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Quantum Mechanics - Prof. Eric R. Bittner - University of Houston

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<strong>Quantum</strong> <strong>Mechanics</strong><br />

Lecture Notes for<br />

Chemistry 6312<br />

<strong>Quantum</strong> Chemistry<br />

<strong>Eric</strong> R. <strong>Bittner</strong><br />

<strong>University</strong> 1<br />

<strong>of</strong> <strong>Houston</strong><br />

Department <strong>of</strong> Chemistry


Lecture Notes on <strong>Quantum</strong> Chemistry<br />

Lecture notes to accompany Chemistry 6321<br />

Copyright @1997-2003, <strong>University</strong> <strong>of</strong> <strong>Houston</strong> and <strong>Eric</strong> R. <strong>Bittner</strong><br />

All Rights Reserved.<br />

August 12, 2003


Contents<br />

0 Introduction 8<br />

0.1 Essentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10<br />

0.2 Problem Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11<br />

0.3 2003 Course Calendar . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13<br />

I Lecture Notes 14<br />

1 Survey <strong>of</strong> Classical <strong>Mechanics</strong> 15<br />

1.1 Newton’s equations <strong>of</strong> motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16<br />

1.1.1 Elementary solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16<br />

1.1.2 Phase plane analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17<br />

1.2 Lagrangian <strong>Mechanics</strong> . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18<br />

1.2.1 The Principle <strong>of</strong> Least Action . . . . . . . . . . . . . . . . . . . . . . . . . 18<br />

1.2.2 Example: 3 dimensional harmonic oscillator in spherical coordinates . . . . 20<br />

1.3 Conservation Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23<br />

1.4 Hamiltonian Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24<br />

1.4.1 Interaction between a charged particle and an electromagnetic field. . . . . 24<br />

1.4.2 Time dependence <strong>of</strong> a dynamical variable . . . . . . . . . . . . . . . . . . . 26<br />

1.4.3 Virial Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26<br />

2 Waves and Wavefunctions 29<br />

2.1 Position and Momentum Representation <strong>of</strong> |ψ〉 . . . . . . . . . . . . . . . . . . . . 29<br />

2.2 The Schrödinger Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31<br />

2.2.1 Gaussian Wavefunctions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32<br />

2.2.2 Evolution <strong>of</strong> ψ(x) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34<br />

2.3 Particle in a Box . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36<br />

2.3.1 Infinite Box . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36<br />

2.3.2 Particle in a finite Box . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38<br />

2.3.3 Scattering states and resonances. . . . . . . . . . . . . . . . . . . . . . . . 40<br />

2.3.4 Application: <strong>Quantum</strong> Dots . . . . . . . . . . . . . . . . . . . . . . . . . . 43<br />

2.4 Tunneling and transmission in a 1D chain . . . . . . . . . . . . . . . . . . . . . . 49<br />

2.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49<br />

2.6 Problems and Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50<br />

1


3 Semi-Classical <strong>Quantum</strong> <strong>Mechanics</strong> 55<br />

3.1 Bohr-Sommerfield quantization . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56<br />

3.2 The WKB Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58<br />

3.2.1 Asymptotic expansion for eigenvalue spectrum . . . . . . . . . . . . . . . . 58<br />

3.2.2 WKB Wavefunction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60<br />

3.2.3 Semi-classical Tunneling and Barrier Penetration . . . . . . . . . . . . . . 62<br />

3.3 Connection Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65<br />

3.4 Scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70<br />

3.4.1 Classical Scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70<br />

3.4.2 Scattering at small deflection angles . . . . . . . . . . . . . . . . . . . . . . 73<br />

3.4.3 <strong>Quantum</strong> treatment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74<br />

3.4.4 Semiclassical evaluation <strong>of</strong> phase shifts . . . . . . . . . . . . . . . . . . . . 75<br />

3.4.5 Resonance Scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78<br />

3.5 Problems and Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78<br />

4 Postulates <strong>of</strong> <strong>Quantum</strong> <strong>Mechanics</strong> 80<br />

4.0.1 The description <strong>of</strong> a physical state: . . . . . . . . . . . . . . . . . . . . . . 85<br />

4.0.2 Description <strong>of</strong> Physical Quantities: . . . . . . . . . . . . . . . . . . . . . . 85<br />

4.0.3 <strong>Quantum</strong> Measurement: . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86<br />

4.0.4 The Principle <strong>of</strong> Spectral Decomposition: . . . . . . . . . . . . . . . . . . . 86<br />

4.0.5 The Superposition Principle . . . . . . . . . . . . . . . . . . . . . . . . . . 87<br />

4.0.6 Reduction <strong>of</strong> the wavepacket: . . . . . . . . . . . . . . . . . . . . . . . . . 89<br />

4.0.7 The temporal evolution <strong>of</strong> the system: . . . . . . . . . . . . . . . . . . . . 90<br />

4.0.8 Dirac <strong>Quantum</strong> Condition . . . . . . . . . . . . . . . . . . . . . . . . . . . 90<br />

4.1 Dirac Notation and Linear Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . 94<br />

4.1.1 Transformations and Representations . . . . . . . . . . . . . . . . . . . . . 94<br />

4.1.2 Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96<br />

4.1.3 Products <strong>of</strong> Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97<br />

4.1.4 Functions Involving Operators . . . . . . . . . . . . . . . . . . . . . . . . . 98<br />

4.2 Constants <strong>of</strong> the Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100<br />

4.3 Bohr Frequency and Selection Rules . . . . . . . . . . . . . . . . . . . . . . . . . . 101<br />

4.4 Example using the particle in a box states . . . . . . . . . . . . . . . . . . . . . . 102<br />

4.5 Time Evolution <strong>of</strong> Wave and Observable . . . . . . . . . . . . . . . . . . . . . . . 103<br />

4.6 “Unstable States” . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104<br />

4.7 Problems and Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106<br />

5 Bound States <strong>of</strong> The Schrödinger Equation 110<br />

5.1 Introduction to Bound States . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110<br />

5.2 The Variational Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112<br />

5.2.1 Variational Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112<br />

5.2.2 Constraints and Lagrange Multipliers . . . . . . . . . . . . . . . . . . . . . 114<br />

5.2.3 Variational method applied to Schrödinger equation . . . . . . . . . . . . . 117<br />

5.2.4 Variational theorems: Rayleigh-Ritz Technique . . . . . . . . . . . . . . . . 118<br />

5.2.5 Variational solution <strong>of</strong> harmonic oscillator ground State . . . . . . . . . . . 119<br />

5.3 The Harmonic Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121<br />

2


5.3.1 Harmonic Oscillators and Nuclear Vibrations . . . . . . . . . . . . . . . . . 124<br />

5.3.2 Classical interpretation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131<br />

5.3.3 Molecular Vibrations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134<br />

5.4 Numerical Solution <strong>of</strong> the Schrödinger Equation . . . . . . . . . . . . . . . . . . . 136<br />

5.4.1 Numerov Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136<br />

5.4.2 Numerical Diagonalization . . . . . . . . . . . . . . . . . . . . . . . . . . . 139<br />

5.5 Problems and Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144<br />

6 <strong>Quantum</strong> <strong>Mechanics</strong> in 3D 152<br />

6.1 <strong>Quantum</strong> Theory <strong>of</strong> Rotations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153<br />

6.2 Eigenvalues <strong>of</strong> the Angular Momentum Operator . . . . . . . . . . . . . . . . . . 157<br />

6.3 Eigenstates <strong>of</strong> L 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158<br />

6.4 Eigenfunctions <strong>of</strong> L 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159<br />

6.5 Addition theorem and matrix elements . . . . . . . . . . . . . . . . . . . . . . . . 162<br />

6.6 Legendre Polynomials and Associated Legendre Polynomials . . . . . . . . . . . . 164<br />

6.7 <strong>Quantum</strong> rotations in a semi-classical context . . . . . . . . . . . . . . . . . . . . 165<br />

6.8 Motion in a central potential: The Hydrogen Atom . . . . . . . . . . . . . . . . . 170<br />

6.8.1 Radial Hydrogenic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . 173<br />

6.9 Spin 1/2 Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173<br />

6.9.1 Theoretical Description . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174<br />

6.9.2 Other Spin Observables . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175<br />

6.9.3 Evolution <strong>of</strong> a state . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175<br />

6.9.4 Larmor Precession . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176<br />

6.10 Problems and Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177<br />

7 Perturbation theory 180<br />

7.1 Perturbation Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180<br />

7.2 Two level systems subject to a perturbation . . . . . . . . . . . . . . . . . . . . . 182<br />

7.2.1 Expansion <strong>of</strong> Energies in terms <strong>of</strong> the coupling . . . . . . . . . . . . . . . 183<br />

7.2.2 Dipole molecule in homogenous electric field . . . . . . . . . . . . . . . . . 184<br />

7.3 Dyson Expansion <strong>of</strong> the Schrödinger Equation . . . . . . . . . . . . . . . . . . . . 188<br />

7.4 Van der Waals forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190<br />

7.4.1 Origin <strong>of</strong> long-ranged attractions between atoms and molecules . . . . . . . 190<br />

7.4.2 Attraction between an atom a conducting surface . . . . . . . . . . . . . . 192<br />

7.5 Perturbations Acting over a Finite amount <strong>of</strong> Time . . . . . . . . . . . . . . . . . 193<br />

7.5.1 General form <strong>of</strong> time-dependent perturbation theory . . . . . . . . . . . . 193<br />

7.5.2 Fermi’s Golden Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194<br />

7.6 Interaction between an atom and light . . . . . . . . . . . . . . . . . . . . . . . . 197<br />

7.6.1 Fields and potentials <strong>of</strong> a light wave . . . . . . . . . . . . . . . . . . . . . 197<br />

7.6.2 Interactions at Low Light Intensity . . . . . . . . . . . . . . . . . . . . . . 198<br />

7.6.3 Photoionization <strong>of</strong> Hydrogen 1s . . . . . . . . . . . . . . . . . . . . . . . . 202<br />

7.6.4 Spontaneous Emission <strong>of</strong> Light . . . . . . . . . . . . . . . . . . . . . . . . 204<br />

7.7 Time-dependent golden rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209<br />

7.7.1 Non-radiative transitions between displaced Harmonic Wells . . . . . . . . 210<br />

7.7.2 Semi-Classical Evaluation . . . . . . . . . . . . . . . . . . . . . . . . . . . 216<br />

3


7.8 Problems and Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219<br />

8 Many Body <strong>Quantum</strong> <strong>Mechanics</strong> 222<br />

8.1 Symmetry with respect to particle Exchange . . . . . . . . . . . . . . . . . . . . . 222<br />

8.2 Matrix Elements <strong>of</strong> Electronic Operators . . . . . . . . . . . . . . . . . . . . . . . 227<br />

8.3 The Hartree-Fock Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . 229<br />

8.3.1 Two electron integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 230<br />

8.3.2 Koopman’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231<br />

8.4 <strong>Quantum</strong> Chemistry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231<br />

8.4.1 The Born-Oppenheimer Approximation . . . . . . . . . . . . . . . . . . . . 232<br />

8.5 Problems and Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241<br />

A Physical Constants and Conversion Factors 247<br />

B Mathematical Results and Techniques to Know and Love 249<br />

B.1 The Dirac Delta Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 249<br />

B.1.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 249<br />

B.1.2 Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 250<br />

B.1.3 Spectral representations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 250<br />

B.2 Coordinate systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253<br />

B.2.1 Cartesian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253<br />

B.2.2 Spherical . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254<br />

B.2.3 Cylindrical . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255<br />

C Mathematica Notebook Pages 256<br />

4


List <strong>of</strong> Figures<br />

1.1 Tangent field for simple pendulum with ω = 1. The superimposed curve is a linear<br />

approximation to the pendulum motion. . . . . . . . . . . . . . . . . . . . . . . . 17<br />

1.2 Vector diagram for motion in a central forces. The particle’s motion is along the<br />

Z axis which lies in the plane <strong>of</strong> the page. . . . . . . . . . . . . . . . . . . . . . . 21<br />

1.3 Screen shot <strong>of</strong> using Mathematica to plot phase-plane for harmonic oscillator.<br />

Here k/m = 1 and our xo = 0.75. . . . . . . . . . . . . . . . . . . . . . . . . . . . 28<br />

2.1 A gaussian wavepacket, ψ(x) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33<br />

2.2 Momentum-space distribution <strong>of</strong> ψ(k). . . . . . . . . . . . . . . . . . . . . . . . . 33<br />

2.3 Go for fixed t as a function <strong>of</strong> x. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35<br />

2.4 Evolution <strong>of</strong> a free particle wavefunction. . . . . . . . . . . . . . . . . . . . . . . 36<br />

2.5 Particle in a box states . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38<br />

2.6 Graphical solution to transendental equations for an electron in a truncated hard<br />

well <strong>of</strong> depth Vo = 10 and width a = 2. The short-dashed blue curve corresponds<br />

to the symmetric case�and the long-dashed blue curve corresponds to the asymetric<br />

case. The red line is<br />

1 − V o/E. Bound state solution are such that the red and<br />

blue curves cross. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41<br />

2.7 Transmission (blue) and Reflection (red) coefficients for an electron scattering over<br />

a square well (V = −40 and a = 1 ). . . . . . . . . . . . . . . . . . . . . . . . . . 42<br />

2.8 Transmission Coefficient for particle passing over a bump. . . . . . . . . . . . . . 43<br />

2.9 Scattering waves for particle passing over a well. . . . . . . . . . . . . . . . . . . . 44<br />

2.10 Argand plot <strong>of</strong> a scattering wavefunction passing over a well. . . . . . . . . . . . . 45<br />

2.11 Density <strong>of</strong> states for a 1-, 2- , and 3- dimensional space. . . . . . . . . . . . . . . . 46<br />

2.12 Density <strong>of</strong> states for a quantum well and quantum wire compared to a 3d space.<br />

Here L = 5 and s = 2 for comparison. . . . . . . . . . . . . . . . . . . . . . . . . . 47<br />

2.13 Spherical Bessel functions, j0, j1, and j1 (red, blue, green) . . . . . . . . . . . . . 48<br />

2.14 Radial wavefuncitons (left column) and corresponding PDFs (right column) for an<br />

electron in a R = 0.5˚Aquantum dot. The upper two correspond to (n, l) = (1, 0)<br />

(solid) and (n, l) = (1, 1) (dashed) while the lower correspond to (n, l) = (2, 0)<br />

(solid) and (n, l) = (2, 1) (dashed) . . . . . . . . . . . . . . . . . . . . . . . . . . . 49<br />

3.1 Eckart Barrier and parabolic approximation <strong>of</strong> the transition state . . . . . . . . . 63<br />

3.2 Airy functions, Ai(y) (red) and Bi(y) (blue) . . . . . . . . . . . . . . . . . . . . . 66<br />

3.3 Bound states in a graviational well . . . . . . . . . . . . . . . . . . . . . . . . . . 69<br />

3.4 Elastic scattering trajectory for classical collision . . . . . . . . . . . . . . . . . . 70<br />

5


3.5 Form <strong>of</strong> the radial wave for repulsive (short dashed) and attractive (long dashed)<br />

potentials. The form for V = 0 is the solid curve for comparison. . . . . . . . . . . 76<br />

4.1 Gaussian distribution function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81<br />

4.2 Combination <strong>of</strong> two distrubitions. . . . . . . . . . . . . . . . . . . . . . . . . . . . 82<br />

4.3 Constructive and destructive interference from electron/two-slit experiment. The<br />

superimposed red and blue curves are P1 and P2 from the classical probabilities . 83<br />

4.4 The diffraction function sin(x)/x . . . . . . . . . . . . . . . . . . . . . . . . . . . 102<br />

5.1 Variational paths between endpoints. . . . . . . . . . . . . . . . . . . . . . . . . . 116<br />

5.2 Hermite Polynomials, Hn up to n = 3. . . . . . . . . . . . . . . . . . . . . . . . . 128<br />

5.3 Harmonic oscillator functions for n = 0 to 3 . . . . . . . . . . . . . . . . . . . . . 132<br />

5.4 <strong>Quantum</strong> and Classical Probability Distribution Functions for Harmonic Oscillator.133<br />

5.5 London-Eyring-Polanyi-Sato (LEPS) empirical potential for the F +H2 → F H+H<br />

chemical reaction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135<br />

5.6 Morse well and harmonic approximation for HF . . . . . . . . . . . . . . . . . . . 136<br />

5.7 Model potential for proton tunneling. . . . . . . . . . . . . . . . . . . . . . . . . . 137<br />

5.8 Double well tunneling states as determined by the Numerov approach. . . . . . . . 138<br />

5.9 Tchebyshev Polynomials for n = 1 − 5 . . . . . . . . . . . . . . . . . . . . . . . . 140<br />

5.10 Ammonia Inversion and Tunneling . . . . . . . . . . . . . . . . . . . . . . . . . . 147<br />

6.1 Vector model for the quantum angular momentum state |jm〉, which is represented<br />

here by the vector j which precesses about the z axis (axis <strong>of</strong> quantzation) with<br />

projection m. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156<br />

6.2 Spherical Harmonic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160<br />

6.3 Classical and <strong>Quantum</strong> Probability Distribution Functions for Angular Momentum.168<br />

7.1 Variation <strong>of</strong> energy level splitting as a function <strong>of</strong> the applied field for an ammonia<br />

molecule in an electric field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185<br />

7.2 Photo-ionization spectrum for hydrogen atom. . . . . . . . . . . . . . . . . . . . . 205<br />

8.1 Various contributions to the H + 2 Hamiltonian. . . . . . . . . . . . . . . . . . . . . 236<br />

8.2 Potential energy surface for H + 2 molecular ion. . . . . . . . . . . . . . . . . . . . . 238<br />

8.3 Three dimensional representations <strong>of</strong> ψ+ and ψ− for the H + 2 molecular ion. . . . . 238<br />

8.4 Setup calculation dialog screen . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243<br />

8.5 HOMO-1, HOMO and LUMO for CH2 = O. . . . . . . . . . . . . . . . . . . . . . 245<br />

8.6 Transition state geometry for H2 + C = O → CH2 = O. The Arrow indicates the<br />

reaction path. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 246<br />

B.1 sin(xa)/πx representation <strong>of</strong> the Dirac δ-function . . . . . . . . . . . . . . . . . . 251<br />

B.2 Gaussian representation <strong>of</strong> δ-function . . . . . . . . . . . . . . . . . . . . . . . . . 251<br />

6


List <strong>of</strong> Tables<br />

3.1 Location <strong>of</strong> nodes for Airy, Ai(x) function. . . . . . . . . . . . . . . . . . . . . . . 68<br />

5.1 Tchebychev polynomials <strong>of</strong> the first type . . . . . . . . . . . . . . . . . . . . . . . 140<br />

5.2 Eigenvalues for double well potential computed via DVR and Numerov approaches 143<br />

6.1 Spherical Harmonics (Condon-Shortley Phase convention. . . . . . . . . . . . . . . 160<br />

6.2 Relation between various notations for Clebsch-Gordan Coefficients in the literature169<br />

8.1 Vibrational Frequencies <strong>of</strong> Formaldehyde . . . . . . . . . . . . . . . . . . . . . . . 244<br />

A.1 Physical Constants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 248<br />

A.2 Atomic Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 248<br />

A.3 Useful orders <strong>of</strong> magnitude . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 248<br />

7


Chapter 0<br />

Introduction<br />

Nothing conveys the impression <strong>of</strong> humungous intellect so much as even the sketchiest<br />

knowledge <strong>of</strong> quantum physics, and since the sketchiest knowledge is all anyone will<br />

ever have, never be shy about holding forth with bags <strong>of</strong> authority about subatomic<br />

particles and the quantum realm without having done any science whatsoever.<br />

Jack Klaff –Bluff Your Way in the <strong>Quantum</strong> Universe<br />

The field <strong>of</strong> quantum chemistry seeks to provide a rigorous description <strong>of</strong> chemical processes<br />

at its most fundamental level. For ordinary chemical processes, the most fundamental and underlying<br />

theory <strong>of</strong> chemistry is given by the time-dependent and time-independent version <strong>of</strong> the<br />

Schrödinger equation. However, simply stating an equation that provides the underlying theory<br />

in now shape or form yields and predictive or interpretive power. In fact, most <strong>of</strong> what we do<br />

in quantum mechanics is to develop a series <strong>of</strong> well posed approximation and physical assumptions<br />

to solve basic equations <strong>of</strong> quantum mechanics. In this course, we will delve deeply into<br />

the underlying physical and mathematical theory. We will learn how to solve some elementary<br />

problems and apply these to not so elementary examples.<br />

As with any course <strong>of</strong> this nature, the content reflects the instructors personal interests in<br />

the field. In this case, the emphasis <strong>of</strong> the course is towards dynamical processes, transitions<br />

between states, and interaction between matter and radiation. More “traditional” quantum<br />

chemistry courses will focus upon electronic structure. In fact, the moniker “quantum chemistry”<br />

typically refers to electronic structure theory. While this is an extremely rich topic, it is my<br />

personal opinion that a deeper understanding <strong>of</strong> dynamical processes provides a broader basis<br />

for understanding chemical processes.<br />

It is assumed from the beginning, that students taking this course have had some exposure<br />

to the fundamental principles <strong>of</strong> quantum theory as applied to chemical systems. This is usually<br />

in the context <strong>of</strong> a physical chemistry course or a separate course in quantum chemistry. I<br />

also assume that students taking this course have had undergraduate level courses in calculus,<br />

differential equations, and have some concepts <strong>of</strong> linear algebra. Students lacking in any <strong>of</strong> these<br />

areas are strongly encouraged to sit through my undergraduate Physical Chemistry II course<br />

(<strong>of</strong>fered in the Spring Semester at the Univ. <strong>of</strong> <strong>Houston</strong>) before attempting this course. This<br />

course is by design and purpose theoretical in nature.<br />

The purpose <strong>of</strong> this course is to provide a solid and mathematically rigorous tour through<br />

modern quantum mechanics. We will begin with simple examples which can be worked out<br />

8


exactly on paper and move on to discuss various approximation schemes. For cases in which<br />

analytical solutions are either too obfuscating or impossible, computer methods will be introduced<br />

using Mathematica. Applications toward chemically relevant topics will be emphasized<br />

throughout.<br />

We will primarily focus upon single particle systems, or systems in which the particles are<br />

distinguishable. Special considerations for systems <strong>of</strong> indistinguishable particles, such as the<br />

electrons in a molecule, will be discussed towards the end <strong>of</strong> the course. The pace <strong>of</strong> the course<br />

is fairly rigorous, with emphasis on solving problems either analytically or using computer.<br />

I also tend to emphasize how to approach a problem from a theoretical viewpoint. As you<br />

will discover rapidly, very few <strong>of</strong> the problem sets in this course are <strong>of</strong> the “look-up the right<br />

formula” type. Rather, you will need to learn to use the various techniques (perturbation theory,<br />

commutation relations, etc...) to solve and work out problems for a variety <strong>of</strong> physical systems.<br />

The lecture notes in this package are really to be regarded as a work in progress and updates<br />

and additions will be posted as they evolve. Lacking is a complete chapter on the Hydrogen<br />

atom and atomic physics and a good overview <strong>of</strong> many body theory. Also, I have not included<br />

a chapter on scattering and other topics as these will be added over the course <strong>of</strong> time. Certain<br />

sections are clearly better than others and will be improved upon over time. Each chapter ends<br />

with a series <strong>of</strong> exercises and suggested problems Some <strong>of</strong> which have detailed solutions. Others,<br />

you should work out on your own. At the end <strong>of</strong> this book are a series <strong>of</strong> Mathematica notebooks<br />

I have written which illustrate various points and perform a variety <strong>of</strong> calculations. These can be<br />

downloaded from my web-site (http://k2.chem.uh.edu/quantum/) and run on any recent version<br />

<strong>of</strong> Mathematica. (≥ v4.n).<br />

It goes entirely without saying (but I will anyway) that these notes come from a wide variety<br />

<strong>of</strong> sources which I have tried to cite where possible.<br />

9


0.1 Essentials<br />

• Instructor: <strong>Pr<strong>of</strong></strong>. <strong>Eric</strong>. R. <strong>Bittner</strong>.<br />

• Office: Fleming 221 J<br />

• Email: bittner@uh.edu<br />

• Phone: -3-2775<br />

• Office Hours: Monday and Thurs. afternoons or by appointment.<br />

• Course Web Page: http://k2.chem.uh.edu/quantum/ Solution sets, course news, class<br />

notes, sample computer routines, etc...will be posted from time to time on this web-page.<br />

• Other Required Text: <strong>Quantum</strong> <strong>Mechanics</strong>, Landau and Lifshitz. This is volume 3 <strong>of</strong><br />

L&L’s classical course in modern physics. Every self-respecting scientist has at least two or<br />

three <strong>of</strong> their books on their book-shelves. This text tends to be pretty terse and uses the<br />

classical phrase it is easy to show... quite a bit (it usually means just the opposite). The<br />

problems are usually worked out in detail and are usually classic applications <strong>of</strong> quantum<br />

theory. This is a land-mark book and contains everything you really need to know to do<br />

quantum mechanics.<br />

• Recommended Texts: I highly recommend that you use a variety <strong>of</strong> books since one<br />

author’s approach to a given topic may be clearer than another’s approach.<br />

– <strong>Quantum</strong> <strong>Mechanics</strong>, Cohen-Tannoudji, et al. This two volume book is very comprehensive<br />

and tends to be rather formal (and formidable) in its approach. The problems<br />

are excellent.<br />

–<br />

– Lectures in <strong>Quantum</strong> <strong>Mechanics</strong>, Gordon Baym. Baym’s book covers a wide range <strong>of</strong><br />

topics in a lecture note style.<br />

– <strong>Quantum</strong> Chemistry, I. Levine. This is usually the first quantum book that chemists<br />

get. I find it to be too wordy and the notation and derivations a bit ponderous. Levine<br />

does not use Dirac notation. However, he does give a good overview <strong>of</strong> elementary<br />

electronic structure theory and some if its important developments. Good for starting<br />

<strong>of</strong>f in electronic structure.<br />

– Modern <strong>Quantum</strong> <strong>Mechanics</strong>, J. J. Sakurai. This is a real classic. Not good for a first<br />

exposure since it assumes a fairly sophisticated understanding <strong>of</strong> quantum mechanics<br />

and mathematics.<br />

– Intermediate <strong>Quantum</strong> <strong>Mechanics</strong>, Hans Bethe and Roman Jackiw. This book is a<br />

great exploration <strong>of</strong> advanced topics in quantum mechanics as illustrated by atomic<br />

systems.<br />

10


– What is <strong>Quantum</strong> <strong>Mechanics</strong>?, Transnational College <strong>of</strong> LEX. OK, this one I found<br />

at Barnes and Noble and it’s more or less a cartoon book. But, it is really good. It<br />

explores the historical development <strong>of</strong> quantum mechanics, has some really interesting<br />

insights into semi-classical and ”old” quantum theory, and presents the study <strong>of</strong><br />

quantum mechanics as a unfolding story. This book I highly recommend if this<br />

is the first time you are taking a course on quantum mechanics.<br />

– <strong>Quantum</strong> <strong>Mechanics</strong> in Chemistry by George Schatz and Mark Ratner. Ratner and<br />

Schatz have more in terms <strong>of</strong> elementary quantum chemistry, emphasizing the use <strong>of</strong><br />

modern quantum chemical computer programs than almost any text I have reviewed.<br />

• Prequisites: Graduate status in chemistry. This course is required for all Physical Chemistry<br />

graduate students. The level <strong>of</strong> the course will be fairly rigorous and I assume that students<br />

have had some exposure to quantum mechanics at the undergraduate level–typically<br />

in Physical Chemistry, and are competent in linear algebra, calculus, and solving elementary<br />

differential equations.<br />

• Tests and Grades: There are no exams in this course, only problem sets and participation<br />

in discussion. This means coming to lecture prepared to ask and answer questions. My<br />

grading policy is pretty simple. If you make an honest effort, do the assigned problems<br />

(mostly correctly), and participate in class, you will be rewarded with at least a B. Of<br />

course this is the formula for success for any course.<br />

0.2 Problem Sets<br />

Your course grade will largely be determined by your performance on these problems as well as<br />

the assigned discussion <strong>of</strong> a particular problem. My philosophy towards problem sets is that this<br />

is the only way to really learn this material. These problems are intentionally challenging, but<br />

not overwhelming, and are paced to correspond to what will be going on in the lecture.<br />

Some ground rules:<br />

1. Due dates are posted on each problem–usually 1 week or 2 weeks after they are assigned.<br />

Late submissions may be turned in up to 1 week later. All problems must be turned in by<br />

December 3. I will not accept any submissions after that date.<br />

2. Handwritten Problems. If I can’t read it, I won’t grade it. Period. Consequently, I strongly<br />

encourage the use <strong>of</strong> word processing s<strong>of</strong>tware for your final submission. Problem solutions<br />

can be submitted electronically as Mathematica, Latex, or PDF files to bittner@uh.edu with<br />

the subject: QUANTUM PROBLEM SET. Do not send me a MSWord file as an email<br />

attachment. I expect some text (written in compete and correct sentences) to explain your<br />

steps where needed and some discussion <strong>of</strong> the results. The computer lab in the basement<br />

<strong>of</strong> Fleming has 20 PCs with copies <strong>of</strong> Mathematica or you can obtain your own license<br />

from the <strong>University</strong> Media Center.<br />

3. Collaborations. You are strongly encouraged to work together and collaborate on problems.<br />

However, simply copying from your fellow student is not an acceptable collaboration.<br />

11


4. These are the only problems you need to turn in. We will have additional exercises–mostly<br />

coming from the lecture. Also, at the end <strong>of</strong> the lectures herein, are a set <strong>of</strong> suggested<br />

problems and exercises to work on. Many <strong>of</strong> these have solutions.<br />

12


0.3 2003 Course Calendar<br />

This is a rough schedule <strong>of</strong> topics we will cover. In essence we will follow the starting from<br />

a basic description <strong>of</strong> quantum wave mechanics and bound states. We will then move onto<br />

the more formal aspects <strong>of</strong> quantum theory: Dirac notation, perturbation theory, variational<br />

theory, and the like. Lastly, we move onto applications: Hydrogen atom, many-electron systems,<br />

semi-classical approximations, and a semi-classical treatment <strong>of</strong> light absorption and emission.<br />

We will also have a recitation session in 221 at 10am Friday morning. The purpose <strong>of</strong> this<br />

will be to specifically discuss the problem sets and other issues.<br />

• 27-August: Course overview: Classical Concepts<br />

• 3-Sept: Finishing Classical <strong>Mechanics</strong>/Elementary <strong>Quantum</strong> concepts<br />

• 8-Sept: Particle in a box and hard wall potentials (Perry?)<br />

• 10 Sept: Tunneling/Density <strong>of</strong> states (Perry)<br />

• 15/17 Bohr-Sommerfield Quantization/Old quantum theory/connection to classical mechanics<br />

(Perry)<br />

• 22/24 Sept: Semiclassical quantum mechanics: WKB Approx. Application to scattering<br />

• 29 Sept/1 Oct. Postulates <strong>of</strong> quantum mechanics: Dirac notation, superposition principle,<br />

simple calculations.<br />

• 6/8 Oct: Bound States: Variational principle, quantum harmonic oscillator<br />

• 13/15 Oct: <strong>Quantum</strong> mechanics in 3D: Angular momentum (Chapt 4.1-4.8)<br />

• 20/22 Oct: Hydrogen atom/Hydrogenic systems/Atomic structure<br />

• 27/29 Oct: Perturbation Theory:<br />

• 3/5 Nov: Time-dependent Perturbation Theory:<br />

• 10/12 Identical Particles/<strong>Quantum</strong> Statistics<br />

• 17/19 Nov: Helium atom, hydrogen ion<br />

• 24/26 Nov: <strong>Quantum</strong> Chemistry<br />

• 3 Dec–Last day to turn in problem sets<br />

• Final Exam: TBA<br />

13


Part I<br />

Lecture Notes<br />

14


Chapter 1<br />

Survey <strong>of</strong> Classical <strong>Mechanics</strong><br />

<strong>Quantum</strong> mechanics is in many ways the cumulation <strong>of</strong> many hundreds <strong>of</strong> years <strong>of</strong> work and<br />

thought about how mechanical things move and behave. Since ancient times, scientists have<br />

wondered about the structure <strong>of</strong> matter and have tried to develop a generalized and underlying<br />

theory which governs how matter moves at all length scales.<br />

For ordinary objects, the rules <strong>of</strong> motion are very simple. By ordinary, I mean objects that<br />

are more or less on the same length and mass scale as you and I, say (conservatively) 10 −7 m to<br />

10 6 m and 10 −25 g to 10 8 g moving less than 20% <strong>of</strong> the speed <strong>of</strong> light. On other words, almost<br />

everything you can see and touch and hold obey what are called “classical” laws <strong>of</strong> motion. The<br />

term “classical” means that that the basic principles <strong>of</strong> this class <strong>of</strong> motion have their foundation<br />

in antiquity. Classical mechanics is a extremely well developed area <strong>of</strong> physics. While you may<br />

think that given that classical mechanics has been studied extensively for hundreds <strong>of</strong> years<br />

there really is little new development in this field, it remains a vital and extremely active area <strong>of</strong><br />

research. Why? Because the majority <strong>of</strong> universe “lives” in a dimensional realm where classical<br />

mechanics is extremely valid. Classical mechanics is the workhorse for atomistic simulations<br />

<strong>of</strong> fluids, proteins, polymers. It provides the basis for understanding chaotic systems. It also<br />

provides a useful foundation <strong>of</strong> many <strong>of</strong> the concepts in quantum mechanics.<br />

<strong>Quantum</strong> mechanics provides a description <strong>of</strong> how matter behaves at very small length and<br />

mass scales: i.e. the realm <strong>of</strong> atoms, molecules, and below. It was developed over the last century<br />

to explain a series <strong>of</strong> experiments on atomic systems that could not be explained using purely<br />

classical treatments. The advent <strong>of</strong> quantum mechanics forced us to look beyond the classical<br />

theories. However, it was not a drastic and complete departure. At some point, the two theories<br />

must correspond so that classical mechanics is the limiting behavior <strong>of</strong> quantum mechanics for<br />

macroscopic objects. Consequently, many <strong>of</strong> the concepts we will study in quantum mechanics<br />

have direct analogs to classical mechanics: momentum, angular momentum, time, potential<br />

energy, kinetic energy, and action.<br />

Much like classical music is in a particular style, classical mechanics is based upon the principle<br />

that the motion <strong>of</strong> a body can be reduced to the motion <strong>of</strong> a point particle with a given mass<br />

m, position x, and velocity v. In this chapter, we will review some <strong>of</strong> the concepts <strong>of</strong> classical<br />

mechanics which are necessary for studying quantum mechanics. We will cast these in form<br />

whereby we can move easily back and forth between classical and quantum mechanics. We will<br />

first discuss Newtonian motion and cast this into the Lagrangian form. We will then discuss the<br />

principle <strong>of</strong> least action and Hamiltonian dynamics and the concept <strong>of</strong> phase space.<br />

15


1.1 Newton’s equations <strong>of</strong> motion<br />

Newton’s Principia set the theoretical basis <strong>of</strong> mathematical mechanics and analysis <strong>of</strong> physical<br />

bodies. The equation that force equals mass times acceleration is the fundamental equation <strong>of</strong><br />

classical mechanics. Stated mathematically<br />

m¨x = f(x) (1.1)<br />

The dots refer to differentiation with respect to time. We will use this notion for time derivatives.<br />

We may also use x ′ or dx/dt as well. So,<br />

¨x = d2x .<br />

dt2 For now we are limiting our selves to one particle moving in one dimension. For motion in<br />

more dimensions, we need to introduce vector components. In cartesian coordinates, Newton’s<br />

equations are<br />

m¨x = fx(x, y, z) (1.2)<br />

m¨y = fy(x, y, z) (1.3)<br />

m¨z = fz(x, y, z) (1.4)<br />

where the force vector � f(x, y, z) has components in all three dimensions and varies with location.<br />

We can also define a position vector, �x = (x, y, z), and velocity vector �v = ( ˙x, ˙y, ˙z). We can also<br />

replace the second-order differential equation with two first order equations.<br />

˙x = vx (1.5)<br />

˙vx = fx/m (1.6)<br />

These, along with the initial conditions, x(0) and v(0) are all that are needed to solve for the<br />

motion <strong>of</strong> a particle with mass m given a force f. We could have chosen two end points as well<br />

and asked, what path must the particle take to get from one point to the next. Let us consider<br />

some elementary solutions.<br />

1.1.1 Elementary solutions<br />

First the case in which f = 0 and ¨x = 0. Thus, v = ˙x = const. So, unless there is an applied<br />

force, the velocity <strong>of</strong> a particle will remain unchanged.<br />

Second, we consider the case <strong>of</strong> a linear force, f = −kx. This is restoring force for a spring<br />

and such force laws are termed Hooke’s law and k is termed the force constant. Our equations<br />

are:<br />

˙x = vx (1.7)<br />

˙vx = −k/mx (1.8)<br />

or ¨x = −(k/m)x. So we want some function which is its own second derivative multiplied by<br />

some number. The cosine and sine functions have this property, so let’s try<br />

x(t) = A cos(at) + B sin(bt).<br />

16


Taking time derivatives<br />

˙x(t) = −aA sin(at) + bB cos(bt);<br />

¨x(t) = −a 2 A cos(at) − b 2 B sin(bt).<br />

�<br />

So we get the required result if a = b = k/m, leaving A and B undetermined. Thus, we need<br />

two initial conditions to specify these coefficients. Let’s pick x(0) = xo �<br />

and v(0) = 0. Thus,<br />

x(0) = A = xo and B = 0. Notice that the term k/m has units <strong>of</strong> angular frequency.<br />

So, our equation <strong>of</strong> motion are<br />

v<br />

3<br />

2<br />

1<br />

0<br />

-1<br />

-2<br />

-3<br />

�<br />

k<br />

ω =<br />

m<br />

x(t) = xo cos(ωt) (1.9)<br />

v(t) = −xoω sin(ωt). (1.10)<br />

-2 p -p 0 p 2 p<br />

-2 p -p 0 p 2 p<br />

x<br />

Figure 1.1: Tangent field for simple pendulum with ω = 1. The superimposed curve is a linear<br />

approximation to the pendulum motion.<br />

1.1.2 Phase plane analysis<br />

Often one can not determine the closed form solution to a given problem and we need to turn<br />

to more approximate methods or even graphical methods. Here, we will look at an extremely<br />

useful way to analyze a system <strong>of</strong> equations by plotting their time-derivatives.<br />

First, let’s look at the oscillator we just studied. We can define a vector s = ( ˙x, ˙v) =<br />

(v, −k/mx) and plot the vector field. Fig. 1.3 shows how to do this in Mathematica. The<br />

17


superimposed curve is one trajectory and the arrows give the “flow” <strong>of</strong> trajectories on the phase<br />

plane.<br />

We can examine more complex behavior using this procedure. For example, the simple<br />

pendulum obeys the equation ¨x = −ω 2 sin x. This can be reduced to two first order equations:<br />

˙x = v and ˙v = −ω 2 sin(x).<br />

We can approximate the motion <strong>of</strong> the pendulum for small displacements by expanding the<br />

pendulum’s force about x = 0,<br />

−ω 2 sin(x) = −ω 2 (x − x3<br />

6<br />

For small x the cubic term is very small, and we have<br />

˙v = −ω 2 x = − k<br />

m x<br />

+ · · ·).<br />

which is the equation for harmonic motion. So, for small initial displacements, we see that the<br />

pendulum oscillates back and forth with an angular frequency ω. For large initial displacements,<br />

xo = π or if we impart some initial velocity on the system vo > 1, the pendulum does not oscillate<br />

back and forth, but undergoes librational motion (spinning!) in one direction or the other.<br />

1.2 Lagrangian <strong>Mechanics</strong><br />

1.2.1 The Principle <strong>of</strong> Least Action<br />

The most general form <strong>of</strong> the law governing the motion <strong>of</strong> a mass is the principle <strong>of</strong> least action<br />

or Hamilton’s principle. The basic idea is that every mechanical system is described by a single<br />

function <strong>of</strong> coordinate, velocity, and time: L(x, ˙x, t) and that the motion <strong>of</strong> the particle is such<br />

that certain conditions are satisfied. That condition is that the time integral <strong>of</strong> this function<br />

S =<br />

� tf<br />

to<br />

L(x, ˙x, t)dt<br />

takes the least possible value give a path that starts at xo at the initial time and ends at xf at<br />

the final time.<br />

Lets take x(t) be function for which S is minimized. This means that S must increase for<br />

any variation about this path, x(t) + δx(t). Since the end points are specified, δx(0) = δx(t) = 0<br />

and the change in S upon replacement <strong>of</strong> x(t) with x(t) + δx(t) is<br />

δS =<br />

� tf<br />

to<br />

L(x + δx, ˙x + δ ˙x, t)dt −<br />

� tf<br />

to<br />

L(x, ˙x, t)dt = 0<br />

This is zero, because S is a minimum. Now, we can expand the integrand in the first term<br />

� �<br />

∂L ∂L<br />

L(x + δx, ˙x + δ ˙x, t) = L(x, ˙x, t) + δx + δ ˙x<br />

∂x ∂ ˙x<br />

Thus, we have<br />

� tf<br />

to<br />

� �<br />

∂L ∂L<br />

δx + δ ˙x dt = 0.<br />

∂x ∂ ˙x<br />

18


Since δ ˙x = dδx/dt and integrating the second term by parts<br />

δS =<br />

�<br />

∂L<br />

δ ˙x δx<br />

�tf to<br />

+<br />

� tf<br />

to<br />

� ∂L<br />

∂x<br />

�<br />

d ∂L<br />

− δxdt = 0<br />

dt ∂ ˙x<br />

The surface term vanishes because <strong>of</strong> the condition imposed above. This leaves the integral. It<br />

too must vanish and the only way for this to happen is if the integrand itself vanishes. Thus we<br />

have the<br />

∂L d ∂L<br />

− = 0<br />

∂x dt ∂ ˙x<br />

L is known as the Lagrangian. Before moving on, we consider the case <strong>of</strong> a free particle.<br />

The Lagrangian in this case must be independent <strong>of</strong> the position <strong>of</strong> the particle since a freely<br />

moving particle defines an inertial frame. Since space is isotropic, L must only depend upon the<br />

magnitude <strong>of</strong> v and not its direction. Hence,<br />

L = L(v 2 ).<br />

Since L is independent <strong>of</strong> x, ∂L/∂x = 0, so the Lagrange equation is<br />

d ∂L<br />

dt ∂v<br />

= 0.<br />

So, ∂L/∂v = const which leads us to conclude that L is quadratic in v. In fact,<br />

which is the kinetic energy for a particle.<br />

L = 1<br />

m v2 ,<br />

T = 1<br />

2 mv2 = 1<br />

2 m ˙x2 .<br />

For a particle moving in a potential field, V , the Lagrangian is given by<br />

L = T − V.<br />

L has units <strong>of</strong> energy and gives the difference between the energy <strong>of</strong> motion and the energy <strong>of</strong><br />

location.<br />

This leads to the equations <strong>of</strong> motion:<br />

d ∂L<br />

dt ∂v<br />

= ∂L<br />

∂x .<br />

Substituting L = T − V , yields<br />

m ˙v = − ∂V<br />

∂x<br />

which is identical to Newton’s equations given above once we identify the force as the minus the<br />

derivative <strong>of</strong> the potential. For the free particle, v = const. Thus,<br />

S =<br />

� tf<br />

to<br />

m<br />

2 v2 dt = m<br />

2 v2 (tf − to).<br />

19


You may be wondering at this point why we needed a new function and derived all this from<br />

some minimization principle. The reason is that for some systems we have constraints on the type<br />

<strong>of</strong> motion they can undertake. For example, there may be bonds, hinges, and other mechanical<br />

hinderances which limit the range <strong>of</strong> motion a given particle can take. The Lagrangian formalism<br />

provides a mechanism for incorporating these extra effects in a consistent and correct way. In<br />

fact we will use this principle later in deriving a variational solution to the Schrodinger equation<br />

by constraining the wavefunction solutions to be orthonormal.<br />

Lastly, it is interesting to note that v 2 = (dl/d) 2 = (dl) 2 /(dt) 2 is the square <strong>of</strong> the element<br />

<strong>of</strong> an arc in a given coordinate system. Thus, within the Lagrangian formalism it is easy to<br />

convert from one coordinate system to another. For example, in cartesian coordinates: dl 2 =<br />

dx 2 + dy 2 + dz 2 . Thus, v 2 = ˙x 2 + ˙y 2 + ˙z 2 . In cylindrical coordinates, dl = dr 2 + r 2 dφ 2 + dz 2 , we<br />

have the Lagrangian<br />

L = 1<br />

2 m( ˙r2 + r 2 ˙ φ 2 + ˙z 2 )<br />

and for spherical coordinates dl 2 = dr 2 + r 2 dθ 2 + r 2 sin 2 θdφ 2 ; hence<br />

L = 1<br />

2 m( ˙r2 + r 2 ˙ θ 2 + r 2 sin 2 θ ˙ φ 2 ).<br />

1.2.2 Example: 3 dimensional harmonic oscillator in spherical coordinates<br />

Here we take the potential energy to be a function <strong>of</strong> r alone (isotropic)<br />

V (r) = kr 2 /2.<br />

Thus, the Lagrangian in cartesian coordinates is<br />

L = m<br />

2 ( ˙x2 + ˙y 2 + ˙z 2 ) + k<br />

2 r2<br />

since r 2 = x 2 + y 2 + z 2 , we could easily solve this problem in cartesian space since<br />

L = m<br />

=<br />

2 ( ˙x2 + ˙y 2 + ˙z 2 ) + k<br />

2 (x2 + y 2 + z 2 ) (1.11)<br />

�<br />

�<br />

�<br />

� m<br />

2 ˙x2 + k<br />

2 x2<br />

+<br />

� m<br />

2 ˙y2 + k<br />

2 y2<br />

+<br />

� m<br />

2 ˙z2 + k<br />

2 z2<br />

(1.12)<br />

and we see that the system is separable into 3 independent oscillators. To convert to spherical<br />

polar coordinates, we use<br />

and the arc length given above.<br />

x = r sin(φ) cos(θ) (1.13)<br />

y = r sin(φ) sin(θ) (1.14)<br />

z = r cos(θ) (1.15)<br />

L = m<br />

2 ( ˙r2 + r 2 ˙ θ 2 + r 2 sin 2 θ ˙ φ 2 ) − k<br />

2 r2<br />

20


The equations <strong>of</strong> motion are<br />

d ∂L<br />

dt ∂ ˙ ∂L<br />

−<br />

φ ∂φ<br />

d ∂L<br />

dt ∂ ˙ ∂L<br />

−<br />

θ ∂θ<br />

d ∂L ∂L<br />

−<br />

dt ∂ ˙r ∂r<br />

= d<br />

dt (mr2 sin 2 θ ˙ φ = 0 (1.16)<br />

= d<br />

dt (mr2 ˙ θ) − mr 2 sin θ cos θ ˙ φ = 0 (1.17)<br />

= d<br />

dt (m ˙r) − mr ˙ θ 2 − mr sin 2 θ ˙ φ 2 + kr = 0 (1.18)<br />

We now prove that the motion <strong>of</strong> a particle in a central force field lies in a plane containing<br />

the origin. The force acting on the particle at any given time is in a direction towards the origin.<br />

Now, place an arbitrary cartesian frame centered about the particle with the z axis parallel to<br />

the direction <strong>of</strong> motion as sketched in Fig. 1.2 Note that the y axis is perpendicular to the plane<br />

<strong>of</strong> the page and hence there is no force component in that direction. Consequently, the motion<br />

<strong>of</strong> the particle is constrained to lie in the zx plane, i.e. the plane <strong>of</strong> the page and there is no<br />

force component which will take the particle out <strong>of</strong> this plane.<br />

Let’s make a change <strong>of</strong> coordinates by rotating the original frame to a new one whereby the<br />

new z ′ is perpendicular to the plane containing the initial position and velocity vectors. In the<br />

sketch above, this new z ′ axis would be perpendicular to the page and would contain the y axis<br />

we placed on the moving particle. In terms <strong>of</strong> these new coordinates, the Lagrangian will have<br />

the same form as before since our initial choice <strong>of</strong> axis was arbitrary. However, now, we have<br />

some additional constraints. Because the motion is now constrained to lie in the x ′ y ′ plane,<br />

θ ′ = π/2 is a constant, and ˙ θ = 0. Thus cos(π/2) = 0 and sin(π/2) = 1 in the equations above.<br />

From the equations for φ we find<br />

d<br />

dt mr2 ˙ φ = 0<br />

or<br />

mr 2 ˙ φ = const = pφ.<br />

This we can put into the r equation<br />

o<br />

d<br />

dt (m ˙r) − mr ˙ φ 2 + kr = 0 (1.19)<br />

d<br />

dt (m ˙r) − p2φ + kr<br />

mr3 = 0 (1.20)<br />

Figure 1.2: Vector diagram for motion in a central forces. The particle’s motion is along the Z<br />

axis which lies in the plane <strong>of</strong> the page.<br />

a<br />

21<br />

F<br />

Z<br />

Y<br />

X


Z'<br />

Y<br />

Z<br />

Y'<br />

where we notice that −p 2 φ/mr 3 is the centrifugal force. Taking the last equation, multiplying by<br />

˙r and then integrating with respect to time gives<br />

i.e.<br />

X'<br />

˙r 2 = − p2 φ<br />

m 2 r 2 − kr2 + b (1.21)<br />

˙r =<br />

Integrating once again with respect to time,<br />

�<br />

t − to =<br />

=<br />

X<br />

− p2 φ<br />

m 2 r 2 − kr2 + b (1.22)<br />

� rdr<br />

�<br />

= 1<br />

�<br />

2<br />

˙r<br />

�<br />

rdr<br />

− p2 φ<br />

m 2 − kr 4 + br 2<br />

dx<br />

√ a + bx + cx 2<br />

(1.23)<br />

(1.24)<br />

(1.25)<br />

where x = r 2 , a = −p 2 φ/m 2 , b is the constant <strong>of</strong> integration, and c = −k This is a standard<br />

integral and we can evaluate it to find<br />

where<br />

r 2 = 1<br />

2ω (b + A sin(ω(t − to))) (1.26)<br />

�<br />

A =<br />

m2 .<br />

What we see then is that r follows an elliptical path in a plane determined by the initial velocity.<br />

b 2 − ω2 p 2 φ<br />

22


This example also illustrates another important point which has tremendous impact on molecular<br />

quantum mechanics, namely, the angular momentum about the axis <strong>of</strong> rotation is conserved.<br />

We can choose any axis we want. In order to avoid confusion, let us define χ as the angular<br />

rotation about the body-fixed Z ′ axis and φ as angular rotation about the original Z axis. So<br />

our conservation equations are<br />

mr 2 ˙χ = pχ<br />

about the Z ′ axis and<br />

mr 2 sin θ ˙ φ = pφ<br />

for some arbitrary fixed Z axis. The angle θ will also have an angular momentum associated<br />

with it pθ = mr 2 ˙ θ, but we do not have an associated conservation principle for this term since it<br />

varies with φ. We can connect pχ with pθ and pφ about the other axis via<br />

Consequently,<br />

pχdχ = pθdθ + pφdφ.<br />

mr 2 ˙χ 2 dχ = mr 2 ( ˙ φ sin θdφ + ˙ θdθ).<br />

Here we see that the the angular momentum vector remains fixed in space in the absence <strong>of</strong><br />

any external forces. Once an object starts spinning, its axis <strong>of</strong> rotation remains pointing in a<br />

given direction unless something acts upon it (torque), in essence in classical mechanics we can<br />

fully specify Lx, Ly, and Lz as constants <strong>of</strong> the motion since d � L/dt = 0. In a later chapter, we<br />

will cover the quantum mechanics <strong>of</strong> rotations in much more detail. In the quantum case, we<br />

will find that one cannot make such a precise specification <strong>of</strong> the angular momentum vector for<br />

systems with low angular momentum. We will, however, recover the classical limit in end as we<br />

consider the limit <strong>of</strong> large angular momenta.<br />

1.3 Conservation Laws<br />

We just encountered one extremely important concept in mechanics, namely, that some quantities<br />

are conserved if there is an underlying symmetry. Next, we consider a conservation law arising<br />

from the homogeneity <strong>of</strong> time. For a closed dynamical system, the Lagrangian does not explicitly<br />

depend upon time. Thus we can write<br />

dL<br />

dt<br />

= ∂L<br />

∂x<br />

∂L<br />

˙x + ¨x (1.27)<br />

∂ ˙x<br />

Replacing ∂L/∂x with Lagrange’s equation, we obtain<br />

dL<br />

dt<br />

=<br />

� �<br />

d ∂L<br />

˙x +<br />

dt ∂ ˙x<br />

∂L<br />

=<br />

¨x<br />

∂ ˙x<br />

(1.28)<br />

d<br />

�<br />

˙x<br />

dt<br />

∂L<br />

�<br />

∂ ˙x<br />

(1.29)<br />

Now, rearranging this a bit,<br />

�<br />

d<br />

˙x<br />

dt<br />

∂L<br />

�<br />

− L = 0. (1.30)<br />

∂ ˙x<br />

23


So, we can take the quantity in the parenthesis to be a constant.<br />

�<br />

E = ˙x ∂L<br />

�<br />

− L = const. (1.31)<br />

∂ ˙x<br />

is an integral <strong>of</strong> the motion. This is the energy <strong>of</strong> the system. Since L can be written in form<br />

L = T − V where T is a quadratic function <strong>of</strong> the velocities, and using Euler’s theorem on<br />

homogeneous functions:<br />

This gives,<br />

˙x ∂L<br />

∂ ˙x<br />

= ˙x∂T<br />

∂ ˙x<br />

E = T + V<br />

= 2T.<br />

which says that the energy <strong>of</strong> the system can be written as the sum <strong>of</strong> two different terms: the<br />

kinetic energy or energy <strong>of</strong> motion and the potential energy or the energy <strong>of</strong> location.<br />

One can also prove that linear momentum is conserved when space is homogeneous. That is,<br />

when we can translate our system some arbitrary amount ɛ and our dynamical quantities must<br />

remain unchanged. We will prove this in the problem sets.<br />

1.4 Hamiltonian Dynamics<br />

Hamiltonian dynamics is a further generalization <strong>of</strong> classical dynamics and provides a crucial link<br />

with quantum mechanics. Hamilton’s function, H, is written in terms <strong>of</strong> the particle’s position<br />

and momentum, H = H(p, q). It is related to the Lagrangian via<br />

Taking the derivative <strong>of</strong> H w.r.t. x<br />

H = ˙xp − L(x, ˙x)<br />

∂H<br />

∂x<br />

= −∂L<br />

∂x<br />

= − ˙p<br />

Differentiation with respect to p gives<br />

∂H<br />

= ˙q.<br />

∂p<br />

These last two equations give the conservation conditions in the Hamiltonian formalism. If H<br />

is independent <strong>of</strong> the position <strong>of</strong> the particle, then the generalized momentum, p is constant in<br />

time. If the potential energy is independent <strong>of</strong> time, the Hamiltonian gives the total energy <strong>of</strong><br />

the system,<br />

H = T + V.<br />

1.4.1 Interaction between a charged particle and an electromagnetic<br />

field.<br />

We consider here a free particle with mass m and charge e in an electromagnetic field. The<br />

Hamiltonian is<br />

H = px ˙x + py ˙y + pz ˙z − L (1.32)<br />

= ˙x ∂L<br />

∂ ˙x<br />

+ ˙y ∂L<br />

∂ ˙y<br />

24<br />

+ ˙z ∂L<br />

∂ ˙z<br />

− L. (1.33)


Our goal is to write this Hamiltonian in terms <strong>of</strong> momenta and coordinates.<br />

For a charged particle in a field, the force acting on the particle is the Lorenz force. Here it<br />

is useful to introduce a vector and scaler potential and to work in cgs units.<br />

�F = e<br />

c �v × (� ∇ × � A) − e ∂<br />

c<br />

� A<br />

∂t − e� ∇φ.<br />

The force in the x direction is given by<br />

Fx = d<br />

�<br />

e<br />

m ˙x = ˙y<br />

dt c<br />

∂Ay<br />

�<br />

∂Az<br />

+ ˙z −<br />

∂x ∂x<br />

e<br />

�<br />

˙y<br />

c<br />

∂Ax<br />

�<br />

∂Ax ∂Ax<br />

+ ˙z + − e<br />

∂y ∂z ∂t<br />

∂φ<br />

∂x<br />

with the remaining components given by cyclic permutation. Since<br />

dAx ∂Ax<br />

∂Ax ∂Ax<br />

= + ˙x∂Ax + ˙y + ˙z<br />

dt ∂t ∂x ∂y ∂z ,<br />

Fx = e<br />

�<br />

+ ˙x<br />

c<br />

∂Ax<br />

�<br />

∂Ax ∂Ax<br />

+ ˙y + ˙z −<br />

∂x ∂y ∂z<br />

e<br />

c �v · � A − eφ.<br />

Based upon this, we find that the Lagrangian is<br />

L = 1 1 1<br />

m ˙x2 m ˙y2<br />

2 2 2 m ˙z2 + e<br />

c �v · � A − eφ<br />

where φ is a velocity independent and static potential.<br />

Continuing on, the Hamiltonian is<br />

H = m<br />

2 ( ˙x2 + ˙y 2 + ˙z 2 ) + eφ (1.34)<br />

= 1<br />

2m ((m ˙x)2 + (m ˙y) 2 + (m ˙y) 2 ) + eφ (1.35)<br />

The velocities, m ˙x, are derived from the Lagrangian via the canonical relation<br />

From this we find,<br />

and the resulting Hamiltonian is<br />

H = 1<br />

�� px −<br />

2m<br />

e<br />

c Ax<br />

p = ∂L<br />

∂ ˙x<br />

m ˙x = px − e<br />

c Ax<br />

m ˙y = py − e<br />

c Ay<br />

m ˙z = pz − e<br />

c Az<br />

� 2<br />

�<br />

+ py − e<br />

c Ay<br />

� 2<br />

�<br />

+ pz − e<br />

c Az<br />

� �<br />

2<br />

+ eφ.<br />

(1.36)<br />

(1.37)<br />

(1.38)<br />

We see here an important concept relating the velocity and the momentum. In the absence <strong>of</strong> a<br />

vector potential, the velocity and the momentum are parallel. However, when a vector potential<br />

is included, the actual velocity <strong>of</strong> a particle is no longer parallel to its momentum and is in fact<br />

deflected by the vector potential.<br />

25


1.4.2 Time dependence <strong>of</strong> a dynamical variable<br />

On <strong>of</strong> the important applications <strong>of</strong> Hamiltonian mechanics is in the dynamical evolution <strong>of</strong> a<br />

variable which depends upon p and q, G(p, q). The total derivative <strong>of</strong> G is<br />

dG<br />

dt<br />

= ∂G<br />

∂t<br />

+ ∂G<br />

∂q<br />

˙q + ∂G<br />

∂p ˙p<br />

From Hamilton’s equations, we have the canonical definitions<br />

Thus,<br />

dG<br />

dt<br />

dG<br />

dt<br />

˙q = ∂H<br />

, ˙p = −∂H<br />

∂p ∂q<br />

∂G<br />

=<br />

∂t<br />

∂G<br />

=<br />

∂t<br />

∂G ∂H<br />

+<br />

∂q ∂p<br />

∂G ∂H<br />

−<br />

∂p ∂q<br />

(1.39)<br />

+ {G, H}, (1.40)<br />

where {A, B} is called the Poisson bracket <strong>of</strong> two dynamical quantities, G and H.<br />

{G, H}, = ∂G ∂H<br />

∂q ∂p<br />

∂G ∂H<br />

−<br />

∂p ∂q<br />

We can also define a linear operator L as generating the Poisson bracket with the Hamiltonian:<br />

LG = 1<br />

{H, G}<br />

i<br />

so that if G does not depend explicitly upon time,<br />

G(t) = exp(iLt)G(0).<br />

where exp(iLt) is the propagator which carried G(0) to G(t).<br />

Also, note that if {G, H} = 0, then dG/dt = 0 so that G is a constant <strong>of</strong> the motion. This<br />

too, along with the construction <strong>of</strong> the Poisson bracket has considerable importance in the realm<br />

<strong>of</strong> quantum mechanics.<br />

1.4.3 Virial Theorem<br />

Finally, we turn our attention to a concept which has played an important role in both quantum<br />

and classical mechanics. Consider a function G that is a product <strong>of</strong> linear momenta and<br />

coordinate,<br />

G = pq.<br />

The time derivative is simply.<br />

G<br />

dt<br />

= q ˙p + p ˙q<br />

26


Now, let’s take a time average <strong>of</strong> both sides <strong>of</strong> this last equation.<br />

�<br />

d<br />

dt pq<br />

�<br />

1<br />

= lim<br />

T →∞<br />

= lim<br />

T →∞<br />

= lim<br />

T →∞<br />

� T<br />

�<br />

d<br />

dt pq<br />

�<br />

dt (1.41)<br />

T 0<br />

�<br />

1 T<br />

d(pq) (1.42)<br />

T 0<br />

1<br />

T ((pq)T − (pq)0) (1.43)<br />

If the trajectories <strong>of</strong> system are bounded, both p and q are periodic in time and are therefore<br />

finite. Thus, the average must vanish as T → ∞ giving<br />

Since p ˙q = 2T and ˙p = −F , we have<br />

〈p ˙q + q ˙p〉 = 0 (1.44)<br />

〈2T 〉 = −〈qF 〉. (1.45)<br />

In cartesian coordinates this leads to<br />

� �<br />

�<br />

〈2T 〉 = − xiFi . (1.46)<br />

i<br />

For a conservative system F = −∇V . Thus, if we have a centro-symmetric potential given<br />

by V = Cr n , it is easy to show that<br />

〈2T 〉 = n〈V 〉.<br />

For the case <strong>of</strong> the Harmonic oscillator, n = 2 and 〈T 〉 = 〈V 〉. So, for example, if we have a<br />

total energy equal to kT in this mode, then 〈T 〉 + 〈V 〉 = kT and 〈T 〉 = 〈V 〉 = kT/2. Moreover,<br />

for the interaction between two opposite charges separated by r, n = −1 and<br />

〈2T 〉 = −〈V 〉.<br />

27


Figure 1.3: Screen shot <strong>of</strong> using Mathematica to plot phase-plane for harmonic oscillator. Here<br />

k/m = 1 and our xo = 0.75.<br />

28


Chapter 2<br />

Waves and Wavefunctions<br />

In the world <strong>of</strong> quantum physics, no phenominon is a phenominon until it is a recorded<br />

phenominon.<br />

– John Archibald Wheler<br />

The physical basis <strong>of</strong> quantum mechanics is<br />

1. That matter, such as electrons, always arrives at a point as a discrete chunk, but that the<br />

probibility <strong>of</strong> finding a chunk at a specified position is like the intensity distribution <strong>of</strong> a<br />

wave.<br />

2. The “quantum state” <strong>of</strong> a system is described by a mathematical object called a “wavefunction”<br />

or state vector and is denoted |ψ〉.<br />

3. The state |ψ〉 can be expanded in terms <strong>of</strong> the basis states <strong>of</strong> a given vector space, {|φi〉}<br />

as<br />

where 〈φi|ψ〉 denotes an inner product <strong>of</strong> the two vectors.<br />

|ψ〉 = �<br />

|φi〉〈φi|ψ〉 (2.1)<br />

i<br />

4. Observable quantities are associated with the expectation value <strong>of</strong> Hermitian operators and<br />

that the eigenvalues <strong>of</strong> such operators are always real.<br />

5. If two operators commute, one can measure the two associated physical quantities simultaneously<br />

to arbitrary precision.<br />

6. The result <strong>of</strong> a physical measurement projects |ψ〉 onto an eigenstate <strong>of</strong> the associated<br />

operator |φn〉 yielding a measured value <strong>of</strong> an with probability |〈φn|ψ〉| 2 .<br />

2.1 Position and Momentum Representation <strong>of</strong> |ψ〉<br />

1 Two common operators which we shall use extensively are the position and momentum operator.<br />

1 The majority <strong>of</strong> this lecture comes from Cohen-Tannoudji Chapter 1, part from Feynman & Hibbs<br />

29


The position operator acts on the state |ψ〉 to give the amplitude <strong>of</strong> the system to be at a<br />

given position:<br />

ˆx|ψ〉 = |x〉〈x|ψ〉 (2.2)<br />

= |x〉ψ(x) (2.3)<br />

We shall call ψ(x) the wavefunction <strong>of</strong> the system since it is the amplitude <strong>of</strong> |ψ〉 at point x. Here<br />

we can see that ψ(x) is an eigenstate <strong>of</strong> the position operator. We also define the momentum<br />

operator ˆp as a derivative operator:<br />

Thus,<br />

ˆp = −i¯h ∂<br />

∂x<br />

(2.4)<br />

ˆpψ(x) = −i¯hψ ′ (x). (2.5)<br />

Note that ψ ′ (x) �= ψ(x), thus an eigenstate <strong>of</strong> the position operator is not also an eigenstate <strong>of</strong><br />

the momentum operator.<br />

We can deduce this also from the fact that ˆx and ˆp do not commute. To see this, first consider<br />

∂<br />

∂x xf(x) = f(x) + xf ′ (x) (2.6)<br />

Thus (using the shorthand ∂x as partial derivative with respect to x.)<br />

[ˆx, ˆp]f(x) = i¯h(x∂xf(x) − ∂x(xf(x))) (2.7)<br />

= −i¯h(xf ′ (x) − f(x) − xf ′ (x)) (2.8)<br />

= i¯hf(x) (2.9)<br />

What are the eigenstates <strong>of</strong> the ˆp operator? To find them, consider the following eigenvalue<br />

equation:<br />

ˆp|φ(k)〉 = k|φ(k)〉 (2.10)<br />

Inserting a complete set <strong>of</strong> position states using the idempotent operator<br />

�<br />

I = |x〉〈x|dx (2.11)<br />

and using the “coordinate” representation <strong>of</strong> the momentum operator, we get<br />

Thus, the solution <strong>of</strong> this is (subject to normalization)<br />

− i¯h∂xφ(k, x) = kφ(k, x) (2.12)<br />

φ(k, x) = C exp(ik/¯h) = 〈x|φ(k)〉 (2.13)<br />

30


We can also use the |φ(k)〉 = |k〉 states as a basis for the state |ψ〉 by writing<br />

�<br />

|ψ〉 = dk|k〉〈k|ψ〉 (2.14)<br />

�<br />

= dk|k〉ψ(k) (2.15)<br />

where ψ(k) is related to ψ(x) via:<br />

�<br />

ψ(k) = 〈k|ψ〉 =<br />

�<br />

dx〈k|x〉〈x|ψ〉 (2.16)<br />

= C dx exp(ikx/¯h)ψ(x). (2.17)<br />

This type <strong>of</strong> integral is called a “Fourier Transfrom”. There are a number <strong>of</strong> ways to define<br />

the normalization C when using this transform, for our purposes at the moment, we’ll set C =<br />

1/ √ 2π¯h so that<br />

and<br />

ψ(x) = 1<br />

�<br />

√<br />

2π¯h<br />

ψ(x) = 1<br />

�<br />

√<br />

2π¯h<br />

dkψ(k) exp(−ikx/¯h) (2.18)<br />

dxψ(x) exp(ikx/¯h). (2.19)<br />

Using this choice <strong>of</strong> normalization, the transform and the inverse transform have symmetric forms<br />

and we only need to remember the sign in the exponential.<br />

2.2 The Schrödinger Equation<br />

Postulate 2.1 The quantum state <strong>of</strong> the system is a solution <strong>of</strong> the Schrödinger equation<br />

i¯h∂t|ψ(t)〉 = H|ψ(t)〉, (2.20)<br />

where H is the quantum mechanical analogue <strong>of</strong> the classical Hamiltonian.<br />

From classical mechanics, H is the sum <strong>of</strong> the kinetic and potential energy <strong>of</strong> a particle,<br />

H = 1<br />

2m p2 + V (x). (2.21)<br />

Thus, using the quantum analogues <strong>of</strong> the classical x and p, the quantum H is<br />

H = 1<br />

2m ˆp2 + V (ˆx). (2.22)<br />

To evaluate V (ˆx) we need a theorem that a function <strong>of</strong> an operator is the function evaluated<br />

at the eigenvalue <strong>of</strong> the operator. The pro<strong>of</strong> is straight forward, Taylor expand the function<br />

about some point, If<br />

V (x) = (V (0) + xV ′ (0) + 1<br />

2 V ′′ (0)x 2 · · ·) (2.23)<br />

31


then<br />

Since for any operator<br />

Thus, we have<br />

V (ˆx) = (V (0) + ˆxV ′ (0) + 1<br />

2 V ′′ (0)ˆx 2 · · ·) (2.24)<br />

So, in coordinate form, the Schrödinger Equation is written as<br />

[ ˆ f, ˆ f p ] = 0∀ p (2.25)<br />

〈x|V (ˆx)|ψ〉 = V (x)ψ(x) (2.26)<br />

i¯h ∂<br />

�<br />

ψ(x, t) = −<br />

∂t ¯h<br />

2m<br />

2.2.1 Gaussian Wavefunctions<br />

∂ 2<br />

+ V (x)<br />

∂x2 �<br />

ψ(x, t) (2.27)<br />

Let’s assume that our initial state is a Gaussian in x with some initial momentum k◦.<br />

ψ(x, 0) =<br />

The momentum representation <strong>of</strong> this is<br />

�<br />

2<br />

πa2 �1/4<br />

exp(ikox) exp(−x 2 /a 2 ) (2.28)<br />

ψ(k, 0) = 1<br />

�<br />

2π¯h<br />

dxe −ikx ψ(x, 0) (2.29)<br />

= (πa) 1/2 e −(k−ko)2a2 /4)<br />

(2.30)<br />

In Fig.2.1, we see a gaussian wavepacket centered about x = 0 with ko = 10 and a = 1.<br />

For now we will use dimensionaless units. The red and blue components correspond to the real<br />

and imaginary components <strong>of</strong> ψ and the black curve is |ψ(x)| 2 . Notice, that the wavefunction is<br />

pretty localized along the x axis.<br />

In the next figure, (Fig. 2.2) we have the momentum distribution <strong>of</strong> the wavefunction, ψ(k, 0).<br />

Again, we have chosen ko = 10. Notice that the center <strong>of</strong> the distribution is shifted about ko.<br />

So, for f(x) = exp(−x 2 /b 2 ), ∆x = b/ √ 2. Thus, when x varies form 0 to ±∆x, f(x) is<br />

diminished by a factor <strong>of</strong> 1/ √ e. (∆x is the RMS deviation <strong>of</strong> f(x).)<br />

For the Gaussian wavepacket:<br />

or<br />

Thus, ∆x∆p = ¯h/2 for the initial wavefunction.<br />

∆x = a/2 (2.31)<br />

∆k = 1/a (2.32)<br />

∆p = ¯h/a (2.33)<br />

32


0.75<br />

0.5<br />

0.25<br />

-3 -2 -1 1 2 3<br />

-0.25<br />

-0.5<br />

-0.75<br />

Figure 2.1: Real (red), imaginary (blue) and absolute value (black) <strong>of</strong> gaussian wavepacket ψ(x)<br />

è<br />

y@kD<br />

2.5<br />

2<br />

1.5<br />

1<br />

0.5<br />

6 8 10 12 14<br />

Figure 2.2: Momentum-space distribution <strong>of</strong> ψ(k).<br />

33<br />

k


2.2.2 Evolution <strong>of</strong> ψ(x)<br />

Now, let’s consider the evolution <strong>of</strong> a free particle. By a “free” particle, we mean a particle<br />

whose potential energy does not change, I.e. we set V (x) = 0 for all x and solve:<br />

i¯h ∂<br />

�<br />

ψ(x, t) = −<br />

∂t ¯h<br />

2m<br />

∂2 ∂x2 �<br />

ψ(x, t) (2.34)<br />

This equation is actually easier to solve in k-space. Taking the Fourier Transform,<br />

Thus, the temporal solution <strong>of</strong> the equation is<br />

i¯h∂tψ(k, t) = k2<br />

ψ(k, t) (2.35)<br />

2m<br />

ψ(k, t) = exp(−ik 2 /(2m)t/¯h)ψ(k, 0). (2.36)<br />

This is subject to some initial function ψ(k, 0). To get the coordinate x-representation <strong>of</strong> the<br />

solution, we can use the FT relations above:<br />

ψ(x, t) =<br />

=<br />

=<br />

=<br />

�<br />

1<br />

√ dkψ(k, t) exp(−ikx) (2.37)<br />

2π¯h<br />

�<br />

dx ′ 〈x| exp(−iˆp 2 /(2m)t/¯h)|x ′ 〉ψ(x ′ , 0) (2.38)<br />

�<br />

�<br />

�<br />

m<br />

2πi¯ht<br />

�<br />

dx ′ exp<br />

� im(x − x ′ ) 2<br />

2¯ht<br />

ψ(x ′ , 0) (2.39)<br />

dx ′ Go(x, x ′ )ψ(x ′ , 0) (2.40)<br />

(homework: derive Go and show that Go is a solution <strong>of</strong> the free particle schrodinger equation<br />

HGo = i∂tGo.) The function Go is called the “free particle propagator” or “Green’s Function”<br />

and tells us the amplitude for a particle to start <strong>of</strong>f at x ′ and end up at another point x at time<br />

t.<br />

The sketch tells me that in order to got far away from the initial point in time t I need to<br />

have a lot <strong>of</strong> energy (wiggles get closer together implies higher Fourier component )<br />

Here we see that the probability to find a particle at the initial point decreases with time.<br />

Since the period <strong>of</strong> oscillation (T ) is the time required to increase the phase by 2π.<br />

2π = mx2 mx2<br />

−<br />

2¯ht 2¯h(t + T )<br />

= mx2<br />

2¯ht2 � �<br />

2 T<br />

1 + T/t<br />

Let ω = 2π/T and take the long time limit t ≫ T , we can estimate<br />

ω ≈ m<br />

2¯h<br />

� �<br />

x 2<br />

t<br />

34<br />

(2.41)<br />

(2.42)<br />

(2.43)


Figure 2.3: Go for fixed t as a function <strong>of</strong> x.<br />

0.4<br />

0.2<br />

-10 -5 5 10<br />

-0.2<br />

-0.4<br />

Since the classical kinetic energy is given by E = m/2v 2 , we obtain<br />

E = ¯hω (2.44)<br />

Thus, the energy <strong>of</strong> the wave is proportional to the period <strong>of</strong> oscillation.<br />

We can evaluate the evolution in x using either the Go we derived above, or by taking the<br />

FT <strong>of</strong> the wavefunction evolving in k-space. Recall that the solution in k-space was<br />

ψ(k, t) = exp(−ik 2 /(2m)t/¯h)ψ(k, 0) (2.45)<br />

Assuming a Gaussian form for ψ(k) as above,<br />

√<br />

a<br />

ψ(x, t) =<br />

(2π) 3/4<br />

�<br />

dke −a2 /4(k−ko) 2<br />

e i(kx−ω(k)t)<br />

where ω(k) is the dispersion relation for a free particle:<br />

Cranking through the integral:<br />

ψ(x, t) =<br />

� 2a 2<br />

π<br />

� 1/4<br />

�<br />

ω(k) = ¯hk2<br />

2m<br />

e iφ<br />

a 4 + 4¯h2 t 2<br />

m 2<br />

� 1/4 e ikox exp<br />

where φ = −θ − ¯hk 2 o/(2m)t and tan 2θ = 2¯ht/(ma 2 ).<br />

Likewise, for the amplitude:<br />

�<br />

(x − ¯hko/mt) 2<br />

�<br />

a 2 + 2i¯ht/m<br />

|ψ(x, t)| 2 �<br />

�<br />

2<br />

1<br />

(x − v◦t)<br />

=<br />

exp −<br />

2π∆x(t) 2 2∆x(t) 2<br />

�<br />

35<br />

(2.46)<br />

(2.47)<br />

(2.48)<br />

(2.49)


Figure 2.4: Evolution <strong>of</strong> a free particle wavefunction. In this case we have given the initial state<br />

a kick in the +x direction. Notice that as the system moves, the center moves at a constant rate<br />

where as the width <strong>of</strong> the packet constantly spreads out over time.<br />

Where I define<br />

∆x(t) = a<br />

�<br />

1 +<br />

2<br />

4¯h2 t2 m2a4 as the time dependent RMS width <strong>of</strong> the wave and the group velocity:<br />

(2.50)<br />

vo = ¯hko<br />

. (2.51)<br />

m<br />

Now, since ∆p = ¯h∆k = ¯h/a is a constant for all time, the uncertainty relation becomes<br />

∆x(t)∆p ≥ ¯h/2 (2.52)<br />

corresponding to the particle’s wavefunction becoming more and more diffuse as it evolves in<br />

time.<br />

2.3 Particle in a Box<br />

2.3.1 Infinite Box<br />

The Mathematica handout shows how one can use Mathematica to set up and solve some simple<br />

problems on the computer. (One good class problem would be to use Mathematica to carry<br />

out the symbolic manipulations for a useful or interesting problem and/or to solve the problem<br />

numerically.)<br />

The potential we’ll work with for this example consists <strong>of</strong> two infinitely steep walls placed<br />

at x = ℓ and x = 0 such that between the two walls, V (x) = 0. Within this region, we seek<br />

solutions to the differential equation<br />

∂ 2 xψ(x) = −2mE/¯h 2 ψ(x). (2.53)<br />

The solutions <strong>of</strong> this are plane waves traveling to the left and to the right,<br />

ψ(x) = A exp(−ikx) + B exp(+ikx) (2.54)<br />

The coefficients A and B we’ll have to determine. k is determined by substitution back into the<br />

differential equation<br />

ψ ′′ (x) = −k 2 ψ(x) (2.55)<br />

Thus, k 2 = 2mE/¯h 2 , or ¯hk = √ 2mE. Let’s work in units in which ¯h = 1 and me = 1. Energy in<br />

these units is the Hartree (≈ 27.eV.) Posted on the web-page is a file (c-header file) which has a<br />

number <strong>of</strong> useful conversion factors.<br />

36


Since ψ(x) must vanish at x = 0 and x = ℓ<br />

A + B = 0 (2.56)<br />

A exp(ikℓ) + B exp(−ikℓ) = 0 (2.57)<br />

We can see immediately that A = −B and that the solutions must correspond to a family <strong>of</strong> sine<br />

functions:<br />

Just a check,<br />

ψ(x) = A sin(nπ/ℓx) (2.58)<br />

ψ(ℓ) = A sin(nπ/ℓℓ) = A sin(nπ) = 0. (2.59)<br />

To obtain the coefficient, we simply require that the wavefunctions be normalized over the range<br />

x = [0, ℓ].<br />

� ℓ<br />

sin(nπx/ℓ)<br />

0<br />

2 dx = ℓ<br />

(2.60)<br />

2<br />

Thus, the normalized solutions are<br />

�<br />

2<br />

ψn(x) = sin(nπ/ℓx) (2.61)<br />

ℓ<br />

The eigenenergies are obtained by applying the Hamiltonian to the wavefunction solution<br />

Thus we can write En as a function <strong>of</strong> n<br />

Enψn(x) = − ¯h2<br />

2m ∂2 xψn(x) (2.62)<br />

= ¯h2 n 2 π 2<br />

2a 2 m ψn(x) (2.63)<br />

En = ¯h2 π2 2a2m n2<br />

(2.64)<br />

for n = 0, 1, 2, .... What about the case where n = 0? Clearly it’s an allowed solution <strong>of</strong><br />

the Schrödinger Equation. However, we also required that the probability to find the particle<br />

anywhere must be 1. Thus, the n = 0 solution cannot be permitted.<br />

Note also that the cosine functions are also allowed solutions. However, the restriction <strong>of</strong><br />

ψ(0) = 0 and ψ(ℓ) = 0 discounts these solutions.<br />

In Fig. 2.5 we show the first few eigenstates for an electron trapped in a well <strong>of</strong> length a = π.<br />

The potential is shown in gray. Notice that the number <strong>of</strong> nodes increases as the energy increases.<br />

In fact, one can determine the state <strong>of</strong> the system by simply counting nodes.<br />

What about orthonormality. We stated that the solution <strong>of</strong> the eigenvalue problem form an<br />

orthonormal basis. In Dirac notation we can write<br />

�<br />

〈ψn|ψm〉 = dx〈ψn|x〉〈x|ψm〉 (2.65)<br />

=<br />

� ℓ<br />

0<br />

� ℓ<br />

dxψ ∗ n(x)ψm(x) (2.66)<br />

= 2<br />

ℓ 0<br />

dx sin(nπx/ℓ) sin(mπx/ℓ) (2.67)<br />

= δnm. (2.68)<br />

37


14<br />

12<br />

10<br />

8<br />

6<br />

4<br />

2<br />

-1 1 2 3 4<br />

Figure 2.5: Particle in a box states<br />

Thus, we can see in fact that these solutions do form a complete set <strong>of</strong> orthogonal states on<br />

the range x = [0, ℓ]. Note that it’s important to specify “on the range...” since clearly the sin<br />

functions are not a set <strong>of</strong> orthogonal functions over the entire x axis.<br />

2.3.2 Particle in a finite Box<br />

Now, suppose our box is finite. That is<br />

V (x) =<br />

�<br />

−Vo if −a < x < a<br />

0 otherwise<br />

(2.69)<br />

Let’s consider the case for E < 0. The case E > 0 will correspond to scattering solutions. In<br />

side the well, the wavefunction oscillates, much like in the previous case.<br />

ψW (x) = A sin(kix) + B cos(kix) (2.70)<br />

where ki comes from the equation for the momentum inside the well<br />

¯hki =<br />

�<br />

2m(En + Vo) (2.71)<br />

We actually have two classes <strong>of</strong> solution, a symmetric solution when A = 0 and an antisymmetric<br />

solution when B = 0.<br />

Outside the well the potential is 0 and we have the solutions<br />

ψO(x) = c1e ρx andc2e −ρx<br />

38<br />

(2.72)


We will choose the coefficients c1 and c2 as to create two cases, ψL and ψR on the left and right<br />

hand sides <strong>of</strong> the well. Also,<br />

¯hρ = √ −2mE (2.73)<br />

Thus, we have three pieces <strong>of</strong> the full solution which we must hook together.<br />

ψL(x) = Ce ρx for x < −a (2.74)<br />

ψR(x) = De −ρx for x > −a (2.75)<br />

(2.76)<br />

ψW (x) = A sin(kix) + B cos(kix)for inside the well (2.77)<br />

To find the coefficients, we need to set up a series <strong>of</strong> simultaneous equations by applying the<br />

conditions that a.) the wavefunction be a continuous function <strong>of</strong> x and that b.) it have continuous<br />

first derivatives with respect to x. Thus, applying the two conditions at the boundaries:<br />

The matching conditions at x = a<br />

The final results are (after the chalk dust settles):<br />

ψL(−a) − ψW (−a) = 0 (2.78)<br />

(2.79)<br />

ψR(a) − ψW (a) = 0 (2.80)<br />

(2.81)<br />

ψ ′ L(−a) − ψ ′ W (−a) = 0 (2.82)<br />

(2.83)<br />

ψ ′ R(a) − ψ ′ W (a) = 0 (2.84)<br />

1. For A = 0. B = D sec(aki)e −aρ and C = D. (Symmetric Solution)<br />

2. For B = 0, A = C csc(aki)e −aρ and C = −D. (Antisymmetric Solution)<br />

So, now we have all the coefficients expressed in terms <strong>of</strong> D, which we can determine by normalization<br />

(if so inclined). We’ll not do that integral, as it is pretty straightforward.<br />

For the energies, we substitute the symmetric and antisymmetric solutions into the Eigenvalue<br />

equation and obtain:<br />

ρ cos(aki) = ki sin(ki) (2.85)<br />

39


or<br />

for the symmetric case and<br />

�<br />

for the anti-symmetric case, or<br />

�<br />

E<br />

Vo − E<br />

ρ<br />

ki<br />

= tan(aki) (2.86)<br />

= tan(a<br />

�<br />

2m(Vo − E)/¯h) (2.87)<br />

ρ sin(aki) = −ki cos(aki) (2.88)<br />

E<br />

Vo − E<br />

ρ<br />

ki<br />

= cot(aki) (2.89)<br />

(2.90)<br />

= cot(a<br />

�<br />

2m(Vo − E)/¯h) (2.91)<br />

Substituting the expressions for ki and ρ into final results for each case we find a set <strong>of</strong><br />

matching conditions: For the symmetric case, eigenvalues occur when ever the two curves<br />

and for the anti-symmetric case,<br />

�<br />

�<br />

1 − Vo/E = tan(a 2m(E − Vo)/¯h) (2.92)<br />

�<br />

�<br />

1 − Vo/E = cot(a 2m(E − Vo)/¯h) (2.93)<br />

These are called “transcendental” equations and closed form solutions are generally impossible<br />

to obtain. Graphical solutions are helpful. In Fig. ?? we show the graphical solution to the<br />

transendental equations for an electron in a Vo = −10 well <strong>of</strong> width a = 2. The black dots<br />

indicate the presence <strong>of</strong> two bound states, one symmetric and one anti-symmetric at E = 2.03<br />

and 3.78 repectively.<br />

2.3.3 Scattering states and resonances.<br />

Now let’s take the same example as above, except look at states for which E > 0. In this case, we<br />

have to consider where the particles are coming from and where they are going. We will assume<br />

that the particles are emitted with precise energy E towards the well from −∞ and travel from<br />

left to right. As in the case above we have three distinct regions,<br />

1. x > −a where ψ(x) = e ik1x + Re −ik1x = ψL(x)<br />

2. −a ≤ x ≤ +a where ψ(x) = Ae −ik2x + Be +ik2x = ψW (x)<br />

40


symêasym<br />

4<br />

3<br />

2<br />

1<br />

-1<br />

-2<br />

-3<br />

-4<br />

2 4 6 8 10 E<br />

Figure 2.6: Graphical solution to transendental equations for an electron in a truncated hard<br />

well <strong>of</strong> depth Vo = 10 and width a = 2. The short-dashed blue curve corresponds to the<br />

symmetric � case and the long-dashed blue curve corresponds to the asymetric case. The red line<br />

is 1 − V o/E. Bound state solution are such that the red and blue curves cross.<br />

3. x > +a where ψ(x) = T e +ik1x = ψR(x)<br />

where k1 = √ �<br />

2mE/¯h is the momentum outside the well, k2 = 2m(E − V )/¯h is the momentum<br />

inside the well, and A, B, T , and R are coefficients we need to determine. We also have the<br />

matching conditions:<br />

ψL(−a) − ψW (−a) = 0<br />

ψ ′ L(−a) − ψ ′ W (−a) = 0<br />

ψR(a) − ψW (a) = 0<br />

ψ ′ R(a) − ψ ′ W (a) = 0<br />

This can be solved by hand, however, Mathematica make it easy. The results are a series <strong>of</strong> rules<br />

which we can use to determine the transmission and reflection coefficients.<br />

T →<br />

A →<br />

B →<br />

R →<br />

−4e−2iak1+2iak2k1k2 −k1 2 + e4iak2k1 2 − 2k1k2 − 2e4iak2k1k2 − k2 2 + e4iak2k2 2 ,<br />

2e−iak1+3iak2k1 (k1 − k2)<br />

−k1 2 + e4iak2k1 2 − 2k1k2 − 2e4iak2k1k2 − k2 2 + e4iak2k2 2 ,<br />

−2e−iak1+iak2k1 (k1 + k2)<br />

−k1 2 + e4iak2k1 2 − 2k1k2 − 2e4iak2k1k2 − k2 2 + e4iak2k2 2 ,<br />

�<br />

−1 + e4iak2 � �<br />

k1 2 − k2 2�<br />

e2iak1 �<br />

−k1 2 + e4iak2k1 2 − 2k1k2 − 2e4iak2k1k2 − k2 2 + e4iak2k2 2�<br />

41


1<br />

0.8<br />

0.6<br />

0.4<br />

0.2<br />

R,T<br />

10 20 30 40<br />

En HhartreeL<br />

Figure 2.7: Transmission (blue) and Reflection (red) coefficients for an electron scattering over<br />

a square well (V = −40 and a = 1 ).<br />

The R and T coefficients are related to the rations <strong>of</strong> the reflected and transimitted flux to<br />

the incoming flux. The current operator is given by<br />

Inserting the wavefunctions above yields:<br />

j(x) = ¯h<br />

2mi (ψ∗ ∇ψ − ψ∇ψ ∗ ) (2.94)<br />

jin = ¯hk1<br />

m<br />

2 ¯hk1R<br />

jref = −<br />

m<br />

2 ¯hk1T<br />

jtrans =<br />

m<br />

Thus, R 2 = −jref/jin and T 2 = jtrans/jin. In Fig. 2.7 we show the transmitted and reflection<br />

coefficients for an electron passing over a well <strong>of</strong> depth V = −40, a = 1 as a function <strong>of</strong> incident<br />

energy, E.<br />

Notice that the transmission and reflection coefficients under go a series oscillations as the<br />

incident energy is increased. These are due to resonance states which lie in the continuum. The<br />

condition for these states is such that an integer number <strong>of</strong> de Broglie wavelength <strong>of</strong> the wave in<br />

the well matches the total length <strong>of</strong> the well.<br />

λ/2 = na<br />

Fig. 2.8,show the transmission coefficient as a function <strong>of</strong> both incident energy and the well<br />

depth and (or height) over a wide range indicating that resonances can occur for both wells and<br />

bumps. Figures 2.9 show various scattering wavefunctions for on an <strong>of</strong>f-resonance cases. Lastly,<br />

Fig. ?? shows an Argand plot <strong>of</strong> both complex components <strong>of</strong> ψ.<br />

42


1<br />

0.95<br />

T<br />

0.9<br />

0.85<br />

10<br />

En<br />

20<br />

30<br />

-10<br />

40<br />

Figure 2.8: Transmission Coefficient for particle passing over a bump. Here we have plotted T as<br />

a function <strong>of</strong> V and incident energy En. The oscillations correspond to resonance states which<br />

occur as the particle passes over the well (for V < 0) or bump V > 0.<br />

2.3.4 Application: <strong>Quantum</strong> Dots<br />

One <strong>of</strong> the most active areas <strong>of</strong> research in s<strong>of</strong>t condensed matter is that <strong>of</strong> designing physical<br />

systems which can confine a quantum state in some controllable way. The idea <strong>of</strong> engineering a<br />

quantum state is extremely appealing and has numerous technological applications from small<br />

logic gates in computers to optically active materials for biomedical applications. The basic<br />

physics <strong>of</strong> these materials is relatively simple and we can use the basic ideas presented in this<br />

chapter. The basic idea is to layer a series <strong>of</strong> materials such that electrons can be trapped in a<br />

geometrically confined region. This can be accomplished by insulator-metal-insulator layers and<br />

etching, creating disclinations in semiconductors, growing semi-conductor or metal clusters, and<br />

so on. A quantum dot can even be a defect site.<br />

We will assume through out that our quantum well contains a single electron so that we can<br />

treat the system as simply as possible. For a square or cubic quantum well, energy levels are<br />

simply those <strong>of</strong> an n-dimensional particle in a box. For example for a three dimensional system,<br />

Enx,ny,nz = ¯h2 π2 ⎛<br />

⎝<br />

2m<br />

� �<br />

nx<br />

2<br />

Lx<br />

+<br />

-5<br />

� �2 ny<br />

Ly<br />

+<br />

0<br />

� nz<br />

Lz<br />

5<br />

V<br />

10<br />

⎞<br />

�2 ⎠ (2.95)<br />

where Lx, Ly, and Lz are the lengths <strong>of</strong> the box and m is the mass <strong>of</strong> an electron.<br />

The density <strong>of</strong> states is the number <strong>of</strong> energy levels per unit energy. If we take the box to be<br />

43


1.5<br />

1<br />

0.5<br />

-10 -5 5 10<br />

-0.5<br />

-1<br />

-1.5<br />

1<br />

0.5<br />

-10 -5 5 10<br />

-0.5<br />

-1<br />

Figure 2.9: Scattering waves for particle passing over a well. In the top graphic, the particle is<br />

partially reflected from the well (V < 0) and in the bottom graphic, the particle passes over the<br />

well with a slightly different energy than above, this time with little reflection.<br />

a cube Lx = Ly = Lz we can relate n to a radius <strong>of</strong> a sphere and write the density <strong>of</strong> states as<br />

ρ(n) = 4π 2 2 dn<br />

n<br />

dE = 4π2n 2<br />

Thus, for a 3D cube, the density <strong>of</strong> states is<br />

�<br />

2 4mL<br />

ρ(n) =<br />

π¯h 2<br />

�<br />

n<br />

� �−1 dE<br />

.<br />

dn<br />

i.e. for a three dimensional cube, the density <strong>of</strong> states increases as n and hence as E 1/2 .<br />

Note that the scaling <strong>of</strong> the density <strong>of</strong> states with energy depends strongly upon the dimen-<br />

44


Im@yD<br />

4<br />

2<br />

0<br />

-2<br />

-4<br />

-10<br />

x<br />

0<br />

10<br />

-5<br />

0<br />

Figure 2.10: Argand plot <strong>of</strong> a scattering wavefunction passing over a well. (Same parameters as<br />

in the top figure in Fig. 2.9).<br />

5<br />

Re@yD<br />

sionality <strong>of</strong> the system. For example in one dimension,<br />

and in two dimensions<br />

ρ(n) = 2mL2<br />

¯h 2 π 2<br />

ρ(n) = const.<br />

The reason for this lies in the way the volume element for linear, circular, and spherical integration<br />

scales with radius n. Thus, measuring the density <strong>of</strong> states tells us not only the size <strong>of</strong> the system,<br />

but also its dimensionality.<br />

We can generalize the results here by realizing that the volume <strong>of</strong> a d dimensional sphere in<br />

k space is given by<br />

Vd = kd π d/2<br />

Γ(1 + d/2)<br />

where Γ(x) is the gamma-function. The total number <strong>of</strong> states per unit volume in a d-dimensional<br />

space is then<br />

nk = 2 1<br />

Vd<br />

2π2 and the density is then the number <strong>of</strong> states per unit energy. The relation between energy and<br />

k is<br />

Ek = ¯h2<br />

2m k2 .<br />

i.e.<br />

√<br />

2Ekm<br />

k =<br />

¯h<br />

45<br />

1<br />

n


3<br />

2.5<br />

2<br />

1.5<br />

1<br />

0.5<br />

DOS<br />

which gives<br />

0.2 0.4 0.6 0.8 1<br />

energy HauL<br />

Figure 2.11: Density <strong>of</strong> states for a 1-, 2- , and 3- dimensional space.<br />

d<br />

d<br />

−2+<br />

2−1+ 2 d π 2<br />

� √ �<br />

m ɛ<br />

d<br />

¯h<br />

ρd(E) =<br />

ɛ Γ(1 + d<br />

2 )<br />

A quantum well is typically constructed so that the system is confined in one dimension<br />

and unconfined in the other two. Thus, a quantum well will typically have discrete state only<br />

in the confined direction. The density <strong>of</strong> states for this system will be identical to that <strong>of</strong> the<br />

3-dimensional system at energies where the k vectors coincide. If we take the thickness to be s,<br />

then the density <strong>of</strong> states for the quantum well is<br />

ρ = L<br />

s ρ2(E)<br />

�<br />

L ρ3(E)<br />

�<br />

Lρ2(E)/s<br />

where ⌊x⌋ is the ”floor” function which means take the largest integer less than x. This is plotted<br />

in Fig. 2.12 and the stair-step DOS is indicative <strong>of</strong> the embedded confined structure.<br />

Next, we consider a quantum wire <strong>of</strong> thickness s along each <strong>of</strong> its 2 confined directions. The<br />

DOS along the unconfined direction is one-dimensional. As above, the total DOS will be identical<br />

46


30<br />

20<br />

10<br />

DOS <strong>Quantum</strong> well vs. 3d body<br />

e HauL<br />

0.005 0.01 0.015 0.02 0.025 0.03<br />

120<br />

100<br />

DOS <strong>Quantum</strong> wire vs. 3d body<br />

80<br />

60<br />

40<br />

20<br />

e HauL<br />

0.05 0.1 0.15 0.2 0.25 0.3<br />

Figure 2.12: Density <strong>of</strong> states for a quantum well and quantum wire compared to a 3d space.<br />

Here L = 5 and s = 2 for comparison.<br />

to the 3D case when the wavevectors coincide. Increasing the radius <strong>of</strong> the wire eventually leads<br />

to the case where the steps decrease and merge into the 3D curve.<br />

� �2<br />

�<br />

2 L L ρ2(E)<br />

ρ = ρ1(E)<br />

s L2 �<br />

ρ2(E)/s<br />

For a spherical dot, we consider the case in which the radius <strong>of</strong> the quantum dot is small<br />

enough to support discrete rather than continuous energy levels. In a later chapter, we will derive<br />

this result in more detail, for now we consider just the results. First, an electron in a spherical<br />

dot obeys the Schrödinger equation:<br />

where ∇ 2 is the Laplacian operator in spherical coordinates<br />

∇ 2 = 1<br />

r<br />

∂ 2<br />

r +<br />

∂r2 1<br />

− ¯h2<br />

2m ∇2 ψ = Eψ (2.96)<br />

r 2 sin θ<br />

∂ ∂<br />

sin θ<br />

∂θ ∂θ +<br />

1<br />

r 2 sin 2 θ<br />

∂2 .<br />

∂φ2 The solution <strong>of</strong> the Schrodinger equation is subject to the boundary condition that for r ≥ R,<br />

ψ(r) = 0, where R is the radius <strong>of</strong> the sphere and are given in terms <strong>of</strong> the spherical Bessel<br />

function, jl(r) and spherical harmonic functions, Ylm.<br />

with energy<br />

ψnlm = 21/2<br />

R3/2 jl(αr/R)<br />

jl+1(α) Ylm(Ω), (2.97)<br />

E = ¯h2 α<br />

2m<br />

2<br />

R2 (2.98)<br />

Note that the spherical Bessel functions (<strong>of</strong> the first kind) are related to the Bessel functions via,<br />

jl(x) =<br />

� π<br />

2x Jl+1/2(x). (2.99)<br />

47


1<br />

0.8<br />

0.6<br />

0.4<br />

0.2<br />

-0.2<br />

j lHxL<br />

5 10 15 20 x<br />

Figure 2.13: Spherical Bessel functions, j0, j1, and j1 (red, blue, green)<br />

The first few <strong>of</strong> these are<br />

sin x<br />

j0(x) = (2.100)<br />

x<br />

sin x cos x<br />

j1(x) = − (2.101)<br />

x2 x<br />

� �<br />

3 1<br />

j2(x) = − sin x −<br />

x3 x<br />

3<br />

cos x (2.102)<br />

x2 jn(x) = (−1) n x n<br />

� �n 1 d<br />

j0(x) (2.103)<br />

x dx<br />

where the last line provides a way to generate jn from j0.<br />

The α’s appearing in the wavefunction and in the energy expression are determined by the<br />

boundary condition that ψ(R) = 0. Thus, for the lowest energy state we require<br />

i.e. α = π. For the next state (l = 1),<br />

j1(α) =<br />

j0(α) = 0, (2.104)<br />

sin α cos α<br />

−<br />

α2 α<br />

= 0. (2.105)<br />

This can be solved to give α = 4.4934. These correspond to where the spherical Bessel functions<br />

pass through zero. The first 6 <strong>of</strong> these are 3.14159, 4.49341, 5.76346, 6.98793, 8.18256, 9.35581.<br />

These correspond to where the first zeros occur and give the condition for the radial quantization,<br />

n = 1 with angular momentum l = 0, 1, 2, 3, 4, 5. There are more zeros, and these correspond to<br />

the case where n > 1.<br />

In the next set <strong>of</strong> figures (Fig. 2.14), we look at the radial wavefunctions for an electron in<br />

a 0.5˚Aquantum dot. First, the case where n = 1, l = 0 and n = 0, l = 1. In both cases, the<br />

wavefunctions vanish at the radius <strong>of</strong> the dot. The radial probability distribution function (PDF)<br />

is given by P = r 2 |ψnl(r)| 2 . Note that increasing the angular momentum l from 0 to 1 causes<br />

the electron’s most probable position to shift outwards. This is due to the centrifugal force due<br />

to the angular motion <strong>of</strong> the electron. For the n, l = (2, 0) and (2, 1) states, we have 1 node in<br />

the system and two peaks in the PDF functions.<br />

48


12<br />

10<br />

5<br />

-5<br />

-10<br />

-15<br />

-20<br />

-25<br />

8<br />

6<br />

4<br />

2<br />

y<br />

y<br />

0.1 0.2 0.3 0.4 0.5 r<br />

0.1 0.2 0.3 0.4 0.5 r<br />

4<br />

3<br />

2<br />

1<br />

P<br />

4<br />

3<br />

2<br />

1<br />

P<br />

0.1 0.2 0.3 0.4 0.5 r<br />

0.1 0.2 0.3 0.4 0.5 r<br />

Figure 2.14: Radial wavefuncitons (left column) and corresponding PDFs (right column) for an<br />

electron in a R = 0.5˚Aquantum dot. The upper two correspond to (n, l) = (1, 0) (solid) and<br />

(n, l) = (1, 1) (dashed) while the lower correspond to (n, l) = (2, 0) (solid) and (n, l) = (2, 1)<br />

(dashed) .<br />

2.4 Tunneling and transmission in a 1D chain<br />

In this example, we are going to generalize the ideas presented here and look at what happens if<br />

we discretize the space in which a particle can move. This happens physically when we consider<br />

what happens when a particle (eg. an electron) can hop from one site to another. If an electron<br />

is on a given site, it has a certain energy ε to be there and it takes energy β to move the electron<br />

from the site to its neighboring site. We can write the Schrödinger equation for this system as<br />

ujε + βuj+1 + βuj−1 = Euj.<br />

for the case where the energy depends upon where the electron is located. If the chain is infinite,<br />

we can write uj = T e ikdj and find that the energy band goes as E = ε + 2β cos(kd) where k is<br />

now the momentum <strong>of</strong> the electron.<br />

2.5 Summary<br />

We’ve covered a lot <strong>of</strong> ground. We now have enough tools at hand to begin to study some physical<br />

systems. The traditional approach to studying quantum mechanics is to progressively a series<br />

<strong>of</strong> differential equations related to physical systems (harmonic oscillators, angular momentum,<br />

49


hydrogen atom, etc...). We will return to those models in a week or so. Next week, we’re going to<br />

look at 2 and 3 level systems using both time dependent and time-independent methods. We’ll<br />

develop a perturbative approach for computing the transition amplitude between states. We will<br />

also look at the decay <strong>of</strong> a state when its coupled to a continuum. These are useful models for<br />

a wide variety <strong>of</strong> phenomena. After this, we will move on to the harmonic oscillator.<br />

2.6 Problems and Exercises<br />

Exercise 2.1 1. Derive the expression for<br />

where ho is the free particle Hamiltonian,<br />

Go(x, x ′ ) = 〈x| exp(−ihot/¯h)|x ′ 〉 (2.106)<br />

ho = − ¯h2 ∂<br />

2m<br />

2<br />

∂x2 2. Show that Go is a solution <strong>of</strong> the free particle Schrödinger Equation<br />

(2.107)<br />

i¯h∂tGo(t) = hoGo(t). (2.108)<br />

Exercise 2.2 Show that the normalization <strong>of</strong> a wavefunction is independent <strong>of</strong> time.<br />

Solution:<br />

i∂t〈ψ(t)|ψ(t)〉 = (i〈 ˙ ψ(t)|)(|ψ(t)〉) + (〈ψ(t)|)(i| ˙ ψ(t)〉) (2.109)<br />

= −〈ψ(t)| ˆ H † |ψ(t)〉 + 〈ψ(t)| ˆ H|ψ(t)〉 (2.110)<br />

= −〈ψ(t)| ˆ H|ψ(t)〉 + 〈ψ(t)| ˆ H|ψ(t)〉 = 0 (2.111)<br />

Exercise 2.3 Compute the bound state solutions (E < 0) for a square well <strong>of</strong> depth Vo where<br />

�<br />

−Vo<br />

V (x) =<br />

0<br />

−a/2 ≤ x ≤ a/2<br />

otherwise<br />

(2.112)<br />

1. How many energy levels are supported by a well <strong>of</strong> width a.<br />

2. Show that a very narrow well can support only 1 bound state, and that this state is an even<br />

function <strong>of</strong> x.<br />

3. Show that the energy <strong>of</strong> the lowest bound state is<br />

4. Show that as<br />

ρ =<br />

E ≈<br />

mV 2<br />

o a 2<br />

2¯h 2<br />

the probability <strong>of</strong> finding the particle inside the well vanishes.<br />

�<br />

(2.113)<br />

− 2mE<br />

2 → 0 (2.114)<br />

¯h<br />

50


Exercise 2.4 Consider a particle with the potential<br />

V (x) =<br />

⎧<br />

⎪⎨<br />

⎪⎩<br />

0 for x > a<br />

−Vo for 0 ≤ x ≤ a<br />

∞ for x < 0<br />

(2.115)<br />

1. Let φ(x) be a stationary state. Show that φ(x) can be extended to give an odd wavefunction<br />

corresponding to a stationary state <strong>of</strong> the symmetric well <strong>of</strong> width 2a (i.e the one studied<br />

above) and depth Vo.<br />

2. Discuss with respect to a and Vo the number <strong>of</strong> bound states and argue that there is always<br />

at least one such state.<br />

3. Now turn your attention toward the E > 0 states <strong>of</strong> the well. Show that the transmission<br />

<strong>of</strong> the particle into the well region vanishes as E → 0 and that the wavefunction is perfectly<br />

reflected <strong>of</strong>f the sudden change in potential at x = a.<br />

Exercise 2.5 Which <strong>of</strong> the following are eigenfunctions <strong>of</strong> the kinetic energy operator<br />

e x , x 2 , x n ,3 cos(2x), sin(x) + cos(x), e −ikx ,<br />

.<br />

Solution Going in order:<br />

1. e x<br />

2. x 2<br />

3. x n<br />

4. 3 cos(2x)<br />

5. sin(x) + cos(x)<br />

6. e −ikx<br />

ˆT = − ¯h2 ∂<br />

2m<br />

2<br />

∂x2 f(x − x ′ � ∞<br />

) = dke −ik(x−x′ ) −ik<br />

e 2 /(2m)<br />

−∞<br />

(2.116)<br />

(2.117)<br />

Exercise 2.6 Which <strong>of</strong> the following would be acceptable one dimensional wavefunctions for a<br />

bound particle (upon normalization): f(x) = e−x , f(x) = e−x2, f(x) = xe−x2, and<br />

Solution In order:<br />

f(x) =<br />

�<br />

e−x2 2e−x2 51<br />

x ≥ 0<br />

x < 0<br />

(2.118)


1. f(x) = e −x<br />

2. f(x) = e −x2<br />

3. f(x) = xe −x2<br />

4.<br />

f(x) =<br />

�<br />

e−x2 2e−x2 x ≥ 0<br />

x < 0<br />

Exercise 2.7 For a one dimensional problem, consider a particle with wavefunction<br />

ψ(x) = N exp(ipox/¯h)<br />

√ x 2 + a 2<br />

where a and po are real constants and N the normalization.<br />

1. Determine N so that ψ(x) is normalized.<br />

� ∞<br />

−∞<br />

Thus ψ(x) is normalized when<br />

dx|ψ(x)| 2 = N 2<br />

� ∞<br />

N =<br />

2 π<br />

= N<br />

a<br />

� a<br />

π<br />

−∞<br />

dx<br />

1<br />

x 2 + a 2<br />

(2.119)<br />

(2.120)<br />

(2.121)<br />

(2.122)<br />

(2.123)<br />

2. The position <strong>of</strong> the particle is measured. What is the probability <strong>of</strong> finding a result between<br />

−a/ √ 3 and +a/ √ 3?<br />

� √<br />

a +a/ 3<br />

π −a/ √ dx|ψ(x)|<br />

3<br />

2 =<br />

� +a/ √ 3<br />

−a/ √ 3<br />

dx<br />

= 1<br />

π tan−1 (x/a)<br />

= 1<br />

3<br />

1<br />

x2 + a2 �<br />

�+a/<br />

�<br />

�<br />

√ 3<br />

−a/ √ 3<br />

3. Compute the mean value <strong>of</strong> a particle which has ψ(x) as its wavefunction.<br />

(2.124)<br />

(2.125)<br />

(2.126)<br />

〈x〉 = a<br />

� ∞ x<br />

dx<br />

π −∞ x2 + a2 (2.127)<br />

= 0 (2.128)<br />

52


Exercise 2.8 Consider the Hamiltonian <strong>of</strong> a particle in a 1 dimensional well given by<br />

H = 1<br />

2m ˆp2 + ˆx 2<br />

where ˆx and ˆp are position and momentum operators. Let |φn〉 be a solution <strong>of</strong><br />

for n = 0, 1, 2, · · ·. Show that<br />

(2.129)<br />

H|φn〉 = En|φn〉 (2.130)<br />

〈φn|ˆp|φm〉 = αnm〈φn|ˆx|φm〉 (2.131)<br />

where αnm is a coefficient depending upon En − Em. Compute αnm. (Hint: you will need to use<br />

the commutation relations <strong>of</strong> [ˆx, H] and [ˆp, H] to get this). Finally, from all this, deduce that<br />

�<br />

m<br />

(En − Em) 2 |φn|ˆx|φm〉| 2 = ¯h2<br />

2m 〈φn|ˆp 2 |φn〉 (2.132)<br />

Exercise 2.9 The state space <strong>of</strong> a certain physical system is three-dimensional. Let |u1〉, |u2〉,<br />

and |u3〉 be an orthonormal basis <strong>of</strong> the space in which the kets |ψ1〉 and |ψ2〉 are defined by<br />

1. Are the states normalized?<br />

|ψ1〉 = 1<br />

√ 2 |u1〉 + i<br />

2 |u2〉 + 1<br />

2 |u3〉 (2.133)<br />

|ψ2〉 = 1<br />

√ 3 |u1〉 + i<br />

√ 3 |u3〉 (2.134)<br />

2. Calculate the matrices, ρ1 and ρ2 representing in the {|ui〉〉 basis, the projection operators<br />

onto |ψ1〉 and |ψ2〉. Verify that these matrices are Hermitian.<br />

Exercise 2.10 Let ψ(r) = ψ(x, y, z) be the normalized wavefunction <strong>of</strong> a particle. Express in<br />

terms <strong>of</strong> ψ(r):<br />

1. A measurement along the x-axis to yield a result between x1 and x2.<br />

2. A measurement <strong>of</strong> momentum component px to yield a result between p1 and p2.<br />

3. Simultaneous measurements <strong>of</strong> x and pz to yield x1 ≤ x ≤ x2 and pz > 0.<br />

4. Simultaneous measurements <strong>of</strong> px, py, and pz, to yield<br />

p1 ≤ px ≤ p2<br />

p3 ≤ py ≤ p4<br />

p5 ≤ pz ≤ p6<br />

(2.135)<br />

(2.136)<br />

(2.137)<br />

(2.138)<br />

(2.139)<br />

Show that this result is equal to the result <strong>of</strong> part 2 when p3, p5 → −∞ and p4, p6 → +∞.<br />

53


Exercise 2.11 Consider a particle <strong>of</strong> mass m whose potential energy is<br />

V (x) = −α(δ(x + l/2) + δ(x − l/2))<br />

1. Calculate the bound states <strong>of</strong> the particle, setting<br />

Show that the possible energies are given by<br />

E = − ¯h2 ρ 2<br />

2m .<br />

e −ρl �<br />

= ± 1 − 2ρ<br />

�<br />

µ<br />

where µ = 2mα/¯h 2 . Give a graphic solution <strong>of</strong> this equation.<br />

(a) The Ground State. Show that the ground state is even about the origin and that it’s<br />

energy, Es is less than the bound state <strong>of</strong> a particle in a single δ-function potential,<br />

−EL. Interpret this physically. Plot the corresponding wavefunction.<br />

(b) Excited State. Show that when l is greater than some value (which you need to determine),<br />

there exists an odd excited state <strong>of</strong> energy EA with energy greater than −EL.<br />

Determine and plot the corresponding wavefunction.<br />

(c) Explain how the preceeding calculations enable us to construct a model for an ionized<br />

diatomic molecule, eg. H + 2 , whose nuclei are separated by l. Plot the energies <strong>of</strong> the<br />

two states as functions <strong>of</strong> l, what happens as l → ∞ and l → 0?<br />

(d) If we take Coulombic repulsion <strong>of</strong> the nuclei into account, what is the total energy<br />

<strong>of</strong> the system? Show that a curve which gives the variation with respect to l <strong>of</strong> the<br />

energies thus obtained enables us to predict in certain cases the existence <strong>of</strong> bound<br />

states <strong>of</strong> H + 2 and to determine the equilibrium bond length.<br />

2. Calculate the reflection and transmission coefficients for this system. Plot R and T as<br />

functions <strong>of</strong> l. Show that resonances occur when l is an integer multiple <strong>of</strong> the de Broglie<br />

wavelength <strong>of</strong> the particle. Why?<br />

54


Chapter 3<br />

Semi-Classical <strong>Quantum</strong> <strong>Mechanics</strong><br />

Good actions ennoble us, and we are the sons <strong>of</strong> our own deeds.<br />

–Miguel de Cervantes<br />

The use <strong>of</strong> classical mechanical analogs for quantum behavour holds a long and proud tradition in<br />

the development and application <strong>of</strong> quantum theory. In Bohr’s original formulation <strong>of</strong> quantum<br />

mechanics to explain the spectra <strong>of</strong> the hydrogen atom, Bohr used purely classical mechanical<br />

notions <strong>of</strong> angular momentum and rotation for the basic theory and imposed a quantization<br />

condition that the angular momentum should come in integer multiples <strong>of</strong> ¯h. Bohr worked under<br />

the assumption that at some point the laws <strong>of</strong> quantum mechanics which govern atoms and<br />

molecules should correspond to the classical mechanical laws <strong>of</strong> ordinary objects like rocks and<br />

stones. Bohr’s Principle <strong>of</strong> Correspondence states that quantum mechanics was not completely<br />

separate from classical mechanics; rather, it incorporates classical theory.<br />

From a computational viewpoint, this is an extremely powerful notion since performing a<br />

classical trajectory calculation (even running 1000’s <strong>of</strong> them) is simpler than a single quantum<br />

calculation <strong>of</strong> a similar dimension. Consequently, the development <strong>of</strong> semi-classical methods has<br />

and remains an important part <strong>of</strong> the development and untilization <strong>of</strong> quantum theory. In fact<br />

even in the most recent issues <strong>of</strong> the Journal <strong>of</strong> Chemical Physics, Phys. Rev. Lett, and other<br />

leading physics and chemical physics journals, one finds new developments and applications <strong>of</strong><br />

this very old idea.<br />

In this chapter we will explore this idea in some detail. The field <strong>of</strong> semi-classical mechanics<br />

is vast and I would recommend the following for more information:<br />

1. Chaos in Classical and <strong>Quantum</strong> <strong>Mechanics</strong>, Martin Gutzwiller (Springer-Verlag, 1990).<br />

Chaos in quantum mechanics is a touchy subject and really has no clear-cut definition that<br />

anyone seems to agree upon. Gutzwiller is one <strong>of</strong> the key figures in sorting all this out. This<br />

is very nice and not too technical monograph on quantum and classical correspondence.<br />

2. Semiclassical Physics, M. Brack and R. Bhaduri (Addison-Wesley, 1997). Very interesting<br />

book, mostly focusing upon many-body applications and Thomas-Fermi approximations.<br />

3. Computer Simulations <strong>of</strong> Liquids, M. P. Allen and D. J. Tildesley (Oxford, 1994). This<br />

book mostly focus upon classical MD methods, but has a nice chapter on quantum methods<br />

which were state <strong>of</strong> the art in 1994. Methods come and methods go.<br />

There are many others, <strong>of</strong> course. These are just the ones on my bookshelf.<br />

55


3.1 Bohr-Sommerfield quantization<br />

Let’s first review Bohr’s original derivation <strong>of</strong> the hydrogen atom. We will go through this a bit<br />

differently than Bohr since we already know part <strong>of</strong> the answer. In the chapter on the Hydrogen<br />

atom we derived the energy levels in terms <strong>of</strong> the principle quantum number, n.<br />

En = − me4<br />

2¯h 2<br />

In Bohr’s correspondence principle, the quantum energy must equal the classical energy. So<br />

for an electron moving about a proton, that energy is inversely proportional to the distance <strong>of</strong><br />

separation. So, we can write<br />

− me4<br />

2¯h 2<br />

1<br />

n 2<br />

1 e2<br />

= −<br />

n2 2r<br />

Now we need to figure out how angular momentum gets pulled into this. For an orbiting body<br />

the centrifugal force which pulls the body outward is counterbalenced by the inward tugs <strong>of</strong> the<br />

centripetal force coming from the attractive Coulomb potential. Thus,<br />

(3.1)<br />

(3.2)<br />

mrω 2 = e2<br />

, (3.3)<br />

r2 where ω is the angular frequency <strong>of</strong> the rotation. Rearranging this a bit, we can plug this into<br />

the RHS <strong>of</strong> Eq. 3.2 and write<br />

− me4<br />

2¯h 2<br />

1<br />

n2 = −mr3 ω2 2r<br />

The numerator now looks amost like the classical definition <strong>of</strong> angular momentum: L = mr 2 ω.<br />

So we can write the last equation as<br />

Solving for L 2 :<br />

− me4<br />

2¯h 2<br />

(3.4)<br />

1 L2<br />

= − . (3.5)<br />

n2 2mr2 L 2 = me4<br />

2¯h 2<br />

2mr2 n2 . (3.6)<br />

Now, we need to pull in another one <strong>of</strong> Bohr’s results for the orbital radius <strong>of</strong> the H-atom:<br />

Plug this into Eq.3.6 and after the dust settles, we find<br />

r = ¯h2<br />

me 2 n2 . (3.7)<br />

L = ¯hn. (3.8)<br />

But, why should electrons be confined to circular orbits? Eq. 3.8 should be applicable to<br />

any closed path the electron should choose to take. If the quantization condition only holds<br />

56


for circular orbits, then the theory itself is in deep trouble. At least that’s what Sommerfield<br />

thought.<br />

The numerical units <strong>of</strong> ¯h are energy times time. That is the unit <strong>of</strong> action in classical<br />

mechanics. In classical mechanics, the action <strong>of</strong> a mechanical system is given by the integral <strong>of</strong><br />

the classical momentum along a classical path:<br />

S =<br />

� x 2<br />

x1<br />

pdx (3.9)<br />

For an orbit, the initial point and the final point must coincide, x1 = x2, so the action integral<br />

must describe some the area circumscribed by a closed loop on the p−x plane called phase-space.<br />

�<br />

S = pdx (3.10)<br />

So, Bohr and Sommerfield’s idea was that the circumscribed area in phase-space was quantized<br />

as well.<br />

As a check, let us consider the harmonic oscillator. The classical energy is given by<br />

E(p, q) = p2 k<br />

+<br />

2m 2 q2 .<br />

This is the equation for an ellipse in phase space since we can re-arrange this to read<br />

1 =<br />

p 2<br />

2mE<br />

= p2 q2<br />

+<br />

a2 b2 + k<br />

2E q2<br />

(3.11)<br />

where a = √ �<br />

2mE and b = 2E/k describe the major and minor axes <strong>of</strong> the ellipse. The area <strong>of</strong><br />

an ellipse is A = πab, so the area circumscribed by a classical trajectory with energy E is<br />

�<br />

S(E) = 2Eπ m/k (3.12)<br />

Since<br />

�<br />

k/m = ω, S = 2πE/ω = E/ν. Finally, since E/ν must be an integer multiple <strong>of</strong> h, the<br />

Bohr-Sommerfield condition for quantization becomes<br />

�<br />

pdx = nh (3.13)<br />

�<br />

where p is the classical momentum for a path <strong>of</strong> energy E, p = 2m(V (x) − E. Taking this a<br />

bit farther, the de Broglie wavelength is p/h, so the Bohr-Sommerfield rule basically states that<br />

stationary energies correspond to classical paths for which there are an integer number <strong>of</strong> de<br />

Broglie wavelengths.<br />

Now, perhaps you can see where the problem with quantum chaos. In classical chaos, chaotic<br />

trajectories never return to their exact staring point in phase-space. They may come close, but<br />

there are no closed orbits. For 1D systems, this is does not occur since the trajectories are the<br />

contours <strong>of</strong> the energy function. For higher dimensions, the dimensionality <strong>of</strong> the system makes<br />

it possible to have extremely complex trajectories which never return to their starting point.<br />

Exercise 3.1 Apply the Bohr-Sommerfield proceedure to determine the stationary energies for<br />

a particle in a box <strong>of</strong> length l.<br />

57


3.2 The WKB Approximation<br />

The original Bohr-Sommerfield idea can be imporoved upon considerably to produce an asymptotic<br />

(¯h → 0) approximation to the Schrödinger wave function. The idea was put forward at about<br />

the same time by three different theoreticians, Brillouin (in Belgium), Kramers (in Netherlands),<br />

and Wentzel (in Germany). Depending upn your point <strong>of</strong> origin, this method is the WKB (US<br />

& Germany), BWK (France, Belgium), JWKB (UK), you get the idea. The original references<br />

are<br />

1. “La mécanique odularatoire de Schrödinger; une méthode générale de résolution par approximations<br />

successives”, L. Brillouin, Comptes rendus (Paris). 183, 24 (1926).<br />

2. “Wellenmechanik und halbzahlige Quantisierung”, H. A. Kramers, Zeitschrift für Physik<br />

39, 828 (1926).<br />

3. “Eine Verallgemeinerung der Quantenbedingungen für die Zwecke der Wellenmechanik”,<br />

Zeitschrift für Physik 38, 518 (1926).<br />

We will first go through how one can use the approach to determine the eigenvalues <strong>of</strong> the<br />

Schrödinger equation via semi-classical methods, then show how one can approximate the actual<br />

wavefunctions themselves.<br />

3.2.1 Asymptotic expansion for eigenvalue spectrum<br />

The WKB proceedure is initiated by writing the solution to the Schödinger equation<br />

as<br />

ψ ′′ + 2m<br />

2 (E − V (x))ψ = 0<br />

¯h<br />

� �<br />

i<br />

ψ(x) = exp<br />

¯h<br />

�<br />

χdx . (3.14)<br />

We will soon discover that χ is the classical momentum <strong>of</strong> the system, but for now, let’s consider<br />

it to be a function <strong>of</strong> the energy <strong>of</strong> the system. Substituting into the Schrodinger equation<br />

produces a new differential equation for χ<br />

If we take ¯h → 0, it follows then that<br />

¯h dχ<br />

i dx = 2m(E − V ) − χ2 . (3.15)<br />

χ = χo =<br />

�<br />

2m(E − V ) = |p| (3.16)<br />

which is the magnitude <strong>of</strong> the classical momentum <strong>of</strong> a particle. So, if we assume that this is<br />

simply the leading order term in a series expansion in ¯h we would have<br />

χ = χo + ¯h<br />

i χ1 +<br />

58<br />

� �2 ¯h<br />

χ2 . . . (3.17)<br />

i


Substituting Eq. 3.17 into<br />

χ = ¯h<br />

i<br />

1<br />

ψ<br />

∂ψ<br />

x<br />

(3.18)<br />

and equating to zero coefficients with different powers <strong>of</strong> ¯h, one obtains equations which determine<br />

the χn corrections in succession:<br />

for n = 1, 2, 3 . . .. For example,<br />

and so forth.<br />

χ2 = − χ2 1 + χ ′ 1<br />

2χo<br />

= − 1<br />

�<br />

2χo<br />

= −<br />

d<br />

dx χn−1<br />

n�<br />

= − χn−mχm<br />

m=0<br />

χ1 = − 1<br />

2<br />

χ ′ o<br />

χo<br />

V ′2<br />

+<br />

16(E − V ) 2<br />

= 1 V<br />

4<br />

′<br />

E − V<br />

V ′2<br />

+<br />

4(E − V ) 2<br />

V ′′<br />

�<br />

.<br />

4(E − V )<br />

5V ′2<br />

32(2m) 1/2 V<br />

−<br />

(E − V ) 5/2 ′′<br />

8(2m) 1/2 (E − V ) 3/2<br />

Exercise 3.2 Verify Eq. 3.19 and derive the first order correction in Eq.3.20.<br />

(3.19)<br />

(3.20)<br />

(3.21)<br />

Now, to use these equations to determine the spectrum, we replace x everywhere by a complex<br />

coordinate z and suppose that V (z) is a regular and analytic function <strong>of</strong> z in any physically<br />

relevant region. Consequently, we can then say that ψ(z) is an analytic function <strong>of</strong> z. So, we<br />

can write the phase integral as<br />

n = 1<br />

�<br />

h<br />

= 1<br />

2πi<br />

C<br />

�<br />

χ(z)dz<br />

C<br />

ψ ′ n(z)<br />

dz (3.22)<br />

ψn(z)<br />

where ψn is the nth discrete stationary solution to the Schrödinger equation and C is some<br />

contour <strong>of</strong> integration on the z plane. If there is a discrete spectrum, we know that the number<br />

<strong>of</strong> zeros, n, in the wavefunction is related to the quantum number corresponding to the n + 1<br />

energy level. So if ψ has no real zeros, this is the ground state wavefunction with energy Eo, one<br />

real zero corresponds to energy level E1 and so forth.<br />

Suppose the contour <strong>of</strong> integration, C is taken such that it include only these zeros and no<br />

others, then we can write<br />

n = 1<br />

�<br />

χodz +<br />

¯h C<br />

1<br />

� �<br />

−¯h χ2dz + . . . (3.23)<br />

2πi c C<br />

59


Each <strong>of</strong> these terms involves E − V in the denominator. At the classical turning points where<br />

V (z) = E, we have poles and we can use the residue theorem to evaluate the integrals. For<br />

example, χ1 has a pole at each turnining point with residue −1/4 at each point, hence,<br />

1<br />

2πi<br />

The next term we evaluate by integration by parts<br />

Hence, we can write<br />

Putting it all together<br />

�<br />

C<br />

�<br />

V ′′<br />

�<br />

dz = −3<br />

(E − V (z)) 3/2 2 C<br />

C<br />

χ2(z)dz =<br />

n + 1/2 = 1<br />

h<br />

−<br />

�<br />

C<br />

χ1dz = − 1<br />

. (3.24)<br />

2<br />

1<br />

32(2m) 1/2<br />

�<br />

C<br />

�<br />

c<br />

�<br />

2m(E − V (z))dz<br />

h<br />

128π 2 (2m) 1/2<br />

�<br />

c<br />

V ′2<br />

dz. (3.25)<br />

(E − V (z)) 5/2<br />

V ′2<br />

dz. (3.26)<br />

(E − V (z)) 5/2<br />

V ′2<br />

dz + . . . (3.27)<br />

(E − V (z)) 5/2<br />

Granted, the above analysis is pretty formal! But, what we have is something new. Notice that<br />

we have an extra 1/2 added here that we did not have in the original Bohr-Sommerfield (BS)<br />

theory. What we have is something even more general. The original BS idea came from the<br />

notion that energies and frequencies were related by integer multiples <strong>of</strong> h. But this is really<br />

only valid for transitions between states. If we go back and ask what happens at n = 0 in<br />

the Bohr-Sommerfield theory, this corresponds to a phase-space ellipse with major and minor<br />

axes both <strong>of</strong> length 0–which violates the Heisenberg Uncertainly rule. This new quantization<br />

condition forces the system to have some lowest energy state with phase-space area 1/2.<br />

Where did this extra 1/2 come from? It originates from the classical turning points where<br />

V (x) = E. Recall that for a 1D system bound by a potential, there are at least two such points.<br />

Each contributes a π/4 to the phase. We will see this more explicitly in the next section.<br />

3.2.2 WKB Wavefunction<br />

Going back to our original wavefunction in Eq. 3.14 and writing<br />

ψ = e iS/¯h<br />

where S is the integral <strong>of</strong> χ, we can derive equations for S:<br />

� �<br />

1 ∂S<br />

−<br />

2m ∂x<br />

i¯h<br />

2m<br />

∂2S + V (x) = E. (3.28)<br />

∂x2 Again, as above, one can seek a series expansion <strong>of</strong> S in powers <strong>of</strong> ¯h. The result is simply the<br />

integral <strong>of</strong> Eq. 3.17.<br />

S = So + ¯h<br />

i S1 + . . . (3.29)<br />

60


If we make the approximation that ¯h = 0 we have the classical Hamilton-Jacobi equation for the<br />

action, S. This, along with the definition <strong>of</strong> the momentum, p = dSo/dx = χo, allows us to make<br />

a very firm contact between quantum mechanics and the motion <strong>of</strong> a classical particle.<br />

Looking at Eq. 3.28, it is clear that the classical approximation is valid when the second term<br />

is very small compared to the first. i.e.<br />

¯h d<br />

dx<br />

� dS<br />

dx<br />

¯h S′′<br />

S ′2<br />

� � �2 dx<br />

dS<br />

≪<br />

≪<br />

1<br />

1<br />

¯h d 1<br />

dx p<br />

≪ 1 (3.30)<br />

where we equate dS/dx = p. Since p is related to the de Broglie wavelength <strong>of</strong> the particle<br />

λ = h/p , the same condition implies that<br />

� �<br />

�<br />

� 1 dλ�<br />

�<br />

� � ≪ 1. (3.31)<br />

�2π<br />

dx �<br />

Thus the semi-classical approximation is only valid when the wavelength <strong>of</strong> the particle as determined<br />

by λ(x) = h/p(x) varies slightly over distances on the order <strong>of</strong> the wavelength itself.<br />

Written another way by noting that the gradiant <strong>of</strong> the momentum is<br />

dp<br />

dx<br />

d �<br />

=<br />

dx<br />

Thus, we can write the classical condition as<br />

2m(E − V (x)) = − m<br />

p<br />

dV<br />

dx .<br />

m¯h|F |/p 3 ≪ 1 (3.32)<br />

Consequently, the semi-classical approximation can only be used in regions where the momentum<br />

is not too small. This is especially important near the classical turning points where p → 0. In<br />

classical mechanics, the particle rolls to a stop at the top <strong>of</strong> the potential hill. When this happens<br />

the de Broglie wavelength heads <strong>of</strong>f to infinity and is certainly not small!<br />

Exercise 3.3 Verify the force condition given by Eq. 3.32.<br />

Going back to the expansion for χ<br />

or equivalently for S1<br />

So,<br />

χ1 = − 1<br />

2<br />

χ ′ o<br />

χo<br />

= 1 V<br />

4<br />

′<br />

E − V<br />

S ′ 1 = − S′′ o p′<br />

= − ′<br />

2S<br />

2p<br />

S1(x) = − 1<br />

log p(x)<br />

2<br />

61<br />

(3.33)<br />

(3.34)


If we stick to regions where the semi-classical condition is met, then the wavefunction becomes<br />

ψ(x) ≈ C1<br />

� �<br />

i<br />

p(x)dx C2 i<br />

p(x)dx<br />

� e ¯h + � e− ¯h<br />

p(x) p(x)<br />

(3.35)<br />

The 1/ √ p prefactor has a remarkably simple interpretation. The probability <strong>of</strong> find the particle<br />

in some region between x and x+dx is given by |ψ| 2 so that the classical probability is essentially<br />

proportional to 1/p. So, the fast the particle is moving, the less likely it is to be found in some<br />

small region <strong>of</strong> space. Conversly, the slower a particle moves, the more likely it is to be found in<br />

that region. So the time spend in a small dx is inversely proportional to the momentum <strong>of</strong> the<br />

particle. We will return to this concept in a bit when we consider the idea <strong>of</strong> time in quantum<br />

mechanics.<br />

The C1 and C2 coefficients are yet to be determined. If we take x = a to be one classical<br />

turning point so that x > a corresponds to the classically inaccessible region where E < V (x),<br />

then the wavefunction in that region must be exponentially damped:<br />

ψ(x) ≈ C<br />

�<br />

|p| exp<br />

�<br />

− 1<br />

� x �<br />

|p(x)|dx<br />

¯h a<br />

To the left <strong>of</strong> x = a, we have a combination <strong>of</strong> incoming and reflected components:<br />

ψ(x) = C1<br />

� �<br />

i a �<br />

√p exp pdx +<br />

¯h x<br />

C2<br />

�<br />

√p exp − i<br />

� a �<br />

pdx<br />

¯h x<br />

3.2.3 Semi-classical Tunneling and Barrier Penetration<br />

(3.36)<br />

(3.37)<br />

Before solving the general problem <strong>of</strong> how to use this in an arbitrary well, let’s consider the case<br />

for tunneling through a potential barrier that has some bumpy top or corresponds to some simple<br />

potential. So, to the left <strong>of</strong> the barrier the wavefunction has incoming and reflected components:<br />

Inside we have<br />

ψB(x) =<br />

and to the right <strong>of</strong> the barrier:<br />

ψL(x) = Ae ikx + Be −ikx . (3.38)<br />

C<br />

�<br />

|p(x)|<br />

�<br />

i<br />

|p|dx<br />

e+ ¯h +<br />

D<br />

�<br />

|p(x)|<br />

�<br />

i<br />

|p|dx<br />

e− ¯h<br />

(3.39)<br />

ψR(x) = F e +ikx . (3.40)<br />

If F is the transmitted amplitude, then the tunneling probability is the ratio <strong>of</strong> the transmitted<br />

probability to the incident probability: T = |F | 2 /|A| 2 . If we assume that the barrier is high or<br />

broad, then C = 0 and we obtain the semi-classical estimate for the tunneling probability:<br />

T ≈ exp<br />

�<br />

− 2<br />

� �<br />

b<br />

|p(x)|dx<br />

¯h a<br />

62<br />

(3.41)


where a and b are the turning points on either side <strong>of</strong> the barrier.<br />

Mathematically, we can “flip the potential upside down” and work in imaginary time. In this<br />

case the action integral becomes<br />

S =<br />

� b<br />

a<br />

�<br />

2m(V (x) − E)dx. (3.42)<br />

So we can think <strong>of</strong> tunneling as motion under the barrier in imaginary time.<br />

There are a number <strong>of</strong> useful applications <strong>of</strong> this formula. Gamow’s theory <strong>of</strong> alpha-decay is<br />

a common example. Another useful application is in the theory <strong>of</strong> reaction rates where we want<br />

to determine tunneling corrections to the rate constant for a particular reaction. Close to the top<br />

<strong>of</strong> the barrier, where tunneling may be important, we can expand the potential and approximate<br />

the peak as an upside down parabola<br />

V (x) ≈ Vo − k<br />

2 x2<br />

where +x represents the product side and −x represents the reactant side. See Fig. 3.1 Set the<br />

zero in energy to be the barrier height, Vo so that any transmission for E < 0 corresponds to<br />

tunneling. 1<br />

0<br />

-0.2<br />

-0.4<br />

-0.6<br />

-0.8<br />

e<br />

-4 -2 2 4 x<br />

Figure 3.1: Eckart Barrier and parabolic approximation <strong>of</strong> the transition state<br />

At sufficiently large distances from the turning point, the motion is purely quasi-classical and<br />

we can write the momentum as<br />

�<br />

p = 2m(E + kx2 /2) ≈ x √ �<br />

mk + E m/k/x (3.43)<br />

and the asymptotic for <strong>of</strong> the Schrödinger wavefunction is<br />

ψ = Ae +iξ2 /2 ξ +iɛ−1/2 + Be −iξ 2 /2 ξ −iɛ−1/2<br />

(3.44)<br />

where A and B are coefficients we need to determine by the matching condition and ξ and ɛ are<br />

dimensionless lengths and energies given by ξ = x(mk/¯h) 1/4 �<br />

, and ɛ = (E/¯h) m/k.<br />

1 The analysis is from Kembel, 1935 as discussed in Landau and Lifshitz, QM<br />

63


The particular case we are interested in is for a particle coming from the left and passing to<br />

the right with the barrier in between. So, the wavefunctions in each <strong>of</strong> these regions must be<br />

and<br />

ψR = Be +iξ2 /2 ξ iɛ−1/2<br />

ψL = e −iξ2 /2 (−ξ) −iɛ−1/2 + Ae +iξ 2 /2 (−ξ) iɛ−1/2<br />

(3.45)<br />

(3.46)<br />

where the first term is the incident wave and the second term is the reflected component. So,<br />

|A| 2 | is the reflection coefficient and |B| 2 is the transmission coefficient normalized so that<br />

|A| 2 + |B| 2 = 1.<br />

Lets move to the complex plane and write a new coordinate, ξ = ρe iφ and consider what happens<br />

as we rotate around in φ and take ρ to be large. Since iξ 2 = ρ 2 (i cos 2φ − sin 2φ), we have<br />

and at φ = π<br />

ψR(φ = 0) = Be iρ2<br />

ρ +iɛ−1/2<br />

ψL(φ = 0) = Ae iρ2<br />

(−ρ) +iɛ−1/2<br />

ψR(φ = π) = Be iρ2<br />

(−ρ) +iɛ−1/2<br />

ψL(φ = π) = Ae iρ2<br />

ρ +iɛ−1/2<br />

So, in otherwords, ψR(φ = π) looks like ψL(φ = 0) when<br />

A = B(e iπ ) iɛ−1/2<br />

(3.47)<br />

(3.48)<br />

So, we have the relation A = −iBe−πɛ . Finally, after we normalize this we get the transmission<br />

coefficient:<br />

T = |B| 2 1<br />

=<br />

1 + e−2πɛ which must hold for any energy. If the energy is large and negative, then<br />

T ≈ e −2πɛ .<br />

Also, we can compute the reflection coefficient for E > 0 as 1 − D,<br />

R =<br />

1<br />

.<br />

1 + e +2πɛ<br />

Exercise 3.4 Verify these last relationships by taking the ψR and ψL, performing the analytic<br />

continuation.<br />

64


This gives us the transmission probabilty as a function <strong>of</strong> incident energy. But, normal<br />

chemical reactions are not done at constant energy, they are done at constant temperature.<br />

To get the thermal transmission coefficient, we need to take a Boltzmann weighted average <strong>of</strong><br />

transmission coefficients<br />

Tth(β) = 1<br />

�<br />

Z<br />

dEe −Eβ T (E) (3.49)<br />

where β = 1/kT and Z is the partition function. If E represents a continuum <strong>of</strong> energy states<br />

then<br />

Tth(β) = − βω¯h(ψ(0) ( βω¯h<br />

4π ) − ψ(0) ( 1 βω¯h<br />

( 4 π<br />

4π<br />

+ 2)))<br />

(3.50)<br />

where ψ (n) (z) is the Polygamma function which is the nth derivative <strong>of</strong> the digamma function,<br />

ψ (0) (z), which is the logarithmic derivative <strong>of</strong> Eulers gamma function, ψ (0) (z) = Γ(z)/Γ(z). 2<br />

3.3 Connection Formulas<br />

In what we have considered thus far, we have assumed that up until the turning point the<br />

wavefunction was well behaved and smooth. We can think <strong>of</strong> the problem as having two domains:<br />

an exterior and an interior. The exterior part we assumed to be simple and the boundary<br />

conditions trivial to impose. The next task is to figure out the matching condition at the turning<br />

point for an arbitrary system. So far what we have are two pieces, ψL and ψR, in the notation<br />

above. What we need is a patch. To do so, we make a linearizing assumption for the force at<br />

the classical turning point:<br />

E − V (x) ≈ Fo(x − a) (3.51)<br />

where Fo = −dV/dx evaluated at x = a. Thus, the phase integral is easy:<br />

�<br />

1 x<br />

pdx =<br />

¯h a<br />

2 �<br />

2mFo(x − a)<br />

3¯h<br />

3/2<br />

(3.52)<br />

But, we can do better than that. We can actually solve the Schrodinger equation for the linear<br />

potential and use the linearized solutions as our patch. The Mathematica Notebook AiryFunctions.nb<br />

goes through the solution <strong>of</strong> the linearized Schrodinger equation<br />

which can be re-written as<br />

with<br />

2 See the Mathematica Book, sec 3.2.10.<br />

− ¯h2 dψ<br />

2m<br />

dx 2 + (E + V ′ )ψ = 0 (3.53)<br />

α =<br />

ψ ′′ = α 3 xψ (3.54)<br />

�<br />

2m<br />

¯h 2 V ′ �1/3<br />

(0) .<br />

65


Absorbing the coefficient into a new variable y, we get Airy’s equation<br />

ψ ′′ (y) = yψ.<br />

The solutions <strong>of</strong> Airy’s equation are Airy functions, Ai(y) and Bi(y) for the regular and irregular<br />

cases. The integral representation <strong>of</strong> the Ai and Bi are<br />

and<br />

Bi(y) = 1<br />

π<br />

Plots <strong>of</strong> these functions are shown in Fig. 3.2.<br />

Ai@yD, Bi@yD<br />

1.5<br />

-10 -8 -6 -4 -2 2<br />

1<br />

0.5<br />

-0.5<br />

Ai(y) = 1<br />

� � �<br />

∞ 3 s<br />

cos + sy ds (3.55)<br />

π 0 3<br />

� �<br />

∞<br />

e<br />

0<br />

−s3 � ��<br />

3<br />

/3+sy s<br />

+ sin + sy ds (3.56)<br />

3<br />

y<br />

Figure 3.2: Airy functions, Ai(y) (red) and Bi(y) (blue)<br />

Since both Ai and Bi are acceptible solutions, we will take a linear combination <strong>of</strong> the two<br />

as our patching function and figure out the coefficients later.<br />

ψP = aAi(αx) + bBi(αx) (3.57)<br />

We now have to determine those coefficients. We need to make two assumptions. One,<br />

that the overlap zones are sufficiently close to the turning point that a linearized potential is<br />

reasonable. Second, the overlap zone is far enough from the turning point (at the origin) that<br />

the WKB approximation is accurate and reliable. You can certainly cook up some potential<br />

for which this will not work, but we will assume it’s reasonable. In the linearized region, the<br />

momentum is<br />

So for +x,<br />

� x<br />

0<br />

p(x) = ¯hα 3/2 (−x) 3/2<br />

(3.58)<br />

|p(x)|dx = 2¯h(αx) 3/2 /3 (3.59)<br />

66


and the WKB wavefunction becomes:<br />

ψR(x) =<br />

D<br />

√ ¯hα 3/4 x 1/4 e−2(αx)3/2 /3 . (3.60)<br />

In order to extend into this region, we will use the asymptotic form <strong>of</strong> the Ai and Bi functions<br />

for y ≫ 0<br />

Ai(y) ≈ e−2y3/2 /3<br />

2 √ πy 1/4<br />

Clearly, the Bi(y) term will not contribute, so b = 0 and<br />

�<br />

4π<br />

a =<br />

α¯h D.<br />

(3.61)<br />

Bi(y) ≈ e+2y3/2 /3<br />

√ . (3.62)<br />

πy1/4 Now, for the other side, we do the same proceedure. Except this time x < 0 so the phase<br />

integral is<br />

� 0<br />

pdx = 2¯h(−αx) 3/2 /3. (3.63)<br />

Thus the WKB wavefunction on the left hand side is<br />

ψL(x) = 1<br />

√ p<br />

=<br />

x<br />

�<br />

Be 2i(−αx)3/2 /3 + Ce −2i(−αx) 3/2 /3 �<br />

1 �<br />

√ Be<br />

¯hα3/4 (−x) 1/4<br />

2i(−αx)3/2 /3 −2i(−αx)<br />

+ Ce 3/2 /3 �<br />

(3.64)<br />

(3.65)<br />

That’s the WKB part, to connect with the patching part, we again use the asymptotic forms for<br />

y ≪ 0 and take only the regular solution,<br />

Ai(y) ≈<br />

≈<br />

1 �<br />

√ sin 2(−y)<br />

π(−y) 1/4 3/2 /3 + π/4 �<br />

1<br />

2i √ π(−y) 1/4<br />

�<br />

e iπ/4 e i2(−y)3/2 /3 −iπ/4 −i2(−y)<br />

− e e 3/2 /3 �<br />

Comparing the WKB wave and the patching wave, we can match term-by-term<br />

a<br />

2i √ π eiπ/4 = B<br />

√<br />

¯hα<br />

−a<br />

2i √ π e−iπ/4 = C<br />

√<br />

¯hα<br />

(3.66)<br />

(3.67)<br />

(3.68)<br />

Since we know a in terms <strong>of</strong> the normalization constant D, B = ie iπ/4 D and C = ie −iπ/4 . This<br />

is the connection! We can write the WKB function across the turning point as<br />

⎧<br />

⎪⎨<br />

ψW KB(x) =<br />

⎪⎩<br />

√2D sin<br />

p(x) � 1<br />

1<br />

√ 2D −<br />

e ¯h<br />

|p(x)|<br />

� 0<br />

¯h x<br />

� 0<br />

x pdx<br />

67<br />

pdx + π/4�<br />

x < 0<br />

x > 0<br />

(3.69)


node xn<br />

1 -2.33811<br />

2 -4.08795<br />

3 -5.52056<br />

4 -6.78671<br />

5 -7.94413<br />

6 -9.02265<br />

7 -10.0402<br />

Table 3.1: Location <strong>of</strong> nodes for Airy, Ai(x) function.<br />

Example: Bound states in the linear potential<br />

Since we worked so hard, we have to use the results. So, consider a model problem for a particle<br />

in a gravitational field. Actually, this problem is not so far fetched since one can prepare trapped<br />

atoms above a parabolic reflector and make a quantum bouncing ball. Here the potential is<br />

V (x) = mgx where m is the particle mass and g is the graviational constant (g = 9.80m/s).<br />

We’ll take the case where the reflector is infinite so that the particle cannot penetrate into it.<br />

The Schrödinger equation for this potential is<br />

− ¯h2<br />

2m ψ′′ + (E − mgx)ψ = 0. (3.70)<br />

The solutions are the Airy Ai(x) functions. Setting, β = mg and c = ¯h 2 /2m, the solutions are<br />

ψ = CAi(<br />

�<br />

− β<br />

c<br />

� 1/3<br />

(x − E/β)) (3.71)<br />

However, there is one caveat: ψ(0) = 0, thus the Airy functions must have their nodes at x = 0.<br />

So we have to systematically shift the Ai(x) function in x until a node lines up at x = 0. The<br />

nodes <strong>of</strong> the Ai(x) function can be determined and the first 7 <strong>of</strong> them are To find the energy<br />

levels, we systematically solve the equation<br />

�<br />

− β<br />

�1/3 En<br />

c β<br />

= xn<br />

So the ground state is where the first node lands at x = 0,<br />

E1 = 2.33811β<br />

(β/c) 1/3<br />

= 2.33811mg<br />

(2m 2 g/¯h 2 ) 1/3<br />

(3.72)<br />

and so on. Of course, we still have to normalize the wavefunction to get the correct energy.<br />

We can make life a bit easier by using the quantization condition derived from the WKB<br />

approximation. Since we require the wavefunction to vanish exactly at x = 0, we have:<br />

�<br />

1 xt<br />

p(x)dx +<br />

¯h 0<br />

π<br />

= nπ. (3.73)<br />

4<br />

68


15<br />

10<br />

5<br />

-5<br />

-10<br />

2 4 6 8 10<br />

Figure 3.3: Bound states in a graviational well<br />

This insures us that the wave vanishes at x = 0, xt in this case is the turning point E = mgxt.<br />

(See Figure 3.3) As a consequence,<br />

Since p(x) =<br />

� xt<br />

0<br />

� xt<br />

0<br />

p(x)dx = (n − 1/4)π<br />

�<br />

2m(En − mgx), The integral can be evaluated<br />

�<br />

2m(E − mghdx = √ ⎛<br />

2 ⎝ 2En<br />

√ �<br />

⎞<br />

Enm 2 m (En − gmxt) (−En + gmxt)<br />

+ ⎠ (3.74)<br />

3gm<br />

3gm<br />

Since xt = En/mg for the classical turning point, the phase intergral becomes<br />

{ 2√2En 2<br />

3g √ } = (n − 1/4)π¯h.<br />

Enm<br />

Solving for En yields the semi-classical approximation for the eigenvalues:<br />

En =<br />

2<br />

g 3 m 1 �<br />

1<br />

2<br />

2� 3<br />

3 (1 − 4 n) (3 π) 3 ¯h 2<br />

3<br />

4 2 1<br />

3<br />

(3.75)<br />

In atomic units, the gravitional constant is g = 1.08563 × 10 −22 bohr/au 2 (Can you guess why<br />

we rarely talk about gravitational effects on molecules?). For n = 0, we get for an electron<br />

E sc<br />

o = 2.014 × 10 −15 hartree or about 12.6 Hz. So, graviational effects on electrons are extremely<br />

tiny compared to the electron’s total energy.<br />

69


3.4 Scattering<br />

b<br />

r<br />

θ<br />

Figure 3.4: Elastic scattering trajectory for classical collision<br />

The collision between two particles plays an important role in the dynamics <strong>of</strong> reactive molecules.<br />

We consider here the collosion between two particles interacting via a central force, V (r). Working<br />

in the center <strong>of</strong> mass frame, we consider the motion <strong>of</strong> a point particle with mass µ with<br />

position vector �r. We will first examine the process in a purely classical context since it is<br />

intuitive and then apply what we know to the quantum and semiclassical case.<br />

3.4.1 Classical Scattering<br />

The angular momentum <strong>of</strong> the particle about the origin is given by<br />

θ c<br />

r c<br />

�L = �r × �p = µ(�r × ˙ �r) (3.76)<br />

we know that angular momentum is a conserved quantity and it is is easy to show that ˙ � L = 0<br />

viz.<br />

˙�L = d<br />

dt �r × �p = (˙ �r × ˙ �r + (�r × ˙ �p). (3.77)<br />

Since ˙r = ˙p/µ, the first term vanishes; likewise, the force vector, ˙ �p = −dV/dr, is along �r so that<br />

the second term vanishes. Thus, L = const meaning that angular momentum is a conserved<br />

quantity during the course <strong>of</strong> the collision.<br />

In cartesian coordinates, the total energy <strong>of</strong> the collision is given by<br />

To convert from cartesian to polar coordinates, we use<br />

E = µ<br />

2 ( ˙x2 + ˙y 2 ) + V. (3.78)<br />

x = r cos θ (3.79)<br />

70<br />

χ


Thus,<br />

where we use the fact that<br />

y = r sin θ (3.80)<br />

˙x = ˙r cos θ − r ˙ θ sin θ (3.81)<br />

˙y = ˙r sin θ + r ˙ θ cos θ. (3.82)<br />

E = mu<br />

2 ˙r2 + V (r) + L2<br />

2µr 2<br />

L = µr 2 ˙ θ 2<br />

(3.83)<br />

(3.84)<br />

where L is the angular momentum. What we see here is that we have two potential contributions.<br />

The first is the physical attraction (or repulsion) between the two scattering bodies. The second is<br />

a purely repulsive centrifugal potential which depends upon the angular momentum and ultimatly<br />

upon the inpact parameters. For cases <strong>of</strong> large impact parameters, this can be the dominant<br />

force. The effective radial force is given by<br />

µ¨r = L2<br />

2r3 ∂V<br />

−<br />

µ ∂r<br />

(3.85)<br />

Again, we note that the centrifugal contribution is always repulsive while the physical interaction<br />

V(r) is typically attractive at long range and repulsive at short ranges.<br />

We can derive the solutions to the scattering motion by integrating the velocity equations for<br />

r and θ<br />

˙r =<br />

� �<br />

2<br />

± E − V (r) −<br />

µ<br />

L2<br />

2µr2 ��1/2 ˙θ = L<br />

µr2 (3.86)<br />

(3.87)<br />

and taking into account the starting conditions for r and θ. In general, we could solve the<br />

equations numerically and obtain the complete scatering path. However, really what we are<br />

interested in is the deflection angle χ since this is what is ultimately observed. So, we integrate<br />

the last two equations and derive θ in terms <strong>of</strong> r.<br />

θ(r) =<br />

� θ<br />

= −<br />

0<br />

� r<br />

� r<br />

dθ = −<br />

∞<br />

L<br />

µr 2<br />

∞<br />

dθ<br />

dr (3.88)<br />

dr<br />

1<br />

�<br />

(2/µ)(E − V − L 2 /2µr 2 )<br />

dr (3.89)<br />

where the collision starts at t = −∞ with r = ∞ and θ = 0. What we want to do is derive this<br />

in terms <strong>of</strong> an impact parameter, b, and scattering angle χ. These are illustrated in Fig. 3.4 and<br />

can be derived from basic kinematic considerations. First, energy is conserved through out, so if<br />

71


we know the asymptotic velocity, v, then E = µv 2 /2. Secondly, angular momentum is conserved,<br />

so L = µ|r × v| = µvb. Thus the integral above becomes<br />

� r<br />

θ(r) = −b<br />

= −<br />

∞<br />

� r<br />

∞<br />

dθ<br />

dr (3.90)<br />

dr<br />

dr<br />

r2 �<br />

1 − V/E − b2 . (3.91)<br />

/r2 Finally, the angle <strong>of</strong> deflection is related to the angle <strong>of</strong> closest approach by 2θc + χ = π; hence,<br />

� ∞<br />

χ = π − 2b<br />

rc<br />

dr<br />

r2 �<br />

1 − V/E − b2 /r2 The radial distance <strong>of</strong> closest approach is determined by<br />

which can be restated as<br />

E = L2<br />

2µr 2 c<br />

b 2 = r 2 c<br />

�<br />

1 −<br />

(3.92)<br />

+ V (rc) (3.93)<br />

�<br />

V (rc)<br />

E<br />

(3.94)<br />

Once we have specified the potential, we can compute the deflection angle using Eq.3.94. If<br />

V (rc) < 0 , then rc < b and we have an attractive potential, if V (rc) > 0 , then rc > b and the<br />

potential is repulsive at the turning point.<br />

If we have a beam <strong>of</strong> particles incident on some scattering center, then collisions will occur<br />

with all possible impact parameter (hence angular momenta) and will give rise to a distribution<br />

in the scattering angles. We can describe this by a differential cross-section. If we have some<br />

incident intensity <strong>of</strong> particles in our beam, Io, which is the incident flux or the number <strong>of</strong> particles<br />

passing a unit area normal to the beam direction per unit time, then the differential cross-section,<br />

I(χ), is defined so that I(χ)dΩ is the number <strong>of</strong> particles per unit time scattered into some solid<br />

angle dΩ divided by the incident flux.<br />

The deflection pattern will be axially symmetric about the incident beam direction due the<br />

spherical symmetry <strong>of</strong> the interaction potential; thus, I(χ) only depends upon the scattering angle.<br />

Thus, dΩ can be constructed by the cones defining χ and χ+dχ, i.e. dΩ = 2π sin χdχ. Even<br />

if the interaction potential is not spherically symmetric, since most molecules are not spherical,<br />

the scattering would be axially symmetric since we would be scattering from a homogeneous distribution<br />

<strong>of</strong> al possible orientations <strong>of</strong> the colliding molecules. Hence any azimuthal dependency<br />

must vanish unless we can orient or the colliding species.<br />

Given an initial velocity v, the fraction <strong>of</strong> the incoming flux with impact parameter between<br />

b and b + db is 2πbdb. These particles will be deflected between χ and χ + dχ if dχ/db > 0 or<br />

between χ and χ − dχ if dχ/db < 0. Thus, I(χ)dΩ = 2πbdb and it follows then that<br />

I(χ) =<br />

b<br />

. (3.95)<br />

sin χ|dχ/db|<br />

72


Thus, once we know χ(b) for a given v, we can get the differential cross-section. The total<br />

cross-section is obtained by integrating<br />

σ = 2π<br />

� π<br />

0<br />

I(χ) sin χdχ. (3.96)<br />

This is a measure <strong>of</strong> the attenuation <strong>of</strong> the incident beam by the scattering target and has the<br />

units <strong>of</strong> area.<br />

3.4.2 Scattering at small deflection angles<br />

Our calculations will be greatly simplified if we consider collisions that result in small deflections<br />

in the forward direction. If we let the initial beam be along the x axis with momentum p, then<br />

the scattered momentum, p ′ will be related to the scattered angle by p ′ sin χ = p ′ y. Taking χ to<br />

be small<br />

χ ≈ p′ y momentum transfer<br />

= . (3.97)<br />

p ′ momentum<br />

Since the time derivative <strong>of</strong> momentum is the force, the momentum transfered perpendicular to<br />

the incident beam is obtained by integrating the perpendicular force<br />

F ′ y = − ∂V<br />

∂y<br />

= −∂V<br />

∂r<br />

∂r<br />

∂y<br />

where we used r 2 = x 2 + y 2 and y ≈ b. Thus we find,<br />

χ =<br />

p ′ y<br />

µ(2E/µ) 1/2<br />

= −b(2µE) −1/2<br />

= −b(2µE) −1/2<br />

� +∞<br />

−∞<br />

� 2E<br />

µ<br />

= −∂V<br />

∂r<br />

∂V dt<br />

∂r r<br />

b<br />

r<br />

� −1/2 � +∞<br />

−∞<br />

∂V<br />

∂r<br />

dx<br />

r<br />

(3.98)<br />

(3.99)<br />

(3.100)<br />

(3.101)<br />

= − b<br />

� ∞ ∂V<br />

E b ∂r (r2 − b 2 ) −1/2 dr (3.102)<br />

where we used x = (2E/µ) 1/2t and x varies from −∞ to +∞ as r goes from −∞ to b and back.<br />

Let us use this in a simple example <strong>of</strong> the V = C/rs potential for s > 0. Substituting V into<br />

the integral above and solving yields<br />

χ = sCπ1/2<br />

2b s E<br />

Γ((s + 1)/2)<br />

. (3.103)<br />

Γ(s/2 + 1)<br />

This indicates that χE ∝ b−s and |dχ/db| = χs/b.<br />

differential cross-section<br />

Thus, we can conclude by deriving the<br />

I(χ) = 1<br />

s χ−(2+2/s)<br />

� �<br />

1/2<br />

2/s<br />

sCπ Γ((s + 1)/2)<br />

2E Γ(s/2 + 1)<br />

(3.104)<br />

for small values <strong>of</strong> the scattering angle. Consequently, a log-log plot <strong>of</strong> the center <strong>of</strong> mass<br />

differential cross-section as a function <strong>of</strong> the scattering angle at fixed energy should give a straight<br />

line with a slope −(2 + 2/s) from which one can determine the value <strong>of</strong> s. For the van der Waals<br />

potential, s = 6 and I(χ) ∝ E −1/3 χ −7/3 .<br />

73


3.4.3 <strong>Quantum</strong> treatment<br />

The quantum mechanical case is a bit more complex. Here we will develop a brief overview<br />

<strong>of</strong> quantum scattering and move onto the semiclassical evaluation. The quantum scattering is<br />

determined by the asymptotic form <strong>of</strong> the wavefunction,<br />

ψ(r, χ) r→∞<br />

�<br />

−→ A e ikz + f(χ)<br />

r eikr<br />

�<br />

(3.105)<br />

where A is some normalization constant and k = 1/λ = µv/¯h is the initial wave vector along<br />

the incident beam direction (χ = 0). The first term represents a plane wave incident upon<br />

the scatterer and the second represents an out going spherical wave. Notice that the outgoing<br />

amplitude is reduced as r increases. This is because the wavefunction spreads as r increases. If<br />

we can collimate the incoming and out-going components, then the scattering amplitude f(χ) is<br />

related to the differential cross-section by<br />

I(χ) = |f(χ)| 2 . (3.106)<br />

What we have is then that asymptotic form <strong>of</strong> the wavefunction carries within it information<br />

about the scattering process. As a result, we do not need to solve the wave equation for all <strong>of</strong><br />

space, we just need to be able to connect the scattering amplitude to the interaction potential.<br />

We do so by expanding the wave as a superposition <strong>of</strong> Legendre polynomials<br />

∞�<br />

ψ(r, χ) = Rl(r)Pl(cos χ). (3.107)<br />

l=0<br />

Rl(r) must remain finite as r = 0. This determines the form <strong>of</strong> the solution.<br />

When V (r) = 0, then ψ = A exp(ikz) and we can expand the exponential in terms <strong>of</strong> spherical<br />

waves<br />

e ikz ∞�<br />

ilπ/2 sin(kr − lπ/2)<br />

= (2l + 1)e Pl(cos χ) (3.108)<br />

l=0<br />

kr<br />

= 1<br />

∞�<br />

(2l + 1)i<br />

2i l=0<br />

l �<br />

i(kr−lπ/2) e<br />

Pl(cos χ)<br />

+<br />

kr<br />

e−i(kr−lπ/2)<br />

�<br />

(3.109)<br />

kr<br />

We can interpret this equation in the following intuitive way: the incident plane wave is equivalent<br />

to an infinite superposition <strong>of</strong> incoming and outgoing spherical waves in which each term<br />

corresponds to a particular angular momentum state with<br />

�<br />

L = ¯h l(l + 1) ≈ ¯h(l + 1/2). (3.110)<br />

From our analysis above, we can relate L to the impact parameter, b,<br />

b = L<br />

µv<br />

l + 1/2<br />

≈ λ. (3.111)<br />

k<br />

In essence the incoming beam is divided into cylindrical zones in which the lth zone contains<br />

particles with impact parameters (and hence angular momenta) between lλ and (l + 1)λ.<br />

74


Exercise 3.5 The impact parameter, b, is treated as continuous; however, in quantum mechanics<br />

we allow only discrete values <strong>of</strong> the angular momentum, l. How will this affect our results, since<br />

b = (l + 1/2)λ from above.<br />

If V (r) is short ranged (i.e. it falls <strong>of</strong>f more rapidly than 1/r for large r), we can derive a<br />

general solution for the asymptotic form<br />

∞�<br />

� � ��<br />

lπ sin(kr − lπ/2 + ηl<br />

ψ(r, χ) −→ (2l + 1) exp i + ηl<br />

Pl(cos χ). (3.112)<br />

2 kr<br />

l=0<br />

The significant difference between this equation and the one above for the V (r) = 0 case is the<br />

addition <strong>of</strong> a phase-shift ηl. This shift only occurs in the outgoing part <strong>of</strong> the wavefunction and so<br />

we conclude that the primary effect <strong>of</strong> a potential in quantum scattering is to introduce a phase<br />

in the asymptotic form <strong>of</strong> the scattering wave. This phase must be a real number and has the<br />

following physical interpretation illustrated in Fig. 3.5 A repulsive potential will cause a decrease<br />

in the relative velocity <strong>of</strong> the particles at small r resulting in a longer de Broglie wavelength. This<br />

causes the wave to be “pushed out” relative to that for V = 0 and the phase shift is negative.<br />

An attractive potential produces a positive phase shift and “pulls” the wavefunction in a bit.<br />

Furthermore, the centrifugal part produces a negative shift <strong>of</strong> −lπ/2.<br />

Comparing the various forms for the asymptotic waves, we can deduce that the scattering<br />

amplitude is given by<br />

f(χ) = 1<br />

2ik<br />

From this, the differential cross-section is<br />

I(χ) = λ 2<br />

�<br />

� ∞� �<br />

� (2l + 1)e<br />

�<br />

iηl<br />

�<br />

�<br />

�<br />

sin(ηl)Pl(cos χ) �<br />

�<br />

∞�<br />

(2l + 1)(e<br />

l=0<br />

2iηl − 1)Pl(cos χ). (3.113)<br />

l=0<br />

2<br />

(3.114)<br />

What we see here is the possibility for interference between different angular momentum components<br />

Moving forward at this point requires some rather sophisticated treatments which we reserve<br />

for a later course. However, we can use the semiclassical methods developed in this chapter to<br />

estimate the phase shifts.<br />

3.4.4 Semiclassical evaluation <strong>of</strong> phase shifts<br />

The exact scattering wave is not so important. What is important is the asymptotic extent <strong>of</strong><br />

the wavefunction since that is the part which carries the information from the scattering center<br />

to the detector. What we want is a measure <strong>of</strong> the shift in phase between a scattering with and<br />

without the potential. From the WKB treatment above, we know that the phase is related to<br />

the classical action along a given path. Thus, in computing the semiclassical phase shifts, we<br />

are really looking at the difference between the classical actions for a system with the potential<br />

switched on and a system with the potential switched <strong>of</strong>f.<br />

η SC<br />

l<br />

= lim<br />

R→∞<br />

� � R<br />

rc<br />

dr<br />

λ(r) −<br />

� R dr<br />

b λ(r)<br />

75<br />

�<br />

(3.115)


1<br />

0.5<br />

-0.5<br />

-1<br />

5 10 15 20 25 30<br />

Figure 3.5: Form <strong>of</strong> the radial wave for repulsive (short dashed) and attractive (long dashed)<br />

potentials. The form for V = 0 is the solid curve for comparison.<br />

R is the radius a sphere about the scattering center and λ(r) is a de Broglie wavelength<br />

λ(r) = ¯h<br />

p<br />

= 1<br />

k(r) =<br />

¯h<br />

µv(1 − V (r) − b 2 /r 2 ) 1/2<br />

(3.116)<br />

associated with the radial motion. Putting this together:<br />

η SC<br />

l = lim<br />

R→∞ k<br />

⎛<br />

� R V (r) b2<br />

⎝ (1 − −<br />

rc E r2 )1/2 � �<br />

R<br />

dr − 1 −<br />

b<br />

b2<br />

r2 =<br />

� ⎞<br />

1/2<br />

dr⎠<br />

⎛<br />

� R<br />

� �<br />

R<br />

lim ⎝ k(r)dr − k 1 −<br />

R→∞ rc<br />

b<br />

(3.117)<br />

b2<br />

r2 � ⎞<br />

1/2<br />

dr⎠<br />

(3.118)<br />

(k is the incoming wave-vector.) The last integral we can evaluate:<br />

� R<br />

k<br />

b<br />

(r 2 − b 2 )) 1/2<br />

r<br />

dr = k<br />

�<br />

(r 2 − b 2 −1 b<br />

) − b cos<br />

r<br />

��<br />

���� R<br />

b<br />

= kR − kbπ<br />

2<br />

(3.119)<br />

Now, to clean things up a bit, we add and subtract an integral over k. (We do this to get rid <strong>of</strong><br />

the R dependence which will cause problems when we take the limit R → ∞.)<br />

η SC<br />

�� R<br />

l = lim<br />

R→∞ rc<br />

� R<br />

=<br />

=<br />

rc<br />

� R<br />

rc<br />

k(r)dr −<br />

� R<br />

rc<br />

kdr +<br />

� R<br />

rc<br />

kdr − (kR − kbπ<br />

2 )<br />

�<br />

(3.120)<br />

(k(r) − k)dr − k(rc − bπ/2) (3.121)<br />

(k(r) − k)dr − krcπ(l + 1/2)/2 (3.122)<br />

This last expression is the standard form <strong>of</strong> the phase shift.<br />

The deflection angle can be determined in a similar way.<br />

�� �<br />

� � �<br />

��<br />

χ = lim<br />

R→∞<br />

π − 2 dθ<br />

actual path<br />

− π − dθ<br />

V =0 path<br />

76<br />

(3.123)


We transform this into an integral over r<br />

⎡<br />

� �<br />

∞ V (r) b2<br />

χ = −2b ⎣ 1 − −<br />

E r2 �−1/2 � �<br />

dr ∞<br />

− 1 −<br />

r2 b<br />

b2<br />

r2 �−1/2 dr<br />

r2 ⎤<br />

⎦ (3.124)<br />

rc<br />

Agreed, this is weird way to express the scattering angle. But let’s keep pushing this forward.<br />

The last integral can be evaluated<br />

� �<br />

∞<br />

1 −<br />

b<br />

b2<br />

r2 �−1/2 �<br />

dr 1 b �<br />

�<br />

= cos−1 �<br />

r2 b r �<br />

∞<br />

b<br />

= − π<br />

. (3.125)<br />

2b<br />

which yields the classical result we obtained before. So, why did we bother? From this we can<br />

derive a simple and useful connection between the classical deflection angle and the rate <strong>of</strong> change<br />

<strong>of</strong> the semiclassical phase shift with angular momentum, dη SC<br />

l /dl. First, recall the Leibnitz rule<br />

for taking derivatives <strong>of</strong> integrals:<br />

�<br />

d b(x)<br />

f(x, y)dy =<br />

dx a(x)<br />

b<br />

dx<br />

Taking the derivative <strong>of</strong> η SC<br />

l<br />

(∂b/∂l)E = b/k, we find that<br />

�<br />

da<br />

b(x)<br />

f(b(x), y) − f(a(x), y) +<br />

dx a(x)<br />

∂f(x, y)<br />

dy (3.126)<br />

∂x<br />

with respect to l, using the last equation a and the relation that<br />

dη SC<br />

l<br />

dl<br />

χ<br />

= . (3.127)<br />

2<br />

Next, we examine the differential cross-section, I(χ). The scattering amplitude<br />

f(χ) = λ<br />

2i<br />

∞�<br />

(2l + 1)e<br />

l=0<br />

2iηlPl(cos χ). (3.128)<br />

where we use λ = 1/k and exclude the singular point where χ = 0 since this contributes nothing<br />

to the total flux.<br />

Now, we need a mathematical identity to take this to the semiclassical limit where the potential<br />

varies slowly with wavelength. What we do is to first relate the Legendre polynomial,<br />

Pl(cos θ) to a zeroth order Bessel function for small values <strong>of</strong> θ (θ ≪ 1).<br />

Pl(cos θ) = J0((l + 1/2)θ). (3.129)<br />

Now, when x = (l + 1/2)θ ≫ 1 (i.e. large angular momentum), we can use the asymptotic<br />

expansion <strong>of</strong> J0(x)<br />

Pulling this together,<br />

Pl(cos θ) →<br />

�<br />

2<br />

π(l + 1/2)θ<br />

�<br />

2<br />

J0(x) →<br />

πx sin<br />

�<br />

x + π<br />

�<br />

. (3.130)<br />

4<br />

�1/2 �<br />

sin ((l + 1/2)θ + π/4) ≈<br />

77<br />

2<br />

π(l + 1/2)<br />

�1/2 sin ((l + 1/2)θ + π/4)<br />

(sin θ) 1/2 (3.131)


for θ(l + 1/2) ≫ 1. Thus, we can write the semi-classical scattering amplitude as<br />

where<br />

f(χ) = −λ<br />

∞�<br />

l=0<br />

� �1/2 (l + 1/2) �<br />

e<br />

2π sin χ<br />

iφ+<br />

+ e iφ−�<br />

(3.132)<br />

φ ± = 2ηl ± (l + 1/2)χ ± π/4. (3.133)<br />

The phases are rapidly oscillating functions <strong>of</strong> l. Consequently, the majority <strong>of</strong> the terms must<br />

cancel and the sum is determined by the ranges <strong>of</strong> l for which either φ + or φ − is extremized.<br />

This implies that the scattering amplitude is determined almost exclusively by phase-shifts which<br />

satisfy<br />

2 dηl<br />

dl<br />

± χ = 0, (3.134)<br />

where the + is for dφ + /dl = 0 and the − is for dφ − /dl = 0. This demonstrates that the only the<br />

phase-shifts corresponding to impact parameter b can contribute significantly to the differential<br />

cross-section in the semi-classical limit. Thus, the classical condition for scattering at a given<br />

deflection angle, χ is that l be large enough for Eq. 3.134 to apply.<br />

3.4.5 Resonance Scattering<br />

3.5 Problems and Exercises<br />

Exercise 3.6 In this problem we will (again) consider the ammonia inversion problem, this time<br />

we will proceed in a semi-classical context.<br />

Recall that the ammonia inversion potential consists <strong>of</strong> two symmetrical potential wells separated<br />

by a barrier. If the barrier was impenetrable, one would find energy levels corresponding to<br />

motion in one well or the other. Since the barrier is not infinite, there can be passage between<br />

wells via tunneling. This causes the otherwise degenerate energy levels to split.<br />

In this problem, we will make life a bit easier by taking<br />

V (x) = α(x 4 − x 2 )<br />

as in the examples in Chapter 5.<br />

Let ψo be the semi-classical wavefunction describing the motion in one well with energy Eo.<br />

Assume that ψo is exponentially damped on both sides <strong>of</strong> the well and that the wavefunction<br />

is normalized so that the integral over ψ 2 o is unity. When tunning is taken into account, the<br />

wavefunctions corresponding to the new energy levels, E1 and E2 are the symmetric and antisymmetric<br />

combinations <strong>of</strong> ψo(x) and ψo(−x)<br />

ψ1 = (ψo(x) + ψo(−x)/ √ 2<br />

ψ2 = (ψo(x) − ψo(−x)/ √ 2<br />

where ψo(−x) can be thought <strong>of</strong> as the contribution from the zeroth order wavefunction in the<br />

other well. In Well 1, ψo(−x) is very small and in well 2, ψo(+x) is very small and the product<br />

ψo(x)ψo(−x) is vanishingly small everywhere. Also, by construction, ψ1 and ψ2 are normalized.<br />

78


1. Assume that ψo and ψ1 are solutions <strong>of</strong> the Schrödinger equations<br />

and<br />

ψ ′′<br />

o + 2m<br />

¯h 2 (Eo − V )ψo = 0<br />

ψ ′′<br />

1 + 2m<br />

¯h 2 (E1 − V )ψ1 = 0,<br />

Multiply the former by ψ1 and the latter by ψo, combine and subtract equivalent terms, and<br />

integrate over x from 0 to ∞ to show that<br />

Perform a similar analysis to show that<br />

E1 − Eo = − ¯h2<br />

m ψo(0)ψ ′ o(0),<br />

E2 − Eo = + ¯h2<br />

m ψo(0)ψ ′ o(0),<br />

2. Show that the unperturbed semiclassical wavefunction is<br />

and<br />

where vo =<br />

ψo(0) =<br />

� ω<br />

2πvo<br />

�<br />

exp − 1<br />

� a �<br />

|p|dx<br />

¯h 0<br />

ψ ′ o(0) = mvo<br />

¯h ψo(0)<br />

�<br />

2(Eo − V (0))/m and a is the classical turning point at Eo = V (a).<br />

3. Combining your results, show that the tunneling splitting is<br />

∆E = ¯hω<br />

π exp<br />

�<br />

− 1<br />

� +a �<br />

|p|dx .<br />

¯h −a<br />

where the integral is taken between classical turning points on either side <strong>of</strong> the barrier.<br />

4. Assuming that the potential in the barrier is an upside-down parabola<br />

what is the tunneling splitting.<br />

V (x) ≈ Vo − kx 2 /2<br />

5. Now, taking α = 0.1 expand the potential about the barrier and compute determine the<br />

harmonic force constant for the upside-down parabola. Using the equations you derived and<br />

compute the tunneling splitting for a proton in this well. How does this compare with the<br />

calculations presented in Chapter 5.<br />

79


Chapter 4<br />

Postulates <strong>of</strong> <strong>Quantum</strong> <strong>Mechanics</strong><br />

When I hear the words “Schrödinger’s cat”, I wish I were able to reach for my gun.<br />

Stephen Hawkings.<br />

The dynamics <strong>of</strong> physical processes at a microscopic level is very much beyond the realm <strong>of</strong><br />

our macroscopic comprehension. In fact, it is difficult to imagine what it is like to move about on<br />

the length and timescales for whcih quantum mechanics is important. However, for molecules,<br />

quantum mechanics is an everyday reality. Thus, in order to understand how molecules move and<br />

behave, we must develop a model <strong>of</strong> that dynamics in terms which we can comprehend. Making<br />

a model means developing a consistent mathematical framework in which the mathematical<br />

operations and constructs mimic the physical processes being studied.<br />

Before moving on to develop the mathematical framework required for quantum mechanics,<br />

let us consider a simple thought experiment. WE could do the experiment, however, we would<br />

have to deal with some additional technical terms, like funding. The experiment I want to<br />

consider goes as follows: Take a machine gun which shoots bullets at a target. It’s not a very<br />

accurate gun, in fact, it sprays bullets randomly in the general direction <strong>of</strong> the target.<br />

The distribution <strong>of</strong> bullets or histogram <strong>of</strong> the amount <strong>of</strong> lead accumulated in the target is<br />

roughly a Gaussian, C exp(−x 2 /a). The probability <strong>of</strong> finding a bullet at x is given by<br />

P (x) = Ce −x2 /a . (4.1)<br />

C here is a normalization factor such that the probability <strong>of</strong> finding a bullet anywhere is 1. i.e.<br />

� ∞<br />

−∞<br />

The probability <strong>of</strong> finding a bullet over a small interval is<br />

� b<br />

a<br />

dxP (x) = 1 (4.2)<br />

dxP (x)〉0. (4.3)<br />

Now suppose we have a bunker with 2 windows between the machine gun and the target such<br />

that the bunker is thick enough that the bullets coming through the windows rattle around a<br />

few times before emerging in random directions. Also, let’s suppose we can “color” the bullets<br />

with some magical (or mundane) means s.t. bullets going through 1 slit are colored “red” and<br />

80


Figure 4.1: Gaussian distribution function<br />

1<br />

0.8<br />

0.6<br />

0.4<br />

0.2<br />

-10 -5 5 10<br />

81


Figure 4.2: Combination <strong>of</strong> two distrubitions.<br />

1<br />

0.8<br />

0.6<br />

0.4<br />

0.2<br />

-10 -5 5 10<br />

bullets going throught the other slit are colored “blue”. Thus the distribution <strong>of</strong> bullets at a<br />

target behind the bunker is now<br />

P12(x) = P1(x) + P2(x) (4.4)<br />

where P1 is the distribution <strong>of</strong> bullets from window 1 (the blue bullets) and P2 the “red” bullets.<br />

Thus, the probability <strong>of</strong> finding a bullet that passed through either 1 or 2 is the sum <strong>of</strong> the<br />

probabilies <strong>of</strong> going through 1 and 2. This is shown in Fig. ??<br />

Now, let’s make an “electron gun” by taking a tungsten filiment heated up so that electrons<br />

boil <strong>of</strong>f and can be accellerated toward a phosphor screen after passing through a metal foil with<br />

a pinhole in the middle We start to see little pin points <strong>of</strong> light flicker on the screen–these are<br />

the individual electron “bullets” crashing into the phosphor.<br />

If we count the number <strong>of</strong> electrons which strike the screen over a period <strong>of</strong> time–just as in<br />

the machine gun experiment, we get a histogram as before. The reason we get a histogram is<br />

slightly different than before. If we make the pin hole smaller, the distribution gets wider. This<br />

is a manifestation <strong>of</strong> the Heisenberg Uncertainty Principle which states:<br />

∆x · δp ≥ ¯h/2 (4.5)<br />

In otherwords, the more I restrict where the electron can be (via the pin hole) the more uncertain<br />

I am about which direction is is going (i.e. its momentum parallel to the foil.) Thus, I wind up<br />

with a distribution <strong>of</strong> momenta leaving the foil.<br />

Now, let’s poke another hole in the foil and consider the distribution <strong>of</strong> electrons on the foil.<br />

Based upon our experience with bullets, we would expect:<br />

P12 = P1 + P2<br />

BUT electrons obey quantum mechanics! And in quantum mechanics we represent a particle<br />

via an amplitude. And one <strong>of</strong> the rules <strong>of</strong> quantum mechanics is that we first add amplitudes<br />

and that probabilities are akin to the intensity <strong>of</strong> the combinded amplitudes. I.e.<br />

P = |ψ1 + ψ2| 2<br />

82<br />

(4.6)<br />

(4.7)


Figure 4.3: Constructive and destructive interference from electron/two-slit experiment. The<br />

superimposed red and blue curves are P1 and P2 from the classical probabilities<br />

1<br />

0.8<br />

0.6<br />

0.4<br />

0.2<br />

-10 -5 5 10<br />

where ψ1 and ψ2 are the complex amplitudes associated with going through hole 1 and hole 2.<br />

Since they are complex numbers,<br />

Thus,<br />

ψ1 = a1 + ib1 = |psi1|e iφ1 (4.8)<br />

ψ2 = a2 + ib2 = |psi2|e iφ2 (4.9)<br />

ψ1 + ψ2 = |ψ1|e iφ1 + |ψ2|e iφ2 (4.10)<br />

|ψ1 + ψ2| 2 = (|ψ1|e iφ1 + |ψ2|e iφ2 )<br />

× (|ψ1|e −iφ1 + |ψ2|e −iφ2 ) (4.11)<br />

P12 = |ψ1| 2 + |ψ2| 2 + 2|ψ1||ψ2| cos(φ1 − φ2)<br />

�<br />

(4.12)<br />

P12 = P1 + P2 + 2<br />

P1P2 cos(φ1 − φ2) (4.13)<br />

In other words, I get the same envelope as before, but it’s modulated by the cos(φ1 − φ2)<br />

“interference” term. This is shown in Fig. 4.3. Here the actual experimental data is shown as a<br />

dashed line and the red and blue curves are the P1 and P2. Just as if a wave <strong>of</strong> electrons struck<br />

the two slits and diffracted (or interfered) with itself. However, we know that electrons come in<br />

definite chunks–we can observe individual specks on the screen–only whole lumps arrive. There<br />

are no fractional electrons.<br />

Conjecture 1 Electrons–being indivisible chunks <strong>of</strong> matter–either go through slit 1 or slit 2.<br />

Assuming Preposition 1 is true, we can divide the electrons into two classes:<br />

1. Those that go through slit 1<br />

83


2. Those that go through slit 2.<br />

We can check this preposition by plugging up hole 1 and we get P2 as the resulting distribution.<br />

Plugging up hole 2, we get P1. Perhaps our preposition is wrong and electrons can be split in<br />

half and half <strong>of</strong> it went through slit 1 and half through slit 2. NO! Perhaps, the electron went<br />

through 1 wound about and went through 2 and through some round-about way made its way<br />

to the screen.<br />

Notice that in the center region <strong>of</strong> P12, P12 > P1+P2, as if closing 1 hole actually decreased the<br />

number <strong>of</strong> electrons going through the other hole. It seems very hard to justify both observations<br />

by proposing that the electrons travel in complicated pathways.<br />

In fact, it is very mysterious. And the more you study quantum mechanics, the more mysterious<br />

it seems. Many ideas have been cooked up which try to get the P12 curve in terms <strong>of</strong><br />

electrons going in complicated paths–all have failed.<br />

Surprisingly, the math is simple (in this case). It’s just adding complex valued amplitudes.<br />

So we conclude the following:<br />

Electrons always arrive in discrete, indivisible chunks–like particles. However, the<br />

probability <strong>of</strong> finding a chunk at a given position is like the distribution <strong>of</strong> the intensity<br />

<strong>of</strong> a wave.<br />

We could conclude that our conjecture is false since P12 �= P1 + P2. This we can test.<br />

Let’s put a laser behind the slits so that an electron going through either slit scatters a bit<br />

<strong>of</strong> light which we can detect. So, we can see flashes <strong>of</strong> light from electrons going through slit<br />

1, flashes <strong>of</strong> light from electrons going through slit 2, but NEVER two flashes at the same<br />

time. Conjecture 1 is true. But if we look at the resulting distribution: we get P12 = P1 + P2.<br />

Measuring which slit the electon passes through destroys the phase information. When we make<br />

a measurement in quantum mechanics, we really disturb the system. There is always the same<br />

amount <strong>of</strong> disturbance because electrons and photons always interact in the same way every time<br />

and produce the same sized effects. These effects “rescatter” the electrons and the phase info is<br />

smeared out.<br />

It is totally impossible to devise an experiment to measure any quantum phenomina without<br />

disturbing the system you’re trying to measure. This is one <strong>of</strong> the most fundimental and perhaps<br />

most disturbing aspects <strong>of</strong> quantum mechanics.<br />

So, once we have accepted the idea that matter comes in discrete bits but that its behavour<br />

is much like that <strong>of</strong> waves, we have to adjust our way <strong>of</strong> thinking about matter and dynamics<br />

away from the classical concepts we are used to dealing with in our ordinary life.<br />

These are the basic building blocks <strong>of</strong> quantum mechanics. Needless to say they are stated in<br />

a rather formal language. However, each postulate has a specific physical reason for its existance.<br />

For any physical theory, we need to be able to say what the system is, how does it move, and<br />

what are the possible outcomes <strong>of</strong> a measurement. These postulates provide a sufficient basis for<br />

the development <strong>of</strong> a consistent theory <strong>of</strong> quantum mechanics.<br />

84


4.0.1 The description <strong>of</strong> a physical state:<br />

The state <strong>of</strong> a physical system at time t is defined by specifying a vector |ψ(t)〉 belonging to a<br />

state space H. We shall assume that this state vector can be normalized to one:<br />

〈ψ|ψ〉 = 1<br />

4.0.2 Description <strong>of</strong> Physical Quantities:<br />

Every measurable physical quantity, A, is described by an operator acting in H; this operator is<br />

an observable.<br />

A consequence <strong>of</strong> this is that any operator related to a physical observable must be Hermitian.<br />

This we can prove. Hermitian means that<br />

〈x|O|y〉 = 〈y|O|x〉 ∗<br />

Thus, if O is a Hermitian operator and 〈O〉 = 〈ψ|O|ψ〉 = λ〈ψ|ψ〉,<br />

Likewise,<br />

If O is Hermitian, we can also write<br />

(4.14)<br />

〈O〉 = 〈x|O|x〉 + 〈x|O|y〉 + 〈y|O|x〉 + 〈y|O|y〉. (4.15)<br />

〈O〉 ∗ = 〈x|O|x〉 ∗ + 〈x|O|y〉 ∗ + 〈y|O|x〉 ∗ + 〈y|O|y〉 ∗<br />

= 〈x|O|x〉 + 〈y|O|x〉 ∗ + 〈x|O|y〉 + 〈y|O|y〉<br />

= 〈O〉 (4.16)<br />

= λ (4.17)<br />

〈ψ|O = λ〈ψ|. (4.18)<br />

which shows that 〈ψ| is an eigenbra <strong>of</strong> O with real eigenvalue λ. Therefore, for an arbitrary ket,<br />

〈ψ|O|φ〉 = λ〈ψ|φ〉 (4.19)<br />

Now, consider eigenvectors <strong>of</strong> a Hermitian operator, |ψ〉 and |φ〉. Obviously we have:<br />

Since O is Hermitian, we also have<br />

Thus, we can write:<br />

O|ψ〉 = λ|ψ〉 (4.20)<br />

O|φ〉 = µ|φ〉 (4.21)<br />

〈ψ|O = λ〈ψ| (4.22)<br />

〈φ|O = µ〈φ| (4.23)<br />

〈φ|O|ψ〉 = λ〈φ|ψ〉 (4.24)<br />

〈φ|O|ψ〉 = µ〈φ|ψ〉 (4.25)<br />

Subtracting the two: (λ − µ)〈φ|ψ〉 = 0. Thus, if λ �= µ, |ψ〉 and |φ〉 must be orthogonal.<br />

85


4.0.3 <strong>Quantum</strong> Measurement:<br />

The only possible result <strong>of</strong> the measurement <strong>of</strong> a physical quantity is one <strong>of</strong> the eigenvalues <strong>of</strong><br />

the corresponding observable. To any physical observable we ascribe an operator, O. The result<br />

<strong>of</strong> a physical measurement must be an eigenvalue, a. With each eigenvalue, there corresponds<br />

an eigenstate <strong>of</strong> O, |φa〉. This function is such that the if the state vector, |ψ(t◦)〉 = |φa〉 where<br />

t◦ corresponds to the time at which the measurement was preformed, O|ψ〉 = a|ψ〉 and the<br />

measurement will yield a.<br />

Suppose the state-function <strong>of</strong> our system is not an eigenfunction <strong>of</strong> the operator we are<br />

interested in. Using the superposition principle, we can write an arbitrary state function as a<br />

linear combination <strong>of</strong> eigenstates <strong>of</strong> O<br />

|ψ(t◦)〉 = �<br />

〈φa|ψ(t◦)〉|φa〉<br />

a<br />

= �<br />

ca|φa〉. (4.26)<br />

a<br />

where the sum is over all eigenstates <strong>of</strong> O. Thus, the probability <strong>of</strong> observing answer a is |ca| 2 .<br />

IF the measurement DOES INDEED YIELD ANSWER a, the wavefunction <strong>of</strong> the system<br />

at an infinitesmimal bit after the measurement must be in an eigenstate <strong>of</strong> O.<br />

4.0.4 The Principle <strong>of</strong> Spectral Decomposition:<br />

|ψ(t + ◦ )〉 = |φa〉. (4.27)<br />

For a non-discrete spectrum: When the physical quantity, A, is measured on a system in a<br />

normalized state |ψ〉, the probability P(an) <strong>of</strong> obtaining the non-degenerate eigenvalue an <strong>of</strong> the<br />

corresponding observable is given by<br />

P(an) = |〈un|ψ〉| 2<br />

where |un〉 is a normalized eigenvector <strong>of</strong> A associated with the eigenvalue an. i.e.<br />

A|un〉 = an|un〉<br />

(4.28)<br />

For a discrete spectrum: the sampe principle applies as in the non-discrete case, except we<br />

sum over all possible degeneracies <strong>of</strong> an<br />

gn�<br />

P(an) = |〈un|ψ〉|<br />

i=1<br />

2<br />

Finally, for the case <strong>of</strong> a continuous spectrum: the probability <strong>of</strong> obtaining a result between<br />

α and α + dα is<br />

dPα = |〈α|ψ〉| 2 dα<br />

86


4.0.5 The Superposition Principle<br />

Let’s formalize the above discussion a bit and write the electron’s state |ψ〉 = a|1〉 + b|2〉 where<br />

|1〉 and |2〉 are “basis states” corresponding to the electron passing through slit 1 or 2. The<br />

coefficients, a and b, are just the complex numbers ψ1 and ψ2 written above. This |ψ〉 is a vector<br />

in a 2-dimensional complex space with unit length since ψ 2 1 + ψ 2 1 = 1. 1<br />

Let us define a Vector Space by defining a set <strong>of</strong> objects {|ψ〉}, an addition rule: |φ〉 =<br />

|ψ〉 + |ψ ′ > which allows us to construct new vectors, and a scaler multiplication rule |φ〉 = a|ψ〉<br />

which scales the length <strong>of</strong> a vector. A non-trivial example <strong>of</strong> a vector space is the x, y plane.<br />

Adding two vectors gives another vector also on the x, y plane and multiplying a vector by a<br />

constant gives another vector pointed in the same direction but with a new length.<br />

The inner product <strong>of</strong> two vectors is written as<br />

〈φ|ψ〉 = (φxφy)<br />

�<br />

ψx<br />

�<br />

(4.29)<br />

=<br />

ψy<br />

φ ∗ xψx + φ ∗ yψy (4.30)<br />

= 〈ψ|φ〉 ∗ . (4.31)<br />

The length <strong>of</strong> a vector is just the inner product <strong>of</strong> the vector with itself, i.e. ψ|ψ〉 = 1 for the<br />

state vector we defined above.<br />

The basis vectors for the slits can be used as a basis for an arbitrary state |ψ〉 by writing it<br />

as a linear combination <strong>of</strong> the basis vectors.<br />

|ψ〉 = ψ1|1〉 + ψ1|1〉 (4.32)<br />

In fact, any vector in the vector space can always be written as a linear combination <strong>of</strong> basis<br />

vectors. This is the superposition principle.<br />

The different ways <strong>of</strong> writing the vector |ψ〉 are termed representations. Often it is easier to<br />

work in one representation than another knowing fully that one can always switch back in forth at<br />

will. Each different basis defines a unique representation. An example <strong>of</strong> a representation are the<br />

unit vectors on the x, y plane. We can also define another orthonormal representation <strong>of</strong> the x, y<br />

plane by introducing the unit vectors |r〉, |θ〉, which define a polar coordinate system. One can<br />

write the vector v = a|x〉 + b|y > as v = √ a 2 + b 2 |r〉 + tan −1 (b/a)|θ〉 or v = r sin θ|x〉 + r cos θ|y〉<br />

and be perfectly correct. Usually experience and insight is the only way to determine a priori<br />

which basis (or representation) best suits the problem at hand.<br />

Transforming between representations is accomplished by first defining an object called an<br />

operator which has the form:<br />

I = �<br />

|i〉〈i|. (4.33)<br />

The sum means “sum over all members <strong>of</strong> a given basis”. For the xy basis,<br />

i<br />

I = |x〉〈x| + |y〉〈y| (4.34)<br />

1 The notation we are introducing here is known as “bra-ket” notation and was invented by Paul Dirac. The<br />

vector |ψ〉 is called a “ket”. The corresponding “bra” is the vector 〈ψ| = (ψ ∗ xψ ∗ y), where the ∗ means complex<br />

conjugation. The notation is quite powerful and we shall use is extensively throughout this course.<br />

87


This operator is called the “idempotent” operator and is similar to multiplying by 1. For example,<br />

We can also write the following:<br />

I|ψ〉 = |1〉〈1|ψ〉 + |2〉〈2|ψ〉 (4.35)<br />

= ψ1|1〉 + ψ2|2〉 (4.36)<br />

= |ψ〉 (4.37)<br />

|ψ〉 = |1〉〈1|ψ〉 + |2〉〈2|ψ〉 (4.38)<br />

The state <strong>of</strong> a system is specified completely by the complex vector |ψ〉 which can be written<br />

as a linear superposition <strong>of</strong> basis vectors spanning a complex vector space (Hilbert space). Inner<br />

products <strong>of</strong> vectors in the space are as defined above and the length <strong>of</strong> any vector in the space<br />

must be finite.<br />

Note, that for state vectors in continuous representations, the inner product relation can be<br />

written as an integral:<br />

�<br />

〈φ|ψ〉 = dqφ ∗ (q)φ(q) (4.39)<br />

and normalization is given by<br />

�<br />

〈ψ|ψ〉 = dq|φ(q)| 2 ≤ ∞. (4.40)<br />

The functions, ψ(q) are termed square integrable because <strong>of</strong> the requirement that the inner<br />

product integral remain finite. The physical motivation for this will become apparent in a<br />

moment when ascribe physical meaning to the mathematical objects we are defining. The class<br />

<strong>of</strong> functions satisfying this requirement are also known as L 2 functions. (L is for Lebesgue,<br />

referring to the class <strong>of</strong> integral.)<br />

The action <strong>of</strong> the laser can also be represented mathematically as an object <strong>of</strong> the form<br />

and<br />

P1 = |1〉〈1|. (4.41)<br />

P2 = |2〉〈2| (4.42)<br />

and note that P1 + P2 = I.<br />

When P1 acts on |ψ〉 it projects out only the |1〉 component <strong>of</strong> |ψ〉<br />

The expectation value <strong>of</strong> an operator is formed by writing:<br />

Let’s evaluate this:<br />

P1|ψ〉 = ψ1|1〉. (4.43)<br />

〈P1〉 = 〈ψ|P1|ψ〉 (4.44)<br />

〈Px〉 = 〈ψ|1〉〈1|ψ〉<br />

= ψ 2 1 (4.45)<br />

88


Similarly for P2.<br />

Part <strong>of</strong> our job is to insure that the operators which we define have physical counterparts.<br />

We defined the projection operator, P1 = |1〉〈1|, knowing that the physical polarization filter<br />

removed all “non” |1〉 components <strong>of</strong> the wave. We could have also written it in another basis,<br />

the math would have been slightly more complex, but the result the same. |ψ1| 2 is a real number<br />

which we presumably could set <strong>of</strong>f to measure in a laboratory.<br />

4.0.6 Reduction <strong>of</strong> the wavepacket:<br />

If a measurement <strong>of</strong> a physical quantity A on the system in the state |ψ〉 yields the result, an,<br />

the state <strong>of</strong> the physical system immediately after the measurement is the normalized projection<br />

Pn|ψ〉 onto the eigen subspace associated with an.<br />

In more plain language, if you observe the system at x, then it is at x. This is perhaps<br />

the most controversial posulate since it implies that the act <strong>of</strong> observing the system somehow<br />

changes the state <strong>of</strong> the system.<br />

Suppose the state-function <strong>of</strong> our system is not an eigenfunction <strong>of</strong> the operator we are<br />

interested in. Using the superposition principle, we can write an arbitrary state function as a<br />

linear combination <strong>of</strong> eigenstates <strong>of</strong> O<br />

|ψ(t◦)〉 = �<br />

〈φa|ψ(t◦)〉|φa〉<br />

a<br />

= �<br />

ca|φa〉. (4.46)<br />

a<br />

where the sum is over all eigenstates <strong>of</strong> O. Thus, the probability <strong>of</strong> observing answer a is |ca| 2 .<br />

IF the measurement DOES INDEED YIELD ANSWER a, the wavefunction <strong>of</strong> the system<br />

at an infinitesmimal bit after the measurement must be in an eigenstate <strong>of</strong> O.<br />

|ψ(t + ◦ )〉 = |φa〉. (4.47)<br />

This is the only postulate which is a bit touchy deals with the reduction <strong>of</strong> the wavepacket<br />

as the result <strong>of</strong> a measurement. On one hand, you could simply accept this as the way one<br />

goes about business and simply state that quantum mechanics is an algorithm for predicting<br />

the outcome <strong>of</strong> experiments and that’s that. It says nothing about the inner workings <strong>of</strong> the<br />

universe. This is what is known as the “Reductionist” view point. In essence, the Reductionist<br />

view point simply wants to know the answer: “How many?”, “How wide?”, “How long?”.<br />

On the other hand, in the Holistic view, quantum mechanics is the underlying physical theory<br />

<strong>of</strong> the universe and say that the process <strong>of</strong> measurement does play an important role in how the<br />

universe works. In otherwords, in the Holist wants the (w)hole picture.<br />

The Reductionist vs. Holist argument has been the subject <strong>of</strong> numerous articles and books<br />

in both the popular and scholarly arenas. We may return to the philosophical discussion, but<br />

for now we will simply take a reductionist view point and first learn to use quantum mechanics<br />

as a way to make physical predictions.<br />

89


4.0.7 The temporal evolution <strong>of</strong> the system:<br />

The time evolution <strong>of</strong> the state vector is given by the Schrödinger equation<br />

i¯h ∂<br />

|ψ(t)〉 = H(t)|ψ(t)〉<br />

∂t<br />

where H(t) is an operator/observable associated withthe total energy <strong>of</strong> the system.<br />

As we shall see, H is the Hamiltonian operator and can be obtained from the classical Hamiltonian<br />

<strong>of</strong> the system.<br />

4.0.8 Dirac <strong>Quantum</strong> Condition<br />

One <strong>of</strong> the crucial aspects <strong>of</strong> any theory is that we need to be able to construct physical observables.<br />

Moreover, we would like to be able to connect the operators and observables in quantum<br />

mechanics to the observables in classical mechanics. At some point there must be a correspondence.<br />

This connection can be made formally by relating what is known as the Poisson bracket<br />

in classical mechanics:<br />

{f(p, q), g(p, q)} = ∂f ∂g ∂g ∂f<br />

−<br />

∂q ∂p ∂q ∂p<br />

which looks a lot like the commutation relation between two linear operators:<br />

(4.48)<br />

[ Â, ˆ B] = Â ˆ B − ˆ BÂ (4.49)<br />

Of course, f(p, q) and g(p, q) are functions over the classical position and momentum <strong>of</strong> the<br />

physical system. For position and momentum, it is easy to show that the classical Poisson<br />

bracket is<br />

{q, p} = 1.<br />

Moreover, the quantum commutation relation between the observable x and p is<br />

[ˆx, ˆp] = i¯h.<br />

Dirac proposed that the two are related and that this relation defines an acceptible set <strong>of</strong><br />

quantum operations.<br />

The quantum mechanical operators ˆ f and ˆg, which in classical theory replace the<br />

classically defined functions f and g, must always be such that the commutator <strong>of</strong> ˆ f<br />

and ˆg corresponds to the Poisson bracket <strong>of</strong> f and g according to<br />

To see how this works, we write the momentum operator as<br />

i¯h{f, g} = [ ˆ f, ˆg] (4.50)<br />

ˆp = ¯h<br />

i<br />

∂<br />

∂x<br />

90<br />

(4.51)


Thus,<br />

Thus,<br />

Let’s see if ˆx and ˆp commute. First <strong>of</strong> all<br />

ˆpψ(x) = ¯h ∂ψ(x)<br />

i ∂x<br />

(4.52)<br />

∂<br />

∂x xf(x) = f(x) + xf ′ (x) (4.53)<br />

[ˆx, ˆp]f(x) = ¯h ∂ ∂<br />

(x f(x) −<br />

i ∂x ∂x xf(x)<br />

= ¯h<br />

i (xf ′ (x) − f(x) − xf ′ (x))<br />

= i¯hf(x) (4.54)<br />

The fact that ˆx and ˆp do not commute has a rather significant consequence:<br />

In other words, if two operators do not commute, one cannot devise and experiment to<br />

simultaneously measure the physical quantities associated with each operator. This in fact limits<br />

the precision in which we can preform any physical measurement.<br />

The principle result <strong>of</strong> the postulates is that the wavefunction or state vector <strong>of</strong> the system<br />

carries all the physical information we can obtain regarding the system and allows us to make<br />

predictions regarding the probable outcomes <strong>of</strong> any experiment. As you may well know, if one<br />

make a series <strong>of</strong> experimental measurements on identically prepared systems, one obtains a<br />

distribution <strong>of</strong> results–usually centered about some peak in the distribution.<br />

When we report data, we usually don’t report the result <strong>of</strong> every single experiment. For<br />

a spectrscopy experiment, we may have made upwards <strong>of</strong> a million individual measurement,<br />

all distributed about some average value. From statistics, we know that the average <strong>of</strong> any<br />

distribution is the expectation value <strong>of</strong> some quantity, in this case x:<br />

for the case <strong>of</strong> a discrete spectra, we would write<br />

�<br />

E(x) = P(x)xdx (4.55)<br />

E[h] = �<br />

n<br />

hnPn<br />

where hn is some value and Pn the number <strong>of</strong> times you got that value normalized so that<br />

�<br />

Pn = 1<br />

n<br />

. In the language above, the hn’s are the possible eigenvalues <strong>of</strong> the h operator.<br />

A similar relation holds in quantum mechanics:<br />

(4.56)<br />

Postulate 4.1 Observable quantities are computed as the expectation value <strong>of</strong> an operator 〈O〉 =<br />

〈ψ|O|ψ〉. The expectation value <strong>of</strong> an operator related to a physical observable must be real.<br />

91


For example, the expectation value <strong>of</strong> ˆx the position operator is computed by the integral<br />

or for the discrete case:<br />

〈x〉 =<br />

� +∞<br />

−∞<br />

ψ ∗ (x)xψ(x)dx.<br />

〈O〉 = �<br />

on|〈n|ψ〉| 2 .<br />

n<br />

Of course, simply reporting the average or expectation values <strong>of</strong> an experiment is not enough,<br />

the data is usually distributed about either side <strong>of</strong> this value. If we assume the distribution is<br />

Gaussian, then we have the position <strong>of</strong> the peak center xo = 〈x〉 as well as the width <strong>of</strong> the<br />

Gaussian σ 2 .<br />

The mean-squared width or uncertainty <strong>of</strong> any measurement can be computed by taking<br />

σ 2 A = 〈(A − 〈A〉)〉.<br />

In statistical mechanics, this the fluctuation about the average <strong>of</strong> some physical quantity, A. In<br />

quantum mechanics, we can push this definition a bit further.<br />

Writing the uncertainty relation as<br />

σ 2 A = 〈(A − 〈A〉)(A − 〈A〉)〉 (4.57)<br />

= 〈ψ|(A − 〈A〉)(A − 〈A〉)|ψ〉 (4.58)<br />

= 〈f|f〉 (4.59)<br />

where the new vector |f〉 is simply short hand for |f〉 = (A − 〈A〉)|ψ〉. Likewise for a different<br />

operator B<br />

σ 2 B = = 〈ψ|(B − 〈B〉)(B − 〈B〉)|ψ〉 (4.60)<br />

We now invoke what is called the Schwartz inequality<br />

= 〈g|g〉. (4.61)<br />

σ 2 Aσ 2 B = 〈f|f〉〈g|g〉 ≥ |〈f|g〉| 2<br />

So if we write 〈f|g〉 as a complex number, then<br />

So we conclude<br />

|〈f|g〉| 2 = |z| 2<br />

σ 2 Aσ 2 B ≥<br />

= ℜ(z) 2 + ℑ(z) 2<br />

≥ ℑ(z) 2 �<br />

1<br />

=<br />

2i (z − z∗ )<br />

� �2<br />

1<br />

≥ (〈f|g〉 − 〈g|f〉)<br />

2i<br />

� �2<br />

1<br />

(〈f|g〉 − 〈g|f〉)<br />

2i<br />

92<br />

�2<br />

(4.62)<br />

(4.63)<br />

(4.64)


Now, we reinsert the definitions <strong>of</strong> |f〉 and |g〉.<br />

Likewise<br />

Combining these results, we obtain<br />

〈f|g〉 = 〈ψ|(A − 〈A〉)(B − 〈B〉)|ψ〉<br />

= 〈ψ|(AB − 〈A〉B − A〈B〉 + 〈A〉〈B〉)|ψ〉<br />

= 〈ψ|AB|ψ〉 − 〈A〉〈ψ|B|ψ〉 − 〈B〉〈ψ|A|ψ〉 + 〈A〉〈B〉<br />

= 〈AB〉 − 〈A〉〈B〉 (4.65)<br />

〈g|f〉 = 〈BA〉 − 〈A〉〈B〉. (4.66)<br />

〈f|g〉 − 〈g|f〉 = 〈AB〉 − 〈BA〉 = 〈AB − BA〉 = 〈[A, B]〉. (4.67)<br />

So we finally can conclude that the general uncertainty product between any two operators is<br />

given by<br />

σ 2 Aσ 2 � �2<br />

1<br />

B = 〈[A, B]〉<br />

(4.68)<br />

2i<br />

This is commonly referred to as the Generalized Uncertainty Principle. What is means is that<br />

for any pair <strong>of</strong> observables whose corresponding operators do not commute there will always be<br />

some uncertainty in making simultaneous measurements. In essence, if you try to measure two<br />

non-commuting properties simultaneously, you cannot have an infinitely precise determination <strong>of</strong><br />

both. A precise determination <strong>of</strong> one implies that you must give up some certainty in the other.<br />

In the language <strong>of</strong> matrices and linear algebra this implies that if two matrices do not commute,<br />

then one can not bring both matrices into diagonal form using the same transformation<br />

matrix. in other words, they do not share a common set <strong>of</strong> eigenvectors. Matrices which do<br />

commute share a common set <strong>of</strong> eigenvectors and the transformation which diagonalizes one will<br />

also diagonalize the other.<br />

Theorem 4.1 If two operators A and B commute and if |ψ〉 is an eigenvector <strong>of</strong> A, then B|ψ〉<br />

is also an eigenvector <strong>of</strong> A with the same eigenvalue.<br />

Pro<strong>of</strong>: If |ψ〉 is an eigenvector <strong>of</strong> A, then A|ψ〉 = a|ψ〉. Thus,<br />

Assuming A and B commute, i.e. [A, B] = AB − BA = 0,<br />

Thus, (B|ψ〉) is an eigenvector <strong>of</strong> A with eigenvalue, a.<br />

BA|ψ〉 = aB|ψ〉 (4.69)<br />

AB|ψ〉 = a(B|ψ〉) (4.70)<br />

Exercise 4.1 1. Show that matrix multiplication is associative, i.e. A(BC) = (AB)C, but<br />

not commutative (in general), i.e. BC �= CB<br />

2. Show that (A + B)(A − B) = A 2 + B 2 only <strong>of</strong> A and B commute.<br />

3. Show that if A and B are both Hermitian matrices, AB + BA and i(AB − BA) are also<br />

Hermitian. Note that Hermitian matrices are defined such that Aij = A ∗ ji where ∗ denotes<br />

complex conjugation.<br />

93


4.1 Dirac Notation and Linear Algebra<br />

Part <strong>of</strong> the difficulty in learning quantum mechanics comes from fact that one must also learn a<br />

new mathematical language. It seems very complex from the start. However, the mathematical<br />

objects which we manipulate actually make life easier. Let’s explore the Dirac notation and the<br />

related mathematics.<br />

We have stated all along that the physical state <strong>of</strong> the system is wholly specified by the<br />

state-vector |ψ〉 and that the probability <strong>of</strong> finding a particle at a given point x is obtained via<br />

|ψ(x)| 2 . Say at some initial time |ψ〉 = |s〉 where s is some point along the x axis. Now, the<br />

amplitude to find the particle at some other point is 〈x|s〉. If something happens between the<br />

two points we write<br />

〈x|operator describing process|s〉 (4.71)<br />

The braket is always read from right to left and we interpret this as the amplitude for“starting<br />

<strong>of</strong>f at s, something happens, and winding up at i”. An example <strong>of</strong> this is the Go function in the<br />

homework. Here, I ask “what is the amplitude for a particle to start <strong>of</strong>f at x and to wind up at<br />

x ′ after some time interval t?”<br />

Another Example: Electrons have an intrinsic angular momentum called “spin” . Accordingly,<br />

they have an associated magnetic moment which causes electrons to align with or against an<br />

imposed magnetic field (eg.this gives rise to ESR). Lets say we have an electron source which<br />

produces spin up and spin down electrons with equal probability. Thus, my initial state is:<br />

|i〉 = a|+〉 + b|−〉 (4.72)<br />

Since I’ve stated that P (a) = P (b), |a| 2 = |b| 2 . Also, since P (a) + P (b) = 1, a = b = 1/ √ 2.<br />

Thus,<br />

|i〉 = 1<br />

√ 2 (|+〉 + |−〉) (4.73)<br />

Let’s say that the spin ups can be separated from the spin down via a magnetic field, B and we<br />

filter <strong>of</strong>f the spin down states. Our new state is |i ′ 〉 and is related to the original state by<br />

4.1.1 Transformations and Representations<br />

〈i ′ |i〉 = a〈+|+〉 + b〈+|−〉 = a. (4.74)<br />

If I know the amplitudes for |ψ〉 in a representation with a basis |i〉 , it is always possible to<br />

find the amplitudes describing the same state in a different basis |µ〉. Note, that the amplitude<br />

between two states will not change. For example:<br />

also<br />

|a〉 = �<br />

|i〉〈i|a〉 (4.75)<br />

i<br />

|a〉 = �<br />

|µ〉〈µ|a〉 (4.76)<br />

µ<br />

94


Therefore,<br />

and<br />

〈µ|a〉 = �<br />

〈µ|i〉〈i|a〉 (4.77)<br />

i<br />

〈i|a〉 = �<br />

〈i|µ〉〈µ|a〉. (4.78)<br />

µ<br />

Thus, the coefficients in |µ〉 are related to the coefficients in |i〉 by 〈µ|i〉 = 〈i|µ〉 ∗ . Thus, we can<br />

define a transformation matrix Sµi as<br />

and a set <strong>of</strong> vectors<br />

Thus, we can see that<br />

Thus,<br />

Now, we can also write<br />

Siµ =<br />

Sµi =<br />

⎡<br />

⎢<br />

⎣<br />

⎡<br />

⎢<br />

⎣<br />

〈µ|i〉 〈µ|j〉 〈µ|k〉<br />

〈ν|i〉 〈ν|j〉 〈ν|k〉<br />

〈λ|i〉 〈λ|j〉 〈λ|k〉<br />

ai =<br />

aµ =<br />

⎡<br />

⎢<br />

⎣<br />

⎡<br />

⎢<br />

⎣<br />

aµ = �<br />

〈i|a〉<br />

〈j|a〉<br />

〈k|a〉<br />

〈µ|a〉<br />

〈ν|a〉<br />

〈λ|a〉<br />

i<br />

⎤<br />

Sµiai<br />

⎤<br />

⎥<br />

⎦ (4.79)<br />

⎥<br />

⎦ (4.80)<br />

⎤<br />

〈µ|i〉 ∗ 〈µ|j〉 ∗ 〈µ|k〉 ∗<br />

〈ν|i〉 〈ν|j〉 〈ν|k〉 ∗<br />

〈λ|i〉 ∗ 〈λ|j〉 ∗ 〈λ|k〉 ∗<br />

ai = �<br />

µ<br />

Siµaµ<br />

⎥<br />

⎦ (4.81)<br />

⎤<br />

⎥<br />

⎦ = S ∗ µi<br />

Since 〈i|µ〉 = 〈µ|i〉 ∗ , S is the Hermitian conjugate <strong>of</strong> S. So we write<br />

S = S †<br />

95<br />

(4.82)<br />

(4.83)<br />

(4.84)<br />

(4.85)<br />

(4.86)


So in short;<br />

thus<br />

S † = (S T ) ∗<br />

(S † )ij = S ∗ ji<br />

and S is called a unitary transformation matrix.<br />

4.1.2 Operators<br />

(4.87)<br />

(4.88)<br />

(4.89)<br />

a = Sa = SSa = S † Sa (4.90)<br />

S † S = 1 (4.91)<br />

A linear operator  maps a vector in space H on to another vector in the same space. We can<br />

write this in a number <strong>of</strong> ways:<br />

or<br />

Linear operators have the property that<br />

|φ〉 Â<br />

↦→ |χ〉 (4.92)<br />

|χ〉 = Â|φ〉 (4.93)<br />

Â(a|φ〉 + b|χ〉) = aÂ|φ〉 + bÂ|χ〉 (4.94)<br />

Since superposition is rigidly enforced in quantum mechanics, all QM operators are linear operators.<br />

The Matrix Representation <strong>of</strong> an operator is obtained by writing<br />

Aij = 〈i| Â|j〉 (4.95)<br />

For example: Say we know the representation <strong>of</strong> A in the |i〉 basis, we can then write<br />

Thus,<br />

�<br />

|χ〉 = Â|φ〉 = Â|i〉〈i|φ〉 = �<br />

|j〉〈j|χ〉 (4.96)<br />

i<br />

j<br />

〈j|χ〉 = �<br />

〈j|A|i〉〈i|φ〉 (4.97)<br />

We can keep going if we want by continuing to insert 1’s where ever we need.<br />

i<br />

96


The matrix A is Hermitian if A = A † . If it is Hermitian, then I can always find a basis |µ〉 in<br />

which it is diagonal, i.e.<br />

So, what is Â|µ〉 ?<br />

Aµν = aµδµν<br />

(4.98)<br />

Â|µ〉 = �<br />

|i〉〈i|A|j〉〈j|µ〉 (4.99)<br />

ij<br />

= �<br />

|i〉Aijδjµ<br />

ij<br />

= �<br />

|i〉Aiµ<br />

i<br />

= �<br />

|i〉aµδiµ<br />

i<br />

(4.100)<br />

(4.101)<br />

(4.102)<br />

(4.103)<br />

(4.104)<br />

(4.105)<br />

(4.106)<br />

= aµ|µ〉 (4.107)<br />

An important example <strong>of</strong> this is the “time-independent” Schroedinger Equation:<br />

which we spend some time in solving above.<br />

Finally, if Â|φ〉 = |χ〉 then 〈φ|A† = 〈χ|.<br />

4.1.3 Products <strong>of</strong> Operators<br />

An operator product is defined as<br />

ˆH|ψ〉 = E|ψ〉 (4.108)<br />

( Â ˆ B)|ψ〉 = Â[ ˆ B|ψ〉] (4.109)<br />

where we operate in order from right to left. We proved that in general the ordering <strong>of</strong> the<br />

operations is important. In other words, we cannot in general write  ˆ B = ˆ BÂ. An example <strong>of</strong><br />

this is the position and momentum operators. We have also defined the “commutator”<br />

[ Â, ˆ B] = Â ˆ B − ˆ BÂ. (4.110)<br />

Let’s now briefly go over how to perform algebraic manipulations using operators and commutators.<br />

These are straightforward to prove<br />

97


1. [ Â, ˆ B] = −[ ˆ B, Â]<br />

2. [ Â, Â] = −[Â, Â] = 0<br />

3. [ Â, ˆ BĈ] = [Â, ˆ B] Ĉ + ˆ B[ Â, Ĉ]<br />

4. [ Â, ˆ B + Ĉ] = [Â, ˆ B] + [ Â, Ĉ]<br />

5. [ Â, [ ˆ B, Ĉ]] + [ ˆ B, [ Ĉ, Â]] + [Ĉ, [Â, ˆ B]] = 0 (Jacobi Identity)<br />

6. [ Â, ˆ B] † = [ † , ˆ B † ]<br />

4.1.4 Functions Involving Operators<br />

Another property <strong>of</strong> linear operators is that the inverse operator always can be found. I.e. if<br />

|χ〉 = Â|φ〉 then there exists another operator ˆ B such that |φ〉 = ˆ B|χ〉. In other words ˆ B = Â−1 .<br />

We also need to know how to evaluate functions <strong>of</strong> operators. Say we have a function, F (z)<br />

which can be expanded as a series<br />

Thus, by analogy<br />

For example, take exp( Â)<br />

thus<br />

exp(x) =<br />

∞�<br />

F (z) = fnz<br />

n=0<br />

n<br />

F ( Â) =<br />

∞�<br />

n=0<br />

(4.111)<br />

∞�<br />

fnÂn . (4.112)<br />

n=0<br />

x n<br />

n! = 1 + x + x2 /2 + · · · (4.113)<br />

exp( Â) =<br />

If  is Hermitian, then F (Â) is also Hermitian. Also, note that<br />

Likewise, if<br />

then<br />

∞�<br />

n=0<br />

 n<br />

n!<br />

[ Â, F (Â)] = 0.<br />

(4.114)<br />

Â|φa〉 = a|φa〉 (4.115)<br />

 n |φa〉 = a n |φa〉. (4.116)<br />

98


Thus, we can show that<br />

F ( Â)|φa〉 = �<br />

n<br />

fn Ân |φa〉 (4.117)<br />

(4.118)<br />

= �<br />

fna n |φa〉 (4.119)<br />

n<br />

(4.120)<br />

= F (a) (4.121)<br />

Note, however, that care must be taken when we evaluate F ( Â + ˆ B) if the two operators<br />

do not commute. We ran into this briefly in breaking up the propagator for the Schroedinger<br />

Equation in the last lecture (Trotter Product). For example,<br />

exp( Â + ˆ B) �= exp( Â) exp( ˆ B) (4.122)<br />

unless [ Â, ˆ B] = 0. One can derive, however, a useful formula (Glauber)<br />

exp( Â + ˆ B) = exp( Â) exp( ˆ B) exp(−[ Â, ˆ B]/2) (4.123)<br />

Exercise 4.2 Let H be the Hamiltonian <strong>of</strong> a physical system and |φn〉 the solution <strong>of</strong><br />

1. For an arbitrary operator, Â, show that<br />

2. Let<br />

(a) Compute [H, ˆp], [H, ˆx], and [H, ˆxˆp].<br />

(b) Show 〈φn|ˆp|φn〉 = 0.<br />

H|φn〉 = En|φn〉 (4.124)<br />

〈φn|[ Â, H]|φn〉 = 0 (4.125)<br />

H = 1<br />

2m ˆp2 + V (ˆx) (4.126)<br />

(c) Establish a relationship between the average <strong>of</strong> the kinetic energy given by<br />

and the average force on a particle given by<br />

Ekin = 〈φn| ˆp2<br />

2m |φn〉 (4.127)<br />

∂V (x)<br />

F = 〈φn|ˆx<br />

∂x |φn〉. (4.128)<br />

Finally, relate the average <strong>of</strong> the potential for a particle in state |φn〉 to the average<br />

kinetic energy.<br />

99


Exercise 4.3 Consider the following Hamiltonian for 1d motion with a potential obeying a simple<br />

power-law<br />

H = p2<br />

+ αxn<br />

2m<br />

where α is a constant and n is an integer. Calculate<br />

(4.129)<br />

〈A〉 = 〈ψ|[xp, H]||ψ〉 (4.130)<br />

and use the result to relate the average potential energy to the average kinetic energy <strong>of</strong> the<br />

system.<br />

4.2 Constants <strong>of</strong> the Motion<br />

In a dynamical system (quantum, classical, or otherwise) a constant <strong>of</strong> the motion is any quantity<br />

such that<br />

For quantum systems, this means that<br />

∂tA = 0. (4.131)<br />

[A, H] = 0 (4.132)<br />

(What’s the equivalent relation for classical systems?) In other words, any quantity which<br />

commutes with H is a constant <strong>of</strong> the motion. Furthermore, for any conservative system (in<br />

which there is no net flow <strong>of</strong> energy to or from the system),<br />

From Eq.4.131, we can write that<br />

[H, H] = 0. (4.133)<br />

∂t〈A〉 = ∂t〈ψ(t)|A|ψ(t)〉 (4.134)<br />

Since [A, H] = 0, we know that if the state |φn〉 is an eigenstate <strong>of</strong> H,<br />

then<br />

H|φn〉 = En|φn〉 (4.135)<br />

A|φn〉 = an|φn〉 (4.136)<br />

The an are <strong>of</strong>ten referred to as “good quantum numbers”. What are some constants <strong>of</strong> the<br />

motion for systems that we have studied thus far? (Bonus: how are constants <strong>of</strong> motion related<br />

to particular symmetries <strong>of</strong> the system?)<br />

A state which is in an eigenstate <strong>of</strong> H it’s also in an eigenstate <strong>of</strong> A. Thus, I can simultaneously<br />

measure quantities associated with H and A. Also, after I measure with A, the system remains<br />

in a the original state.<br />

100


4.3 Bohr Frequency and Selection Rules<br />

What if I have another operator, B, which does not commute with H? What is 〈B(t)〉? This we<br />

can compute by first writing<br />

|ψ(t)〉 = �<br />

cne −iEnt/¯h |φn〉. (4.137)<br />

Then<br />

= �<br />

Let’s define the “Bohr Frequency” as ωnm = (En − Em)/¯h.<br />

〈B(t)〉 = �<br />

n<br />

n<br />

〈B(t)〉 = 〈ψ|B|ψ(t)〉 (4.138)<br />

cnc ∗ me −i(En−Em)t/¯h 〈φm|B|φn〉. (4.139)<br />

n<br />

cnc ∗ me −iωnmt 〈φm|B|φn〉. (4.140)<br />

Now, the observed expectation value <strong>of</strong> B oscillates in time at number <strong>of</strong> frequencies corresponding<br />

to the energy differences between the stationary states. The matrix elements Bnm =<br />

〈φm|B|φn〉 do not change with time. Neither do the coefficients, {cn}. Thus, let’s write<br />

B(ω) = cnc ∗ m〈φm|B|φn〉δ(ω − ωnm) (4.141)<br />

and transform the discrete sum into an continuous integral<br />

〈B(t)〉 = 1<br />

� ∞<br />

e<br />

2π 0<br />

−iωt B(ω) (4.142)<br />

where B(ω) is the power spectral <strong>of</strong> B. In other words, say I monitor < B(t) > with my<br />

instrument for a long period <strong>of</strong> time, then take the Fourier Transform <strong>of</strong> the time-series. I get<br />

the power-spectrum. What is the power spectrum for a set <strong>of</strong> discrete frequencies: If I observe<br />

the time-sequence for an infinite amount <strong>of</strong> time, I will get a series <strong>of</strong> discretely spaced sticks<br />

along the frequency axis at precisely the energy difference between the n and m states. The<br />

intensity is related to the probability <strong>of</strong> making a transition from n to m under the influence<br />

<strong>of</strong> B. Certainly, some transitions will not be allowed because 〈φn|B|φm〉 = 0. These are the<br />

“selection rules”.<br />

We now prove an important result regarding the integrated intensity <strong>of</strong> a series <strong>of</strong> transitions:<br />

Exercise 4.4 Prove the Thomas-Reiche-Kuhn sum rule: 2<br />

� 2m|xn0| 2<br />

n<br />

¯h 2 (En − Eo) = 1 (4.143)<br />

where the sum is over a compete set <strong>of</strong> states, |ψn〉 <strong>of</strong> energy En <strong>of</strong> a particle <strong>of</strong> mass m which<br />

moves in a potential; |ψo〉 represents a bound state, and xn0 = 〈ψn|x|ψo〉. (Hint: use the commutator<br />

identity: [x, [x, H]] = ¯h 2 /m)<br />

2 This is perhaps one <strong>of</strong> the most important results <strong>of</strong> quantum mechanics since it is gives the total spectral<br />

intensity for a series <strong>of</strong> transitions. c.f Bethe and Jackiw for a great description <strong>of</strong> sum-rules.<br />

101


Figure 4.4: The diffraction function sin(x)/x<br />

1<br />

0.8<br />

0.6<br />

0.4<br />

0.2<br />

-20 -10 10 20<br />

-0.2<br />

4.4 Example using the particle in a box states<br />

What are the constants <strong>of</strong> motion for a particle in a box?<br />

Recall that the energy levels and wavefunctions for this system are<br />

φn(x) =<br />

2ma2 (4.144)<br />

�<br />

2<br />

a sin(nπ x) (4.145)<br />

a<br />

En = n2 π 2 ¯h 2<br />

Say our system in in the nth state. What’s the probability <strong>of</strong> measuring the momentum and<br />

obtaining a result between p and p + dp?<br />

where<br />

Pn(p)dp = |φn(p)| 2 dp (4.146)<br />

�<br />

�<br />

1 a 2<br />

φn(p) = √ dx<br />

2π¯h 0 a sin(nπx/a)e−ipx/¯h<br />

(4.147)<br />

= 1<br />

�<br />

i(nπ/a−p/¯h)a 1 e − 1<br />

√<br />

2i 2π¯h i(nπ/a − p/¯h) − e−i(nπ/a−p/¯h)a �<br />

− 1<br />

(4.148)<br />

−i(nπ/a − p/¯h)<br />

= 1<br />

�<br />

a<br />

2i π¯h exp i(nπ/2 − pa/(2¯h))[F (p − nπ¯h/2) + (−1)n+1F (p + nπ¯h/a)] (4.149)<br />

Where the F (p) are “diffraction functions”<br />

F (p) = sin(pa/(2¯h))<br />

pa/(2¯h)<br />

(4.150)<br />

Note that the width 4π¯h/a does not change as I change n. Nor does the amplitude. However,<br />

note that (F (x + n) ± FF (x − n)) 2 is always an even function <strong>of</strong> x. Thus, we can say<br />

〈p〉n =<br />

� +∞<br />

−∞<br />

Pn(p)pdp = 0 (4.151)<br />

102


We can also compute:<br />

〈p 2 〉 = ¯h 2<br />

� � �<br />

∞ �<br />

�∂φn(x)<br />

�2<br />

�<br />

dx � �<br />

0 � ∂x �<br />

= ¯h 2<br />

� a<br />

0<br />

=<br />

2<br />

a<br />

(4.152)<br />

� �<br />

nπ 2<br />

cos(nπx/a)dx (4.153)<br />

a<br />

� nπ¯h<br />

a<br />

� 2<br />

Thus, the RMS deviation <strong>of</strong> the momentum:<br />

∆pn =<br />

= 2mEn<br />

�<br />

〈p 2 〉n − 〈p〉 2 n = nπ¯h<br />

a<br />

(4.154)<br />

(4.155)<br />

Thus, as n increases, the relative accuracy at which we can measure p increases due the fact<br />

that we can resolve the wavefunction into two distinct peaks corresponding to the particle either<br />

going to the left or to the right. ∆p increases due to the fact that the two possible choices for<br />

the measurement are becoming farther and farther apart and hence reflects the distance between<br />

the two most likely values.<br />

4.5 Time Evolution <strong>of</strong> Wave and Observable<br />

Now, suppose we put our system into a superposition <strong>of</strong> box-states:<br />

|ψ(0)〉 = 1<br />

√ 2 (|φ1〉 + |φ2〉) (4.156)<br />

What is the time evolution <strong>of</strong> this state? We know the eigen-energies, so we can immediately<br />

write:<br />

|ψ(t)〉 = 1<br />

√ 2 (exp(−iE1t/¯h)|φ1〉 + exp(−iE2t/¯h)|φ2〉) (4.157)<br />

Let’s factor out a common phase factor <strong>of</strong> e −iE1t/¯h and write this as<br />

|ψ(t)〉 ∝ 1<br />

√ 2 (|φ1〉 + exp(−i(E2 − E1)t/¯h)|φ2〉) (4.158)<br />

and call (E2 − E1)/¯h = ω21 the Bohr frequency.<br />

where<br />

|ψ(t)〉 ∝ 1<br />

√ 2 (|φ1〉 + exp(−iω21t)|φ2〉) (4.159)<br />

ω21 = 3π2¯h . (4.160)<br />

2ma2 103


The phase factor is relatively unimportant and cancels out when I make a measurement. Eg.<br />

the prob. density:<br />

|ψ(x, t)| 2 = |〈x|ψ(t)〉| 2<br />

(4.161)<br />

(4.162)<br />

= 1<br />

2 φ2 1(x) + 1<br />

2 φ2 2(x) + φ1(x)φ2(x) cos(ω21t) (4.163)<br />

Now, let’s compute 〈x(t)〉 for the two state system. To do so, let’s first define x ′ = x − a/2<br />

as the center <strong>of</strong> the well to make the integrals easier. The first two are easy:<br />

〈φ1|x ′ |φ1〉 ∝<br />

〈φ2|x ′ |φ2〉 ∝<br />

which we can do by symmetry. Thus,<br />

Thus,<br />

� a<br />

0<br />

� a<br />

0<br />

dx(x − a/2) sin 2 (πx/a) = 0 (4.164)<br />

dx(x − a/2) sin 2 (2πx/a) = 0 (4.165)<br />

〈x ′ (t)〉 = Re{e −iω21t 〈φ1|x ′ |φ2〉} (4.166)<br />

〈φ1|x ′ |φ2〉 = 〈φ1|x|φ2〉 − (a/2)〈φ1|φ2〉 (4.167)<br />

= 2<br />

� a<br />

dxx sin(πx/a) sin(2πx/a) (4.168)<br />

a 0<br />

= − 16a<br />

9π2 (4.169)<br />

〈x(t)〉 = a 16a<br />

−<br />

2 9π2 cos(ω21t) (4.170)<br />

Compare this to the classical trajectory. Also, what about 〈E(t)〉?<br />

4.6 “Unstable States”<br />

So far in this course, we have been talking about systems which are totally isolated from the<br />

rest <strong>of</strong> the universe. In these systems, there is no influx or efflux <strong>of</strong> energy and all our dynamics<br />

are governed by the three principle postulates I mentioned a the start <strong>of</strong> the lecture. IN essence,<br />

if at t = 0 I prepare the system in an eigenstate <strong>of</strong> H, then for all times later, it’s still in that<br />

state (to within a phase factor). Thus, in a strictly conservative system, a system prepared in<br />

an eigenstate <strong>of</strong> H will remain in an eigenstate forever.<br />

However, this is not exactly what is observed in nature. We know from experience that atoms<br />

and molecules, if prepared in an excited state (say via the absorption <strong>of</strong> a photon) can relax<br />

104


ack to the ground state or some lower state via the emission <strong>of</strong> a photon or a series <strong>of</strong> photons.<br />

Thus, these eigenstates are “unstable”.<br />

What’s wrong here? The problem is not so much to do with what is wrong with our description<br />

<strong>of</strong> the isolated system, it has to do with full description is not included. A isolated atom or<br />

molecule can still interact with the electro-magnetic field (unless we do some tricky confinement<br />

experiments). Thus, there is always some interaction with an outside “environment”. Thus,<br />

while it is totally correct to describe the evolution <strong>of</strong> the global system in terms <strong>of</strong> some global<br />

“atom” + “environment” Hamiltonian, it it NOT totally rigorous to construct a Hamiltonian<br />

which describes only part <strong>of</strong> the story. But, as the great <strong>Pr<strong>of</strong></strong>. Karl Freed (at U. Chicago) once<br />

told me “Too much rigor makes rigor mortis”.<br />

Thankfully, the coupling between an atom and the electromagnetic field is pretty weak. Each<br />

photon emission probability is weighted by the fine-structure constant, α ≈ 1/137. Thus a 2<br />

photon process is weighted by α 2 . Thus, the isolated system approximation is pretty good. Also,<br />

we can pretty much say that most photon emission processes occur as single photon events.<br />

Let’s play a bit “fast and loose” with this idea. We know from experience that if we prepare<br />

the system in an excited state at t = 0, the probability <strong>of</strong> finding it still in the excited state at<br />

some time t later, is<br />

P (t) = e −t/τ<br />

(4.171)<br />

where τ is some time constant which we’ll take as the lifetime <strong>of</strong> the state. One way to “prove”<br />

this relation is to go back to Problem Set 0. Let’s say we have a large number <strong>of</strong> identical systems<br />

N , each prepared in the excited state at t = 0. At time t, there are<br />

N(t) = N e −t/τ<br />

(4.172)<br />

systems in the excited state. Between time t and t + dt a certain number, dn(t) will leave the<br />

excited state via photon emission.<br />

Thus,<br />

dn(t) = N(t) − N(t + dt) = − dN(t)<br />

dt = N(t)dt<br />

dt τ<br />

dn(t)<br />

N(t)<br />

= dt<br />

τ<br />

Thus, 1/τ is the probability per unit time for leaving the unstable state.<br />

The Avg. time a system spends in the unstable state is given by:<br />

1<br />

τ<br />

(4.173)<br />

(4.174)<br />

� ∞<br />

dtte<br />

0<br />

−t/τ = τ (4.175)<br />

For a stable state P (t) = 1 thus, τ → ∞.<br />

The time a system spends in the state is independent <strong>of</strong> its history. This is a characteristic <strong>of</strong><br />

an unstable state. (Also has to do with the fact that the various systems involved to not interact<br />

with each other. )<br />

105


Finally, according to the time-energy uncertainty relation:<br />

∆Eτ ≈ ¯h. (4.176)<br />

Thus, an unstable system has an intrinsic “energy width” associated with the finite time the<br />

systems spends in the state.<br />

For a stable state:<br />

and<br />

for real energies.<br />

What if I instead write: E ′ n = En − i¯hγn/¯h? Then<br />

Thus,<br />

|ψ(t)〉 = e −iEnt/¯h |φn〉 (4.177)<br />

Pn(t) = |e −iEnt/¯h | 2 = 1 (4.178)<br />

Pn(t) = |e −iEnt/¯h e −γn/2t | 2 = e −γnt<br />

γn = 1/τn<br />

(4.179)<br />

(4.180)<br />

is the “Energy Width” <strong>of</strong> the unstable state.<br />

The surprising part <strong>of</strong> all this is that in order to include dissipative effects (photon emission,<br />

etc..) the Eigenvalues <strong>of</strong> H become complex. In other words, the system now evolves under a<br />

non-hermitian Hamiltonian! Recall the evolution operator for an isolated system:<br />

U(t) = e −iHt/¯h (4.181)<br />

(4.182)<br />

U † (t) = e iHt/¯h (4.183)<br />

where the first is the forward evolution <strong>of</strong> the system and the second corresponds to the backwards<br />

evolution <strong>of</strong> the system. Thus, Unitarity is thus related to the time-reversal symmetry<br />

<strong>of</strong> conservative systems. The inclusion <strong>of</strong> an “environment” breaks the intrinsic time-reversal<br />

symmetry <strong>of</strong> an isolated system.<br />

4.7 Problems and Exercises<br />

Exercise 4.5 Find the eigenvalues and eigenvectors <strong>of</strong> the matrix:<br />

⎛<br />

⎜<br />

M = ⎜<br />

⎝<br />

0 0 0 1<br />

0 0 1 0<br />

0 1 0 0<br />

1 0 0 0<br />

106<br />

⎞<br />

⎟<br />

⎠<br />

(4.184)


Solution: You can either do this the hard way by solving the secular determinant and then<br />

finding the eigenvectors by Gramm-Schmidt orthogonalization, or realize that since M = M −1<br />

and M = M † , M is a unitary matrix, this, its eigenvalues can only be ± 1. Furthermore, since<br />

the trace <strong>of</strong> M is 0, the sum <strong>of</strong> the eigenvalues must be 0 as well. Thus, λ = (1, 1, −1, −1) are<br />

the eigenvalues. To get the eigenvectors, consider the following. Let φmu be an eigenvector <strong>of</strong><br />

M, thus,<br />

⎛<br />

⎜<br />

φµ = ⎜<br />

⎝<br />

Since Mφµ = λµφmu, x1 = λµx4 and x2 = λµx3 Thus, 4 eigenvectors are<br />

for λ = (−1, −1, 1, 1).<br />

⎛<br />

⎜<br />

⎝<br />

−1<br />

0<br />

0<br />

1<br />

⎞<br />

⎛<br />

⎟<br />

⎠ ,<br />

⎜<br />

⎝<br />

0<br />

1<br />

−1<br />

0<br />

⎞<br />

x1<br />

x2<br />

x3<br />

x4<br />

⎛<br />

⎟<br />

⎠ ,<br />

⎜<br />

⎝<br />

Exercise 4.6 Let λi be the eigenvalues <strong>of</strong> the matrix:<br />

calculate the sums:<br />

and<br />

H =<br />

⎛<br />

⎜<br />

⎝<br />

⎞<br />

⎟ . (4.185)<br />

⎠<br />

1<br />

0<br />

0<br />

1<br />

⎞<br />

2 −1 −3<br />

−1 1 2<br />

−3 2 3<br />

3�<br />

λi<br />

i<br />

3�<br />

λ<br />

i<br />

2 i<br />

⎛<br />

⎟<br />

⎠ ,<br />

⎜<br />

⎝<br />

⎞<br />

0<br />

1<br />

1<br />

0<br />

⎞<br />

⎟<br />

⎠<br />

(4.186)<br />

⎟<br />

⎠ (4.187)<br />

Hint: use the fact that the trace <strong>of</strong> a matrix is invariant to choice representation.<br />

Solution: Using the hint,<br />

and<br />

(4.188)<br />

(4.189)<br />

trH = �<br />

λi = �<br />

Hii = 2 + 3 + 1 = 6 (4.190)<br />

i<br />

i<br />

�<br />

λ<br />

i<br />

2 i = �<br />

HijHji =<br />

ij<br />

�<br />

H<br />

ij<br />

2 ij = 42 (4.191)<br />

107


Exercise 4.7 1. Let |φn〉 be the eigenstates <strong>of</strong> the Hamiltonian, H <strong>of</strong> some arbitrary system<br />

which form a discrete, orthonormal basis.<br />

Define the operator, Unm as<br />

(a) Calculate the adjoint U † nm <strong>of</strong> Unm.<br />

(b) Calculate the commutator, [H, Unm].<br />

(c) Prove: UmnU † pq = δnqUmp<br />

H|φn〉 = En|φn〉.<br />

Unm = |φn〉〈φm|.<br />

(d) For an arbitrary operator, A, prove that<br />

〈φn|[A, H]|φn〉 = 0.<br />

(e) Now consider some arbitrary one dimensional problem for a particle <strong>of</strong> mass m and<br />

potential V (x). For here on, let<br />

H = p2<br />

+ V (x).<br />

2m<br />

i. In terms <strong>of</strong> p, x, and V (x), compute: [H, p], [H, x], and [H, xp].<br />

ii. Show that 〈φn|p|φn〉 = 0.<br />

iii. Establish a relation between the average value <strong>of</strong> the kinetic energy <strong>of</strong> a state<br />

(f) Show that<br />

〈T 〉 = 〈φn| p2<br />

2m |φn〉<br />

and<br />

〈φn|x dV<br />

dx |φn〉.<br />

The average potential energy in the state φn is<br />

〈V 〉 = 〈φn|V |φn〉,<br />

find a relation between 〈V 〉 and 〈T 〉 when V (x) = Vox λ for λ = 2, 4, 6, . . ..<br />

〈φn|p|φm〉 = α〈φn|x|φm〉<br />

where α is some constant which depends upon En − Em. Calculate α, (hint, consider<br />

the commutator [x, H] which you computed above.<br />

(g) Deduce the following sum-rule for the linear -response function.<br />

〈φ0|[x, [H, x]]|φ0〉 = 2 �<br />

(En − E0)|〈φ0|x|φn〉| 2<br />

n>0<br />

Here |φ0〉 is the ground state <strong>of</strong> the system. Give a physical interpretation <strong>of</strong> this last<br />

result.<br />

108


Exercise 4.8 For this section, consider the following 5 × 5 matrix:<br />

⎛<br />

⎜<br />

H = ⎜<br />

⎝<br />

0 0 0 0 1<br />

0 0 0 1 0<br />

0 0 1 0 0<br />

0 1 0 0 0<br />

1 0 0 0 0<br />

⎞<br />

⎟<br />

⎠<br />

(4.192)<br />

1. Using Mathematica determine the eigenvalues, λj, and eigenvectors, φn, <strong>of</strong> H using the<br />

Eigensystem[] command. Determine the eigenvalues only by solving the secular determinant.<br />

|H − Iλ| = 0<br />

Compare the computational effort required to perform both calculations. Note: in entering<br />

H into Mathematica, enter the numbers as real numbers rather than as integers (i.e. 1.0<br />

vs 1 ).<br />

2. Show that the column matrix <strong>of</strong> the eigenvectors <strong>of</strong> H,<br />

T = {φ1, . . . , φ5},<br />

provides a unitary transformation <strong>of</strong> H between the original basis and the eigenvector basis.<br />

T † HT = Λ<br />

where Λ is the diagonal matrix <strong>of</strong> the eigenvalues λj. i.e. Λij = λiδij.<br />

3. Show that the trace <strong>of</strong> a matrix is invarient to representation.<br />

4. First, without using Mathematica, compute: T r(H 2 ). Now check your result with Mathematica.<br />

109


Chapter 5<br />

Bound States <strong>of</strong> The Schrödinger<br />

Equation<br />

A #2 pencil and a dream can take you anywhere.<br />

– Joyce A. Myers<br />

Thus far we have introduced a series <strong>of</strong> postulates and discussed some <strong>of</strong> their physical implications.<br />

We have introduced a powerful notation (Dirac Notation) and have been studying how<br />

we describe dynamics at the atomic and molecular level where ¯h is not a small number, but is<br />

<strong>of</strong> order unity. We now move to a topic which will serve as the bulk <strong>of</strong> our course, the study<br />

<strong>of</strong> stationary systems for various physical systems. We shall start with some general principles,<br />

(most <strong>of</strong> which we have seen already), and then tackle the following systems in roughly this<br />

order:<br />

1. Harmonic Oscillators: Molecular vibrational spectroscopy, phonons, photons, equilibrium<br />

quantum dynamics.<br />

2. Angular Momentum: Spin systems, molecular rotations.<br />

3. Hydrogen Atom: Hydrogenic Systems, basis for atomic theory<br />

5.1 Introduction to Bound States<br />

Before moving on to these systems, let’s first consider what is meant by a “bound state”. Say<br />

we have a potential well which has an arbitrary shape except that at x = ±a, V (x) = 0 and<br />

remains so in either direction. Also, in the range <strong>of</strong> −a ≤ x ≤ a, V (x) < 0. The Schrodinger<br />

Equation for the stationary states is:<br />

� −¯h 2<br />

2m<br />

∂2 �<br />

+ V (x)<br />

∂x2 φn(x) = Enφn(x) (5.1)<br />

Rather than solve this exactly (which we can not do since we haven’t specified more about<br />

V ) let’s examine the topology <strong>of</strong> the allowed bound state solutions. As we have done with the<br />

square well cases, let’s cut the x axis into three domains: Domain 1 for x < −a, Domain 2 for<br />

−a ≤ x ≤ a, Domain 3 for x > a. What are the matching conditions that must be met?<br />

110


For Domain 1 we have:<br />

For Domain 2 we have:<br />

For Domain 3 we have:<br />

∂2 ∂x2 φI(x) = − 2mE<br />

¯h 2 φI(x) ⇒ (+)φI(x) (5.2)<br />

∂ 2<br />

∂x 2 φII(x) =<br />

2m(V (x) − E)<br />

¯h 2 φII(x) ⇒ (−)φII(x) (5.3)<br />

∂2 ∂x2 φIII(x) = − 2m(E)<br />

¯h 2 φIII(x) ⇒ (+)φIII(x) (5.4)<br />

At the rightmost end <strong>of</strong> each equation, the (±) indicates the sign <strong>of</strong> the second derivative <strong>of</strong><br />

the wavefunction. (i.e. the curvature must have the same or opposite sign as the function itself.)<br />

For the + curvature functions, the wavefunctions curve away from the x-axis. For − curvature,<br />

the wavefunctions are curved towards the x-axis.<br />

Therefore, we can conclude that for regions outside the well, the solutions behave much like<br />

exponentials and within the well, the behave like superpositions <strong>of</strong> sine and cosine functions.<br />

Thus, we adopt the asymptotic solution that<br />

and<br />

φ(x) ≈ exp(+αx)for x < a as x → −∞ (5.5)<br />

φ(x) ≈ exp(−αx)for x > a as x → +∞ (5.6)<br />

Finally, in the well region, φ(x) oscillates about the x-axis. We can try to obtain a more complete<br />

solution by combining the solutions that we know. To do so, we must find solutions which are<br />

both continuous functions <strong>of</strong> x and have continuous first derivatives <strong>of</strong> x.<br />

Say we pick an arbitrary energy, E, and seek a solution at this energy. Define the righthand<br />

part <strong>of</strong> the solution to within a multiplicative factor, then the left hand solution is then a<br />

complicated function <strong>of</strong> the exact potential curve and can be written as<br />

φIII(x) = B(E)e +ρx + B ′ (E)e −ρx<br />

where B(E) and B ′ (E) are both real functions <strong>of</strong> E and depend upon the potential function.<br />

Since the solutions must be L 2 , the only appropriate bound states are those for which B(E) = 0.<br />

Any other value <strong>of</strong> B(E) leads to diverging solutions.<br />

Thus we make the following observations concerning bound states:<br />

1. They have negative energy.<br />

2. They vanish exponentially outside the potential well and oscillate within.<br />

(5.7)<br />

3. They form a discrete spectrum as the result <strong>of</strong> the boundary conditions imposed by the<br />

potential.<br />

111


5.2 The Variational Principle<br />

Often the interaction potential is so complicated that an exact solution is not possible. This<br />

is <strong>of</strong>ten the case in molecular problems in which the potential energy surface is a complicated<br />

multidimensional function which we know only at a few points connected by some interpolation<br />

function. We can, however, make some approximations. The method, we shall use is the<br />

“Variational method”.<br />

5.2.1 Variational Calculus<br />

The basic principle <strong>of</strong> the variational method lies at the heart <strong>of</strong> most physical principles. The<br />

idea is that we represent a quantity as a stationary integral<br />

J =<br />

� x2<br />

x1<br />

f(y, yx, x)dx (5.8)<br />

where f(y, yx, x) is some known function which depened upon three variables, which are also<br />

functions <strong>of</strong> x, y(x), yx = dy/dx, and x itself. The dependency <strong>of</strong> y on x uis generally unknown.<br />

This means that while we have fixed the end-points <strong>of</strong> the integral, the path that we actually<br />

take between the endpoints is not known.<br />

Picking different paths leads to different values <strong>of</strong> J. However ever certain paths will minimize,<br />

maximize, or find the saddle points <strong>of</strong> J. For most cases <strong>of</strong> physical interest, its the extrema<br />

that we are interested. Lets say that there is one path, yo(t) which minimizes J (See Fig. 5.1).<br />

If we distort that path slightly, we get another path y(x) which is not too unlike yo(x) and we<br />

will write it as y(x) = yo(x) + η(x) where η(x1) = η(x2) = 0 so that the two paths meet at the<br />

terminal points. If η(x) differs from 0 only over a small region, we can write the new path as<br />

and the variation from the minima as<br />

y(x, α) = yo(x) + αη(x)<br />

δy = y(x, α) − yo(x, 0) = αη(x).<br />

Since yo is the path which minimizes J, and y(x, α) is some other path, then J is also a<br />

function <strong>of</strong> α.<br />

� x2<br />

J(α) = f(y(x, α), y ′ (x, α), x)dx<br />

x1<br />

and will be minimized when � �<br />

∂J<br />

= 0<br />

∂α α=0<br />

Because J depends upon α, we can examine the α dependence <strong>of</strong> the integral<br />

Since<br />

� � � �<br />

∂J<br />

x2 ∂f ∂y ∂f<br />

=<br />

+<br />

∂α<br />

x1 ∂y ∂α ∂y α=0<br />

′<br />

∂y ′<br />

�<br />

dx<br />

∂α<br />

∂y<br />

∂α<br />

= η(x)<br />

112


and<br />

∂y ′<br />

∂α<br />

= ∂η<br />

∂x<br />

we have � � � �<br />

∂J<br />

x2 ∂f ∂f<br />

= η(x) +<br />

∂α<br />

x1 ∂y ∂y α=0<br />

′<br />

�<br />

∂η<br />

dx.<br />

∂x<br />

Now, we need to integrate the second term by parts to get η as a common factor. Remember<br />

integration by parts? �<br />

�<br />

udv = vu − vdu<br />

From this<br />

� x2<br />

x1<br />

�<br />

∂f<br />

∂y ′<br />

�<br />

∂η<br />

dx = η(x)<br />

∂x<br />

∂f<br />

�<br />

�<br />

�<br />

�<br />

∂x �<br />

x2<br />

x1<br />

−<br />

� x2<br />

x1<br />

η(x) d ∂f<br />

dx<br />

dx ∂y ′<br />

The boundaty term vanishes since η vanishes at the end points. So putting it all together and<br />

setting it equal to zero:<br />

� �<br />

x2 ∂f d ∂f<br />

η(x) −<br />

∂y dx ∂y ′<br />

�<br />

dx. = 0<br />

x1<br />

We’re not done yet, since we still have to evaluate this. Notice that α has disappeared from<br />

the expression. In effect, we can take an arbitrary variation and still find the desired path tha<br />

minimizes J. Since η(x) is arbitrary subject to the boundary conditions, we can make it have the<br />

same sign as the remaining part <strong>of</strong> the integrand so that the integrand is always non-negative.<br />

Thus, the only way for the integral to vanish is if the bracketed term is zero everywhere.<br />

�<br />

∂f d ∂f<br />

−<br />

∂y dx ∂y ′<br />

�<br />

= 0 (5.9)<br />

This is known as the Euler equation and it has an enormous number <strong>of</strong> applications. Perhaps<br />

the simplest is the pro<strong>of</strong> that the shortest distance between two points is a straight line (or<br />

on a curved space, a geodesic). The straightline distance between two points on the xy plane<br />

is s = √ x 2 + y 2 and the differential element <strong>of</strong> distance is ds =<br />

Thus, we can write a distance along some line in the xy plane as<br />

J =<br />

� x2y2<br />

x1y1<br />

ds =<br />

� x2y2<br />

x1y1<br />

�<br />

1 + y 2 xdx.<br />

�<br />

(dx) 2 + (dy) 2 = �<br />

1 + y 2 xdx.<br />

If we knew y(x) then J would be the arclength or path-length along the function y(x) between<br />

two points. Sort <strong>of</strong> like, how many steps you would take along a trail between two points. The<br />

trail may be curvy or straight and there is certainly a single trail which is the shortest. So,<br />

setting<br />

f(y, yx, x) =<br />

and substituting it into the Euler equation one gets<br />

d ∂f<br />

dx ∂yx<br />

�<br />

1 + y 2 x<br />

= − d 1<br />

�<br />

dx 1 + y2 x<br />

= 0. (5.10)<br />

113


So, the only way for this to be true is if<br />

1<br />

�<br />

1 + y2 = constant. (5.11)<br />

x<br />

Solving for yx produces a second constant: yx = a, which immediatly yields that y(x) = ax + b.<br />

In other words, its a straight line! Not too surprising.<br />

An important application <strong>of</strong> this principle is when the integrand f is the classical Lagrangian<br />

for a mechanical system. The Lagrangian is related to Hamiltonian and is defined as the difference<br />

between the kinetic and potential energy.<br />

L = T − V (5.12)<br />

where as H is the sum <strong>of</strong> T +V . Rather than taking x as the independent variable, we take time,<br />

t, and position and velocity oa a particle as the dependent variables. The statement <strong>of</strong> δJ = 0<br />

is a mathematical statement <strong>of</strong> Hamilton’s principle <strong>of</strong> least action<br />

� t2<br />

δ L(x, ˙x, t)dt = 0. (5.13)<br />

t1<br />

In essence, Hamilton’s principle asserts that the motion <strong>of</strong> the system from one point to another<br />

is along a path which minimizes the integral <strong>of</strong> the Lagrangian. The equations <strong>of</strong> motion for that<br />

path come from the Euler-Lagrange equations,<br />

So if we write the Lagrangian as<br />

d ∂L ∂L<br />

−<br />

dt ∂ ˙x ∂x<br />

and substitute this into the Euler-Lagarange equation, we get<br />

which is Newton’s law <strong>of</strong> motion: F = ma.<br />

= 0. (5.14)<br />

L = 1<br />

2 m ˙x2 − V (x) (5.15)<br />

m¨x = − ∂V<br />

∂x<br />

5.2.2 Constraints and Lagrange Multipliers<br />

(5.16)<br />

Before we can apply this principle to a quantum mechanical problem, we need to ask our selves<br />

what happens if there is a constraint on the system which exclues certain values or paths so that<br />

not all the η’s may be varied arbitrarily? Typically, we can write the constraint as<br />

φi(y, x) = 0 (5.17)<br />

For example, for a bead on a wire we need to constrain the path to always lie on the wire or for<br />

a pendulum, the path must lie on in a hemisphere defined by the length <strong>of</strong> the pendulum from<br />

114


the pivot point. In any case, the general proceedure is to introduce another function, λi(x) and<br />

integrate<br />

so that<br />

� x2<br />

x1<br />

λi(x)φi(y, x)dx = 0 (5.18)<br />

� x2<br />

δ λi(x)φi(y, x)dx = 0 (5.19)<br />

x1<br />

as well. In fact it turns out that the λi(x) can be even be taken to be a constant, λi for this<br />

whole proceedure to work.<br />

Regardless <strong>of</strong> the case, we can always write the new stationary integral as<br />

�<br />

δ<br />

(f(y, yx, x) + �<br />

λiφi(y, x))dx = 0. (5.20)<br />

i<br />

The multiplying constants are called Lagrange Mulipliers. In your statistical mechanics course,<br />

these will occur when you minimize various thermodynamic functions subject to the various<br />

extensive constraints, such as total number <strong>of</strong> particles in the system, the average energy or<br />

temperature, and so on.<br />

In a sence, we have redefined the original function or Lagrangian to incorporate the constraint<br />

into the dynamics. So, in the presence <strong>of</strong> a constraint, the Euler-Lagrange equations become<br />

d ∂L ∂L<br />

−<br />

dt ∂ ˙x ∂x<br />

= �<br />

i<br />

∂φi<br />

∂x λi<br />

(5.21)<br />

where the term on the right hand side <strong>of</strong> the equation represents a force due to the constraint.<br />

The next issue is that we still need to be able to determine the λi Lagrange multipliers.<br />

115


Figure 5.1: Variational paths between endpoints. The thick line is the stationary path, yo(x)<br />

and the dashed blue curves are variations y(x, α) = yo(x) + αη(x).<br />

fHxL<br />

2<br />

1<br />

-1.5 -1 -0.5 0.5 1 1.5 x<br />

-1<br />

-2<br />

116


5.2.3 Variational method applied to Schrödinger equation<br />

The goal <strong>of</strong> all this is to develop a procedure for computing the ground state <strong>of</strong> some quantum<br />

mechanical system. What this means is that we want to minimize the energy <strong>of</strong> the system<br />

with respect to arbitrary variations in the state function subject to the constraint that the state<br />

function is normalized (i.e. number <strong>of</strong> particles remains fixed). This means we want to construct<br />

the variation:<br />

δ〈ψ|H|ψ〉 = 0 (5.22)<br />

with the constraint 〈ψ|ψ〉 = 0.<br />

In the coordinate representation, the integral involves taking the expectation value <strong>of</strong> the<br />

kinetic energy operator...which is a second derivative operator. That form is not too convenient<br />

for our purposes since it will allow us to write Eq.5.22 in a form suitable for the Euler-Lagrange<br />

equations. But, we can integrate by parts!<br />

�<br />

ψ ∗ ∂2 �<br />

ψ ∂ψ �<br />

�<br />

dx = ψ∗ �<br />

∂x2 ∂x � −<br />

� � ∂ψ∗ � � �<br />

∂ψ<br />

dx (5.23)<br />

∂x ∂x<br />

Assuming that the wavefunction vanishes at the limits <strong>of</strong> the integration, the surface term vanishes<br />

leaving only the second term. We can now write the energy expectation value in terms<br />

<strong>of</strong> two dependent variables, ∇ψ and ψ. OK, they’re functions, but we can still treat them as<br />

dependent variables just like we treated the y(x) ′ s above.<br />

�<br />

E =<br />

� ¯h 2<br />

2m (∇ψ∗ )(∇ψ) + V ψ ∗ �<br />

ψ dx (5.24)<br />

Adding on the constraint and defining the Lagrangian as<br />

L =<br />

� ¯h 2<br />

2m (∇ψ∗ )(∇ψ) + V ψ ∗ ψ<br />

we can substitute this into the Euler-Lagrange equations<br />

This produces the result<br />

�<br />

− λψ ∗ ψ, (5.25)<br />

∂L ∂L<br />

− ∇∂x = 0. (5.26)<br />

∂ψ∗ ∂(∇ψ∗ (V − λ)ψ = ¯h2<br />

2m ∇2 ψ, (5.27)<br />

which we immediately recognize as the Schrödinger equation.<br />

While this may be a rather academic result, it gives us the key to recognize that we can<br />

make an expansion <strong>of</strong> ψ in an arbitrary basis and take variations with respect to the coeffients<br />

<strong>of</strong> that basis to find the lowest energy state. This is the basis <strong>of</strong> a number <strong>of</strong> powerful numerical<br />

methods used solve the Schrödinger equation for extremely complicated systems.<br />

117


5.2.4 Variational theorems: Rayleigh-Ritz Technique<br />

We now discuss two important theorems:<br />

Theorem 5.1 The expectation value <strong>of</strong> the Hamiltonian is stationary in the neighborhood <strong>of</strong> its<br />

eigenstates.<br />

To demonstrate this, let |ψ〉 be a state in which we compute the expectation value <strong>of</strong> H. Also,<br />

let’s modify the state just a bit and write<br />

Expectation values are computed as<br />

|ψ〉 → |ψ〉 + |δψ〉. (5.28)<br />

〈H〉 = 〈ψ|H|ψ〉<br />

〈ψ|ψ〉<br />

(where we assume arb. normalization). In other words<br />

Now, insert the variation<br />

or<br />

(5.29)<br />

〈ψ|ψ〉〈H〉 = 〈ψ|H|ψ〉 (5.30)<br />

〈ψ|ψ〉δ〈H〉 + 〈δψ|ψ〉〈H〉 + 〈ψ|δψ〉〈H〉 = 〈δψ|H|ψ〉 + 〈ψ|H|δψ〉 (5.31)<br />

〈ψ|ψ〉δ〈H〉 = 〈δψ|H − 〈H〉|ψ〉 − 〈ψ|H − 〈H〉|δψ〉 (5.32)<br />

If the expectation value is to be stationary, then δ〈H〉 = 0. Thus the RHS must vanish for an<br />

arbitrary variation in the wavefunction. Let’s pick<br />

Thus,<br />

|δψ〉 = ɛ|ψ〉. (5.33)<br />

(H − 〈H〉)|ψ〉 = 0 (5.34)<br />

That is to say that |ψ〉 is an eigenstate <strong>of</strong> H. Thus proving the theorem.<br />

The second theorem goes:<br />

Theorem 5.2 The Expectation value <strong>of</strong> the Hamiltonian in an arbitrary state is greater than or<br />

equal to the ground-state energy.<br />

The pro<strong>of</strong> goes as this: Assume that H has a discrete spectrum <strong>of</strong> states (which we demonstrated<br />

that it must) such that<br />

H|n〉 = En|n〉 (5.35)<br />

118


Thus, we can expand any state |ψ〉 as<br />

Consequently<br />

and<br />

Thus, (assuming that |ψ〉 is normalized)<br />

〈H〉 = �<br />

En|cn| 2 ≥ �<br />

Eo|cn| 2 ≥ Eo<br />

|ψ〉 = �<br />

cn|n〉. (5.36)<br />

n<br />

〈ψ|ψ〉 = �<br />

|cn| 2 , (5.37)<br />

n<br />

〈ψ|H|ψ〉 = �<br />

|cn| 2 En. (5.38)<br />

n<br />

n<br />

n<br />

(5.39)<br />

quid est demonstrato.<br />

Using these two theorems, we can estimate the ground state energy and wavefunctions for a<br />

variery <strong>of</strong> systems. Let’s first look at the Harmonic Oscillator.<br />

Exercise 5.1 Use the variational principle to estimate the ground-state energy <strong>of</strong> a particle in<br />

the potential<br />

�<br />

Cx x > 0<br />

V (x) =<br />

(5.40)<br />

+∞ x ≤ 0<br />

Use xe −ax as the trial function.<br />

5.2.5 Variational solution <strong>of</strong> harmonic oscillator ground State<br />

The Schrödinger Equation for the Harmonic Osc. (HO) is<br />

Take as a trial function,<br />

�<br />

− ¯h2 ∂<br />

2m<br />

2<br />

∂x<br />

2 + k2<br />

2 x2<br />

�<br />

φ(x) − Eφ(x) = 0 (5.41)<br />

φ(x) = exp(−λx 2 ) (5.42)<br />

where λ is a positive, non-zero constant to be determined. The variational principle states that<br />

the energy reaches a minimum<br />

∂〈H〉<br />

∂λ<br />

when φ(x) is the ground state solution. Let us first derive 〈H〉(λ).<br />

〈H〉(λ) = 〈φ|H|φ〉<br />

〈φ|φ〉<br />

= 0. (5.43)<br />

119<br />

(5.44)


To evaluate this, we break the problem into a series <strong>of</strong> integrals:<br />

and<br />

Putting it all together:<br />

〈φ|p 2 � ∞<br />

|φ〉 =<br />

� ∞<br />

〈φ|φ〉 = dx|φ(x)| 2 =<br />

−∞<br />

� π<br />

2λ<br />

(5.45)<br />

dxφ<br />

−∞<br />

′′ (x)φ(x) = −2λ〈φ|φ〉 + 4λ 2 〈φ|x 2 |φ〉 (5.46)<br />

< φ|x 2 � ∞<br />

|φ〉 =<br />

Taking the derivative with respect to λ:<br />

Thus,<br />

−∞<br />

dxx 2 |φ(x)| 2 = 1<br />

〈φ|φ〉. (5.47)<br />

4λ<br />

〈φ|H|φ〉<br />

〈φ|φ〉 =<br />

�<br />

− ¯h2<br />

� �<br />

�<br />

2 1<br />

−2λ + 4λ +<br />

2m<br />

4λ<br />

k 1<br />

2 4λ<br />

〈φ|H|φ〉<br />

〈φ|φ〉 =<br />

� �<br />

2<br />

¯h<br />

λ +<br />

2m<br />

k<br />

8λ<br />

∂〈H〉<br />

∂λ<br />

Since only positive values <strong>of</strong> λ are allowed.<br />

= ¯h2<br />

2m<br />

λ = ±<br />

λ =<br />

(5.48)<br />

(5.49)<br />

k<br />

− = 0 (5.50)<br />

8λ2 √ mk<br />

2¯h<br />

√ mk<br />

2¯h<br />

Using this we can calculate the ground state energy by substituting λ back into 〈H〉(λ).<br />

〈H〉(λ) =<br />

Now, define the angular frequency: ω =<br />

√ �<br />

2<br />

mk ¯h<br />

2¯h 2m<br />

�<br />

k/m.<br />

k 4¯h<br />

+<br />

8<br />

2<br />

�<br />

=<br />

mk<br />

¯h<br />

�<br />

k<br />

2 m<br />

(5.51)<br />

(5.52)<br />

(5.53)<br />

〈H〉(λ) = ¯h<br />

ω (5.54)<br />

2<br />

which ( as we can easily prove) is the ground state energy <strong>of</strong> the harmonic oscillator.<br />

Furthermore, we can write the HO ground state wavefunction as<br />

120


φo(x) =<br />

φo(x) =<br />

1<br />

φo(x) = � φ(x) (5.55)<br />

〈φ|φ〉<br />

� �1/4 � √<br />

2λ<br />

mk<br />

exp −<br />

π<br />

2¯h x2<br />

�<br />

�√ mk<br />

¯hπ<br />

�1/4 � √<br />

mk<br />

exp −<br />

2¯h x2<br />

�<br />

(5.56)<br />

(5.57)<br />

To compute the “error” in our estimate, let’s substitute the variational solution back into the<br />

Schrodinger equation:<br />

�<br />

− ¯h2 ∂<br />

2m<br />

2<br />

∂x<br />

2 + k2<br />

2 x2<br />

�<br />

φo(x) = − ¯h2<br />

2m φ′′<br />

o(x) + k2<br />

2 φo(x) (5.58)<br />

− ¯h2<br />

2m φ′′ o(x) + k2<br />

2 φo(x) = − ¯h2<br />

� √<br />

2 kmx − ¯h km<br />

2m ¯h 2<br />

�<br />

φo(x) + k<br />

2 x2φo(x) (5.59)<br />

− ¯h2<br />

2m φ′′ o(x) + k2<br />

2 φo(x) = ¯h<br />

�<br />

k<br />

2 m φo(x) (5.60)<br />

Thus, φo(x) is in fact the correct ground state wavefunction for this system. If it were not<br />

the correct function, we could re-introduce the solution as a new trial function, re-compute the<br />

energy, etc... and iterate through until we either find a solution, or run out <strong>of</strong> patience! (Usually<br />

it’s the latter than the former.)<br />

5.3 The Harmonic Oscillator<br />

Now that we have the HO ground state and the HO ground state energy, let us derive the whole<br />

HO energy spectrum. To do so, we introduce “dimensionless” quantities: X and P related to<br />

the physical position and momentum by<br />

X = ( mω<br />

2¯h )1/2 x (5.61)<br />

1<br />

P = (<br />

2¯hmω )1/2p (5.62)<br />

This will save us from carrying around a bunch <strong>of</strong> coefficients. In these units, the HO Hamiltonian<br />

is<br />

H = ¯hω(P 2 + X 2 ). (5.63)<br />

121


The X and P obey the canonical comutational relation:<br />

We can also write the following:<br />

Thus, I can construct the commutator:<br />

[X, P ] = 1 i<br />

[x, p] =<br />

2¯h 2<br />

(5.64)<br />

(X + iP )(X − iP ) = X 2 + P 2 + 1/2 (5.65)<br />

(X − iP )(X + iP ) = X 2 + P 2 − 1/2. (5.66)<br />

[(X + iP ), (X − iP )] = (X + iP )(X − iP ) − (X − iP )(X + iP )<br />

Let’s define the following two operators:<br />

Therefore, the a and a † commute as<br />

= 1/2 + 1/2<br />

Let’s write H in terms <strong>of</strong> the a and a † operators:<br />

or in terms <strong>of</strong> the a and a † operators:<br />

= 1 (5.67)<br />

a = (X + iP ) (5.68)<br />

a † = (X + iP ) † = (X − iP ). (5.69)<br />

[a, a † ] = 1 (5.70)<br />

H = ¯hω(X 2 + P 2 ) = ¯hω(X − iP )(X + iP ) + ¯hω/2 (5.71)<br />

H = ¯hω(a † a + 1/2) (5.72)<br />

Now, consider that |φn〉 is the nth eigenstate <strong>of</strong> H. Thus, we write:<br />

¯hω(a † a + 1/2)|φn〉 = En|φn〉 (5.73)<br />

What happens when I multiply the whole equation by a? Thus, we write:<br />

Now, since aa † − a † a = 1,<br />

a¯hω(a † a + 1/2)|φn〉 = aEn|φn〉 (5.74)<br />

¯hω(aa † + 1/2)(a|φn〉) = En(a|φn〉) (5.75)<br />

¯hω(a † a + 1/2 − 1)(a|φn〉) = En(a|φn〉) (5.76)<br />

In other words, a|φn〉 is an eigenstate <strong>of</strong> H with energy E = En − ¯hω.<br />

122


Since<br />

What happends if I do the same procedure, this time using a † ? Thus, we write:<br />

we have<br />

we can write<br />

Thus,<br />

or,<br />

a † ¯hω(a † a + 1/2)|φn〉 = a † En|φn〉 (5.77)<br />

[a, a † ] = aa † − a † a (5.78)<br />

a † a = aa † − 1 (5.79)<br />

a † a † a = a † (aa † − 1) (5.80)<br />

= (a † a − 1)a † . (5.81)<br />

a † ¯hω(a † a + 1/2)|φn〉 = ¯hω((a † a − 1 + 1/2)a † )|φn〉 (5.82)<br />

¯hω(a † a − 1/2)(a † |φn〉) = En(a † |φn〉). (5.83)<br />

Thus, a † |φn〉 is an eigenstate <strong>of</strong> H with energy E = En + ¯hω.<br />

Since a † and a act on harmonic oscillator eigenstates to give eigenstates with one more or one<br />

less ¯hω quanta <strong>of</strong> energy, these are termed “creation” and “annihilation” operators since they<br />

act to create additional quata <strong>of</strong> excitation or decrease the number <strong>of</strong> quanta <strong>of</strong> excitation in<br />

the system. Using these operators, we can effectively “ladder” our way up the energy scale and<br />

determine any eigenstate one we know just one.<br />

Well, we know the ground state solution. That we got via the variational calculation. What<br />

happens when I apply a † to the φo(x) we derived above? In coordinate form:<br />

(X − iP ) φo(x) =<br />

X acting on φo is:<br />

iP acting on φo is<br />

=<br />

� �mω<br />

2¯h<br />

� �mω<br />

2¯h<br />

Xφo(x) =<br />

�1/2 � �1/2<br />

�<br />

1 ∂<br />

x +<br />

φo(x)<br />

2mω¯h ∂x<br />

�1/2 � �1/2 1 ∂<br />

x +<br />

2mω¯h ∂x<br />

� �mω<br />

¯hπ<br />

(5.84)<br />

�1/4 mω<br />

(−x2 e 2¯h ) (5.85)<br />

� mω<br />

2¯h xφo(x) (5.86)<br />

�<br />

iP φo(x) = −¯h<br />

1 ∂<br />

mω2¯h ∂x φo(x) (5.87)<br />

123


After cleaning things up:<br />

iP φo(x) =<br />

�<br />

mω<br />

−x<br />

2¯h φo(x) (5.88)<br />

= −Xφo(x) (5.89)<br />

(X − iP ) φo(x) =<br />

=<br />

2Xφo(x)<br />

�<br />

mω<br />

2<br />

2¯h<br />

(5.90)<br />

xφo(x) (5.91)<br />

=<br />

=<br />

2Xφo(x)<br />

�<br />

mω<br />

2<br />

2¯h<br />

(5.92)<br />

x<br />

� �<br />

mω 1/4 � �<br />

2 mω<br />

exp −x<br />

2¯h<br />

2¯h<br />

(5.93)<br />

5.3.1 Harmonic Oscillators and Nuclear Vibrations<br />

We introduced one <strong>of</strong> the most important applications <strong>of</strong> quantum mechanics...the solution <strong>of</strong><br />

the Schrödinger equation for harmonic systems. These are systems in which the amplitude <strong>of</strong><br />

motion is either small enough so that the physical potential energy operator can be expanded<br />

about its minimum to second order in the displacement from the minima. When we do so, the<br />

Hamiltonian can be written in the form<br />

H = ¯hω(P 2 + X 2 ) (5.94)<br />

where P and X are dimensionless operators related to the physical momentum and position<br />

operators via<br />

�<br />

mω<br />

X = x (5.95)<br />

2¯h<br />

and<br />

�<br />

P =<br />

1<br />

p.<br />

2¯hmω<br />

(5.96)<br />

We also used the variational method to deduce the ground state wavefunction and demonstrated<br />

that the spectrum <strong>of</strong> H is a series <strong>of</strong> levels separated by ¯hω and that the ground-state energy is<br />

¯hω/2 above the energy minimum <strong>of</strong> the potential.<br />

We also defined a new set <strong>of</strong> operators by taking linear combinations <strong>of</strong> the X and P .<br />

a = X + iP (5.97)<br />

a † = X − iP. (5.98)<br />

We also showed that the commutation relation for these operators is<br />

[a, a † ] = 1. (5.99)<br />

These operators are non-hermitian operators, and hence, do not correspond to a physical observable.<br />

However, we demonstrated that when a acts on a eigenstate <strong>of</strong> H, it produces another<br />

124


eigenstate withe energy En − ¯hω. Also, a † acting on an eigenstate <strong>of</strong> H produces another eigenstate<br />

with energy En + ¯hω. Thus,we called a the destruction or annihilation operator since it<br />

removes a quanta <strong>of</strong> excitation from the system and a † the creation operator since it adds a<br />

quanta <strong>of</strong> excitation to the system. We also wrote H using these operators as<br />

H = ¯hω(a † a + 1<br />

) (5.100)<br />

2<br />

Finally, ω is the angular frequency <strong>of</strong> the classical harmonic motion, as obtained via Hooke’s<br />

law:<br />

¨x = − k<br />

x. (5.101)<br />

m<br />

Solving this produces<br />

and<br />

x(t) = xo sin(ωt + φ) (5.102)<br />

p(t) = po cos(ωt + φ). (5.103)<br />

Thus, the classical motion in the x, p phase space traces out the circumference <strong>of</strong> a circle every<br />

1/ω regardless <strong>of</strong> the initial amplitude.<br />

The great advantage <strong>of</strong> using the a, and a † operators is that they we can replace a differential<br />

equation with an algebraic equation. Furthermore, since we can represent any Hermitian operator<br />

acting on the HO states as a combination <strong>of</strong> the creation/annihilation operators, we can replace<br />

a potentially complicated series <strong>of</strong> differentiations, integrations, etc... with simple algebraic<br />

manipulations. We just have to remember a few simple rules regarding the commutation <strong>of</strong> the<br />

two operators. Two operators which we may want to construct are:<br />

and<br />

• position operator: � �1/2 2¯h (a mω<br />

† + a)<br />

• momentum operator: i � �1/2 ¯hmω (a 2<br />

† − a).<br />

Another important operator is<br />

N = a † a. (5.104)<br />

H = ¯hω(N + 1/2). (5.105)<br />

Since [H, N] = 0, eigenvalues <strong>of</strong> N are “good quantum numbers” and N is a constant <strong>of</strong> the<br />

motion. Also, since<br />

then if<br />

H|φn〉 = En|φn〉 = ¯hω(N + 1/2)|φn〉 (5.106)<br />

N|φn〉 = n|φn〉, (5.107)<br />

then n must be an integer n = 0, 1, 2, · · · corresponding to the number <strong>of</strong> quanta <strong>of</strong> excitation in<br />

the state. This gets the name “Number Operator”.<br />

Some useful relations (that you should prove )<br />

125


1. [N, a] = [a † a, a] = −a<br />

2. [N, a † ] = [a † a, a † ] = a †<br />

To summarize, we have the following relations using the a and a † operators:<br />

1. a|φn〉 = √ n|φn−1〉<br />

2. a † |φn〉 = √ n + 1|φn+1〉<br />

3. 〈φn|a = √ n + 1〈φn+1 = (a|φn〉) †<br />

4. 〈φn|a † = √ n + 1〈φn−1|<br />

5. N|φn〉 = n|φn〉<br />

6. 〈φn|N = n〈φn|<br />

Using the second <strong>of</strong> these relations we can write<br />

which can be iterated back to the ground state to produce<br />

This is the “generating relation” for the eigenstates.<br />

Now, let’s look at x and p acting on |φn〉.<br />

Also,<br />

|φn+1〉 = a†<br />

√ n + 1 |φn〉 (5.108)<br />

|φn〉 = (a† ) n<br />

√ n! |φo〉 (5.109)<br />

�<br />

x|φn〉 =<br />

¯h<br />

2mω (a† + a)|φn〉 (5.110)<br />

�<br />

¯h<br />

=<br />

2mω (√n + 1|φn+1〉 + √ n|φn−1〉) (5.111)<br />

= i<br />

�<br />

p|φn〉 = i<br />

�<br />

m¯hω<br />

2 (a† − a)|φn〉 (5.112)<br />

m¯hω<br />

2 (√ n + 1|φn+1〉 − √ n|φn−1〉) (5.113)<br />

Thus,the matrix elements <strong>of</strong> x and p in the HO basis are:<br />

�<br />

¯h �√<br />

〈φm|x|φn〉 = n + 1δm,n+1 +<br />

2mω<br />

√ �<br />

nδm,n−1<br />

126<br />

(5.114)


〈φm|p|φn〉 = i<br />

�<br />

mω¯h<br />

2<br />

�√ n + 1δm,n+1 − √ nδm,n−1<br />

The harmonic oscillator wavefunctions can be obtained by solving the equation:<br />

�<br />

(5.115)<br />

〈x|a|φo〉 = (X + iP )φo(x) = 0 (5.116)<br />

� mω<br />

The solution <strong>of</strong> this first order differential equation is easy:<br />

¯h<br />

�<br />

∂<br />

x + φo(x) = 0 (5.117)<br />

∂x<br />

φo(x) = c exp(− mω<br />

2 x2 ) (5.118)<br />

where c is a constant <strong>of</strong> integration which we can obtain via normalization:<br />

Doing the integration produces:<br />

�<br />

φo(x) =<br />

dx|φo(x)| 2 = 1 (5.119)<br />

� �<br />

mω 1/4<br />

mω<br />

−<br />

e 2¯h<br />

¯hπ<br />

x2<br />

127<br />

(5.120)


Figure 5.2: Hermite Polynomials, Hn up to n = 3.<br />

HnHxL<br />

10<br />

7.5<br />

5<br />

2.5<br />

-3 -2 -1 1 2 3 x<br />

-2.5<br />

-5<br />

-7.5<br />

-10<br />

128


Since we know that a † acting on |φo〉 gives the next eigenstate, we can write<br />

φ1(x) =<br />

� mω<br />

¯h<br />

�<br />

∂<br />

x − φo(x) (5.121)<br />

∂x<br />

Finally, using the generating relation, we can write<br />

φn(x) = 1<br />

� �n mω ∂<br />

√ x − φo(x). (5.122)<br />

n! ¯h ∂x<br />

Lastly, we have the “recursion relations” which generates the next solution one step higher or<br />

lower in energy given any other solution.<br />

� �<br />

1 mω ∂<br />

φn+1(x) = √ x − φn(x) (5.123)<br />

n + 1 ¯h ∂x<br />

and<br />

φn−1(x) = 1<br />

� �<br />

mω ∂<br />

√ x + φn(x). (5.124)<br />

n ¯h ∂x<br />

These are the recursion relationships for a class <strong>of</strong> polynomials called Hermite polynomials, after<br />

the 19th French mathematician who studied such functions. These are also termed “Gauss-<br />

Hermite” and form a set <strong>of</strong> orthogonal polynomials. The first few Hermite Polynomials, Hn(x)<br />

are {1, 2 x, −2 + 4 x 2 , −12 x + 8 x 3 , 12 − 48 x 2 + 16 x 4 } for n = 0 to 4. Some <strong>of</strong> these are plotted<br />

in Fig. 5.2<br />

The functions themselves are defined by the generating function<br />

g(x, t) = e −t2 ∞�<br />

+2tx<br />

=<br />

n=0<br />

Hn(x) tn<br />

. (5.125)<br />

n!<br />

Differentiating the generating function n times and setting t = 0 produces the nth Hermite<br />

polynomial<br />

Hn(x) = dn<br />

�<br />

�<br />

�<br />

g(x, t) �<br />

dtn � = (−1)n dn x2<br />

e e−x2<br />

(5.126)<br />

dxn Another useful relation is the Fourier transform relation:<br />

�<br />

1 ∞<br />

√<br />

2π −∞<br />

e itx e −x2 /2 Hn(x)dx = −i n e −t2 /2 Hn(t) (5.127)<br />

which is useful in generating the momentum space representation <strong>of</strong> the harmonic oscillator<br />

functions. Also, from the generating function, we can arrive at the recurrence relation:<br />

and<br />

Hn+1 = 2xHn − 2nHn−1<br />

(5.128)<br />

H ′ n(x) = 2nHn−1(x). (5.129)<br />

129


Consequently, the hermite polynomials are solutions <strong>of</strong> the second-order differental equation:<br />

H ′′<br />

n − 2xH ′ n + 2nHn = 0 (5.130)<br />

which is not self-adjoint! To put this into self-adjoint form, we multiply by the weighting function<br />

w = e−x2, which leads to the orthogonality integral<br />

� ∞<br />

Hn(x)Hm(x)e −x2<br />

dx = δnm. (5.131)<br />

−∞<br />

For the harmonic oscillator functions, we absorb the weighting function into the wavefunction<br />

itself<br />

ψn(x) = e −x2 /2 Hn(x).<br />

When we substitute this function into the differential equation for Hn we get<br />

ψ ′′<br />

n + (2n + 1 − x 2 )ψn = 0. (5.132)<br />

To normalize the functions, we first multipy g by itself and then multiply by w<br />

e −x2<br />

e −s2 +2sx −t<br />

e 2 �<br />

+2tx<br />

=<br />

mn<br />

e −x2<br />

Hn(x)Hm(x) smtn n!m!<br />

(5.133)<br />

When we integrate over −∞ to ∞ the cross terms drop out by orthogonality and we are left<br />

with<br />

� (st)<br />

n=0<br />

n � ∞<br />

e<br />

n!n! −∞<br />

−x2<br />

H 2 n(x)dx =<br />

� ∞<br />

e<br />

−∞<br />

−x2−s2 +2sx−t2 =<br />

+2xt<br />

dx<br />

� ∞<br />

e<br />

−∞<br />

−(x−s−t)2<br />

dx<br />

= π 1/2 e 2st = � 2<br />

n=0<br />

n (st) n<br />

.<br />

n!<br />

(5.134)<br />

Equating like powers <strong>of</strong> st we obtain,<br />

� ∞<br />

e −x2<br />

H 2 n(x)dx = 2 n π 1/2 n!. (5.135)<br />

When we apply this technology to the SHO, the solutions are<br />

where z = αx and<br />

A few gratuitous solutions:<br />

−∞<br />

ψn(z) = 2 −n/2 π −1/4 (n!) −1/2 e −z2<br />

Hn(z) (5.136)<br />

φ1(x) =<br />

φ2(x) =<br />

� 4<br />

π<br />

Fig. 5.3 shows the first 4 <strong>of</strong> these functions.<br />

α 2 = mω<br />

¯h .<br />

� � �<br />

mω 3 1/4<br />

x exp(−<br />

¯h<br />

1<br />

2 mωx2 ) (5.137)<br />

� �<br />

mω 1/4 �<br />

2<br />

4π¯h<br />

mω<br />

¯h x2 �<br />

− 1 exp(− 1<br />

2 mωx2 ) (5.138)<br />

130


5.3.2 Classical interpretation.<br />

In Fig. 5.3 are a few <strong>of</strong> the lowest energy states for the harmonic oscillator. Notice that as the<br />

quantum number increases the amplitude <strong>of</strong> the wavefunction is pushed more and more towards<br />

larger values <strong>of</strong> ±x. This becomes more pronounced when we look at the actual probability<br />

distribution functions, |ψn(x)| 2 | for the same 4 states as shown in Fig 5.4.<br />

Here, in blue are the actual quantum distributions for the ground-state through n = 3. In<br />

gray are the classical probability distrubutions for the corresponding energies. The gray curves<br />

tell us the probabilty per unit length <strong>of</strong> finding the classical particle at some point x and any<br />

point in time. This is inversely proportional to how long a particle spends at a given point...i.e.<br />

Pc(x) ∝ 1/v(x). Since E = mv 2 /2 + V (x),<br />

and<br />

For the Harmonic Oscillator:<br />

v(x) =<br />

�<br />

2(E − V (x))/m<br />

�<br />

m<br />

P (x) ∝<br />

2(E − V (x))<br />

�<br />

m<br />

Pn(x) ∝<br />

2(¯hω(n + 1/2) − kx2 /2) .<br />

Notice that the denominator goes to zero at the classical turning points, in other words,<br />

the particle comes to a dead-stop at the turning point and consequently we have the greatest<br />

likelyhood <strong>of</strong> finding the particle in these regions. Likewise in the quantum case, as we increase<br />

the quantum number, the quantum distrubution function becomes more and more like its classical<br />

counterpart. This is shown in the last four frames <strong>of</strong> Fig. 5.4 where we have the same plots as in<br />

Fig. 5.4, except we look at much higher quantum numbers. For the last case, where n = 19 the<br />

classical and quantum distributions are nearly identical. This is an example <strong>of</strong> the correspondence<br />

principle. As the quantum number increases, we expect the quantum system to look more and<br />

more like its classical counter part.<br />

131


1<br />

0.8<br />

0.6<br />

0.4<br />

0.2<br />

-4 -2 2 4<br />

2<br />

1<br />

-4 -2 2 4<br />

-1<br />

-2<br />

Figure 5.3: Harmonic oscillator functions for n = 0 to 3<br />

1<br />

0.5<br />

-4 -2 2 4<br />

-0.5<br />

-1<br />

4<br />

2<br />

-4 -2 2 4<br />

-2<br />

-4<br />

132


Figure 5.4: <strong>Quantum</strong> and Classical Probability Distribution Functions for Harmonic Oscillator<br />

for n = 0, 1, 2, 3, 4, 5, 9, 14, 19<br />

2.5<br />

2<br />

1.5<br />

1<br />

0.5<br />

-4 -2 2 4<br />

1.5<br />

1.25<br />

1<br />

0.75<br />

0.5<br />

0.25<br />

-4 -2 2 4<br />

0.8<br />

0.6<br />

0.4<br />

0.2<br />

-7.5 -5 -2.5 2.5 5 7.5<br />

0.6<br />

0.5<br />

0.4<br />

0.3<br />

0.2<br />

0.1<br />

-7.5 -5 -2.5 2.5 5 7.5<br />

1.75<br />

1.5<br />

1.25<br />

1<br />

0.75<br />

0.5<br />

0.25<br />

-4 -2 2 4<br />

1.4<br />

1.2<br />

1<br />

0.8<br />

0.6<br />

0.4<br />

0.2<br />

-4 -2 2 4<br />

0.7<br />

0.6<br />

0.5<br />

0.4<br />

0.3<br />

0.2<br />

0.1<br />

-7.5 -5 -2.5 2.5 5 7.5<br />

0.6<br />

0.5<br />

0.4<br />

0.3<br />

0.2<br />

0.1<br />

-7.5 -5 -2.5 2.5 5 7.5<br />

133


5.3.3 Molecular Vibrations<br />

The fully quantum mechanical treatment <strong>of</strong> both the electronic and nuclear dynamics <strong>of</strong> even a<br />

diatomic molecule is a complicated affair. The reason for this is that we are forced to find the<br />

stationary states for a potentially large number <strong>of</strong> particles–all <strong>of</strong> which interact, constrained<br />

by a number <strong>of</strong> symmetry relations (such as the fact that no two electrons can be in the same<br />

state at the same time.) In general, the exact solution <strong>of</strong> a many-body problem (such as this)<br />

is impossible. (In fact, believe it it is rigorously impossible for even three classically interacting<br />

particles..although many have tried. ) However, the mass <strong>of</strong> the electron is on the order <strong>of</strong> 10 3 to<br />

10 4 times smaller than the mass <strong>of</strong> a typical nuclei. Thus, the typical velocities <strong>of</strong> the electrons<br />

is much larger than the typical nuclear velocities. We can then assume that the electronic cloud<br />

surrounding the nuclei will respond instantaneously to small and slow changes to the nuclear<br />

positions. Thus, to a very good approximation, we can separate the nuclear motion from the<br />

electonic motion. This separation <strong>of</strong> the nuclear and electronic motion is called the Born-<br />

Oppenheimer Approximation or the Adiabatic Approximation. This approximation is<br />

one <strong>of</strong> the MOST important concepts in chemical physics and is covered in more detail in Section<br />

8.4.1.<br />

Fundimental notion is that the nuclear motion <strong>of</strong> a molecule occurs in the average field <strong>of</strong><br />

the electrons. In other words, the electronic charge distribution acts as an extremely complex<br />

multi-dimensional potential energy surface which governs the motion and dynamics <strong>of</strong> the atoms<br />

in a molecule. Consequently, since chemistry is the science <strong>of</strong> chemical structure, changes, and<br />

dynamics, nearly all chemical reactions can be described in terms <strong>of</strong> nuclear motion on one (or<br />

more) potential energy surface. In Fig. ?? is the London-Eyring-Polanyi-Sato (LEPS) [1]surface<br />

for the F + H2 → HF + H reaction using the Mukerman V set <strong>of</strong> parameters.[3] The LEPS<br />

surface is an empirical potential energy surface based upon the London-Heitler valance bond<br />

theory. Highly accurate potential functions are typically obtained by performing high level ab<br />

initio electronic structure calculations sampling over numerous configurations <strong>of</strong> the molecule.[2]<br />

For diatomic molecules, the nuclear stretching potential can be approximated as a Morse<br />

potential curve<br />

V (r) = De(1 − e −α(r−req ) 2 − De<br />

(5.139)<br />

where De is the dissociation energy, α sets the range <strong>of</strong> the potential, and req is the equilibrium<br />

bond length. The Morse potential for HF is shown in Fig. 5.6 and is parameterized by De =<br />

591.1kcal/mol, α = 2.2189˚A −1 , and req = 0.917˚A.<br />

Close to the very bottom <strong>of</strong> the potential well, where r − re is small, the potential is nearly<br />

harmonic and we can replace the nuclear SE with the HO equation by simply writing that the<br />

angular frequancy is<br />

�<br />

V<br />

ω =<br />

′′ (re)<br />

(5.140)<br />

m<br />

So, measuring the vibrational spectrum <strong>of</strong> the well will give us the curvature <strong>of</strong> the well since<br />

(En − Em)/¯h is always an integer multiple <strong>of</strong> ω for harmonic systems. The red curve in Fig. 5.6<br />

is a parabolic approximation for the bottom <strong>of</strong> the well.<br />

V (r) = De(−1 + α 2 (r − re) 2 /2 + . . .) (5.141)<br />

134


Figure 5.5: London-Eyring-Polanyi-Sato (LEPS) empirical potential for the F + H2 → F H + H<br />

chemical reaction<br />

3.5<br />

3<br />

2.5<br />

2<br />

1.5<br />

1<br />

0.5<br />

rHH<br />

4<br />

0.5 1 1.5 2 2.5 3 3.5 4<br />

135<br />

rFH


V HkCalêmolL<br />

600<br />

400<br />

200<br />

-200<br />

-400<br />

-600<br />

Figure 5.6: Morse well and harmonic approximation for HF<br />

1 2 3 4<br />

Clearly, Deα2 �<br />

is the force constant. So the harmonic frequency for the well is ω = k/µ, where<br />

µ is the reduced mass, µ = m1m2/(m1 + m2) and one would expect that the vibrational energy<br />

levels would be evenly spaced according to a harmonic progression. Deviations from this are due<br />

to anharmonic effects introduced by the inclusion <strong>of</strong> higher order terms in the Taylor expansion<br />

<strong>of</strong> the well. As one might expect, the harmonic expansion provides a descent estimate <strong>of</strong> the<br />

potential energy surface close to the equilibrium geometry.<br />

5.4 Numerical Solution <strong>of</strong> the Schrödinger Equation<br />

5.4.1 Numerov Method<br />

Clearly, finding bound state soutions for the Schrödinger equation is an important task. Unfortunately,<br />

we can only solve a few systems exactly. For the vast majority <strong>of</strong> system which<br />

we cannot handle exactly, we need to turn to approximate means to finde solutions. In later<br />

chapters, we will examine variational methods and perturbative methods. Here, we will look at<br />

a very simple scheme based upon propagating a trial solution at a given point. This methods is<br />

called the ”Numerov” approach after the Russian astronomer who developed the method. It can<br />

be used to solve any differential equation <strong>of</strong> the form:<br />

f ′′ (r) = f(r)u(r) (5.142)<br />

where u(r) is simply a function <strong>of</strong> r and f ′′ is the second derivative <strong>of</strong> f(r) with respect to r and<br />

which is the solution we are looking for. For the Schrödinger equation we would write:<br />

ψ ′′ = 2m<br />

2 (V (r) − E)ψ (5.143)<br />

¯h<br />

136<br />

rFH


Figure 5.7: Model potential for proton tunneling.<br />

V Hcm<br />

6000<br />

-1L 4000<br />

2000<br />

-1 -0.5 0.5 1<br />

-2000<br />

-4000<br />

x HbohrL<br />

The basic proceedure is to expand the second derivative as a finite difference so that if we know<br />

the solution at one point, xn, we can get the solution at some small point a distance h away,<br />

xn+1 = xn + h.<br />

f[n + 1] = f[n] + hf ′ [n] + h2<br />

2! f ′′ [n] + h3<br />

3! f ′′′ [n] + h4<br />

4! f (4) [n] . . . (5.144)<br />

f[n − 1] = f[n] − hf ′ [n] + h2<br />

2! f ′′ [n] − h3<br />

3! f ′′′ [n] + h4<br />

4! f (4) [n] (5.145)<br />

If we combine these two equations and solve for f[n + 1] we get a result:<br />

f[n + 1] = −f[n − 1] + 2f[n] + f ′′ [n]h 2 + h4<br />

12 f (4) [n] + O[h 6 ] (5.146)<br />

Since f is a solution to the Schrödinger equation, f ′′ = Gf where<br />

G = 2m<br />

2 (V − E)<br />

¯h<br />

we can get the second derivative <strong>of</strong> f very easily. However, for the higher order terms, we have<br />

to work a bit harder. So let’s expand f ′′<br />

f ′′ [n + 1] = −2f ′′ [n] − f ′′ [n − 1] + s 2 f (4) [n]<br />

and truncate at order h 6 . Now, solving for f (4) [n] and substituting f ′′ = Gf we get<br />

f[n + 1] =<br />

�<br />

2f[n] − f[n − 1] + h2 (G[n − 1]f[n − 1] + 10G[n]f[n])�<br />

12<br />

�<br />

1 − h2<br />

(5.147)<br />

G[n + 1]�<br />

12<br />

which is the working equation.<br />

Here we take a case <strong>of</strong> proton tunneling in a double well potential. The potential in this case<br />

is the V (x) = α(x 4 − x 2 ) function shown in Fig. 5.7. Here we have taken the parameter α = 0.1<br />

137


1<br />

0.5<br />

-2 -1 1 2<br />

-0.5<br />

-1<br />

-2 -1 1 2<br />

Figure 5.8: Double well tunneling states as determined by the Numerov approach. On the left<br />

is the approximate lowest energy (symmetric) state with no nodes and on the rigth is the next<br />

lowest (antisymmetric) state with a single node. The fact that the wavefunctions are heading <strong>of</strong>f<br />

towards infinity indicated the introduction <strong>of</strong> an additional node coming in from x = ∞.<br />

and m = 1836 as the proton and use atomic units throughout. Also show in Fig. 5.7 are effective<br />

harmonic oscillator wells for each side <strong>of</strong> the barrier. Notice that the harmonic approximation<br />

is pretty crude since the harminic well tends to over estimate the steepness <strong>of</strong> the inner portion<br />

and under estimate the steepness <strong>of</strong> the outer portions. Nonetheless, we can use the harminic<br />

oscillator ground states in each well as starting points.<br />

To use the Numerov method, one starts by guessing an initial energy, E, and then propagating<br />

a trial solution to the Schrödinger equation. The Curve you obtain is in fact a solution to the<br />

equation, but it will ususally not obey the correct boundary conditions. For bound states, the<br />

boundary condition is that ψ must vanish exponentially outside the well. So, we initialize the<br />

method by forcing ψ[1] to be exactly 0 and ψ[2] to be some small number. The exact values<br />

really make no difference. If we are <strong>of</strong>f by a bit, the Numerov wave will diverge towards ±∞ as x<br />

increases. As we close in on a physically acceptible solution, the Numerov solution will begin to<br />

exhibit the correct asymptotic behavour for a while before diverging. We know we have hit upon<br />

an eigenstate when the divergence goes from +∞ to −∞ or vice versa, signeling the presence<br />

<strong>of</strong> an additional node in the wavefunction. The proceedure then is to back up in energy a bit,<br />

change the energy step and gradually narrow in on the exact energy. In Figs. 5.8a and 5.8b are<br />

the results <strong>of</strong> a Numerov search for the lowest two states in the double well potential. One at<br />

-3946.59cm −1 and the other at -3943.75cm −1 . Notice that the lowest energy state is symmetric<br />

about the origin and the next state is anti-symmetric about the origin. Also in both cases, the<br />

Numerov function diverges since we are not precisely at a stationary solution <strong>of</strong> the Schrödinger<br />

equation...but we are within 0.01cm −1 <strong>of</strong> the true eigenvalue.<br />

The advantage <strong>of</strong> the Numerov method is that it is really easy to code. In fact you can<br />

even code it in Excell. Another advantage is that for radial scattering problems, the out going<br />

boundary conditions occur naturally, making it a method <strong>of</strong> choice for simple scattering problems.<br />

In the Mathematica notebooks, I show how one can use the Numerov method to compute the<br />

scattering phase shifts and locate resonance for atomic collisions. The disadvantage is that you<br />

have to search by hand for the eigenvalues which can be extremely tedious.<br />

138<br />

3<br />

2<br />

1<br />

-1


5.4.2 Numerical Diagonalization<br />

A more general approach is based upon the variational principle (which we will discuss later)<br />

and the use <strong>of</strong> matrix representations. If we express the Hamiltonian operator in matrix form in<br />

some suitable basis, then the eigenfunctions <strong>of</strong> H can also be expressed as linear combinations<br />

<strong>of</strong> those basis functions, subject to the constraint that the eigenfunctions be orthonormal. So,<br />

what we do is write:<br />

〈φn|H|φm〉 = Hnm<br />

and<br />

|ψj〉 = �<br />

〈φn|ψj〉|φn〉<br />

n<br />

The 〈φn|ψj〉 coefficients are also elements <strong>of</strong> a matrix, Tnj which transforms a vector in the φ<br />

basis to the ψ basis. Conseqently, there is a one-to-one relation between the number <strong>of</strong> basis<br />

functions in the ψ basis and the basis functions in the φ basis.<br />

If |ψn〉 is an eigenstate <strong>of</strong> H, then<br />

Multiplying by 〈φm| and resolving the identity,<br />

Thus,<br />

or in more compact form<br />

H|ψj〉 = Ej|ψj〉.<br />

�<br />

〈φm|H|φn〉〈φn|ψj〉 = 〈φm|ψj〉Ej<br />

n<br />

�<br />

HmnTnj<br />

n<br />

= EjTmj (5.148)<br />

�<br />

TmjHmnTnj = Ej<br />

mn<br />

T † HT = E Î<br />

(5.149)<br />

where Î is the identity matrix. In otherwords, the T -matrix is simply the matrix which brings<br />

H to diagonal form.<br />

Diagonalizing a matrix by hand is very tedeous for anything beyond a 3×3 matrix. Since this<br />

is an extremely common numerical task, there are some very powerful numerical diagonalization<br />

routines available. Most <strong>of</strong> the common ones are in the Lapack package and are included as part<br />

<strong>of</strong> the Mathematica kernel. So, all we need to do is to pick a basis, cast our Hamiltonian into<br />

that basis, truncate the basis (usually determined by some energy cut-<strong>of</strong>f) and diagonalize away.<br />

Usually the diagonalization part is the most time consuming. Of course you have to be prudent<br />

in choosing your basis.<br />

A useful set <strong>of</strong> basis functions are the trigonmetric forms <strong>of</strong> the Tchebychev polynomials. 1<br />

These are a set <strong>of</strong> orthogonal functions which obey the following recurrence relation<br />

Tn+1(x) − 2xTn(x) + Tn−1(x) = 0 (5.150)<br />

139


1<br />

0.75<br />

0.5<br />

0.25<br />

-1 -0.5 0.5 1<br />

-0.25<br />

-0.5<br />

-0.75<br />

-1<br />

Figure 5.9: Tchebyshev Polynomials for n = 1 − 5<br />

Table 5.1: Tchebychev polynomials <strong>of</strong> the first type<br />

To = 1<br />

T1 = x<br />

T2 = 2x 2 − 1<br />

T3 = 4x 3 − 3x<br />

T4 = 8x 4 − 8x 2 − 1<br />

T5 = 16x 5 − 20x 3 + 5x<br />

140


Table 5.1 lists a the first few <strong>of</strong> these polynomials as functions <strong>of</strong> x and a few <strong>of</strong> these are plotted<br />

in Fig. 5.9<br />

It is important to realize that these functions are orthogonal on a finite range and that<br />

integrals over these functions must include a weighting function w(x) = 1/ √ 1 − x 2 . The orthogonality<br />

relation for the Tn polynomials is<br />

� +1<br />

−1<br />

⎧<br />

⎪⎨ 0 m �= m<br />

π<br />

Tm(x)Tn(x)w(x)dx = m = n �= 0<br />

⎪⎩<br />

2<br />

π m = n = 0<br />

(5.151)<br />

Arfkin’s Mathematical Methods for Physicists has a pretty complete overview <strong>of</strong> these special<br />

functions as well as many others. As usual, These are encorporated into the kernel <strong>of</strong> Mathematica<br />

and the Mathematica book and on-line help pages has some useful information regarding<br />

these functions as well as a plethera <strong>of</strong> other functions.<br />

From the recurrence relation it is easy to show that the Tn(x) polynomials satisfy the differential<br />

equation:<br />

(1 − x 2 )T ′′<br />

n − xT ′ x + n 2 Tn = 0 (5.152)<br />

If we make a change <strong>of</strong> variables from x = cos(θ) and dx = − sin θdθ, then the differential<br />

equation reads<br />

dTn<br />

dθ + n2 Tn = 0 (5.153)<br />

This is a harmonic oscillator and has solutions sin nθ and cos nθ. From the boundary conditions<br />

we have two linearly independent solutions<br />

and<br />

The normalization condition then becomes:<br />

and<br />

� +1<br />

−1<br />

� +1<br />

−1<br />

Tn = cos nθ = cos n(arccosx)<br />

Vn = sin nθ.<br />

Tm(x)Tn(x)w(x)dx =<br />

Vm(x)Vn(x)w(x)dx =<br />

� π<br />

0<br />

� π/2<br />

−π/2<br />

cos(mθ) cos(nθ)dθ (5.154)<br />

sin(mθ) sin(nθ)dθ (5.155)<br />

which is precisely the normalization integral we perform for the particle in a box state assuming<br />

the width <strong>of</strong> the box was π. For more generic applications, we can scale θ and its range to any<br />

range.<br />

1 There are at least 10 ways to spell Tchebychev’s last name Tchebychev, Tchebyshev, Chebyshev are the<br />

most common, as well as Tchebysheff, Tchebycheff, Chebysheff, Chevychef, . . .<br />

141


The way we use this is to use the φn = N sin nx basis functions as a finite basis and truncate<br />

any expansion in this basis at some point. For example, since we are usually interested in low<br />

lying energy states, setting an energy cut-<strong>of</strong>f to basis is exactly equivalent to keeping only the<br />

lowest ncut states. The kinetic energy part <strong>of</strong> the Hamiltonian is diagonal in this basis, so we get<br />

that part for free. However, the potential energy part is not diagonal in the φn = N sin nx basis,<br />

so we have to compute its matrix elements:<br />

�<br />

Vnm = φn(x)V (x)φm(x)dx (5.156)<br />

To calculate this integral, let us first realize that [V, x] = 0, so the eigenstates <strong>of</strong> x are also<br />

eigenstates <strong>of</strong> the potential. Taking matrix elements in the finite basis,<br />

xnm = N 2<br />

�<br />

φn(x)xφm(x)dx,<br />

and diagonalizing it yields a finite set <strong>of</strong> ”position” eigenvalues, {xi} and a transformation for<br />

converting between the ”position representation” and the ”basis representation”,<br />

Tin = 〈xi|φn〉,<br />

which is simply a matrix <strong>of</strong> the basis functions evaluated at each <strong>of</strong> the eigenvalues. The special<br />

set <strong>of</strong> points defined by the eigenvalues <strong>of</strong> the position operator are the Gaussian quadrature<br />

points over some finite range.<br />

This proceedure termed the ”discrete variable representation” was developed by Light and<br />

coworkers in the 80’s and is a very powerful way to generate coordinate representations <strong>of</strong> Hamiltonian<br />

matrixes. Any matrix in the basis representation (termed the FBR for finite basis representation)<br />

can be transformed to the discrete variable representation (DVR) via the transformation<br />

matrix T . Moreover, there is a 1-1 correspondency between the number <strong>of</strong> DVR points and the<br />

number <strong>of</strong> FBR basis functions. Here we have used only the Tchebychev functions. One can<br />

generate DVRs for any set <strong>of</strong> orthogonal polynomial function. The Mathematica code below generates<br />

the required transformations, the points, the eigenvalues <strong>of</strong> the second-derivative operator,<br />

and a set <strong>of</strong> quadrature weights for the Tchebychev sine functions over a specified range:<br />

dv2fb[DVR_, T_] := T.DVR.Transpose[T];<br />

fb2dv[FBR_, T_] := Transpose[T].FBR.T;<br />

tcheby[npts_, xmin_, xmax_] := Module[{pts, fb, del},<br />

del = xmax - xmin;<br />

pts = Table[i*del*(1/(npts + 1)) + xmin, {i, npts}] // N;<br />

fbrke = Table[(i*(Pi/del))^2, {i, npts}] // N;<br />

w = Table[del/(npts + 1), {i, npts}] // N;<br />

T = Table[<br />

Sqrt[2.0/(npts + 1)]*Sin[(i*j)*Pi/(npts + 1)],<br />

{i, npts}, {j, npts}] // N;<br />

Return[{pts, T, fbrke, w}]<br />

]<br />

To use this, we first define a potential surface, set up the Hamiltonian matrix, and simply<br />

diagonalize. For this example, we will take the same double well system described above and<br />

compare results and timings.<br />

142


V[x_] := a*(x^4 - x^2);<br />

cmm = 8064*27.3;<br />

params = {a -> 0.1, m -> 1836};<br />

{x, T, K, w} = tcheby[100, -1.3, 1.3];<br />

Kdvr = (fb2dv[DiagonalMatrix[K], T]*m)/2 /. params;<br />

Vdvr = DiagonalMatrix[V[x]] /. params;<br />

Hdvr = Kdvr + Vdvr;<br />

tt = Timing[{w, psi} = Transpose[<br />

Sort[Transpose[Eigensystem[Hdvr]]]]];<br />

Print[tt]<br />

(Select[w*cmm , (# < 3000) &]) // TableForm<br />

This code sets up the DVR points x, the transformation T and the FBR eigenvalues K using<br />

the tcheby[n,xmin,xmax] Mathematica module defined above. We then generate the kinetic<br />

energy matrix in the DVR using the transformation<br />

and form the DVR Hamiltonian<br />

KDV R = T † KF BRT<br />

HDV R = KDV R + VDV R.<br />

The eigenvalues and eigenvectors are computed via the Eigensystem[] routine. These are then<br />

sorted according to their energy. Finally we print out only those states with energy less than<br />

3000 cm −1 and check how long it took. On my 300 MHz G3 laptop, this took 0.3333 seconds to<br />

complete. The first few <strong>of</strong> these are shown in Table 5.2 below. For comparison, each Numerov<br />

iteration took roughly 1 second for each trial function. Even then, the eigenvalues we found are<br />

probabily not as accurate as those computed here.<br />

Table 5.2: Eigenvalues for double well potential computed via DVR and Numerov approaches<br />

i ωi (cm −1 ) Numerov<br />

1 -3946.574 -3946.59<br />

2 -3943.7354 -3943.75<br />

3 -1247.0974<br />

4 -1093.5204<br />

5 591.366<br />

6 1617.424<br />

143


5.5 Problems and Exercises<br />

Exercise 5.2 Consider a harmonic oscillator <strong>of</strong> mass m and angular frequency ω. At time<br />

t = 0, the state <strong>of</strong> this system is given by<br />

|ψ(0)〉 = �<br />

cn|φn〉 (5.157)<br />

where the states |φn〉 are stationary states with energy En = (n + 1/2)¯hω.<br />

n<br />

1. What is the probability, P , that at a measurement <strong>of</strong> the energy <strong>of</strong> the oscillator at some<br />

later time will yield a result greater than 2¯hω. When P = 0, what are the non-zero coefficients,<br />

cn?<br />

2. From now on, let only co and c1 be non zero. Write the normalization condition for |ψ(0)〉<br />

and the mean value 〈H〉 <strong>of</strong> the energy in terms <strong>of</strong> co and c1. With the additional requirement<br />

that 〈H〉 = ¯hω, calculate |co| 2 and |c1| 2 .<br />

3. As the normalized state vector |ψ〉 is defined only to within an arbitrary global phase factor,<br />

as can fix this factor by setting co to be real and positive. We set c1 = |c1|e iφ . We assume<br />

also that 〈H〉 = ¯hω and show that<br />

Calculate φ.<br />

〈x〉 = 1<br />

�<br />

2<br />

¯h<br />

.<br />

mω<br />

(5.158)<br />

4. With |ψ〉 so determined, write |ψ(t)〉 for t > 0 and calculate the value <strong>of</strong> φ at time t.<br />

Deduce the mean <strong>of</strong> 〈x〉(t) <strong>of</strong> the position at time t.<br />

Exercise 5.3 Find 〈x〉, 〈p〉, 〈x 2 〉 and 〈p 2 〉 for the ground state <strong>of</strong> a simple harmonic oscillator.<br />

What is the uncertainty relation for the ground state.<br />

Exercise 5.4 In this problem we consider the the interaction between molecule adsorbed on a<br />

surface and the surface phonons. Represent the vibrational motion <strong>of</strong> the molecule (with reduced<br />

mass µ) as harmonic with force constant K<br />

and the coupling to the phonons as<br />

Ho = −¯h2<br />

2µ<br />

∂2 K<br />

+<br />

∂x2 2 x2<br />

(5.159)<br />

H ′ = −x �<br />

Vk cos(Ωkt) (5.160)<br />

k<br />

where Vk is the coupling between the molecule and phonon <strong>of</strong> wavevector k and frequency Ωk.<br />

144


1. Express the total Hamiltonian as a displaced harmonic well. What happens to the well as<br />

a function <strong>of</strong> time?<br />

2. What is the Golden-Rule transition rate between the ground state and the nth excited state<br />

<strong>of</strong> the system due to phonon interactions? Are there any restrictions as to which final state<br />

can be reached? Which phonons are responsible for this process?<br />

3. From now on, let the perturbing force be constant in time<br />

H ′ = x �<br />

k<br />

Vk<br />

(5.161)<br />

where Vk is the interaction with a phonon with wavevector k. Use the lowest order level<br />

<strong>of</strong> perturbation theory necessary to construct the transition probability between the ground<br />

state and the second-excited state.<br />

Exercise 5.5 Let<br />

Show that the Harmonic Oscillator Hamiltonian is<br />

X = ( mω<br />

2¯h )1/2 x (5.162)<br />

1<br />

P = (<br />

2¯hmω )1/2p (5.163)<br />

H = ¯hω(P 2 + X 2 ) (5.164)<br />

Now, define the operator: a † = X − iP . Show that a † acting on the harmonic oscillator ground<br />

state is also an eigenstate <strong>of</strong> H. What is the energy <strong>of</strong> this state? Use a † to define a generating<br />

relationship for all the eigenstates <strong>of</strong> H.<br />

Exercise 5.6 Show that if one expands an arbitrary potential, V (x) about its minimum at xmin,<br />

and neglects terms <strong>of</strong> order x 3 and above, one always obtains a harmonic well. Show that a<br />

harmonic oscillator subject to a linear perturbation can be expressed as an unperturbed harmonic<br />

oscillator shifted from the origin.<br />

Exercise 5.7 Consider the one-dimensional Schrödinger equation with potential<br />

�<br />

m<br />

V (x) = 2 ω2x2 x > 0<br />

+∞ x ≤ 0<br />

Find the energy eigenvalues and wavefunctions.<br />

(5.165)<br />

Exercise 5.8 An electron is contained inside a hard sphere <strong>of</strong> radius R. The radial components<br />

<strong>of</strong> the lowest S and P state wavefunctions are approximately<br />

ψS(x) ≈ sin(kr)<br />

kr<br />

ψP (x) ≈ cos(kr)<br />

kr<br />

− sin(kr)<br />

(kr)<br />

145<br />

(5.166)<br />

∂ψS(kr)<br />

= . (5.167)<br />

2 ∂(kr)


1. What boundary conditions must each state obey?<br />

2. Using E = k 2 ¯h 2 /(2m) and the above boundary conditions, what are the energies <strong>of</strong> each<br />

state?<br />

3. What is the pressure exerted on the surface <strong>of</strong> the sphere if the electron is in the a.) S state,<br />

b.) the P state. (Hint, recall from thermodynamics: dW = P dV = −(dE(R)/dR)dR.)<br />

4. For a solvated e − in water, the S to P energy gap is about 1.7 eV. Estimate the size <strong>of</strong><br />

the the hard-sphere radius for the aqueous electron. If the ground state is fully solvated,<br />

the pressure <strong>of</strong> the solvent on the electron must equal the pressure <strong>of</strong> the electron on the<br />

solvent. What happens to the system when the electron is excited to the P -state from the<br />

equilibrated S state? What happens to the energy gap between the S and P as a result <strong>of</strong><br />

this?<br />

Exercise 5.9 A particle moves in a three dimensional potential well <strong>of</strong> the form:<br />

V (x) =<br />

�<br />

∞ z 2 > a 2<br />

mω 2<br />

2 (x2 + y 2 ), otherwise<br />

Obtain an equation for the eigenvalues and the associated eigenfunctions.<br />

(5.168)<br />

Exercise 5.10 A particle moving in one-dimension has a ground state wavefunction (not-normalized)<br />

<strong>of</strong> the form:<br />

ψo(x) = e −α4 x 4 /4<br />

(5.169)<br />

where α is a real constant with eigenvalue Eo = ¯h 2 α 2 /m. Determine the potential in which the<br />

particle moves. (You do not have to determine the normalization.)<br />

Exercise 5.11 A two dimensional oscillator has the Hamiltonian<br />

H = 1<br />

2 (p2 x + p 2 y) + 1<br />

2 (1 + δxy)(x2 + y 2 ) (5.170)<br />

where ¯h = 1 and δ


Figure 5.10: Ammonia Inversion and Tunneling<br />

1. Using the Spartan electronic structure package (or any other one you have access to), build<br />

a model <strong>of</strong> NH3, and determine its ground state geometry using various levels <strong>of</strong> ab initio<br />

theory. Make a table <strong>of</strong> N − H bond lengths and θ = � H − N − H bond angles for the<br />

equilibrium geometries as a function <strong>of</strong> at least 2 or 3 different basis sets. Looking in the<br />

literature, find experimental values for the equilibrium configuration. Which method comes<br />

closest to the experimental values? Which method has the lowest energy for its equilibrium<br />

configuration.<br />

2. Using the method which you deamed best in part 1, repeat the calculations you performed<br />

above by systematically constraining the H − N − H bond angle to sample configurations<br />

around the equilibrium configuration and up to the planar D3h configuration. Note, it may<br />

be best to constrain two H − N − H angles and then optimize the bond lengths. Sample<br />

enough points on either side <strong>of</strong> the minimum to get a descent potential curve. This is your<br />

Born-Oppenheimer potential as a function <strong>of</strong> θ.<br />

3. Defining the orgin <strong>of</strong> a coordinate system to be the θ = 120 o D3h point on the surface, fit<br />

your ab initio data to the ”W”-potential<br />

V (x) = αx 2 + βx 4<br />

What are the theoretical values <strong>of</strong> α and β?<br />

4. We will now use perturbation theory to compute the tunneling dynamics.<br />

(a) Show that the points <strong>of</strong> minimum potential energy are at<br />

� �1/2 α<br />

xmin = ±<br />

2β<br />

(5.171)<br />

and that the energy difference between the top <strong>of</strong> the barrier and the minimum energy<br />

is given by<br />

V = V (0) − V (xmin) (5.172)<br />

= α2<br />

4β<br />

147<br />

(5.173)


(b) We first will consider the barrier to be infinitely high so that we can expand the potential<br />

function around each xmin. Show that by truncating the Taylor series expansion<br />

above the (x − xmin) 2 terms that the potential for the left and right hand sides are<br />

given by<br />

VL = 2α (x + xmin) 2 − V<br />

and<br />

VR = 2α (x − xmin) 2 − V.<br />

What are the vibrational energy levels for each well?<br />

(c) The wavefunctions for the lowest energy states in each well are given by<br />

with<br />

ψ(x) = γ1/2<br />

exp[−γ2<br />

π1/4 2 (x ± xmin) 2 ]<br />

γ =<br />

� (4µα) 1/2<br />

¯h<br />

� 1/2<br />

.<br />

The energy levels for both sides are degenerate in the limit that the barrier height is<br />

infinite. The total ground state wavefunction for this case is<br />

Ψ(x) =<br />

�<br />

ψL(x)<br />

ψR(x)<br />

However, as the barrier height decreases, the degenerate states begin to mix causing<br />

the energy levels to split. Define the ”high barrier” hamiltonian as<br />

for x < 0 and<br />

�<br />

H = − ¯h2 ∂<br />

2µ<br />

2<br />

+ VL(x)<br />

∂x<br />

H = − ¯h2 ∂<br />

2µ<br />

2<br />

+ VR(x)<br />

∂x<br />

for x > 0. Calculate the matrix elements <strong>of</strong> H which mix the two degenerate left and<br />

right hand ground state wavefunctions: i.e.<br />

where<br />

〈Ψ|H|Ψ〉 =<br />

�<br />

.<br />

HRR HLR<br />

HRL HLL<br />

HRR = 〈ψR|H|ψR〉<br />

, with similar definitions for HRL, HLL and HLR. Obtain numerical values <strong>of</strong> each<br />

matrix element using the values <strong>of</strong> α and β you determined above (in cm −1 ). Use the<br />

mass <strong>of</strong> a H atom for the reduced mass µ.<br />

148<br />


(d) Since the ψL and ψR basis functions are non-orthogonal, you will need to consider the<br />

overlap matrix, S, when computing the eigenvalues <strong>of</strong> H. The eigenvalues for this<br />

system can be determined by solving the secular equation<br />

�<br />

�<br />

�<br />

�<br />

�<br />

α − λ β − λS<br />

β − λS α − λ<br />

�<br />

�<br />

�<br />

� = 0 (5.174)<br />

�<br />

where α = HRR = HLL and β = HLR = HRL (not to be confused with the potential<br />

parameters above). Using Eq. 5.174, solve for λ and determine the energy splitting<br />

in the ground state as a function the unperturbed harmonic frequency and the barrier<br />

height, V . Calculate this splitting using the parameters you computed above. What is<br />

the tunneling frequency? The experimental results is ∆E = 0.794cm −12 .<br />

Exercise 5.14 Consider a system in which the Lagrangian is given by<br />

L(qi, ˙qi) = T (qi, ˙qi) − V (qi) (5.175)<br />

where we assume T is quadratic in the velocities. The potential is independent <strong>of</strong> the velocity<br />

and neither T nor V carry any explicit time dependency. Show that<br />

⎛<br />

d<br />

⎝<br />

dt<br />

�<br />

⎞<br />

∂L<br />

˙qj − L⎠<br />

= 0<br />

∂ ˙qj<br />

j<br />

The constant quantity in the (. . .) defines a Hamiltonian, H. Show that under the assumed<br />

conditions, H = T + V<br />

Exercise 5.15 The Fermat principle in optics states that a light ray will follow the path, y(x)<br />

which minimizes its optical length, S, through a media<br />

S =<br />

� x2,y2<br />

x1,y1<br />

n(y, x)ds<br />

where n is the index <strong>of</strong> refraction. For y2 = y1 = 1 and −x1 = x2 = 1 find the ray-path for<br />

1. n = exp(y)<br />

2. n = a(y − yo) for y > yo<br />

Make plots <strong>of</strong> each <strong>of</strong> these trajectories.<br />

Exercise 5.16 In a quantum mechanical system there are gi distinct quantum states between<br />

energy Ei and Ei + dEi. In this problem we will use the variational principle and Lagrange<br />

multipliers to determine how ni particles are distributed amongst these states subject to the constraints<br />

1. The number <strong>of</strong> particles is fixed:<br />

n = �<br />

2 From Molecular Structure and Dynamics, by W. Flygare, (Prentice Hall, 1978)<br />

149<br />

i<br />

ni


2. the total energy is fixed �<br />

niEi = E<br />

We consider two cases:<br />

i<br />

1. For identical particles obeying the Pauli exclusion principle, the probability <strong>of</strong> a given configuration<br />

is<br />

WF D = �<br />

i<br />

gi<br />

ni!(gi − ni)!<br />

Show that maximizing WF D subject to the constraints above leads to<br />

ni =<br />

gi<br />

e λ1+λ2Ei + 1<br />

(5.176)<br />

with the Lagrange multipliers λ1 = −Eo/kT and λ2 = 1/kT . Hint: try working with the<br />

log W and use Sterling’s approximation in the limit <strong>of</strong> a large number <strong>of</strong> particles .<br />

2. In this case we still consider identical particles, but relax the restriction on the fixed number<br />

<strong>of</strong> particles in a given state. The probability for a given distribution is then<br />

WBE = �<br />

i<br />

(ni + gi − 1)!<br />

.<br />

ni!(gi − 1)!<br />

Show that by minimizing WBE subject to the constraints above leads to the occupation<br />

numbers:<br />

gi<br />

ni =<br />

eλ1+λ2Ei − 1<br />

where again, the Lagrange multipliers are λ1 = −Eo/kT and λ2 = 1/kT . This yields the<br />

Bose-Einstein statistics. Note: assume that gi ≫ 1<br />

3. Photons satisfy the Bose-Einstein distribution and the constraint that the total energy is<br />

constant. However, there is no constrain regarding the total number <strong>of</strong> photons. Show that<br />

by eliminating the fixed number constraint leads to the foregoing result with λ1 = 0.<br />

150


Bibliography<br />

[1] C. A. Parr and D. G. Trular, J. Phys. Chem. 75, 1844 (1971).<br />

[2] H. F. Schaefer III, J. Phys. Chem. 89, 5336 (1985).<br />

[3] P. A. Whitlock and J. T. Muckermann, J. Chem. Phys. 61, 4624 (1974).<br />

151


Chapter 6<br />

<strong>Quantum</strong> <strong>Mechanics</strong> in 3D<br />

In the next few lectures, we will focus upon one particular symmetry, the isotropy <strong>of</strong> free space.<br />

As a collection <strong>of</strong> particles rotates about an arbitrary axis, the Hamiltonian does not change. If<br />

the Hamoltonian does in fact depend explicitly upon the choice <strong>of</strong> axis, the system is “gauged”,<br />

meaning all measurements will depend upon how we set up the coordinate frame. A Hamiltonian<br />

with a potential function which depends only upon the coordinates, e.g. V = f(x, y, z), is gauge<br />

invarient, meaning any measurement that I make will not depend upon my choice <strong>of</strong> reference<br />

frame. On the other hand, if our Hamiltonian contains terms which couple one reference frame<br />

to another (as in the case <strong>of</strong> non-rigid body rotations), we have to be careful in how we select<br />

the “gauge”. While this sounds like a fairly specialized case, it turns out that many ordinary<br />

phenimina depend upon this, eg. figure skaters, falling cats, floppy molecules. We focus upon<br />

rigid body rotations first.<br />

For further insight and information into the quantum mechanics <strong>of</strong> angular momentum, I<br />

recommend the following texts and references:<br />

1. Theory <strong>of</strong> Atomic Structure, E. Condon and G. Shortley. This is the classical book on<br />

atomic physics and theory <strong>of</strong> atomic spectroscopy and has inspired generations since it<br />

came out in 1935.<br />

2. Angular Momentum–understanding spatial aspects in chemistry and physics, R. N. Zare.<br />

This book is the text for the second-semester quantum mechanics at Stanford taught by<br />

Zare (when he’s not out looking for Martians). It’s a great book with loads <strong>of</strong> examples in<br />

spectroscopy.<br />

3. <strong>Quantum</strong> Theory <strong>of</strong> Angular Momentum, D. A. Varshalovich, A. Moskalev, and V. Khersonskii.<br />

Not to much physics in this book, but if you need to know some relation between<br />

Wigner-D functions and Racah coefficients, or how to derive 12j symbols, this book is for<br />

you.<br />

First, we need to look at what happens to a Hamiltonian under rotation. In order to show that<br />

H is invariant to any rotations, we need only to show that it is invarient under an infinitesimal<br />

rotation.<br />

152


6.1 <strong>Quantum</strong> Theory <strong>of</strong> Rotations<br />

Let δ � φ by a vector <strong>of</strong> a small rotation equal in magnitude to the angle δφ directed long an<br />

arbitrary axis. Rotating the system by δ � φ changes the direction vectors �rα by δ�rα.<br />

δ�rα = δ � φ × �rα<br />

Note that the × denotes the vector “cross” product. Since we will be using cross-products<br />

through out these lectures, we pause to review the operation.<br />

A cross product between two vectors is computed as<br />

�c = �a × �b �<br />

�<br />

� î ˆj kˆ<br />

�<br />

= �<br />

�<br />

�<br />

ai aj ak<br />

bi bj bk<br />

�<br />

�<br />

�<br />

�<br />

�<br />

�<br />

�<br />

= î(ajbk − bjak) − ˆj(aibk − biak) + ˆ k(aibj − biaj)<br />

(6.1)<br />

= ɛijkajbk (6.2)<br />

Where ɛijk is the Levi-Cevita symbol or the “anti-symmetric unit tensor” defined as<br />

⎧<br />

⎪⎨ 0 if any <strong>of</strong> the indices are the same<br />

ɛijk = 1 for even permuations <strong>of</strong> the indices<br />

⎪⎩<br />

−1 for odd permutations <strong>of</strong> the indices<br />

(Note that we also have assumed a “summation convention” where by we sum over all repeated<br />

indices. Some elementary properties are ɛiklɛikm = δlm and ɛiklɛikl = 6.)<br />

So, an arbitrary function ψ(r1, r2, · · ·) is transformed by the rotation into:<br />

ψ1(r1 + δr1, r2 + δr2, · · ·) = ψ(r1, r2, · · ·) + �<br />

δra · � ∇ψa<br />

Thus, we conclude, that the operator<br />

= ψ(r1, r2, · · ·) + �<br />

δ� φ × ra · � ∇aψa<br />

=<br />

1 + δ� φ · �<br />

�ra × � ∇a<br />

a<br />

�<br />

1 + δ� φ · �<br />

�ra × � �<br />

∇a<br />

is the operator for an infintesimal rotation <strong>of</strong> a system <strong>of</strong> particles. Since δφ is a constant, we<br />

can show that this operator commutes with the Hamiltonian<br />

�<br />

�<br />

[ �ra × � �<br />

∇a , H] = 0 (6.6)<br />

a<br />

This implies then a particular conservation law related to the isotropy <strong>of</strong> space. This is <strong>of</strong> course<br />

angular momentum so that<br />

�<br />

�<br />

�ra × � �<br />

∇a<br />

(6.7)<br />

a<br />

153<br />

a<br />

a<br />

a<br />

ψa<br />

(6.3)<br />

(6.4)<br />

(6.5)


must be at least proportional to the angular momentum operator, L. The exact relation is<br />

which is much like its classical counterpart<br />

¯hL = �r × �p = −i¯h�r × � ∇ (6.8)<br />

L = 1<br />

�r × �v. (6.9)<br />

m<br />

The operator is <strong>of</strong> course a vector quantity, meaning that is has direction. The components <strong>of</strong><br />

the angular momentum vector are:<br />

¯hLx = ypz − zpy<br />

¯hLy = zpx − zpz<br />

¯hLz = xpy − ypx<br />

¯hLi = ɛijkxjpk<br />

(6.10)<br />

(6.11)<br />

(6.12)<br />

(6.13)<br />

For a system in a external field, antgular momentum is in general not conserved. However, if<br />

the field posesses spherical symmetry about a central point, all directions in space are equivalent<br />

and angular momentum about this point is conserved. Likewise, in an axially symmetric field,<br />

motion about the axis is conserved. In fact all the conservation laws which apply in classical<br />

mechanics have quantum mechanical analogues.<br />

We now move on to compute the commutation rules between the Li operators and the x and<br />

p operators First we note:<br />

In short hand:<br />

[Lx, x] = [Ly, y] = [Lz, z] = 0 (6.14)<br />

[Lx, y] = 1<br />

¯h ((ypz − zpy)y − y(ypz − zpy)) = − z<br />

¯h [py, y] = iz (6.15)<br />

[Li, xk] = iɛiklxl<br />

We need also to know how the various components commute with one another:<br />

¯h[Lx, Ly] = Lx(zpx − xpz) − (zpx − xpz)Lx<br />

(6.16)<br />

(6.17)<br />

= (Lxz − zLx)px − x(Lxpz − pzLx) (6.18)<br />

= −iypx + ixpy<br />

154<br />

(6.19)


Which we can summarize as<br />

= i¯hLz<br />

[Ly, Lz] = iLx<br />

[Lz, Lx] = iLy<br />

[Lx, Ly] = iLz<br />

[Li, Lj] = iɛijkLk<br />

Now, denote the square <strong>of</strong> the modulus <strong>of</strong> the total angular momentum by L 2 , where<br />

L 2 = L 2 x + L 2 y + L 2 z<br />

Notice that this operator commutes with all the other Lj operators,<br />

For example:<br />

also,<br />

Thus,<br />

(6.20)<br />

(6.21)<br />

(6.22)<br />

(6.23)<br />

(6.24)<br />

(6.25)<br />

[L 2 , Lx] = [L 2 , Ly] = [L 2 , Lz] = 0 (6.26)<br />

[L 2 x, Lz] = Lx[Lx, Lz] + [Lx, Lz]Lx = −i(LxLy + LyLx) (6.27)<br />

[L 2 y, Lz] = i(LxLy + LyLx) (6.28)<br />

[L 2 , Lz] = 0 (6.29)<br />

Thus, I can measure L 2 and Lz simultaneously. (Actually I can measure L 2 and any one component<br />

Lk simultaneously. However, we ueually pick this one as the z axis to make the math<br />

easier, as we shall soon see.)<br />

A consequence <strong>of</strong> the fact that Lx, Ly, and Lz do not commute is that the angular momentum<br />

vector � L can never lie exactly along the z axis (or exactly along any other<br />

�<br />

axis for that matter).<br />

We can interpret this in a classical context as a vector <strong>of</strong> length |L| = ¯h L(L + 1) with the Lz<br />

component being ¯hm. The vector is then constrained to lie in a cone as shown in Fig. ??. We<br />

will take up this model at the end <strong>of</strong> this chapter in the semi-classical context.<br />

It is also convienent to write Lx and Ly as a linear combination<br />

L+ = Lx + iLyL− = Lx − iLy<br />

(Recall what we did for Harmonic oscillators?) It’s easy to see that<br />

[L+, L−] = 2Lz<br />

155<br />

(6.30)<br />

(6.31)


Figure 6.1: Vector model for the quantum angular momentum state |jm〉, which is represented<br />

here by the vector j which precesses about the z axis (axis <strong>of</strong> quantzation) with projection m.<br />

Z<br />

Y<br />

Likewise:<br />

m<br />

θ<br />

|j|=(j (j + 1))<br />

1/2<br />

X<br />

[Lz, L+] = L+<br />

[Lz, L−] = −L−<br />

L 2 = L+L− + L 2 z − Lz = L−L+ + L 2 z + Lz<br />

(6.32)<br />

(6.33)<br />

(6.34)<br />

We now give some frequently used expressions for the angular momentum operator for a<br />

single particle in spherical polar coordinates (SP). In SP coordinates,<br />

It’s easy and straightforward to demonstrate that<br />

and<br />

L± = e ±φ<br />

x = r sin θ cos φ (6.35)<br />

y = r sin θ sin φ (6.36)<br />

z = r cos θ (6.37)<br />

Lz = −i ∂<br />

∂φ<br />

�<br />

± ∂<br />

�<br />

∂<br />

+ i cot θ<br />

∂θ ∂φ<br />

156<br />

(6.38)<br />

(6.39)


Thus,<br />

L 2 = − 1<br />

�<br />

1 ∂<br />

sin θ sin θ<br />

2<br />

�<br />

∂ ∂<br />

+ sin θ<br />

∂φ2 ∂θ ∂θ<br />

which is the angular part <strong>of</strong> the Laplacian in SP coordinates.<br />

∇ 2 1<br />

=<br />

r2 �<br />

sin θ<br />

sin θ<br />

∂ ∂ ∂ ∂ 1 ∂<br />

r2 + sin θ +<br />

∂r ∂r ∂θ ∂θ sin θ<br />

2<br />

∂φ2 �<br />

= 1<br />

r 2<br />

∂ ∂<br />

r2<br />

∂r ∂r<br />

1<br />

− L2<br />

r2 In other words, the kinetic energy operator in SP coordinates is<br />

− ¯h2<br />

2m ∇2 = − ¯h2<br />

�<br />

1<br />

2m r2 �<br />

∂ ∂ 1<br />

r2 − L2<br />

∂r ∂r r2 6.2 Eigenvalues <strong>of</strong> the Angular Momentum Operator<br />

Using the SP form<br />

(6.40)<br />

(6.41)<br />

(6.42)<br />

(6.43)<br />

Lzψ = i ∂ψ<br />

∂φ = lzψ (6.44)<br />

Thus, we conclude that ψ = f(r, θ)e ilzφ . This must be single valued and thus periodic in φ with<br />

period 2π. Thus,<br />

Thus, we write the azimuthal solutions as<br />

which are orthonormal functions:<br />

� 2π<br />

0<br />

lz = m = 0, ±1, ±2, · · · (6.45)<br />

Φm(φ) = 1<br />

√ 2π e imφ<br />

(6.46)<br />

Φ ∗ m(φ)Φm ′(φ)dφ = δmm ′ (6.47)<br />

In a centrally symmetric case, stationary states which differ only in their m quantum number<br />

must have the same energy.<br />

We now look for the eigenvalues and eigenfunctions <strong>of</strong> the L 2 operator belonging to a set <strong>of</strong><br />

degenerate energy levels distinguished only by m. Since the +z−axis is physically equivalent to<br />

the −z−axis, for every +m there must be a −m. Let L denote the greatest possible m for a<br />

given L 2 eigenstate. This upper limit must exist because <strong>of</strong> the fact that L 2 − L 2 z = L 2 x + L 2 y is a<br />

operator for an essentially positive quantity. Thus, its eigenvalues cannot be negative. We now<br />

apply LzL± to ψm.<br />

Lz(L±ψm) = (Lz ± 1)(L±ψm) = (m ± 1)(L±ψm) (6.48)<br />

157


(note: we used [Lz, L±] = ±L± ) Thus, L±ψm is an engenfunction <strong>of</strong> Lz with eigenvalue m ± 1.<br />

i.e.<br />

ψm+1 ∝ L+ψm<br />

ψm−1 ∝ L−ψm<br />

If m = l then, we must have L+ψl = 0. Thus,<br />

(6.49)<br />

(6.50)<br />

L−L+ψl = (L 2 − L 2 z − Lz)ψl = 0 (6.51)<br />

L 2 ψl = (L 2 z + Lz)ψl = l(l + 1)ψl<br />

(6.52)<br />

Thus, the eigenvalues <strong>of</strong> L 2 operator are l(l + 1) for l any positive integer (including 0). For a<br />

given value <strong>of</strong> l, the component Lz can take values<br />

l, l − 1, · · · , 0, −l (6.53)<br />

or 2l + 1 different values. Thus an energy level with angular momentum l has 2l + 1 degenerate<br />

states.<br />

6.3 Eigenstates <strong>of</strong> L 2<br />

Since l ansd m are the good quantum numbers, we’ll denote the eigenstates <strong>of</strong> L 2 as<br />

This we will <strong>of</strong>ten write in short hand after specifying l as<br />

Since L 2 = L+L− + L 2 z − Lz, we have<br />

Also, note that<br />

thus we have<br />

〈m|L 2 |m〉 = m 2 − m − �<br />

L 2 |lm〉 = l(l + 1)|lm〉. (6.54)<br />

L 2 |m〉 = l(l + 1)|m〉. (6.55)<br />

m ′<br />

〈m|L+|m ′ 〉〈m ′ |L−|m〉 = l(l + 1) (6.56)<br />

〈m − 1|L−|m〉 = 〈m|L+|m − 1〉 ∗ , (6.57)<br />

|〈m|L+|m − 1〉| 2 = l(l + 1) − m(m − 1) (6.58)<br />

Choosing the phase (Condon and Shortly phase convention) so that<br />

〈m|L+|m − 1〉 =<br />

〈m − 1|L−|m〉 = 〈m|L+|m − 1〉 (6.59)<br />

�<br />

�<br />

l(l + 1) − m(m − 1) = (l + m)(l − m + 1) (6.60)<br />

Using this relation, we note that<br />

〈m|Lx|m − 1〉 = 〈m − 1|Lx|m〉 = 1�<br />

(l + m)(l − m + 1) (6.61)<br />

2<br />

〈m|Ly|m − 1〉 = 〈m − 1|Lx|m〉 = −i�<br />

(l + m)(l − m + 1) (6.62)<br />

2<br />

Thus, the diagonal elements <strong>of</strong> Lx and Ly are zero in states with definite values <strong>of</strong> 〈Lz〉 = m.<br />

158


6.4 Eigenfunctions <strong>of</strong> L 2<br />

The wavefunction <strong>of</strong> a particle is not entirely determined when l and m are presribed. We still<br />

need to specify the radial component. Thus, all the angular momentum operators (in SP coords)<br />

contain and explicit r dependency. For the time, we’ll take r to be fixed and denote the angular<br />

momentum eigenfunctions in SP coordinates as Ylm(θ, φ) with normalization<br />

�<br />

|Ylm(θ, φ)| 2 dΩ (6.63)<br />

where dΩ = sin θdθdφ = d(cos θ)dφ and the integral is over all solid angles. Since we can<br />

determine common eigenfunctions for L 2 and Lz, there must be a separation <strong>of</strong> variables, θ and<br />

φ, so we seek solutions <strong>of</strong> the form:<br />

The normalization requirement is that<br />

and I require<br />

I thus seek solution <strong>of</strong><br />

� 1<br />

sin θ<br />

i.e.<br />

Ylm(θ, φ) = Φm(φ)Θlm(θ) (6.64)<br />

� π<br />

0<br />

� 2π � π<br />

0<br />

0<br />

∂ ∂ 1<br />

sin θ +<br />

∂θ ∂θ sin2 θ<br />

|Θlm(θ)| 2 sin θdθ = 1 (6.65)<br />

Y ∗<br />

l ′ m ′YlmdΩ = δll ′δmm ′. (6.66)<br />

∂2 ∂φ2 �<br />

ψ + l(l + 1)ψ = 0 (6.67)<br />

�<br />

1 ∂ ∂ m2<br />

sin θ −<br />

sin θ ∂θ ∂θ sin2 �<br />

+ l(l + 1) Θlm(θ) = 0 (6.68)<br />

θ<br />

which is well known from the theory <strong>of</strong> spherical harmonics.<br />

Θlm(θ) = (−1) m i l<br />

�<br />

�<br />

�<br />

� (2l + 1)(l − m)!<br />

P<br />

2(l − m)!<br />

m<br />

l (cos θ) (6.69)<br />

for m > 0. Where P m<br />

l are associated Legendre Polynomials. For m < 0 we get<br />

Θl,−|m| = (−1) m Θl,|m|<br />

(6.70)<br />

Thus, the angular momentum eigenfunctions are the spherical harmonics, normalized so that<br />

the matrix relations defined above hold true. The complete expression is<br />

Ylm = (−1) (m+|m|)/2 i l<br />

� 2l + 1<br />

4π<br />

�1/2 (l − |m|)!<br />

P<br />

(l + |m|)!<br />

|m|<br />

l (cos θ)e imφ<br />

159<br />

(6.71)


Table 6.1: Spherical Harmonics (Condon-Shortley Phase convention.<br />

1<br />

Y00 = √<br />

4π<br />

� �1/2 3<br />

Y1,0 = cos(θ)<br />

4π<br />

� �1/2 3<br />

Y1,±1 = ∓ sin(θ)e<br />

8π<br />

±iφ<br />

�<br />

5<br />

Y2,±2 = 3<br />

96π sin2 θe ∓iφ<br />

�<br />

5<br />

Y2,±1 = ∓3 sin θ cos θeiφ<br />

24π<br />

�<br />

�<br />

5 3<br />

Y2,0 =<br />

4π 2 cos2 θ − 1<br />

�<br />

2<br />

These can also be generated by the SphericalHarmonicY[l,m,θ,φ] function in Mathematica.<br />

Figure 6.2: Spherical Harmonic Functions for up to l = 2. The color indicates the phase <strong>of</strong> the<br />

function.<br />

160


For the case <strong>of</strong> m = 0,<br />

Yl0 = i l<br />

� 2l + 1<br />

4π<br />

� 1/2<br />

Pl(cos θ) (6.72)<br />

Other useful relations are in cartesian form, obtained by using the relations<br />

cos θ = z<br />

, (6.73)<br />

r<br />

sin θ cos φ = x<br />

, (6.74)<br />

r<br />

and<br />

sin θ sin φ = y<br />

. (6.75)<br />

r<br />

Y1,0 =<br />

Y1,1 =<br />

� �1/2 3 z<br />

4π r<br />

� �1/2 3 x + iy<br />

8π r<br />

� �1/2 3 x − iy<br />

Y1,−1 =<br />

8π r<br />

The orthogonality integral <strong>of</strong> the Ylm functions is given by<br />

� 2π � π<br />

Another useful relation is that<br />

0<br />

0<br />

(6.76)<br />

(6.77)<br />

(6.78)<br />

Y ∗<br />

lm(θ, φ)Yl ′ m ′(θ, φ) sin θdθdφ = δll ′δmm ′. (6.79)<br />

Yl,−m = (−1) m Y ∗<br />

lm. (6.80)<br />

This relation is useful in deriving real-valued combinations <strong>of</strong> the spherical harmonic functions.<br />

Exercise 6.1 Demonstrate the following:<br />

1. [L+, L 2 ] = 0<br />

2. [L−, L 2 ] = 0<br />

Exercise 6.2 Derive the following relations<br />

�<br />

�<br />

�<br />

ψl,m(θ, φ) = � (l + m)!<br />

(2l!)(l − m)! (L−) l−m ψl,l(θ, φ)<br />

and<br />

�<br />

�<br />

�<br />

ψl,m(θ, φ) = � (l − m)!<br />

(2l!)(l + m)! (L+) l+m ψl,−l(θ, φ)<br />

where ψl,m = Yl,m are eigenstates <strong>of</strong> the L 2 operator.<br />

161


6.5 Addition theorem and matrix elements<br />

In the quantum mechanics <strong>of</strong> rotations, we will come across integrals <strong>of</strong> the general form<br />

�<br />

Y ∗<br />

l1m1 Yl2m2Yl3m3dΩ<br />

or �<br />

Y ∗<br />

l1m1 Pl2Yl3m3dΩ<br />

in computing matrix elements between angular momentum states. For example, we may be<br />

asked to compute the matrix elements for dipole induced transitions between rotational states <strong>of</strong><br />

a spherical molecule or between different orbital angular momentum states <strong>of</strong> an atom. In either<br />

case, we need to evaluate an integral/matrix element <strong>of</strong> the form<br />

�<br />

〈l1m1|z|l2m2〉 =<br />

�<br />

Realizing that z = r cos θ = r 4π/3Y10(θ, φ), Eq. 6.87 becomes<br />

〈l1m1|z|l2m2〉 =<br />

�<br />

4π<br />

3 r<br />

�<br />

Y ∗<br />

l1m1 zYl2m2dΩ (6.81)<br />

Y ∗<br />

l1m1 Y10Yl2m2dΩ (6.82)<br />

Integrals <strong>of</strong> this form can be evaluated by group theoretical analysis and involves the introduction<br />

<strong>of</strong> Clebsch-Gordan coefficients, CLM 1<br />

l1m1l2m2 which are tabulated in various places or can be<br />

computed using Mathematica. In short, some basic rules will always apply.<br />

1. The integral will vanish unless the vector sum <strong>of</strong> the angular momenta sums to zero.<br />

i.e.|l1 − l3| ≤ l2 ≤ (l1 + l3). This is the “triangle” rule and basically means you have to be<br />

able make a triangle with length <strong>of</strong> each side being l1, l2, and l3.<br />

2. The integral will vanish unless m2 + m3 = m1. This reflects the conservation <strong>of</strong> the z<br />

component <strong>of</strong> the angular momentum.<br />

3. The integral vanishes unless l1 + l2 + l3 is an even integer. This is a parity conservation<br />

law.<br />

So the general proceedure for performing any calculation involving spherical harmonics is to first<br />

check if the matrix element violates any <strong>of</strong> the three symmetry rules, if so, then the answer is 0<br />

and you’re done. 2<br />

To actually perform the integration, we first write the product <strong>of</strong> two <strong>of</strong> the Ylm’s as a<br />

Clebsch-Gordan expansion:<br />

Yl1m1Yl2m2 = �<br />

�<br />

�<br />

�<br />

� (2l1 + 1)(2l2 + 1)<br />

4π(2L + 1) CL0<br />

LM<br />

l10l20C LM<br />

l1m1l2m2 YLM. (6.83)<br />

1 Our notation is based upon Varshalovich’s book. There at least 13 different notations that I know <strong>of</strong> for<br />

expressing these coefficients which I list in a table at the end <strong>of</strong> this chapter.<br />

2 In Mathematica, the Clebsch-Gordan coefficients are computed using the function<br />

ClebschGordan[j1,m1,j2,m2,j,m] for the decomposition <strong>of</strong> |jm〉 in to |j1, m1〉 and |j2, m2〉.<br />

162


We can use this to write<br />

�<br />

Y ∗<br />

lmYl1m2Yl2m2dΩ = �<br />

�<br />

�<br />

�<br />

� (2l1 + 1)(2l2 + 1)<br />

4π(2L + 1) CL0<br />

LM<br />

�<br />

�<br />

�<br />

= �<br />

�<br />

LM<br />

(2l1 + 1)(2l2 + 1)<br />

4π(2L + 1) CL0<br />

�<br />

�<br />

�<br />

= � (2l1 + 1)(2l2 + 1)<br />

4π(2l + 1)<br />

l10l20C LM<br />

�<br />

l1m1l2m2<br />

l10l20C LM<br />

l1m1l2m2 δlLδmM<br />

C l0<br />

l10l20C lm<br />

l1m1l2m2<br />

Y ∗<br />

lmYLMdΩ<br />

(6.84)<br />

In fact, the expansion we have done above for the product <strong>of</strong> two spherical harmonics can be<br />

inverted to yield the decomposition <strong>of</strong> one angular momentum state into a pair <strong>of</strong> coupled angular<br />

momentum states, such as would be the case for combining the orbital angular momentum <strong>of</strong> a<br />

particle with, say, its spin angular momentum. In Dirac notation, this becomes pretty apparent<br />

|LM〉 = �<br />

〈l1m1l2m2|LM〉|l1m1l2m2〉 (6.85)<br />

m1m2<br />

where the state |l1m1l2m2〉 is the product <strong>of</strong> two angular momentum states |l1m1〉 and |l2m2〉.<br />

The expansion coefficients are the Clebsch-Gordan coefficients<br />

C LM<br />

l1m1l2m2 = 〈l1m1l2m2|LM〉 (6.86)<br />

Now, let’s go back the problem <strong>of</strong> computing the dipole transition matrix element between<br />

two angular momentum states in Eq. 6.87. The integral we wish to evaluate is<br />

�<br />

〈l1m1|z|l2m2〉 =<br />

Y ∗<br />

l1m1 zYl2m2dΩ (6.87)<br />

and we noted that z was related to the Y10 spherical harmonic. So the integral over the angular<br />

coordinates involves:<br />

�<br />

Y ∗<br />

l1m1 Y10Yl2m2dΩ. (6.88)<br />

First, we evaluate which matrix elements are going to be permitted by symmetry.<br />

1. Clearly, by the triangle inequality, |l1 − l2| = 1. In other words, we change the angular<br />

momentum quantum number by only ±1.<br />

2. Also, by the second criteria, m1 = m2<br />

3. Finally, by the third criteria: l1 + l2 + 1 must be even, which again implies that l1 and l2<br />

differ by 1.<br />

Thus the integral becomes<br />

�<br />

Y ∗<br />

l+1,mY10Ylm =<br />

�<br />

�<br />

�<br />

� (2l + 1)(2 + 1)<br />

4π(2l + 3) Cl+1,0 l010 C1m l+1,ml0<br />

163<br />

(6.89)


From tables,<br />

So<br />

Thus,<br />

�<br />

C l+1,0<br />

l010<br />

C 1m<br />

l+1,ml0 = −<br />

= −<br />

� √ �<br />

2 (1 + l)<br />

√ √<br />

2 + 2 l 3 + 2 l<br />

�√ √ √ �<br />

2 1 + l − m 1 + l + m<br />

√ √<br />

2 + 2 l 3 + 2 l<br />

C l+1,0<br />

l010 C1m<br />

l+1,ml0 = 2 (1 + l) √ 1 + l − m √ 1 + l + m<br />

(2 + 2 l) (3 + 2 l)<br />

Y ∗<br />

l+1,mY10YlmdΩ =<br />

� �<br />

�<br />

3 �<br />

� (l + m + 1)(l − m + 1)<br />

4π (2l + 1)(2l + 3)<br />

Finally, we can construct the matrix element for dipole-transitions as<br />

(6.90)<br />

�<br />

�<br />

�<br />

〈l1m1|z|l2m2〉 = r�<br />

(l + m + 1)(l − m + 1)<br />

δl1±1,l2δm1,m2. (6.91)<br />

(2l + 1)(2l + 3)<br />

Physically, this make sense because a photon carries a single quanta <strong>of</strong> angular momentum. So<br />

in order for molecule or atom to emit or absorb a photon, its angular momentum can only change<br />

by ±1.<br />

Exercise 6.3 Verify the following relations<br />

�<br />

�<br />

Y ∗<br />

l+1,m+1Y11Ylmdω =<br />

Y ∗<br />

l−1,m−1Y11Ylmdω = −<br />

�<br />

� �<br />

�<br />

3 �<br />

� (l + m + 1)(l + m + 2)<br />

8π 2l + 1)(2l + 3)<br />

� �<br />

�<br />

3 �<br />

� (l − m)(l − m − 1)<br />

8π 2l − 1)(2l + 1)<br />

Y ∗<br />

lmY00YlmdΩ = 1<br />

√ 4π<br />

(6.92)<br />

(6.93)<br />

(6.94)<br />

6.6 Legendre Polynomials and Associated Legendre Polynomials<br />

Ordinary Legendre polynomials are generated by<br />

Pl(cos θ) = 1<br />

2ll! d l<br />

(d cos θ) l (cos2 θ − 1) l<br />

164<br />

(6.95)


i.e. (x = cos θ)<br />

and satisfy<br />

Pl(x) = 1<br />

2l ∂<br />

l!<br />

l<br />

∂x l (x2 − 1) l<br />

(6.96)<br />

� �<br />

1 ∂ ∂<br />

sin θ + l(l + 1) Pl = 0 (6.97)<br />

sin θ ∂θ ∂θ<br />

The Associated Legendre Polynomials are derived from the Legendre Polynomials via<br />

P m<br />

l (cos θ) = sin m θ<br />

∂m (∂ cos θ) m Pl(cos θ) (6.98)<br />

6.7 <strong>Quantum</strong> rotations in a semi-classical context<br />

Earlier we established the fact that the angular momentum vector can never exactly lie on a<br />

single spatial axis. By convention we take the quantization axis to be the z axis, but this is<br />

arbitrary and we can pick any axis as the quantization axis, it is just that picking the z axis<br />

make the mathematics much simpler. Furthermore, we established that the maximum length<br />

the angular momentum vector can have along the z axis is the eigenvalue <strong>of</strong> Lz �<br />

when m = l,<br />

so 〈lz〉 = l which is less than l(l + 1). Note, however, we can write the eigenvalue <strong>of</strong> L2 as<br />

l 2 (1+1/l 2 ) . As l becomes very large the eigenvalue <strong>of</strong> Lz and the eigenvalue <strong>of</strong> L 2 become nearly<br />

identical. The 1/l term is in a sense a quantum mechanical effect resulting from the uncertainty<br />

in determining the precise direction <strong>of</strong> � L.<br />

We can develop a more quantative model for this by examining both the uncertainty product<br />

and the semi-classical limit <strong>of</strong> the angular momentum distribution function. First, recall, that if<br />

we have an observable, A then the spread in the measurements <strong>of</strong> A is given by the varience.<br />

∆A 2 = 〈(A − 〈A〉) 2 〉 = 〈A 2 〉 − 〈A〉 2 . (6.99)<br />

In any representation in which A is diagonal, ∆A 2 = 0 and we can determine A to any level <strong>of</strong><br />

precision. But if we look at the sum <strong>of</strong> the variances <strong>of</strong> lx and ly we see<br />

∆L 2 x + ∆L 2 y = l(l + 1) − m 2 . (6.100)<br />

So for a fixed value <strong>of</strong> l and m, the sum <strong>of</strong> the two variences is constant and reaches its minimum<br />

when |m| = l corresponding to the case when the vector points as close to the ±z axis as it<br />

possible can. The conclusion we reach is that the angular momentum vector lies somewhere in a<br />

cone in which the apex half-angle, θ satisfies the relation<br />

m<br />

cos θ = �<br />

(6.101)<br />

l(l + 1)<br />

which we can varify geometrically. So as l becomes very large the denominator becomes for m = l<br />

l<br />

�<br />

l(l + 1) =<br />

1<br />

� → 1 (6.102)<br />

1 1 + 1/l2 165


and θ = 0,cooresponding to the case in which the angular momentum vector lies perfectly along<br />

the z axis.<br />

Exercise 6.4 Prove Eq. 6.100 by writing 〈L 2 〉 = 〈L 2 x〉 + 〈L 2 y〉 + 〈L 2 z〉.<br />

To develop this further, let’s look at the asymptotic behavour <strong>of</strong> the Spherical Harmonics at<br />

large values <strong>of</strong> angular momentum. The angular part <strong>of</strong> the Spherical Harmonic function satisfies<br />

�<br />

1 ∂ ∂<br />

m2<br />

sin θ + l(l + 1) −<br />

sin θ ∂θ ∂θ sin2 �<br />

Θlm = 0 (6.103)<br />

θ<br />

For m = 0 this reduces to the differential equation for the Legendre polynomials<br />

� �<br />

2 ∂ ∂<br />

+ cot θ + l(l + 1) Pl(cos θ) = 0 (6.104)<br />

∂θ2 ∂θ<br />

If we make the substitution<br />

Pl(cos θ) =<br />

χlθ<br />

(sin θ) 1/2<br />

(6.105)<br />

then we wind up with a similar equation for χl(θ)<br />

�<br />

2 ∂<br />

∂θ2 + (l + 1/2)2 + csc2 �<br />

θ<br />

χl = 0. (6.106)<br />

4<br />

For very large l, the l + 1/2 term dominates and we can ignore the cscθ term everywhere except<br />

for angles close to θ = 0 or θ = π. If we do so then our differential equation becomes<br />

� �<br />

2 ∂<br />

+ (l + 1/2)2 χl = 0, (6.107)<br />

∂θ2 which has the solution<br />

χl(θ) = Al sin ((l + 1/2)θ + α) (6.108)<br />

where Al and α are constants we need to determine from the boundary conditions <strong>of</strong> the problem.<br />

For large l and for θ ≫ l −1 and π − θ ≫ l −1 one obtains<br />

Similarly,<br />

Ylo(θ, φ) ≈<br />

sin((l + 1/2)θ + α)<br />

Pl(cos θ) ≈ Al<br />

(sin θ) 1/2 . (6.109)<br />

� l + 1/2<br />

2π<br />

� 1/2<br />

so that the angular probability distribution is<br />

� �<br />

l + 1/2<br />

|Ylo| 2 =<br />

2π<br />

A 2 l<br />

sin((l + 1/2)θ + α)<br />

Al<br />

(sin θ) 1/2 . (6.110)<br />

sin2 ((l + 1/2)θ + α)<br />

. (6.111)<br />

sin θ<br />

166


When l is very large the sin 2 ((l + 1/2)θ) factor is extremely oscillatory and we can replace it<br />

by its average value <strong>of</strong> 1/2. Then, if we require the integral <strong>of</strong> our approximation for |Yl0| 2 to be<br />

normalized, one obtains<br />

|Yl0| 2 =<br />

1<br />

2π 2 sin(θ)<br />

(6.112)<br />

which holds for large values <strong>of</strong> l and all values <strong>of</strong> θ except for theta = 0 or θ = π.<br />

We can also recover this result from a purely classical model. In classical mechanics, the<br />

particle moves in a circular orbit in a plane perpendicular to the angular momentum vector. For<br />

m = 0 this vector lies in the xy plane and we will define θ as the angle between the particle and<br />

the z axis, φ as the azimuthal angle <strong>of</strong> the angular momentum vector in the xy plane. Since the<br />

particles speed is uniform, its distribution in θ is uniform. Thus the probability <strong>of</strong> finding the<br />

particle at any instant in time between θ and θ + dθ is dθ/π. Furthermore, we have not specified<br />

the azimuthal angle, so we assume that the probability distribution is also uniform over φ and<br />

the angular probability dθ/π must be smeared over some band on the uniform sphere defined by<br />

the angles θ and θ + dθ. The area <strong>of</strong> this band is 2π sin θdθ. Thus, we can define the “classical<br />

estimate” <strong>of</strong> the as a probability per unit area<br />

P (θ) = dθ<br />

π<br />

1<br />

2π sin θ =<br />

1<br />

2π 2 sin θ<br />

which is in agreement with the estimate we made above.<br />

For m �= 0 we have to work a bit harder since the angular momentum vector is tilted out <strong>of</strong><br />

the plane. For this we define two new angles γ which is the azimuthal rotation <strong>of</strong> the particle’s<br />

position about the L vector and α with is constrained by the length <strong>of</strong> the angular momentum<br />

vector and its projection onto the z axis.<br />

m m<br />

cos α = � ≈<br />

l(l + 1 l<br />

The analysis is identical as before with the addition <strong>of</strong> the fact that the probability in γ (taken<br />

to be uniform) is spread over a zone 2π sin θdθ. Thus the probability <strong>of</strong> finding the particle with<br />

some angle θ is<br />

P (θ) = dγ 1<br />

dθ 2π2 sin θ .<br />

Since γ is the dihedral angle between the plane containing z and l and the plane containing<br />

l and r (the particle’s position vector), we can relate γ to θ and α by<br />

Thus,<br />

cos θ = cos α cos π π<br />

+ sin α sin cos γ = sin α cos γ<br />

2 2<br />

sin θdθ = sin α sin γdγ.<br />

This allows us to generalize our probability distribution to any value <strong>of</strong> m<br />

|Ylm(θ, φ)| 2 =<br />

=<br />

1<br />

2π2 sin α sin γ<br />

1<br />

2π 2 (sin 2 α − cos 2 θ) 1/2<br />

167<br />

(6.113)<br />

(6.114)


Figure 6.3: Classical and <strong>Quantum</strong> Probability Distribution Functions for Angular Momentum.<br />

1<br />

0.8<br />

0.6<br />

0.4<br />

0.2<br />

1<br />

0.8<br />

0.6<br />

0.4<br />

0.2<br />

1<br />

0.8<br />

0.6<br />

0.4<br />

0.2<br />

»Y4,0» 2<br />

0.5 1 1.5 2 2.5 3<br />

»Y4,2» 2<br />

0.5 1 1.5 2 2.5 3<br />

»Y4,4» 2<br />

0.5 1 1.5 2 2.5 3<br />

1.5<br />

1.25<br />

1<br />

0.8<br />

0.6<br />

0.4<br />

0.2<br />

1<br />

0.8<br />

0.6<br />

0.4<br />

0.2<br />

1<br />

0.75<br />

0.5<br />

0.25<br />

»Y10,0» 2<br />

0.5 1 1.5 2 2.5 3<br />

»Y10,2» 2<br />

0.5 1 1.5 2 2.5 3<br />

»Y10,10» 2<br />

0.5 1 1.5 2 2.5 3<br />

which holds so long as sin 2 α > cos 2 θ. This corresponds to the spatial region (π/2 − α) <<br />

θ < π/2 + α. Outside this region, the distribution blows up and corresponds to the classically<br />

forbidden region.<br />

In Fig. 6.3 we compare the results <strong>of</strong> our semi-classical model with the exact results for l = 4<br />

and l = 10. All in all we do pretty well with a semi-classical model, we do miss some <strong>of</strong> the<br />

wiggles and the distribution is sharp close to the boundaries, but the generic features are all<br />

there.<br />

168


Table 6.2: Relation between various notations for Clebsch-Gordan Coefficients in the literature<br />

Symbol Author<br />

C jm<br />

j1m1j2m2 Varshalovich a<br />

S j1j2<br />

jm1jm2 Wignerb Ajj1j2 mm1m2 Eckartc Cm1m j<br />

2<br />

Van der Weardend (j1j2m1m2|j1j2jm) Condon and Shortley e<br />

C j1j2<br />

jm (m1m2) Fock f<br />

X(j, m, j1, j2, m1) Boys g<br />

C(jm; m1m2) Blatt and Weisskopfh Cj1j2j m1m2m Beidenharni C(j1j2j, m1m2) Rosej �<br />

�<br />

j1 j2 j<br />

m1 m2 m<br />

Yutsis and Bandzaitisk 〈j1m1j2m2|(j1j2)jm〉 Fanol a.) D. A. Varschalovich, et al. <strong>Quantum</strong> Theory <strong>of</strong> Angular Momentum, (World Scientific, 1988).<br />

b.) E. Wigner, Group theory, (Academic Press, 1959).<br />

c.) C. Eckart “The application <strong>of</strong> group theory to the quantum dynamics <strong>of</strong> monatomic systems”,<br />

Rev. Mod. Phys. 2, 305 (1930).<br />

d.) B. L. Van der Waerden, Die gruppentheorische methode in der quantenmechanik, (Springer,<br />

1932).<br />

e.)E. Condon and G. Shortley, Theory <strong>of</strong> Atomic Spectra, (Cambridge, 1932).<br />

f.) V. A. Fock, “ New Deduction <strong>of</strong> the Vector Model”, JETP 10,383 (1940).<br />

g.)S. F. Boys, “Electronic wave functions IV”, Proc. Roy. Soc., London, A207, 181 (1951).<br />

h.) J. M. Blatt and V. F. Weisskopf, Theoretical Nuclear Physics, (McGraw-Hill, 1952).<br />

i.) L. C. Beidenharn, ”Tables <strong>of</strong> Racah Coefficients”, ONRL-1098 (1952).<br />

j.) M. E. Rose, Multipole Fields, (Wiley 1955).<br />

k.) A. P. Yusis and A. A. Bandzaitit, The Theory <strong>of</strong> Angular Momentum in Quanutm <strong>Mechanics</strong>,<br />

(Mintus, Vilinus, 1965).<br />

l.) U. Fano, “Statistical matrix techniques and their application to the directional correlation <strong>of</strong><br />

radiation,” US Nat’l Bureau <strong>of</strong> Standards, Report 1214 (1951).<br />

169


6.8 Motion in a central potential: The Hydrogen Atom<br />

(under development)<br />

The solution <strong>of</strong> the Schrödinger equation for the hydrogen atom was perhaps the most significant<br />

developments in quantum theory. Since it is one <strong>of</strong> the few problems in nature in which we<br />

can derive an exact solution to the equations <strong>of</strong> motion, it deserves special attention and focus.<br />

Perhaps more importantly, the hydrogen atomic orbitals form the basis <strong>of</strong> atomic physics and<br />

quantum chemistry.<br />

The potential energy function between the proton and the electron is the centrosymmetric<br />

Coulombic potential<br />

V (r) = − Ze2<br />

r .<br />

Since the potential is centrosymmetric and has no angular dependency the Hydrogen atom Hamiltonian<br />

separates in to radial and angular components.<br />

H = − ¯h2<br />

2µ<br />

� 1<br />

r 2<br />

∂ ∂ L2<br />

r2 −<br />

∂r ∂r ¯h 2 r2 �<br />

+ e2<br />

r<br />

(6.115)<br />

where L 2 is the angular momentum operator we all know and love by now and µ is the reduced<br />

mass <strong>of</strong> the electron/proton system<br />

µ = memp<br />

me + mp<br />

≈ me = 1<br />

Since [H, L] = 0, angular momentum and one component <strong>of</strong> the angular momentum must be<br />

constants <strong>of</strong> the motion. Since there are three separable degrees <strong>of</strong> freedom, we have one other<br />

constant <strong>of</strong> motion which must correspond to the radial motion. As a consequence, the hydrogen<br />

wavefunction is separable into radial and angular components<br />

ψnlm = Rnl(r)Ylm(θ, φ). (6.116)<br />

Using the Hamiltonian in Eq. 6.115 and this wavefunction, the radial Schrödinger equation reads<br />

(in atomic units)<br />

�<br />

− ¯h2<br />

�<br />

1<br />

2 r2 ∂ ∂ l(l + 1)<br />

r2 −<br />

∂r ∂r r2 �<br />

− 1<br />

�<br />

Rnl(R) = ERnl(r) (6.117)<br />

r<br />

At this point, we introduce atomic units to make the notation more compact and drastically<br />

simplify calculations. In atomic units, ¯h = 1 and e = 1. A list <strong>of</strong> conversions for energy, length,<br />

etc. to SI units is listed in the appendix. The motivation is so that all <strong>of</strong> our numbers are <strong>of</strong><br />

order 1.<br />

The kinetic energy term can be rearranged a bit<br />

and the radial equation written as<br />

�<br />

− ¯h2<br />

�<br />

2 ∂ 2 ∂<br />

+<br />

2 ∂r2 r ∂r<br />

1<br />

r2 ∂ ∂<br />

r2<br />

∂r ∂r<br />

= ∂2<br />

∂r<br />

2 ∂<br />

+ 2 r ∂r<br />

(6.118)<br />

l(l + 1)<br />

−<br />

r2 �<br />

− 1<br />

�<br />

Rnl(R) = ERnl(r) (6.119)<br />

r<br />

170


To solve this equation, we first have to figure out what approximate form the wavefunction must<br />

have. For large values <strong>of</strong> r, the 1/r terms disappear and the asymptotic equation is<br />

or<br />

− ¯h2<br />

2<br />

∂ 2<br />

∂r 2 Rnl(R) = ERnl(r) (6.120)<br />

∂ 2 R<br />

∂r 2 = α2 R (6.121)<br />

where α = −2mE/¯h 2 . This differential equation we have seen before for the free particle, so the<br />

solution must have the same form. Except in this case, the function is real. Furthermore, for<br />

bound states with E < 0 the radial solution must go to zero as r → ∞, so <strong>of</strong> the two possible<br />

asymptotic solutions, the exponentially damped term is the correct one.<br />

R(r) ≡ e −αr<br />

(6.122)<br />

Now, we have to check if this is a solution everywhere. So, we take the asymptotic solution and<br />

plug it into the complete equation:<br />

Eliminating e −αr<br />

α 2 e −αr + 2<br />

r (−αe−αr ) + 2m<br />

¯h 2<br />

� �<br />

2 e<br />

+ E e<br />

r −αr = 0. (6.123)<br />

�<br />

α 2 + 2mE<br />

¯h 2<br />

�<br />

+ 1<br />

� �<br />

2 2me<br />

2 − 2α = 0 (6.124)<br />

r ¯h<br />

For the solution to hold everywhere, it must also hold at r = 0, so two conditions must be met<br />

which we defined above, and<br />

α 2 = −2mE/¯h 2<br />

� 2me 2<br />

2 − 2α<br />

¯h<br />

�<br />

(6.125)<br />

= 0. (6.126)<br />

If these conditions are met, then e −αr is a solution. This last equation also sets the length scale<br />

<strong>of</strong> the system since<br />

α = me 2 /¯h 2 = 1/ao<br />

(6.127)<br />

where ao is the Bohr radius. In atomic units, ao = 1. Likewise, the energy can be determined:<br />

E = − ¯h2<br />

2ma 2 o<br />

= − ¯h2<br />

me 2<br />

e 2<br />

2ao<br />

In atomic units the ground states energy is E = −1/2hartree.<br />

171<br />

= − e2<br />

. (6.128)<br />

2ao


Finally, we have to normalize R<br />

�<br />

d 3 re −2αr � ∞<br />

= 4π r<br />

0<br />

2 e −2αr dr (6.129)<br />

The angular normalization can be absorbed into the spherical harmonic term in the total wavefunction<br />

since Y00 = 1/ √ 4π. So, the ground state wavefunction is<br />

ψn00 = Ne −r/ao Y00<br />

(6.130)<br />

The radial integral can be evaluated using Leibnitz’ theorem for differentiation <strong>of</strong> a definite<br />

integral<br />

Thus,<br />

�<br />

∂ b<br />

� b<br />

f(β, x)dx =<br />

∂β a<br />

a<br />

� ∞<br />

r<br />

0<br />

2 e −βr dr =<br />

Exercise 6.5 Generalize this result to show that<br />

� ∞<br />

0<br />

� ∞<br />

0<br />

= ∂2<br />

∂β 2<br />

= − ∂2<br />

∂β 2<br />

= 2<br />

β 3<br />

∂f(β, x)<br />

dx (6.131)<br />

∂β<br />

∂ 2<br />

∂β2 e−βrdr � ∞<br />

0<br />

1<br />

β<br />

r n e −βr dr = n!<br />

β n+1<br />

e −βr dr<br />

Thus, using this result and putting it all together, the normalized radial wavefunction is<br />

(6.132)<br />

(6.133)<br />

� �3/2 1<br />

R10 = 2 e<br />

ao<br />

−r/ao . (6.134)<br />

For the higher energy states, we examine what happens at r → 0. Using a similar analysis<br />

as above, one can show that close in, the radial solution must behave like a polynomial<br />

which leads to a general solution<br />

R ≡ r l+1<br />

R = r l+1 e −αr<br />

∞�<br />

asr<br />

s=0<br />

s .<br />

172


The proceedure is to substitute this back into the Schrodinger equation and evaluate term by<br />

term. In the end one finds that the energies <strong>of</strong> the bound states are (in atomic units)<br />

En = − 1<br />

2n 2<br />

and the radial wavefunctions<br />

� �l � �<br />

2r 2<br />

Rnl =<br />

nao nao<br />

� (n − l − 1)!<br />

2n((n + l)!) 3<br />

�<br />

e −r/nao L 2l+1<br />

� �<br />

2r<br />

n+1<br />

nao<br />

where the L b a are the associated Laguerre polynomials.<br />

6.8.1 Radial Hydrogenic Functions<br />

(6.135)<br />

The radial wavefunctions for nuclei with atomic number Z are modified hydrogenic wavefunctions<br />

with the Bohr radius scaled by Z. I.e a = ao/Z. The energy for 1 electron about a nucleus with<br />

Z protons is<br />

Some radial wavefunctions are<br />

6.9 Spin 1/2 Systems<br />

En = − Z2<br />

n 2<br />

1<br />

2ao<br />

= − Z2<br />

n2 Ry<br />

2<br />

(6.136)<br />

� �3/2<br />

Z<br />

R1s = 2 e<br />

ao<br />

−Zr/ao (6.137)<br />

R2s = 1<br />

� �3/2 �<br />

Z<br />

√ 1 −<br />

2 ao<br />

Zr<br />

�<br />

e<br />

2ao<br />

−Zr/2ao (6.138)<br />

R2p = 1<br />

2 √ � �5/2<br />

Z<br />

re<br />

6 ao<br />

−Zr/2ao (6.139)<br />

In this section we are going to illustrate the various postulates and concepts we have been<br />

developing over the past few weeks. Rather than choosing as examples problems which are<br />

pedagogic (such as the particle in a box and its variations) or or chosen for theor mathematical<br />

simplicity, we are going to focus upon systems which are physically important. We are going to<br />

examine, with out much theoretical introduction, the case in which the state space is limited to<br />

two states. The quantum mechanical behaviour <strong>of</strong> these systems can be varified experimentally<br />

and, in fact, were and still are used to test various assumptions regarding quantum behaviour.<br />

Recall from undergraduate chemistry that particles, such as the electron, proton, and so forth,<br />

possess an intrinsic angular momentum, � S, called spin. This is a property which has no analogue<br />

in classical mechanics. Without going in to all the details <strong>of</strong> angular momentum and how it gets<br />

quantized (don’t worry, it’s a coming event!) we are going to look at a spin 1/2 system, such as<br />

173


a neutral paramagnetic Ag atom in its ground electronic state. We are going to dispense with<br />

treating the other variables, the nuclear position and momentum,the motion <strong>of</strong> the electrons,<br />

etc... and focus only upon the spin states <strong>of</strong> the system.<br />

The paramagnetic Ag atoms possess an electronic magnetic moment, � M. This magnetic<br />

moment can couple to an externally applied magnetic field, � B, resulting on a net force being<br />

applied to the atom. The potential energy in for this is<br />

W = − � M. � B. (6.140)<br />

We take this without further pro<strong>of</strong>. We also take without pro<strong>of</strong> that the magnetic moment and<br />

the intrinsic angular momentum are proportional.<br />

�M = γ � S (6.141)<br />

The proportionality constant is the gyromagnetic ratio <strong>of</strong> the level under consideration. When<br />

the atoms traverse through the magnetic field, they are deflected according to how their angular<br />

momentum vector is oriented with the applied field.<br />

Also, the total moment relative to the center <strong>of</strong> the atom is<br />

�F = � ∇( � M. � B) (6.142)<br />

� Γ = � M × � B. (6.143)<br />

Thus, the time evolution <strong>of</strong> the angular momentum <strong>of</strong> the particle is<br />

that it to say<br />

∂<br />

∂t � S = � Γ (6.144)<br />

∂<br />

∂t � S = γ � S × � B. (6.145)<br />

Thus, the velocity <strong>of</strong> the angular momentum is perpendicular to � S and the angular momentum<br />

vector acts like a gyroscope.<br />

We can also show that for a homogeneous field the force acts parallel to z and is proportional<br />

to Mz. Thus, the atoms are deflected according to how their angular momentum vector is oriented<br />

with respect to the z axis. Experimentally, we get two distributions. Meaning that measurement<br />

<strong>of</strong> Mz can give rise to two possible results.<br />

6.9.1 Theoretical Description<br />

We associate an observable, Sz, with the experimental observations. This has 2 eigenvalues, at<br />

±¯h/2 We shall assume that the two are not degenerate. We also write the eigenvectors <strong>of</strong> Sz as<br />

|±〉 corresponding to<br />

Sz|+〉 = + ¯h<br />

|+〉 (6.146)<br />

2<br />

174


with<br />

and<br />

The closure, or idempotent relation is thus<br />

with<br />

The most general state vector is<br />

Sz|−〉 = + ¯h<br />

|−〉 (6.147)<br />

2<br />

〈+|+〉 = 〈−|−〉 = 1 (6.148)<br />

〈+|−〉 = 0. (6.149)<br />

|+〉〈+| + |−〉〈−| = 1. (6.150)<br />

|ψ〉 = α|+〉 + β|−〉 (6.151)<br />

|α| 2 + |β| 2 = 1. (6.152)<br />

In the |±〉 basis, the matrix representation <strong>of</strong> Sz is diagonal and is written as<br />

Sz = ¯h<br />

�<br />

2<br />

6.9.2 Other Spin Observables<br />

1 0<br />

0 −1<br />

We can also measure Sx and Sy. In the |±〉 basis these are written as<br />

and<br />

Sx = ¯h<br />

�<br />

2<br />

Sy = ¯h<br />

�<br />

2<br />

0 1<br />

1 0<br />

0 i<br />

−i 0<br />

You can verify that the eigenvalues <strong>of</strong> each <strong>of</strong> these are ±¯h/2.<br />

6.9.3 Evolution <strong>of</strong> a state<br />

The Hamiltonian for a spin 1/2 particle in a B-field is given by<br />

�<br />

�<br />

�<br />

(6.153)<br />

(6.154)<br />

(6.155)<br />

H = −γ|B|Sz. (6.156)<br />

175


Where B is the magnitude <strong>of</strong> the field. This operator is time-independent, thus, we can solve the<br />

Schrodinger Equation and see that the eigenvectors <strong>of</strong> H are also the eigenvectors <strong>of</strong> Sz. (This<br />

the eigenvalues <strong>of</strong> Sz are “good quantum numbers”.) Let’s write ω = −γ|B| so that<br />

H|+〉 = + ¯hω<br />

|+〉 (6.157)<br />

2<br />

H|−〉 = − ¯hω<br />

|−〉 (6.158)<br />

2<br />

Therefore there are two energy levels, E± = ±¯hω/2. The separation is proportional to the<br />

magnetic field. They define a single “Bohr Frequency”.<br />

6.9.4 Larmor Precession<br />

Using the |±〉 states, we can write any arb. angular momentum state as<br />

|ψ(0)〉 = cos( θ<br />

2 )e−iφ/2 |+〉 + sin( θ<br />

2 )e+iφ/2 |−〉 (6.159)<br />

where θ and φ are polar coordinate angles specifing the directrion <strong>of</strong> the angular momentum<br />

vector at a given time. The time evolution under H is<br />

|ψ(0)〉 = cos( θ<br />

2 )e−iφ/2 e −iE+t/¯h |+〉 + sin( θ<br />

2 )e+iφ/2 e −iEmt/¯h |−〉, (6.160)<br />

or, using the values <strong>of</strong> E+ and E−<br />

|ψ(0)〉 = cos( θ<br />

2 )e−i(φ+ωt)/2 |+〉 + sin( θ<br />

2 )e+i(φ+ωt)/2 |−〉 (6.161)<br />

In other words, I can write<br />

θ(t) = θ (6.162)<br />

φ(t) = φ + ωt. (6.163)<br />

This corresponds to the precession <strong>of</strong> the angular momentum vector about the z axis at an angular<br />

frequency <strong>of</strong> ω. More over, the expectation values <strong>of</strong> Sz, Sy, and Sx can also be computed:<br />

〈Sz(t)〉 = ¯h/2 cos(θ) (6.164)<br />

〈Sx(t)〉 = ¯h/2 sin(θ/2) cos(φ + ωt) (6.165)<br />

〈Sy(t)〉 = ¯h/2 sin(θ/2) sin(φ + ωt) (6.166)<br />

Finally, what are the “populations” <strong>of</strong> the |±〉 states as a function <strong>of</strong> time?<br />

|〈+|ψ(t)〉| 2 = cos 2 (θ/2) (6.167)<br />

|〈−|ψ(t)〉| 2 = sin 2 (θ/2) (6.168)<br />

Thus, the populations do not change, neither does the normalization <strong>of</strong> the state.<br />

176


6.10 Problems and Exercises<br />

Exercise 6.6 A molecule (A) with orbital angular momentum S = 3/2 decomposes into two<br />

products: product (B) with orbital angular momentum 1/2 and product (C) with orbital angular<br />

momentum 0. We place ourselves in the rest frame <strong>of</strong> A) and angular momentum is conserved<br />

throughout.<br />

A 3/2 → B 1/2 + C 0<br />

(6.169)<br />

1. What values can be taken on by the relative orbital angular momentum <strong>of</strong> the two final<br />

products? Show that there is only one possible value <strong>of</strong> the parity <strong>of</strong> the relative orbital<br />

state is fixed. Would this result remain the same if the spin <strong>of</strong> A was 3/2?<br />

2. Assume that A is initially in the spin state characterized by the eigenvalue ma¯h <strong>of</strong> its spin<br />

component along the z-axis. We know that the final orbital state has a definite parity. Is it<br />

possible to determine this parity by measuring the probabilities <strong>of</strong> finding B in either state<br />

|+〉 or in state |−〉?<br />

Exercise 6.7 The quadrupole moment <strong>of</strong> a charge distribution, ρ(r), is given by<br />

Qij = 1<br />

�<br />

e<br />

3(xixj − δijr 2 )ρ(r)d 3 r (6.170)<br />

where the total charge e = � d3rρ(r). The quantum mechanical equivalent <strong>of</strong> this can be written<br />

in terms <strong>of</strong> the angular momentum operators as<br />

Qij = 1<br />

�<br />

r<br />

e<br />

2<br />

�<br />

3<br />

2 (JiJj + JjJi) − δijJ 2<br />

�<br />

ρ(r)d 3 r (6.171)<br />

The quadrupole moment <strong>of</strong> a stationary state |n, j〉, where n are other non-angular momentum<br />

quantum numbers <strong>of</strong> the system, is given by the expectation value <strong>of</strong> Qzz in the state in which<br />

m = j.<br />

1. Evaluate<br />

in terms <strong>of</strong> j and 〈r 2 〉 = 〈nj|r 2 |nj〉.<br />

Qo = 〈Qzz〉 = 〈njm = j|Qzz|njm = j〉 (6.172)<br />

2. Can a proton (j = 1/2) have a quadrupole moment? What a bout a deuteron (j = 1)?<br />

3. Evaluate the matrix element<br />

What transitions are induced by this operator?<br />

〈njm|Qxy|nj ′ m ′ 〉 (6.173)<br />

4. The quantum mechanical expression <strong>of</strong> the dipole moment is<br />

po = 〈njm = j| r<br />

e Jz|njm = j〉 (6.174)<br />

Can an eigenstate <strong>of</strong> a Hamiltonian with a centrally symmetric potential have an electric<br />

dipole moment?<br />

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Exercise 6.8 The σx matrix is given by<br />

prove that<br />

σx =<br />

where α is a constant and I is the unit matrix.<br />

�<br />

0 1<br />

1 0<br />

�<br />

, (6.175)<br />

exp(iασx) = I cos(α) + iσx sin(α) (6.176)<br />

Solution: To solve this you need to expand the exponential. To order α 4 this is<br />

e iασx = I + iασx − α2<br />

2 σ2 x − iα3<br />

3! σ3 + α4<br />

4! σ4 x + · · · (6.177)<br />

Also, note that σx.σx = I, thus, σ2n x = I and σ2n+1 x = σx. Collect all the real terms and all the<br />

imaginary terms:<br />

e iασx =<br />

�<br />

I + I α2<br />

2<br />

+ I α4<br />

4!<br />

+ · · ·<br />

These are the series expansions for cos and sin.<br />

�<br />

+ iσx<br />

�<br />

α − −i α3<br />

3!<br />

+ · · ·<br />

�<br />

(6.178)<br />

e iασx = I cos(α) + iσx sin(α) (6.179)<br />

Exercise 6.9 Because <strong>of</strong> the interaction between the proton and the electron in the ground state<br />

<strong>of</strong> the Hydrogen atom, the atom has hyperfine structure. The energy matrix is <strong>of</strong> the form:<br />

in the basis defined by<br />

⎛<br />

⎜<br />

H = ⎜<br />

⎝<br />

A 0 0 0<br />

0 −A 2A 0<br />

0 2A −A 0<br />

0 0 0 A<br />

⎞<br />

⎟<br />

⎠<br />

(6.180)<br />

|1〉 = |e+, p+〉 (6.181)<br />

|2〉 = |e+, p−〉 (6.182)<br />

|3〉 = |e−, p+〉 (6.183)<br />

|4〉 = |e−, p−〉 (6.184)<br />

where the notation e+ means that the electron’s spin is along the +Z-axis, and e− has the spin<br />

pointed along the −Z axis. i.e. |e+, p+〉 is the state in which both the electron spin and proton<br />

spin is along the +Z axis.<br />

1. Find the energy <strong>of</strong> the stationary states and sketch an energy level diagram relating the<br />

energies and the coupling.<br />

178


2. Express the stationary states as linear combinations <strong>of</strong> the basis states.<br />

3. A magnetic field <strong>of</strong> strength B applied in the +Z direction and couples the |e+, p+〉 and<br />

|e−, p−〉 states. Write the new Hamiltonian matrix in the |e±, p±〉 basis. What happens to<br />

the energy levels <strong>of</strong> the stationary states as a result <strong>of</strong> the coupling? Add this information<br />

to the energy level diagram you sketched in part 1.<br />

Exercise 6.10 Consider a spin 1/2 particle with magnetic moment � M = γ � S. The spin space is<br />

spanned by the basis <strong>of</strong> |+〉 and |−〉 vectors, which are eigenvectors <strong>of</strong> Sz with eigenvalues ±¯h/2.<br />

At time t = 0, the state <strong>of</strong> the system is given by<br />

|ψ(0)〉 = |+〉<br />

1. If the observable Sx is measured at time t = 0, what results can be found and with what<br />

probabilities?<br />

2. Taking |ψ(0)〉 as the initial state, we apply a magnetic field parallel to the y axis with<br />

strength Bo. Calculate the state <strong>of</strong> the system at some later time t in the {|±〉} basis.<br />

3. Plot as a function <strong>of</strong> time the expectation values fo the observables Sx, Sy, and Sz. What<br />

are the values and probabilities? Is there a relation between Bo and t for the result <strong>of</strong> one<br />

<strong>of</strong> the measurements to be certain? Give a physical interpretation <strong>of</strong> this condition.<br />

4. Again, consider the same initial state, this time at t = 0, we measure Sy and find +¯h/2<br />

What is the state vector |ψ(0 + )〉 immediately after this measurement?<br />

5. Now we take |ψ(0 + )〉 and apply a uniform time-dependent field parallel to the z-axis. The<br />

Hamiltonian operator <strong>of</strong> the spin is then given by<br />

H(t) = ω(t)Sz<br />

Assume that prior to t = 0, ω(t) = 0 and for t > 0 increases linearly from 0 to ωo at time<br />

t = T . Show that for 0 ≤ t ≤ T , the state vector can be written as<br />

|ψ(t)〉 = 1 �<br />

√ e<br />

2<br />

iθ(t) |+〉 + ie −iθ(t) |−〉 �<br />

where θ(t) is a real function <strong>of</strong> t (which you need to determine).<br />

6. Finally, at time t = τ > T , we measure Sy. What results can we find and with what<br />

probabilities? Determine the relation which must exist between ωo and T in order for us to<br />

be sure <strong>of</strong> the result. Give the physical interpretation.<br />

179


Chapter 7<br />

Perturbation theory<br />

If you perturbate to much, you will go blind.<br />

– T. A. Albright<br />

In previous lectures, we discusses how , say through application <strong>of</strong> and external driving force,<br />

the stationary states <strong>of</strong> a molecule or other quantum mechanical system can be come coupled<br />

so that the system can make transitions from one state to another. We can write the transition<br />

amplitude exactly as<br />

G(i → j, t) = 〈j| exp(−iH(tj − ti))/¯h)|i〉 (7.1)<br />

where H is the full Hamiltonian <strong>of</strong> the uncoupled system plus the applied perturbation. Thus, G<br />

tells us the amplitude for the system prepared in state |i〉 at time ti and evolve under the applied<br />

Hamiltonian for some time tj − ti and be found in state |j〉. In general this is a complicated<br />

quantity to calculate. Often, the coupling is very complex. In fact, we can only exactly determine<br />

G for a few systems: linearly driven harmonic oscillators, coupled two level systems to name the<br />

more important ones.<br />

In today’s lecture and following lectures, we shall develop a series <strong>of</strong> well defined and systematic<br />

approximations which are widely used in all applications <strong>of</strong> quantum mechanics. We start<br />

with a general solution <strong>of</strong> the time-independent Schrödinger equation in terms and eventually<br />

expand the solution to infinite order. We will then look at what happens if we have a perturbation<br />

or coupling which depends explicitly upon time and derive perhaps the most important<br />

rule in quantum mechanics which is called: “Fermi’s Golden Rule”. 1<br />

7.1 Perturbation Theory<br />

In most cases, it is simply impossible to obtain the exact solution to the Schrödinger equation.<br />

In fact, the vast majority <strong>of</strong> problems which are <strong>of</strong> physical interest can not be resolved exactly<br />

and one is forced to make a series <strong>of</strong> well posed approximations. The simplest approximation is<br />

to say that the system we want to solve looks a lot like a much simpler system which we can<br />

1 During a seminar, the speaker mentioned Fermi’s Golden Rule. <strong>Pr<strong>of</strong></strong>. Wenzel raised his arm and in German<br />

spiked English chided the speaker that it was in fact HIS golden rule!<br />

180


solve with some additional complexity (which hopefully is quite small). In other words we want<br />

to be able to write our total Hamiltonian as<br />

H = Ho + V<br />

where Ho represents that part <strong>of</strong> the problem we can solve exactly and V some extra part which<br />

we cannot. This we take as a correction or perturbation to the exact problem.<br />

Perturbation theory can be formuated in a variery <strong>of</strong> ways, we begin with what is typically<br />

termed Rayleigh-Schrödinger perturbation theory. This is the typical approach and used most<br />

commonly. Let Ho|φn〉 = Wn|φn〉 and (Ho + λV )|ψ〉 = En|ψ〉 be the Schrödinger equations for<br />

the uncoupled and perturbed systems. In what follows, we take λ as a small parameter and<br />

expand the exact energy in terms <strong>of</strong> this parameter. Clearly, we write En as a function <strong>of</strong> λ and<br />

write:<br />

En(λ) = E (0)<br />

n + λE (1)<br />

n + λ 2 E (2)<br />

n . . . (7.2)<br />

Likewise, we can expand the exact wavefunction in terms <strong>of</strong> λ<br />

|ψn〉 = |ψ (0)<br />

n 〉 + λ|ψ (1)<br />

n 〉 + λ 2 |ψ (2)<br />

n 〉 . . . (7.3)<br />

Since we require that |ψ〉 be a solution <strong>of</strong> the exact Hamiltonian with energy En, then<br />

H|ψ〉 = (Ho + λV ) �<br />

|ψ (0)<br />

n 〉 + λ|ψ (1)<br />

= �<br />

E (0)<br />

n + λE (1)<br />

n + λ 2 E (2)<br />

n . . .<br />

Now, we collect terms order by order in λ<br />

• λ 0 : Ho|ψ (0)<br />

n 〉 = E (0)<br />

n |ψ (0)<br />

n 〉<br />

• λ 1 : Ho|ψ (1)<br />

n 〉 + V |ψ (0)<br />

n 〉 = E (0)<br />

n |ψ (1) 〉 + E (1)<br />

n |ψ (0)<br />

n 〉<br />

n 〉 + λ 2 |ψ (2)<br />

� �<br />

|ψ (0)<br />

n 〉 + λ|ψ (1)<br />

n 〉 + λ 2 |ψ (2)<br />

n 〉 . . . �<br />

• λ 2 : Ho|ψ (2)<br />

n 〉 + V |ψ (1) 〉 = E (0)<br />

n |ψ (2)<br />

n 〉 + E (1)<br />

n |ψ (1)<br />

n 〉 + E (2)<br />

n |ψ (0)<br />

n 〉<br />

and so on.<br />

n 〉 . . . �<br />

The λ 0 problem is just the unperturbed problem we can solve. Taking the λ 1 terms and<br />

multiplying by 〈ψ (0)<br />

n | we obtain:<br />

(7.4)<br />

(7.5)<br />

〈ψ (0)<br />

n |Ho|ψ (0)<br />

n 〉 + 〈ψ (0)<br />

n |V |ψ (0) 〉 = E (0)<br />

n 〈ψ (0)<br />

n |ψ (1)<br />

n 〉 + E (1)<br />

n 〈ψ (0)<br />

n |ψ (0)<br />

n 〉 (7.6)<br />

In other words, we obtain the 1st order correction for the nth eigenstate:<br />

E (1)<br />

n = 〈ψ (0)<br />

n |V |ψ (0) 〉.<br />

Note to obtain this we assumed that 〈ψ (1)<br />

n |ψ (0)<br />

n 〉 = 0. This is easy to check by performing a<br />

similar calculation, except by multiplying by 〈ψ (0)<br />

m | for m �= n and noting that 〈ψ (0)<br />

n |ψ (0)<br />

m 〉 = 0 are<br />

orthogonal state.<br />

〈ψ (0)<br />

m |Ho|ψ (0)<br />

n 〉 + 〈ψ (0)<br />

m |V |ψ (0) 〉 = E (0)<br />

n 〈ψ (0)<br />

m |ψ (1)<br />

n 〉 (7.7)<br />

181


Rearranging things a bit, one obtains an expression for the overlap between the unperturbed and<br />

perturbed states:<br />

〈ψ (0)<br />

m |ψ (1)<br />

n 〉 = 〈ψ(0)<br />

E (0)<br />

m |V |ψ (0)<br />

n 〉<br />

n − E (0)<br />

m<br />

Now, we use the resolution <strong>of</strong> the identity to project the perturbed state onto the unperturbed<br />

states:<br />

|ψ (1)<br />

n 〉 = �<br />

|ψ (0)<br />

m 〉〈ψ (0)<br />

m |ψ (1)<br />

n 〉<br />

m<br />

= �<br />

m�=n<br />

〈ψ (0)<br />

m |V |ψ (0)<br />

E (0)<br />

n − E (0)<br />

m<br />

n 〉<br />

(7.8)<br />

|ψ (0)<br />

m 〉 (7.9)<br />

where we explictly exclude the n = m term to avoid the singularity. Thus, the first-order<br />

correction to the wavefunction is<br />

|ψn〉 ≈ |ψ (0)<br />

n 〉 + �<br />

This also justifies our assumption above.<br />

m�=n<br />

〈ψ (0)<br />

m |V |ψ (0)<br />

E (0)<br />

n − E (0)<br />

m<br />

n 〉<br />

|ψ (0)<br />

m 〉. (7.10)<br />

7.2 Two level systems subject to a perturbation<br />

Let’s say that in the |±〉 basis our total Hamiltonian is given by<br />

In matrix form:<br />

H = ωSz + V Sx. (7.11)<br />

H =<br />

�<br />

ω V<br />

V −ω<br />

Diagonalization <strong>of</strong> the matrix is easy, the eigenvalues are<br />

We can also determine the eigenvectors:<br />

where<br />

�<br />

(7.12)<br />

E+ = √ ω 2 + V 2 (7.13)<br />

E− = − √ ω 2 + V 2 (7.14)<br />

|φ+〉 = cos(θ/2)|+〉 + sin(θ/2)|−〉 (7.15)<br />

|φ−〉 = − sin(θ/2)|+〉 + cos(θ/2)|−〉 (7.16)<br />

|V |<br />

tan θ = (7.17)<br />

ω<br />

For constant coupling, the energy gap ω between the coupled states determines how the states<br />

are mixed as the result <strong>of</strong> the coupling.<br />

plot splitting as a function <strong>of</strong> unperturbed energy gap<br />

182


7.2.1 Expansion <strong>of</strong> Energies in terms <strong>of</strong> the coupling<br />

We can expand the exact equations for E± in terms <strong>of</strong> the coupling assuming that the coupling<br />

is small compared to ω. To leading order in the coupling:<br />

E+ = ω(1 + 1<br />

� �<br />

�<br />

�|V<br />

| �2<br />

�<br />

� � · · ·) (7.18)<br />

2 � ω �<br />

E− = ω(1 − 1<br />

� �<br />

�<br />

�|V<br />

| �2<br />

�<br />

� � · · ·) (7.19)<br />

2 � ω �<br />

On the otherhand, where the two unperturbed states are identical, we can not do this expansion<br />

and<br />

and<br />

E+ = |V | (7.20)<br />

E− = −|V | (7.21)<br />

We can do the same trick on the wavefunctions: When ω ≪ |V | (strong coupling) , θ ≈ π/2,<br />

Thus,<br />

|ψ+〉 = 1<br />

√ 2 (|+〉 + |−〉) (7.22)<br />

|ψ−〉 = 1<br />

√ 2 (−|+〉 + |−〉). (7.23)<br />

In the weak coupling region, we have to first order in the coupling:<br />

|ψ+〉 = (|+〉 +<br />

|ψ−〉 = (|−〉 +<br />

|V |<br />

|−〉) (7.24)<br />

ω<br />

|V |<br />

|+〉). (7.25)<br />

ω<br />

In other words, in the weak coupling region, the perturbed states look a lot like the unperturbed<br />

states. Where as in the regions <strong>of</strong> strong mixing they are a combination <strong>of</strong> the unperturbed<br />

states.<br />

183


7.2.2 Dipole molecule in homogenous electric field<br />

Here we take the example <strong>of</strong> ammonia inversion in the presence <strong>of</strong> an electric field. From the<br />

problem sets, we know that the NH3 molecule can tunnel between two equivalent C3v configurations<br />

and that as a result <strong>of</strong> the coupling between the two configurations, the unperturbed<br />

energy levels Eo are split by an energy A. Defining the unperturbed states as |1〉 and |2〉 we can<br />

define the tunneling Hamiltonian as:<br />

or in terms <strong>of</strong> Pauli matrices:<br />

H =<br />

�<br />

Eo −A<br />

−A Eo<br />

�<br />

H = Eoσo − Aσx<br />

Taking ψ to be the solution <strong>of</strong> the time-dependent Schrödinger equation<br />

H|ψ(t)〉 = i¯h| ˙ ψ〉<br />

we can insert the identity |1〉〈1| + |2〉〈2| = 1 and re-write this as<br />

i¯h ˙c1 = Eoc1 − Ac2<br />

i¯h ˙c2 = Eoc2 − Ac1<br />

(7.26)<br />

(7.27)<br />

(7.28)<br />

where c1 = 〈1|ψ〉 and c2 = 〈2|ψ〉. are the projections <strong>of</strong> the time-evolving wavefunction onto the<br />

two basis states. Taking these last two equations and adding and subtracting them from each<br />

other yields two new equations for the time-evolution:<br />

i¯h ˙c+ = (Eo − A)c+<br />

i¯h ˙c− = (Eo + A)c−<br />

where c± = c1 ± c2 (we’ll normalize this later). These two new equations are easy to solve,<br />

Thus,<br />

�<br />

i<br />

c±(t) = A± exp<br />

¯h (Eo<br />

�<br />

∓ A)t .<br />

(7.29)<br />

(7.30)<br />

c1(t) = 1 �<br />

eiEot/¯h A+e<br />

2 −iAt/¯h + A−e +iAt/¯h�<br />

and<br />

c2(t) = 1 �<br />

eiEot/¯h A+e<br />

2 −iAt/¯h − A−e +iAt/¯h�<br />

.<br />

Now we have to specify an initial condition. Let’s take c1(0) = 1 and c2(0) = 0 corresponding to<br />

the system starting <strong>of</strong>f in the |1〉 state. For this initial condition, A+ = A− = 1 and<br />

and<br />

c1(t) = e iEot/¯h cos(At/¯h)<br />

c2(t) = e iEot/¯h sin(At/¯h).<br />

184


So that the time evolution <strong>of</strong> the state vector is given by<br />

|ψ(t)〉 = e iEot/¯h [|1〉 cos(At/¯h) + |2〉 sin(At/¯h)]<br />

So, left alone, the molecule will oscillate between the two configurations at the tunneling frequency,<br />

A/¯h.<br />

Now, we apply an electric field. When the dipole moment <strong>of</strong> the molecule is aligned parallel<br />

with the field, the molecule is in a lower energy configuration, whereas for the anti-parrallel case,<br />

the system is in a higher energy configuration. Denote the contribution to the Hamiltonian from<br />

the electric field as:<br />

H ′ = µeEσz<br />

The total Hamiltonian in the {|1〉, |2〉} basis is thus<br />

Solving the eigenvalue problem:<br />

we find two eigenvalues:<br />

H =<br />

�<br />

Eo + µeE −A<br />

−A Eo − µeE<br />

λ± = Eo ±<br />

|H − λI| = 0<br />

�<br />

A 2 + µ 2 eE 2 .<br />

These are the exact eigenvalues.<br />

In Fig. 7.1 we show the variation <strong>of</strong> the energy levels as a function <strong>of</strong> the field strength.<br />

�<br />

(7.31)<br />

Figure 7.1: Variation <strong>of</strong> energy level splitting as a function <strong>of</strong> the applied field for an ammonia<br />

molecule in an electric field<br />

Weak field limit<br />

If µeE/A ≪ 1, then we can use the binomial expansion<br />

√ 1 + x 2 ≈ 1 + x 2 /2 + . . .<br />

to write<br />

�<br />

A2 + µ 2 eE 2 �<br />

= A 1 +<br />

≈ A<br />

�<br />

1 + 1<br />

2<br />

�<br />

µeE<br />

A<br />

�<br />

µeE<br />

A<br />

�2 � 1<br />

/2<br />

�2 �<br />

(7.32)<br />

Thus in the weak field limit, the system can still tunnel between configurations and the energy<br />

splitting are given by<br />

E± ≈ (Eo ∓ A) ∓ µ2 eE 2<br />

A<br />

185


To understand this a bit further, let us use perturbation theory in which the tunneling<br />

dominates and treat the external field as a perturbing force. The unperturbed hamiltonian can<br />

be diagonalized by taking symmetric and anti-symmetric combinations <strong>of</strong> the |1〉 and |2〉 basis<br />

functions. This is exactly what we did above with the time-dependent coefficients. Here the<br />

stationary states are<br />

|±〉 = 1<br />

√ (|1〉 ± |2〉)<br />

2<br />

with energies E± = Eo ∓ A. So that in the |±〉 basis, the unperturbed Hamiltonian becomes:<br />

H =<br />

�<br />

Eo − A 0<br />

0 Eo + A<br />

The first order correction to the ground state energy is given by<br />

E (1) = E (0) + 〈+|H ′ |+〉<br />

To compute 〈+|H ′ |+〉 we need to transform H ′ from the {|1〉, |2〉} uncoupled basis to the new |±〉<br />

coupled basis. This is accomplished by inserting the identity on either side <strong>of</strong> H ′ and collecting<br />

terms:<br />

〈+|H ′ |+〉 = 〈+|(|1〉 < 1| + |2〉〈2|)H ′ (|1〉 < 1| + |2〉〈2|)〉 (7.33)<br />

= 1<br />

2 (〈1| + 〈2|)H′ (|1〉 + |2〉) (7.34)<br />

= 0 (7.35)<br />

Likewise for 〈−|H ′ |−〉 = 0. Thus, the first order correction vanish. However, since 〈+|H ′ |−〉 =<br />

µeE does not vanish, we can use second order perturbation theory to find the energy correction.<br />

W (2)<br />

+ = �<br />

H ′ miH ′ im<br />

m�=i<br />

Ei − Em<br />

�<br />

= 〈+|H′ |−〉〈−|H ′ |+〉<br />

E (0)<br />

+ − E (0)<br />

−<br />

(µeE)<br />

=<br />

2<br />

Eo −