm(θ) . (8.56)

We now change variables toω = cos θ , (8.57)

and write the derivative with respect to θ as

∂θ=∂ω

∂θ

∂ω= − sin θ

∂ω

4We can use the separation of variables, i.e. Y`m(θ, φ) = Φ(φ) Θ(θ) and Eq. (8.51) to write the φdependence as

Φ(φ) = eimφ .

Here m has to be an integer for Φ(φ) = Φ(φ+ 2π), i.e the function is single valued.

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144 CHAPTER 8. CENTRAL FORCE PROBLEM I

or∂

∂ω= − 1

sin θ

∂θ.

The equation for Pλm can now be written as− ∂

∂ω

(− sin2 θ

∂ω

)+ λ− m2

sin2 θ

Pλm(θ) = 0

or d

((1− ω2)

d

)+

(λ− m2

1− ω2

)Pλm(ω) = 0 ,

since sin2 θ = 1− cos2 θ = 1− ω2. Therefore, we have

(1− ω2)d2P

dω2− 2ω

dP

dω+

(λ− m2

1− ω2

)P = 0 . (8.58)

This is known as the Legendre differential equation, and can be solved by the methods ofseries. For λ = `(` + 1) with ` an integer and |m| ≤ `, the solutions can be written interms of a polynomial known as Associate Legendre function. These functions are givenby5

Pm` (ω) = (1− ω2)m/2

dm

dωmP`(ω)

=1

2``!(1− ω2)m/2

d`+m

dω`+m(ω2 − 1)` . (8.59)

In the above expression we have introduced the functions P`(ω) = P 0` (ω) which are known

as the Legendre polynomials. These polynomials satisfy the orthogonality relation

+1∫−1

dω Pm` (ω)Pm

`′ (ω) =2

2`+ 1

(`+m)!

(`−m)!δ``′ . (8.60)

They also satisfy the recursion relations

(2`+ 1)ωPm` = (`+ 1−m)Pm

`+1 + (`+m)Pm`−1

(1− ω2)d

dωPm` = −`ωPm

` + (`+m)Pm`−1 (8.61)

= (`+ 1)ωPm` − (`+ 1−m)Pm

`+1

5 We have introduced the notation that for λ = `(`+ 1)

Pm` (ω) ≡ Pλm(ω) .

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8.2. ANGULAR MOMENTUM IN QUANTUM MECHANICS 145

Some special properties of these polynomials are

P`(1) = 1 ; P`(−1) = (−)`

Pm` (1) = Pm

` (−1) = 0 for m 6= 0

P0(ω) = 1 ; P1(ω) = ω

P2(ω) =1

2(3ω2 − 1) ; P 1

1 (ω) =√

1− ω2

P 12 (ω) = 3ω

√1− ω2 ; P 2

2 (ω) = 3(1− ω2) .

We now combine the solution for the φ and θ dependence to get the spherical harmonics,Y`m(θ, φ), given by

Y`m(θ, φ) = (−)m

(2`+ 1)

(`−m)!

(`+m)!

12

Pm` (cos θ) eimφ . (8.62)

This function, which is an eigenstate of L2 and Lz, is normalized to the extent that

+1∫−1

d(cos θ)

2π∫0

dφ Y ∗`m(θ, φ)Y`′m′(θ, φ) = δ``′ δmm′ . (8.63)

The lowest order spherical harmonics are:

Y00(θ, φ) =1√4π

; Y10(θ, φ) =

√3

4πcos θ

Y1±1(θ, φ) = ∓√

3

8πsin θ e±iφ ; Y20(θ, φ) =

√5

16π(3 cos2 θ − 1)

Y2±1(θ, φ) = ∓√

15

8πsin θ cos θ e±iφ ; Y2±2(θ, φ) =

1

4

√15

2πsin2 θ e±2iφ .

We thus have established that the components of the angular momentum ~L do not com-mute. In particular,

[Li, Lj] = ihεijkLk . (8.64)

However[L2, Li] = 0 , (8.65)

which allows us to write eigenstates of L2 and Lz as

L2 Y`m(θ, φ) = h2`(`+ 1)Y`m(θ, φ)

(8.66)

Lz Y`m(θ, φ) = hmY`m(θ, φ) ,

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146 CHAPTER 8. CENTRAL FORCE PROBLEM I

with the eigenstates Y`m(θ, φ) being the spherical harmonics.

Here we observe that the magnitude of the angular momentum,√〈L2〉 is given by

∫dΩY ∗`m(θ, φ)L2 Y`m(θ, φ)

12

= h√`(`+ 1)

where the integration is over all direction in space, i.e., dΩ = dφ dθ sin θ. On the otherhand, the maximum projection of ~L along the z-axis is given by∫

dΩY ∗``(θ, φ)Lz Y``(θ, φ) = h` < h√`(`+ 1) ,

i.e., the projection of the angular momentum along the z-axis is always less than themagnitude of the angular momentum. This suggests that the angular momentum vectorcannot point along the z-axis. If we force ~L to be along the z-axis this would mean itscomponents along the x and y axis are identically zero. This is not possible because the x-and y-components of the angular momentum do not commute, and from the uncertaintyprinciple we know that we can not determine both the x- and y-components of the angularmomentum to any desired accuracy, e.g. both being zero. This suggests that the angularmomentum vector is precessing about the z-axis, and that this precession is purely aquantum mechanical effect resulting from the uncertainty principle,

8.3 Rotational Motion

In this section we will take advantage of the quantization of the angular momentumto examine the rotation of a diatomic molecule or a deformed nucleus. For excitationof a diatomic molecule that are small compared to the vibrational excitation hω (seeSec. 7.2), the distance between the atoms of a molecule are fixed, and the molecule canhave rotational energy as if it is a rigid object. Classically, this rotational energy is thekinetic energy of rotation given by

1

2Iω2 , (8.67)

where I is the moment of inertia for the rigid body about the axis of rotation, and ω isthe angular velocity of rotation. We can write this kinetic energy in terms of the classicalangular momentum ~L as

Hrot =L2

2I. (8.68)

Since the kinetic energy of rotation is also the total energy, we have defined this totalrotational energy as Hrot.

We now have to quantize this classical Hamiltonian for rotational motion. From theresults of the last section, we know that the angular momentum in quantum mechanics

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8.3. ROTATIONAL MOTION 147

is given by Eq. (8.38), and in particular L2 is the differential operator in Eq. (8.50). Thismeans we can write the Hamiltonian for a quantized rigid body as

Hrot = − h2

2I

1

sin θ

∂θ

(sin θ

∂θ

)+

1

sin2 θ

∂2

∂φ2

. (8.69)

and the corresponding Schrodinger equation for this quantized rigid body is

Hrot ψrot = E ψrot . (8.70)

But we know that the spherical harmonics Y`m are the eigenstates of L2. We thereforecan establish that

ψrot = Y`m(θ, φ) , (8.71)

where the angles θ and φ give the orientation of the solid in space.6 The correspondingeigenstates or possible energy of rotation is given by

E` =h2

2I`(`+ 1) . (8.72)

Thus the energy spectrum for rotation is completely determined by the moment of inertiaof the molecule.

To test the validity of this model for the rotational spectrum of a diatomic molecule,we consider the molecules H2 and D2. Since in both cases the electronic wave functionis the same, we expect the distance between the two nuclei in the two molecules to bethe same. Assuming the ground state energy of both molecules to be zero, and the firstexcited state to correspond to ` = 2, then the ratio of the energy of the first excited statefor the two molecules is given by

E2(H2)

E2(D2)=I(D2)

I(H2)(8.73)

Taking the moment of inertia of the molecule about an axis through the center of massand perpendicular to the line joining the two nuclei in the molecule, we have

I(H2) =1

2mpR

20 and I(D2) =

1

2mdR

20 = mpR

20 . (8.74)

where R0 is the separation between the two atoms in the molecule, i.e. the bond length.Therefore, the ratio of the energy of the first excited state of the two molecules is

E2(H2)

E2(D2)= 2 . (8.75)

6For the most general solid body, we need three angles to determine the orientation of the body inspace. As a result, the motion of the most general solid object has a wave function that depends on threeangles and not two. However, for diatomic molecules, where there is an axis of symmetry along a linejoining the two nuclei, we only need two angles and therefore the wave function is a function of these twoangles.

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148 CHAPTER 8. CENTRAL FORCE PROBLEM I

We can compare this result with the experimental ratio, which is 1.995. In a similarmanner if we calculate the ratio of the moment of inertia of HD and H2 molecules, thenthe model gives a value of 4/3, while experiment gives 1.3286. Here again we have goodagreement, suggesting that our simple model for the rotational motion of a diatomicmolecule is very good.

In the discussion above, we took the first excited state of H2 to correspond to ` = 2.The fact that there is no ` = 1 or for that matter no odd ` state is a result of the fact thatthe H2 molecule has a symmetry about a plane going through the center of mass of themolecule and perpendicular to the axis joining the two nuclei in the molecule. This samesymmetry is present in O2 and N2, but is not satisfied for HD or NaCl. As a result, thelatter molecules will have states with ` being both even and odd.

This same model has been extended to deformed nuclei which also exhibit a rotationalspectrum similar to that observed in molecules.

8.4 The Radial Equation

Let us now consider potentials that are a function of the radial variables r, i.e. V (~r) =V (r). These potentials are known as central potentials, and have spherical symmetry, i.e.,they are invariant under rotation in three dimensions. The Hamiltonian for this class ofpotentials is of the form

H = T + V (r) , (8.76)

where the kinetic energy, T , is given as

T =p2

2µ= Tr +

L2

2µr2, (8.77)

which can be written in terms of the radial variable r and the angular momentum operatorsquare L2 as given in Eq. (8.31). Since the angular momentum operator is a function of

the angular variables θ and φ, then [Tr, ~L] = 0. We also have shown that [L2, ~L] = 0, thus

the kinetic energy T always commutes with the angular momentum ~L. If in addition thepotential is central, i.e.

V (~r ) = V (r) ,

then the Hamiltonian H commutes with the angular momentum ~L, i.e.

[H, ~L ] = 0 . (8.78)

This result implies that the eigenstates of H can also be eigenstates of the angular mo-mentum. But since the components of the angular momentum do not commute, i.e.[Lx, Ly] 6= 0, we cannot find an eigenstate of ~L, but we can find eigenstates of L2 and

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8.4. THE RADIAL EQUATION 149

Lz. Thus the solution of the Schrodinger equation for a central potential can also be aneigenstate of L2 and Lz, i.e.

H ψn`m = En`m ψn`m ( 8.79a)

L2 ψn`m = h2`(`+ 1)ψn`m ( 8.79b)

Lz ψn`m = hmψn`m , ( 8.79c)

where n, ` and m are quantum numbers with ` being the angular momentum and m, theprojection of the angular momentum along the z-axis. Here, n is an additional quantumnumber to be determined later. We will find that the definition of n in terms of theenergy will depend on the choice of the central potential. Because the Hamiltonian canbe divided into two parts, one depending on r, the other on θ and φ, the wave functioncan be written as a product, i.e.

ψn`m(~r ) = Rn`(r)Y`m(θ, φ) (8.80)

with the radial function Rn`(r) satisfying the equation

− h2

1

r2

d

dr

(r2 d

dr

)− `(`+ 1)

r2

R`(r) + V (r)R`(r) = E R`(r) . (8.81)

We now write the radial wave function as

R`(r) =u`(r)

r, (8.82)

and then we have thatdR`

dr=

1

r

du`dr− u`(r)

r2

and

d

dr

(r2dR`

dr

)=

d

dr

(rdu`dr

)− du`

dr

= rd2u`dr2

.

This allows us to write the radial Schrodinger equation for u`(r) as

− h2

d2u`dr2

+

(V (r) +

h2

`(`+ 1)

r2

)u`(r) = Eu`(r) (8.83)

or, in terms of an ‘effective potential’, as

− h2

d2u`dr2

+ Veff (r)u`(r) = Eu`(r) , (8.84)

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150 CHAPTER 8. CENTRAL FORCE PROBLEM I

where this angular momentum dependent effective potential is given in terms of the orig-inal potential V (r) as:

Veff (r) = V (r) +h2

`(`+ 1)

r2. (8.85)

This is illustrated in Figure 8.2, where we have assumed V (r) is a square well, and theresultant effective potential is represented by the dashed curve. We observe that Eq. (8.84)is identical in form to the one dimensional Schrodinger equation we considered in previouschapters, the only differences being:

1. We now have r ≥ 0 only.

2. The potential V (r) is replaced by Veff (r) which depends on `, the angular momen-tum.

3. Near the origin of the wave function u(r) → 0 as r → 0, otherwise R(r) will beinfinite at the origin. In this way we have insured the radial probability density|R(r)|2 is finite at r = 0. Thus at r = 0 we have a situation similar to the onedimensional case with V (x)→ +∞ for x ≤ 0.

V(r)

l(l+1)

Veff (r)

r 2

Figure 8.2: A comparison of the ‘effective potential’ Veff and the square well V (r).

We observe here that for r > a the effective potential is repulsive for ` 6= 0. Thisrepulsion is due to an angular momentum barrier which keeps the particle away from theorigin if the particle is outside the well, or possibly traps the particle if the particle isinside the well. This potential is illustrated in the Figure 8.2.

8.5 Problems

1. Use the tensor notation developed in this chapter to prove that

∇× ( ~A× ~B) = ~A(∇ · ~B)− ~B(∇ · ~A) + ( ~B · ∇) ~A− ( ~A · ∇) ~B .

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8.5. PROBLEMS 151

2. Given the change of variables from rectangular to spherical polar coordinates (seeEq. (8.46)), show that the Laplacian in spherical polar coordinates is given by

∇2 =1

r2

∂r

(r2 ∂

∂r

)+

1

r2 sin θ

∂θ

(sin θ

∂θ

)+

1

r2 sin2 θ

∂2

∂φ2.

3. The generating function for the Legendre polynomial is given by

1√1− 2ws+ s2

=∞∑`=0

P`(w) s` .

Use this generating function to evaluate

+1∫−1

dw P`(w)P`′(w) .

4. Using the definition of the components of the angular momentum Lx, Ly, Lz, andthe commutation relation of the coordinate with the canonical momentum, i.e.,[ri, pj] = ihδij, show that:

(a) The commutation relation of the component of the angular momentum opera-tor with the component of the position and momentum operator are

[Li, rj] = ihεijkrk and [Li, pj] = ihεijkpk

(b) The commutation relation of the component of the angular momentum are;

[Li, Lj] = LiLj − LjLi = ihεijkLk

and [Lz, L

2]

= 0 ,

whereL2 = L2

x + L2y + L2

z .

5. A rigid system rotates freely about the z-axis with moment of inertia I. By express-ing the energy of the system in terms of the angular momentum, Lz, show that thepossible energy levels of the system are

Em =h2m2

2I, m = 0, 1, 2, . . .

with eigenfunctionsum(φ) = e±imφ ,

where φ is the angle specifying the orientation of the system in the x− y plane.

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152 CHAPTER 8. CENTRAL FORCE PROBLEM I

6. A diatomic molecule, e.g. H2, rotates about its center of mass and in a planeperpendicular to the z-axis. The Hamiltonian for this system is given by

H =1

2IL2z ,

where I is the moment of inertia of the molecule.

(a) Assuming the mass of each of the atoms in the molecule is m, and the distancebetween the atoms in a, what is the moment of inertia of the molecule?

(b) Write the above Hamiltonian in spherical polar coordinates.

(c) What are the solutions of the Schrodinger equation for this Hamiltonian?

(d) What is the rotational energy spectrum for this diatomic molecule?

(e) What is the ration of the energies of the first excited state of H2 and D2

molecules, given that the mass of D is twice the mass of H, and the separationbetween the nuclei in the two molecules is the same?

7. The rotational energy of a diatomic molecule may be obtained by regarding themolecule as a rigid system consisting of two point particles a fixed distance apart.This rigid system is free to rotate about its center of mass. By expressing the totalenergy in terms of the moment of inertia I, and the total angular momentum, showthat the rotational energy levels are

E` =h

2I`(`+ 1), ` = 0, 1, 2, . . .

and that the level specified by ` has (2`+ 1) different eigenstates, corresponding tothe possible values of the z-component of the angular momentum

Lz = hm m = 0,±1,±2 . . . ,±` .

8. Given the operator

L− = Lx − iLy ;

show that L−Y11(θ, φ) is proportional to Y10(θ, φ), i.e.

L−Y11(θ, φ) = aY10(θ, φ) .

Determine the constant a.

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8.5. PROBLEMS 153

9. Show that

|Y11(θ, φ)|2 + |Y10(θ, φ)|2 + |Y1−1(θ, φ)|2 =3

4πP1(cos θ) .

This is a particular case of the theorem, which states that

∑m

Y ∗`m(θ, φ)Y`m(θ, φ) =2`+ 1

4πP`(cos θ) .

This result shows that when all m states corresponding to a given ` are occupied,the system has spherical symmetry.

10. The parity operator P is defined by the relation

P ψ(~r) = ψ(−~r) .

Show thatP Y`m(θ, φ) = (−1)` Y`m(θ, φ) .

Show that the solutions of the Schrodinger equation for central potentials havedefinite parity.

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154 CHAPTER 8. CENTRAL FORCE PROBLEM I

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Chapter 9

Central Force Problem II

Abstract : This chapter is devoted to the solution of the radial Schrodingerequation for three different spherically symmetric potentials whose solutionsare used to discuss approximation methods in atomic, molecular and nuclearproblems in Chapter13.

In the last Chapter we started with the simple model of the hydrogen atom and showedthat in general the problem of two particles interacting via a potential that depends onthe relative distance between them can be divided into two parts. The angular part of thewave function, which was the same for all central potential, and a radial wave functionthat depended on the form of the potential. In the last chapter we considered in detailsthe angular part of the Schrodinger equation for central potential. In this Chapter weturn to the radial part of the Schrodinger equation and start by considering the case ofa square well in three-dimensions, then proceed to the harmonic oscillator, and finallythe Coulomb problem which was the initial motivation for considering the Schrodingerequation for central potentials. In the process we introduce the special functions neededfor scattering theory, and which form the basis for modeling more complex systems.

9.1 The Three-Dimensional Square Well

In the last Chapter we showed that the solution to the Schrodinger equation for a centralpotential can be reduced to solving the differential equation (8.81) for the radial wavefunction R`(r). To solve this radial equation we need to specify the potential V (r). Inthe present section we will consider the case of a simple square well potential given by

V (r) =

−V0 for r < a

0 for r > a, (9.1)

155

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156 CHAPTER 9. CENTRAL FORCE PROBLEM II

where a is the radius of the well. For this potential, the radial equation for r < a takesthe form

d2R`

dr2+

2

r

dR`

dr+

(α2 − `(`+ 1)

r2

)R`(r) = 0 , (9.2)

where

α2 =2m

h2 (E + V0) . (9.3)

If we now define ρ = αr, the radial equation reduces to

d2R`

dρ2+

2

ρ

dR`

dρ+

(1− `(`+ 1)

ρ2

)R`(ρ) = 0 . (9.4)

This equation is known as the spherical Bessel equation. To relate it to the ordinaryBessel’s differential equation. i.e.,

d2Jpdρ2

+1

ρ

dJpdρ

+

(1− p2

ρ2

)Jp(ρ) = 0 , (9.5)

we have to write the radial wave function as

R`(ρ) = ρ−1/2 J(ρ) .

Substituting this in Eq. (9.4), we find that J(ρ) satisfies the equation

d2J

dρ2+

1

ρ

dJ

dρ+

(1−

(`+ 12)2

ρ2

)J(ρ) = 0 (9.6)

which is basically Eq. (9.5) for p = ` + 1/2. We therefore can write the radial wavefunction in terms of the half-integer Bessel function as

R`(ρ) =

√π

2ρJp(ρ) p = ±

(`+

1

2

). (9.7)

We now introduce the spherical Bessel function, j`(ρ) as

j`(ρ) =

√π

2ρJ`+ 1

2(ρ) (9.8)

and the spherical Neumann function, n`(ρ) as

n`(ρ) = (−1)`√π

2ρJ−`− 1

2(ρ) , (9.9)

Alternatively, we can define the spherical Hankel function of the first kind h(+)` (ρ), and

the second kind h(−)` (ρ) as

h(±)` (ρ) = n`(ρ)± i j`(ρ) . (9.10)

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9.1. THE THREE-DIMENSIONAL SQUARE WELL 157

From the above results, we may deduce that there are two independent solutions to theradial equation. These are j`(ρ) and n`(ρ), or the Hankel functions h

(±)` . The existence of

two independent solutions is a consequence of the fact that we have a second order lineardifferential equation. Here, j` and n` are real solutions and are the convenient choice forthe application of the boundary condition at r = 0. On the other hand, h

(±)` are complex

solutions, which we will use to impose the boundary condition for r → ∞. For the caseof ` = 0 and 1, these functions take the form of 1

j0(ρ) =sin ρ

ρ, j1(ρ) =

sin ρ

ρ2− cos ρ

ρ,

n0(ρ) =cos ρ

ρ, n1(ρ) =

cos ρ

ρ2+

sin ρ

ρ, (9.11)

h(±)0 (ρ) =

e±iρ

ρ, h

(±)1 (ρ) =

(1

ρ2∓ i

ρ

)e±iρ .

The asymptotic forms of these functions for ρ→∞ are

j`(ρ) → 1

ρsin

(ρ− π`

2

)

n`(ρ) → 1

ρcos

(ρ− π`

2

)(9.12)

h(±)` (ρ) → 1

ρ

[1± i`(`+ 1)

2ρ− · · ·

]exp

[±i(ρ− π`

2)

].

On the other hand the behavior of these functions as ρ→ 0 are

j`(ρ) → ρ`

(2`+ 1)!!

[1− ρ2

2(2`+ 3)+ · · ·

](9.13)

n`(ρ) → (2`+ 1)!!

(2`+ 1)

1

ρ`+1

[1 +

ρ2

2(2`+ 1)+ · · ·

],

where (2`+ 1)!! = (2`+ 1)(2`− 1)(2`− 3) · · · 1.Having established the asymptotic behavior of the two independent solutions of Eq. (9.4),

we can write the most general solution for r < a as

R`(ρ) = Aj`(ρ) +Bn`(ρ) for r < a .

1For more detailed properties of Bessel functions and other functions encountered in this chapter seeAbramowitz and Stegun [24]

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158 CHAPTER 9. CENTRAL FORCE PROBLEM II

Using the boundary condition that R`(ρ) should be finite as ρ → 0 and the results ofEq. (9.13), we conclude that B = 0, and therefore

R`(ρ) = Aj`(ρ) for r < a . (9.14)

On the other hand, for r > a, the radial Schrodinger equation takes the form

d2R`

dr2+

2

r

dR`

dr+

(β2 − `(`+ 1)

r2

)R`(r) = 0 (9.15)

with

β2 =2mE

h2 . (9.16)

For bound states, i.e., E < 0, we have β2 = −2m|E|/h2. We now define ρ = βr ≡ iγrwhere γ is real, and our general solution is a linear combination of j`(ρ) and n`(ρ), or

h(+)` (ρ) and h

(−)` (ρ). The latter choice will turn out to be more convenient for imposing the

boundary condition for r →∞. We therefore write the solution of the radial Schrodingerequation for r > a and E < 0 as

R`(r) = A′h(+)` (ρ) +B′h

(−)` (ρ) for r > a . (9.17)

Here ρ is pure imaginary for bound states. Considering the asymptotic behavior of h(±)` (ρ)

(Eq. (9.12)), and the boundary condition that R`(r) be finite as r → ∞, we find thatB′ = 0. We therefore have for r > a that

R`(r) = A′ h(+)` (ρ)

→ A′

iγre−γr−iπ`/2 =

C

re−γr for r →∞ . (9.18)

We now can write the radial wave function for a square well potential as

R`(r) =

Aj`(αr) for r < a

A′h(+)` (iγr) for r > a

. (9.19)

The constants A, A′ and the bound state energy E are determined by the requirement thatthe wave function R`(r) and its derivative be continuous at r = a, and the normalizationcondition

∞∫0

dr r2R∗` (r)R`(r) = 1 (9.20)

be satisfied.

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9.2. THE THREE-DIMENSIONAL HARMONIC OSCILLATOR 159

9.2 The Three-Dimensional Harmonic Oscillator

The harmonic oscillator in three-dimensions is very important because of its simplicity andthe fact that the solution is a good approximation to many physically interesting problems.For the present, we are going to use the harmonic oscillator potential to illustrate thesolution of the radial Schrodinger equation using series solution. In particular, we wantto illustrate how the discrete eigenvalues arise in a bound state problem as a result of theboundary conditions imposed on the solution of the Schrodinger equation.

The potential for the three dimensional harmonic oscillator is given by

V (r) =1

2mω2r2 . (9.21)

The radial Schrodinger equation can be written ( see Eq. (8.83) ) as(− h2

2m

d2

dr2+h2

2m

`(`+ 1)

r2+

1

2mω2r2 − E

)u`(r) = 0 , (9.22)

where u`(r) = r R`(r) and m is the mass of the particle in the potential well. To simplifythe differential equation, we define

β =

√mω

h, ξ = βr , and ε = E/hω .

With these definitions we can write Eq. (9.22) as(d2

dξ2− `(`+ 1)

ξ2− ξ2 + 2ε

)u`(ξ) = 0 . (9.23)

We can solve this differential equation by series solution. However, before we proceedto implement the series solution, let us extract the asymptotic solution for r → 0 andr → ∞. In this way we will find that the final series solution corresponds to a knownpolynomial when the boundary conditions are applied.

1. For ξ → 0, Eq. (9.23) can be approximated by the equation(d2

dξ2− `(`+ 1)

ξ2

)u` = 0 , (9.24)

since the terms −ξ2 and 2ε are small in comparison to `(`+ 1)/ξ2. The solutions ofEq. (9.24) are

u` = ξ`+1 or ξ−` . (9.25)

Since the radial wave function has to be finite at the origin, i.e. r → 0, the secondsolution in Eq. (9.25) is not acceptable because u` →∞ as ξ → 0, and the solution

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160 CHAPTER 9. CENTRAL FORCE PROBLEM II

would be singular at the origin. From the above analysis we can conclude that thesolution to Eq. (9.23) is given by

u`(ξ)→ ξ`+1 for ξ → 0 .

Note, this result is valid for all potentials that satisfy the condition that V (r) →r−2+δ for r → 0 and δ > 0.

2. For ξ → ∞. In this case we can drop terms like `(` + 1)/ξ2 and 2ε, since they aresmall in comparison to the term proportional to ξ2. The asymptotic equation nowis given by (

d2

dξ2− ξ2

)u` = 0 . (9.26)

Taking z = ξ2, this equation can be rewritten as(4z

d2

dz2+ 2

d

dz− z

)u` = 0 , (9.27)

or for large z the differential equation can be approximated by(d2

dz2− 1

4

)u` = 0 . (9.28)

This equation has a known solution which is

u` = e−z/2 = e−ξ2/2 . (9.29)

From these results we can conclude the u`(ξ) which is a solution of Eq. (9.23) has thefollowing asymptotic behavior

u`(ξ) =

ξ`+1 for ξ → 0

e−ξ2/2 for ξ →∞

. (9.30)

We therefore could write u`(ξ) as

u`(ξ) = ξ`+1 e−ξ2/2 f`(ξ) . (9.31)

In this way we hope f`(ξ) to be a simple function of ξ. Substituting u`(ξ) from Eq. (9.31)into Eq. (9.23), we get an equation for f`(ξ) of the form

d2f`dξ2

+ 2

(`+ 1

ξ− ξ

)df`dξ

+ [2ε− (2`+ 3)] f`(ξ) = 0 . (9.32)

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9.2. THE THREE-DIMENSIONAL HARMONIC OSCILLATOR 161

If we now define L(η) = f`(ξ) where η = ξ2, then L(η) is a solution to the differentialequation

ηd2L

dη2+[(`+

3

2)− η

]dL

dη+[ε

2− 1

2

(`+

3

2

)]L(η) = 0 . (9.33)

This is a special case of Laplace’s equation(zd2

dz2+ (β − z)

d

dz− α

)F = 0 . (9.34)

This equation has a solution known as a confluent hypergeometric function which can bewritten as a series of the form

F (α|β|z) =∞∑p=0

Γ(α + p) Γ(β)

Γ(β + p) Γ(α)

zp

p!, (9.35)

where Γ(a) is a Γ-function. Comparing Eqs. (9.33) and (9.34), we can write the seriessolution for L(η) in the form

L(η) = F (b|a|η) =∞∑p=0

Γ(b+ p)

Γ(a+ p)

Γ(a)

Γ(b)

ηp

p!, (9.36)

where

b =1

2

(`+

3

2

)− ε

2, a = `+

3

2. (9.37)

For large η, the confluent hypergeometric function has the asymptotic behavior that

F (b|a|η)→ Γ(a)

Γ(b)ηb− a eη =

Γ(a)

Γ(b)(βr)2(b− a) eβ

2r2for η →∞ . (9.38)

If we combine this result with that of Eq. (9.31), we observe that u`(ξ) ∝ eβ2r2/2, i.e.,

the wave function goes to infinity as r →∞ which violates the boundary condition. Theonly way this situation can be avoided, is for the series solution in Eq. (9.36) to becomea finite polynomial. This can be achieved by taking

b =1

2

(`+

3

2

)− ε

2= −n , (9.39)

where n is a positive integer. Under this condition the series solution becomes

L(η) =∞∑p=0

Γ(p− n)

Γ(p+ a)

Γ(a)

Γ(−n)

ηp

p!. (9.40)

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162 CHAPTER 9. CENTRAL FORCE PROBLEM II

But now we have that2

Γ(p− n)

Γ(−n)=

0 for p > n

(−1)p n!(n− p)! for p ≤ n

. (9.41)

This result, which turns L(η) to a polynomial of degree n, is due to the fact that thegamma function, Γ(n), has simple poles for values of n which are negative integers. Wenow can write L(η) as

L(η) =n∑p=0

(−1)pΓ(`+ 3

2

)Γ(`+ p+ 3

2

) ( np

)ηp . (9.42)

This polynomial solution of Eq. (9.33) is known as the associated Laguerre polynomial.Combining the results of Eqs. (9.31) and (9.42), we can write the solution of the radialSchrodinger equation for the harmonic oscillator potential as

Rn`(r) = Nn`(β) r` e−β2r2/2

n∑p=0

(−1)p(np

)Γ(`+ 3

2

)Γ(`+ p+ 3

2

) (β2r2)p

, (9.43)

with

β =

√mω

h

and the normalization constant Nn`(β) is given by

N2n`(β) =

2`−n+2 (2`+ 2n+ 1)!! β2`+3

√π n! [(2`+ 1)!!]2

. (9.44)

Here we observe that had we not taken the asymptotic behavior of Rn`(r) for r → 0 andr →∞ out, we would have not obtained a simple polynomial solution for the remainingwave function. In particular, a series solution for Rn`(r) would have been an infiniteseries because it would have included a power series expansion for e−β

2r2/2. Furthermore,it would have been very difficult to impose the boundary condition for r →∞.

Let us now examine the condition under which L(η) became a polynomial, which wasthe condition to guarantee a finite solution for r →∞. From Eq. (9.39), we have

ε = 2n+ `+3

2

2To prove this result, we make use of the reflection property of the Γ-function, i.e.

Γ(z)Γ(−z) = − π

z sinπz,

and the fact that the residue of Γ(z) at z = −n is (−1)n/n! (see problem 1).

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9.3. THE COULOMB POTENTIAL 163

Table 9.1: The lowest eigenvalues and corresponding radial wave function for the Har-monic Oscillator in three-dimension. Also included are spectroscopic names for thesestates.

n ` En` Rn`/Nn` Name

0 0 32hω e−β

2r2/2 0s

0 1 52hω r e−β

2r2/2 0p

1 0 72hω

(1− 2

3β2r2

)e−β

2r2/2 1s

0 2 72hω r2 e−β

2r2/2 0d

or

En` = hωε = hω(

2n+ `+3

2

). (9.45)

Thus we have found that the condition that the wave function be finite as r →∞ quantizesthe energy levels of the Harmonic Oscillator.

In the Table 9.1 we have the eigenvalues and eigenfunctions of the lowest energyeigenstates with their spectroscopic notation.

9.3 The Coulomb Potential

We started the last Chapter by considering the simplest model for the hydrogen atomas a negative charge electron attracted to a positively charged proton by the Coulombpotential. We have already determined the angular part of the wave function to be thespherical harmonics. In this section we solve the radial equation for a hydrogen like atomwhich consists of an electron in the Coulomb field of a nucleus of charge Ze, i.e.

Vc(r) = −Ze2

r. (9.46)

In this case the radial equation for u`(r) is given byd2

dr2− `(`+ 1)

r2+

2µZe2

h2r+

2µE

h2

u`(r) = 0 . (9.47)

Since we want to consider bound states only, we have E < 0. To simplify our differentialequation for u`(ρ), we define new parameters for the energy and strength of the potential

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164 CHAPTER 9. CENTRAL FORCE PROBLEM II

as:

κ2 = −2µE

h2 and β =µZe2

h2κ. (9.48)

Our differential equation for u`(ρ) now reduces to(d2

dr2− `(`+ 1)

r2+

2βκ

r− κ2

)u`(r) = 0 . (9.49)

Changing the variable in the differential equation to ρ = 2κr, we get(d2

dρ2− `(`+ 1)

ρ2+β

ρ− 1

4

)u`(ρ) = 0 . (9.50)

Before we attempt a general solution of this differential equation, let us follow the pro-cedure implemented in the last section and examine the asymptotic solutions for ρ → 0and ρ → ∞. For ρ → 0, the dominant term is −`(` + 1)/ρ2 and the equation has theasymptotic form

d2

dρ2− `(`+ 1)

ρ2

u`(ρ) = 0 . (9.51)

This equation has one of two possible solutions:

u` ∼ ρ`+1 , or u` ∼ ρ−` .

The requirement that u`(ρ) be finite, and in particular zero at ρ = 0, demands that weonly consider the solution

u`(ρ) ∼ ρ`+1 . (9.52)

For the case of ρ→∞, our radial Schrodinger equation for u`(ρ) reduces to(d2

dρ2− 1

4

)u`(ρ) = 0 . (9.53)

Here again two solutions are possible,

u`(ρ) ∼ eρ/2 , or u`(ρ) ∼ e−ρ/2 .

The boundary condition requirement that the wave function for a bound state be zero asr →∞, implies that the only solution that is acceptable is

u`(ρ) ∼ e−ρ/2 . (9.54)

We now can build the two asymptotic solutions into u(r) by taking

u`(ρ) = ρ`+1e−ρ/2f`(ρ) . (9.55)

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9.3. THE COULOMB POTENTIAL 165

This procedure, we hope, will imply that f`(ρ) is a simple function like a polynomial. Tofind the differential equation that f`(ρ) satisfies, we make use of the Schrodinger Eq. (9.50)and the fact that

du`dρ

= (`+ 1)ρ`e−ρ/2f`(ρ)− 1

2ρ`+1e−ρ/2f`(ρ) + ρ`+1e−ρ/2

df`dρ

,

with the second derivative of u`(ρ) given by

d2u`dρ2

= `(`+ 1)ρ`−1e−ρ/2f`(ρ)− (`+ 1)ρ`e−ρ/2f`(ρ) + 2(`+ 1)ρ`e−ρ/2df`dρ

+1

4ρ`+1e−ρ/2f`(ρ)− ρ`+1e−ρ/2

df`dρ

+ ρ`+1e−ρ/2d2f`dρ2

.

This result for the second derivative we substitute in Eq. (9.50) to write a second orderdifferential equation for the function f`(ρ) that is of the form

ρd2f`dρ2

+ [2(`+ 1)− ρ]df`dρ− [(`+ 1)− β]f`(ρ) = 0 . (9.56)

If we attempt a series solution for this equation3, i.e.,

f`(ρ) =∞∑s=0

as ρs , (9.57)

thendf

dρ=∞∑s=0

s as ρs−1 =

∞∑s=0

(s+ 1) as+1 ρs ,

where the second expression results from the fact that the first term in the sum is zeroand can be dropped in conjunction with a change in the sum index s→ s+1. In a similarmanner, we have

d2f`dρ2

=∞∑s=0

s (s− 1) as ρs−2 =

∞∑s=0

(s+ 2)(s+ 1) as+2 ρs .

Using these results, our differential equation, Eq. (9.56), becomes

∞∑s=0

(s+ 2)(s+ 1)as+2ρ

s+1 + 2(`+ 1)(s+ 1)as+1ρs − (s+ 1)as+1ρ

s+1

− [(`+ 1)− β]asρs = 0 . (9.58)

3This procedure is identical to that employed in the last chapter to derive the wave function for a onedimensional harmonic oscillator.

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166 CHAPTER 9. CENTRAL FORCE PROBLEM II

Since this sum is zero for any value of ρ, we expect the coefficients of ρ p,p = 0, 1, · · · tobe zero, i.e.,

ρ0 : 2(`+ 1)a1 = (`+ 1− β)a0

therefore a1 =`+ 1− β2(`+ 1)

a0

ρ1 : 2a2 + 4(`+ 1)a2 = a1 + [(`+ 1)− β]a1

therefore a2 =`+ 2− β

2[2(`+ 1) + 1]a1

......

ρs : s(s+ 1)as+1 + 2(`+ 1)(s+ 1)as+1 = sas + [(`+ 1)− β]as

therefore as+1 =[s+ `+ 1− β]

(s+ 1)[2(`+ 1) + s]as . (9.59)

From the above recursion relation for the coefficients as, we have that

as+1

as∼ 1

sfor large s .

Here again, as for the case of the one dimensional harmonic oscillator discussed in thelast chapter, if we don’t terminate the series, the solution for large ρ will not satisfy theboundary condition that the wave function should go to zero as ρ → ∞. For the seriessolution, with the coefficients given by Eq. (9.59), to terminate we should have

s+ `+ 1− β = 0 . (9.60)

This can be achieved by taking β to be a positive integer, i.e. β = n ≥ 1. Making use ofthe definition of β as given in Eq. (9.50), we have

β = n =µZe2

h2κ.

We can solve this equation for κ, i.e.,

κ =µZe2

h2n=

Z

anor

κ2 = −2µE

h2 =µ2Z2e4

h4n2=

Z2

a2n2,

where a = h2/µe2 is the Bohr radius. Therefore the energy is given by

E = −µZ2e4

2h2n2. (9.61)

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9.3. THE COULOMB POTENTIAL 167

Table 9.2: The energy and radial wave function for the hydrogen atom (Z = 1). Theenergies are given in terms of Rydbergs R∞, with R∞ = 13.6 eV. Here ρ = 2r/na, wherea is the Bohr radius and is given by a = h2/µe2.

name n ` En Rn`(r)

1s 1 0 R∞ 2(

1a

)3/2e−ρ/2

2s 2 0 14R∞

(12a

)3/2(2− ρ) e−ρ/2

2p 2 1 14R∞

1√3

(12a

)3/2ρ e−ρ/2

Defining the fine structure constant, α as

α =e2

hc=

1

137, (9.62)

the energy is then given in terms of the fine structure constant as

En = −µc2

2

(Zα)2

n2≡ −R∞Z

2

n2, (9.63)

where the Rydberg, R∞ = µe4/2h2 = 13.6 eV. In Table 9.2 we list the values for thelowest states in this model of the hydrogen atoms. Also included are the spectroscopicname and wave function for each state. Here we observe that for a given n, all stateswith ` ≤ n have the same energy. This is a result of taking the potential between thenucleus and the electron to be the Coulomb potential only, and is due to a symmetryof the Hamiltonian that is not obvious4. In addition, for each ` we can have (2` + 1)different projections of the angular momentum, i.e., we have a (2`+ 1) degeneracy. Thisdegeneracy is a result of the rotational symmetry of the Hamiltonian. Although theseresults are a good approximation for the hydrogen spectrum, it is by no means the fullstory. We need to include the magnetic interaction, relativistic effects, and the effect ofpolarization of the vacuum if we are to give a full description of the hydrogen spectrum.Some of these effects will be considered in later chapters.

4The Coulomb spectrum was first calculated by Pauli [25] on the basis that the Hamiltonian has O(4)symmetry. For a detailed discussion of dynamic symmetry and application to the energy spectrum ofHamiltonians see Schiff [26].

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168 CHAPTER 9. CENTRAL FORCE PROBLEM II

We now turn to the wave function for the Coulomb potential in terms of the seriessolution. For β = n the coefficients in the series are related by the recursion relations

as+1 =[s+ `+ 1− n]

(s+ 1)[2(`+ 1) + s]as . (9.64)

The series solution for f(ρ) can now be written as

f(ρ) = a0

1 +

a1

a0

ρ+a2

a0

ρ2 +a3

a0

ρ3 + · · ·

= a0

1 +

a

bρ+

a(a+ 1)

b(b+ 1)

ρ2

2!+a(a+ 1)(a+ 2)

b(b+ 1)(b+ 2)

ρ3

3!+ · · ·

. (9.65)

where a = ` + 1 − n and b = 2` + 2. The above infinite series is an expansion of theconfluent hypergeometric function, which is defined as:

1F1(a, b; ρ) = 1 +a

bρ+

a(a+ 1)

b(b+ 1)

ρ2

2!+a(a+ 1)(a+ 2)

b(b+ 1)(b+ 2)

ρ3

3!+ · · · . (9.66)

For the case when a = ` + 1 − n is a negative integer, the infinite series becomes apolynomial known as the Laguerre Polynomial, which is defined in this case as

L2`+1n−`−1(ρ) ≡ (n+ `)!

(n− `− 1)! (2`+ 1)!1F1(−n+ `+ 1, 2`+ 2; ρ) (9.67)

The full solution of the Schrodinger equation for the Coulomb Hamiltonian can now bewritten as

Rn`(r) = Nn`ρ`e−ρ/2 L2`+1

n−`−1(ρ) with ρ = 2κr =2Z

anr . (9.68)

where a is the Bohr radius given by a = h2/µe2, and Z the charge on the nucleus. Thenormalization of the wave function, Nn` is given by5

N2n` =

[(2Z

na

)3 (n− `− 1)!

2n[(n+ `)!]3

]. (9.69)

The radial solutions tabulated in Table 9.1 are plotted in Figure 9.1.

5In determining the normalization of the wave function, i.e.∞∫

0

dr r2 [Rn`(r)]2,

we have made use of the fact that∞∫

0

dρ ρ2`+2 e−ρ[L2`+1n−`−1(ρ)

]2=

2n[(n+ `)!]3

(n− `− 1)!.

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9.3. THE COULOMB POTENTIAL 169

2 4 6 8 10

0.5

1.0

1.5

ξ

10 x R21

R10

R20

Rnl

Figure 9.1: The radial wave function for the lowest three states of the Coulomb potential.

The wave functions given in Figure 9.1 cannot be measured in coordinate space, butthe momentum space wave function has been measured. To compare our results withexperiment, we need to Fourier transform this coordinate space wave function. For theground state, the momentum space wave function is given by

φ100(~k) =1

(2π)3/2

∫d3r ei

~k·~r ψ100(~r) . (9.70)

This integral can be performed in spherical polar coordinates. Making use of the fact that

ψ100(~r) =1√4π

R10(r)

where R10 is given in Table 9.1, we get

φ100(~k) =1

(2π)3/2

2

a3/2√

∫d3r ei

~k·~r e−r/a

=1

(2π)3/2

2√4πa3/2

∞∫0

drr2 e−r/a 2π

+1∫−1

dx eikrx .

In writing the second line we have taken d3r = r2dr dφ dx, where x = cos θ with θ the anglebetween ~k and ~r. The φ integration gives a factor of 2π. Performing the x integration weget

φ100(~k) =1

(2π)3/2,

√4π

a3/2

1

ik

∞∫0

drr e−r/a[eikr − e−ikr] .

To perform the r integration we define ξ = 1/a. This change of variable allows us to writeour integral as

φ100(~k) =1

(2π)3/2

2√

a3/2

1

k

∞∫0

drr e−ξr sin kr

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170 CHAPTER 9. CENTRAL FORCE PROBLEM II

= − 1

(2π)3/2

2√

a3/2

1

k

∂ξ

∞∫0

dr e−ξr sin kr .

But we have that∞∫0

dr e−ξr sin kr =k

ξ2 + k2.

This allows us to write the wave function for the ground state of the Coulomb Hamiltonianas

φ100(~k) =2√

2

π

a3/2

(1 + k2a2)2. (9.71)

The probability of finding the electron with momentum ~p = h~k is given by the square ofthe magnitude of the momentum space wave function, i.e.,

|φ100(~k)|2 =8

π2

a3

(1 + k2a2)4. (9.72)

This quantity has been recently measured. In Figure 9.1 we compare our results withthe experimental results, and the agreement could not be better. This illustrates the factthat a direct measurement of the wave function is possible.

Figure 9.2: Comparison of the momentum wave function for the ground state of Hydrogenwith the results of (e, 2e) experiment on hydrogen by B. Lohmann and E. Weigold [27].

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9.4. PROBLEMS 171

9.4 Problems

1. Making use of the fact that the Γ-function can be written as

Γ(z) =1

z

∞∏l=1

(1 +

1

l

)z (1 +

z

l

)−1

to show that the residue of the Γ-function for negative integers, n, is (−1)n/n!, i.e.,

limz→−n

(z + n)Γ(z) =(−1)n

n!for n positive integer

2. Obtain an approximate analytic expression for the energy levels in a square wellpotential with depth V0 and radius a, when V0a

2 is slightly greater than π2h2/8m.Take the orbital angular momentum to be zero.

3. If the ground state of a particle in a square well is just barely bound, show that thewell depth V0 and radius a are related to the binding energy by the expansion

2µV0a2

h2 =π2

4+ 2η +

(1 +

4

π2

)η2 + · · ·

where

η =

√−2µEa

h.

Here, E is the energy of the ground state and µ is the mass of the particle.

4. Solve the radial Schrodinger equation for ` = 0 and

V (r) = −V0 e−r/a .

Change variables from r toz = e−r/2a

and show that Bessel’s equation results. What boundary conditions are to be im-posed on the solution U0(r) as a function of z? How can these be used to determinethe energy levels in this potential?

5. Solve the radial Schrodinger equation for the potential

V (r) =A

r2+Br2 .

Consider only the discrete spectrum.

Hint: Compare with the three-dimensional harmonic oscillator in spherical coordi-nates.

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172 CHAPTER 9. CENTRAL FORCE PROBLEM II

6. To introduce electromagnetic interaction into quantum systems, we make use of theminimal coupling principle which involves the substitution

~p→ ~p− e

c~A ,

where ~A is the vector potential and c the velocity of light. This procedure gives, foran electron in an external vector potential ~A(~r), the Schrodinger equation

1

2m

(h

i∇− e

c~A(~r)

)·(h

i∇− e

c~A(~r)

)ψ(~r) = Eψ(~r) .

Simplify the equation for the case where ~A(~r) describes a uniform magnetic fieldpointing in the z-direction.

(a) Show that the Hamiltonian can be written as

H = − h2

2m∇2 − e

2mcB0Lz +

e2B20

8mc2(x2 + y2) .

(b) Show that in cylindrical coordinates this Hamiltonian is the sum of two Hamil-tonian with one corresponding to a two-dimensional harmonic oscillator.

(c) What is the energy spectrum for this Hamiltonian?

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Chapter 10

Scattering by a Central Potential

Abstract : With the central role played by scattering in any measurements on aquantum system, in this chapter we introduce the concepts in scattering theorywith application to the simple square well problem as well as the Coulombproblem.

Most scattering experiments consist of a beam of particles incident on a stationarytarget, and a detector to measure the number of scattered particles in a given directionper unit time. In practice, the beam has many particles in it, but we will neglect anyinteraction between the particles. Similarly, the target has many particles in a confinedregion of space and again we will assume the interaction between the target particles canbe ignored. Although we should be considering a wave packet incident on the target, itcan be shown1 that we can get the same result for the scattering amplitude assuming theincident beam is described by a plane wave. In this way we reduce the complexity ofthe algebra in the formulation of the scattering problem. The plane incident wave withmomentum ~ki is then given, up to a normalization, by

ψinc(~r) ∝ ei~ki·~r . (10.1)

On the other hand, the scattered particles will be originating at the target and moving ina spherical wave. The amplitude of this scattered wave might depend on the scatteringangles. However, because of the cylindrical symmetry about the incident beam, the onlyangle dependence is the angle between the incident beam direction, ki, and the scatteredparticle’s direction of momentum kf . Here, the subscripts i and f refer to the initial andfinal states. We now can write the scattered wave in terms of a spherical wave, eikr/r, as

ψscat ∝ f(ki, ki · kf )eikf r

r. (10.2)

1For a full treatment of scattering using wave packets consult Goldberger and Watson [28].

173

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174 CHAPTER 10. SCATTERING BY A CENTRAL POTENTIAL

For elastic scattering, i.e. when the initial energy of the incident particle is equal to thefinal energy of the scattered particle, we have that |~ki| = |~kf | = k,2 and we can write thetime independent wave function outside the interaction region as

ψ(~r) ∝ ei~ki·~r + f(k, θ)

eikr

r, (10.3)

where cos θ = ki · kf and f(k, θ) is the amplitude of the scattered wave or the scatteringamplitude. This wave function is illustrated in Figure 10.1.

r

e i k . re i k r

Figure 10.1: Illustration of the asymptotic wave function for a scattering problem. It con-sists of an incident plane wave and an outgoing spherical wave to represent the scatteredparticles.

In the next two sections we will relate the probability for scattering in a given directionto the coefficient of the outgoing spherical wave f(k, θ). We will then relate the amplitudef(k, θ) to the solution of the Schrodinger equation. In this way we establish the relationbetween the wave function and the quantities we measure in any scattering experiment.In the final section of this chapter we will consider the problem of scattering by a Coulombpotential, which will require special treatment because the interaction in this case has aninfinite range.

10.1 The Cross Section

In any scattering experiment, we measure the number of particles scattered at a givenangle θ. If we divide the number of scattered particles per unit time by the flux of incidentparticles, we get the probability for a particle scattering at a given angle. This is referred

2This definition of elastic scattering assumes we are in the center of mass where ~ki and ~kf are therelative initial and final momentum (see Sec. 2.2).

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10.1. THE CROSS SECTION 175

to as the differential cross section for that angle. In this section we derive the relationbetween the cross section and the scattering amplitude f(k, θ).

The time dependent Schrodinger equation for the potential V (r) is given by

ih∂Ψ

∂t=

(− h

2

2µ∇2 + V (r)

)Ψ(~r, t) , (10.4)

where µ is the reduced mass (see Eq. (8.12)) of the incident and target particles. Thecomplex conjugate solution, Ψ∗(~r, t) satisfies the equation

− ih∂Ψ∗

∂t=

(− h

2

2µ∇2 + V (r)

)Ψ∗(~r, t) . (10.5)

Multiplying Eq. (10.4) by Ψ∗ from the left, and Eq. (10.5) by Ψ from the right andsubtracting, we get

ih∂

∂t(Ψ∗Ψ) = − h

2

(Ψ∗∇2Ψ− (∇2Ψ∗)Ψ

)= −∇ · h

2

2µ( Ψ∗∇Ψ− (∇Ψ∗)Ψ ) . (10.6)

If we now define the density, ρ = Ψ∗Ψ, then Eq. (10.6) can be written as

∂ρ

∂t+∇ ·~j = 0 , (10.7)

where the current ~j is given by

~j =h

2µi(Ψ∗∇Ψ− (∇Ψ∗)Ψ)

=h

2µi(ψ∗∇ψ − (∇ψ∗)ψ) . (10.8)

Here, we have dropped the time dependence in the wave function since we are not dealingwith wave packets. In any scattering experiment the scattered particle is measured at adistance r which is large compared to the range of the interaction. This means that thedetector is in the asymptotic region where the wave function is given by

ψ(~r) = ei~ki·~r + f(k, θ)

eikr

r. (10.9)

Using Eq. (10.9) in Eq, (10.8), we get, after some algebra that involves writing ∇ in

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176 CHAPTER 10. SCATTERING BY A CENTRAL POTENTIAL

spherical polar coordinates,3 an expression of the form

~j =h~kiµ

+hk

µ

err2|f(k, θ)|2

+eikr(1−cos θ) [· · ·] + e−ikr(1−cos θ) [· · ·] . (10.10)

Since the detector is never put in the forward direction, i.e. θ = 0, and it subtends afinite solid angle, we need to integrate over the angular range of the detector, i.e.,∫

dθ dφ sin θ g(θ, φ) e±ikr(1−cos θ) ,

where g(θ, φ) is a smooth function that depends on the detector’s properties. For kr 1,we are integrating a highly oscillatory function, and the integral is zero according to theRiemann-Labegue lemma. We thus have for the current

~j =h~kiµ

+h k

µ

err2|f(k, θ)|2 . (10.11)

The first term represents the incident beam of particles, and is present even if there wasno scattering target. The radial flux of scattered particles is then given by

~j · er =hk

µ

|f(k, θ)|r2

. (10.12)

D

θ

d A = r 2 d Ω

k i

Figure 10.2: Illustration of the scattering angle θ, and the solid angle dΩ, subtended bythe detector D.

The number of particles crossing the area that subtends a solid angle dΩ is given by(see Figure 10.2)

~j · er dA =hk

µ|f(k, θ)|2 dΩ . (10.13)

3In spherical polar coordinates we have

∇ = er∂

∂r+ eθ

1r

∂θ+ eφ

1r sin θ

∂φ.

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10.2. KINEMATICS 177

The differential cross section is the flux of scattered particles per unit area divided by theincident flux, hk/µ. Therefore we have that

dσ = |f(k, θ)|2 dΩ

and we havedσ

dΩ= |f(k, θ)|2 . (10.14)

This in fact is the result we would have expected from the construction of Eq. (10.3) forthe asymptotic wave function. We now have to relate the scattering amplitude f(k, θ) tothe solution of the Schrodinger equation. One important observation we can make at thisstage is that the cross section measured experimentally depends on the form of the wavefunction for large r, i.e., outside the interaction region.

10.2 Kinematics

Consider the scattering of two particles where the potential between the particles is afunction of the relative distance between the particles, i.e. V (|~r1−~r2|). The Hamiltonianfor such a system is given by

H =p2

1

2m1

+p2

2

2m2

+ V (|~r1 − ~r2|) , (10.15)

where p21 = −h2∇2

1 and p22 = −h2∇2

2. We introduce relative and center of mass coordinatesand momenta (see Eqs. (8.6) and (8.7)), i.e.

~r = ~r1 − ~r2 and ~R =m1~r1 +m2~r2

m1 +m2

(10.16)

~p =m2~p1 −m1~p2

m1 +m2

and ~P = ~p1 + ~p2 .

In terms of these new variables the Hamiltonian takes the form

H =P 2

2M+p2

2µ+ V (r) , (10.17)

whereM = m1 +m2 and µ =

m1m2

m1 +m2

. (10.18)

Here, µ is the reduced mass, and M the total mass. In the center of mass we have ~P = 0,and the two-particle Hamiltonian reduces to the one-particle Hamiltonian with a reducedmass µ, i.e.

H =p2

2µ+ V (r) . (10.19)

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178 CHAPTER 10. SCATTERING BY A CENTRAL POTENTIAL

We thus have reduced the two-particle problem to a one particle problem of mass µscattering from a potential V (r). The Schrodinger equation in the two-particle center ofmass is (

− h2

2µ∇2 + V (r)

)ψ(~r) = Eψ(~r) . (10.20)

10.3 The Square Well Potential

Before we proceed to a general discussion of the solution of Eq. (10.20), let us considerthe solution of Eq. (10.20) for the case of a simple square well of radius a, i.e.,

V (r) =

−V0 for r < a

0 for r > a. (10.21)

In the last chapter, we considered the bound state (i.e. E < 0) problem for this potential.We now have to consider the case of E > 0, i.e. the scattering problem. For r < a, thesolution of the radial equation is the same as for the bound state (see Eq. (9.19)), whichis

R`(r) = Aj`(αr) with α2 =2µ

h2 (E + V0) for r < a . (10.22)

For r > a, we have a linear combination of the two solutions of the radial equation, i.e.,

R`(r) = B j`(kr) + C n`(kr) , (10.23)

orR`(r) = B′ h

(+)` (kr) + C ′ h

(−)` (kr) , (10.24)

where k2 = 2µE/h2. Since we are not considering bound states, the wave function R`(r)does not go to zero as r →∞, otherwise, the current at the detector would be zero. Thismeans neither B nor C (B′ or C ′) is zero. In this case either combination is valid and wewill use both at different times.

To get a feeling for the form of the scattering wave function, let us take ` = 0. In thiscase we have that

R0(r) =u0(r)

r, (10.25)

with

u0(r) =

A sinαr for r < a

B sin kr + C cos kr for r > a. (10.26)

An alternative way of writing the solution for r > a is

u0(r) = A′ sin(kr + δ) for r > a . (10.27)

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10.3. THE SQUARE WELL POTENTIAL 179

ψ ( r )

r

V = 0

V ≠ 0

Figure 10.3: A comparison the radial scattering wave function in the presence and absenceof a potential V .

Note that both forms for the solution for r > a have two constants to be adjustedby the boundary condition which requires that the wave function and its derivative becontinuous at r = a. Before we determine these constants, let us examine the form of thewave function for the case of V0 = 0, and V0 6= 0 at large distances, i.e. for r > a. InFigure 10.3, we sketch both wave functions. We observe that for V0 6= 0 the wave lengthfor r < a is smaller than is the case for r > a. However, for V0 = 0, the wave length isthe same for all r. Thus the presence of the potential, shifts the wave function for r > arelative to the wave function for the case V0 = 0 by an amount δ. In other words, theconstant δ introduced in Eq. (10.27) depends on the parameters of the potential, in thiscase V0 and a. The inverse might also be possible, i.e., if we measure δ we might be ableto determine the parameters of the potential. The δ introduced in Eq. (10.27) is calledthe scattering Phase Shift.

To guarantee that the wave function and its derivative are continuous at r = a, wetake

A sinαa = A′ sin(ka+ δ)

αA cosαa = kA′ cos(ka+ δ) .

Therefore, to determine δ, we take the ratio of the above two equations. This gives usthe result that

α cotαa = k cot(ka+ δ)

= kcot ka cot δ − 1

cot δ + cot ka. (10.28)

Solving this equation for cot δ we get

k cot δ =α cotαa+ k tan ka

1− αk

cotαa tan ka. (10.29)

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180 CHAPTER 10. SCATTERING BY A CENTRAL POTENTIAL

Having determined the phase shift δ we can determine one of the other two constants(A or A′). This leaves one overall multiplicative constant to be determined. For boundstates, this constant was determined by the requirement that the wave function be nor-malized. However, the scattering wave function is not normalizable in the same manner asthe bound state, because the normalization integral is mathematically not well defined.4

This is a consequence of the fact that the wave function does not go to zero as r →∞.

θk

iz

Figure 10.4: The scattering plane in the two-body center of mass. The incident beam isalong the positive z-axis.

10.4 The Scattering Amplitude

Having considered the simple problem of S-wave (i.e. ` = 0) scattering by a square well,we now turn to the more general problem of two-particle scattering. In the two-bodycenter of mass, we have two particles with opposite momenta, initially along the z-axis(see Figure 10.4). Since the scattering takes place in a plane, we can eliminate the φdependence in this problem. Furthermore, the initial momentum defines a direction inspace, thus braking the isotropy of the space. This implies that the angular momentumof the system is not fixed as was the case for the bound state problem, but dependson the momentum in the initial state. In fact, the angular momentum, classically, isperpendicular to the scattering plane. In general, the wave function is a linear combinationof many such angular momenta. As we have chosen our z-axis to be along the directionof the incident momentum and therefore in the scattering plane, we expect the projectionof our angular momentum along the z-axis to be zero. This means that the wave functionfor a given angular momentum is of the form

R`(r)Y`0(θφ) . (10.30)

But

Y`0(θ, φ) =

√2`+ 1

4πP`(cos θ) , (10.31)

4We will show in a later chapter that the scattering wave function has a δ-function normalization.

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10.4. THE SCATTERING AMPLITUDE 181

where cos θ = r · ki. We now can write the general form of the wave function as

ψ(~r) =1

(2π)3/2

∑`

i`(2`+ 1)ψ`(r)P`(cos θ) . (10.32)

The choice of the factor of (2π)−3/2 i` (2`+1) is for later convenience in the normalization.At this stage we would like to point out that this choice for the normalization is not unique.

Since P`(cos θ) is related to the spherical harmonics, it is an eigenstate of the angularmomentum operator square, L2. Making use of the orthogonality of the P`, we can showthat the radial Schrodinger equation for ψ`(r) is given by

d2ψ`dr2

+2

r

dψ`dr− `(`+ 1)

r2ψ` + [k2 − U(r)]ψ`(r) = 0 , (10.33)

where

k2 =2µE

h2 and U(r) =2µ

h2 V (r) . (10.34)

We now assume that the potential satisfies the condition that; U(r)→ 0 faster than r−1.In this case the radial equation for large r is given by

d2ψ`dr2

+2

r

dψ`dr− `(`+ 1)

r2ψ`(r) + k2ψ`(r) = 0 (10.35)

which is the spherical Bessel’s equation.

Note: Our assumption that U(r)→ 0 faster than r−1 excludes the Coulombpotential. From this point on we restrict ourselves to the class of potentialsthat satisfy the above condition. We will examine the Coulomb potential asa special case at the end of this chapter.

Before we write the general solution to Eq. (10.33), let us rewrite the solution to thesquare well potential in terms of the spherical Bessel function. We have from Eq. (10.25)and (10.27) that, for ` = 0,

ψ0(r) =A′

rsin(kr + δ) =

A′

r[ sin kr cos δ + cos kr sin δ ]

= kA′ [ cos δ j0(kr) + sin δ n0(kr) ] . (10.36)

This gives us the idea of writing the general solution to Eq. (10.35), for any angularmomentum, `, as

ψ`(r)→ A` [ cos δ` j`(kr) + sin δ` n`(kr) ] for r →∞ . (10.37)

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182 CHAPTER 10. SCATTERING BY A CENTRAL POTENTIAL

Since for r →∞ we have that (see Eq. (9.12))

j`(ρ) → 1

ρsin

(ρ− π`

2

)(10.38)

n`(ρ) → 1

ρcos

(ρ− π`

2

),

then

ψ`(r)→A`kr

sin

(kr − π`

2+ δ`

)for r →∞ . (10.39)

Here, A` and δ` are the two constants to be determined by the normalization of the wavefunction and the continuity of the logarithmic derivative.5 We now would like to makeuse of Eqs. (10.39) and (10.32) to write the total asymptotic wave function in the formgiven by Eq. (10.3). In this way we hope to relate the scattering amplitude f(k, θ) tothe phase shifts δ`. To achieve this result we first write sin(kr − π`/2 + δ`) in terms ofexponentials, to get

ψ`(r) →A`kr

1

2i

[ei(kr−

π`2

+δ`) − e−i(kr−π`2

+δ`)]

=A`

2ikre−iδ`

[−e−i(kr−

π`2 ) + e2iδ` ei(kr−

π`2 )]

=A`

2ikre−iδ`

[2i sin

(kr − π`

2

)+(e2iδ` − 1

)ei(kr−

π`2 )]. (10.40)

Using Eq. (10.39) we get the asymptotic radial wave function for a given ` to be of theform

ψ`(r)→ A`e−iδ`

(j`(kr) + (−i)` 1

2ik

(e2iδ` − 1

) eikrr

). (10.41)

Using this result in Eq. (10.32), and the fact that A` = eiδ` , we get the scattering wavefunction for r →∞ to be

ψ(~r)→ 1

(2π)3/2

(ei~k·~r + f(k, θ)

eikr

r

), (10.42)

5The logarithmic derivative is given by

d

drlogψ`(r) =

1ψ`(r)

dψ`(r)dr

which is identical to the condition given in Eq. (10.28) for the square well with ` = 0.

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10.4. THE SCATTERING AMPLITUDE 183

whereei~k·~r =

∑`

i` (2`+ 1) j`(kr)P`(cos θ) , (10.43)

and

f(k, θ) =1

k

∑`

(2`+ 1)

(e2iδ` − 1

2i

)P`(cos θ) , (10.44)

where cos θ = k · r. Now since the direction of the final momentum ~kf is the radial

direction, i.e., r = kf , we have cos θ = ki · kf , with ~ki being the initial momentum of theparticles in the beam.

In Eq. (10.44), we have established the relation between the scattering amplitude andthe phase shifts. In this way we have completed the relation between the experimentallymeasured cross section and the wave function which is a solution to the Schrodingerequation. Finally, we can write Eq. (10.44) as

f(k, θ) =1

k

∑`

(2`+ 1) f`(k)P`(cos θ) , (10.45)

where f`(k), the partial wave amplitude, is given by

f`(k) =1

2i

(e2iδ` − 1

)= eiδ` sin δ` . (10.46)

In general, this partial wave amplitude is a complex number, i.e. it has a magnitude anda phase. With the help of Eq. (10.14) we can write the differential cross section in termsof the partial wave amplitude as

dΩ=

1

k2

∣∣∣ ∑`

(2`+ 1) f`(k)P`(cos θ)∣∣∣2 . (10.47)

This gives the probability for an incident particle, with momentum k and energy E = k2

2µ,

to be scattered in the direction defined by the angle θ. To get the total cross section, i.e.the probability of scattering in any direction, we have to integrate the differential crosssection over the 4π solid angle, i.e.

σT =∫ (

)dΩ

=1

k2

∑`

∑`′

(2`+ 1)(2`′ + 1) f`(k)f ∗`′(k)∫dΩP`(cos θ)P`′(cos θ) .

Using the orthogonality of the Legendre polynomials (i.e. Eq. (8.60)), we get

σT =4π

k2

∑`

(2`+ 1) |f`(k)|2 =4π

k2

∑`

(2`+ 1) sin2 δ` . (10.48)

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184 CHAPTER 10. SCATTERING BY A CENTRAL POTENTIAL

Comparing Eqs. (10.47) and (10.48), we observe that the differential cross section has moreinformation about the scattering amplitude than the total cross section. To illustrate this,consider the case when only two partial waves are important, the ` = 0 and 1. In thiscase we have for the differential cross section

dΩ=

1

k2

(|f0|2 + 9|f1|2 cos2 θ + 3(f0f

∗1 + f ∗0 f1) cos θ

), (10.49)

while for the total cross section we have

σT =4π

k2

(|f0|2 + 3|f1|2

). (10.50)

Thus we see that the differential cross section will give us information on the relativephase of the ` = 0 and ` = 1 amplitudes. Here, we note that the in the event of S-wavescattering only, the differential cross section is angle independent, i.e. the differentialcross section is isotropic.

10.5 The Optical Theorem

This theorem, is a special case of a more general theorem that sets a nonlinear constrainton the scattering amplitude called unitarity. In particular it gives a relation between thescattering amplitude in the forward direction, f(k, θ = 0), and the total cross section,σT . It results from the condition that probability should be conserved in the scatteringprocess.

From Eq. (10.45), we have that the forward scattering amplitude is given by

f(k, θ = 0) =1

k

∑`

(2`+ 1) f`(k)P`(1)

=1

k

∑`

(2`+ 1) f`(k) (10.51)

since P`(1) = 1. Taking the imaginary part of this equation we get

Im [f(k, 0)] =1

k

∑`

(2`+ 1) Im [f`(k)] . (10.52)

But we have from Eq. (10.46) that for real phase shifts

Im [f`(k)] = sin2 δ` . (10.53)

Therefore, we can write

Im [f(k, 0)] =1

k

∑`

(2`+ 1) sin2 δ` . (10.54)

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10.6. THE PHASE SHIFTS FOR TWO-BODY SCATTERING 185

Making use of Eq. (10.48), we can write

Im [f(k, 0)] =k

4πσT . (10.55)

This result is commonly known as the Optical Theorem, and relates the forward scatteringamplitude to the total cross section. Considering the fact that the total cross section isproportional to the scattering amplitude squared, then Eq. (10.55) imposes a non-linearconstraint on the scattering amplitude, f(k, θ).

To illustrate the relation between unitarity and the optical theorem, we recall fromEq. (10.46) that the partial wave scattering amplitude is given by

f`(k) =1

2i

(e2iδ` − 1

)≡ 1

2i(S`(k)− 1) , (10.56)

where S`(k) is the partial wave S-matrix element. Solving Eq. (10.52) for S`(k), we get

S`(k) = 1 + 2if`(k) . (10.57)

Unitarity is the result of the fact that the S-matrix is unitary, i.e.,

S† S = I , (10.58)

which is obviously the case for S`(k) if the the phase shifts δ` are real. Also, the unitarityof S`(k) gives us the result

S†` (k)S`(k) = 1 , (10.59)

orImf`(k) = |f`(k)|2 . (10.60)

This non-linear relation for the partial wave scattering amplitude is identical to the opticaltheorem. We will see in a later chapter on formal scattering theory, Chapter 14, that thisresult can be derived from the Schrodinger equation directly.

10.6 The Phase Shifts for Two-Body Scattering

So far we have determined the cross section in terms of the scattering amplitude or thephase shifts. However, the phase shifts are constants used in writing the asymptoticform of the wave function. In the case of a square well potential these phase shifts weredetermined by matching the logarithmic derivative of the radial wave function at the wellradius r = a. In general, we can follow the same procedure and integrate the differentialequation from the origin to a large enough radial distance r, which is larger than the

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186 CHAPTER 10. SCATTERING BY A CENTRAL POTENTIAL

range of the potential, and then match the logarithmic derivative, as calculated fromthe asymptotic solution and the solution we get by integrating the radial Schrodingerequation. This is achieved by taking r0 to be such that r0 a, where a is the rangeof the potential, and solving the radial Schrodinger equation for ψ`(r) for r < r0 andcalculating the logarithmic derivative of ψ`(r), i.e.,

γ =d

drlogψ`(r)

∣∣∣r=r0−ε

, (10.61)

where ε is infinitesimal. On the other hand, for r > r0, we have the asymptotic solution

ψ`(r) = cos δ` ( j`(kr) + tan δ` n`(kr) )

and its derivativedψ`dr

= k cos δ` ( j′`(kr) + tan δ` n′`(kr) )

where j′` and n′` are the derivative of the spherical Bessel and Neumann functions. Wenow can write the logarithmic derivative for r > r0 as

d

drlogψ`(r)

∣∣∣r=r0+ε

= kj′`(kr0) + tan δ` n

′`(kr0)

j`(kr0) + tan δ` n`(kr0)

= γ . (10.62)

This equation can be solved for the phase shift, or tan δ`, to give

tan δ` =k j′`(kr0)− γ j`(kr0)

γ n`(kr0)− k n′`(kr0). (10.63)

Here, we observe that given γ, we can determine the phase shift δ`.

Note: Here γ is a function of r0, and r0 should be chosen large enough sothat δ` or tan δ` is independent of r0.

From the above results we can study the behavior of δ` at low energies, i.e. k → 0. Forkr0 `, we can write the spherical Bessel and Neumann functions and their derivativeas

j`(kr0)→ (kr0)`

(2`+ 1)!!and n`(kr0)→ (2`− 1)!!

(kr0)`+1

(10.64)

j′`(kr0)→ − `(kr0)`−1

(2`+ 1)!!and n′`(kr0)→ (`+ 1)(2`− 1)!!

(kr0)`+2.

We then can write the phase shift for kr0 ` as

tan δ` →(kr0)2`+1

(2`+ 1)!! (2`− 1)!!

`− γr0

`+ 1 + γr0

, (10.65)

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10.7. COULOMB SCATTERING 187

or

tan δ` ∝ k2`+1 for kr0 ` . (10.66)

From this result we may deduce that for small wave number k, i.e. low energy, sin δ` ∝k2`+1. If we now write the cross section as

σT =4π

k2

∑`

(2`+ 1) sin2 δ` ≡∑`

σ` , (10.67)

then the partial wave cross section, σ`, is given by

σ` =4π

k2(2`+ 1) sin2 δ`

→ (const.) k4` for kr0 ` . (10.68)

From this result we may conclude that for k → 0 we have

σ` =

const. for ` = 00 for ` 6= 0

. (10.69)

Thus at low energies, we expect the ` = 0 partial wave to dominate the cross section.

10.7 Coulomb Scattering

The analysis in this chapter has so far been restricted to finite range potentials. Thisexcludes the Coulomb potential which is considered to be infinite in range. Because ofthe central role played by the Coulomb potential in both atomic and molecular physics,and the fact that all accelerators produce beams of charged particles which are scatteredby targets often made of charged particles (e.g. nuclei), a discussion of scattering theorythat excludes the Coulomb problem is incomplete. The aim of this section is to derivethe amplitude for the scattering of two charged particles, and from that, extract theRutherford cross section.

Consider for the present, the scattering of two charged particles with charges Ze andZ ′e′, with the Coulomb potential between the particles given by

V (~r) =ZZ ′ee′

r. (10.70)

The Schrodinger equation for this potential is then given by(− h

2

2µ∇2 +

ZZ ′ee′

r

)ψ(~r ) = E ψ(~r ) , (10.71)

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188 CHAPTER 10. SCATTERING BY A CENTRAL POTENTIAL

where µ is the reduced mass of the two particles. We can rewrite this equation, aftermultiplication by −2µ

h2 , as (∇2 + k2 − 2γk

r

)ψ(~r ) = 0 , (10.72)

where

k2 =2µE

h2 and γ =µZZ ′ee′

h2k. (10.73)

We are going to consider the solution of this equation for E > 0, i.e., we would like toderive the scattering wave function ψ(~r ), and from its asymptotic behavior extract thescattering amplitude.

For any scattering experiment, we can define the direction of the incident beam to bethe z-axis. The presence of this preferred direction in space breaks the symmetry (thatspace is isotropic) and it gives us a problem which has cylindrical symmetry. As a resultof this new symmetry we expect the wave function and the scattering amplitude to beindependent of the angle φ, as illustrated in Eq. (10.32). Furthermore, the wave functionasymptotically should have an incident beam given up to a normalization by

eikz ,

and a scattered beam that is proportional to a spherical out going wave, i.e.,

eikr

r.

This suggests that the solution to Eq. (10.72) can be written as

ψ(~r ) = eikz g(r − z) ,

and excludes the possibility of having a function of the form

ψ(~r) = eikz g(r + z) ,

since the latter leads to an incoming spherical wave given by

e−ikr

r.

The fact that the boundary condition on this scattering problem requires the total wavefunction to be a function of z and r−z, suggests that the optimum choice for a coordinatesystem for solving the Schrodinger equation is the parabolic coordinate given by

ξ = r − z = r(1− cos θ)

η = r + z = r(1 + cos θ) (10.74)

φ = φ .

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10.7. COULOMB SCATTERING 189

In this new coordinate system, our total scattering wave function is a product of anincident plane wave and a function of ξ, i.e.,

ψ(~r) = eik(η−ξ)/2 g(ξ) . (10.75)

The Laplacian, ∇2, in this coordinate system is given by

∇2 =4

ξ + η

∂ξ

(ξ∂

∂ξ

)+

∂η

(η∂

∂η

)+

1

ξη

∂2

∂φ2. (10.76)

We now can write the Schrodinger equation in this coordinate system as a partial differ-ential equation in two variables, since we have no φ dependence. Furthermore, for thewave function with the structure given in Eq. (10.75), we have

∇2ψ(~r) =4 eik(η−xi)/2

η + ξ

[ξd2g

dξ2+ (1− ikξ) dg

dξ− k2

4(η + ξ) g

]. (10.77)

This allows us to rewrite Eq. (10.72) for g(ξ) as

ξd2g

dξ2+ (1− ikξ) dg

dξ− γkg(ξ) = 0 , (10.78)

which is the Confluent Hypergeometric equation, with the solution given by

g(ξ) = F (−iγ|1|ikξ) , (10.79)

where the Confluent Hypergeometric function F (α|β|z) is given in Eq. (8.35) as an infiniteseries. We thus can write the total wave function for the scattering of two charged particlesas

ψ(~r) = eikz g(r − z)

= eikz F (−iγ|1|ik(r − z)) . (10.80)

To get the asymptotic form of this wave function, and thus determine the scatteringamplitude, we need to know the behavior of F (α|β|ρ) for large ρ. We have that6

F (α|β|ρ)→ Γ(β)

(−ρ)−α

Γ(β − α)+ρα−β eρ

Γ(α)

as |ρ| → ∞ . (10.81)

With this result for the asymptotic behavior of the Confluent Hypergeometric equation,we can write the total scattering wave function for r →∞ as

ψ(~r) = eikz F (−iγ|1|ikξ)

→ eπγ/2

Γ(1 + iγ)

ei(kz+γ log k(r−z)) +

Γ(1 + iγ)

iΓ(−iγ)

e−iγ log sin2 θ/2

2k sin2 θ/2

ei(kr−γ log 2kr

r

≡ eπγ/2

Γ(1 + iγ)

[ei(kz+γ log k(r−z)) + f(k, θ)

ei(kr−γ log 2kr)

r

], (10.82)

6See Abramowitz and Stegun [24] Eq. 13.5.1

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190 CHAPTER 10. SCATTERING BY A CENTRAL POTENTIAL

where the scattering amplitude f(k, θ) is given by 7

f(k, θ) = −γ Γ(1 + iγ)

Γ(1− iγ)

e−iγ log sin2 θ/2

2k sin2 θ/2

= − γ

2k sin2 θ/2ei(η0−γ log sin2 θ/2) , (10.83)

whereη0 = arg Γ(1 + iγ) . (10.84)

The corresponding differential cross section is then given by

dΩ=

γ2

4k2 sin4 θ/2. (10.85)

This cross section is commonly known as the Rutherford cross section. Here we observethat:

1. The differential cross section for the scattering of two charged particles, dσdΩ→ ∞

as θ → 0. This in fact is the case both for electron scattering on atoms and protonscattering off a nucleus.

2. The total cross section is also infinity. This is a result of the fact that the Coulombpotential has an infinite range.

3. The asymptotic wave function as presented in Eq. (10.82) consists of an incidentplane wave and an outgoing spherical scattering wave. However, if we comparethis result with the equation we got for finite range potentials, i.e., Eq. (10.42), weobserve that both the plane wave and the scattering wave are modified by a factorproportional to γ, which is basically the product of the charges on the two particles.This distortion of the incident plane wave and the scattered spherical wave are theresult of the infinite range of the Coulomb potential.

10.8 Problems

1. Calculate the S-wave phase shift (i.e. ` = 0) for neutron proton scattering at centerof mass energies of 5 and 10 MeV given that the potential between a neutron and

7In writing this result we have made use of the fact that

(1− cos θ) = 2 sin2 θ/2 ,

and−z Γ(−z) = Γ(1− z) .

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10.8. PROBLEMS 191

proton is a square well of radius r0 = 2.51 fm. and the depth is V0 = 17.8 MeV.Calculate the corresponding cross section.

Note: 1 fm. = 10−13 cm, and h2/2µ = 41.47 MeV fm2.

2. Calculate the cross section for scattering off a hard sphere of radius R at very lowenergies, i.e. in the limit where kR 1.

Hint: A hard sphere can be represented by the potential

V (r) =

+∞ r ≤ R0 r > R

.

3. Show that for P -wave (` = 1) scattering by an attractive square well potential ofradius a and depth V0, the phase shifts satisfy the equation

α2 [ka cot(ka+ δ1)− 1] = k2 (αa cotαa− 1) ,

where k2 = 2µh2E and α2 = 2µ

h2 (E + V0). For neutron proton scattering we can take

V0 = 36.2 MeV. and a = 2.02 fm. Taking h2

2µ= 41.47 MeV fm2, calculate the phase

shift at E = 10 MeV.

4. Using the orthogonality of the Legendre polynomials, show that if the scatteringwave function is written as

ψ(~r) =1

(2π)3/2

∑`

i`(2`+ 1)ψ`(r)P`(cos θ) ,

then ψ`(r) satisfies the radial Schrodinger equation.

5. Show that for complex phase shifts the total cross section as calculated from the Op-tical Theorem is larger than the total elastic cross section obtained from integratingthe differential cross section.

Hint: To prove the above you can restrict your argument to one partial wave, e.g.` = 0.

6. The amplitude of S-wave scattering at at low energies is given by

f0 = − ak

1 + iak − 12arek2

where a and re are real constants, and k is the momentum.

(a) Show that this amplitude satisfies unitarity.

(b) Show that the corresponding cross section goes to a constant as the energygoes to zero, i.e. k → 0.

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192 CHAPTER 10. SCATTERING BY A CENTRAL POTENTIAL

(c) Considering your result for part (b), can you give the constant a a physicalmeaning?

7. Prove that a simple statement equivalent to the Optical Theorem for S-waves is

Im(

1

f0

)= −1 .

8. The P -wave amplitude for positive pion proton (i.e. π+p) scattering is given by

f1(k) = Λk3k4 − iΛk3 +

(2β2 − 3

2Λβ)k2 +

(β4 − 1

2Λβ3

)−1

where β = 5.3344 fm−1 and Λ = 10.337 fm−1.

(a) Calculate the phase shift δ1(k) as a function of k for 0 < k < 2.5 fm−1. Plotthe phase shifts.

(b) Calculate the total cross section for P -wave π+p scattering as a function ofenergy.

(c) What happens to the cross section when the phase shift is π/2?

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Chapter 11

Matrix Formulation of QuantumMechanics

So far in our study of quantum mechanics we have come across a number of quantitiesthat can be measured, e.g. momentum, energy and angular momentum. For each of thesequantities or observables, we can introduce an operator. Thus in Table 11.1 below wehave a number of observables and the corresponding operators. The question is, whatproperties should these and other operators that correspond to observables, satisfy. In thischapter we will develop the properties of these operators, the states of the system underobservation, and there interrelation. In the process we develop a theoretical frameworkof quantum mechanics that is more general than just the solution of the Schrodingerequation. In the following we first examine the properties of operators corresponding toobservables and the eigenvalues and eigenfunctions of these operator. This is followedwith an introduction of Dirac notation which is used to detail the unitary transformationrelating the different representation of the operators and their eigenfunctions. Finallywe consider the three representations most often used, and these are the coordinate,momentum and angular momentum. The latter will allow us to introduce spin 1/2 basisfor the representation of the angular momentum operator.

11.1 Operators and Observables

Let us consider the operator F which corresponds to a quantity F that can be measured.If we have a system in a state ψα, then the average value of the observable F is

〈F 〉 =∫ψ∗α F ψα ≡ (ψα, F ψα) . (11.1)

For example, if we need to know the average momentum of an electron in the ground, i.e.0s, state of hydrogen, then we need to calculate

〈~p 〉 =∫d3r ψ∗100(~r) [−ih∇] ψ100(~r) ≡ (ψ100, −ih∇ψ100) . (11.2)

193

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194 CHAPTER 11. MATRIX FORMULATION OF QUANTUM MECHANICS

Table 11.1: Commonly encountered observables and the corresponding operators

Physical Observable Operator

~r ~rCoordinate

x, y, z x, y, z

~p −ih∇Momentum

px, py, pz −ih ∂∂x, −ih ∂

∂y, −ih ∂

∂z

~L = ~r × ~p ~L = −ih~r ×∇Angular Lx = ypz − zpy Lx = −ih

(y ∂∂z− z ∂

∂y

)Momentum Ly = zpx − xpz Ly = −ih

(z ∂∂x− x ∂

∂z

)Lz = xpy − ypx Lz = −ih

(x ∂∂y− y ∂

∂x

)

Energy H = p2

2m + V (~r) H = − h2

2m ∇2 + V (~r)

In general the action of the operator F on the state ψα gives a new state ψ′α, i.e.,

ψ′α = F ψα . (11.3)

To satisfy the condition of addition of probability amplitudes, i.e. superposition of states,F should satisfy the two conditions

F (ψ1 + ψ2) = F ψ1 + F ψ2 (11.4)

andF a ψα = a F ψα , (11.5)

where a is any complex number. These two conditions define a linear operator. We willfind that most operators that correspond to physically measurable quantities are linearoperators. One exception is the time reversal operator which is anti-linear, i.e.,

F a ψα = a∗ F ψα , (11.6)

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11.1. OPERATORS AND OBSERVABLES 195

where a∗ is the complex conjugate of a.For the operator F to correspond to a measurable quantity, the average value of F

should be real, i.e.〈F 〉 = 〈F 〉∗

or (ψα, F ψα

)=(ψα, F ψα

)∗=(F ψα, ψα

). (11.7)

In writing the above result we have made use of the definition of the bracket in terms ofan integral over the wave function as defined in Eq. (11.1). If we define a matrix elementof F , Fαβ, as

Fαβ ≡(ψα, F ψβ

), (11.8)

then the generalization of Eq. (11.7) is1

Fαβ = F ∗βα

or (ψα, F ψβ

)=

(ψβ, F ψα

)∗=(F ψα, ψβ

)≡

(ψα, F

† ψβ). (11.9)

This means the operator F should be self-adjoint or Hermitian, i.e.,

F = F † =(F T

)∗, (11.10)

where F T is the transpose of F . Here, if we recall the fact that Hermitian matriceshave real eigenvalues, then taking Fαβ as a matrix with Fαβ = F ∗βα guarantees that the

eigenvalues of F are real. We therefore can make the general statement:

In quantum mechanics all operators are linear and Hermitianif they are to correspond to physical observables.

textbfExample: Consider the momentum operator

~p = −ih∇ .

To show that ~p is a linear operator, we have to show that it satisfies the twoconditions for a linear operator, i.e. Eqs. (11.4) and (11.5) are satisfied. Thefirst condition is satisfied because

p (ψ1 + ψ2) = −ih∇ (ψ1 + ψ2) = −ih∇ψ1 − ih∇ψ2

= pψ1 + pψ2 ,

1This corresponds to the matrix Fαβ being Hermitian.

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196 CHAPTER 11. MATRIX FORMULATION OF QUANTUM MECHANICS

while the second condition is valid since

paψα = −ih∇aψα = a(−ih∇)ψα = apψα ,

for any complex constant a. We therefore have established that p is a linearoperator. To show that p is a Hermitian operator, we have to show that

(ψα, p ψβ) = (ψβ, p ψα)∗ .

But we have that

(ψα, p ψβ) = −ih∫d3r ψ∗α(~r )∇ψβ(~r ) .

Integrating by parts, we get

(ψα, p ψβ) = ih∫d3r [∇ψα(~r )]∗ ψβ(~r )

=∫d3r [−ih∇ψα(~r )]∗ ψβ(~r )

=∫

d3r ψ∗β(~r ) [−ih∇ψα(~r )]∗

= (ψβ, p ψα)∗ .

Therefore, the momentum operator p = −ih∇ is a linear and Hermitian op-erator.

Consider the case when the operator F acts on a state ψα and gives the state ψ′α whichis proportional to ψα, i.e.

Fψα = ψ′α = Fαψα , (11.11)

where Fα is the constant of proportionality. If we compare this result to the equivalentmatrix equation, i.e., if F was a matrix rather than an operator, then we would referto Fα as the eigenvalue and ψα as the eigenstate or eigenfunction. For example, for thethree-dimensional harmonic oscillator, we had

H ψn`m = En` ψn`m ,

where En` is the eigenvalue and ψn`m is the eigenfunction of H.To show that if F is Hermitian, then Fα is real, we write Eq. (11.11) as

F ψα = Fα ψα , (11.12)

and then the complex conjugate equation with α replaced by β is given by

F ∗ ψ∗β = F ∗β ψ∗β . (11.13)

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11.1. OPERATORS AND OBSERVABLES 197

We now multiply Eq. (11.12) by ψ∗β from the left and Eq. (11.13) by ψα from the right,and then integrate the two equations. If we now subtract one equation from the other weget (

Fα − F ∗β) ∫

d3r ψ∗β(~r )ψα(~r ) =∫d3r ψ∗β(~r ) F ψα(~r )−

∫d3r (F ∗ ψ∗β(~r ))ψα(~r )

= Fβα − F ∗αβ .

Since F is Hermitian, Fαβ = F ∗βα, the above equation reduces to

(Fα − F ∗β )∫d3r ψ∗β(~r )ψα(~r ) = 0 . (11.14)

For α = β since∫d3r |ψα|2 6= 0, except for the uninteresting case of ψα(~r ) = 0, we

therefore haveFα = F ∗α . (11.15)

We can now state that:

The eigenvalues of Hermitian operators are real.

On the other hand for Fα 6= Fβ we have that2

∫d3r ψ∗β(~r )ψα(~r ) = (ψβ, ψα) = 0 for α 6= β . (11.16)

which implies that:

The eigenstates of a Hermitian operator are orthogonal.

If the states ψα correspond to a bound state, the wave function ψα(r)→ 0 as r →∞,and the wave function is normalizable. In that case we have

(ψα, ψβ) = δαβ , (11.17)

and we can state that:

The eigenstates of a Hermitian operator are orthonormal.

2In the event that α 6= β and Fα = Fβ , then ψα and ψβ are not orthogonal, and we need to use theSchmidt orthogonalization procedure.

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198 CHAPTER 11. MATRIX FORMULATION OF QUANTUM MECHANICS

Another property of the eigenstates of Hermitian operators is that they form a com-plete set of states, i.e., if we have any state Ψ that is normalizable, then we can write

Ψ(x) =∑α

aα ψα(x) . (11.18)

Using the orthogonality of the ψα’s, i.e. Eq. (11.17), we get

(ψβ, Ψ) =∑α

aα (ψβ, ψα) =∑α

aα δαβ = aβ .

Therefore,

aβ = (ψβ, Ψ) . (11.19)

The statement of completeness of the eigenstates ψα is represented by∑α

ψα(x)ψ∗α(x′) = δ(x− x′) . (11.20)

To prove this result we expand the δ-function δ(x− x′), in terms of our basis states as

δ(x− x′) =∑α

aα(x′)ψα(x) .

To determine the coefficient aα we make use of the orthonormality condition, i.e. Eq. (11.17),to get

aα(x′) =∫dxψ∗α(x) δ(x′ − x) = ψ∗α(x′) .

This proves the result of Eq. (11.20), which is a statement of completeness of the eigen-states ψα.

Summary:

1. For every observable there is a linear Hermitian operator.

2. The eigenvalues of this Hermitian operator are real.

3. The eigenstates of the Hermitian operator are orthonormal, i.e.

(ψα, ψβ) = δαβ

4. The eigenstates of the Hermitian operator form a complete set, i.e.∑α

ψα(x)ψ∗α(x′) = δ(x− x′) .

In the above discussion, we have ignored two facts:

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11.1. OPERATORS AND OBSERVABLES 199

1. In some problems we have more than one eigenstate with the same energy, e.g. ifwe consider the Hamiltonian for the Coulomb problem

H = − h2

2m∇2 − Ze2

r

then the eigenstates of H are ψn`m and

H ψn`m = En ψn`m ,

i.e., the energy of the system is independent of ` and m. A similar situation arisesfor all central potentials in that the eigenvalues are independent of m. In these casesit is not obvious that states with different m are orthogonal. We have guaranteedthe orthogonality of the eigenstates of H by taking ψn`m ∝ Y`m(θ, φ), i.e., that thestate ψn`m be also an eigenstate of the total angular momentum square, L2, and itsz-component, Lz.

2. In most problems in quantum mechanics the eigenvalues of the Hamiltonian forma discreet as well as continuous spectrum. The continuous part of the spectrumcorresponds to the scattering states. For such states, ψα(r) does not go to zero asr →∞ and the eigenstates are not normalizable, e.g., for the free Hamiltonian,

H = − h2

2m∇2 ,

the eigenstates are

φ~p(~r) =1

(2πh)3/2ei~k·~r

where ~p = h~k and the normalization taken to be

(φ~p, φ~p′) = δ(~p− ~p′) ,

which is infinite for ~p = ~p′. One either accepts this δ−function normalization, orputs the system in a box and at the end of the calculation takes the limit as thevolume of the box goes to ∞.

With this extended basis that includes scattering as well as bound states, Eqs. (11.18)and (11.20) become

Ψ(~r ) =∑α

aα ψα(~r ) +∫d3k a(~k )ψ~k(~r ) , (11.21)

and ∑α

ψα(~r )ψ∗α(~r ′) +∫d3k ψ~k(~r )ψ∗~k(~r

′) = δ(~r − ~r ′) . (11.22)

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200 CHAPTER 11. MATRIX FORMULATION OF QUANTUM MECHANICS

Let us now show that if two observables can be measured simultaneously, then the op-erators corresponding to these observables commute. If two observables can be measuredat the same time, then we have one eigenstate that is an eigenstate of both operatorscorresponding to the two observables, i.e.

Fψα = Fαψα and Mψα = Mαψα , (11.23)

where F and M are the two operators corresponding to the two observables, and ψα isthe eigenstate of both operators. Multiplying the first of the equations in Eq. (11.23) byM , and the second equation by F we get

MFψα = MFαψα = FαMψα = FαMαψα

F Mψα = FMαψα = MαFψα = MαFαψα .

Subtracting one equation from the other, we get(MF − F M

)ψα = (FαMα −MαFα)ψα = 0 .

For any state Ψ we haveΨ =

∑α

aα ψα

and then (MF − F M

)Ψ =

∑α

aα(MF − F M

)ψα = 0 .

Since this is true for any Ψ, it follows that(MF − F M

)=[M, F

]= 0 . (11.24)

Thus the operators corresponding to two observables that can be measured simultaneouslycommute.

The inverse of the above result can be easily shown; i.e., if[M, F

]= 0, then

Fψα = Fαψα and Mψα = Mαψα . (11.25)

To prove that the statement in Eq. (11.25) is valid, given Eq. (11.24), we assume that

Mψα = Mαψα .

We then multiply by F from the left to get

F Mψα = FMαψα = MαFψα .

Using the fact that F and M commute, i.e. F M = MF , we can write the above equationas

M(Fψα

)= Mα

(Fψα

),

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11.1. OPERATORS AND OBSERVABLES 201

i.e. Fψα is an eigenstate of M with eigenvalue Mα. This implies that

Fψα ∝ ψα .

In particular, we can define the constant of proportionality to be Fα such that

Fψα = Fαψα .

This proves the result that if two operators commute, then we have a state that is aneigenstate of both operators, and this corresponds to the fact that the observables corre-sponding to the two operators can be measured simultaneously.

Finally, we will prove that if [K, F

]= iM , (11.26)

then

〈(∆F )2〉 〈(∆K)2〉 ≥ 1

4〈M〉2 (11.27)

with

∆F = F − 〈F 〉∆K = K − 〈K〉 (11.28)

〈K〉 ≡(Φ, K Φ

).

This basically says that if two operators do not commute, then there is an uncertainty re-lation between the corresponding measurements, i.e. we cannot measure both observablesto any desired degree of accuracy.

Proof : Take a stateΨ = φ+ λψ λ real (11.29)

then

0 ≤ (Ψ, Ψ) = (φ+ λψ, φ+ λψ)

= (φ, φ) + λ (φ, ψ) + (ψ, φ)+ λ2 (ψ, ψ) . (11.30)

The right hand side of this equation can be considered as a quadratic in λ.For the case when this quadratic is equal to zero, we have λ given by

λ =1

2 (ψ, ψ)

[−(φ, ψ) + (ψ, φ)±

[(φ, ψ) + (ψ, φ)]2 − 4 (ψ, ψ) (φ, φ)

1/2].

(11.31)The requirement that the r.h.s. of Eq. (11.30) be ‘greater than’ zero forany real λ can only be satisfied if the quantity in the square root bracketin Eq. (11.31) is negative, i.e.,

(ψ, ψ) (φ, φ) ≥ 1

4| (φ, ψ) + (ψ, φ) |2 . (11.32)

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202 CHAPTER 11. MATRIX FORMULATION OF QUANTUM MECHANICS

The equality sign in Eq. (11.32) corresponds to the equal sign in Eq. (11.30).

To prove Eq. (11.27), we take

φ =(F − 〈F 〉

)Φ = ∆F Φ and ψ = i

(K − 〈K〉

)Φ = i∆K Φ . (11.33)

Then

(φ, φ) = 〈(∆F )2〉 =(Φ,(F − 〈F 〉

) (F − 〈F 〉

)Φ)

(11.34)

(ψ, ψ) = 〈(∆K)2〉 =(Φ,(K − 〈K〉

) (K − 〈K〉

)Φ)

while

(φ, ψ) = i(Φ,[F K − F 〈K〉 − 〈F 〉K − 〈F 〉 〈K〉

]Φ)

(11.35)

(ψ, φ) = −i(Φ,[KF − K〈F 〉 − 〈K〉F − 〈K〉 〈F 〉

]Φ)

and then

(φ, ψ) + (ψ, φ) = i(Φ,[F K − KF

]Φ)

=(Φ, M Φ

)= 〈M〉 . (11.36)

Using the results of Eqs. (11.34) and (11.36) in Eq. (11.32), we get the resultin Eq. (11.27).

11.2 Dirac Notation

In the last section we showed that:

1. For each observable there is a linear Hermitian operator F .

2. For each such operator there is a set of eigenstates ψα, and eigenvalues Fα such that

Fψα = Fαψα .

3. The eigenvalues Fα are real and the eigenstates form a complete and orthonormalset, i.e.

(ψα, ψβ) = δαβ .

Any state Ψ can now be written as

Ψ =∑α

aα ψα with aα = (ψα, Ψ) ,

or ∑α

ψα ψ∗α = I ,

where I is the unit operator.

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11.2. DIRAC NOTATION 203

Let us compare the above properties of ψα with the basis vectors used in three-dimension (3-D). In 3-D we have that our basis vectors are e1, e2, and e3. These basisvectors satisfy the condition that

ei · ej = δij ,

i.e., they are orthonormal. Any vector ~V can be written as

~V =3∑i=1

viei with vi = ei · ~V .

Comparing the properties of ei with those of ψα, we may deduce:

1. The expression (ψα, ψβ) is equivalent to the scalar product, i.e.,

(ψα, ψβ) =∫d3r ψ∗α(~r )ψβ(~r )

is equivalent to the 3-D vector product

~V · ~U =∑i

Vi Ui .

However, unlike the case in three-dimensional space where the basis states are unitvectors ei (i = 1, 2, 3), in quantum mechanics we need both the space of basis statesψα (α = 1, 2, · · ·), and the adjoint space ψ∗α (α = 1, 2, · · ·). To distinguish betweenthe two spaces, Dirac introduced the notation

ψα ≡ |α〉 called ket (11.37)

andψ∗α ≡ 〈α| called bra (11.38)

so that the scalar product is given by the bra|ket, i.e.,

(ψα, ψβ) ≡ 〈α|β〉 = δαβ , (11.39)

which is a statement of the orthonormality of the basis states.

2. In most problems in 3-D we introduce a set of basis vectors e1, e2, and e3 in termsof which we can write any vector, e.g., e1, e2, and e3 can be the unit vectors in thex, y, and z direction, or they can be the unit vectors in the r, θ, and φ directionif we are working in spherical polar coordinates. In a similar manner in quantummechanics we take a Hermitian operator and take its eigenstates to form a basis interms of which we can write any state. Thus we can take the eigenstates of F todefine our basis states, i.e.,

Fψα = Fαψα =⇒ F |α〉 = Fα|α〉 , (11.40)

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204 CHAPTER 11. MATRIX FORMULATION OF QUANTUM MECHANICS

and now we can write the state Ψ = | · · ·〉 in terms of our basis as

Ψ =∑α

aα ψα =⇒ | · · ·〉 =∑α

|α〉 aα , (11.41)

withaα = (ψα, Ψ) =⇒ aα = 〈α| · · ·〉 . (11.42)

In this notation a statement of completeness of our basis states is given by∑α

ψα ψ∗α =

∑α

|α〉〈α| = 1 . (11.43)

To illustrate the simplicity of the Dirac notation, consider the state |a〉 and the set ofbasis vectors |α〉 (α = 1, 2, · · ·). Making use of the completeness and orthonormality ofthe basis states, we can write

|a〉 =

(∑α

|α〉〈α|)|a〉 =

∑α

|α〉〈α|a〉 . (11.44)

This is basically the expansion of the state |a〉 in terms of the basis |α〉. In a similarmanner we have

〈a| = 〈a|(∑

α

|α〉〈α|)

=∑α

〈a|α〉〈α| . (11.45)

Taking into consideration the fact that 〈a| = (|a〉)†, it is clear that

〈a|α〉 = (〈α|a〉)† . (11.46)

|2>

|1>

|a>

<1|a>

Figure 11.1: The projection of the vector |a〉 on to the unit vector |1〉.

Let us now turn our attention to the physical interpretation of the scalar product 〈α|a〉.If the state |a〉 is a vector in the space where the basis vectors are |α〉 (α = 1, 2, · · ·), then

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11.2. DIRAC NOTATION 205

〈α|a〉 is the projection of the state vector |a〉 along the |α〉 axis (see Figure 11.2). We nowcan define a projection operator which when acting on a state vector |a〉 will give us theprojection of |a〉 along that axis. Such a projection operator is

Pα = |α〉〈α| . (11.47)

When Pα acts on |a〉 we getPα |a〉 = |α〉〈α|a〉 ,

which is the component of |a〉 along |α〉 times the unit vector along |α〉. Thus Pα|a〉 is avector along the |α〉-axis with a magnitude equal to the projection of |a〉 along |α〉. Someof the properties of the projection operator are

PαPβ = |α〉〈α|β〉〈β| = |α〉 δαβ 〈β|= Pα δαβ , (11.48)

and ∑α

Pα =∑α

|α〉〈α| = I . (11.49)

So far, we have assumed that if we make a measurement for which there is an operatorF when the system is in the state |α〉 = ψα then the resultant of the measurement is Fα,i.e.

F |α〉 = Fα|α〉 .Suppose we prepare the system in a state |a〉 which is not an eigenstate of the operator Fwhose corresponding observable we want to measure. For example, consider the measure-ment of the position of the electron in a hydrogen atom. The electron in the hydrogenatom is in an eigenstate of the Hamiltonian (i.e. energy), but is not in an eigenstate ofposition. We first write the state |a〉 in terms of the eigenstate of F , i.e.

|a〉 =∑α

|α〉〈α|a〉

and then we operate on |a〉 with the operator F corresponding to the measurement wewant to perform, i.e.

F |a〉 =∑α

F |α〉〈α|a〉 =∑α

Fα|α〉〈α|a〉 .

Note, the operation of F on the state |a〉 gives a new state |a′〉, so that

|a′〉 =∑α

Fα|α〉〈α|a〉 =∑α

|α〉Fα〈α|a〉 .

Here |a′〉 6= (const.)|a〉. To determine the average value of the operator F , we calculate

〈a|F |a〉 = 〈a|a′〉 =∑α

〈a|α〉Fα〈α|a〉

=∑α

|〈α|a〉|2 Fα . (11.50)

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206 CHAPTER 11. MATRIX FORMULATION OF QUANTUM MECHANICS

This implies that |〈α|a〉|2 is the probability of finding the system described by the state|a〉 in the quantum state that is an eigenstate of the operator F . Therefore

〈α|a〉 = probability amplitude of getting Fα for the measurement of

the observable F on a system in a state |a〉.

If |a〉 is an eigenstate of the operator A, i.e.

A|a〉 = Aa|a〉

then

〈a|α〉 = probability amplitude of getting Aa for the measurement of the

observable A on a system in state |α〉.

| α 1 >

| α 2 >

< α 2 | a 1 >

| a 1 >

< a 1 | α 1 >

| α 1 >

| a 1 >

| a 2 >

Figure 11.2: The projection of the state |a1〉 on to the basis |α1〉 left, and the projectionof the state |α1〉 on the basis |a1〉 on the right.

Note that〈a|α〉 = (〈α|a〉)† .

11.3 Representation of Operators

Having established the fact that we can find a set of basis states |α〉 (α = 1, 2, · · ·), suchthat ∑

α

|α〉〈α| = I ,

we now can write operators in a format that allows us to extract numbers from quantumtheory that can be compared with experiment. For example, consider the operator A

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11.3. REPRESENTATION OF OPERATORS 207

in whose eigenvalues and eigenstates we are interested. We can write the operator A interms of the basis states |α〉 (α = 1, 2, · · ·) as

A = IAI =∑α

∑α′|α〉〈α|A|α′〉〈α′| . (11.51)

It is clear that if we know all the matrix elements of the operator A, i.e. 〈α|A|α′〉, thenwe know the operator A.

In most problems in quantum mechanics when we perform a measurement, the systemis in an eigenstate of the operator corresponding to the quantity being measured. Forexample, when we measure the energy of an electron in the hydrogen atom, the electronis in an eigenstate of the Hamiltonian, and is described by the wave function which is asolution of the Schrodinger equation. That means given the operator A we need to findthe state |a〉 such that

A|a〉 = Aa|a〉 , (11.52)

or ∑α

A|α〉〈α|a〉 = Aa|a〉

and therefore ∑α

〈α′|A|α〉〈α|a〉 = Aa〈α′|a〉 . (11.53)

Thus given the operator A, we can determine the matrix elements 〈α′|A|α〉 and then solvethe above equation for 〈α|a〉. In Eq. (11.53) we have a standard eigenvalue problem, andthe equation can be written as∑

α

〈α′|A|α〉 − Aaδα′α

〈α|a〉 = 0 . (11.54)

For this equation to have solutions, we require that the determinant of coefficient be zero,i.e.

det〈α′|A|α〉 − Aaδα′α

= 0 . (11.55)

This determines the eigenvalues. Since A = A†, i.e., A is Hermitian then(〈α′|A|α〉

)†= 〈α|A†|α′〉 = 〈α|A|α′〉 , (11.56)

and the matrix 〈α′|A|α〉 is Hermitian. If we make our basis states |α〉 finite in dimension,i.e. α = 1, 2, · · · , N , then the solution of Eqs. (11.54) and (11.55) reduces to that ofsolving N homogeneous algebraic equations. Furthermore, since the matrix 〈α′|A|α〉 isHermitian, the eigenvalues are real. Having determined the eigenvalues from Eq. (11.55),we can determine the eigenvectors from Eq. (11.54). From these eigenvectors we canconstruct the matrix U defined as

Uaα = 〈a|α〉 . (11.57)

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208 CHAPTER 11. MATRIX FORMULATION OF QUANTUM MECHANICS

Then since the eigenvectors are orthogonal[U U †

]aa′

=∑α

〈a|α〉〈α|a′〉 = δaa′ (11.58)

and the matrix U is unitary. We also have∑αα′〈a|α〉〈α|A|α′〉〈α′|a′〉 = 〈a|A|a′〉 = Aa δaa′

or [UAU †

]aa′

= Aa δaa′ , (11.59)

i.e., UAU † is diagonal with its diagonal elements the eigenvalues of A.Suppose we have two operators A and B which are linear and Hermitian. We then

can construct two different sets of basis states given by

A|a〉 = Aa|a〉 and B|b〉 = Bb|b〉 . (11.60)

If we have a general state vector | · · ·〉 which is written in terms of the basis of theeigenstates of the operator A, i.e. we have 〈a| · · ·〉, how can we write this general statevector | · · ·〉 in terms of the eigenstates of the operator B? In other words, what is therelationship between 〈b| · · ·〉 and 〈a| · · ·〉? Using the completeness and orthonormality ofthe two basis, we have

〈b| · · ·〉 =∑a

〈b|a〉〈a| · · ·〉 =∑a

Uba〈a| · · ·〉 , (11.61)

while〈a| · · ·〉 =

∑b

〈a|b〉〈b| · · ·〉 =∑b

U∗ba〈b| · · ·〉 . (11.62)

Thus the matrix U can be used to relate 〈a| · · ·〉 and 〈b| · · ·〉.In a similar way, if we have an operator F in terms of one set of basis states, we can

write it in terms of another set of basis states. This is achieved by making use of thecompleteness of the states which allows us to write

〈a|F |a′〉 =∑bb′〈a|b〉〈b|F |b′〉〈b′|a′〉 (11.63)

with the inverse relation given by

〈b|F |b′〉 =∑aa′〈b|a〉〈a|F |a′〉〈a′|b′〉 . (11.64)

The above two equations can be written in matrix form as

F a = U †F bU and F b = U F aU † . (11.65)

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11.3. REPRESENTATION OF OPERATORS 209

It is clear from these relations between the matrix elements of F in the two basis states,that U must satisfy the condition

U U † = U †U = I , (11.66)

i.e.,U † = U−1 . (11.67)

In other words, the matrix U is unitary. From this it follows that the scalar product isindependent of the basis, i.e. if

|αu〉 = U |α〉 and |βu〉 = U |β〉 (11.68)

then〈αu|βu〉 = 〈α|U †U |β〉 = 〈α|β〉 , (11.69)

i.e., the scalar product is the same in the transformed basis as in the original basis.

Definitions:

1. The basis in terms of which a state vector and operator are written is referred to asthe representation.

2. The change of basis through a unitary matrix or operator is referred to as a changeof representation.

11.3.1 The Coordinate Representation

So far the representation we have been using is the coordinate representation (or r-representation), i.e, the basis states are the eigenstates of the position operator

~r|~r 〉 = ~r |~r 〉 (11.70)

with the orthonormality and completeness of the states |~r 〉 given by

〈~r |~r ′〉 = δ (~r − ~r ′) and∫d3r |~r 〉〈~r | = 1 . (11.71)

Here we have replaced the sum by an integral since ~r takes on continuous values.To illustrate some operators in this representation let us consider the momentum

operator in one dimension. We have that the position and momentum operator satisfythe commutation relation

xp− px = ih . (11.72)

If we take the matrix element of this equation in coordinate representation, we get

〈x| (xp− px) |x′〉 = ih〈x|x′〉 = ihδ(x− x′) (11.73)

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210 CHAPTER 11. MATRIX FORMULATION OF QUANTUM MECHANICS

but we have thatx|x〉 = x|x〉 (11.74)

so that Eq. (11.73) can be written as

(x− x′)〈x|p|x′〉 = ihδ(x− x′) = −ih(x− x′)δ′(x− x′) , (11.75)

where δ′(x) = ddxδ(x). The second equality can be justified by recalling that the Dirac

δ-function always appears in an integral, and an integration by parts allows us to write3

f ′(x) δ(x− a) = −f(x) δ′(x− a) . (11.76)

We now can write Eq. (11.75) as

〈x|p|x′〉 = −ih ddxδ(x− x′) = −ih d

dx〈x|x′〉 = ih

d

dx′〈x|x′〉

or

p|x′〉 = ihd

dx′|x′〉 . (11.77)

For the momentum operator acting on a general state | · · ·〉, we have, using the complete-ness of our basis states,

p| · · ·〉 =∫p |x〉 dx 〈x| · · ·〉 = ih

∫ d

dx|x〉 dx 〈x| · · ·〉

= −ih∫|x〉 dx d

dx〈x| · · ·〉 , (11.78)

where the final expression is a result of an integration by parts and the assumption thatthe state 〈x| · · ·〉 goes to zero for x → ±∞. In general, we have for a function of themomentum operator that4

3

+∞∫−∞

dx δ(x− x′) =

+∞∫−∞

dx

[d

dx(x− x′)

]δ(x− x′)

= −+∞∫−∞

dx (x− x′) δ′(x− x′)

4The function of an operator f(p) is only defined in terms of a power series in the operator p, i.e.,

f(p) = a0 + a1p+ a2p2 + · · · .

This is analogues to the definition of a function of a matrix, e.g.,

e−p = 1− p+12!p2 − 1

3!p3 + · · · ,

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11.3. REPRESENTATION OF OPERATORS 211

f(p) | · · ·〉 =∫dx |x〉 f

(−ih d

dx

)〈x| · · ·〉 . (11.79)

In writing this result we have made use of the fact that

p→(−ih d

dx

)and p2 →

(−ih d

dx

)2

.

If we now have a function of the momentum and position operators, i.e. A(p, x), and wewant to solve the equation

A (p, x) |a〉 = Aa |a〉 , (11.80)

we first have to write |a〉 in terms of our basis states, i.e.

|a〉 =∫dx |x〉 〈x|a〉

so that

A |a〉 =∫dx |x〉A

(−ih d

dx, x

)〈x|a〉 . (11.81)

We now use the orthogonality of our basis states (i.e. 〈x|x′〉 = δ(x − x′)) to writeEq. (11.80) in the coordinate representation as

A

(−ih d

dx, x

)〈x|a〉 = Aa 〈x|a〉 . (11.82)

In a similar manner, we have the coordinate representation in three-dimensions to corre-spond to

~r → ~r , ~p→ −ih∇ , and |a〉 → 〈~r |a〉 . (11.83)

An example of Eq. (11.82) is the case where A is the Hamiltonian operator which is afunction of the momentum and position operator, in which case our operator equation isgiven by

H(~p, ~r

)|Eα〉 = Eα |Eα〉 . (11.84)

In coordinate representation this equation reduces to the standard Schrodinger equationas we are familiar with it, i.e. in Dirac notation it takes the form

H (−ih∇, ~r ) 〈~r |Eα〉 = Eα 〈~r |Eα〉 . (11.85)

Here the wave function in coordinate space is written as 〈~r |Eα〉.

where p2 = pp and p3 = ppp, · · · . In the event that f(p) acts on an eigenstate of p, i.e., p|pn 〉 = pn|pn 〉then

f(p) |pn 〉 = f(pn) |pn 〉 .

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212 CHAPTER 11. MATRIX FORMULATION OF QUANTUM MECHANICS

11.3.2 The Momentum Representation

If we take for our basis the eigenstates of the momentum operator ~p, then we have

~p |~p 〉 = ~p |~p 〉 , 〈~p |~p ′〉 = δ (~p− ~p ′) and∫|~p 〉 d3p 〈~p | = I . (11.86)

Proceeding in a similar manner as in the coordinate representation, we can write anyfunction of the momentum operator acting on a general state as

f(~p ) | · · ·〉 =∫d3p |~p 〉 f(~p ) 〈~p | · · ·〉 . (11.87)

To determine the coordinate operator in momentum representation, we consider againthe matrix element of the commutation relation of the coordinate and the momentum inone-dimension, as given in Eq. (11.72), in momentum representation, i.e.

〈p| (xp− px) |p′〉 = ih 〈p|p′〉 = ih δ(p− p′) .

Using Eq. (11.86), we have

(p− p′) 〈p|x|p′〉 = −ih δ(p− p′) = ih (p− p′) δ′(p− p′) .

Therefore, we have for the position operator, the matrix element

〈p|x|p′〉 = ihd

dp〈p|p′〉 = −ih d

dp′〈p|p′〉 .

This allows us to write

x |p′〉 = −ih d

dp′|p′〉 . (11.88)

Now, any function of the position operator, when acting on an eigenstate of the momentumoperator, gives

f(x) |p〉 = f

(−ih d

dp

)|p〉 ,

and for the most general state | · · ·〉, we have

f(x) | · · ·〉 =∫dp |p〉 f

(ihd

dp

)〈p| · · ·〉 . (11.89)

We are now in a position to write our standard eigenvalue problem for the operator A(~p, ~r )in momentum space. Given that

A(~p, ~r ) |a〉 = Aa |a〉 ,

we can write this equation in momentum representation as

A(~p, ih∇p) 〈~p |a〉 = Aa 〈~p |a〉 . (11.90)

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11.3. REPRESENTATION OF OPERATORS 213

Example 1: As a first example, let us consider the case when the operatorA(~p, ~r ) is just the position operator, i.e.

A(~p, ~r ) = ~r . (11.91)

Then Eq. (11.90) takes the form

ih∇p 〈~p |~r 〉 = ~r 〈~p |~r 〉 .

The solution to this first order differential equation is given by

〈~p |~r 〉 =1

(2πh)3/2e−i

~k·~r with ~k =~p

h. (11.92)

This is the eigenstate of the position operator in momentum representation.The eigenstate of the momentum operator in coordinate representation is〈~r |~p 〉, and is given by

〈~r |~p 〉 = (〈~p |~r 〉)† =1

(2πh)3/2ei~k·~r . (11.93)

Example 2: For our second example, we consider the case when A is theHamiltonian operator, i.e.

H(~p, ~r ) =1

2mp2 + V (~r, ~p ) , (11.94)

We have taken the potential to be as general as possible and thus a function ofboth the position and the momentum operator. In coordinate representation,H takes the form

〈~r |H|~r ′〉 =1

2m〈~r |p2|~r ′〉+ 〈~r |V |~r ′〉

= − h2

2m∇2r′ δ(~r − ~r ′) + 〈~r |V |~r ′〉 . (11.95)

The δ-function in the first term tells us that the operator p2 is diagonal in thecoordinate representation.

The Schrodinger equation now takes the form of

H |E〉 = E |E〉

or, in coordinate representation it reduces to

〈~r |H|E〉 = E 〈~r |E〉 .

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214 CHAPTER 11. MATRIX FORMULATION OF QUANTUM MECHANICS

Introducing a complete set of eigenstates of the position operator ~r, we get∫d3r′ 〈~r |H|~r ′〉〈~r ′|E〉 = E 〈~r |E〉

Using the expression for the Hamiltonian in coordinate representation as given in Eq. (11.95),we get ∫

d3r′(− h2

2m∇2r′

)δ(~r − ~r ′) 〈~r ′|E〉+

∫d3r′ 〈~r |V |~r ′〉〈~r ′|E〉 = E 〈~r |E〉 .

Integrating the first term on the right hand side of the above equation by parts, twice,we get

− h2

2m∇2r 〈~r |E〉+

∫d3r′ 〈~r |V |~r ′〉〈~r ′|E〉 = E 〈~r |E〉 . (11.96)

For most of the potentials we have considered V is a function of the position operatoronly, and in that case

〈~r |V (~r )|~r ′〉 = V (~r ) δ(~r − ~r ′) , (11.97)

and Eq. (11.96) reduces to the ordinary Schrodinger equation, i.e.− h2

2m∇2r + V (~r )

〈~r |E〉 = E 〈~r |E〉 . (11.98)

In other words, 〈~r |E〉 is the usual wave function encountered before.In momentum representation, the Schrodinger equation takes the form

〈~p |H|E〉 = E 〈~p |E〉 ,or on introducing a complete set of eigenstates of the momentum operator, we get∫

d3p′ 〈~p |H|~p ′〉〈~p ′|E〉 = E 〈~p |E〉 .

But we have for the Hamiltonian in momentum space that

〈~p |H|~p ′〉 =1

2mp2 δ(~p− ~p ′) + 〈~p |V |~p ′〉 . (11.99)

This allows us to write the Schrodinger equation in momentum representation as(E − p2

2m

)〈~p |E〉 =

∫d3p′ 〈~p |V |~p ′〉〈~p ′|E〉 . (11.100)

We will see when we come to formal scattering theory, that in some cases it is advan-tageous to work in momentum space to get the scattering amplitude. In these cases wewill need to start with the Schrodinger equation in momentum representation. In caseswhen the potential is not diagonal in coordinate representation, the Schrodinger equationin coordinate representation is an integrodifferential equation. On the other hand, inmomentum representation we have an integral equation that is easier to solve. Thus foreach problem we have a representation in which the Schrodinger equation is simplest tosolve, and we should take advantage of this convenience.

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11.3. REPRESENTATION OF OPERATORS 215

11.3.3 Angular Momentum Representation

In Chapter 1, we showed that if ~L is the angular momentum operator then[L2, Li

]= 0 for i = 1, 2, 3

(11.101)

[Li, Lj] = ih∑k

εijkLk

where the totally antisymmetric tensor εijk is defined as

εijk =

+1 If (i, j, k) is an even permutation of (1,2,3).−1 If (i.j.k) is an odd permutation of (1,2,3).0 If any two of (i, j, k) are equal.

. (11.102)

Since L2 and L3 commute, we can construct simultaneous eigenstates of L2 and L3, i.e.

L2 Y`m(θ, φ) = h2`(`+ 1)Y`m(θ, φ)

(11.103)

L3 Y`m(θ, φ) = hmY`m(θ, φ)

where Y`m turns out to be the usual spherical harmonics as defined in Chapter 1.In the present section we would like to generalize this result using the matrix formu-

lation. Consider the vector operator ~J with components J1, J2 and J3 which satisfy a setof commutation relations identical to the angular momentum operator, i.e.[

Ji, Jj]

= ihεijkJk . (11.104)

Here, we make use of the convention that there is a sum over the repeated index, whichin the case of Eq. (11.104) is the index k. Taking

J2 =3∑i=1

J2i , (11.105)

we can show, using Eq. (11.104), that

[J2, Jk

]=

3∑i=1

[J2i , Jk

]= 0 . (11.106)

We now define two new operators

J+ = J1 + iJ2 J− = J1 − iJ2 , (11.107)

and then using the commutation relation, Eq. (11.104), we can show that[J3, J±

]= ±hJ±

[J+, J−

]= 2hJ3 . (11.108)

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216 CHAPTER 11. MATRIX FORMULATION OF QUANTUM MECHANICS

The commutation relation in Eqs. (11.104) and (11.108) are a special case of a Lie Algebra.Since J2 and J3 are two commuting operators, we have states that are simultaneouseigenstates of both J2 and J3, i.e.

J2 |β,m〉 = h2β|β,m〉(11.109)

J3 |β,m〉 = hm |β,m〉 .

The problem now is to determine the eigenvalues of J2 and J3, i.e. β and m. Since J3 is a

component of ~J , we expect that for a given value of β, the eigenvalue m has a maximumvalue, mmax, and a minimum value, mmin. From Eq. (11.109) we have that

J+ J3 |β,m〉 = hm J+ |β,m〉 .

But from the commutation relation in Eq. (11.108), we have that

J+J3 = J3J+ − hJ+

and therefore

J3J+ |β,m〉 = h(m+ 1)J+ |β,m〉 . (11.110)

This means that J+ |β,m〉 is an eigenstate of J3 with eigenvalue h(m + 1) except whenm = mmax in which case

J+ |β,mmax〉 = 0 . (11.111)

In a similar manner, we have that

J−J3 |β,m〉 = hm J− |β,m〉

and using the commutation relation given in Eq. (11.108), we get

J3J− |β,m〉 = h(m− 1) J− |β,m〉 . (11.112)

In other words, J− |β,m〉 is an eigenstate of J3 with eigenvalue h(m − 1), except whenm = mmin in which case

J− |β,mmin〉 = 0 . (11.113)

We thus have established that J+ (J− ) increases (decreases) m by one. In this way, givenone state |β,m〉, we can generate all the states with different m and a fixed value of β bythe action of J+ or J−. Using the definition of J+ and J−, we can write

J−J+ =(J1 − iJ2

) (J1 + iJ2

)= J2 − J2

3 − hJ3 , (11.114)

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11.3. REPRESENTATION OF OPERATORS 217

and then

J− J+ |β,mmax〉 = h2(β −m2

max −mmax

)|β,mmax〉 = 0 .

Assuming the state |β,mmax〉 6= 0, then the quantity in brackets should be zero and thiscan be satisfied if

β = mmax (mmax + 1) . (11.115)

On the other hand, we have that

J+ J− = J2 − J33 + hJ3

and the action of this operator on the state |β,mmin〉 gives

J+ J− |β,mmin〉 = h2(β −m2

min +mmin

)|β,mmin〉 = 0 .

From this we get that

β = m2min −mmin . (11.116)

Comparing Eqs. (11.115) and (11.116), we get

mmax (mmax + 1)−m2min +mmin = 0 .

This can be written as

(mmax +mmin) (mmax −mmin + 1) = 0

and has two possible solutions: mmax = −mmin and mmax −mmin = −1. Since mmax ≥mmin, the only possible solution is

mmax = −mmin . (11.117)

Since there are as many values of m greater than zero as there are less than zero mmax−mmin = 2j. On the other hand since m changes by intervals of one then j = 0, 1

2, 1, · · ·,

i.e.,

mmax −mmin = 2j with j = 0,1

2, 1,

3

2, · · · . (11.118)

In other words, m has the range of values

− j ≤ m ≤ j with (2j + 1) possible values. (11.119)

This means that mmax = j, and the eigenvalue corresponding to the operator J2, i.e. β,is given by

β = j(j + 1) , (11.120)

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218 CHAPTER 11. MATRIX FORMULATION OF QUANTUM MECHANICS

and we can write our eigenvalue problem for the operators J2 and J3 as

J2 |j,m〉 = h2j(j + 1) |j,m〉(11.121)

J3 |j,m〉 = hm |j,m〉 .

Here, we observe that we have solved the above eigenvalue problem to the extent ofdetermining the eigenvalues, the number of eigenstates and how they are related, withoutsolving any differential equations. However, we have had to make use of the commutationrelation of the operators associated with this problem. This procedure can be appliedto a number of problems, and we will use this method to solve the harmonic oscillatorproblem in Chapter 13 when we consider the occupation number representation.

Although we have demonstrated that the operators J+ (J−) will increase (decrease)the value of m, we need to determine the normalization of the new state generated bythese two operators. To determine this normalization let us first consider the action ofJ− on the state with maximum m, i.e.

|j, j − 1〉 = Cj,j−1 J− |j, j〉 . (11.122)

We then can write the general state as

|j,m〉 = Cj,m(J−)j−m

|j, j〉 , (11.123)

where Cj,m is a normalization constant to be determined. We now can write

|j,m〉 = Cj,m J−

(J−)j−m−1

|j, j〉

=Cj,mCj,m+1

J− |j,m+ 1〉 , (11.124)

and

〈j,m| =(Cj,mCj,m+1

)∗〈j,m+ 1| J+ , (11.125)

since(J−)∗

= J+. The requirement that |j,m〉 be normalized is then

〈j,m|j,m〉 = 1 =

∣∣∣∣∣ Cj,mCj,m+1

∣∣∣∣∣2

〈j,m+ 1|J+ J−|j,m+ 1〉

=

∣∣∣∣∣ Cj,mCj,m+1

∣∣∣∣∣2

〈j,m+ 1|(J2 − J2

3 + hJ3

)|j,m+ 1〉

=

∣∣∣∣∣ Cj,mCj,m+1

∣∣∣∣∣2

h2 (j(j + 1)−m(m+ 1)) ,

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11.4. SPIN 12

PARTICLES 219

and therefore ∣∣∣∣∣Cj,m+1

Cj,m

∣∣∣∣∣ = h√j(j + 1)−m(m+ 1) .

We then can write

J− |j,m〉 = h√j(j + 1)−m(m− 1) |j,m− 1〉 . (11.126)

Similarly, we can show that

J+ |j,m〉 = h√j(j + 1)−m(m+ 1) |j,m+ 1〉 . (11.127)

The above results are valid for j = ` (where ` is an integer) and in this case the states|`,m〉 are the spherical harmonics, which in coordinate representation are given by

〈θ, φ | `,m〉 = Y`m(θ, φ) . (11.128)

For the case of j = 12, 3

2, · · · the eigenstates of J2 and J3 can be considered as spinor

representation, and we will discuss the case of j = 12

in the next section. The above basis

of eigenstates of the two commuting operators J2 and J3, with J2 = J21 + J2

2 + J23 and

the Ji satisfying the commutation relation given in Eq. (11.104), form a basis in terms ofwhich we can write any eigenstate of a linear Hermitian operator.

11.4 Spin 12 Particles

As an example of the usefulness of the angular momentum, let us consider the case ofj = 1/2, and as a first step calculate the matrices

〈j,m|Ji|j,m′〉 .

To determine these matrices, we recall that we have determined the result of the actionof J± on the state |j,m〉, and we have that

J1 =1

2

(J+ + J−

)J2 =

1

2i

(J+ − J−

).

We then have, using the orthogonality of the states |j,m〉, that

〈j,m|J1|j,m′〉 =h

2

√j(j + 1)−m(m− 1) δm,m′+1

+√j(j + 1)−m(m+ 1) δm,m′−1

. (11.129)

For j = 1/2, m = ±1/2 and the above expression is the element of a 2× 2 matrix σ1, i.e.

〈j,m|J1|j,m′〉 =h

2[σ1]mm′ ,

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220 CHAPTER 11. MATRIX FORMULATION OF QUANTUM MECHANICS

where σ1 is given as

σ1 =

(0 11 0

). (11.130)

In a similar manner, the matrix elements of J2 are given by

〈j,m|J2|j,m′〉 =h

2i

√j(j + 1)−m(m− 1) δm,m′+1

−√j(j + 1)−m(m+ 1) δm,m′−1

, (11.131)

and for j = 1/2 this reduces to the matrix element of σ2, i.e

〈j,m|J2|j,m′〉 =h

2[σ2]mm′

where the matrix σ2 is given as

σ2 =

(0 −ii 0

). (11.132)

Finally, we have for the matrix elements of J3 for j = 1/2 as the elements of the matrixσ3, i.e.

〈j,m|J3|j,m′〉 =h

2[σ3]mm′

where the matrix σ2 is

σ3 =

(1 00 −1

). (11.133)

Here, σ1, σ2 and σ3 are known as the Pauli spin matrices, and in conjunction with theunit 2× 2 matrix form a complete basis for writing any 2× 2 matrix (see Problem 3). Wealso have

〈j,m|J2|j,m′〉 =h2

4[σ1σ1 + σ2σ2 + σ3σ3 ]mm′

with

[σ1σ1 + σ2σ2 + σ3σ3 ] = 3

(1 00 1

). (11.134)

The last step follows from the fact that σ2i = I. From the commutation relation of the

Ji, Eq. (11.104), we can show that the Pauli spin matrices σi satisfy the commutationrelation

[σi, σj] = 2i∑k

εijk σk . (11.135)

They also satisfy the anti-commutation relation

σi, σj ≡ σiσj + σjσi = 0 for i 6= j .

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11.4. SPIN 12

PARTICLES 221

orσi, σj = 2 δij (11.136)

In this case the states |j,m〉 for j = 12

turn out to be two component spinors. To show

this we consider the eigenvalue equation for the operator (or matrix) S3 = J3 = h2σ3, i.e.

S3

(uv

)= ± h

2

(uv

)or (

1 00 −1

) (uv

)= ±

(uv

).

After matrix multiplication on the right hand side this reduces to(u−v

)= ±

(uv

).

Therefore, the eigenstate corresponding to the eigenvalue of + h2

is

(u0

), while the eigen-

state corresponding to the eigenvalue − h2

is

(0v

). After normalization, we get the two

eigenstates to be

α ≡(

10

)and β ≡

(01

). (11.137)

These column matrices of length two are referred to as two-component Pauli spinors.5

In general, the eigenstates of Si are a linear combination of α and β, e.g. if we considerthe case of i = 3 then, as observed above, α and β are the eigenstates of S3. To determinethe eigenstates of S1 and S2, we consider the operator

S1 cosφ+ S2 sinφ ,

and then for φ = 0 we will get the eigenstates of S1, while for φ = π2

we will get theeigenstates of S2. The eigenvalue equation in this case is given by

(S1 cosφ+ S2 sinφ)

(uv

)=h

(uv

).

5Although spinors are different from the standard wave functions we have encountered in coordinateor momentum space, we can recast the Pauli spinor to look like an eigenstate of S3 with eigenvalues of± h2 , i.e.

S3 |m〉 = mh |m〉 with m = ±12.

We can write the state |m〉 in a ‘coordinate’ representation, i.e. 〈ξ|m〉, where now the coordinate cantake on two values ξ = ± 1

2 . In this case〈ξ|m〉 = δξm .

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222 CHAPTER 11. MATRIX FORMULATION OF QUANTUM MECHANICS

Making use of the explicit matrix representation of S1 and S2, we get

h

2

(0 e−iφ

eiφ 0

) (uv

)=h

(uv

).

This gives us two algebraic equations of the form

e−iφ v = λu and eiφ u = λ v

which have a solution forλ = ± 1 .

For λ = 1, the eigenstates are

1√2

(e−iφ/2

eiφ/2

)=

1√2

e−iφ/2 α + eiφ/2 β

, (11.138)

while for λ = −1, we have

1√2

(e−iφ/2

−eiφ/2

)=

1√2

e−iφ/2 α− eiφ/2 β

. (11.139)

Thus for φ = 0 the eigenstates of S1 corresponding to λ = ±1 are

1√2

(α± β) =1√2

(1±1

), (11.140)

while for φ = π2

the eigenstates of S2 are

eiπ/4√2

(α± iβ) =eiπ/4√

2

(1±i

). (11.141)

Here we note that for φ → φ + 2π, the solution changes sign and in fact the solution isinvariant under the transformation φ→ φ + 4π. This is a property of the spin 1/2 wavefunction which we will discuss in more detail when considering the relation of symmetryto conservation laws in the next chapter.

There are many particles in nature with spin 1/2, e.g. the electron, proton, quarks,· · ·. Associated with the spin there is a magnetic moment, which in the case of the electronis given by

~M = − eg

2mc~S = − ehg

4mc~σ . (11.142)

Here, e, and m are the charge and mass of the electron, while c is the velocity of lightand g is the gyromagnetic ratio, which is approximately 2, and more specifically

g = 2.0023192 .

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11.4. SPIN 12

PARTICLES 223

For an electron in a constant magnetic field ~B, the Hamiltonian for the interaction of themagnetic moment and magnetic field is given by

H = − ~M · ~B =egh

4mc~σ · ~B . (11.143)

Since the Hamiltonian, H, is a 2× 2 matrix in spin space, then the wave function is givenby a two component spinor of the form

ψ(t) =

(u(t)v(t)

),

and the time dependent Schrodinger equation becomes

ihdψ

dt=egh

4mc~σ · ~B ψ(t) .

Since the Hamiltonian has no space or time dependence, we expect the wave function ψto be independent of the position of the electron, and its time dependence to be of theform

ψ(t) =

(u(t)v(t)

)= e−iωt

(α+

α−

).

We now can write the time independent Schrodinger equation as

(α+

α−

)=egh

4mc~σ · ~B

(α+

α−

).

Taking the direction of the magnetic field to be along the 3−axis, we get

(α+

α−

)=eghB

4mc

(1 00 −1

) (α+

α−

)=eghB

4mc

(α+

−α−

).

The solutions to this eigenvalue problem are

ω =egB

4mcwith

(α+

α−

)=

(10

)= α

and

ω = − egB

4mcwith

(α+

α−

)=

(01

)= β .

We now can write the general solution as

ψ(t) = ae−iωt(

10

)+ beiωt

(01

)

=

(ae−iωt

beiωt

)with ω =

egB

4mc.

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224 CHAPTER 11. MATRIX FORMULATION OF QUANTUM MECHANICS

Let us consider the case when the spin in the initial state (i.e. t=0) is an eigenstateof S1, i.e.

ψ(0) =

(ab

)=

1√2

(11

)or

S1 ψ(0) =h

2ψ(0)

The solution at a later time t is then given by

ψ(t) =1√2

(e−iωt

eiωt

).

The average value of the spin, along the 1-axis, at time t is given by

〈S1〉 = 〈ψ(t)|S1|ψ(t)〉

=h

2

1√2

(eiωt e−iωt

) ( 0 11 0

)1√2

(e−iωt

eiωt

)

=h

2cos 2ωt .

On the other hand, the average value of the spin along the 2-axis is given by

〈S2〉 = 〈ψ(t)|S2|ψ(t)〉

=h

2

1√2

(eiωt e−iωt

) ( 0 −ii 0

)1√2

(e−iωt

eiωt

)

=h

2sin 2ωt .

Combining the above two results we observe that as a function of time the spin is pre-cessing about the 3-axis and the frequency of precession is given by

2w =egB

2mc≈ eB

mc.

11.5 Problems

1. The vector ~r in real three-dimensional space is subject to the transformation

~r ′ = A ~r = ~a× ~r

where ~a is a given fixed vector. Show that A is a linear operator which satisfies theequation

A3 + a2A = 0 with a = |~a|

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11.5. PROBLEMS 225

2. Construct the matrix of a linear transformation in real two-dimensional space whichdoubles the length of every vector drawn from the origin, and rotate it through apositive angle of 45. Show that this matrix satisfies the equation A4 = −16.

3. Show that every 2× 2 matrix can be written in the form

a0I + a1σ1 + a2σ2 + a3σ3

where the ai are complex numbers, and the σi are given by

σ1 =

(0 11 0

)σ2 =

(0 −ii 0

)σ3 =

(1 00 −1

).

In the above I is the unit matrix.

4. Given the matrix

A =

2

√2/3 0√

2/3 2√

1/3

0√

1/3 2

.

(a) Show that the equation (x, Ax) = 1 represents an ellipsoid in real three-dimensional space, and find the length of the principle axes of the ellipsoid.

(b) What are the direction cosines of the principle axis?

(c) Construct the unitary matrix U which diagonalizes A.

(d) Show by direct matrix multiplication that UAU−1 is a diagonal matrix.

5. The characteristic equation for a matrix A has the simple form

det aij − αδij = 0

only if the basis of the representation is diagonal. Write down the eigenvalue equa-tion

A x = λx

using an arbitrary basis |yi|, and show that the characteristic equation assumes theform

det aij − λ∆ij = 0

where

aij =(yi, A yj

)and ∆ij = (yi, yj) .

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226 CHAPTER 11. MATRIX FORMULATION OF QUANTUM MECHANICS

6. Show that the set of all polynomials of a degree smaller than N in a real variableu can be regarded as a linear vector space of N dimensions. Let the scalar productin this space be defined as

(x, y) =

1∫−1

du x∗(u) y(u) .

Prove that the operator

A = − d

du(1− u2)

d

du

is Hermitian.

7. Construct the characteristic equation for the operator A of the previous problem inthe special case of N = 3. Show that it has the eigenvalues 0, 2, 6. What are thecorresponding eigenvectors?

8. Show that for an operator A,

trA

=∑α

〈α|A|α〉

is independent of the choice of basis |α〉.

9. Let |u〉 and |v〉 be two vectors of finite norm. Show that

tr |u〉〈v| = 〈v|u〉tr |v〉〈u| = 〈u|v〉

10. Let H be a positive definite Hermitian operator.

(a) Show that for any |u〉 and |v〉

|〈u|H|v〉|2 ≤ 〈u|H|u〉 〈v|H|v〉 ,

and that the equality 〈u|H|u〉 = 0 necessarily implies that H|u〉 = 0.

(b) Show that tr H ≥ 0, and that the equality implies that H = 0.

11. Solve the eigenvalue problem~σ · n χλ = λχλ

where n is a unit vector given by

n = (sin θ cosφ, sin θ sinφ, cos θ)

and ~σ is the Pauli spin matrix.

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11.5. PROBLEMS 227

(a) Find the states χµ.

(b) Write the states χµ in terms of the two component spinors for spin up, α, andspin down, β.

(c) construct the projection operator

Pλ = |χλ〉 〈χλ| .

(d) If n = (1, 0, 0), what is the projection operator Pµ?

12. Show that it is impossible to construct a nonvanishing 2× 2 matrix which anticom-mutes with all three Pauli matrices.

13. Prove that for any two vectors ~A and ~B, we have(~σ · ~A

) (~σ · ~B

)= ~A · ~B + i~σ ·

(~A× ~B

)where ~σ is the Pauli spin matrix.

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228 CHAPTER 11. MATRIX FORMULATION OF QUANTUM MECHANICS

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Chapter 12

Symmetry and Conservation

We have seen that for a central potential, the Hamiltonian commutes with the angularmomentum. This means we can write the wave function for the system to be an eigenstateof both the energy and angular momentum. In particular, this will allow us to write thewave function in terms of a radial part R`(r), and an angle dependent part, given by thespherical harmonics Y`m(θ, φ), i.e.,

ψ(~r) = R`(r)Y`m(r) .

This in turn reduces the Schrodinger equation to a second order linear differential equationin the radial variable r. What we have accomplished, is to use the symmetry of the systemto divide the problem into two parts: (a) The geometry given by the spherical harmonicsY`m(θ, φ), (b) The dynamics, as given by R`(r).

What I want to discuss in this chapter is how we can make use of the symmetry of theproblem to separate the geometry from the dynamics. Here , it is important to rememberthat the geometry has all the symmetries and thus the conservation laws are built in. Wewill consider three symmetries in detail:

1. Symmetry under translation in space and time.

2. Symmetry under rotation.

3. Symmetry under time reversal.

All of these symmetries are associated with space-time. There are other symmetries innature which are also associated with conservation laws. These are often referred to asinternal symmetries and are as important as the space-time symmetries discussed in thischapter. However, since we have not encountered them in any of the problems we haveconsidered, we will postpone the discussion of these symmetries until later chapters wherewe will need to introduce them.

229

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230 CHAPTER 12. SYMMETRY AND CONSERVATION

12.1 Translation in Space and Time

We start by considering the simplest symmetry and the corresponding conservation law.Let us consider a system described by the wave function ψα(x) ≡ 〈x|α〉, in one spacedimension. If we now displace the entire physical system by a distance λ, then the stateof the system after displacement is given by ψα′(x) ≡ 〈x|α′〉. This transformation isillustrated in Figure 12.1

x xx + λ x + λx x

Ψα (x) Ψα' (x)

Figure 12.1: An illustration of the translation of the wave function ψα along the x-axisby a distance λ.

From the above figure it is clear that

ψα′(x) = ψα(x− λ) . (12.1)

This process of displacing the physical system by a distance λ can be represented by anoperator D(λ) such that

|α′〉 = D(λ) |α〉 . (12.2)

The problem is how to determine D(λ). If we expand the right hand side of Eq. (12.1) ina Taylor series about the point x, we get;

ψα′(x) = 〈x|α′〉 = ψα(x)− λ d

dxψα(x) +

λ2

2!

d2

dx2ψα(x)− · · ·

=

1− λ d

dx+λ2

2!

d2

dx2+ · · ·

ψα(x) (12.3)

and therefore,

ψα′(x) = exp

−λ d

dx

ψα(x)

= 〈x| exp

−iλhpx

|α〉 , (12.4)

where px is the momentum operator (i.e. px = −ih ddx

). We now can write

|α′〉 = exp

−iλhpx

|α〉 , (12.5)

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12.1. TRANSLATION IN SPACE AND TIME 231

with the operator D(λ) given by

D(λ) = exp

−iλhpx

. (12.6)

The generalization of this result to three-dimensions is now straightforward. Thedisplacement of the system is now defined by a direction and a magnitude, i.e.,

λ→ ~λ ,

and the momentum in one dimension is replaced by the momentum vector, i.e.,

p→ ~p .

The displacement operator is now given by;

D(~λ) = exp− ih~λ · ~p

. (12.7)

The statement of Eq. (12.1) implies that the system does not change under the operationof translation, i.e., the system is invariant under translation in space.

To study the corresponding conservation law, we take the time derivative of Eq. (12.2),i.e.,

ihd

dt|α′〉 = ih

d

dtD(~λ) |α〉 = ihD(~λ)

d

dt|α〉 . (12.8)

If the state of the system satisfies the Schrodinger equation then

ihd

dt|α〉 = H |α〉 , (12.9)

where H is the Hamiltonian for the system. This allows us to write Eq. (12.8) as

ihd

dt|α′〉 = D(~λ)H |α〉

= D(~λ)H D−1(~λ) |α′〉 . (12.10)

Here, D−1(~λ) is the inverse operator for translation (i.e. |α〉 = D−1(~λ) |α′〉) and is givenby,

D−1(~λ) = expi

h~λ · ~p

=

[D(~λ)

]†, (12.11)

i.e. D(~λ) is a unitary operator.

For the system to be invariant under translation, the states |α〉 and |α′〉 = D(~λ) |α〉should satisfy the same Schrodinger equation, i.e., if

ihd

dt|α〉 = H |α〉

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232 CHAPTER 12. SYMMETRY AND CONSERVATION

then

ihd

dt|α′〉 = H |α′〉 .

Making use of this result in Eq. (12.10) we get,

H |α′〉 = D(~λ)HD−1(~λ) |α′〉

or [H,D(~λ)

]= 0 . (12.12)

Therefore, for a system to be invariant under the transformation D(~λ), the Hamiltonian

for that system should commute with D(~λ). We note at this point that the only property

of D(~λ) that we have made use of is that it has an inverse. If D(~λ) is the operator fortranslation in space then the commutation relation in Eq. (12.12) is equivalent to

[H, ~p ] = 0 , (12.13)

i.e., the Hamiltonian commutes with the momentum operator, which in turn impliesthat one can construct simultaneous eigenstates of the Hamiltonian and the momentumoperator. One such Hamiltonian is the free particle Hamiltonian, i.e.,

H =p2

2m,

for which it is obvious that [H, ~p ] = 0, and the eigenstates of H are also eigenstates of ~p,i.e.,

H |~k 〉 =h2k2

2m|~k 〉 and ~p |~k 〉 = h~k |~k 〉 .

The second of these two equations can be written in coordinate representation as

−ih∇r 〈~r |~k 〉 = hk 〈~r |~k 〉

with the solution of this first order differential equation given by,

〈~r |~k 〉 =1

(2π)3/2ei~k·~r . (12.14)

Thus in this case the eigenstates of the momentum operator are also eigenstates of theHamiltonian. Also these eigenstates are such that the operator D(~λ ) is diagonal in thisbasis, i.e.,

〈~k |D(~λ)|~k′ 〉 = δ(~k − ~k ′) ei~λ·~k . (12.15)

The matrix elements of the operator D(~λ ) are referred to as the representation of thegroup of translation. We will come back to this concept later when we are consideringrotations.

Before we proceed to a discussion of other symmetries, let us consider some of thegeneral properties of D(~λ ).

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12.1. TRANSLATION IN SPACE AND TIME 233

1. If we first translate the system by a displacement ~λ1, and then by ~λ2, we can write

|α′〉 = D(~λ1) |α〉 ,

and

|α′′〉 = D(~λ2) |α′〉 = D(~λ2)D(~λ1) |α〉 .

But

|α′′〉 = D(~λ1 + ~λ2) |α〉 ,

therefore

D(~λ2)D(~λ1) = D(~λ1 + ~λ2)

i.e., the product of two translations is another translation.

2. We can define an identity transformation D(0) such that;

D(0)D(~λ ) = D(~λ )D(0) = D(~λ ) .

3. For every translation D(~λ ), there is an inverse D−1(~λ ) such that,

D(~λ )D−1(~λ ) = D−1(~λ )D(~λ ) = D(~λ ) .

4. For three successive transformations we have,

D(~λ3)D(~λ2)D(~λ1)

=D(~λ3)D(~λ2)

D(~λ1) .

The above four properties are those of a group, and D(~λ ) for a given ~λ is an elementof this group. If in addition the elements of the group satisfy the condition that;

5. The order in which we perform the transformation is not important, i.e.,

D(~λ1)D(~λ2) = D(~λ2)D(~λ1) ,

then the group is said to be an Abelian group.

6. If the elements of the group D(~λ ) have a parameter ~λ that takes on continuousvalues, as in the case of translation, then the group is a continuous group or Liegroup.

Having established that the operators of translation in space D(~λ ) form a group,let us turn to the symmetry of the system under translation in time. Given the state

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234 CHAPTER 12. SYMMETRY AND CONSERVATION

ψα(t) = 〈t|α〉, if we displace this system in time by an interval τ , then we get the stateψα′(t) and

ψα′(t) = ψα(t− τ)

=

1− τ d

dt+τ 2

2!

d2

dt2+ · · ·

ψα(t)

= D(τ)ψα(t) .

But from the Schrodinger equation we have

ihd

dtψα(t) = H ψα(t) or

d

dt⇒ − i

hH

so that the operator of translation in time is

D(τ) = expi

hHτ

. (12.16)

We have established the fact that for a system which is invariant under translation themomentum operator commutes with the Hamiltonian for that system. Thus at timet = 0 we can construct a state that is an eigenstate of both the Hamiltonian H and themomentum operator ~p. Since the operator that translates the system in time is a functionof the Hamiltonian H, the state of the system at time t > 0 will have the same momentumas the state at time t = 0. In other words if

Hψn,~k(t = 0) = Enψn,~k(t = 0) and ~pψn,~k(t = 0) = h~kψn,~k(t = 0) ,

thenψ(t) = D(t)ψ(0)

~pψ(t) = ~pD(t)ψ(0)

= D(t) ~pψ(0) since [H, ~p] = 0

= h~k D(t)ψ(0)

= h~kψ(t) .

In other words the momentum is the same at t > 0 as it was at t = 0, i.e., the momentumis a constant of the motion, or momentum is conserved.

If the Hamiltonian is time dependent, then the operator for translation in time is givenby

D(τ) = exp

−τ d

dt

, (12.17)

which does not commute with H, and the system is not invariant under translation intime.

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12.2. ROTATION IN THREE-DIMENSIONS 235

12.2 Rotation in Three-Dimensions

So far we have considered the translation of the system in space and time. We found thatthe operators that generate translation form a group, and that the group is an Abeliangroup. We now turn our attention to the rotation of the system in three-dimensions. Thisis the first case of non-trivial symmetry because the group associated with rotation inthree-dimensions is not Abelian. The structure of the resultant group has many elementsin common with other symmetry groups which we will encounter in later chapters of thisbook.

ε

r

r'

ε (n × r)

n

θ

r Sin θ = |n x r|

Figure 12.2: The rotation of the vector ~r by the infinitesimal angle ε about the axis n.

Consider the rotation of the system by an amount ε in the positive sense about a unitvector n. Then every point labeled ~r in the old system moves to a new point ~r ′ in thenew system. For ε small, we can write (see Figure 12.2)

~r ′ = ~r + ε (n× ~r) . ( 12.18a)

or~r = ~r ′ − ε (n× ~r ′) . ( 12.18b)

If n is along the z-axis, then the transformation can be written as x′

y′

z′

=

1 −ε 0ε 1 00 0 1

xyz

, (12.19)

or in general this transformation takes the form

~r ′ = Pε~r . (12.20)

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236 CHAPTER 12. SYMMETRY AND CONSERVATION

For finite rotation the matrix Pθ is given by

Pθ =

cos θ − sin θ 0sin θ cos θ 0

0 0 1

. (12.21)

We note here that the operator rotates the vector ~r by the angle θ leaving the coordinatesystem fixed.

We now turn our attention to the state of the physical system in order to examinewhat happens to the state under rotation. We take the system to be initially in a stateψα(~r ) = 〈~r |ψα〉. After rotation, the system is in a state ψα′(~r ′) = 〈~r ′|ψα′〉. These twostates are related by the operator R, i.e.,

|ψα′〉 = R|ψα〉 . (12.22)

Here the operator R is analogous to the translation operator D(λ) we encountered in thelast section.

r

r'

nΨα'

Ψα

Figure 12.3: The rotation of the state ψα about the axes n which gives the state ψα′ .

From Figure 12.3, we observe that if we had a set of axes that rotated with the state,then the new state in the rotated coordinate would be functionally identical to the originalstate in the old coordinate system. In other words

ψα(~r ) = ψα′(~r ′ ) . (12.23)

Making use of the inverse of the transformation in Eq. (12.18b), i.e. ~r = ~r ′ − ε (n× ~r ′ ),we can write the above expression as

ψα(~r ) = ψα (~r ′ − εn× ~r ′ ) . (12.24)

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12.2. ROTATION IN THREE-DIMENSIONS 237

For an infinitesimal rotation, ε 1, we have

ψα(~r ′ − εn× ~r ′ ) = ψα(~r ′ )− ε(n× ~r ′ ) · ∇ψα(~r ′ )

= ψα(~r ′ )− εn · (~r ′ ×∇)ψα(~r ′ )

= [1− εn · (~r ′ ×∇)]ψα(~r ′) .

Since the momentum, ~p, is proportional to the gradient operator ∇, we can replace theoperator ~r ′ ×∇ by the angular momentum, i.e.,

~r ′ ×∇ =i

h~L .

With this result in hand we can write the rotated state as

ψα′(~r ′) ≡ 〈~r ′|ψα′〉 =[1− i

h~ε · ~L

]ψα(~r ′)

= 〈~r ′|R|ψα〉 , (12.25)

where

R(~ε ) = 1− i

h~ε · ~L , (12.26)

with ~ε defined to be a vector along the axis of rotation n with magnitude equal to ε, i.e.,~ε = εn.

So far we have only considered infinitesimal rotation. For finite rotation we have toconsider n infinitesimal rotations of magnitude ε, and then take the limit as n→∞ ε→ 0with nε→ θ. Thus the rotation operator for finite rotation is given by

R(~θ ) = limε→0

limn→∞

[1− i

h~ε · ~L

]n= exp

[− ih~θ · ~L

], (12.27)

and the rotated state is given by

ψα′(~r ) = R(~θ )ψα(~r ) . (12.28)

Here R(~θ ) is a unitary operator that rotates the system by an angle θ about the axis θ.

The rotation operators R(~θ ) form a group in that

1. The product of two rotations is a third rotation, i.e.,

R(~θ1)R(~θ2) = R(~θ3) .

2. There exists a unit operator, R(0) = 1, such that

R(~θ )R(0) = R(0)R(~θ ) = R(~θ ) .

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238 CHAPTER 12. SYMMETRY AND CONSERVATION

3. There exists an inverse operator R−1(~θ ) = R(−~θ ) such that

R(~θ ) R−1(~θ ) = R(0) .

4. The associative law is satisfied, i.e.,

R(~θ1)R(~θ2)R(~θ3)

=R(~θ1)R(~θ2)

R(~θ3) .

In addition to the above properties, we have two additional properties.

5. In three-dimensions the product of two rotations depends on the order in which therotations have been carried out, i.e.,

R(~θ1)R(~θ2) 6= R(~θ2)R(~θ1) ,

i.e., the group is a non-Abelian group.

6. The elements of the group are characterized by a parameter θ that can take oncontinuous values. Thus, R(~θ ) forms a Lie Group.

In the above definition of R(~θ), ~L = L1, L2, L3 are the generators of the group andthey satisfy the commutation relation

[Li, Lj] = ihεijkLk . (12.29)

This commutation relation defines the Lie Algebra for the rotation group R(3).To relate the invariance under rotation with the corresponding conservation law, con-

sider the fact that|ψα′〉 = R(~θ ) |ψα〉 (12.30)

and

ihd

dt|ψα〉 = H |ψα〉 , (12.31)

We then have that

ihd

dt|ψα′〉 = ih

d

dtR(~θ )|ψα〉

= ihR(~θ )d

dt|ψα〉

= R(~θ )H |ψα〉= R(~θ )HR−1(~θ )|ψα′〉 . (12.32)

For the system to be invariant under rotation, both |ψα〉 and |ψα′〉 must satisfy the sameequation, i.e., if

ihd

dt|ψα〉 = H |ψα〉 ,

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12.2. ROTATION IN THREE-DIMENSIONS 239

then

ihd

dt|ψα′〉 = H |ψα′〉 .

For this to be true, we require that

R(~θ )HR−1(~θ ) = H ,

or [R(~θ ), H

]= 0 . (12.33)

Since the rotation operator R(~θ ) is defined in terms of the angular momentum operator~L, then for the system to be invariant under rotation the angular momentum operatorshould commute with the Hamiltonian, i.e.,[

H, ~L]

= 0 , (12.34)

i.e., the angular momentum of the system must be a constant of the motion.To use the elements of the group R(~θ ) to transform the system from one set of

coordinates to another, we need to write R(~θ ) in some representation, e.g.,

〈α|R(~θ )|β〉 = Dαβ(~θ ) . (12.35)

These matrices are the representation of the rotation group. The group representation issaid to be reducible if we can find a unitary transformation on the basis states |α〉, |β〉, · · ·that will make D(~θ ) block diagonal. On the other hand, the representation is said to beirreducible if there is no such unitary transformation. For the group R(3), the representa-tion is irreducible if the basis states are taken to be eigenstates of the angular momentumoperator square, L2, and the 3-component of the angular momentum, L3, i.e.,

L2|`,m〉 = h2`(`+ 1)|`,m〉(12.36)

L3|`,m〉 = hm|`,m〉 .

To write the representation of the rotation operator in this basis, we recall that ~L actingon the state |`,m〉 changes m but not `, i.e.

〈`′,m′|~L|`,m〉 = δ`′` 〈`,m′|~L|`,m〉 .

This result follows from the fact that ~L is proportional to L3 and L±. The rotationoperator, given in Eq.(12.27), has matrix elements given by

〈`′,m′|R(~θ )|`,m〉 = δ`′` 〈`,m′|R(~θ )|`,m〉= δ`′`

[D`]m′m

, (12.37)

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240 CHAPTER 12. SYMMETRY AND CONSERVATION

where

D` = exp− ih~θ · ~L`

. (12.38)

Here ~L` is the (2`+ 1)× (2`+ 1) matrix representation of the operator ~L, i.e.,[~L`]m′m

= 〈`,m′|~L|`,m〉 .

By always specifying the irreducible representation of a group, we have uniquely de-fined the dimensionality of the representation matrices. However, to specify the repre-sentation, we need to define the most general rotation possible. This can be achieved interms of the three Euler angles which define the orientation of a body in space. We willdefine our Euler angles by the follows rotations:

1. A rotate by angle α about the z-axis.

2. A rotate by an angle β about the new y-axis.

3. A rotate by an angle γ about the new z-axis.

These rotations should be performed in the order in which they are listed, i.e.,

|ψα′〉 = e−ihγLz′′ e−

ihβLy′ e−

ihαLz |ψα〉 = R(α, β, γ) |ψα〉 . (12.39)

This operator is rather difficult to work with as it stands. This is mainly due to thefact that it is written in terms of three different coordinate systems, and to calculate thecorresponding representation is impossible with the operator in its present form. Buta careful examination of these rotations reveals that a rotation about the new y-axis isequivalent to a rotation about the old z-axis by an angle of −α, followed by the rotationabout the old y-axis by an angle β and then a rotation by an angle α about the old zaxis, i.e.

e−ihβLy′ = e−

ihαLz e−

ihβLy e

ihαLz . (12.40)

In a similar manner we can write the rotation about the z′′-axis in terms of rotation aboutthe original set of coordinates. This procedure allows us to rewrite the operator R(α, β, γ)in terms or the original coordinates as

R(α, β, γ) = e−ihαLz e−

ihβLy e−

ihγLz , (12.41)

and now the matrix elements of this operator are simple to calculate. Thus the irreduciblerepresentation of the rotation group results from taking the matrix elements of R(α, β, γ)between the eigenstates of L2 and Lz, i.e.,

〈`′,m′|R(α, β, γ)|`,m〉 = δ`′` 〈`,m′|R(α, β, γ)|`,m〉

≡ δ`′` D`m′m(α, β, γ) , (12.42)

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12.2. ROTATION IN THREE-DIMENSIONS 241

with

D`m′m(α, β, γ) = e−iαm′ 〈`,m′|e−

ihβLy |`,m〉 e−iγm

≡ e−iαm′d`m′m(β) e−iγm , (12.43)

where

d`m′m(β) = (−1)m′−m

[(`+m′)! (`−m′)!(`+m)! (`−m)!

]1/2 (cos

β

2

)m′+m (sin

β

2

)m′−m

× P(m′−m,m′+m)`−m′ (cos β) . (12.44)

Here, P(α,β)` are the Jacobi polynomials.

Let us now consider the transformation of the state |`,m〉 to the state |`,m〉′, where

|`,m〉′ = R(α, β, γ) |`,m〉=

∑`′m′|`′,m′〉 〈`′,m′|R(α, β, γ)|`,m〉

=∑m′|`,m′〉D`m′m(α, β, γ) . (12.45)

In coordinate representation, this reduces to

〈r|`,m〉′ =∑m′〈r|`,m′〉D`m′m(α, β, γ) .

But we have that〈r|`,m〉′ = 〈P−1

R r|`,m〉 = 〈r′|`,m〉and therefore, we can write the transformation of the spherical harmonics associated withthe system that is being rotated, as

Y`m(r′) =∑m′Y`m′(r)D`m′m(α, β, γ) . (12.46)

In the above discussion we have restricted our analysis to rotation of systems with nospin, i.e., the angular momentum is orbital angular momentum. In that case ` = integer,and therefore a rotation of the system by an angle of 2π about the z-axis is given by

ψ′ = e−ih

2πLz ψ = e−i2mπ ψ = ψ for m = integer .

If we now extend the above analysis of rotation to include spin, and in general to includesystems with total angular momentum ~J , then the rotation operator can be written as

R(~θ ) = exp[− ih~θ · ~J

]. (12.47)

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242 CHAPTER 12. SYMMETRY AND CONSERVATION

However, we now have for the case of spin 1/2 that a rotation of the system by 2π doesnot lead to the original state. In fact, to reach the original state, we have to rotate thesystem by 4π, since

χ′ = e−ih

2πJz χ = e−2mπ χ = −χ for m = ±1

2

while

χ′ = e−ih

4πJz χ = e−4mπ χ = χ for m = ±1

2.

We therefore have to enlarge our group of rotations to include the half integer angularmomentum. This new group is called the covering group, and is identical to the groupSU(2), the special unitary group in two-dimensions.

We now can write the representation of SU(2) as

Djm′m(α, β, γ) = e−iαm′djm′m(β) e−iγm , (12.48)

where

djm′m(β) = 〈j,m′|e−ihβJy |j,m〉

=e−

ihβJjym′m

. (12.49)

As an example, let us consider the evaluation of d12m′m(β). This is given by

d12m′m(β) = 〈 1

2,m′|e−

ihβJy | 1

2,m〉 .

But for j = 1/2 we have Jy = h2σy = h

2

(0 −ii 0

), therefore

d12 (β) = e−

i2βσy

= 1− i

2βσy +

1

2!

(i

2βσy

)2

− 1

3!

(i

2βσy

)3

+ · · · .

But we have that (σy)2 = 1. This allows us to write d

12 (β) as

d12 (β) = 1− i

2βσy +

1

2

(i

2β)2

1− 1

3!σy

(i

2β)3

+ · · · .

This series can be regrouped to give

d12 (β) = 1

1− 1

2!

2

)2

+ · · ·

− iσyβ2 − 1

3!

2

)3

+ · · ·

= 1 cos

β

2− iσy sin

β

2

=

(1 00 1

)cos

β

2− i

(0 −ii 0

)sin

β

2.

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12.3. ADDITION OF ANGULAR MOMENTUM 243

Therefore

d12 (β) =

cos β2− sin β

2

sin β2

cos β2

. (12.50)

In particular d12 (π) when acting on spin up,

(10

), gives spin down,

(01

), i.e.,

d12 (π)

(10

)=

(0 −11 0

) (10

)=

(01

).

12.3 Addition of Angular Momentum

We have established in the last section that in the presence of rotational symmetry, thetotal angular momentum of the system is conserved, i.e. the angular momentum is aconstant of the motion, and the total angular momentum of the system commutes withthe Hamiltonian. This can be stated as[

J2, H]

= 0 and [Jz, H] = 0 .

Because we have three commuting operators, the eigenstates of the system, i.e., the eigen-states of the Hamiltonian H, can also be eigenstates of the angular momentum squareJ2, and its component along the z-axis, Jz. To take full advantage of the symmetry ofthe problem, we need to write the eigenstates of the system in terms of the eigenstates ofJ2 and Jz. In this way we separate the geometry in the problem from the dynamics, aswas done for a particle in a central potential in Chapter 1 when we wrote the total wavefunction in terms of a radial wave function R`(r) and the angular function Y`m(θ, φ). Infact the spherical harmonic Y`m(θ, φ) was an eigenstate of the total angular momentumand its z-component. We also observed that for a system with spherical symmetry, thedependence on the potential, i.e. the dynamics, was only present in the radial equation.The geometry of the problem, which is very important and needs to be taken into con-sideration exactly, was in the spherical harmonics. Here, it should be emphasized thatany approximation in the geometry will automatically lead to a violation of the symmetry(conservation law), which in this case is spherical symmetry (angular momentum).

For a system with more than one particle and with spherical symmetry, we need toconstruct the states of total angular momentum. These will be given in terms of the basisstates of the individual particle’s angular momentum states. In general, for a system withn particles with angular momentum ~Ji, i = 1, · · · , n, the total angular momentum is givenby

~J = ~J1 + ~J2 + · · ·+ ~Jn . (12.51)

A similar problem arises when we consider a system of one particle with spin ~S and orbitalangular momentum ~L. In this case the total angular momentum is given by

~J = ~L+ ~S . (12.52)

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244 CHAPTER 12. SYMMETRY AND CONSERVATION

In this section, we will consider the problem of constructing eigenstates of the totalangular momentum when ~J is the sum of two angular momenta, e.g. n = 2 in Eq. (12.51).This incorporates the case when one of the angular momenta is the spin of the particlewhile the other is the orbital angular momentum of that particle. The reason why we neednot make a distinction between spin and orbital angular momentum is due to the factthat they both satisfy the same algebra, the algebra of SU(2). For two angular momenta,since they are a function of different variables, we have that[

~J1, ~J2

]= 0 . (12.53)

This means we can take as our basis, the eigenstates of the four commuting operators

J21 , J1,z , J2

2 , J2,z .

These basis states are|j1,m1; j2,m2〉 = |j1,m1〉 |j2,m2〉 , (12.54)

with

J21 |j1,m1〉 = h2j1(j1 + 1) |j1,m1〉 ( 12.55a)

J1,z |j1,m1〉 = hm1 |j1,m1〉 ( 12.55b)

J22 |j2,m2〉 = h2j2(j2 + 1) |j2,m2〉 ( 12.55c)

J2,z |j2,m2〉 = hm2 |j2,m2〉 . ( 12.55d)

The disadvantage of this choice of basis states (i.e. |j1,m1; j2,m2〉) is that the statesare not eigenstates of the total angular momentum J2 which is obvious from the fact that

J2 =(~J1 + ~J2

)·(~J1 + ~J2

)= J2

1 + J22 + 2 ~J1 · ~J2

= J21 + J2

2 + 2 (J1,xJ2,x + J1,yJ2,y + J1,zJ2,z)

and |j1,m1; j2,m2〉 is not an eigenstate of Ji,x and Ji,y. Also, in this basis, the repre-sentations of the rotation group are reducible, which implies that there exists a unitarytransformation to another basis in which the representations of the rotation group areirreducible, i.e. block diagonal.

An alternative set of basis are the eigenstates of the operators

J21 , J2

2 , J2 , Jz .

We can show that these operators commute among themselves, and their eigenstates aregiven by

J2i |(j1j2)jm〉 = h2ji(ji + 1) |(j1j2)jm〉 i = 1, 2 ( 12.56a)

J2 |(j1j2)jm〉 = h2j(j + 1) |(j1j2)jm〉 ( 12.56b)

Jz |(j1j2)jm〉 = hm |(j1j2)jm〉 . ( 12.56c)

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12.3. ADDITION OF ANGULAR MOMENTUM 245

This basis has the advantage that the set of quantum numbers j1j2jm are constantsof motion because the corresponding operators commute with the Hamiltonian. Also, inthis basis, the representations of the rotation group are irreducible. The problem nowis to find the unitary transformation that will allow us to write the states |(j1j2)jm〉 interms of the known states |j1,m1; j2,m2〉, i.e.,

|(j1j2)jm〉 =∑j′1m

′1

j′2m′2

|j′1,m′1; j′2m′2〉 〈j′1,m′1; j′2m

′2|(j1j2)jm〉 . (12.57)

The elements of this unitary transformation are the brackets 〈j′1,m′1; j′2m′2|(j1j2)jm〉. Be-

fore we proceed to determine these brackets, let us determine some of their properties.We have that

J21 |(j1j2)jm〉 = h2j1(j1 + 1) |(j1j2)jm〉 , (12.58)

and〈j′1,m′1; j′2,m

′2| J2

1 = h2j′1(j′1 + 1) 〈j′1,m′1; j′2,m′2| . (12.59)

We now multiply Eq. (12.57) by 〈j′1,m′1; j′2,m′2| and Eq. (12.58) by |(j1j2)jm〉 and subtract

to geth2 j1(j1 + 1)− j′1(j′1 + 1) 〈j′1,m′1; j′2,m

′2|(j1j2)jm〉 = 0 .

Therefore for j1 6= j′1 we have

〈j′1,m′1; j′2,m′2|(j1j2)jm〉 = 0

or〈j′1,m′1; j′2,m

′2|(j1j2)jm〉 = δj′1j1 〈j1,m

′1; j′2,m

′2|(j1j2)jm〉 .

In a similar way we can show that for j′2 6= j2, the brackets that form the elements of theunitary transformation are zero. We now can write Eq. (12.57) as

|(j1j2)jm〉 =∑m1m2

|j1,m1; j2,m2〉 〈j1,m1; j2,m2|(j1j2)jm〉 . (12.60)

We now can further reduce this sum by noting that

Jz = J1,z + J2,z .

We then can writeJz |(j1j2)jm〉 = hm |(j1j2)jm〉 , (12.61)

and〈j1,m1; j2,m2| (J1,x + J2,z) = h (m1 +m2) 〈j1,m1; j2,m2| . (12.62)

Proceeding as before and multiplying Eq. (12.61) by 〈j1,m1; j2,m2| and Eq. (12.62) by|(j1j2)jm〉 and subtracting we get

h m− (m1 +m2) 〈j1,m1; j2,m2|(j1j2)jm〉 = 0

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246 CHAPTER 12. SYMMETRY AND CONSERVATION

and therefore

〈j1,m1; j2,m2|(j1j2)jm〉 = δm,m1+m2 (j1m1j2m2|j1j2jm) . (12.63)

We now can write the unitary transformation from one basis to the other as

|(j1j2)jm〉 =∑

m1m2

m1 +m2 = m

|j1,m1; j2,m2〉 (j1m1j2m2|j1j2jm) . (12.64)

The coefficient (j1m1j2m2|j1j2jm) is known as the Clebsch-Gordan coefficient. Here wenote that in the ket on the right hand side of Eq. (12.64), we have put brackets aroundthe quantum numbers j1 and j2. This is to indicate that these two angular momenta areadded to get a total angular momentum j with projection m. The inverse transformationis then given by

|j1,m1; j2,m2〉 =∑jm

m = m1 +m2

|(j1j2)jm〉 (j1m1j2m2|j1j2jm) . (12.65)

Since the transformation from one basis to the next is unitary, and the states in eachbasis are orthonormal, these coefficients satisfy the orthogonality relation∑

jm

(j1m1j2m2|j1j2jm) (j1m′1j2m

′2|j1j2jm) = δm1m′

1δm2m′

2, (12.66)

and ∑m1m2

(j1m1j2m2|j1j2jm) (j1m1j2m2|j1j2j′m′) = δjj′ δmm′ . (12.67)

The values of these coefficients can be obtained from standard codes available on mostcomputers.

Finally, we should note that since ~J is the vector sum of ~J1 and ~J2, that the corre-sponding quantum numbers satisfy the relation

|j1 − j2| ≤ j ≤ |j1 + j2| .

Furthermore, since ~J satisfy the same algebra as the angular momentum operator, wehave that

−j ≤ m ≤ j .

Example: The simplest example we can consider is that of two particles of spin 1/2, i.e.,

~S = ~s1 + ~s2 .

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12.4. SPACE INVERSION AND PARITY 247

In this case

|( 12

12)SM〉 =

∑m1=± 1

2

( 12m1

12M −m1| 12 1

2SM) | 1

2m1〉 | 12 M −m1〉 .

For S = 1 and M = +1, m1 must be + 12

and we have

|( 12

12)S = 1M = +1〉 = ( 1

212

12

12| 1

212

1 1) | 12

12〉 | 1

212〉

= | 12

12〉 | 1

212〉 = α(1)α(2)

In this case the normalization of basis required that ( 12

12

12

12| 1

212

1 1) = 1. Similarly, forS = 1 and M = −1 we have that m1 = − 1

2and

|( 12

12)S = 1M = −1〉 = β(1) β(2) .

Here, α and β correspond to Pauli spinors with spin up and down respectively, while α(1)indicates that particle 1 is in a state with spin up, i.e. along the positive z-axis.

For M = 0, we have two possible values for the total spin, i.e. S = 1 or 0. In this casem1 = ± 1

2and we have two terms in the sum. For S = 1 we have

|( 12

12)S = 1M = 0〉 =

1√2α(1) β(2) + α(2) β(1) ,

while for S = 0, we have

|( 12

12)S = 0M = 0〉 =

1√2α(1)β(2)− α(2) β(1) .

We observe here that all the S = 1 (Spin Triplet) states have a symmetric wave functionunder the exchange of the particles coordinates. On the other hand, the S = 0 (SpinSinglet) state is antisymmetric under this exchange. We will come back to this problemat a later stage when we consider the symmetry of the wave function under permutationof coordinates.

12.4 Space Inversion and Parity

Until 1957, it was always assumed that the observables for any system will not changeif we consider the system to have gone under the transformation ~r → −~r. However,it was realized in 1957 that there has been no experimental test of this symmetry forcertain class of interaction commonly known as the Weak Interaction. The validity of thissymmetry corresponds to the question of getting the same result from an experiment andits mirror image. At the suggestion of T. D. Lee and C. N. Yang three experiments werecarried out which demonstrated that in Weak Interactions this symmetry is violated at

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248 CHAPTER 12. SYMMETRY AND CONSERVATION

the level of 100%. However, for all other interactions, space inversion is a good symmetry,and the corresponding conserved quantity is commonly referred to as parity. Thus thestates of any system, where the governing interaction is not the Weak Interaction, parityis one of the quantum numbers that labels the states, and in any transition parity will beconserved.

Let us consider the action of space inversion to be represented by the operator P . Wethen have that

P ψ(~r) = ψ(−~r) . (12.68)

This operator has the property that when acting on a state twice, it will give the originalstate, i.e.,

P2 ψα(~r) = P ψα(−~r) = ψα(~r) . (12.69)

This special property of the operator P2, having eigenvalue one, implies that the operatorP could have eigenvalues ±1, i.e.,

P ψα(~r) = ±ψα(~r) . (12.70)

This in turn means that we can label the state ψα by the eigenvalue corresponding to theoperator P .

We now examine the condition under which space inversion is a valid symmetry andthus parity is a conserved quantum number. Let us consider the case when the state ψαis a solution of the time dependent Schrodinger equation, i.e.,

ih∂ψα∂t

= H ψα . (12.71)

Then the action of the space inversion operator on this equation gives

ihP ∂ψα∂t

= P H ψα . (12.72)

Taking P ψα = ψ′α and therefore ψα = P−1 ψ′α, we can write Eq. (12.72) as

ih∂ψ′α∂t

= PHP−1ψ′α . (12.73)

For the theory to be invariant under space inversion, the equation of motion, i.e., theSchrodinger equation, should maintain its form under the transformation. In this casethis implies that the Hamiltonian H will commute with the parity operator P , i.e.,

[P , H] = 0 . (12.74)

We thus may conclude that for systems for which the Hamiltonian commutes with theparity operator, the eigenstates of the system can be labeled by the parity quantumnumber which can take a value of ±1.

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12.5. TIME REVERSAL 249

12.5 Time Reversal

We now turn to time reversal which involves the reversal in the direction of time, i.e. itis equivalent to asking the following question. When running a film backwards, are thephysical phenomenon observed as valid as those observed when the film runs forward?Although not valid at the macroscopic level, this seems to be true at the microscopiclevel. To examine this symmetry, first consider the state ψα that describes our system.Corresponding to this state then, there is a state ψ′α that describes the time reversedsystem. In this time reversed system all momenta and angular momenta point in theopposite direction to the original system, i.e., under time reversal ~p→ −~p and ~L→ −~L.We now introduce an operator T such that

T ψα = ψ′α . (12.75)

To the best of our knowledge T is a symmetry operation that is valid for most systems,i.e., if ψα is an eigenstate of the Hamiltonian H, then ψ′α = T ψα is also an eigenstate ofH.

We first examine the general property of the operator T by considering a state ψα attime t = 0 to be given by

ψα =∑k

aαk φk ,

where φk is an eigenstate of the Hamiltonian H with an eigenvalue Ek, i.e.,

H φk = Ek φk .

We now would like to follow the time development of this system along two paths:

1. Let the state ψα propagate in time from t = 0 to time t = t1. Then the state att = t1 is given by

ψ′α = e−iHt1/h ψα =∑k

aαk e−iEkt1/h φk at t = t1 .

We next apply the time reversal operator to this state, i.e., we act on the state withthe time reversal operator T to get

ψ′′α = T ψ′α =∑k

aαk e−iEkt1/h T φk at t = −t1 .

In writing this result we have assumed T is a linear operator.

2. Now we can get to the same final state by first applying the time reversal operatorto get

ψ′′′α =∑k

aαk T φk at t = 0 ,

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250 CHAPTER 12. SYMMETRY AND CONSERVATION

and then by letting the state propagate in time to t = −t1, we get

ψ′′α = eiHt1/h ψ′′′α =∑k

aαk eiEkt1/h T φk at t = −t1 .

To get this result, we have taken [H, T ] = 0.

Clearly the final states obtained along the two different paths are different, yet theyboth should describe the time reversed system at time t = −t1. This suggests that wehave made an error in the above analysis. We observe that if the operator T had theproperty that for any complex constant c

T cΨ = c∗ T Ψ

where Ψ is any state vector, then we would have had the same result for our final stateirrespective of the path followed. In fact, with this new definition of the time reversaloperator, our two paths give

1. For the first path

ψ′′α = T e−iHt1/h ψα= T

∑k

aαk e−iEkt1/hφk

=∑k

a∗αk eiEkt1/h T φk .

2. While for the second path we have

ψ′′α = eiHt1/h T ψα= eiHt1/h

∑k

a∗αk T φk

=∑k

a∗αk eiEkt1/h T φk .

We now have the same result for the state at time t = −t1 irrespective of the path followed.From the above property of the time reversal operator we may conclude that this operatoris in fact not linear but is referred to as an anti-linear operator.

In fact we can show from the Schrodinger equation that the operator T has to beanti-linear. The time dependent Schrodinger equation is given by

ih∂ψα∂t

= H ψα .

If under time reversal t→ −t, then the Schrodinger equation for the time reversed stateis given by

−ih∂ψ′α

∂t= H ψ′α

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12.5. TIME REVERSAL 251

But this is the Schrodinger equation for the state ψ′α = ψ∗α. If we operate with T on theSchrodinger equation, we get

−ih∂T ψα∂t

= T H ψα .

If [T , H] = 0, then the above equation becomes

−ih∂T ψα∂t

= H T ψα

and we have the state T ψα satisfying the same equation as the state ψ∗α. Thus if T ψα =ψ∗α, then time reversal is a symmetry of the system.

In general, the time reversal operator is of the form

T = K U , (12.76)

where K is complex conjugation operator while U is linear operator. Furthermore, if|〈α|β〉| is to remain the same under time reversal, then U must be unitary, i.e., U−1 = U †,and in the case of T = K U , is an anti-unitary operator.

For particles with spin zero, we want to choose T and thus U such that

T ~r T −1 = ~r and T ~p T −1 = −~p , (12.77)

i.e., the time reversal operator changes the direction of the momentum not the coordinate.In coordinate representation, we have that ~r is real while ~p is pure imaginary (~p = −ih∇),so that the simplest choice for T is

T = K , (12.78)

where K is the complex conjugation operator.For particles with spin, we also want to change the direction of the spin and angular

momentum, i.e., in addition to the conditions in Eq. (12.77) we have

T ~S T −1 = −~S and T ~J T −1 = − ~J , (12.79)

where ~J is the angular momentum of the system. As before, if we specify the representa-tion of the spin, we can write the Pauli matrices as

σx =

(0 11 0

), σy =

(0 −ii 0

), σz =

(1 00 −1

).

In this case ~r, σx, σz are real, while ~p and σy are pure imaginary. Thus for T ~S T −1 = −~Sand T = K U , we require that

U Sx U−1 = −Sx and U Sz U

−1 = −Sz . (12.80)

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252 CHAPTER 12. SYMMETRY AND CONSERVATION

We therefore need to choose U such that it reverses the direction of the x and z axis forthe spin matrices, but not the y-axis. This can be achieved by a rotation about the y-axisby an angle π. In this case the rotation operator is

U = e−iπSy/h

and the time reversal operator is

T = K U = e−iπSy/hK . (12.81)

12.6 Isospin and the Pauli Principle

Soon after the neutron was discovered in 1932, Heisenberg suggested that in the absenceof electromagnetic interaction, the neutron and the proton are two states of the samesystem. In fact, the mass of the proton is almost the same as the mass of the neutronsince

mp = 938.2796± 0.0027 MeV./c2 mn = 939.5731± 0.0027 MeV./c2 , (12.82)

where c is the velocity of light in vacuum. Furthermore, the interaction between theneutron and proton is approximately the same as the force between two protons when theCoulomb repulsion is subtracted. This equivalence of the proton and neutron is equivalentto a degeneracy, and therefore, a possible symmetry. In this case the symmetry is notassociated with either space or time, but is an internal symmetry. The fact that we havea doublet of states suggested to Heisenberg that he should introduce the same algebrathat described spin, because with spin we have also a doublet corresponding to spin upand spin down. Thus isospin is the algebra that describes protons and neutrons to whichwe give the generic name nucleon so that

proton = nucleon with isospin up

neutron = nucleon with isospin down

We now can adapt the algebra of spin to isospin by introducing the isospin operators

τ1 =

(0 11 0

)τ2 =

(0 −ii 0

)τ3 =

(1 00 −1

). (12.83)

which are the Pauli operators that we used for spin. Since we know the algebra of theseoperators, we may adapt the result we derived for spin angular momentum to isospin,and in particular we would have

1

2(τ1 + iτ2) n = p

1

2(τ1 − iτ2) p = n , (12.84)

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12.6. ISOSPIN AND THE PAULI PRINCIPLE 253

where n and p are the states corresponding to a neutron and proton, i.e.,

n =

(01

)p =

(10

). (12.85)

Furthermore, we can use the same Clebsch-Gordan coefficients we used for angular mo-mentum to construct states of total isospin. In fact, since isospin is a good symmetry ofthe nuclear Hamiltonian, the total isospin operator commutes with the Hamiltonian, andto that extent the eigenstates of total isospin are the proper basis to be working in, as wasthe case with angular momentum where we needed to construct states of total angularmomentum. For the two nucleon system, the states of total isospin consist of a triplet ofstates with isospin one which are

|( 12

12)1 + 1〉 = p(1) p(2)

|( 12

12)1 − 1〉 = n(1)n(2)

|( 12

12)1 0〉 =

1√2

[p(1)n(2) + n(1) p(2)] , ( 12.86a)

while for the isospin singlet we have

|( 12

12)0 0〉 =

1√2

[p(1)n(2)− n(1) p(2)] . ( 12.86b)

As expected, these states are identical to those we encountered for spin 12.3.

Table 12.1: The partial waves that are allowed by the Pauli exclusion principle and theexperiment required to determine the phase shifts in that partial wave.

` S T Spectroscopic Notation Experiment0 0 1 1S0 p− p

1 0 3S1 p− p and n− p1 0 0 1P1 p− p and n− p

1 1 3P0, 3P1, 3P2 p− p2 0 1 1D2 p− p

1 0 3D1, 3D2, 3D3 p− p and n− p

The wave function for the two nucleon system is now a product of the space, spin andisospin, i.e.,

Ψ(1, 2) = ψ(space) φ(spin) χ(isospin) . (12.87)

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254 CHAPTER 12. SYMMETRY AND CONSERVATION

Because our two nucleon system now consists of two identical Fermions,1 their wavefunction should be antisymmetric, i.e.,

Ψ(1, 2) = −Ψ(2, 1) . (12.88)

This basically states that two Fermions cannot be at the same position in space and atthe same time have the same quantum numbers. This statement is known as the Pauliexclusion principle. Since each part of the wave function of the two nucleons can have adefinite symmetry under the exchange of the coordinates of the two nucleons, it is clearthat only certain combinations are allowed according to this Pauli exclusion principle.From the spin and isospin wave functions we know that the singlet is antisymmetric whilethe triplet is symmetric. In other words, under the interchange of coordinates we have

φ(2, 1) = (−1)1+S φ(1, 2) χ(2, 1) = (−1)1+Tχ(1, 2) , (12.89)

where S and T are the total spin and isospin of the two nucleon system. On the otherhand, the space part of the wave function is labeled by the orbital angular momentum `and is a function of the relative coordinates of the two nucleons. Then under exchange ofcoordinates, the space part of the wave function satisfies

ψ(2, 1) = (−1)` ψ(1, 2) . (12.90)

Combining Eqs. (12.89) and (12.90), we can see that the total wave function has thesymmetry

Ψ(2, 1) = (−1)`+S+T Ψ(1, 2) . (12.91)

To satisfy the Pauli exclusion principle, we need to have `+S+T to be an odd integer. InTable 12.1 we present the states that are allowed, and neutron-proton (n− p) or proton-proton (p− p) experiments that have to be done to get information about the scatteringamplitude in that partial wave. We have used the spectroscopic notation

2S+1`J

where for ` = 0, 1, 2, · · · we have S, P, D · · ·. Here, J is the total angular momentum.

12.7 Problems

1. The Hamiltonian of a rigid rotator (i.e. quantum mechanical football) in an externalmagnetic field pointing along the z-axis is given by

H =L2

2I+B Lz ,

1Fermions are particles with half integer spin, and are called by this name because they obey Fermi-Dirac statistics. In contrast, Bosons are particles with integer spin that satisfy Bose-Einstein statistics.As you will see in the course on Thermal Physics, the statistic of spin 1

2 particles is different from thestatistic of integer spin particles.

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12.7. PROBLEMS 255

where I is the moment of inertia of the rotator, and B is a constant. Here ~L is theangular momentum operator. Determine the energy and wave function of the rigidrotator.

2. Consider the infinitesimal rotation operator to be of the form

R(ε, n) =

1 + εn · ~E.

Evaluate:

(a) First, a rotation about the y-axis followed by a rotation about the x-axis, i.e.,

R(ε, nx)R(ε, ny)ψα(~r).

(b) Second a rotation about the x-axis followed by a rotation about the y-axis, i.e.,

R(ε, ny)R(ε, nx)ψα(~r).

Using the definition of rotation about n by an angle ε to be

ψα′(~r ′) = ψα(~r ′ − εn× ~r ′ ) = R(ε, n)ψα(~r ′) ,

determine the commutation relation for Ex, Ey, Ez and relate them to the commu-tation relation of the angular momentum.

3. Calculate the matrices〈jm|Ji|jm′〉 i = 1, 2, 3

for j = 1. Determine the rotation operator

〈jm|R(π

2, e2)|jm′〉 = exp

− ih

π

2J j2

,

where e2 is a unit vector in the 2 direction, for j = 1 as a 3× 3 matrix.

4. Prove the following relations for the irreducible representation of the rotation group∑µ

D`mµ(α, β, γ)D`∗m′µ(α, β, γ) = δmm′

and〈`m|R−1(α, β, γ)|`m′〉 =

[D`m′m(α, β, γ)

]∗.

5. The positronium is a bound state of an electron and a positron (both are spin 1/2particles). In the ground state, i.e., n = 1, ` = 0, the main term in the Hamiltonian,besides the Coulomb attraction is

V1 = γ ~σ1 · ~σ2 .

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256 CHAPTER 12. SYMMETRY AND CONSERVATION

(a) Calculate the Clebsch-Gordan coefficients for the coupling of spin 1/2 with spin1/2 by diagonalizing the above interaction in the basis |s1 = 1

2,m1; s2 = 1

2,m2〉.

(b) Show that|(s1s2)SM〉 = (−1)S+1 |(s2s1)SM〉 ,

if s1 = s2 = 12.

6. Consider the reactionπ− + d→ n+ n .

where π− is the negative pion, d is the deuteron, and n is the neutron. Suppose theπ− has spin 0 and negative intrinsic parity and was captured from an s-orbit.

(a) What possible states would the two neutrons be in?

(b) What is the intrinsic parity of the pion?

Hint: Use conservation of angular momentum, parity and the Pauli exclusion prin-ciple.

7. Show that

D`m0(α, β, γ) = (−1)m

√4π

2`+ 1Y ∗`m(β, α) .

Hint Note the Jacobi polynomial P(m,m)`−m (x) is given by

P(m,m)`−m (x) = (−2)m

`!

(`−m)!(1− x2)−m/2P−m` (x) .

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Chapter 13

Approximation Methods for BoundStates

In most problems in quantum mechanics we are given a Hamiltonian H for which weneed to determine the eigenvalues and the corresponding eigenstates. For most physi-cally interesting systems, the Hamiltonian is complex enough that we cannot solve theSchrodinger equation

H |ψ〉 = E |ψ〉 (13.1)

in a closed form. In this case, to get any information about the energy spectrum we needto resort to approximation methods. Two such methods that are often used for both thebound state and scattering problem are:

1. Perturbation theory.

2. Variational method.

In the present chapter, we will consider both of these methods within the context of boundstate problems.

13.1 Perturbation Theory

This method is most useful when we can write the Hamiltonian for the system underconsideration as the sum of two parts, i.e.,

H = H0 +H1 , (13.2)

where H0 is chosen such that the corresponding Schrodinger equation can be solved ex-actly, i.e., we can solve the equation

H0|φn〉 = εn|φn〉 , (13.3)

257

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258 CHAPTER 13. APPROXIMATION METHODS FOR BOUND STATES

and at the same time H1 is small enough that the contribution to the energy and wavefunction from H1 is relatively small. In particular, the aim of perturbation theory is towrite a series solution for both the eigenvalues and eigenstates of H. The lowest term inthe series would be the eigenvalues and eigenstates of H0. The rest of the terms in theexpansion would be a power series in matrix elements of H1 with respect to the eigenstatesof H0.

Let us first consider the Schrodinger equation for H, i.e.,

H|ψn〉 = En|ψn〉 . (13.4)

With the help of Eq. (13.2), this can be written as

(En −H0 ) |ψn〉 = H1 |ψn〉 , (13.5)

We now multiply, from the left, by the state 〈φn| to get

〈φn| (En −H0) |ψn〉 = 〈φn|H1|ψn〉 .

Making use of the fact that 〈φn| is an eigenstate of H0, we can solve this equation for theenergy, En, to get

En = εn +〈φn|H1|ψn〉〈φn|ψn〉

= εn + 〈φn|H1|ψn〉 , (13.6)

where we have taken the normalization of the wave functions to be such that

〈φ|ψn〉 = 1 . (13.7)

At the end of this section we will show that this normalization is consistent with ourformulation. From Eq. (13.6) it is clear that a series solution for |ψn〉 in powers of H1

will give us a corresponding series for the eigenvalues En, and provided H1 is small, thisseries will converge. To get a power series for the eigenstates |ψn〉, we write Eq. (13.5) as

(En −∆−H0) |ψn〉 = (H1 −∆) |ψn〉 ,

where the constant ∆, to be determined at a later stage, is chosen such that the operator(En−∆−H0) does not have any zero eigenvalues. We then can rewrite the above equationas

|ψn〉 =1

En −∆−H0

(H1 −∆) |ψn〉 . (13.8)

Making use of the completeness of the eigenstates of H0, i.e.,∑m

|φm〉 〈φm| = 1 ,

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13.1. PERTURBATION THEORY 259

we can write Eq. (13.8) as

|ψn〉 =∑m

|φm〉 〈φm|1

En −∆−H0

(H1 −∆) |ψn〉

= |φn〉 〈φn|1

En −∆−H0

(H1 −∆) |ψn〉

+∑m 6=n|φm〉 〈φm|

1

En −∆−H0

(H1 −∆) |ψn〉 . (13.9)

We now introduce projection operators Pn and Qn defined in terms of the states |φn〉 as

Pn ≡ |φn〉 〈φn| , (13.10)

and

Qn = 1− Pn=

∑m 6=n|φm〉 〈φm| . (13.11)

In terms of these projection operators we can write the eigenstate of the full Hamiltonianas

|ψn〉 = |φn〉1

En −∆− εn(〈φn|H1|ψn〉 −∆) +Qn

1

En −∆−H0

(H1 −∆) |ψn〉

= |φn〉+Qn1

En −∆−H0

(H1 −∆) |ψn〉 , (13.12)

where we have made use of Eq. (13.6) to simplify the first term on the right hand sideof the equation. In Eq. (13.12), we have an integral equation that is equivalent to theSchrodinger equation. We will examine such equations in the next chapter when dealingwith scattering theory.

This result for |ψn〉 can be iterated to give a power series in H1 that is of the form

|ψn〉 = |φn〉+Qn1

En −∆−H0

(H1 −∆) |φn〉

+Qn1

En −∆−H0

(H1 −∆) Qn1

En −∆−H0

(H1 −∆) |φn〉

+ · · · . (13.13)

This series can be formally written as

|ψn〉 =∞∑l=0

Qn

1

En −∆−H0

(H1 −∆)l|φn〉 . (13.14)

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260 CHAPTER 13. APPROXIMATION METHODS FOR BOUND STATES

At this stage we still have to decide on the value of the constant ∆. Here we observethat the eigenvalues En, which are yet to be determined, appear on the right hand sideof Eq. (13.14). To overcome this Gordian knot, we take ∆ to be

∆ = En − εn . (13.15)

In this case the power series in H1 for the wave function becomes

|ψn〉 =∞∑l=0

Qn

1

εn −H0

(H1 + εn − En)l|φn〉

= |φn〉+∑m 6=n|φm〉

1

εn − εm〈φm|H1|φn〉+ · · · . (13.16)

Although En still appears on the right hand side of the above equation, we will see thatin fact its first appearance is in the term O((〈H1〉)3) and higher. With this result at handwe can use Eq. (13.6) to write a power series in H1 for the energy En as

En = εn +∞∑l=0

〈φn|H1

Qn

1

εn −H0

(H1 + εn − En)l|φn〉

= εn + 〈φn|H1|φn〉+∑m6=n〈φn|H1

|φm〉〈φm|εn − εm

H1 |φn〉+ · · ·

= εn + 〈φn|H1|φn〉+∑m6=n

|〈φn|H1|φm〉|2

εn − εm+ · · · . (13.17)

As we expected, our choice of ∆ gives us a power series in which the first three termsdepend only on the eigenvalues and eigenstates of the un-perturbed Hamiltonian H0. Theterm O((〈H1〉)3) will require a knowledge of En, and for that, we use the value determinedby the first three terms in the series. This procedure can also be followed for the higherorder terms in the series. For the ground state, i.e. E0, the sign of the first order correctionis determined by the perturbing potential H1, while the sign of the second order correctionis always negative irrespective of whether H1 is attractive or repulsive. In most practicalproblems, only the first few terms in the series are calculated unless there is a compellingreason for including the higher order terms due to the presence of special effects such ascollective behavior as might be the case in a many-body system. In these cases we usuallysum a certain sub-set of the infinite series to get the effects we are interested in. We willsee one example when considering an infinite Fermi system with potentials that have aninfinite short range repulsive behavior.

Finally, we need to prove that the normalization we chose in Eq. (13.7) is not violated.To prove this we multiply Eq. (13.16) by 〈φn| from the left to get,

〈φn|ψn〉 = 〈φn|φn〉+∞∑l=1

〈φn|Qn

1

εn −H0

(H1 + εn − En)l|φn〉

= 〈φn|φn〉 = 1 , (13.18)

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13.2. THE HELIUM ATOM IN PERTURBATION THEORY 261

where to write the second line we made use of the fact that 〈φn|Qn = 0 because of theorthonormality of the eigenstate of H0.

13.2 The Helium Atom in Perturbation Theory

As a first example of perturbation theory we will consider the ground state energy of thehelium (He) atom. In this case we will consider the He atom to consist of two spinlesselectrons in motion about an infinitely heavy nucleus.1 The Hamiltonian for the He atom,in units2 in which h = me = e = 1, can be written as

H = H0 +H1 . (13.19)

| r1 - r2 |→ →

r2→r1

Figure 13.1: Definition of the coordinates for the helium atom.

The un-perturbed Hamiltonian includes the kinetic energy of the two electrons and theattraction due to the nucleus, i.e., H0 is the sum of two hydrogen-like Hamiltonians, onefor each of the electrons, i.e.,

H0 = −1

2

(∇2

1 +∇22

)− Z

(1

r1

+1

r2

)=

(−1

2∇2

1 −Z

r1

)+(−1

2∇2

2 −Z

r2

). (13.20)

The perturbation in this case is the Coulomb repulsion between the two electrons, i.e.,

H1 =1

|~r1 − ~r2|. (13.21)

Since the un-perturbed Hamiltonian H0 is written as the sum of the Hamiltonian for thetwo electrons, then the eigenstates of H0 are the product of two eigenstates, one for each

1The ratio of the mass of the electron to the mass of the He nucleus is ≈ 8× 104.2These units in which the Bohr radius a = 1 are commonly used by Atomic Physicists (see Appendix).

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262 CHAPTER 13. APPROXIMATION METHODS FOR BOUND STATES

of the electrons, i.e.

〈~r1;~r2|n1`1m1; n2, `2m2〉 = 〈~r1|n1`1m1〉 〈~r2|n2`2m2〉= φn1`1m1(~r1)φn2`2m2(~r2) , (13.22)

where φn`m(~r ) is the wave function of hydrogen atom with charge Z = 2 on the nucleus.For the ground state we have the two electrons in the lowest energy state of the hydrogenicatom. In this case the un-perturbed wave function is given by (see Table 9.2 with a = 1)

〈~r1;~r2|100; 100〉 =Z3

πe−Z(r1+r2) . (13.23)

With this wave function we can calculate the first order correction to the energy, ∆E(1),which is given by

∆E(1) = 〈100; 100|H1|100; 100〉

=∫d3r1 d

3r2|φ100(~r1)|2 |φ100(~r2)|2

|~r1 − ~r2|. (13.24)

To evaluate this integral, we need to expand |~r1 − ~r2|−1 in terms of spherical harmonicsas3

|~r1 − ~r2|−1 =∑`

(r`<r`+1>

)P`(cos θ)

=∑`m

2`+ 1

(r`<r`+1>

)Y`m(r1)Y ∗`m(r2) , (13.25)

where θ is the angle between the vectors ~r1 and ~r2, and

(r`<r`+1>

)=

r`1r`+1

2

for r1 < r2

r`2r`+1

1

for r2 < r1

. (13.26)

With this expansion, and using the orthogonality of the spherical harmonics, we can writethis first order correction to the energy as4

∆E(1) = 16Z6

∞∫0

dr1 r21 e−2Zr1

1

r1

r1∫0

dr2 r22 e−2Zr2 +

∞∫r1

dr2 r2 e−2Zr2

3For a proof of this result see J. D. Jackson Classical Electrodynamics J. Wiley & Sons (1962).4To evaluate the integrals in Eq. (13.27), we have made use of the fact that∫

dxxm eax = eaxm∑k=0

(−1)km!xm−k

(m− k)!ak+1.

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13.3. ANHARMONIC LINEAR OSCILLATOR 263

=5

8Z . (13.27)

Therefore, the ground state energy of the helium atom to first order in perturbation theoryis given by

E0 = −Z2

2− Z2

2+

5

8Z = −Z

(Z − 5

8

). (13.28)

As expected, the repulsion between the electrons reduces the binding energy of the system.The ionization energy, i.e., the energy required to remove one electron from the atom, isnow given by

I = −E0(He) + E0(He+) = −−Z

(Z − 5

8

)+Z2

2

=Z

2

(Z − 5

4

)=

3

4for Z = 2 . (13.29)

To get this ionization energy in eV, we have to multiply the above result by 2R =27.2116 eV. This gives us the ionization of He to be I = 20.4087 eV, which is to becompared with the experimental value of 24.6401 eV.

13.3 Anharmonic Linear Oscillator

In many problems in physics we are interested in the development of a system around anequilibrium state. In all such systems the lowest order approximation to the behavior ofthe system can be described in terms of a simple harmonic oscillator. This, as we willsee, describes the vibrational states of both molecules and nuclei. The advantage of usinga harmonic oscillator is the fact that we can write an exact solution to the harmonicoscillator as we demonstrated in Chapter 1. In the present section, we would like toconsider problems that require an extension of the analysis beyond the simple harmonicoscillator and include the first anharmonic term. In the process we will re-derive thesolution of the harmonic oscillator in occupation number representation.

Let us consider a molecule consisting of two atoms. The potential between the twoatoms is presented graphically in Figure 13.2. In the ground state, the two atoms inthe molecule have an equilibrium separation r ≈ R. If we are interested in the low lyingstates of the molecule, then these correspond to the small oscillation about the equilibriumseparation. In this case we can write the potential between the two atoms in terms of aTaylor expansion about the equilibrium separation R, i.e.,

V (r) = V (R) + (r −R)

(dV

dr

)r=R

+1

2!(r −R)2

(d2V

dr2

)r=R

+1

3!(r −R)3

(d3V

dr3

)r=R

+ · · · . (13.30)

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264 CHAPTER 13. APPROXIMATION METHODS FOR BOUND STATES

V(r)

rR

Figure 13.2: The potential between two atoms. Here R is the equilibrium distance.

Because the potential has a minimum at r = R, the second term on the right handside is zero since

(dVdr

)r=R

= 0. Furthermore, we can take the zero of our energy to be

V (R). In this way, the lowest order approximation to the potential between two atomsis represented by a harmonic oscillator provided the excitation above the ground state isnot large. Thus, the lowest order Hamiltonian can be written as

H0 =p2

2m+

1

2mω2x2 , (13.31)

where we have taken x = (r−R), and considered the radial variable as a one-dimensionalproblem. The frequency of the oscillator can be traced back to the second derivative ofthe potential, i.e.,

ω2 =1

m

(d2V

dr2

)r=R

This Hamiltonian gives a spectrum which is of the form

Eν = hω(ν +

1

2

). (13.32)

As a first correction to the above Hamiltonian, we have a term that is cubic in thedisplacement from equilibrium, i.e.,

H1 =mω2

2bx3 , (13.33)

where the strength of this perturbation is proportional to the third derivative, i.e.,

mω2

2b=

1

3!

(d3V

dr3

)r=R

.

To determine the effect of this anharmonic term on the vibrational spectrum, we need todetermine the eigenstates of H0. Then for H1 small, we can use perturbation theory tocalculate the shift in energy due to the anharmonicity.

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13.3. ANHARMONIC LINEAR OSCILLATOR 265

13.3.1 Occupation Number Representation

Although we have solved the Schrodinger equation for the one-dimensional harmonicoscillator in an earlier chapter, we propose here to get the eigenvalues and eigenstates ofthe harmonic oscillator using the same algebraic method used to find the representationof the angular momenta in Chapter 3. Let us define the operator a and a† as

a =i√

2mhω(p− imωx) a† = − i√

2mhω(p+ imωx) , (13.34)

where x and p are the position and momentum of the particle. We now can write theHamiltonian for the one-dimensional harmonic oscillator in terms of a and a† as

H0 = hω(a†a+

1

2

). (13.35)

Making use of the commutation relation of the position and momentum operators, i.e.[x, p] = ih, we can write the commutation relation between the new operators a and a† as[

a, a†]

= 1 [a, a] = 0 =[a†, a†

]. (13.36)

With these commutation relations we can determine the commutation relations of a anda† with the Hamiltonian H0 to be[

H0, a†]

= hω a† [H0, a] = −hω a . (13.37)

Since the Hamiltonian is a linear Hermitian operator, we can assume that it has eigenstates|ν〉 such that

H0 |ν〉 = εν |ν〉 , (13.38)

and then using the commutation relation of the Hamiltonian with the operator a†, we canwrite [

H0, a†]|ν〉 = H0 a

† |ν〉 − a†H0 |ν〉 = hω a† |ν〉 .

This equation can be written as an eigenvalue problem of the form

H(a† |ν〉

)= (εν + hω)

(a† |ν〉

). (13.39)

In other words, if the state |ν〉 has energy εν , then the state a† |ν〉 has energy (εν + hω).This suggests that the operator a†, acting on a state with a given energy, gives us a state ofhigher energy, and the energy of the new state is always hω greater than the original state.Thus, the operator a† seems to have a similar property to the operator J+ encountered inour discussion of the representation of the angular momentum in Chapter 3. In a similarmanner we can show, using the commutation of H0 with a, that

H0 (a |ν〉) = (εν − hω) |ν〉 . (13.40)

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266 CHAPTER 13. APPROXIMATION METHODS FOR BOUND STATES

Thus, the operators a and a† lower or raise the energy of a state by an amount hω. In thisway, given one state of the system and its corresponding energy, these operators allowus to generate all the other states and their corresponding energies. This situation isidentical to that encountered in Chapter 3, where the operators J+ and J− raised andlowered the projection of the angular momentum along the z-axis by h.

Let us now assume that there is a lowest energy state denoted by |0〉. 5 Since bydefinition it is the lowest energy state, we have that

a |0〉 = 0 . (13.41)

This lowest energy state is an eigenstate of the Hamiltonian with eigenvalue hω/2 since

H0 |0〉 =1

2hω |0〉 . (13.42)

We now can generate all the other states by the repeated application of the operator a†.Thus we have

|1〉 ≡ N1a†|0〉 and H0|1〉 = hω

(1 +

1

2

)|1〉

|2〉 ≡ N2

(a†)2|0〉 and H0|2〉 = hω

(2 +

1

2

)|2〉

......

|ν〉 ≡ Nν

(a†)ν|0〉 and H0|ν〉 = hω

(ν +

1

2

)|ν〉 ,

where Nν is the normalization which is to be determined. We now have that the state |ν〉has energy εν with

εν = hω(ν +

1

2

). (13.43)

This result is identical to that obtained by solving the Schrodinger equation in coordinaterepresentation. To determine the normalization of our states, let us write the state |ν〉 as

|ν〉 = Nν

(a†)ν|0〉

= Nν a†(a†)ν−1

|0〉

=Nν

Nν−1

a†|ν − 1〉 .

Then since the state |ν〉 is normalized, we have

1 = 〈ν|ν〉 =∣∣∣∣ Nν

Nν−1

∣∣∣∣2〈ν − 1|aa†|ν − 1〉

=∣∣∣∣ Nν

Nν−1

∣∣∣∣2〈ν − 1|

1 + a†a|ν − 1〉 .

5All operators in quantum mechanics should have at least one bound on their spectrum, i.e. either anupper or lower bound. For the Hamiltonian operator we make sure it has a lower bound on its spectrum.

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13.3. ANHARMONIC LINEAR OSCILLATOR 267

We now can write the operator a†a in terms of the Hamiltonian H0, and then make useof the fact that the state |ν − 1〉 is an eigenstate of the Hamiltonian with eigenvaluehω(ν − 1/2), to write the above expression as

1 =∣∣∣∣ Nν

Nν−1

∣∣∣∣2 ν .We thus have that

Nν−1

=1√ν

which allows us to write the state |ν〉 as

|ν〉 =1√νa†|ν − 1〉 . (13.44)

In a similar manner, we can show that

|ν − 1〉 =1√νa|ν〉 . (13.45)

Finally, we can write the state |ν〉 in terms of the lowest energy state as

|ν〉 =1√ν!

(a†)ν|0〉 . (13.46)

In this way we have generated all the eigenstates and eigenvalues of the Hamiltonianby writing H0 in terms of the operators a and a† and making use of the algebra ofthe commutation relation of these operators. Here we note that since the position andmomentum operators can be written in terms of a and a†, then any operator that is afunction of x and p can also be written in terms of a and a†. This in turn implies thatthe space we have defined is closed under the action of any operator that is a functionof x and p, and to that extent the eigenstates |ν〉 form a complete set. We will see ina later chapter how we can use this approach to quantize vibrational motion of nucleiand molecules. One final observation is the fact that the Hamiltonian in this case playsthe same role as Jz played in the angular momentum case. This algebraic method canbe used to solve a number of other problems, the most famous being the solution of theSchrodinger equation for the Coulomb potential. In fact it was first solved using thisalgebraic method by Pauli.6

6W. Pauli, Z. Phys. 36, 336 (1926); See also L. I. Schiff Quantum Mechanics McGraw-Hill, 3rd Editionp.238 (1968).

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268 CHAPTER 13. APPROXIMATION METHODS FOR BOUND STATES

13.3.2 Lowest Order Contribution

Having determined the eigenenergies and corresponding eigenstates of H0, we are in aposition to examine the contribution of the anharmonic term to the energy and wavefunction of the vibrational motion of a molecule. To calculate the matrix elements of theperturbing Hamiltonian H1, we need to write this perturbation in terms of the operatorsa and a†. We have that

a+ a† =2mω√2mhω

x ,

and therefore

x =

√h

2mω

(a+ a†

). (13.47)

Using this expression for x, we can calculate the matrix element of x between the eigen-states of H0, i.e.

〈ν ′|x|ν〉 =

√h

2mω

〈ν ′|a|ν〉+ 〈ν ′|a†|ν〉

=

√h

2mω

√ν〈ν ′|ν − 1〉+

√ν + 1〈ν ′|ν + 1〉

=

√h

2mω

√ν δν′,ν−1 +

√ν + 1 δν′,ν+1

. (13.48)

We observe at this stage that if the perturbing Hamiltonian was proportional to x, therewould be no first order contribution to the energy from this perturbation, and we wouldneed to go to second order to get a shift in the energy. Furthermore, this shift wouldlower the energy of the ground state as pointed out previously. With this result as anillustration of how we may determine the contribution of the perturbation to the energyand wave function of the system, we can proceed to write H1 in terms of our raising andlowering operators as

H1 =mω2

2bx3 = hωγ(a+ a†) (a+ a†) (a+ a†)

= hωγ

3aa†a+ 3a†aa† + a3 + (a†)3, (13.49)

where we have used the commutation relations of the operators a and a† to write thesecond line of Eq. (13.49). Here the constant γ is given by

γ =1

4b

√h

2mω, (13.50)

From the structure of H1, as given in Eq. (13.49) in term of the creation and annihilationoperators a† and a, it is clear that the contribution to the energy of the anharmonic

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13.3. ANHARMONIC LINEAR OSCILLATOR 269

oscillator, in first order perturbation theory, is zero, i.e., 〈ν|H1|ν〉 = 0. Thus the lowestorder correction to the energy comes in second order perturbation theory, and is of theform ∑

ν ′

ν ′ 6= ν

|〈ν ′|H1|ν〉|2

εν − εν′.

In this sum the only non-zero matrix elements are;

〈ν + 1|H1|ν〉 = 3hωγ〈ν + 1|a†aa†|ν〉 = 3hωγ(ν + 1)3/2 ( 13.51a)

〈ν − 1|H1|ν〉 = 3hωγ〈ν − 1|aa†a|ν〉 = 3hωγν3/2 ( 13.51b)

〈ν + 3|H1|ν〉 = hωγ〈ν + 3|(a†)3|ν〉 = hωγ [(ν + 1)(ν + 2)(ν + 3)]1/2 ( 13.51c)

〈ν − 3|H1|ν〉 = hωγ〈ν − 3|a3|ν〉 = hωγ [ν(ν − 1)(ν − 2)]1/2 . ( 13.51d)

This means that our energy is given by

Eν = hω(ν +

1

2

)+

∑ν ′

ν ′ 6= ν

|〈ν|H1|ν ′〉|2

hω(ν − ν ′). (13.52)

Using Eq. (13.51), we can write Eq. (13.52) as

Eν = hω(ν +

1

2

)+ hωγ2

1

3|〈ν − 3|a3|ν〉|2 + 9|〈ν − 1|aa†a|ν〉|2

−9|〈ν + 1|a†aa†|ν〉|2 − 1

3|〈ν + 3|(a†)3|ν〉|2

= hω

(ν +

1

2

)+ hωγ2

1

3ν(ν − 1)(ν − 2) + 9ν3 − 9(ν + 1)3

−1

3(ν + 1)(ν + 2)(ν + 3)

= hω

(ν +

1

2

)− hωγ2

(30ν2 + 30ν + 11

). (13.53)

This is the energy of the anharmonic oscillator to second order in perturbation theory. Wenote here that this result is valid for ν ≥ 3. For ν < 3, we have to take into considerationthe fact that the state |ν〉 is zero for ν < 0 in our calculation of the matrix elements of H1,e.g., for the calculation of the shift in the ground state energy ν = 0, and the contributionto the energy shift, in second order perturbation theory, of the terms proportional to a3

and aa†a in Eq. (13.49) are zero.The wave function to the same order in perturbation theory can be written as

|ψν〉 = |φν〉+∑ν ′

ν ′ 6= ν

|φν′〉 〈ν′|H1|ν〉

hω(ν − ν ′)

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270 CHAPTER 13. APPROXIMATION METHODS FOR BOUND STATES

= |φν〉+ γ|φν−3〉

1

3

√ν(ν − 1)(ν − 2) + |φν−1〉3ν3/2

−|φν+1〉3(ν + 1)3/2 − |φν+3〉1

3

√(ν + 1)(ν + 2)(ν + 3)

. (13.54)

V(r)

r

Figure 13.3: A comparison of the harmonic oscillator potential with the potential betweentwo atoms.

Let us see if this result agrees with what we intuitively expect. If we take the Hamiltonianfor the two atoms to be H0, then we have taken a harmonic oscillator about r = R, whereR is the equilibrium distance between the two atoms. However, the potential V (r) iswider than the harmonic oscillator, and thus the addition of the contribution of H1 to theenergy should lower the eigenstates, which is exactly what we derived, and is illustratedin Figure 13.3.

13.4 Van der Waal’s Potential

Let us consider the force between two hydrogen like atoms, i.e., two atoms each havingone valence electron which plays an active role in giving rise to the potential betweenthe two atoms. If the distance between the centers of the two atoms is much larger thanthe distance of each electron from its nucleus, i.e. R r1 and r2, then the energy ofthe system as a function of R will be the interatomic potential. The Hamiltonian for thesystem can now be written as

H = H0 +H1 , (13.55)

where the un-perturbed Hamiltonian, in units of h = m = e = 1, is given by

H0 = −1

2

(∇2

1 +∇22

)−(

1

r1

+1

r2

), (13.56)

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13.4. VAN DER WAAL’S POTENTIAL 271

A

BR - r1

→ →

R→

r2→

r1→

Figure 13.4: Definition of the coordinates for the hydrogen molecule.

i.e., it is the sum of the Hamiltonians for the two individual atoms, while the perturbingHamiltonian is given by

H1 =1

R+

1

r12

− 1

r1B

− 1

r2A

. (13.57)

This includes the interaction of the electron and nucleus of one atom with the electronand nucleus of the other atom (see Figure 13.4). Since we are interested in the long rangebehavior of the potential between the two atoms, i.e., when R r1 and r2, we will writethe perturbing Hamiltonian in terms of R, r1, and r2 as

H1 =1

R+

1

|~R− ~r1 + ~r2|− 1

|~R− ~r1|− 1

|~R + ~r2|

=1

R

1 +

[1 +

2R · (~r2 − ~r1)

R+

(~r2 − ~r1)2

R2

]−1/2

−[1− 2

R · ~r1

R+r2

1

R2

]−1/2

−[1 + 2

R · ~r2

R+r2

2

R2

]−1/2 , (13.58)

and then take the limit, as the distance between the two atoms becomes larger than thesize of the individual atoms, i.e., R r1 and r2. To get the first non-zero term in theseries in ri

R, i = 1, 2 we expand the terms in square bracket to get7

H1 ≈ −1

R3

[3(R · ~r1

) (R · ~r2

)− (~r1 · ~r2)

]. (13.59)

The first term on the right hand side corresponds to the interaction of two dipoles. In factwhat is happening is that the lowest order interaction corresponds to the situation when

7Here we have made use of the fact that

(1 + x)−1/2 = 1− x

2+

3x2

8− 5x3

16+ · · · ,

to derive Eq. (13.59).

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272 CHAPTER 13. APPROXIMATION METHODS FOR BOUND STATES

each atom gets polarized by the field of the other atom. This results in a configurationwhere we have two dipoles a distance R apart. We can now proceed to calculate theperturbation due to H1 as given in Eq. (13.59), since we know the eigenstates of H0,which are hydrogen-like wave functions. However, to get the R dependence of the potentialbetween the two atoms, let us replace the two atoms by two one-dimensional harmonicoscillators a distance R apart. This will give us the correct R dependence, and at thesame time simplify the computation. Taking the vector ~R to be along the x-axis, ourperturbing Hamiltonian becomes

H1 = − 2

R3x1 x2 , (13.60)

while the un-perturbed Hamiltonian H0 is modeled to be

H0 =

(p2

1

2m+

1

2mω2

1x21

)+

(p2

2

2m+

1

2mω2

2x22

). (13.61)

The eigenstates of H0 can be written as (see previous section),

|ν1; ν2〉 = |ν1〉 |ν2〉 . (13.62)

This factorization is possible only because the Hamiltonian H0 is the sum of two harmonicoscillator Hamiltonians with no interaction between them. If we take ω1 = ω2 = ω thenthe eigenvalues of H0 are

H0|ν1; ν2〉 = hω (ν1 + ν2 + 1) |ν1; ν2〉 ≡ εν1ν2|ν1; ν2〉 . (13.63)

Taking the two atoms to be in their ground state, i.e. ν1 = ν2 = 0, the contribution tothe energy from first order perturbation theory is given by

〈0; 0|H1|0; 0〉 = − 2

R3〈0|x1|0〉 〈0|x2|0〉 = 0 . (13.64)

Thus the lowest order correction to the ground state energy comes from second orderperturbation theory, and this is given by

∆E(2) =∑ν1ν2

〈0; 0|H1|ν1; ν2〉 〈ν1; ν2|H1|0; 0〉ε00 − εν1ν2

=(

2

R3

)2 ∑ν1ν2

|〈0|x1|ν1〉|2 |〈0|x2|ν2〉|2

ε00 − εν1ν2

= −(

2

R3

)2 ∑ν1ν2

|〈0|x1|ν1〉|2 |〈0|x2|ν2〉|2

hω(ν1 + ν2). (13.65)

Making use of the matrix element of x as given in Eq. (13.48), we can reduce the sum tojust one term corresponding to ν1 = ν2 = 1. This gives us

∆E(2) = −1

2hω

(1

mω2

)2 1

R6= − A

R6. (13.66)

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13.5. THE VARIATIONAL METHOD 273

The above result gives the correct R dependence of the potential between two atoms. Toget a more accurate determination of A, we need to use an atomic wave function for theelectrons rather than the harmonic oscillator wave function. However, in the atomic case,the sum in the calculation of ∆E(2) is infinite and needs to be truncated.

13.5 The Variational Method

An alternative approximation to perturbation theory is the variational method, and likeperturbation theory it has its advantages and disadvantages. Its main advantages are:

1. The method does not require that we write the Hamiltonian as the sum of two parts,e.g., H = H0 +H1, with H1 small and the eigenvalues and eigenstates of H0 known.

2. In many problems the variational results are better than the result we get withperturbation theory with little additional effort.

On the other hand, the disadvantages of the variational method are:

1. The computation can be very difficult if one requires a good result.

2. The starting point and success of the calculation depends to a large extent on howclever we can be. This last point becomes clear when we discuss the method.

Theorem: The ground state energy E0 of a system with a Hamiltonian Hsatisfies the relation

E0 ≤ 〈ψ0|H|ψ0〉 , (13.67)

where ψ0 is any function that is normalized, i.e.,

〈ψ0|ψ0〉 = 1 . (13.68)

Proof: Let us introduce the eigenstates of H as |φn〉, i.e.,

H|φn〉 = En|φn〉 . (13.69)

Since the states |φn〉 form a complete set of states, we can expand |ψ0〉 interms of the eigenstates of H, i.e.,

|ψ0〉 =∑n

an|φn〉 . (13.70)

Then the requirement that |ψ0〉 be normalized means

〈ψ0|ψ0〉 = 1 =∑nm

a∗n am〈φn|φm〉

=∑n

|an|2 . (13.71)

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274 CHAPTER 13. APPROXIMATION METHODS FOR BOUND STATES

We now can write the matrix element of H as

〈ψ0|H|ψ0〉 =∑nm

a∗n 〈φn|H|φm〉 am

=∑n

|an|2En . (13.72)

If we take E0 < E1 < E2 · · ·, then we have that

〈ψ0|H|ψ0〉 =∑n

En |an|2

≥∑n

E0 |an|2 = E0 . (13.73)

Therefore we have that

E0 ≤ 〈ψ0|H|ψ0〉 . (13.74)

The fact that E0 ≤ 〈ψ0|H|ψ0〉 means that if we choose a |ψ0〉 such that 〈ψ0|H|ψ0〉 is aminimum, then we have an upper bound on E0, and as we improve our wave function|ψ0〉, we get better bounds on E0.

To achieve this upper bound, we take ψ0 to be a function of a set of parameters, e.g.

|ψ0〉 = |ψ0(α, β, γ, · · ·)〉 . (13.75)

We then define the function J(α, β, γ, · · ·) as

J(α, β, γ, · · ·) ≡ 〈ψ0|H|ψ0〉 . (13.76)

We now can find the minimum of the function J(α, β, γ, · · ·) with respect to the parametersα, β, γ, · · · by taking

∂J

∂α= 0 ,

∂J

∂β= 0 , · · · . (13.77)

This method of getting the upper bound on the ground state energy E0 is known as Ritzvariational method.

So far we have considered the ground state energy of the system described by theHamiltonian H. We can in principle extend the method to determine bounds on theexcited states of the system. Thus to calculate the first excited state, we construct afunction ψ1 such that it is orthogonal to the ground state ψ0, i.e.,

〈ψ1|ψ0〉 = 0 , (13.78)

and this wave function is normalized so that

〈ψ1|ψ1〉 = 1 . (13.79)

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13.6. THE HELIUM ATOM - VARIATIONAL METHOD 275

Then we can calculate an upper bound on the first excited state as

E1 ≤ 〈ψ1|H|ψ1〉 . (13.80)

Here we note that in calculating an upper bound on a given state, we have to make surethat the symmetry of our trial function ψn is the same as the eigenstate corresponding tothe eigenvalue we are calculating the upper bound on. For example, if the ground stateis a state of zero angular momentum and positive parity, then our trial function shouldcorrespond to a positive parity state with zero angular momentum. If we ignore this andtake a wave function with negative parity, we would be calculating an upper bound onthe lowest negative parity state of the system which could be an excited state and not aground state.

13.6 The Helium Atom - Variational Method

To illustrate the use of the variational method, and at the same time compare the methodof perturbation theory with the variational approach, we consider the helium atom fora second time. Here again, we take the helium atom as a system of two electrons withno spin in order to avoid the problem of angular momentum addition. In this case weexpect the wave function for the two electrons to be the product of two hydrogen-likewave functions with the added feature that the electron does not see the full charge ofthe nucleus because of the screening by the other electron. This means we will take thecharge in the trial wave function to be a variational parameter. In other words, the trialfunction for the ground state is taken as

ψ0(~r1, ~r2) =β3

πe−β(r1+r2) . (13.81)

This wave function, which is properly normalized, assumes that the two electrons are in astate with angular momentum zero, i.e., s-state. Furthermore, each electron sees a chargeβ on the nucleus. To optimize our wave function we have to minimize

J(β) = 〈ψ0|H|ψ0〉 , (13.82)

with respect to β. Here, H is the full Hamiltonian for the helium atom as given byEqs. (13.19) - (13.21). To minimize J(β) we write it as the sum of three terms, i.e.,

J(β) = J1(β) + J2(β) + J3(β) , (13.83)

where J1(β) is the kinetic energy of the system and is given by

J1(β) = −1

2

∫d3r1 d

3r2 ψ∗0(~r1, ~r2)

(∇2

1 +∇22

)ψ0(~r1, ~r2)

= β2 . (13.84)

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276 CHAPTER 13. APPROXIMATION METHODS FOR BOUND STATES

To perform this integral we have made use of the fact that the trial wave function has noangle dependence, and therefore if we take ∇2

i in spherical coordinates, the only contri-bution comes from the radial part of ∇2

i which is given by

1

r2i

∂ri

(r2i

∂ri

).

This result allows us to do the angle integration in a trivial manner since the integrandis angle independent. The radial integral then is a standard integral to be found in anytable of integrals.8 Here, we observe that the kinetic energy is as usual a positive quantity.

The second term on the right hand side of Eq. (13.83) corresponds to the attractionof the electrons by the nucleus. This integral is given by

J2(β) = −Z∫d3r1 d

3r2 ψ∗0(~r1, ~r2)

(1

r1

+1

r2

)ψ0(~r1, ~r2)

= −2Zβ . (13.85)

Here again the integrand has no angle dependence allowing us to do the angle integration,while the radial integrals are standard integrals. This term, as we would expect, is negativebecause of the attraction between the electrons and the nucleus.

Finally, the third term on the right hand side of Eq. (13.83) gives the contribution dueto the repulsion between the two electrons, and therefore should be positive. It is givenby

J3(β) =∫d3r1 d

3r2 ψ∗0(~r1, ~r2)

1

|~r1 − ~r2|ψ0(~r1, ~r2)

=5

8β . (13.86)

This integral was evaluated previously and given in Eq. (13.27). We now combine theresults of Eqs. (13.84) to (13.86) to write

J(β) = β2 − 2Zβ +5

8β . (13.87)

To determine the minimum of this function with respect to β, we differentiate it and setthe derivative to zero, i.e.

∂J

∂β= 2(β − Z) +

5

8= 0

8We have that∞∫

0

xn e−ax dx =Γ(n+ 1)an+1

(n > −1, a > 0) .

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13.6. THE HELIUM ATOM - VARIATIONAL METHOD 277

or

β = Z − 5

16. (13.88)

This is the effective charge on the nucleus that the electron sees, and as expected it isalways less than the full charge on the nucleus. With this value of β we now can calculatethe upper bound on the energy to be

E0 ≤ EU = J(β)|β=Z− 516

= −(Z2 − 5

8Z +

25

256

). (13.89)

If we compare this result with the result of first order perturbation theory E(1)0 as given

in Eq. (13.28), we find that

EU = E(1)0 −

25

256

indicating that the variational result gives a better upper bound on the ground stateenergy. Since the wave function used in the two cases is of the same form, the advantageof the variational method is the use of the effective charge β = Z − 5

16in the variational

calculation.The ionization energy of He is now given by

I = −EU(He) + EU(He+)

=1

2

(Z2 − 5

4+

25

128

). (13.90)

Table 13.1: Ionization energy in atomic units for two electron systems.

Atom or Ion Experimental Perturbation Variationalresults theory method

He 0.9055 0.75 0.85Li+ 2.7798 2.62 2.72

Be++ 5.655 5.5 5.60C++++ 14.4070 14.25 14.35

In Table 13.1 we present the energy needed to remove an electron from a two electronatom. Here we compare the result of experiment, perturbation theory, and the variationalmethod. We find that, in general, the variational method gives a result that is in betteragreement with experiment than the result of first order perturbation theory. The result ofperturbation theory can be improved if we include the higher order terms in the expansion,

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278 CHAPTER 13. APPROXIMATION METHODS FOR BOUND STATES

but then we have a more difficult computational problem. We also can improve thevariational result by taking a better trial wave function. In particular, in the aboveanalysis we made use of a product wave function that was very simple and had only oneparameter. We could take as a trial wave function the sum of such product, e.g.,

ψ0(~r1, ~r2) =∑nn′

Ann′φn00(~r1, β)φn′00(~r2, β) , (13.91)

where Ann′ are variational parameters adjusted to minimize the upper bound on theenergy, and φn00(~r, β) are hydrogenic wave functions with an effective charge β on thenucleus, with β as a parameter. Another direction we can take to improve our trial wavefunction is to deviate from the product ansates by multiplying the wave function we usedby a function of the relative coordinates of the two electrons. All these changes can,in principle, improve the upper bound on the ground state energy at the cost of moreparameters and in general, more complicated calculations.

13.7 Degenerate Perturbation Theory

In our discussion of perturbation theory we have implicitly assumed that the eigenstates ofthe un-perturbed HamiltonianH0 all have different energies, i.e., there are no degeneracies.In particular, we have assumed that there are no degeneracies that are removed by theperturbation H1. On the other hand, all problems with spherical symmetry give us a wavefunction that is proportional to the spherical harmonics, i.e., ψ(~r) = R`(r)Y`m(r), andthe corresponding eigenenergies do not depend on the quantum number m. This impliesthat for systems with spherical symmetry we have at least a degeneracy of (2` + 1). Toappreciate the problem with perturbation theory in the presence of this degeneracy, let usconsider the case when we have two eigenstates of H0, |a〉 and |b〉, with the same energyεa = εb = ε. When we turn on the perturbation one of two things can take place:

1. If 〈a|H1|b〉 = 0, then the perturbation may remove the degeneracy to the extentthat to first order in perturbation theory we have

Ea = εa + 〈a|H1|a〉 = ε+ 〈a|H1|a〉Eb = εb + 〈b|H1|b〉 = ε+ 〈b|H1|b〉

with Ea 6= Eb. Here if we could turn on the interaction, H1, by slowly increasingits strength (e.g. H1 → λH1, with λ ≤ 1, and change λ from zero to 1), we couldsee that the degenerate states |a〉 and |b〉 slowly develop to have different energies.This is illustrated in Figure 13.5. In this case we don’t have a problem in using theresults of perturbation theory presented in section 5.1.

2. In this case we have that H1 is not diagonal in our basis, i.e.,

〈a|H1|b〉 6= 0 ,

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13.7. DEGENERATE PERTURBATION THEORY 279

H0 H0 + H1

| a > , | b >| b >

| a >

Figure 13.5: The splitting in the case when the perturbation H1 is diagonal in the basisconsidered.

and we get the situation where the states |a〉, and |b〉 are split by the perturbationinto two different states |A〉 and |B〉, such that

|A〉 = α |a〉+ β |b〉|B〉 = γ |a〉+ δ |b〉

H0

| a > , | b >

H0 + H1

| A > = α | a > + β | b >

| B > = γ | a > + δ | b >

Figure 13.6: The splitting in the case when the perturbation H1 is not diagonal in thebasis considered.

In this case the perturbation is turned on suddenly, and we have the states |A〉 and|B〉 as soon as the perturbation is present irrespective of its strength. To avoid that,we have to choose our basis, which are eigenstates of H0, such that H1 is diagonalin this basis, i.e., we have to diagonalize the matrix(

〈a|H1|a〉 〈a|H1|b〉〈b|H1|a〉 〈b|H1|b〉

).

Furthermore, since |a〉 and |b〉 are eigenstates of H0 with the same eigenvalue ε,then |A〉 and |B〉 are also eigenstates of H0 with eigenvalue ε.

We may then conclude from the above example that after dividing the Hamiltonian intothe un-perturbed part H0, and the perturbation H1, the next important decision is tochoose the basis for the eigenstates of H0 such that the perturbing Hamiltonian H1 isdiagonal in this basis. If that is not possible, then we need to actually diagonalize ourHamiltonian in the basis we have chosen. This diagonalization gives us not only the eigen-values including the perturbation to first order, but also gives us a new set of eigenstatesto use in calculating the higher order terms in the perturbation series.

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280 CHAPTER 13. APPROXIMATION METHODS FOR BOUND STATES

13.8 The Spin-Orbit Interaction

As an example of a physical problem where we have a degenerate basis for the eigenstatesof H0, and where the perturbation removes part of this degeneracy, let us consider thecontribution of the spin-orbit interaction to the energy spectrum of the hydrogen atom.

If the electron in the hydrogen atom was stationary relative to the proton, then the onlyinteraction would be the Coulomb force. However, since the electron is moving relativeto the proton, then by sitting on the electron we observe a magnetic field generated bythe motion of the proton. This magnetic field is given by9

~B = e(µ0

)~r × ~vr3

, (13.92)

where µ0 is the permeability of free space, v is the velocity of the proton relative to theelectron, and e is the charge on the proton. This expression can be rearranged so that

~B =e

m

(µ0

)~r × ~pr3

=e

m

µ0

~L

r3. (13.93)

However, we have thate2

r3=

1

r

d

dr

(−e

2

r

)=

4πε0r

dVcdr

,

where ε0 is the permittivity of free space, and Vc is the Coulomb potential between theproton and electron. We now can write the magnetic field as

~B =µ0ε0me

1

r

dVcdr

~L =1

mec2

1

r

dVcdr

~L . (13.94)

where c = 1√µ0ε0

is the velocity of light in vacuum. The interaction between the magnetic

moment of the electron, ~M , and the magnetic field generated by the proton will be aninteraction present in the hydrogen atom in addition to the Coulomb interaction. Thisadditional interaction is given by

VLS = − ~M · ~B=

e

m~S · ~B

=1

m2c2

1

r

dVcdr

(~S · ~L

), (13.95)

9We have used MKSA system of units in the derivation of the spin-orbit interaction so that the studentcan make use of results derived in a standard course in electromagnetic theory.

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13.8. THE SPIN-ORBIT INTERACTION 281

where ~S is the spin of the electron. This derivation, which is non-relativistic, is off bya factor of two when compared to the relativistic result, which is the correct one to use,i.e., the spin orbit interaction is given by

VLS =1

2m2c2

1

r

dVcdr

~L · ~S . (13.96)

With this result in hand we can write the Hamiltonian for the hydrogen atom as

H = H0 +H1 , (13.97)

where the un-perturbed Hamiltonian is the one discussed in Chapter 1 and referred to asthe Coulomb Hamiltonian, i.e.,

H0 = − h2

2m∇2 − e2

r, (13.98)

and the perturbation H1 = VLS, with Vc given by

Vc = −e2

r. (13.99)

Since this Hamiltonian for the hydrogen atom now includes the spin of the electron, thecorresponding eigenstate of H0 will need to include the spin of the electron if we are to useit in perturbation theory to calculate the correction to the spectrum due to the spin-orbitinteraction. In other words, the wave function is of the form

〈~r; ξ|n`m`; sms〉 = Rn`(r)Y`m`(r)χsms(ξ) , (13.100)

where χsms(ξ) is the spin wave function with ξ playing the role of the coordinate.10 Sincethis wave function is an eigenstate of H0 with the corresponding eigenvalue depending onn but not ` m` or the spin quantum numbers, we have in general more than one eigenstatefor a given eigenvalue, i.e., we have degenerate eigenstates of H0. This requires that weexamine our basis states to make sure that the perturbing Hamiltonian H1 is diagonal inthis basis. Since H1 involves the operator ~L · ~S, and we know that

~L · ~S =1

2(L+S− + L−S+) + LzSz

and the operators L± change the quantum number m`, we can conclude that in this basis,H1 is not diagonal. We now have a choice of either diagonalizing this Hamiltonian in

10The spin wave function χsms(ξ) is nothing more than the Pauli spinors encountered in Chapters 3

and 4, and can be written asχsms

(ξ) = δmsξ ,

where ξ can take on two values + 12 and − 1

2 .

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282 CHAPTER 13. APPROXIMATION METHODS FOR BOUND STATES

the basis given in Eq. (13.100), or finding an alternative basis set of eigenstates of H0.Here, we can resort to our discussion of symmetry and conservation where we found thatrotational invariance implies that the total angular momentum is a constant of the motion.This implies that the total angular momentum commutes with the Hamiltonian. In otherwords, the Hamiltonian is diagonal in a basis in which the total angular momentum isa good quantum number. The basis we have in Eq. (13.100) is related to the basis in

which the total angular momentum ~J ( ~J = ~L + ~S) is a good quantum number by aunitary transformation, the elements of this transformation matrix being the Clebsch-Gordan coefficients of the group SU(2) (see Chapter 4 Section 3). In other words, thebasis for treating the spin-orbit interaction in perturbation theory are the eigenstates ofthe operators S2, L2, J2, Jz as well as the Hamiltonian. These states can be written as

〈~r, ξ|n(`s)jmj〉 = Rn`(r)Y(`s)jmj(r, ξ) , (13.101)

with

Y(`s)jmj(r, ξ) =∑m`ms

(`m` sms|`s jmj)Y`m`(r)χsms(ξ) . (13.102)

To show that in this basis the perturbation H1 is diagonal, we need to write the operator~L · ~S in terms of the set of operators (J2, L2, S2). To achieve this we consider J2 as

J2 = (~L+ ~S) · (~L+ ~S) = L2 + S2 + 2~L · ~S ,

and then solving for ~L · ~S, we get

~L · ~S =1

2

(J2 − L2 − S2

). (13.103)

Thus, H1 is diagonal in this basis, and the matrix elements of ~L · ~S in this new basis aregiven by

〈(`s)jmj|2~L · ~S|(`s)jmj〉 = h2[j(j + 1)− `(`+ 1)− 3

4

]

= h2

` for j = `+ 1

2

−(`+ 1) for j = `− 12

. (13.104)

With this result we can write the first order correction to the energy due to the spin-orbitinteraction to be

∆E(1) = 〈n(`s)jmj|H1|n(`s)jmj〉

=⟨n`∣∣∣ 1

2m2c2

1

r

dVcdr

∣∣∣n`⟩ 〈(`s)jmj|~L · ~S|(`s)jmj〉 , (13.105)

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13.8. THE SPIN-ORBIT INTERACTION 283

where

⟨n`∣∣∣ 1

2m2c2

1

r

dVcdr

∣∣∣n`⟩ =1

2m2c2

∞∫0

dr r2Rn`(r)

(1

r

dVcdr

)Rn`(r)

≡ 2

h2 αn` > 0 . (13.106)

Since the radial integral is positive, and making use of the matrix elements of the ~L · ~Sas given by Eq. (13.104), we may conclude that the effect of the perturbation is to lowerthe energy of the state with j = `− 1

2while at the same time increase the energy of the

state with j = ` + 12. This is illustrated in Figure 13.7 for the case of n = 3, where we

demonstrate how the the spin-orbit interaction removes the degeneracy between stateswith different j. However, there is still the degeneracy that is the result of sphericalsymmetry which requires that each state with total angular momentum j has a degeneracyof (2j + 1). To remove this degeneracy we need to break the isotropy of space. In thecase of atoms, this can be achieved by placing the atom in a magnetic or electric field (seeproblems at the end of the chapter.)

3S 3P 3D

degeneracy

6

4

2

2

4

3D5/2

3P3/2

3S1/2

3P1/2

3D3/2

Figure 13.7: The splitting of the n = 3 levels as a result of the spin-orbit interaction.

In the above discussion we have only considered the correction to the energy levelsof the hydrogen atom in lowest order perturbation theory. The second and higher ordercorrection require non-diagonal matrix elements of H1. Since H1 is diagonal in spin s,orbital angular momentum `, and total angular momentum j, the only non-zero matrixelements are those with a different principle quantum number n, and in this case theradial matrix elements are small and differ in energy substantially.

Had we attempted to diagonalize our Hamiltonian in the basis given in Eq. (13.100),the result of the diagonalization would have been the states which are eigenstates of(S2, L2, J2, Jz), and the unitary matrix of transformation would have had for its elementsthe Clebsch-Gordan coefficients of the group SU(2) (see problem 2 at the end of thischapter).

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284 CHAPTER 13. APPROXIMATION METHODS FOR BOUND STATES

13.9 The Zeeman Effect

In the last section we found, by examining the symmetry of the system, that we couldchoose a basis in which the Hamiltonian is diagonal. In this way we avoided the problemof diagonalizing the Hamiltonian in the basis given in Eq. (13.100). In the present sectionwe will consider the problem where the choice of basis states for perturbation is not assimple.

Let us consider the case when the isotropy of space is broken by introducing an externalmagnetic field. In particular, let us place a hydrogenic atom in a uniform magnetic field.In this case the total angular momentum of the atom is no more a constant of the motion.However, if we take our constant magnetic field to be pointing along the z-axis, i.e., wehave a uniform field in the x − y plane, then we have rotational invariance for thoserotations about the z-axis. This implies that

R(z, θ) = exp[− ihθLz

]commutes with the Hamiltonian, or that

[Lz, H] = 0 .

In other words, the z-component of the angular momentum is the constant of motion. Inthis case the basis we used to calculate the contribution of the spin-orbit interaction, i.e.,the eigenstates of L2, S2, J2, Jz are not the proper basis, and we will need to diagonalizeour Hamiltonian in order to determine the correct basis states for perturbation theory.

Let us consider the case when both the spin-orbit interaction and the interaction ofthe atom with a constant magnetic field along the z-direction are present. In this casethe perturbing Hamiltonian is given by11

11The Zeeman Hamiltonian is given by

HZ = − ~M · ~B ,

where ~M is the magnetic moment of the electron. There are two contributions to this magnetic moment,one is from the spin, i.e.,

~MS = − e

mc~S ,

while the other is due to the velocity of the electron, which gives rise to a current that produces a magneticmoment

~ML =12c

∫d3r ~r × ~J(~r)

= − e

2mc~L .

To write this final result, we have taken the current, corresponding to an electron with velocity v, to be~J(~r) = −evδ(~r − ~r0). We now can write the Zeeman Hamiltonian as

HZ = − ~M · ~B = ( ~ML + ~MS) · ~B =e

2mc

(~L+ 2~S

)· ~B ,

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13.9. THE ZEEMAN EFFECT 285

H1 =1

2m2c2

1

r

dVcdr

~L · ~S +e

2mc

(~L+ 2~S

)· ~B , (13.107)

where ~B is the magnetic field. This second term is known as the Zeeman term namedafter the Dutch physicist who first observed the line splitting in the spectrum when thesource of the radiation was placed in a magnetic field. Here we observe that the firstterm is diagonal in the basis of eigenstates of S2, L2, J2, Jz. On the other hand, for amagnetic field pointing along the z-axis, the second term is diagonal in the basis whichare eigenstates of L2, Lz, S

2, Sz. In other words, neither bases we are familiar with aresatisfactory for perturbation theory, and we need to diagonalize H1 in one of the two bases.Let us take as our basis the eigenstates of the total angular momentum, i.e., |(`s)jm〉.We then have

〈n, (`s)jmj|H1|n, (`′s)j′m′j〉 =

(h

2mc

)2 ⟨n`∣∣∣1r

dVcdr

∣∣∣n`⟩×[j(j + 1)− `(`+ 1)− 3

4

]δ``′ δjj′ δmjm′

j

+eB0

2mc〈(`s)jmj|Jz + Sz|(`′s)j′m′j〉 , (13.108)

where we have taken the magnetic field to be along the z-axis, i.e. ~B = B0z, and madeuse of the fact that Jz = Lz +Sz. In writing the above result for the matrix element of H1

we have made use of the fact that the spin of the electron s is 12

and can not be changedso the matrix element is diagonal in s. To determine the second term on the right handside of the above equation, we need to calculate the matrix element

〈(`s)jmj|Sz|(`′s)j′m′j〉 .

To calculate this matrix element, we write the states of good total angular momentum interms of eigenstates of L2, Lz, S

2, Sz as

1

h〈(`s)jmj|Sz|(`′s)j′m′j〉 =

1

h

∑m`ms

∑m′`m′s

(`m` sms|`sjmj) (`′m′` sm′s|`′sj′m′j)

×〈`m`|`′m′`〉 〈sms|Sz|sm′s〉

= δ``′ δmjm′j

∑m`ms

ms (`m` sms|`sjmj) (`m` sms|`sj′mj)

= δ``′ δmjm′j

∑ms=± 1

2

ms (`m−ms sms|`sjmj)

× (`m−ms sms|`sj′mj) (13.109)

where ~B is the external magnetic field. The term involving ~L · ~B can be derived using the idea of minimalelectromagnetic coupling which involves the substitution ~p→ ~p− e

c~A, where ~A is the vector potential.

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286 CHAPTER 13. APPROXIMATION METHODS FOR BOUND STATES

Table 13.2: Table of Clebsch-Gordan coefficients (`m` sms|` s j mj) for the addition ofangular momentum ` to spin s = 1

2.

ms = 12

ms = − 12

j = `+ 12

√`+mj+

12

2`+1

√`−mj+ 1

2

2`+1

j = `− 12−√

`−mj+ 12

2`+1

√`+mj+

12

2`+1

From this result we can conclude that the Zeeman Hamiltonian is diagonal in the quantumnumber corresponding to the projection of the angular momentum along the z axis, i.e.,Jz. This result is expected considering the fact that the system is invariant under rotationabout the z-axis.

For the maximum value of mj corresponding to mj = ` + 12

we have that j = j′ =` + 1

2. In this case there is only one term in the sum in Eq. (13.109) with the Clebsch-

Gordan involved being one. In this case the perturbing Hamiltonian given in Eq. (13.107)is diagonal in our basis and therefore no diagonalization is required to calculate thecorrection to the energy which is given by

∆E(1) = 〈n, (`s)j = `+ 12; mj = ±(`+ 1

2)|H1|n, (`s)j = `+ 1

2; mj = ±(`+ 1

2)〉

= αn` [`± (`+ 1)x] , (13.110)

where αn` is defined in Eq. (13.106), and x is given by

x =β

2αn`, (13.111)

with β given in terms of the magnetic field by the relation

β =eB0h

mc. (13.112)

To illustrate the result we have so far, let us consider the 1S1/2 level of Hydrogen. In thiscase ` = 0 and

∆E(1) = ±β2

= ±eB0h

2mc.

Thus the 1S1/2 level is split into two levels, one corresponding to mj = + 12, the other is

for mj = − 12.

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13.9. THE ZEEMAN EFFECT 287

For all other values of mj the sum in Eq. (13.109) has only two terms and can bewritten as

1

h〈(`s)jmj|Sz|(`s)j′mj〉 =

1

2

[(`m− 1

2s 1

2|`sjmj) (`m− 1

2s 1

2|`sj′mj)

−(`m+ 12s− 1

2|`sjmj) (`m+ 1

2s− 1

2|`sj′mj)

].(13.113)

We now make use of Table 13.2 for the Clebsch-Gordan coefficient to calculate the matrixelements of Sz. This gives the 2× 2 matrix in Table 13.3 which has the matrix elementsof Sz.

Table 13.3: The matrix 1h〈(`s)jmj|Sz|(`s)j′mj〉.

j = `− 12

j = `+ 12

j = `− 12

− mj2`+1

√`(`+1)+ 1

4−m2

j

2`+1

j = `+ 12−

√`(`+1)+ 1

4−m2

j

2`+1

mj2`+1

Using the results in Table 13.3 we can write the matrix element of the perturbingHamiltonian H1 as

〈n, (`s)jmj|H1|n, (`s)j′mj〉 .

This 2× 2 matrix is tabulated in Table 13.4

Table 13.4: The matrix 〈n, (`s)jmj|H1|n, (`s)j′mj〉

j = `− 12

j = `+ 12

j = `− 12−αn` (`+ 1) + β `mj

2`+1−β

2

√`(`+1)+ 1

4−m2

j

2`+1

j = `+ 12

−β2

√`(`+1)+ 1

4−m2

j

2`+1αnl `+ βmj(`+1)

2`+1

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288 CHAPTER 13. APPROXIMATION METHODS FOR BOUND STATES

To diagonalize the perturbation H1 in this basis we need to solve the equation

det∣∣∣〈n, (`s)jmj|H1|n, (`s)j′mj〉 − λ δjj′

∣∣∣ = 0 . (13.114)

This gives a quadratic equation in λ with a solution given by

λ = αn`

mjx− 12±[(`+ 1

2)2 +mjx+

x2

4

] 12

≡ ∆E(1) , (13.115)

For a weak magnetic field, i.e., x 1 the above expression for ∆E(1) reduces to

∆E(1) ≈ αn`

mjx− 12± (`+ 1

2)

[1 +

mjx

(`+ 12)2

] 12

≈ αn`

mjx− 1

2± (`+ 1

2)

[1 +

mjx

2(`+ 12)2

]

= αn`

`+mjx

(1 + 1

2`+1

)−(`+ 1) +mjx

(1− 1

2`+1

) Zeeman Effect . (13.116)

Here we observe that the splitting in the hydrogen spectrum due to the spin-orbit inter-action is further split by the presence of the magnetic field, which is what we expectedconsidering the fact that we have broken the symmetry, that space is isotropic, by intro-ducing a magnetic field. For small magnetic field, i.e., x 1, the splitting is such thatthe spin orbit effect is still recognized (see Figure 5.7). On the other hand for strongmagnetic fields, i.e., x 1, we have that[

(`+ 12)2 +mjx+

x2

4

] 12

=[(`+ 1

2)2 −m2

j + (mj + 12x)2

] 12

= (mj + 12x)

[1 +

(`+ 12)2 −m2

j

(mj + 12x)2

] 12

≈ (mj + 12x) .

With this result in hand, we can write the correction to the energy in first order pertur-bation theory in a strong magnetic field as

∆E(1) ≈ αn`

(mj − 1

2) + (mj + 1

2)x

−(mj + 12) + (mj − 1

2)x

Paschen-Back Effect . (13.117)

Here the splitting due to the magnetic field is so large that the level spacing has noresemblance to the original spin-orbit splitting. In Figure 13.8 we sketch the levels forthe P -state, i.e., ` = 1 and indicate how the spacing changes as we increase the magneticfield or x.

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13.10. PROBLEMS 289

0 1 2 3

l = 1

x

P3/2

P1/2

mj = +3/2

mj = +1/2

mj = -1/2

mj = +1/2

mj = -1/2

mj = -3/2

Figure 13.8: The splitting of the ` = 1 levels as a result of the spin-orbit interaction anda magnetic field. In particular we demonstrate how the spectrum changes with x whichis proportional to the magnetic field.

13.10 Problems

1. To first order in perturbation theory, calculate the correction to the ground state ofthe hydrogen-like atom due to the finite size of the nucleus. For simplicity, assumethat the nucleus is spherical, of radius R, and that its charge Ze is uniformlydistributed throughout its volume.

Hint: Use Gauss’s law to determine the potential due to a uniformly charged sphereas a function of r, the distance from the center of the sphere.

2. The positronium is a bound state of an electron and a positron (both spin 12

par-ticles). In the ground state, i.e. n = 1, ` = 0, the main term in the Hamiltonian,besides the Coulomb attraction is

H1 = γ~σ1 · ~σ2 .

(a) Calculate the Clebsch-Gordan coefficients for the coupling of spin 12

with spin12

by diagonalizing the above interaction, H1, in the basis

|s1 =1

2m1; s2 =

1

2m2〉 .

(b) Show that

|(s1s2)SM〉 = (−1)S+1 |(s2s1)SM〉 ,

if s1 = s2 = 12.

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290 CHAPTER 13. APPROXIMATION METHODS FOR BOUND STATES

(c) Calculate the matrix elements of H1 in this basis, and thus predict, to firstorder in perturbation theory, the energy splitting of the spin singlet and spintriplet states.

(d) If we put the positronium atom, in the 1S state, in an external magnetic fieldB along the z-axis, we get an additional term in the Hamiltonian of the form

Vz =eB

µc(S1z − S2z)

where S1z is the projection of the spin of the electron along the z-axis, whileS2z is the projection of the spin of the positron along the z-axis. Calculate theeigenvalues and eigenstates of the total Hamiltonian, and classify the statesaccording to the quantum numbers associated with the constants of motion.

3. Consider a charged one-dimensional harmonic oscillator in a constant external elec-tric field E. The potential energy of such a charged particle in an electric field isgiven by −eEx, where x and e are the position and the charge of the particle.

(a) Find the ground state energy of this system to second order in perturbationtheory.

(b) Show that for this problem the Schrodinger equation can be solved exactly.

(c) Compare the result of the exact solution with that of perturbation theory.

4. Consider the nucleus 3He to consist of a deuteron of spin one, and a proton of spin12

in relative angular momentum zero.

(a) What are the possible spin states of 3He? Write these states in terms of the spinstates of the deuteron and proton using the table of Clebsch-Gordan coefficientsgiven in Table 13.2.

(b) If the interaction between the deuteron and proton is

H1 = V0~Sd · ~Sp ,

where ~Sd is the spin of the deuteron and ~Sp the spin of the proton. Whatshould the sign of V0 be, for the ground state spin of 3He, to be 1

2.

5. The Hamiltonian for a system is given as

H = − h2

2m∇2 − Ae−ar .

Using the trial functionψ(r) = B e−βr ,

calculate the upper bound on the ground state energy.

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13.10. PROBLEMS 291

6. Given a one-dimensional harmonic oscillator with Hamiltonian

H = − h2

2m

d2

dx2+

1

2mω2x2 .

Assume a trial wave function

ψ0(x, a) = Ae−ax2/2 with A =

(a

π

)1/4

,

for the ground state.

(a) Calculate the ground state energy using the above trial function. Compareyour result with the exact eigenvalue.

(b) Find a trial function for the first excited state, and calculate the energy of thatstate.

7. Calculate the upper bound on the ground state energy of the Coulomb Hamiltonianusing the trial functions

ψ0(r, a) = N e−ar2

, ψ0(r, a) =N

r2 + a2and ψ0(r, a) = N re−ar .

Compare your results with the exact answer.

8. Using the variational method, find an approximate energy and wave function forthe 2S state of the hydrogen atom.

9. A trial function Ψ differs from an eigenfunction φE,

HφE = EφE,

by a small amount, so that Ψ = φE + εψ, where φE and ψ are both normalized andε 1. Show that the upper bound on the energy 〈H 〉 differs from E only by aterm of order ε2, and find the term.

Note the function Ψ is not normalized.

10. Calculate the splitting in the n = 3 levels of hydrogen due to the spin-orbit andZeeman interaction with constant magnetic field in the z-direction of magnitude B0.

11. If the general form of a spin-orbit interaction for a particle of mass m and spin ~Smoving in a potential V (r) is

VLS =1

2m2c2

1

r

dV

dr~L · ~S ,

what is the effect of this interaction on the spectrum of the three-dimensional har-monic oscillator?

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292 CHAPTER 13. APPROXIMATION METHODS FOR BOUND STATES

12. Calculate the perturbation of the first two energy levels of a hydrogen atom placedin a constant electric field ~E along the z-axis. (This effect is known as the Starkeffect).

13. The normal Zeeman effect involves the interaction of the magnetic moment of theelectron due to its angular momentum with an external magnetic field, i.e., the spinof the electron is neglected. In this case the Hamiltonian for the hydrogen atom ina magnetic field ~B is given by

H =1

2∇2 − 1

r+

1

2~L · ~B ,

in atomic units.

(a) Calculate the splitting of the n = 2 levels in hydrogen, to first order in per-turbation theory, for the case when the magnetic field is in the z-direction, i.e.~B = B0z.

(b) How many spectral lines does the n = 2 to n = 1 transition split into?

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Chapter 14

Scattering Theory; Revisited

In Chapter 2 we showed how we can calculate the cross section for two particle scatteringin terms of the scattering amplitude f(k, θ), and how this scattering amplitude is relatedto the phase shifts δ`. In particular, we found that we needed to solve the Schrodingerequation for the wave function ψ`(r) for all r and then extract the phase shifts, knowingthat for r →∞

ψ`(r)→1

krsin(kr − 1

2π`+ δ`) .

In other words, the asymptotic wave function determines the scattering amplitude and,therefore cross section, yet we need to calculate the wave function for all r. Since thewave function is not the observable we measure, we would like to set up an equation forthe scattering amplitude f(k, θ) which is the observable.

In this chapter we will first derive an equation, the Lippmann-Schwinger equation,which can be solved for the scattering amplitude. We will find that this equation allowsus not only to derive the Optical Theorem, but also to consider perturbation expansionfor the scattering amplitude in the form of the Born series. Finally, we will consider aspecial class of potentials, called separable potentials, where we can study the propertiesof the scattering amplitude analytically.

14.1 Formal Theory of Scattering

To derive an equation for the scattering amplitude f(k, θ), we need to convert the Schrodingerequation plus boundary condition into an integral equation which incorporates theseboundary conditions. Consider the Hamiltonian

H = H0 + V , (14.1)

where H0 is taken to be the kinetic energy in the center of mass, i.e., H0 = p2

2µwith µ the

reduced mass of the system given by

µ =m1m2

m1 +m2

.

293

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294 CHAPTER 14. SCATTERING THEORY; REVISITED

Here m1 and m2 are the masses of the two particles in the collision. In Eq. (14.1), V isthe interaction between the two particles. We also introduce the eigenstates of H0 to be

H0|φ~k〉 = E|φ~k〉 , (14.2)

where E = h2k2

2µ. These states are labeled by the initial momentum ~p = h~k. In coordinate

representation, the state |φ~k〉 is a plane wave, i.e.,

〈~r |φ~k〉 =1

(2π)3/2ei~k·~r . (14.3)

We note here that this state is identical to the eigenstate of the momentum operatoras given in Eq. (11.93), i.e. 〈~r |~k〉 = 〈~r |φ~k〉. The Schrodinger equation for the fullHamiltonian H can now be written as

(E −H0) |ψ〉 = V |ψ〉 . (14.4)

This can be thought of as an inhomogeneous differential equation if we take it in coordinaterepresentation, and in this case, the solution can be written as the sum of a particularsolution plus a solution to the homogeneous equation. The particular solution is given by

|ψ〉 =1

E −H0

V |ψ〉 . (14.5)

In writing the operator (E −H0)−1, we have assumed that the operator (E −H0) has nozero eigenvalues. In fact, as we will find, the operator (E−H0) does have zero eigenvaluesand therefore its inverse which appears in Eq. (14.5) is singular. We will show next how theboundary conditions in the Schrodinger equation are used to overcome these singularities.The general solution of Eq. (14.4) is then given by

|ψ~k〉 = |φ~k〉+1

E −H0

V |ψ~k〉 . (14.6)

We have labeled our solution |ψ~k〉 by the vector ~k to indicate that the momentum of the

incident beam is h~k. This in turn means that the energy of the incident beam in thecenter of mass is given by E = h2k2

2µ.

To examine the singularities of (E −H0)−1, we write Eq. (14.6) in coordinate repre-sentation as

〈~r |ψ~k〉 = 〈~r |φ~k〉+∫d3r 〈~r |(E −H0)−1|~r ′〉 〈~r ′|V |ψ~k〉 . (14.7)

We are now in a position to examine the singularities of the Green’s function.1 Makinguse of the fact that the complete set of eigenstates of the momentum operator are also

1This Green’s function is identical to the Green’s function encountered in the theory of differentialequations. After all, we are solving the Schrodinger equation which is a second order differential equation.

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14.1. FORMAL THEORY OF SCATTERING 295

eigenstates of H0, we can write the Green’s function as

〈~r |(E −H0)−1|~r ′〉 =∫d3k 〈~r |(E −H0)−1|~k 〉 〈~k |~r ′〉

=∫d3k〈~r |~k 〉 〈~k |~r ′〉E − h2k2

, (14.8)

since the eigenvalue of H0 is h2k2

2µ. Taking the energy E, which is the energy of the initial

incident particle in the two-body center mass, to be related to the incident momentum,hk0, then the integral in Eq. (14.8) can be written as

〈~r |(E −H0)−1|~r ′〉 =1

(2π)3

h2

∫d3k

ei~k·(~r−~r ′)

k20 − k2

=1

(2π)3

h2

∞∫0

dkk2

k20 − k2

π∫0

dθ sin θ eik|~r−~r′| cos θ

2π∫0

=1

(2π)2

h2

∞∫0

dkk2

k20 − k2

eik|~r−~r′| − e−ik|~r−~r ′|

ik|~r − ~r ′|

=1

i(2π)2

h2

1

|~r − ~r ′|

+∞∫−∞

dk keik|~r−~r

′|

k20 − k2

. (14.9)

However this integral is not defined for k20 > 0, (i.e. E > 0) because the integrand has a

pole at k = ±k0, which is on the integration path. To overcome this problem, we need togo around these poles. This can be achieved by taking2

k20 → k2

0 ± iε , (14.10)

where ε is an infinitesimal quantity. We then perform the integral and take the limit ofε→ 0. As we will see, the choice of sign in k2

0± iε will determine the boundary condition.Let us take the positive sign, i.e., k2

0 + iε, for the present. Then our Green’s function hasthe integral representation given by

〈~r |(E −H0)−1|~r ′〉 =1

i(2π)2

h2

1

|~r − ~r ′|

+∞∫−∞

dk keik|~r−~r

′|

k20 + iε− k2

. (14.11)

The integrand now has two poles at

k = ±(k0 + iε) .

2This is one choice of integration path. Other choices will entail taking the integration path below orabove both poles. These correspond to other boundary conditions.

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296 CHAPTER 14. SCATTERING THEORY; REVISITED

We now can perform the integral in Eq. (14.11) by making use of Cauchy’s Theorem. Fork in the upper half of the complex k-plane, i.e., Im(k) > 0, the integrand has the propertythat it goes to zero as |k| → ∞. This condition allows us to convert the integral over theinterval −∞ to +∞ to an integral along a contour C in Figure 14.1 where the semi-circlecorresponds to |k| → ∞. Because the integrand is zero on the infinite semi-circle it doesnot contribute to the value of the Green’s function. This allows us to write the Green’sfunction as

k

C

k0 + iε

-k0 - iε

Figure 14.1: The contour of integration used to calculate the Green’s function with anoutgoing spherical wave.

〈~r |(E −H0)−1|~r ′〉 =1

i(2π)2

h2

1

|~r − ~r ′|

∮C

dk k eik|~r−~r′|

(k0 + iε− k)(k0 + iε+ k)

= − 1

h2

eik0|~r−~r ′|

|~r − ~r ′|, (14.12)

where we have made use of Cauchy’s Residue Theorem3 and have taken the limit ε→ 0.Had we chosen k2

0 → k20 − iε, the poles in the integrand would have been at

k = ±(k0 − iε) .

In this case the contour C would have enclosed the pole at k = −k0 + iε, and we wouldget the the factor of

e−ik0|~r−~r ′|

3Cauchy’s theorem allows us to write the integral over a closed contour in terms of the sum over theresidues of all the poles in the integrand that are inside the contour, i.e.,∮

C

dk f(k) = 2πi∑n

Res[f(k)] ,

where n runs over all the poles inside the contour C.

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14.1. FORMAL THEORY OF SCATTERING 297

in Eq. (14.12) instead ofeik0|~r−~r ′| .

This we will see corresponds to spherical waves which are moving towards the scatteringcenter, while the Green’s function in Eq. (14.12) corresponds to spherical outgoing waves.These two Green’s functions can now be written as

〈~r | (E± −H0)−1|~r ′〉 = limε→0〈~r | (E ± iε−H0)−1|~r ′〉

= ∓ 1

h2

e±ik0|~r−~r ′|

|~r − ~r ′|. (14.13)

With this result in hand we can write the scattering wave function, in coordinate rep-resentation, by substituting the Green’s function 〈~r | (E+ − H0)−1|~r ′〉 in Eq. (14.7) toget

〈~r |ψ(+)~k0〉 = 〈~r |φ~k0

〉 − 1

h2

∫d3r′

eik0|~r−~r ′|

|~r − ~r ′|〈~r ′|V |ψ(+)

~k0〉 , (14.14)

where (+) superscript on the wave function indicates that the boundary condition cor-responds to an outgoing spherical wave. To extract the scattering amplitude out of thiswave function we need to take the limit as r → ∞ to determine the coefficient of theoutgoing spherical wave and compare that equation with Eq. (10.3) to determine thescattering amplitude. To take the limit as r →∞ we make use of the fact that

|~r − ~r ′| =√r2 + r′2 − 2~r · ~r ′

= r

1 +

(r′

r

)2

− 2~r · ~r ′

r2

12

→ r − r · ~r ′ for r →∞ ,

where r is a unit vector in the ~r direction. We now can write the asymptotic wave function(i.e., r →∞) as

〈~r |ψ(+)~k0〉 → 1

(2π)3/2ei~k0·~r − 1

h2

eik0r

r

∫d3r′ e−ik0r·~r ′ 〈~r |V |ψ(+)

~k0〉 . (14.15)

Since the unit vector r is in the same direction as the final momentum of the scatteredparticle, and the magnitude of the initial and final momentum are the same due to energyconservation, we can take the final momentum to be ~kf = k0r. Making use of this resultand the fact that

〈φ~kf |~r′〉 =

1

(2π)3/2e−i

~kf ·~r ′,

we can write the asymptotic wave function as

〈~r |ψ(+)~ki〉 → 1

(2π)3/2

ei~ki·~r − 4π2µ

h2 〈φ~kf |V |ψ(+)~ki〉 e

ikr

r

. (14.16)

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298 CHAPTER 14. SCATTERING THEORY; REVISITED

In writing the above scattering wave function, we have labeled the initial momentum, i.e.,the momentum of the incident beam, by ~ki, while the final momentum is labeled as ~kf .Conservation of energy then requires that the magnitude of the initial and final momentumbe equal if the scattering is elastic, i.e., |~ki| = |~kf | ≡ k. Comparing Eqs. (14.16) and (10.3)allows us to write the scattering amplitude in terms of the matrix elements of the potentialas

f(k, θ) = −4π2µ

h2 〈φ~kf |V |ψ(+)~ki〉 , (14.17)

where cos θ = kf · ki. Since the scattering amplitude is basically an observable, we willintroduce a corresponding operator which we will call the T -matrix, and which is definedby the relation

V |ψ(+)~ki〉 ≡ T (E+)|φ~ki〉 , (14.18)

where the boundary condition is now specified by labeling the T -matrix with E+ = E+iε.We now can write the scattering amplitude in terms of the T -matrix as

f(k, θ) = −4π2µ

h2 〈φ~kf |T (E+)|φ~ki〉 ( 14.19a)

= −4π2µ

h2 〈~kf |T (E+)|~ki〉 . ( 14.19b)

Therefore, the problem of finding an equation for the scattering amplitude has beenreduced to the problem of finding an equation for the T -matrix. Making use of thedefinition of the T -matrix as given in Eq. (14.18), we can write the scattering wavefunction given in Eq. (14.6) as

|ψ(+)~ki〉 = |φ~ki〉+G0(E+)T (E+)|φ~ki〉 , (14.20)

where G0(E+) = (E+ − H0)−1 is the Green’s function in operator form. This Green’sfunction is often referred to as the free-particle Green’s function since it is the Green’sfunction for the Schrodinger equation in the absence of any interaction. We now multiplythis equation from the left by the potential V , to get

V |ψ(+)~ki〉 = V |φ~ki〉+ V G0(E+)T (E+)|φ~ki〉 . (14.21)

Applying the definition of the T -matrix, Eq. (14.18), to the left hand side of Eq. (14.21)gives us the equation

T (E+)|φ~ki〉 = V |φ~ki〉+ V G0(E+)T (E+)|φ~ki〉 . (14.22)

Since this is valid for any state |φ~ki〉, we can write an operator equation for the T -matrixwhich is of the form

T (E+) = V + V G0(E+)T (E+) . (14.23)

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14.1. FORMAL THEORY OF SCATTERING 299

We have taken the operator T to be a function of the energy E+ since the Green’s functionG0(E+) is a function of the energy, and we need to know the Green’s function to determinethe T -matrix. Also, by specifying the energy in the form of E+, we have specified theboundary condition that we have spherically outgoing waves. Equation (14.23) is knownas the Lippmann-Schwinger equation, and is equivalent to the Schrodinger equation in-cluding the boundary conditions. The solution of this equation will give us the scatteringamplitude directly and thus the cross section.

To solve Eq. (14.23), we need to write the equation in a given representation. Inthis case the natural representation is the momentum representation4 which gives us〈~k |T (E+)|~k ′〉 and then the scattering amplitude, as measured in elastic scattering, isgiven by

〈~k |T (E+)|~k ′〉 with |~k| = |~k ′| and E =h2k2

2µ. (14.24)

The Lippmann-Schwinger equation, Eq. (14.23), can be written in momentum space as

〈~k |T (E+)|~k ′〉 = 〈~k |V |~k ′〉+ 〈~k |V G0(E+)T (E+)|~k ′〉

= 〈~k |V |~k ′〉+∫d3k′′

〈~k |V |~k ′′〉 〈~k ′′|T (E+)|~k ′〉E + iε− h2k′′2

. (14.25)

In writing the second line of the above equation we have made use of the fact thatthe eigenstates of the momentum operator are complete, and that these states are alsoeigenstates of H0 with eigenvalue h2k′′2

2µ. In Eq. (14.25), we have an integral equation in

three-dimensions which is very difficult to solve as it stands. To reduce the dimensionalityof the equation, we need to partial wave expand the matrix elements of the potential Vand the T -matrix T (E+). This expansion involves the introduction of eigenstates of thetotal angular momentum and its projection along the z-axis. In the case of the scatteringof spin-less particles, the total angular momentum is the orbital angular momentum. Forthe potential V , the partial wave expansion involves writing the matrix elements of V as5

〈~k|V |~k ′〉 = 〈k; k|V |k′; k′〉=

∑`m

〈k|`m〉 〈`m, k|V |k′, `m〉 〈`m|k′〉

4Note that the eigenstates of the momentum operator and the free Hamiltonian H0 are identical withour choice of normalization, i.e.,

|φ~k〉 = |~k 〉 .

5In writing Eq. (14.26), we are making use of the notation

|~k 〉 = |k; k〉 = |k〉 |k〉

to separate the radial from the angular variables.

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300 CHAPTER 14. SCATTERING THEORY; REVISITED

≡∑`m

〈k|`m〉 〈k|V`|k′〉 〈`m|k′〉 . (14.26)

In writing the first line of the above equation we have separated the radial from theangular part of the momentum eigenstates. We have also taken into consideration thatthe potential is spherically symmetric by taking the matrix elements of V to be diagonalin ` and independent of m, i.e., since [H,L2] = [H,Lz] = 0, we have

〈`m, k|V |k′, `′m′〉 = δ``′ δmm′ 〈k|V`|k′〉 . (14.27)

The independence of the matrix elements from m is a result of the fact that the radialwave functions for a spherically symmetric potential are independent of m. In Eq. (14.26)the bracket 〈k|`m〉 is nothing other than the spherical harmonic, i.e.,

〈k|`m〉 = Y`m(k) .

In other words the expansion in Eq. (14.26) is similar to that in Eq. (10.45) if we takeinto consideration that the Legendre polynomial can be written in terms of sphericalharmonics using the addition theorem, i.e.,

∑m

Y`m(k)Y ∗`m(k′) =2`+ 1

4πP`(cos θ) . (14.28)

In this case we can write the potential in momentum representation as

〈~k |V |~k ′〉 =1

∑`

(2`+ 1)V`(k, k′)P`(cos θ) , (14.29)

with

V`(k, k′) = 〈k|V`|k′〉 , (14.30)

and cos θ = k · k′. In a similar manner we can write a partial wave expansion for theT -matrix which is of the form

〈~k |T (E+)|~k ′〉 =∑`m

〈k|`m〉 〈k|T`(E+)|k′〉 〈`m|k′〉

=1

∑`

(2`+ 1)T`(k, k′;E+)P`(cos θ) . (14.31)

With these partial wave expansions for the potential V and the T -matrix T (E+), andusing the orthogonality of the spherical harmonics, i.e.,∫

dk 〈`m|k〉 〈k|`′m′〉 = δ``′ δmm′ , (14.32)

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14.1. FORMAL THEORY OF SCATTERING 301

we can write the Lippmann-Schwinger equation as a one dimensional integral equation ofthe form

〈k|T`(E+)|k′〉 = 〈k|V`|k′〉+

∞∫0

dk′′k′′2〈k|V`|k′′〉 〈k′′|T`(E+)|k′〉

E + iε− h2k′′2

, ( 14.33a)

or

T`(k, k′;E+) = V`(k, k

′) +

∞∫0

dk′′k′′2 V`(k, k

′′)

E + iε− h2k′′2

T`(k′′, k′;E+) . ( 14.33b)

This equation can be solved numerically on any present day computer. This is achievedby replacing the integral by a sum and in this way we turn the integral equation into aset of linear algebraic equations. In converting the above integral equation into a set ofalgebraic equations we should note that the energy E and the initial momentum k′ areparameters and play no role in the solution of the equation. In fact, to get the physicalamplitude we should take

E =h2k2

0

2µand k′ = k0

where k0 is referred to as the on-shell momentum.By comparing the partial wave expansion of the scattering amplitude, f(k, θ), as given

in Eq. (10.45) with the expansion of the T -matrix as given in Eq. (14.31), and makinguse of Eq. (14.19), we can write the partial wave scattering amplitude f`(k) in terms ofthe partial wave T -matrix T`(k0, k0;E+) as

f`(k0) =1

2i

(e2iδ` − 1

)= −πµk0

h2 T`(k0, k0;E+)

≡ −πµk0

h2 T`(k0) . (14.34)

To solve Eq. (14.33) for the scattering amplitude, we need to determine the inputpartial wave potential V`(k, k

′), and the Green’s function G0(E+) in momentum repre-sentation. To determine the partial wave potential we need to first write the potential inmomentum representation in terms of the potential in coordinate space as we have usedit in previous chapters. All the potentials we have encountered to date are diagonal incoordinate representation, i.e.,

〈~r |V |~r ′〉 = δ(~r − ~r ′)V (~r) . (14.35)

These potentials are known as local potentials in contrast to non-local potentials that arenot diagonal in coordinate representation. For local potentials, we have in momentumspace

〈~k |V |~k ′〉 =∫d3r d3r′ 〈~k |~r 〉 〈~r |V |~r ′〉 〈~r ′|~k ′〉

=∫d3r 〈~k |~r 〉V (~r) 〈~r |~k ′〉 . (14.36)

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302 CHAPTER 14. SCATTERING THEORY; REVISITED

Making use of the partial wave expansion of the eigenstates of the momentum operator,i.e.,

〈~r |~k 〉 =1

(2π)3/2ei~k·~r

=

√2

π

∑`m

i` j`(kr)Y`m(r)Y ∗`m(k) , (14.37)

and the orthogonality of the spherical harmonics in Eq. (14.36) allows us to write thepotential in momentum space as

〈~k |V |~k ′〉 =∑`m

Y`m(k)V`(k, k′)Y ∗`m(k′) , (14.38)

where

V`(k, k′) =

2

π

∞∫0

dr r2

∞∫0

dr′ r′2 j`(kr)V`(r, r′) j`(k

′r′) ( 14.39a)

=2

π

∞∫0

dr r2 j`(kr)V`(r) j`(k′r) . ( 14.39b)

Here, Eq. (14.39b) is to be used for the special case when the potential is local andcentral, in which case V`(r) = V (r). Thus for any local central potential, Eq. (14.39b)defines the momentum space partial wave potential for use in the integral equation givenin Eq. (14.33). The solution of this Lippmann-Schwinger equation then gives us thescattering amplitude for a given angular momentum `. To get the differential cross section,we need to solve Eq. (14.33) for all partial waves that are important. For finite rangepotentials, the maximum value of ` can be estimated by taking the classical argumentthat the maximum angular momentum is the cross product of the incident momentum,times the maximum impact parameter, which in this case is the range of the potential,i.e., `max = r0k, where r0 is the range of the potential.6

14.2 The Born Approximation

The solution of the Lippmann-Schwinger equation, Eq. (14.33), is often not simple toget, and one may need to resort to approximation methods. One approximation oftenused at high energies or weak potentials in both atomic and nuclear physics is the Bornapproximation. This is the first term in a series similar to the perturbation series we

6Here, we should note that for the case of the Coulomb potential, where the range of the interactionis infinite, the partial wave sum should be examined carefully.

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14.2. THE BORN APPROXIMATION 303

developed in the last chapter. To get the series, we iterate Eq. (14.23) to get, in operatorform, the Born series given by

T (E±) = V + V G0(E±)V + V G0(E±)V G0(E±)V + · · · . (14.40)

This is a power series in the potential, and therefore for a potential that is weak, theseries converges. Here, as in the case of perturbation theory for bound states, we candefine H0 to include some of the interaction and in this way, the remaining interaction(H − H0) is weak enough for the Born series to converge. However, in this case, to getthe amplitude or T -matrix, we need to take the matrix element of (H −H0) with respectto the solution of the Schrodinger equation with the Hamiltonian H0. Also the Green’sfunction G0(E±) has to be replaced by the Green’s function for the Hamiltonian H0 whichincludes the interaction. This procedure has been implemented for Coulomb plus shortrange potential, where we know the solution for the Coulomb Hamiltonian analytically.

Another condition under which the above Born series converges is when the energy Eis high.7In that case the series is again a power series in E−1, and the first few terms ofthe series give a good approximation.

The Born approximation is used when the first term in the Born series is taken to bethe amplitude for scattering. For this case, we have

〈~k |T (E+)|~k ′〉 ≈ 〈~k |V |~k ′〉 ≡ 〈~k |TB|~k ′〉 . (14.41)

For the case of a local central potential, the Born approximation reduces to

〈~kf |TB|~ki〉 = 〈φ~kf |TB|φ~ki〉

=∫d3r φ∗~kf V (r)φ~ki

=1

(2π)3

∫d3r ei~q·~r V (r) , (14.42)

where ~q = ~ki − ~kf is the momentum transfer in the scattering.As an example of the application of the Born approximation, let us consider the

problem of scattering by a Coulomb potential, and in particular the potential due to thenucleus with charge Ze, i.e.

VC(r) = −Ze2

r. (14.43)

Here the Born approximation is given by

〈φ~kf |TB|φ~ki〉 = − Ze2

(2π)3

∫d3r ei~q·~r

1

r

7By high energy we mean high, relative to the depth of the potential V , or the energy of bound statesin the potential V . For example, for electron scattering in atomic physics, high is more than 103 eV,while for proton scattering in nuclear physics, high is more than 103 MeV.

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304 CHAPTER 14. SCATTERING THEORY; REVISITED

= − Ze2

(2π)2

∞∫0

dr r

+1∫−1

dx eiqrx

= − Ze2

(2π)2

2

q

∞∫0

dr sin qr . (14.44)

This integral, as it stands is not well defined. However, if we replace the Coulomb potentialby the screened Coulomb potential, i.e.,

V (r) = −Ze2 e−ar

r,

the integral becomes well defined, and then we can take the limit as a → 0 to get theBorn amplitude for the Coulomb potential, i.e.8

〈φ~kf |TB|φ~ki〉 = −Ze2

2π2

1

qlima→0

∞∫0

dr e−ar sin qr

= −Ze2

2π2

1

qlima→0

q

q2 + a2

= −Ze2

2π2

1

q2, (14.45)

where the momentum transfer square is given by

q2 = 2k2(1− cos θ) , (14.46)

and cos θ = kf · ki, and |~kf | = |~ki| = k. The scattering amplitude can now be written as

f(k, θ) =Ze2µ

h2

1

k2(1− cos θ). (14.47)

This scattering amplitude is real, and to that extent does not satisfy unitarity. This istrue for the Born approximation for any real potential. With this result, we can calculatethe differential cross section, which is given by

dΩ= |f(k, θ)|2

=Z2e4µ2

h4

1

k4(1− cos θ)2

=Z2e4

16E2 sin4(θ/2), (14.48)

8We have made use of the integral∞∫

0

dx e−ax sin qx =q

a2 + q2

to get the Born amplitude for Coulomb scattering.

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14.3. ELECTRON ATOM SCATTERING 305

which is the Rutherford cross section we got in Eq. (10.85) This cross section goes toinfinity when cos θ = 1, i.e., for θ = 0 and θ = π. In fact, the total cross section is infinity.This is the result of the fact that the Coulomb potential is infinite in range. In nature, wealways have screened potentials in the sense that proton-proton scattering is a problem oftwo charge particles when the two hydrogen atoms overlap. For large distances we haveneutral hydrogen atoms. Although the Born amplitude for the Coulomb potential is realand does not satisfy unitarity, the cross section we get is the exact classical Rutherfordcross section.

14.3 Electron Atom scattering

As a second application of the Born approximation, let us consider electron atom scat-tering, and in particular, electron scattering from hydrogen. This approximation is verygood at high energies where the second term in the Born series, which goes as 1

Ebecomes

negligible. The potential that describes the interaction between the incident electron andthe target atom is the sum of two terms. The first term describes the interaction of theelectron with the point nucleus, is attractive, and of the form

V1(r) = −Zr, (14.49)

where Z is the charge on the nucleus. Here we are using atomic units. The secondterm corresponds to the interaction of the incident electron with the charge distributionresulting from the bound electrons in the atom. This is of the from

V2(~r ) =∫d3r′

ρ(~r ′)

|~r − ~r ′|, (14.50)

where ρ(~r ′) is the charge distribution of the bound electrons and is given by

ρ(~r ) =Z∑n=1

|φn(~r )|2 . (14.51)

Here, we note that the sum runs over the Z electrons each in a state φn(~r ), n = 1, · · · , Z.For the case of electron hydrogen scattering, Z = 1 and the bound state electron wave

function can be taken as the ground state of the hydrogen atom, i.e., φ100(~r ). In this caseV2 is given by

V2(~r ) =∫d3r′

|φ100(~r ′)|2

|~r − ~r ′|, (14.52)

with

φ100(~r ) = R10(r)Y00(r) =1√4π

R10(r) , (14.53)

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306 CHAPTER 14. SCATTERING THEORY; REVISITED

and the radial wave function is given in Sec. 9.3, which in atomic units is given by

R10(r) = 2 e−r . (14.54)

We now can write the total potential for electron hydrogen scattering as

V (~r ) = −1

r+∫d3r′

|φ100(~r ′)|2

|~r − ~r ′|. (14.55)

To calculate the second term on the right hand side of the above expression for thepotential, we need to expand the factor of |~r− ~r ′|−1 in terms of spherical harmonics (seeSec. 8.2), i.e.

1

|~r − ~r ′|= 4π

∑`m

,1

2`+ 1

r`<r`+1>

Y`m(r)Y`m(r′) . (14.56)

This allows us to write the potential as

V (~r ) = −1

r+∑`m

2`+ 1

∞∫0

dr′ r′2|R10(r)|2

r`<r`+1>

Y`m(r)∫dr Y`m(r′) . (14.57)

Making use of the orthogonality of the spherical harmonics give one term in the sumcorresponding to ` = m = 0. This reduces the potential to just a radial integral of theform

V (~r ) = −1

r+

∞∫0

dr′ r′2|R10(r′)|2

r>. (14.58)

At this stage we note that the potential we have is central, i.e., it is a function of r = |~r|.To evaluate the integral we need to first divide the domain of integration into two partscorresponding to r′ < r and r′ > r which determines the value of r>. We then make useof the fact that the radial wave function is normalized to write our potential as

V (r) =

∞∫r

dr′ r′2(

1

r′− 1

r

)|R10(r′)|2 . (14.59)

To simplify this result we need to calculate the integral using the radial wave function forthe ground state of hydrogen. this gives9

V (r) = 4

∞∫r

dr′ r′2(

1

r′− 1

r

)e−2r′

= −(1 + r)e−2r

r. (14.60)

9To calculate the radial integral we have made use of the fact that∫dxxm e−ax = −e−ax

m∑l=0

m!xm−l

(m− l)! al+1.

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14.3. ELECTRON ATOM SCATTERING 307

Here we observe that the range of the potential is determined by the charge distributiondue to the bound electron. This we expect that the incident electron sees a neutral atom,and therefore no potential energy until we are the electron gets to the charge distributionof the atom.

We now turn to the calculation of the Born amplitude for the above potential. In theBorn approximation the T -matrix is given by

〈φ~kf |TB|φ~ki〉 =1

(2π)3

∫d3r e−i~q·~r V (r)

=2

(2π)2

∞∫0

dr r2 V (r)sin qr

qr. (14.61)

Making use of the fact that

∞∫0

dxxn e−βx sin qx = (−1)n∂n

∂βn

(q

q2 + β2

),

we can evaluate the above integral and write the Born T -matrix as

〈φ~kf |TB|φ~ki〉 = − 1

2π2

q2 + 8

(q2 + 4)2, (14.62)

where the momentum transfer q is given in term of the center of mass scattering angleand the momentum of the incident electron by

q2 = 2k2(1− cos θ) . (14.63)

The scattering amplitude and cross section can then be written as

f(k, θ) = 2q2 + 8

(q2 + 4)2

= 22k2(1− cos θ) + 8

[2k2(1− cos θ) + 4]2(14.64)

and

= |f(k, θ)|2

= 4

2k2(1− cos θ) + 8

[2k2(1− cos θ) + 4]2

2

. (14.65)

This cross section is in units of Bohr radii square, and illustrated in Fig 14.2

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308 CHAPTER 14. SCATTERING THEORY; REVISITED

25 50 75 100 125 150 175

-3

-2

-1

0

1

2

log(

)dσ

/dΩ

θ

Figure 14.2: Born cross section for electron hydrogen scattering at 100 eV.

14.4 Unitarity of the T-matrix

In Chapter 10, we demonstrated that the scattering amplitude f(k, θ) satisfied the OpticalTheorem, and as a result, the S-matrix in a given partial wave is unitary, i.e., it isrepresented by a phase. In this section we will derive this unitarity relation from theLippmann-Schwinger equation.

In operator from, the Lippmann-Schwinger equation is given by

T (E±) = V + V G0(E±)T (E±) . (14.66)

We now multiply this equation from the left by V −1 and from the right by T−1(E±). Thisgives us the equation

T−1(E±) = V −1 −G0(E±). (14.67)

If we consider T (z) as an analytic function in the complex z-plane, then

T−1(E+)− T−1(E−) = −[G0(E+)−G0(E−)

]≡ −∆G0(E) . (14.68)

We have that

G0(E+) =1

E + iε−H0

=(E −H0)

(E −H0)2 + ε2− iε

(E −H0)2 + ε2.

But10

limε→0(+)

ε

(E −H0)2 + ε2= π δ(E −H0) . (14.69)

10For a proof of this result see I. M. Gel’fand and G. E. Shilov Generalized Functions, Vol. 1, Sec. 2.4,Academic Press (1964).

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14.4. UNITARITY OF THE T-MATRIX 309

Using this result we can write

G0(E±) =P

E −H0

∓ iπδ(E −H0) , (14.70)

where the P indicates that we should calculate the principle value integral when theoperator is written in terms of the eigenstates of H0. Making use of Eq. (14.70), we canwrite ∆G0 as

∆G0(E) = −2iπ δ(E −H0) . (14.71)

We now can write Eq. (14.68) as

T−1(E−)[T (E+)− T (E−)

]T−1(E+) = −2iπ δ(E −H0) ,

and therefore

∆T (E) ≡ T (E+)− T (E−) = −2iπ T (E−) δ(E −H0)T (E+) , ( 14.72a)

orImT (E) = −π T (E−) δ(E −H0)T (E+) . ( 14.72b)

This is the operator form of the unitarity equation, and when taken in momentumrepresentation it gives the off-shell unitarity equation if the initial and final momenta aredifferent and not related to the energy E. The diagonal elements of this operator willgive us the optical theorem, i.e.,

〈~kf |ImT (E)|~ki〉 = −π 〈~kf |T (E−) δ(E −H0)T (E+)|~ki〉 . (14.73)

We now introduce a complete set of eigenstates of the momentum operator, which are alsoeigenstates of H0, next to the δ-function on the left hand side of Eq. (14.73). This allowsus to do the radial integral and get the intermediate momentum to have a magnitude k0

such that E =h2k2

0

2µ. We thus can write

〈~kf |ImT (E)|~ki〉 = −π∫d3k 〈~kf |T (E−)|~k〉 δ

(E − h2k2

)〈~k|T (E+)|~ki〉

= −π∫dk0 〈~kf |T (E−)|~k0〉

µk0

h2 〈~k0|T (E+)|~ki〉 . (14.74)

If we now take ~kf = ~ki, i.e., the forward direction, then the above unitarity equationreduces to

〈~ki|ImT (E)|~ki〉 = −πµk0

h2

∫dk0 |〈~ki|T (E)|~k0〉|2 . (14.75)

Taking |~ki| = k0, the T -matrices in the above expression are then on-shell and we can writethis equation in terms of the scattering amplitude f(k, θ). This is achieved by multiplying

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310 CHAPTER 14. SCATTERING THEORY; REVISITED

the above unitarity equation by the factor −4π2µh2 , and making use of Eq. (14.19) to relate

the T -matrix to the scattering amplitude. This gives us the result

−4π2µ

h2 〈~ki|ImT (E)|~ki〉 =4π2µ

h2

πµk0

h2

∫dk0 |〈~ki|T (E)|~k0〉|2 ,

or

Imf(k0, 0) =k0

4πσT . (14.76)

This result is identical to the Optical Theorem derived in Section 10.5.

14.5 The T -matrix for Separable Potentials

In Section 14.1 we showed that the Schrodinger equation plus the boundary conditioncan be replaced by an integral equation commonly known as the Lippmann-Schwingerequation. After partial wave expansion, this equation can be written as

〈k|T`(E+)|k′〉 = 〈k|V`|k′〉+

∞∫0

dk′′ k′′2〈k|V`|k′′〉 〈k′′|T`(E+)|k′〉

E+ − h2k′′2

. (14.77)

In this section we will solve this equation for a special class of potentials called separablepotentials. These have the special feature that the Lippmann-Schwinger equation canbe solved analytically because the kernel of the integral equation is separable. Becausewe can get an analytic solution to our equations, we will be able to present a generaldiscussion of the properties of the scattering amplitude that are valid in general for alarge class of potentials.

Before we proceed with our discussion, we will introduce some definitions that willsimplify the algebra without obscuring the physics. These are:

• We will take h2

2µ= 1.

• We will restrict our discussion to S-wave scattering, i.e., ` = 0, and therefore dropany reference to the angular momentum.

• Matrix elements of operators are understood in the sense that

〈φ|O|ψ〉 ≡∞∫0

dk k2 φ∗(k)O(k)ψ(k) . (14.78)

This is a result of the fact that we will be considering one partial wave only.

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14.5. THE T -MATRIX FOR SEPARABLE POTENTIALS 311

With these definitions we can write the operator form of the partial wave Lippmann-Schwinger equation as

T`(E+) = V` + V`G0(E+)T`(E

+) , ( 14.79a)

or with no reference to ` as

T (E+) = V + V G0(E+)T (E+) . ( 14.79b)

The separable potential we are considering is given in momentum representation forthe `th partial wave as

〈k|V`|k′〉 = 〈k|g`〉λ` 〈g`|k′〉 ( 14.80a)

= g`(k)λ` g`(k′) , ( 14.80b)

or in operator form with no reference to ` as

V = |g〉λ 〈g| . ( 14.80c)

Here, the function g`(k) is referred to as the form factor. We now can write the Lippmann-Schwinger equation, Eq. (14.79b), for this potential as

T (E+) = |g〉λ 〈g|+ |g〉λ 〈g|G0(E+)T (E+)

= |g〉λ〈g|+ 〈g|G0(E+)T (E+)

≡ |g〉λ 〈F| , (14.81)

where the bra 〈F| is defined in terms of the T -matrix as

〈F| = 〈g|+ 〈g|G0(E+)T (E+) . (14.82)

Making use of Eq. (14.81) we can write

〈F| = 〈g|+ 〈g|G0(E+)|g〉λ 〈F| . (14.83)

If we now solve this equation for 〈F| and substitute in Eq. (14.64), we get the result thatthe T -matrix has the form

T (E+) = |g〉 τ(E+) 〈g| , (14.84)

where

τ(E+) =λ−1 − 〈g|G0(E+)|g〉

−1

=

λ−1 −∞∫0

dk k2 [g(k)]2

E+ − k2

−1

. (14.85)

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312 CHAPTER 14. SCATTERING THEORY; REVISITED

In momentum representation this T -matrix takes the form

〈k|T (E+)|k′〉 = 〈k|g〉 τ(E+) 〈g|k′〉 , ( 14.86a)

orT (k, k′;E+) = g(k) τ(E+) g(k′) . ( 14.86b)

To establish the relation between the wave function for bound and scattering states andthe T -matrix, let us now consider the Schrodinger equation for the bound state problemusing this separable potential. In operator form, the Schrodinger equation can be writtenas

(E −H0)|ψ〉 = V |ψ〉 . (14.87)

Here again, we are considering the equation for a given angular momentum `. For boundstates, E < 0. Since the eigenvalues of H0, the kinetic energy in the center of mass, ispositive, we have no problems dividing by the operator (E − H0). Also, the equation(E −H0)|φ〉 = 0 has no solution. This allows us to write the solution of Eq. (14.87) forE < 0 as

|ψ〉 =1

E −H0

V |ψ〉 = G0(E)V |ψ〉 . (14.88)

Since E < 0 and (E − H0)−1 is not singular, we need not introduce the +iε required inSec. 14.1. In Eq. (14.88), we have a homogeneous integral equation, and if we replacethe integral by a sum, the integral equation will be converted to a set of homogeneousalgebraic equations that have a solution only for certain values of E, i.e., we have aneigenvalue problem.

For the separable potential considered above, Eq. (14.88) reduces to

|ψ〉 = G0(E)|g〉λ 〈g|ψ〉 for E < 0 . (14.89)

We note here, that 〈g|ψ〉 is a matrix element of the unit operator in the sense of Eq. (14.78),and therefore a constant that can be incorporated into the normalization of the wavefunction. In momentum representation, this wave function takes the simple form

〈k|ψ〉 = 〈k|(E −H0)−1|g〉λ 〈g|ψ〉

=1

E − k2〈k|g〉λ 〈g|ψ〉 , ( 14.90a)

or in functional form

ψ(k) = N g(k)

E − k2. ( 14.90b)

Here, N = λ 〈g|ψ〉 is the normalization of the wave function. To determine this normal-ization we need to determine 〈g|ψ〉. This we get by multiplying Eq. (14.89) from the leftby 〈g|. This gives us the result that

〈g|ψ〉 = 〈g|G0(E)|g〉λ 〈g|ψ〉 ,

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14.5. THE T -MATRIX FOR SEPARABLE POTENTIALS 313

or λ−1 − 〈g|G0(E)|g〉

λ 〈g|ψ〉 = 0 . (14.91)

For this equation to be satisfied, we either have

〈g|ψ〉 = 0 ,

which corresponds to the normalization being zero, i.e., zero wave function, or we havethat

λ−1 − 〈g|G0(E)|g〉

= 0 . (14.92)

This condition will determine the energy (E < 0) at which we have a bound state. If wenow compare this condition with Eq. (14.85), we find that the energy at which we havea bound state, the T -matrix has a singularity. To determine the form of this singularity,let us take the energy of the bound state to be E = B < 0. Then

λ−1 − 〈g|G0(B)|g〉

= 0 , (14.93)

but we have that

G0(E) =1

E −H0

=1

B −H0

(B −H0)1

E −H0

=1

B −H0

− 1

B −H0

(E −B)1

E −H0

.

This equation can be written as

G0(E) = G0(B)− (E −B)G0(B)G0(E) . (14.94)

With this result in hand we can write

〈g|G0(E)|g〉 = 〈g|G0(B)|g〉 − (E −B) 〈g|G0(B)G0(E)|g〉 .

Making use of this result and Eq. (14.85), we can write τ(E) as

τ(E) =1

(E −B) 〈g|G0(B)G0(E)|g〉. (14.95)

Therefore, at E ≈ B the T -matrix has a simple pole in the complex energy plane alongthe negative real axis. Having established the energy at which we have a bound state, wenow can write the wave function given in Eq. (14.90) as

|ψB〉 =N

B −H0

|g〉 , (14.96)

which in momentum space takes the simple form

ψB(~k) = N g(~k)

B − k2. (14.97)

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314 CHAPTER 14. SCATTERING THEORY; REVISITED

The normalization N is now determined by the condition that

1 = 〈ψB|ψB〉 = 〈g|G0(B)G0(B)|g〉N 2

and thereforeN = 〈g|G0(B)G0(B)|g〉 −

12 . (14.98)

Comparing this result for the normalization with Eq. (14.95), we observe that N 2 is theresidue of τ(E) at E = B, and the residue of the T -matrix is given by

|g〉N 2 〈g| ,

which is related to the square of the bound state wave function.We now turn to the scattering wave function which is given in Eq. (14.6). In momen-

tum space, this wave function takes the form

〈~k |ψ(+)~ki〉 = 〈~k |~ki〉+ 〈~k |G0(E+)V |ψ(+)

~ki〉

= δ(~k − ~ki) +1

E+ − k2〈~k |V |ψ(+)

~ki〉 . (14.99)

But we have from the definition of the T -matrix in Eq. (14.18) that

〈~k |V |ψ(+)~ki〉 = 〈~k |T (E+)|~ki〉

= 〈~k |g〉 τ(E+) 〈g|~ki〉 . (14.100)

With this result we can write the scattering wave function in momentum space as

〈~k |ψ(+)~ki〉 = δ(~k − ~ki) +

〈~k |g〉E+ − k2

τ(E+) 〈g|~ki〉 , ( 14.101a)

or in functional form this wave function is given by

ψ(+)~ki

(~k) = δ(~k − ~ki) +g(~k)

E+ − k2τ(E+) g(~k) . ( 14.101b)

Here we observe that the scattering wave function is related to the T -matrix. However,this T -matrix has the initial and final momentum as independent variables since ~ki is theinitial momentum of the incident beam and is related to the energy E = |~ki|2, while the

momentum ~k is a variable whose presence is the result of the fact that we are consider-ing the scattering wave function in momentum representation. This T -matrix that is afunction of two variables ~ki and ~k is known as the half-off-shell T -matrix.

To get the scattering wave function in coordinate space, we need to transform thewave function, i.e.,

〈~r |ψ(+)~ki〉 =

∫d3k 〈~r |~k〉 〈~k |ψ~ki〉

= 〈~r |~ki〉+∫d3k〈~r |~k〉 〈~k |g〉E+ − k2

τ(E+) 〈g|~ki〉 . ( 14.102a)

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14.5. THE T -MATRIX FOR SEPARABLE POTENTIALS 315

This equation can be written in functional form as

ψ(+)~ki

(~r) =1

(2π)3/2

ei~ki·~r +∫d3k ei

~k·~r g(~k)

E+ − k2τ(E+) g(~k)

, ( 14.102b)

where the form factors g(~k) are given by

〈~k |g〉 = g(~k) =1√4π

g(k) =1√4π〈k|g〉 . (14.103)

The factor of 1√4π

in the above equation is due to the fact that we have restricted our

analysis to S-wave, i.e., ` = 0, and Y00(k) = 1√4π

.All of the above results are valid, in general, for any potential even though we have

only derived the equation for a separable potential. Also, the extension to all angularmomentum proceeds in the same identical steps with the additional feature that thepotential and T -matrix have to be expanded in partial wave form. Finally, we shouldemphasis that the determination of the T -matrix, the bound state wave function, and thescattering wave function have been reduced to the evaluation of the integral

〈g|G0(E+)|g〉 =

∞∫0

dk k2 [g(k)]2

E − k2, (14.104)

which is required in determining the function τ(E).To get explicit expressions for the wave function and T -matrix, we need to specify the

form factor g(k). For S-waves, the simplest form factor we can have is

g(k) =1

k2 + β2. (14.105)

where β is a parameter that is referred to as the range of the interaction. For this formfactor, the integral in Eq. (14.104) is given by

〈g|G0(E+)|g〉 =π

2

1

(k20 + β2)2

(k2

0 − β2

2β− ik0

), (14.106)

where E = k20. Making use of this result in Eq. (14.85) we can determine τ(E+) which is

used in Eq. (14.86) to give us the on-shell T -matrix as

T (k0) = g(k0) τ(E+) g(k0)

=

λ−1

(k2

0 + β2)2− π

2

(k2

0 − β2

2β− ik0

)−1

. (14.107)

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316 CHAPTER 14. SCATTERING THEORY; REVISITED

This is the inverse of a second order polynomial whose zeros could give rise to boundstates if the corresponding energy was negative. To determine the phase shifts for thispotential, we make use of Eq. (14.34) to write the partial wave amplitude as11

f(k0) = eiδ sin δ = −πk0

2T (k0) . (14.108)

We now can write the phase shifts as

cot δ =Re T (k0)

ImT (k0),

and therefore

k0 cot δ = −β2

(1 +

4β3

πλ

)+k2

0

(1− 8β3

πλ

)− 2

πλk4

0 . (14.109)

To determine the scattering wave function in coordinate space, we need to calculate theintegral

∫d3k ei

~k·~r g(~k)

E+ − k2=

√π

ir

∞∫0

dk k

(eikr − e−ikr

)(k2

0 + iε− k2) (k2 + β2)

=

√π

ir

∞∫−∞

dkk eikr

(k20 + iε− k2) (k2 + β2)

. (14.110)

k

Ci β

-i β

k

0 + iε

-k0 - iε

Figure 14.3: The contour of integration used to calculate the scattering wave function incoordinate representation.

This integral can be evaluated by closing the contour of integration in the upper half ofthe complex k-plane, and then by using Cauchy’s theorem to write the integral in termsof the residues at the poles inside the contour. In this case the integrand has four poles in

11We have modified Eq. (14.34) to take into account the fact that in this section we have taken 2µh2 = 1.

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14.5. THE T -MATRIX FOR SEPARABLE POTENTIALS 317

the complex k-plane (see Figure 14.3), and only two of the poles are inside the contour.These poles are at k = iβ and k = k0 + iε. The pole at k = iβ comes from the form factorand its position depends on the parameters of the potential and in particular the rangeof the interaction. On the other hand, the pole at k = k0 + iε is from the free-particleGreen’s function, and its position depends only on the energy of the incident beam inthe center of mass. In other words, this second pole is independent of the form of thepotential. Using the residue theorem to calculate the integral in Eq. (14.110), we get∫

d3k ei~k·~r g(~k)

E+ − k2=

2π3/2

r

k eikr

(k20 + iε− k2) (k + iβ)

∣∣∣∣∣k=iβ

− k eikr

(k0 + iε+ k) (k2 + β2)

∣∣∣∣∣k=k0+iε

= π3/2

(e−βr − eik0r

r

)1

k20 + β2

. (14.111)

We now can write the scattering wave function as

〈~r |ψ(+)~ki〉 = 〈~r |~ki〉 −

√π

2

(eikf r − e−βr

r

)〈~kf |g〉 τ(E+) 〈g|~ki〉

=1

(2π)3/2

ei~ki·~r − 2π2

(eikf r − e−βr

r

)〈~kf |T (E+)|~ki〉

,( 14.112a)

where we have taken k0 = |~ki| = |~kf |. This equation can be written in functional form as

ψ(+)~ki

(~r) =1

(2π)3/2

ei~ki·~r − 2π2

(eikf r − e−βr

r

)T (~kf , ~ki;E

+)

. ( 14.112b)

This scattering wave function, which is valid for all ~r, consists as the sum of three parts:

1. The first part is the incident plane wave with momentum ~ki and given by

ei~ki·~r .

The amplitude of this wave is unity.

2. The second term is a spherical outgoing wave with momentum ~kf pointing in theradial direction. This part is given by

−4π2 T (~kf , ~ki, E+)eikf r

r.

The amplitude of this wave is the scattering amplitude given by12

f(k0, θ) = −4π2 T (~kf , ~ki;E+) ,

with cos θ = kf · ki.12Recall that we have taken h2

2µ = 1 when comparing this result with Eq. (14.19).

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318 CHAPTER 14. SCATTERING THEORY; REVISITED

3. Finally, the scattering wave function has a component that is decaying because itis proportional to

e−βr

r.

The rate of decay of this component depends on the parameter of the potentialβ. This is why we referred to this parameter as the range of the potential. Herewe note that the longer the range of the potential the further we have to go fromthe scattering center before we get the asymptotic solution that has the scatteringamplitude and therefore the phase shifts.

We next turn to the bound state wave function for the form factor given in Eq. (14.105).In momentum space this wave function is of the form

〈~k |ψB〉 = N 〈~k |g〉

B − k2

= − N√4π

g(k)

k2 + α2

= − N√4π

1

(k2 + α2)(k2 + β2), (14.113)

where α2 = −B. In coordinate space this wave function takes the form

〈~r |ψB〉 =∫d3k 〈~r |~k 〉 〈~k |ψB〉

=1

(2π)3/2

∫d3k ei

~k·~r 〈~k |ψB〉 . (14.114)

Making use of Eq. (14.113), we get after doing the angular integration

〈~r |ψB〉 = − N√2

1

2πi

1

r

+∞∫−∞

dkk eikr

(k2 + α2) (k2 + β2). (14.115)

This integral can be converted, as was the case for the scattering wave function, to acontour integral by closing the contour in the upper half k-plane. In this case there aretwo poles inside the contour. One pole at k = iβ comes from the form factor and itsposition is the same as in the scattering wave function case. The second pole at k = iαcomes from the Green’s function. This pole was on the positive real axis for the scatteringcase, and has in a continuous manner moved to the positive imaginary axis as we changeour energy from a positive value for the scattering case to a negative value for the boundstate. Using the residue theorem we can perform the integration to get

〈~r |ψB〉 = − N√2

1

2(β2 − α2)

(e−αr − e−βr

r

). (14.116)

This bound state wave function, up to a normalization, is an analytic continuation of thespherically outgoing wave part of the scattering wave function.

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14.6. SPIN DEPENDENT SCATTERING 319

14.6 Spin Dependent Scattering

Since the two nucleons have spin and can be in a total spin singlet and triplet, we canconsider the scattering amplitude to consist of an amplitude for scattering in spin singletand an amplitude for scattering in spin triplet, i.e., we can write our total amplitude as

f(k, θ) = Ps fs(k, θ) + Pt ft(k, θ) , (14.117)

where Ps and Pt are the spin singlet and spin triplet projection operators. Making useof the facts that ~S = ~s1 + ~s2 and and that the nucleon spin can be written in terms ofthe Pauli spin matrices as ~s = 1

2h~σ, we can determine the matrix elements of ~σ(1) · ~σ(2)

between states of total spin to be

〈~σ(1) · ~σ(2)〉 =

−3 for S = 0+1 for S = 1

. (14.118)

With this result in hand, we can write the spin single and spin triplet projection operatorsas

Ps =1

4[1− ~σ(1) · ~σ(2)] ( 14.119a)

Ps =1

4[3 + ~σ(1) · ~σ(2)] . ( 14.119b)

As we would expect, the projection operators satisfy the condition Ps + Pt = 1. Thusif the singlet amplitude was the same as the triplet amplitude, i.e., fs = ft, then wewouldn’t have any spin dependence in the interaction.

The cross section for scattering in this case will depend on our measurement of thedirection of the spin in the initial and final state. If we carry out no measurement onthe orientation of the spin in the initial and final state, then we need to average over allpossible orientations of the spin of the two particles in the initial state, and sum over allpossible orientations of the spin of the final two particles, i.e., the differential cross willbe section

dΩ=

1

(2s1 + 1)(2s2 + 1)

∑m1m2

∑m3 m4

|〈m1,m2|f(k, θ)|m3,m4〉|2 , (14.120)

where m1 and m2 are the projection of the spin s1 and s2 of the two particles in the initialstate, and m3 and m4 are the projection of the spin of the particles in the final states.This result is valid for the scattering of any two particles of spin s1 and s2 which producestwo particles with spin s3 and s4. For the case of two nucleon scattering, this equationreduces to

dΩ=

1

4

∑m1 m2

∑m3m4

|〈m1,m2|f(k, θ)|m3,m4〉|2 . (14.121)

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320 CHAPTER 14. SCATTERING THEORY; REVISITED

If we couple the spin in the initial and final states to states of total spin, and take intoconsideration the fact that spin is a good quantum number so that the matrix elementsof the amplitude are diagonal in the total spin, we can write

〈m1,m2|f(k, θ)|m3,m4〉 =∑SMS

( 12m1, 1

2m2| 12 1

2SMS) ( 1

2m3, 1

2m4| 12 1

2SMS)

×〈S|f(k, θ)|S〉 . (14.122)

Using the orthogonality of the Clebsch-Gordan coefficients Eqs. (12.66) and (12.67), wecan write the cross section as

dΩ=

1

4

∑SMS

|〈S|f(k, θ)|S〉|2

=3

4|ft(k.θ)|2 +

1

4|fs(k, θ)|2 . (14.123)

We note again that if ft = fs, then our expression for the differential cross section reducesto the one for spinless particles, which what we expect.

14.7 Effective Range Parameters

In Sec.refSec.14.2, we considered the Born approximation which can have good conver-gence at high energies since it can be thought of as a power series in E−1. We now turn toan approximation that is valid at low energies. For low energies, the scattering amplitudeis dominated by S-wave scattering, i.e. ` = 0, and the scattering amplitude can be writtenas

f(k, θ) =1

k

∑` (2`+ 1) eiδ` sin δ` P`(cosθ)

≈ 1

keiδ0 sin δ0

=1

k cot δ − ik, (14.124)

where in the last line we dropped the subscript 0 on the δ with the understanding that wewill be considering S-wave scattering only. Since we are considering low energy scattering,we should be able to expand k cot δ as a power series in the energy or k2, i.e.,

k cot δ = −1

a+

1

2rek

2 + · · · . (14.125)

This series is known as the effective range expansion, with a being the scattering lengthand re the effective range.

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14.7. EFFECTIVE RANGE PARAMETERS 321

To develop this expansion, we consider the radial Schrodinger equation at two energiesE1 and E2, i.e.,

d2u1

dr2+

h2 [E1 − V (r)]u1 = 0 , ( 14.126a)

andd2u2

dr2+

h2 [E2 − V (r)]u2 = 0 , ( 14.126b)

where V (r) is the potential. If we multiply Eq. (14.126a) by u2 and Eq. (14.126b) byu1 and subtract, we get

u2d2u1

dr2− u1

d2u2

dr2=(k2

2 − k21

)u2(r)u1(r) . (14.127)

We now integrate this equation to getr0∫

0

dr

u2d2u1

dr2− u1

d2u2

dr2

=(k2

2 − k21

) r0∫0

dr u2(r)u1(r) .

If we now integrate the left hand side of the equation by parts, i.e.,r0∫

0

dr

u2d2u1

dr2− u1

d2u2

dr2

=

(u2(r)

du1

dr− u1(r)

du2

dr

) ∣∣∣∣∣r0

0

we get (u2(r)

du1

dr− u1(r)

du2

dr

) ∣∣∣∣∣r0

0

=(k2

2 − k21

) r0∫0

dr u2(r)u1(r) . (14.128)

If r0 is taken to be the range of the potential, then for r > r0 the radial wave functiontakes the form

ui(r) ∝ sin(kir + δi) for i = 1, 2 . (14.129)

We now define the function vi(r) as

vi(r) ≡sin(kir + δi)

sin δifor i = 1, 2 . (14.130)

Since the function vi(r) satisfies the Schrodinger equation, it also satisfies the equation(v2(r)

dv1

dr− v1(r)

dv2

dr

) ∣∣∣∣r00

=(k2

2 − k21

) r0∫0

dr v1(r) v2(r) . (14.131)

If u(r) is chosen such that u(r) = v(r) for r ≥ r0, and since u(0) = 0 from the boundarycondition at the origin, subtracting Eq. (14.128) from Eq. (14.131) and making use of thefact that vi(0) = 1, we get(

dv2

dr− dv1

dr

) ∣∣∣∣∣r=0

=(k2

2 − k21

) ∞∫0

dr [v1(r) v2(r)− u1(r)u2(r)] . (14.132)

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322 CHAPTER 14. SCATTERING THEORY; REVISITED

But we have thatdvidr

∣∣∣∣∣r=0

= k cot δi , (14.133)

and therefore

k2 cot δ2 = k1 cot δ1 +(k2

2 − k21

) ∞∫0

dr [v1(r) v2(r)− u1(r)u2(r)] . (14.134)

Taking k1 → 0 and k2 → k, we get

k cot δ = −1

a+

1

2k2ρ(0, E) , (14.135)

where

1

2ρ(E1, E2) =

∞∫0

dr [v1(r) v2(r)− u1(r)u2(r)]

=

r0∫0

dr [v1(r) v2(r)− u1(r)u2(r)] . (14.136)

For low energies, the wave function u(r) is not very sensitive function of E. Thus, we candefine the effective range as

re = ρ(0, 0) , (14.137)

and we have

k cot δ = −1

a+

1

2rek

2 + · · · . (14.138)

If we now go through the above procedure with E2 = −B, where B is the binding energyof the two-body system, then

v2(r) = e−αr where α2 =2µB

h2 , (14.139)

and thendv2

dr

∣∣∣∣∣r=0

= −α . (14.140)

In this way we can write the binding energy in terms of the effective range parameters,i.e.,

− α = −1

a− 1

2α2ρ(0, B)

≈ −1

a− 1

2α2ρ(0, 0) = −1

a− 1

2α2re . (14.141)

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14.7. EFFECTIVE RANGE PARAMETERS 323

In other words we can use this effective range expansion to determine the behavior of thesystem in the neighborhood of the origin in the energy variable. We will next apply thiseffective range expansion to neutron-proton and proton-proton scattering.

With the above results, we are in a position to write the phase shifts in terms of thecross section measured experimentally. Since we are considering low energy scattering,i.e. E < 10 MeV. and therefore k < 1

2, the only important partial wave is the ` = 0,

and we can use effective range theory as a parameterization of the S-wave phase shifts ork cot δ, i.e.,

k cot δ = −1

a+

1

2k2re − Pk4r3

e + · · · , (14.142)

where

a = Scattering length

re = Effective range

P = Shape parameter .

Since we have that

limk→0

f(k, θ) = −a

then the differential cross section at zero energy is given by

dΩ= a2

and the total cross section is

limE→0

σT = 4πa2 . (14.143)

Thus the scattering length is a measure of the cross section at zero energy.We now turn to the determination of the scattering length for the singlet and triplet.

First, let us consider the scattering of neutrons from molecular hydrogen. Molecularhydrogen is a mixture of

ortho-hydrogen parallel spin for the two protons. ↑↑para-hydrogen antiparallel spin for the two protons. ↑↓ .

The para-hydrogen is a lower energy state than the ortho-hydrogen, and therefore as welower the temperature of molecular hydrogen more of the molecules drop into the lowerenergy state of para-hydrogen until at a temperature of ≈ 20 K, molecular hydrogenis predominantly para-hydrogen. We can take advantage of this property of molecularhydrogen to carry out experiments in which neutrons are scattered at low energy fromthe two protons in the molecule coherently and in this way determine the singlet andtriplet scattering length.

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324 CHAPTER 14. SCATTERING THEORY; REVISITED

The scattering amplitude in Eq. (14.117) can be written with the help of Eq. (14.119)as

f(k, θ) =1

4(3ft + fs) +

1

4(ft − fs) ~σ(1) · ~σ(2) . (14.144)

At very low energies we have that f ≈ −a and we can write an effective scattering lengthin terms of the singlet and triplet scattering length, i.e.,

a =1

4(3as + at) +

1

4(at − as) ~σn · ~σp

=1

4(3as + at) +

1

4(at − as)~sn · ~sp , (14.145)

where ~sn and ~sp are the spin operators for the neutron and proton respectively. For lowenergy scattering of neutrons from molecular hydrogen, the scattering off the two protonsin the molecule is coherent, and in that case we can write the effective scattering lengthas

aH2 =1

2(3at + as) + (at − as) ~sn · (~sp1 + ~sp2)

=1

2(3at + as) + (at − as) ~sn · ~SH2 . (14.146)

The cross section involves squaring this and averaging over spins to get 13

a2 =1

4(3at + as)

2 +1

4(at − as)2 SH2(SH2 + 1)

=

14

(3at + as)2 + 1

2(at − as)2 for ortho-hydrogen SH = 1

14

(3at + as)2 for para-hydrogen SH = 0

. (14.147)

13 The squaring of the effective scattering length a gives an expression that includes the operator~sn · ~SH2 and its square. The average over spin of this operator is zero, i.e.

〈~sn · ~SH2〉 = 0 .

On the other hand, we have that(~sn · ~SH2

)2

=14

(~σn · ~SH2

) (~σn · ~SH2

)=

14

[~SH2 · ~SH2 + i~σn ·

(~SH2 × ~SH2

)]=

14

[~SH2 · ~SH2 − ~σn · ~SH2

].

Averaging this over spin, we get zero from the second term on the right hand side, while the first termgives SH(SH + 1)/4.

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14.7. EFFECTIVE RANGE PARAMETERS 325

From experiment, we have that

σpara ≈ 4 barns ( 14.148a)

σortho ≈ 125 barns ( 14.148b)

where 1 barn = 10−24 cm2. Comparing these cross sections with the expression givenin Eq. (14.147), we may conclude that as < 0 if at > 0 and vice versa. We now candetermine the sign of at from the knowledge that the triplet ` = 0 channel has a boundstate, the deuteron, with a binding energy is 2.2246 MeV. Since this is a small binding onthe nuclear scale, we could use the effective range expansion given in Eq. (14.141), i.e.,

α =1

a+

1

2α2re (14.149)

to determine the scattering length. For a binding energy of 2.2246 MeV, we have thatα = 0.23, and therefore taking the effective range to be of the order of magnitude of therange of the nucleon-nucleon interaction, i.e. re ≈ 1.4 fm, we get the triplet scatteringlength to be

at ≈ 5.15 fm .

Since the triplet scattering length is positive, then the singlet scattering length must benegative. To determine the magnitude of singlet scattering length, we make use of thefact that according to our estimate the triplet total cross section is given by

σt ≈ 4πa2t ≈ 3.0 barnes .

But the experimental total cross section is given by

σT =1

4σs +

3

4σt ≈ 20. barns ,

and thereforeσs = 4σT − 3σt ≈ 71 barns.

This corresponds to a singlet scattering length of

as ≈ −24 fm.

A more accurate analysis of the data gives us the triplet and singlet scattering length at

at = 5.414± 0.005 fm as = −23.719± 0.013 fm . ( 14.150a)

To get the effective range, we need to get measurements of the cross section at low energyand analyze the data using the effective range expansion. This gives for the effectiveranges for the triplet and singlet channel to be

ret = 1.75± 0.005 fm res = 2.76± 0.05 fm . ( 14.150b)

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326 CHAPTER 14. SCATTERING THEORY; REVISITED

When comparing the zero energy cross section for the singlet (71 barns) and triplet (3barns), we find that the singlet cross section is much larger than the triplet cross section.To see if this is due to a bound state or resonance near the zero of the energy, we use thescattering length and effective range in Eq. (14.133) to determine the binding energy ofthat state, i.e. we solve the equation

αs =1

as+

1

2α2sres .

This equation is a quadratic in αs with two solutions corresponding to

αs = 0.04 fm and 1.39 fm .

Clearly, αs = 1.39 fm is not a solution, since there is no such nucleus with spin zero.However, the large cross section at zero energy suggests that there is a virtual singletstate at

αs = −0.05 fm .

The wrong sign for αs in our analysis is due to the fact that the singlet cross section σsdoes not determine the sign of the scattering length.

14.8 Problems

1. The potential between a proton and a neutron is due to the exchange of a pion.This potential is given by

V (r) = V0e−mπr

r,

where mπ is the mass of the pion and V0 is a constant.

(a) Calculate the Born amplitude for proton-neutron scattering.

(b) Write the differential cross section for proton-neutron scattering, in Born ap-proximation, as function of the scattering angle.

2. The potential for electron-hydrogen scattering can be written as

V (r) = −e2

r+∫d3r′|φ100(r′)|2 e2

|~r − ~r ′|

(a) Using the expression

1

|~r − ~r ′|= 4π

∑`m

1

2`+ 1

r`<r`+1>

Y`m(r)Y ∗`m(r′)

simplify the potential V (r) taking for the wave function φ100(r) the groundstate of the hydrogen atom.

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14.8. PROBLEMS 327

(b) Calculate the differential cross section in the Born Approximation as a functionof the angle at intervals of 10 for 0 < θ < 60, and incident electrons of energy1 KeV.

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328 CHAPTER 14. SCATTERING THEORY; REVISITED

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Appendix A

Complex Analysis in a Nutshell

In this appendix I would like to very briefly summarize those features of complex analysisoften encountered in a course on quantum mechanics:

1. The Fourier transforms of the wave function from coordinate to momentum space,and vice versa, often requires the use of Cauchy’s theorem and the analytic structureof integrand in the complex plane.

2. The determination of the Green’s function for differential equation, and in par-ticular the free Green’s function, requires the evaluation of integrals that can bedetermined with the help of Cauchy’s theorem. This in terms allows the inclusionof the boundary condition of the differential equation in the Green’s function.

3. The study of scattering theory, the amplitude for scattering is a complex function ofthe energy. The structure of this function often reflects the physics of the scatteringprocess.

In all three cases, we find a sound understanding of complex analysis, and in particularthe use of Cauchy’s theorem, plays a central role in understanding the mathematicalframework, as well as the physical phenomena.

A.1 Functions of a Complex Variable

We can represent a complex variable either in rectangular coordinates as z = x+ iy, or inpolar coordinates as z = reiθ. These two representations of a complex number are relatedby

r =√x2 + y2 tan θ =

y

x. (A.1)

We now can define a function of a complex variable as f(z), e.g.,

f(z) = z2 = x2 − y2 + 2ixy = r2 e2iθ

= fR(x, y) + ifI(x, y) (A.2)

329

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330 APPENDIX A. COMPLEX ANALYSIS IN A NUTSHELL

z f

a

b

cA

B

C

Figure A.1: The complex z-plane and the corresponding f -plane, where f(z) = z2. Thepoints a, b and c in the z-plane are mapped into the points A, B and C in the f -plane.

wherefR(x, y) = x2 − y2 = r2 cos 2θ fI(x, y) = 2xy = r2 sin 2θ (A.3)

In other words for every point (x, y) in the complex z-plane we have a point (fR, fI) inthe complex f -plane, see Figure A.1 below. However, we observe in this example thatthe points in the upper half of the z-plane are mapped onto the full f -plane. Similarly,the points in the lower half of the z-plane are mapped onto the full f -plane. In fact, thepoints +z and −z are mapped onto the same point in the f -plane.

The inverse transformation

f(z) = z1/2 = r1/2 eiθ/2 (A.4)

has the problem that for a given z we have two possible values of f(z), i.e. the functionis multi-valued. In fact for every value of (x, y) we have the two values of f(z) = (fR, fI)corresponding to θ/2 and (θ + 2π)/2. In this case the z-plane is mapped onto the upperhalf of the f -plane OR the lower half of the f -plane, see Figure A.2.

There are many other mappings that we could consider, e.g.

f(z) = log(z) = log(reiθ) = log r + iθ . (A.5)

In this case, replacing θ → θ+2nπ does not change z but changes f(z). In fact we shouldwrite

fn(z) = log(z) = log r + i(θ + 2nπ) . (A.6)

In this case for a given value of z we can generate an infinite number of possible valuesof f(z) = log z by changing the integer n.

To overcome this problem of the multi-valued nature of mappings, to allow for thefunction f(z) to change in a continuous manner when z is varied continuously, Riemannsuggested the following procedure, whereby he assigned the same values of f(z) corre-sponding to different values of z to different complex planes. To illustrate this, consider

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A.1. FUNCTIONS OF A COMPLEX VARIABLE 331

z f

a

b

c

A

B

C

C'A'

B'

Figure A.2: The complex z-plane and the corresponding f -plane, where f(z) = z1/2. Thepoints a, b and c in the z-plane are mapped into the points A, B and C, or the points A′,B′ and C ′.

the function

f(z) = z2 .

We could assign the first complex f -plane to those values of z corresponding to the upperhalf of the z-plane, i.e. Im(z) > 0, while the second complex f -plane could correspondto those values of z with Im(z) < 0. If we now write z in polar coordinates, i.e. z = reiθ,then for r > 0 as we increase θ from 0 to π/4 to π/2 to 3π/4 to π, the point f(z) in thef -plane proceeds from the first quadrant to the second, third and then fourth quadrantof that plane. Now as we continue to vary the point z by changing θ from θ = π − ε toθ = π + ε with ε 1, we move from the fourth quadrant of the first f -plane, to the firstquadrant of the second f -plane. In this way we get a continuous change in f(z) as wechange z. To get back onto the first f -plane we need change θ from π to 3π/2 to 2π. Thischange in θ from π to 3π/2 to 2π will allow us to scan the second f -plane. As we proceedfrom θ = 2π − ε to θ = 2π + ε, which is the same as θ = ε, we return from the secondf -plane to the first f -plane. Thus we have two planes that are continuously joined. Theseplanes are referred to as Riemann sheets and we say that the function f(z) = z2 has atwo Riemann sheet structure. The point r = 0 is a problem, in that it either belongs toboth sheets, or the function f(z) is not continuous at this point. This point is known asa branch point. The lines along which the two planes join is known as a branch cut. Inthe above analysis and Figure A.3 this branch cut is taken to be along the positive realaxis. However, the branch cut could be from the point z = 0 in any radial direction tothe infinity circle.

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332 APPENDIX A. COMPLEX ANALYSIS IN A NUTSHELL

z f

a

b

cA

B

C

Figure A.3: The complex z-plane and the corresponding f -plane, where f(z) = z2. Inthe f -plane we have a branch point at the origin and a branch cut along the positive realaxis.

A.2 Analytic Functions

In the last section we considered the function of a complex variable f(z) that can bewritten as

f(z) = fR(x, y) + ifI(x, y) , (A.7)

where both fR(x, y) and fI(x, y) are real functions of (x, y). We now define the derivativeof this function of a complex variable as

df

dz= lim

h→0

f(z + h)− f(z)

h, (A.8)

where h is a complex number. Unlike a function of a real variable, there are an infinitenumber of ways we can approach the point z in the complex z-plane. The derivative ofthe function f(z) along each direction could be different, in which case the concept of aderivative is not meaningful. Thus for the derivative to be meaningful, we have to makesure that the above definition of a derivative does not depend on the direction. This canbe achieved by requiring that the derivative of f(z) be the same with respect to the twodirections, x and iy, in the complex z-plane. We have that the derivative with respect tothe Re(z) = x is

∂f

∂x=∂fR∂x

+ i∂fI∂x

, (A.9)

while the derivative with respect to the Im(z) = iy is

1

i

∂f

∂y=

1

i

(∂fR∂y

+ i∂fI∂y

)=∂fI∂y− i ∂fR

∂y. (A.10)

For these two derivatives to be identical we require that,

∂fR∂x

=∂fI∂y

∂fI∂x

= −∂fR∂y

. (A.11)

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A.3. CAUCHY’S THEOREM 333

These two conditions are known as the Cauchy-Riemann condition. If these conditionsare satisfied, we can say that the function f(z) is differentiable, and if the function f(z)is differentiable at z it is said to be an analytic function at z. If the function f(z) is notdifferentiable at z, then the function has a singularity at z.

If the function f(z) is analytic for all z except for z = a, then the point z = a isan isolated singularity. If now the function f(z) behaves like (z − a)−n with the integern > 0, for z in the neighborhood of a, then the singularity of f(z) is an nth-order pole,and for n = 1 we have a simple pole.

A.3 Cauchy’s Theorem

We now turn to the integral of a complex function f(z). For functions of a real variable,the integral was defined on a segment of a line, e.g. the integral

b∫a

dx f(x) (A.12)

is over the interval a < x < b. On the other hand for functions of complex variablez, the integral is from one point in the z-plane (x1, y1) to another point in the z-plane(x2, y2) along a specified path. Thus the integral of a function of a complex variable canbe written as ∫

Cdz f(z) =

∫C

(dx+ idy) [fR(x, y) + ifI(x, y)] , (A.13)

where C specifies the path of integration in the complex z-plane. If the path is a closedpath specified by C, we write the integral as∮

Cdz f(z) . (A.14)

If the function f(z) is analytic in a domain D, and the contour C is in this domainand is piecewise continuous, see Figure A.4, then∮

Cdz f(z) = 0 . (A.15)

This is known as the Cauchy Integral Theorem. This theorem has strong implications tothe extent that we can deform the integration path C without changing the value of theintegral, provided the function f(z) is analytic over the domain in which the contourof integration resides. However, we can not deform the contour of integration to includenew poles or singularities of the function f(z) without changing the value of the integral.The simplest case of an integral over a contour which includes a singularity is the casewhere the singularity is a simple pole, i.e.∮

Cdz

1

z − a,

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334 APPENDIX A. COMPLEX ANALYSIS IN A NUTSHELL

D

C

Figure A.4: The function f(z) is analytic over the domain D, the circle labeled C is thecontour of integration. This contour is in the domain D.

where the point z = a is inside the contour C. We now can change the contour C to bea circle of radius ε centered about the pole at z = a. The integral in this case can bewritten as an integral over an angle θ. This can be achieved by defining z−a = εeiθ, withdz = iεeiθ dθ, so that the integral over the contour C can be written as

∮Cdz

1

z − a= i

2π∫0

dθ = 2πi . (A.16)

In the event that we have a higher order pole we have

∮Cdz

1

(z − a)n= i

2π∫0

dθ (εeiθ)1−n = iε1−n2π∫0

[cos(n− 1)θ − i sin(n− 1)θ] = 0 for n 6= 1 .

(A.17)We now consider the more general case where we have a function f(z) that is analyticover the domain D, with the contour C also in this domain. Then we have that∮

Cdz

f(z)

z − a= 2πif(a) . (A.18)

To prove this result we collapse the contour to a circle of radius ε 1. In this way thefunction f(z) can be taken out of the integral and given the value f(a). Now with thehelp of Eq. (A.17) we have established the result in Eq. (A.18). Eq. (A.18) is a specialcase of a more general result where we can write the value of a function f(z) in terms ofa contour integral that includes the point z, i.e.

f(z) =1

2πi

∮Cdz′

f(z′)

z′ − z(A.19)

This result is often used in physics to determine the value of a function at a point z inthe complex plane given the measured values of the function along the contour C.

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A.4. TAYLOR AND LAURENT SERIES 335

A.4 Taylor and Laurent Series

For functions of a real variable we can expand a function f(x) about a point x = a as aTaylor series of the form

f(x) = f(a) + (x− a)df

dx

∣∣∣∣∣x=a

+ · · ·+ (x− a)n

n!

dn

dxn

∣∣∣∣∣x=a

+ · · · . (A.20)

This series converges for |x−a| < 1. We can carry over this Taylor expansion to functionsof a complex variable z by writing

f(z) =∞∑n=0

cn(z − a)n . (A.21)

This series has a radius of convergence that is given by |z − a| < 1. Since the right handside of this expansion is analytic, we are restricted to the application of this expansionover domains where the function f(z) is analytic. To extend this expansion to includefunctions that have singularities at a point a, with this singularity being either an nth

order pole or essential singularity, we introduce the Laurent series, i.e.,

f(z) =+∞∑

n=−∞cn (z − a)n . (A.22)

Then the coefficient cm can be determined by multiplying Eq. (A.22) by (z − a)−(m+1)

and integrating the resultant expression over a closed contour C that includes the pointZ = a, i.e.

∮Cdz

f(z)

(z − a)m+1=

+∞∑n=−∞

cn

∮Cdz (z − a)n−m−1

=m−1∑n=−∞

cn

∮C

(z − a)n−m−1 + cm

∮C

dz

z − a

++∞∑

n=m+1

cn

∮Cdz (z − a)n−m−1 . (A.23)

The first integral on the right hand side is zero according to Eq. (A.17), while the lastintegral on the right hand side is zero because the integrand is analytic in the domainthat includes the contour C. Using Eq. (A.16) for the remaining integrals on the righthand side we get

cm =1

2πi

∮Cdz

f(z)

(z − a)m+1. (A.24)

This allows us to determine the coefficients of the Laurent series.

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336 APPENDIX A. COMPLEX ANALYSIS IN A NUTSHELL

A.5 Residue Theorem

Let us now consider the Laurent series for the function f(z). The integral of this functionover a closed contour is given by∮

Cdz f(z) =

+∞∑n=−∞

cn

∮Cdz (z − z0)n

= c−1

∮C

dz

z − z0

= 2πic−1 . (A.25)

We now can define the residue of the function f(z) as

Res[f(z0)] = c−1 =1

2πi

∮Cdz f(z) . (A.26)

This assumes that f(z) has only one simple pole inside the contour C. In the event thatwe have several poles inside the contour C, we can write the Residue Theorem as∮

Cdz f(z) = 2πi

n∑i=1

Res[f(zi)] , (A.27)

where zi, i = 1, . . . , n are the positions of poles of f(z) inside the contour C. This ResidueTheorem is very useful in calculating Fourier transform and the determination of theGreen’s function for the Schrodinger equations and the Laplace’s equation in Electromag-netic theory.

For further reading on the subject and examples of the application of complex analysisto physical problems you can consider any book on Mathematical Physics or ComplexAnalysis, e.g.

1. C. W. Wong, Introduction to Mathematical Physics: Methods & Concepts, OxfordUniversity Press (1991).

2. G. Arfken, Mathematical Methods for Physicists, Academic Press (1985).

3. M. L. Boas, Mathematical Method for Physical Sciences, Wiley (1983).

A.6 Problems

1. Use the Residue Theorem to calculate the following integrals;

(a)

+∞∫−∞

dx1

(x2 + b2)2

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A.6. PROBLEMS 337

(b)

+∞∫−∞

dxeikx

(x2 + b2)2

(c)

+∞∫−∞

dxcos kx

x2 + a2

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338 APPENDIX A. COMPLEX ANALYSIS IN A NUTSHELL

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Appendix B

Atomic Units

In most fields of physics the units are chosen such that all observables are of the orderone. This facilitates most computational problems by avoiding under flows and over flowson modern computers. In Atomic physics the most commonly choice of units are atomicunits in which h = e = me = 1. To see how these units come about, let us considerthe Schrodinger equation for the Coulomb problem as a model of the Hydrogen atom. Inunits where length is measured in nm (1 nm = 10−9m) or fm (1 fm = 10−15m), whileenergy is measured in eV or MeV , we have[

− h2

2me

∇2 − e2

r

]ψ = E ψ . (B.1)

In these units hc = 197.3nmeV = 197.3 fmMeV . Then the Bohr radius is given by

aB =h2

mee2= 0.0529nm , (B.2)

while the Rydberg, R, is given by

R =mee

4

2h2 = 13.6058 eV. . (B.3)

We now can write the Schrodinger equation as[−1

2∇2 − me e

2

h2

1

r

]ψ =

meE

h2 ψ , (B.4)

or [−1

2∇2 −− 1

aB r

]ψ =

meE

h2 ψ . (B.5)

Multiply this equation by a2B to get[

−a2B

2∇2 − aB

r

]ψ =

(h2

mee2

)2meE

h2 ψ =E

2Rψ . (B.6)

339

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340 APPENDIX B. ATOMIC UNITS

If we now measure lengths in units of aB and energies in units of 2R, then our equationreduces to [

−1

2∇2 − 1

r

]ψ = E ψ . (B.7)

This is equivalent to takingh = e = me = 1 (B.8)

in the Schrodinger equation. In these units the ground state of the Hydrogen atom is −12.

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Appendix C

Numerical Solution of theSchrodinger Equation

To calculate the phase shifts for a general central potential V (r), we need to numericallysolve the Schrodinger equation, Eq. (10.33), for the radial wave function ψ`(r). This isachieved by integrating the differential equation from the origin up to a point r0 beyondthe range of the potential, V (r), and then by matching this numerical solution, and itsfirst derivative, to the asymptotic solution, which is given in Eq. (10.39). In this sectionwe present the numerical procedure for integrating the Schrodinger equation for the radialwave function.

If we write the radial wave function, ψ`(r), as

ψ`(r) =u`(r)

r(C.1)

then the Schrodinger equation for u`(r) can be written as a second order differentialequation without a first order derivative, of the form

d2u`dr2

+ w(r)u`(r) = 0 , (C.2)

or

u′′` (r) + w(r)u`(r) = 0 , (C.3)

where

w(r) =2µ

h2 [E − V (r)]− `(`+ 1)

r2. (C.4)

Here µ is the reduced mass, E is the energy, and ` the angular momentum.To solve the above differential equation numerically, we are going to take advantage

of the fact that the equation does not include terms proportional to the first derivative,u′`(r). This will improve the numerical accuracy of the procedure. In general, to solve a

341

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342APPENDIX C. NUMERICAL SOLUTION OF THE SCHRODINGER EQUATION

differential equation numerically, we convert the equation to a difference equation, andthen solve the difference equation. This involves writing the first derivative as

du

dr= lim

h→0

u(r + h)− u(r − h)

2h

≈ u(r + h)− u(r − h)

2h, (C.5)

where h is taken to be small. Here we note that on most computers the value of thefunctions u(r+h) and u(r−h) are known to a finite number of significant figures, e.g. wehave seven significant figures in single precision on most 32 bit computers. This meansthat for small h, when we take the difference ∆u = u(r + h)− u(r − h) in Eq. (C.5), welose accuracy as a result of the reduction in the number of significant figures to which thedifference ∆u is known. On the other hand, for h large, the above approximation to thederivative has an error that depends on h. To determine this error let us write a Taylorseries expansion for u(r ± h) about the point r, i.e.

u(r + h) = u+ hu(1) +h2

2!u(2) + · · · =

∞∑n=0

hn

n!u(n) (C.6)

u(r − h) = u− hu(1) +h2

2!u(2) + · · · =

∞∑n=0

(−1)nhn

n!u(n) , (C.7)

where u(n) is the nth derivative of the function u calculated at r. In particular u(0) = u(r).We now can write, the first derivative by using Eq. (C.5), as

1

2[u(r + h)− u(r − h) ] = hu(1) +

h3

3!u(3) + · · · . (C.8)

This gives us an error of the order of h2

3!u(3) in the first derivative. In a similar manner

we can write the second derivative as

d2u

dr2=

1h

[u(r + h)− u(r)]− 1h

[u(r)− u(r − h)]

h

=1

h2[u(r + h) + u(r − h)− 2u(r) ]

= u(2) +h2

12u(4) + · · · . (C.9)

Here again, we have an error of the order of h2. To improve our accuracy when replacingthe derivative by a difference, without reducing the value of h, we have to define thederivative so that the error is proportional to a higher power of h than Eqs. (C.8) and(C.9) provide us with. This is possible for the Schrodinger equation because we have a

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343

second order differential equation with no first order derivative. To see how this can beachieved, let us consider the relation

1

2[u(r + h) + u(r − h) ] = u+

h2

2!u(2) +

h4

4!u(4) +

h6

6!u(6) + · · · . (C.10)

If we differentiate this equation twice, or take an expansion similar to Eqs. (C.6) and(C.7) for u′′(r), we get

1

2[u′′(r + h) + u′′(r − h) ] = u(2) +

h2

2!u(4) +

h4

4!u(6) + · · · . (C.11)

We now can eliminate the error of the order of h4 by multiplying Eq. (C.11) by h2

12and

subtracting the result from Eq. (C.10) to get

1

2

[(u(r + h)− h2

12u′′(r + h)

)+

(u(r − h)− h2

12u′′(r − h)

)]

= u(r) +5h2

12u(2) − h6

480u(6) + · · · (C.12)

Using the differential equation, Eq. (C.2), we can write

h2

12u′′(x) = −h

2

12w(x)u(x) ≡ −t(x)u(x) . (C.13)

This in turn allows us to write Eq. (C.12) as

( 1 + t(r + h) ) u(r + h) = − ( 1 + t(r − h) ) u(r − h)

+ ( 2− 10 t(r) ) u(r)− h6

240u(6) + · · · . (C.14)

The error in this case is of the order of h6, rather than h2. This allows us to consider amuch larger value of h yet maintain the accuracy needed for the calculation. Finally wecan solve Eq. (C.14) for u(r + h) in terms of u(r) and u(r − h) to get

u(r + h) =(2− 10 t(r)) u(r)− (1 + t(r − h)) u(r − h)

1 + t(r + h). (C.15)

With this result in hand we can calculate the wave function at (r + h), given the wavefunction at (r − h) and r. In other words, we need to know the wave function at twopoints a distance h apart to know the wave function for all values of r. From Eq. (C.1)we know that u`(r) at r = 0 has to be zero for the radial wave function ψ`(r) to be finiteat the origin. Thus by taking

u(0) = 0 and u(h) = const. , (C.16)

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344APPENDIX C. NUMERICAL SOLUTION OF THE SCHRODINGER EQUATION

we can calculate the wave function at u(2h). We then can use our knowledge of the wavefunction at h and 2h to calculate the wave function at r = 3h. In this way we can integratethe Schrodinger equation to determine the wave function for all r. This procedure givesus the wave function up to the constant used for u(h). For the bound state problem,the value of this constant is determined by the fact that the wave function has to benormalized. For the case when u(r) is the scattering wave function, to determine thephase shifts, we need to calculate the logarithmic derivative at some large value of r. Inthis case the logarithmic derivative is independent of the constant used for u(h).

This leads us to the problem of how to calculate the first derivative of the wave functionat large r to the same degree of accuracy used to integrate the differential equation. Hereagain we resort to the same procedure used above by calculating the difference

1

2[u(r + h)− u(r − h)] = hu(1) +

h3

3!u(3) +

h5

5!u(5) + · · · . (C.17)

To eliminate the error term proportional to u(3), we write the similar expansion for thesecond derivative, i.e.,

1

2[u′′(r + h)− u′′(r − h)] = hu(3) +

h3

3!u(5) +

h5

5!u(7) + · · · . (C.18)

We now multiply Eq. (C.18) by h2

3!and subtract the result from Eq. (C.17) to get

1

2

[(u(r + h)− h2

3!u′′(r + h)

)+

(u(r − h)− h2

3!u′′(r − h)

)]

= hu(1) − 7h5

360u(5) + · · · . (C.19)

In this way the error in our first derivative is of the order of h4,1 rather than h2. Makinguse of the differential equation, Eq. (C.2), to replace u′′(r± h) by −w(r± h)u(r± h), weget for the first derivative at r to be

u(1) =1

2h

[(1 +

h2

3!w(r + h)

)u(r + h)−

(1 +

h2

3!w(r − h)

)u(r − h)

]. (C.20)

The error in this derivative is of the order if h4 u(5) which is comparable to the errorwe had in integrating the differential equation. We now can calculate the logarithmicderivative at r = r0 from the wave function at r = r0±h, which we have from integratingthe differential equation starting at the origin and finishing at r = r0 + h.

1This can be further improved so that the error in the first derivative is of the order of h9, seeJ. M. Blatt, Journal of Computational Physics, 1, 382 (1967).

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C.1. PROBLEMS 345

C.1 Problems

1. Calculate the S-wave phase shifts as a function of energy for the potential

V (r) = V0 e−β r2

,

using the numerical procedure described above. For α−α scattering the parametersof the potential are

V0 = −0.62178 fm−1 and β = 0.22 fm−2 .

All masses and energies are in units of fm−1. To convert units of MeV. to fm−1, wedivide energies and mass by hc = 197.327 Mev fm.

(a) Show that in the above problem the phase shifts are not sensitive to the valueof r0, provided r0 is greater than the range of the potential.

(b) Plot the wave function u(r) for S-wave scattering for the potential in Problem1 of Chapter 10. Can you estimate the range of the potential from the wavefunction? Does your result agree with the value of r0 you got in Problem 1?

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346APPENDIX C. NUMERICAL SOLUTION OF THE SCHRODINGER EQUATION

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Bibliography

[1] W. Heisenberg, Z. Physik 43, 127 (1927). 1

[2] W. Heisenberg, Physics and Beyond. 1

[3] James Gleick, CHAOS 0 Making a New Science, Viking (1987). 1

[4] J. C. Maxwell, Treatise on Electricity and Magnetism, 3rd ed., 2 vols., reprinted byDover, New York (1954). 1.1.2

[5] M. Planck, Ann. Physik, 4, 553 (1901). 1.2.1

[6] A. Einstein, Ann. Physik, 17, 132 (1905). 1.2.2

[7] A. H. Compton, Phys. Rev., 21, 715 (1923); 22, 409 (1923). 1.2.3

[8] Louis de Broglie, Phil. Mag. 47,446 (1924); Ann. Phys. 3, 22 (1925). 1.3

[9] C. J. Davisson and L. H. Germer, Phys. Rev. 30, 705 (1927). 1.3

[10] J. J. Balmer, Ann. Physik, 25, 80 (1885). 1.4

[11] E. Rutherford, Phil. Mag. 21, 669 (1911). 1.4

[12] N. Bohr, Phil. Mag., 26, 1 (1913). 1.4

[13] R.P. Feynman, Rev. Mod. Phys. 20, 267 (1948). 2.2

[14] J. B. J. Fourier Theory Analytique de la Chaleur (1822); English translation in Ana-lyitc Theory of Heat, Dover, New York (1955). 3.1

[15] E. Schrodinger, Ann. der Phys. 79, 361 (1926); English translation in Collected paperson wave mechanics by E. Schrodinger, trans. J. Shearer and W. Deans, Blackie,Glasgow (1928). 3.4

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347

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348 BIBLIOGRAPHY

[18] Henri Becquerel, “Sur les radiations emises par phosphorecene”. Comptes Rendue,122, 420-421 (1896). 6.1

[19] G. Gamow, Z. Phys. 51, 204 (1928). 6.1, 2

[20] E. U. Condon and R. W. Gurney, Nature 122, 439 (1928) and Phys. Rev. 33, 127(1929). 6.1, 2

[21] Felix Bloch, “Uber die Quantenmechanik der Electronen in Kristallgittern”, Z.Physik, 52, 555-600 (1928). 6.6.1

[22] R. de L. Kronig and W. G. Penney, Proc. Roy. Soc. (London) A 130, 499 (1931).6.6.2

[23] J.E. Lennard-Jones, “On the determination of molecular fields”, Proc. R. Soc. LondonA 106 463-477 (1924). 7.1

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[25] W. Pauli, Z. Physik 36, 336 (1926). 4

[26] Leonard I. Schiff, Quantum Mechanics 3rd Ed.McGraw-Hill, New York (1968) p236.4

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[28] Marvin L. Goldberger and Kenneth M. Watson, Collision Theory, John Wiley &Sons, New York (1964). 1

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Index

349