Download Quantum Chemistry - Eric R. Bittner

Quantum Chemistry
Lecture notes to accompany Chemistry 6321
Copyright @1997-2001, University of Houston and Eric R. Bittner
All Rights Reserved.
November 21, 2001
Contents
0 Introduction
7
0.1 Essentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
0.2 Problem Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
0.3 Course Calendar . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
I
Lecture Notes
12
1 Waves and Wavefunctions
1.1 Position and Momentum Representation of
1.2 The Schrödinger Equation . . . . . . . . .
1.2.1 Gaussian Wavefunctions . . . . . .
1.2.2 Evolution of ψ(x) . . . . . . . . . .
1.3 Particle in a Box . . . . . . . . . . . . . .
1.3.1 Infinite Box . . . . . . . . . . . . .
1.3.2 Particle in a finite Box . . . . . . .
1.3.3 Scattering states and resonances. .
1.4 Summary . . . . . . . . . . . . . . . . . .
1.5 Problems and Exercises . . . . . . . . . . .
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2 Postulates of Quantum Mechanics
2.0.1 The description of a physical state: . . .
2.0.2 Description of Physical Quantities: . . .
2.0.3 Quantum Measurement: . . . . . . . . .
2.0.4 The Principle of Spectral Decomposition:
2.0.5 The Superposition Principle . . . . . . .
2.0.6 Reduction of the wavepacket: . . . . . .
2.0.7 The temporal evolution of the system: .
2.0.8 Dirac Quantum Condition . . . . . . . .
2.1 Dirac Notation and Linear Algebra . . . . . . .
2.1.1 Transformations and Representations . .
2.1.2 Operators . . . . . . . . . . . . . . . . .
2.1.3 Products of Operators . . . . . . . . . .
2.1.4 Functions Involving Operators . . . . . .
2.2 Constants of the Motion . . . . . . . . . . . . .
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2.3
2.4
2.5
2.6
2.7
Bohr Frequency and Selection Rules . . . .
Example using the particle in a box states
Time Evolution of Wave and Observable .
“Unstable States” . . . . . . . . . . . . . .
Problems and Exercises . . . . . . . . . . .
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56
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3 Bound States of The Schrödinger Equation
3.1 Introduction to Bound States . . . . . . . . . . . . . . . . . .
3.2 The Variational Principle . . . . . . . . . . . . . . . . . . . . .
3.2.1 Variational Calculus . . . . . . . . . . . . . . . . . . .
3.2.2 Constraints and Lagrange Multipliers . . . . . . . . . .
3.2.3 Variational method applied to Schrödinger equation . .
3.2.4 Variational theorems: Rayleigh-Ritz Technique . . . . .
3.2.5 Variational solution of harmonic oscillator ground State
3.3 The Harmonic Oscillator . . . . . . . . . . . . . . . . . . . . .
3.3.1 Harmonic Oscillators and Nuclear Vibrations . . . . . .
3.3.2 Classical interpretation. . . . . . . . . . . . . . . . . .
3.3.3 Molecular Vibrations . . . . . . . . . . . . . . . . . . .
3.4 Numerical Solution of the Schrödinger Equation . . . . . . . .
3.4.1 Numerov Method . . . . . . . . . . . . . . . . . . . . .
3.4.2 Numerical Diagonalization . . . . . . . . . . . . . . . .
3.5 Problems and Exercises . . . . . . . . . . . . . . . . . . . . . .
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107
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. 113
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4 Quantum Mechanics in 3D
4.1 Quantum Theory of Rotations . . . . . . . . . . . . . . . . .
4.2 Eigenvalues of the Angular Momentum Operator . . . . . .
4.3 Eigenstates of L2 . . . . . . . . . . . . . . . . . . . . . . . .
4.4 Eigenfunctions of L2 . . . . . . . . . . . . . . . . . . . . . .
4.5 Addition theorem and matrix elements . . . . . . . . . . . .
4.6 Legendre Polynomials and Associated Legendre Polynomials
4.7 Quantum rotations in a semi-classical context . . . . . . . .
4.8 Motion in a central potential: The Hydrogen Atom . . . . .
4.8.1 Radial Hydrogenic Functions . . . . . . . . . . . . . .
4.9 Spin 1/2 Systems . . . . . . . . . . . . . . . . . . . . . . . .
4.9.1 Theoretical Description . . . . . . . . . . . . . . . . .
4.9.2 Other Spin Observables . . . . . . . . . . . . . . . .
4.9.3 Evolution of a state . . . . . . . . . . . . . . . . . . .
4.9.4 Larmor Precession . . . . . . . . . . . . . . . . . . .
4.10 Problems and Exercises . . . . . . . . . . . . . . . . . . . . .
5 Perturbation theory
5.1 Perturbation Theory . . . . . . . . . . . . . . . . . . . . . .
5.2 Two level systems subject to a perturbation . . . . . . . . .
5.2.1 Expansion of Energies in terms of the coupling . . .
5.2.2 Dipole molecule in homogenous electric field . . . . .
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5.3 Dyson Expansion of the Schrödinger Equation . . . . . . . . . . . . . .
5.4 Van der Waals forces . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.4.1 Origin of long-ranged attractions between atoms and molecules .
5.4.2 Attraction between an atom a conducting surface . . . . . . . .
5.5 Perturbations Acting over a Finite amount of Time . . . . . . . . . . .
5.5.1 General form of time-dependent perturbation theory . . . . . .
5.5.2 Fermi’s Golden Rule . . . . . . . . . . . . . . . . . . . . . . . .
5.6 Interaction between an atom and light . . . . . . . . . . . . . . . . . .
5.6.1 Fields and potentials of a light wave . . . . . . . . . . . . . . .
5.6.2 Interactions at Low Light Intensity . . . . . . . . . . . . . . . .
5.6.3 Photoionization of Hydrogen 1s . . . . . . . . . . . . . . . . . .
5.6.4 Spontaneous Emission of Light . . . . . . . . . . . . . . . . . .
5.7 Time-dependent golden rule . . . . . . . . . . . . . . . . . . . . . . . .
5.7.1 Non-radiative transitions between displaced Harmonic Wells . .
5.7.2 Semi-Classical Evaluation . . . . . . . . . . . . . . . . . . . . .
5.8 Problems and Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . .
6 Semi-Classical Quantum Mechanics
6.1 Bohr-Sommerfield quantization . . . . . . . . . . . . . .
6.2 The WKB Approximation . . . . . . . . . . . . . . . . .
6.2.1 Asymptotic expansion for eigenvalue spectrum . .
6.2.2 WKB Wavefunction . . . . . . . . . . . . . . . .
6.2.3 Semi-classical Tunneling and Barrier Penetration
6.3 Connection Formulas . . . . . . . . . . . . . . . . . . . .
6.4 Problems and Exercises . . . . . . . . . . . . . . . . . . .
7 Many Body Quantum Mechanics
7.1 Symmetry with respect to particle Exchange .
7.2 Matrix Elements of Electronic Operators . . .
7.3 The Hartree-Fock Approximation . . . . . . .
7.3.1 Two electron integrals . . . . . . . . .
7.3.2 Koopman’s Theorem . . . . . . . . . .
7.4 Quantum Chemistry . . . . . . . . . . . . . .
7.4.1 The Born-Oppenheimer Approximation
7.5 Problems and Exercises . . . . . . . . . . . . .
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List of Figures
1.1 A gaussian wavepacket, ψ(x) . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Momentum-space distribution of ψ(k). . . . . . . . . . . . . . . . . . . . . . . .
1.3 Go for fixed t as a function of x. . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.4 Evolution of a free particle wavefunction. . . . . . . . . . . . . . . . . . . . . .
1.5 Particle in a box states . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.6 Graphical solution to transendental equations for an electron in a truncated hard
well of depth Vo = 10 and width a = 2. The short-dashed blue curve corresponds
to the symmetric caseqand the long-dashed blue curve corresponds to the asymetric
case. The red line is 1 − V o/E. Bound state solution are such that the red and
blue curves cross. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.7 Transmission (blue) and Reflection (red) coefficients for an electron scattering over
a square well (V = −40 and a = 1 ). . . . . . . . . . . . . . . . . . . . . . . . .
1.8 Transmission Coefficient for particle passing over a bump. . . . . . . . . . . . .
1.9 Scattering waves for particle passing over a well. . . . . . . . . . . . . . . . . . .
1.10 Argand plot of a scattering wavefunction passing over a well. . . . . . . . . . . .
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2.1
2.2
2.3
. 36
. 37
2.4
Gaussian distribution function . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Combination of two distrubitions. . . . . . . . . . . . . . . . . . . . . . . . . . .
Constructive and destructive interference from electron/two-slit experiment. The
superimposed red and blue curves are P1 and P2 from the classical probabilities
The diffraction function sin(x)/x . . . . . . . . . . . . . . . . . . . . . . . . . .
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. 57
3.1
3.2
3.3
3.4
3.5
Variational paths between endpoints. . . . . . . . . . . . . . . . . . . . . . . . . . 71
Hermite Polynomials, Hn up to n = 3. . . . . . . . . . . . . . . . . . . . . . . . . 83
Harmonic oscillator functions for n = 0 to 3 . . . . . . . . . . . . . . . . . . . . . 87
Quantum and Classical Probability Distribution Functions for Harmonic Oscillator. 88
London-Eyring-Polanyi-Sato (LEPS) empirical potential for the F +H2 → F H+H
chemical reaction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90
3.6 Morse well and harmonic approximation for HF . . . . . . . . . . . . . . . . . . . 91
3.7 Model potential for proton tunneling. . . . . . . . . . . . . . . . . . . . . . . . . . 92
3.8 Double well tunneling states as determined by the Numerov approach. . . . . . . . 93
3.9 Tchebyshev Polynomials for n = 1 − 5 . . . . . . . . . . . . . . . . . . . . . . . . 95
3.10 Ammonia Inversion and Tunneling . . . . . . . . . . . . . . . . . . . . . . . . . . 102
4.1
4.2
Vector Model of angular momentum . . . . . . . . . . . . . . . . . . . . . . . . . . 111
Spherical Harmonic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116
4
4.3
Classical and Quantum Probability Distribution Functions for Angular Momentum.124
5.1
5.2
Variation of energy level splitting as a function of the applied field for an ammonia
molecule in an electric field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141
Photo-ionization spectrum for hydrogen atom. . . . . . . . . . . . . . . . . . . . . 161
6.1
6.2
6.3
Eckart Barrier and parabolic approximation of the transition state . . . . . . . . . 185
Airy functions, Ai(y) (red) and Bi(y) (blue) . . . . . . . . . . . . . . . . . . . . . 188
Bound states in a graviational well . . . . . . . . . . . . . . . . . . . . . . . . . . 191
7.1
7.2
7.3
Various contributions to the H2+ Hamiltonian. . . . . . . . . . . . . . . . . . . . . 208
Potential energy surface for H2+ molecular ion. . . . . . . . . . . . . . . . . . . . . 210
Three dimensional representations of ψ+ and ψ− for the H2+ molecular ion. . . . . 210
5
List of Tables
3.1 Tchebychev polynomials of the first type . . . . . . . . . . . . . . . . . . . . . . . 95
3.2 Eigenvalues for double well potential computed via DVR and Numerov approaches 98
4.1 Spherical Harmonics (Condon-Shortley Phase convention. . . . . . . . . . . . . . . 116
4.2 Relation between various notations for Clebsch-Gordan Coefficients in the literature125
6.1
Location of nodes for Airy, Ai(x) function. . . . . . . . . . . . . . . . . . . . . . . 190
6
Chapter 0
Introduction
Nothing conveys the impression of humungous intellect so much as even the sketchiest
knowledge of quantum physics, and since the sketchiest knowledge is all anyone will
ever have, never be shy about holding forth with bags of authority about subatomic
particles and the quantum realm without having done any science whatsoever.
Jack Klaff –Bluff Your Way in the Quantum Universe
The field of quantum chemistry seeks to provide a rigorous description of chemical processes
at its most fundamental level. For ordinary chemical processes, the most fundamental and underlying theory of chemistry is given by the time-dependent and time-independent version of the
Schrödinger equation. However, simply stating an equation that provides the underlying theory
in now shape or form yields and predictive or interpretive power. In fact, most of what we do
in quantum mechanics is to develop a series of well posed approximation and physical assumptions to solve basic equations of quantum mechanics. In this course, we will delve deeply into
the underlying physical and mathematical theory. We will learn how to solve some elementary
problems and apply these to not so elementary examples.
As with any course of this nature, the content reflects the instructors personal interests in
the field. In this case, the emphasis of the course is towards dynamical processes, transitions
between states, and interaction between matter and radiation. More “traditional” quantum
chemistry courses will focus upon electronic structure. In fact, the moniker “quantum chemistry”
typically refers to electronic structure theory. While this is an extremely rich topic, it is my
personal opinion that a deeper understanding of dynamical processes provides a broader basis
for understanding chemical processes.
The purpose of this course is to provide a solid and mathematically rigorous tour through
modern quantum mechanics. We will begin with simple examples which can be worked out
exactly on paper and move on to discuss various approximation schemes. For cases in which
analytical solutions are either too obfuscating or impossible, computer methods will be introduced using Mathematica. Applications toward chemically relevant topics will be emphasized
throughout.
We will primarily focus upon single particle systems, or systems in which the particles are
distinguishable. Special considerations for systems of indistinguishable particles, such as the
electrons in a molecule, will be discussed towards the end of the course. The pace of the course
is fairly rigorous, with emphasis on solving problems either analytically or using computer.
7
The lecture notes in this package are really to be regarded as a work in progress and updates
and additions will be posted as they evolve. Lacking is a chapter on the Hydrogen atom and
atomic physics and a good overview of many body theory. Also, I have not included a chapter
on scattering. These will be added over the course of time. Certain sections are clearly better
than others and will be improved upon over time. Each chapter ends with a series of exercises
and suggested problems Some of which have detailed solutions. Others, you should work out on
your own. At the end of this book are a series of Mathematica notebooks I have written which
illustrate various points and perform a variety of calculations. These can be downloaded from
my web-site (http://k2.chem.uh.edu/quantum/) and run on any recent version of Mathematica.
(≥ v3.0).
It goes entirely without saying (but I will anyway) that these notes come from a wide variety
of sources which I have tried to cite where possible.
8
0.1
Essentials
• Instructor: Prof. Eric. R. Bittner.
• Office: Fleming 221 J
• Email: [email protected]
• Phone: -3-2775
• Office Hours: Monday and Thurs. afternoons or by appointment.
• Course Web Page: http://k2.chem.uh.edu/quantum/ Solution sets, course news, class
notes, sample computer routines, etc...will be posted from time to time on this web-page.
• Required Text: Introduction to Quantum Mechanics, by David Griffiths. The course
will roughly follow the material in this text.
• Recommended Texts:
– Quantum Mechanics, Cohen-Tannoudji, et al. This two volume book is very comprehensive and tends to be rather formal (and formidable) in its approach. The problems
are excellent.
– Quantum Mechanics, Landau and Lifshitz. This is volume 3 of L&L’s classical course
in modern physics. No self respecting scientist doesn’t have at least two or three of
their books at hand. This text tends to be pretty terse and uses the classical phrase
it is easy to show... quite a bit. The problems are usually worked out in detail and
are usually classic applications of quantum theory. This is a real land mark book, but
not a good book for starting off.
– Lectures in Quantum Mechanics, Gordon Baym. Baym’s book covers a wide range of
topics in a lecture note style.
– Quantum Chemistry, I. Levine. This is usually the first quantum book that chemists
get. I find it to be too wordy and the notation and derivations a bit ponderous. Levine
does not use Dirac notation. However, he does give a good overview of elementary
electronic structure theory and some if its important developments. Good for starting
off in electronic structure.
– Modern Quantum Mechanics, J. J. Sakurai. This is a real classic. Not good for a first
exposure since it assumes a fairly sophisticated understanding of quantum mechanics
and mathematics.
– Intermediate Quantum Mechanics, Hans Bethe and Roman Jackiw. This book is a
great exploration of advanced topics in quantum mechanics as illustrated by atomic
systems.
– What is Quantum Mechanics?, Transnational College of LEX. OK, this one I found
at Barnes and Noble and it’s more or less a cartoon book. But, it is really good. It
explores the historical development of quantum mechanics, has some really interesting insights into semi-classical and ”old” quantum theory, and presents the study of
quantum mechanics as a unfolding story.
9
– Quantum Mechanics in Chemistry by George Schatz and Mark Ratner.
I highly recommend that you use a variety of books since one author’s approach to a given
topic may be clearer than another’s approach.
• Prequisites: Graduate status in chemistry. This course is required for all Physical Chemistry students. The level of the course will be fairly rigorous and I assume that students
have had some exposure to quantum mechanics at the undergraduate level–typically in
Physical Chemistry, and are competent in linear algebra, calculus, and solving elementary
differential equations.
• Tests and Grades: There are no exams in this course, only problem sets and participation
in discussion. Students will be required to make a brief presentation of at least one problem
set over the course of the the semester.
0.2
Problem Sets
Your course grade will largely be determined by your performance on these problems as well as
the assigned discussion of a particular problem. My philosophy towards problem sets is that this
is the only way to really learn this material. These problems are intentionally challenging, but
not overwhelming, and are paced to correspond to what will be going on in the lecture.
Some ground rules:
1. Due dates are posted on each problem–usually 1 week or 2 weeks after they are assigned.
Late submissions may be turned in up to 1 week later. All problems must be turned in by
December 3. I will not accept any submissions after that date.
2. Each student will be required to present their solution to one problem (or part of one
problem) in class as assigned. This should be in the form of a 15 to 20 minutes discussion
showing how to work the problem. For lengthy derivations, students are encouraged to
provide copies to the class for reference.
3. Handwritten Problems. If I can’t read it, I won’t grade it. Period. Consequently, I strongly
encourage the use of word processing software for your final submission. Problem solutions
can be submitted electronically as Mathematica, Latex, or PDF files to [email protected] with
the subject: QUANTUM PROBLEM SET. Do not send me a MSWord file as an email
attachment. I expect some text (written in compete and correct sentences) to explain your
steps where needed and some discussion of the results. The computer lab in the basement
of Fleming has 20 PCs with copies of Mathematica or you can obtain your own license
from the University Media Center.
4. Collaborations. You are strongly encouraged to work together and collaborate on problems.
However, simply copying from your fellow student is not an acceptable collaboration.
5. These are the only problems you need to turn in. We will have additional exercises–mostly
coming from the lecture. Also, at the end of the lectures herein, are a set of suggested
problems and exercises to work on. Many of these have solutions.
10
0.3
Course Calendar
This is a rough schedule of topics we will cover. In essence we will follow the order in Griffith’s
book starting from a basic description of quantum wave mechanics and bound states. We will
then move onto the more formal aspects of quantum theory: Dirac notation, perturbation theory,
variational theory, and the like. Lastly, we move onto applications: Hydrogen atom, manyelectron systems, semi-classical approximations, and a semi-classical treatment of light absorption
and emission.
Corresponding chapters in Griffiths (G) and the lecture notes (L) are indicated in (). Eg. G:
1; L:1 means, read chapter 1 from Griffiths and these notes.
• 22-August: Course overview and review of elementary results
• 27-August: Optional Lecture (ACS MEETING)
• 29-August: Waves, wavefunctions. (G:1, 2.2,2.5,2.6; L:1)
• 3-Sept: No Lecture (Labor Day)
• 5-Sept: Overview of Postulates/review of linear algebra (G:3,L:2)
• 10/12 Sept Postulates: Dirac notation, superposition principle
• 17/19Sept: Bound States: Variational principle, harmonic oscillator (G: 2.3,7.1;L3,3.2,3.3)
• 24/26 Sept: Quantum mechanics in 3D: Angular momentum (G:4, L4.1-4.8)
• 1/3 Oct: Hydrogen atom
• 8/10 Oct: Perturbation Theory: (G:6, L:5.1-5.4)
• 15/17 Oct: Time-dependent Perturbation Theory: (G:9, L5.5-5.7)
• 22/24 Oct: Semi-classical approximations (G: 8, L:)
• 29/31 Oct Finish Semiclassical/Intro Quantum Statistics (G:5, L:7.1)
• 5/7 Nov Identical Particles/Quantum Statistics (G:5, L:7.1)
• 12/14 Nov: Helium atom, hydrogen ion(L:7.2)
• 19/21 Nov: Quantum Chemistry (L:7.3,7.4)
• 26/28 Nov: Scattering Theory (time permitting)
• 3 Dec–Last day to turn in problem sets
11
Part I
Lecture Notes
12
Chapter 1
Waves and Wavefunctions
In the world of quantum physics, no phenominon is a phenominon until it is a recorded
phenominon.
– John Archibald Wheler
The physical basis of quantum mechanics is
1. That matter, such as electrons, always arrives at a point as a discrete chunk, but that the
probibility of finding a chunk at a specified position is like the intensity distribution of a
wave.
2. The “quantum state” of a system is described by a mathematical object called a “wavefunction” or state vector and is denoted |ψi.
3. The state |ψi can be expanded in terms of the basis states of a given vector space, {|φi i}
as
|ψi =
X
|φi ihφi |ψi
(1.1)
i
where hφi |ψi denotes an inner product of the two vectors.
4. Observable quantities are associated with the expectation value of Hermitian operators and
that the eigenvalues of such operators are always real.
5. If two operators commute, one can measure the two associated physical quantities simultaneously to arbitrary precision.
6. The result of a physical measurement projects |ψi onto an eigenstate of the associated
operator |φn i yielding a measured value of an with probability |hφn |ψi|2 .
1.1
1
Position and Momentum Representation of |ψi
Two common operators which we shall use extensively are the position and momentum operator.
1
The majority of this lecture comes from Cohen-Tannoudji Chapter 1, part from Feynman & Hibbs
13
The position operator acts on the state |ψi to give the amplitude of the system to be at a
given position:
x̂|ψi = |xihx|ψi
= |xiψ(x)
(1.2)
(1.3)
We shall call ψ(x) the wavefunction of the system since it is the amplitude of |ψi at point x. Here
we can see that ψ(x) is an eigenstate of the position operator. We also define the momentum
operator p̂ as a derivative operator:
p̂ = −ih̄
∂
∂x
(1.4)
Thus,
p̂ψ(x) = −ih̄ψ 0 (x).
(1.5)
Note that ψ 0 (x) 6= ψ(x), thus an eigenstate of the position operator is not also an eigenstate of
the momentum operator.
We can deduce this also from the fact that x̂ and p̂ do not commute. To see this, first consider
∂
xf (x) = f (x) + xf 0 (x)
∂x
(1.6)
Thus (using the shorthand ∂x as partial derivative with respect to x.)
[x̂, p̂]f (x) = ih̄(x∂x f (x) − ∂x (xf (x)))
= −ih̄(xf 0 (x) − f (x) − xf 0 (x))
= ih̄f (x)
(1.7)
(1.8)
(1.9)
What are the eigenstates of the p̂ operator? To find them, consider the following eigenvalue
equation:
p̂|φ(k)i = k|φ(k)i
(1.10)
Inserting a complete set of position states using the idempotent operator
I=
Z
|xihx|dx
(1.11)
and using the “coordinate” representation of the momentum operator, we get
− ih̄∂x φ(k, x) = kφ(k, x)
(1.12)
Thus, the solution of this is (subject to normalization)
φ(k, x) = C exp(ik/h̄) = hx|φ(k)i
14
(1.13)
We can also use the |φ(k)i = |ki states as a basis for the state |ψi by writing
|ψi =
=
Z
Z
dk|kihk|ψi
(1.14)
dk|kiψ(k)
(1.15)
where ψ(k) is related to ψ(x) via:
ψ(k) = hk|ψi =
= C
Z
Z
dxhk|xihx|ψi
(1.16)
dx exp(ikx/h̄)ψ(x).
(1.17)
This type of integral is called a “Fourier Transfrom”. There are a number of ways to define
the√normalization C when using this transform, for our purposes at the moment, we’ll set C =
1/ 2πh̄ so that
ψ(x) = √
1 Z
dkψ(k) exp(−ikx/h̄)
2πh̄
(1.18)
1 Z
dxψ(x) exp(ikx/h̄).
2πh̄
(1.19)
and
ψ(x) = √
Using this choice of normalization, the transform and the inverse transform have symmetric forms
and we only need to remember the sign in the exponential.
1.2
The Schrödinger Equation
Postulate 1.1 The quantum state of the system is a solution of the Schrödinger equation
ih̄∂t |ψ(t)i = H|ψ(t)i,
(1.20)
where H is the quantum mechanical analogue of the classical Hamiltonian.
From classical mechanics, H is the sum of the kinetic and potential energy of a particle,
H=
1 2
p + V (x).
2m
(1.21)
Thus, using the quantum analogues of the classical x and p, the quantum H is
H=
1 2
p̂ + V (x̂).
2m
(1.22)
To evaluate V (x̂) we need a theorem that a function of an operator is the function evaluated
at the eigenvalue of the operator. The proof is straight forward, Taylor expand the function
about some point, If
1
V (x) = (V (0) + xV 0 (0) + V 00 (0)x2 · · ·)
2
15
(1.23)
then
1
V (x̂) = (V (0) + x̂V 0 (0) + V 00 (0)x̂2 · · ·)
2
(1.24)
[fˆ, fˆp ] = 0∀ p
(1.25)
hx|V (x̂)|ψi = V (x)ψ(x)
(1.26)
Since for any operator
Thus, we have
So, in coordinate form, the Schrödinger Equation is written as
∂
h̄ ∂ 2
+ V (x) ψ(x, t)
ih̄ ψ(x, t) = −
∂t
2m ∂x2
!
1.2.1
(1.27)
Gaussian Wavefunctions
Let’s assume that our initial state is a Gaussian in x with some initial momentum k◦ .
2
ψ(x, 0) =
πa2
1/4
exp(iko x) exp(−x2 /a2 )
(1.28)
The momentum representation of this is
1 Z
ψ(k, 0) =
dxe−ikx ψ(x, 0)
2πh̄
2 2
= (πa)1/2 e−(k−ko ) a /4)
(1.29)
(1.30)
In Fig.1.1, we see a gaussian wavepacket centered about x = 0 with ko = 10 and a = 1.
For now we will use dimensionaless units. The red and blue components correspond to the real
and imaginary components of ψ and the black curve is |ψ(x)|2 . Notice, that the wavefunction is
pretty localized along the x axis.
In the next figure, (Fig. 1.2) we have the momentum distribution of the wavefunction, ψ(k, 0).
Again, we have chosen ko = 10. Notice that√the center of the distribution is shifted about ko .
So, for f (x) = exp(−x2√
/b2 ), ∆x = b/ 2. Thus, when x varies form 0 to ±∆x, f (x) is
diminished by a factor of 1/ e. (∆x is the RMS deviation of f (x).)
For the Gaussian wavepacket:
∆x = a/2
∆k = 1/a
(1.31)
(1.32)
∆p = h̄/a
(1.33)
or
Thus, ∆x∆p = h̄/2 for the initial wavefunction.
16
0.75
0.5
0.25
-3
-2
-1
1
2
3
-0.25
-0.5
-0.75
Figure 1.1: Real (red), imaginary (blue) and absolute value (black) of gaussian wavepacket ψ(x)
è
y@kD
2.5
2
1.5
1
0.5
k
6
8
10
12
14
Figure 1.2: Momentum-space distribution of ψ(k).
17
1.2.2
Evolution of ψ(x)
Now, let’s consider the evolution of a free particle. By a “free” particle, we mean a particle
whose potential energy does not change, I.e. we set V (x) = 0 for all x and solve:
∂
h̄ ∂ 2
ih̄ ψ(x, t) = −
ψ(x, t)
∂t
2m ∂x2
!
(1.34)
This equation is actually easier to solve in k-space. Taking the Fourier Transform,
ih̄∂t ψ(k, t) =
k2
ψ(k, t)
2m
(1.35)
Thus, the temporal solution of the equation is
ψ(k, t) = exp(−ik 2 /(2m)t/h̄)ψ(k, 0).
(1.36)
This is subject to some initial function ψ(k, 0). To get the coordinate x-representation of the
solution, we can use the FT relations above:
1 Z
ψ(x, t) = √
dkψ(k, t) exp(−ikx)
2πh̄
Z
(1.37)
=
dx0 hx| exp(−ip̂2 /(2m)t/h̄)|x0 iψ(x0 , 0)
(1.38)
=
m Z
im(x − x0 )2
dx0 exp
ψ(x0 , 0)
2πih̄t
2h̄t
(1.39)
dx0 Go (x, x0 )ψ(x0 , 0)
(1.40)
r
=
Z
!
(homework: derive Go and show that Go is a solution of the free particle schrodinger equation
HGo = i∂t Go .) The function Go is called the “free particle propagator” or “Green’s Function”
and tells us the amplitude for a particle to start off at x0 and end up at another point x at time
t.
The sketch tells me that in order to got far away from the initial point in time t I need to
have a lot of energy (wiggles get closer together implies higher Fourier component )
Here we see that the probability to find a particle at the initial point decreases with time.
Since the period of oscillation (T ) is the time required to increase the phase by 2π.
mx2
mx2
−
2h̄t
2h̄(t + T )
!
2
mx
T2
=
2h̄t2 1 + T /t
2π =
(1.41)
(1.42)
Let ω = 2π/T and take the long time limit t T , we can estimate
2
m x
ω≈
2h̄ t
18
(1.43)
Figure 1.3: Go for fixed t as a function of x.
0.4
0.2
-10
-5
5
10
-0.2
-0.4
Since the classical kinetic energy is given by E = m/2v 2 , we obtain
E = h̄ω
(1.44)
Thus, the energy of the wave is proportional to the period of oscillation.
We can evaluate the evolution in x using either the Go we derived above, or by taking the
FT of the wavefunction evolving in k-space. Recall that the solution in k-space was
ψ(k, t) = exp(−ik 2 /(2m)t/h̄)ψ(k, 0)
(1.45)
Assuming a Gaussian form for ψ(k) as above,
√
a Z
2
2
dke−a /4(k−ko ) ei(kx−ω(k)t)
ψ(x, t) =
3/4
(2π)
(1.46)
where ω(k) is the dispersion relation for a free particle:
ω(k) =
h̄k 2
2m
(1.47)
Cranking through the integral:
ψ(x, t) =
2a2
π
!1/4
eiφ
(x − h̄ko /mt)2
iko x
e
exp
2 2 1/4
a2 + 2ih̄t/m
a4 + 4h̄m2t
"
#
(1.48)
where φ = −θ − h̄ko2 /(2m)t and tan 2θ = 2h̄t/(ma2 ).
Likewise, for the amplitude:
s
2
|ψ(x, t)| =
1
(x − v◦ t)2
exp
−
2π∆x(t)2
2∆x(t)2
"
19
#
(1.49)
Figure 1.4: Evolution of a free particle wavefunction. In this case we have given the initial state
a kick in the +x direction. Notice that as the system moves, the center moves at a constant rate
where as the width of the packet constantly spreads out over time.
Where I define
s
a
4h̄2 t2
∆x(t) =
1+ 2 4
2
ma
(1.50)
as the time dependent RMS width of the wave and the group velocity:
vo =
h̄ko
.
m
(1.51)
Now, since ∆p = h̄∆k = h̄/a is a constant for all time, the uncertainty relation becomes
∆x(t)∆p ≥ h̄/2
(1.52)
corresponding to the particle’s wavefunction becoming more and more diffuse as it evolves in
time.
1.3
1.3.1
Particle in a Box
Infinite Box
The Mathematica handout shows how one can use Mathematica to set up and solve some simple
problems on the computer. (One good class problem would be to use Mathematica to carry
out the symbolic manipulations for a useful or interesting problem and/or to solve the problem
numerically.)
The potential we’ll work with for this example consists of two infinitely steep walls placed
at x = ` and x = 0 such that between the two walls, V (x) = 0. Within this region, we seek
solutions to the differential equation
∂x2 ψ(x) = −2mE/h̄2 ψ(x).
(1.53)
The solutions of this are plane waves traveling to the left and to the right,
ψ(x) = A exp(−ikx) + B exp(+ikx)
(1.54)
The coefficients A and B we’ll have to determine. k is determined by substitution back into the
differential equation
ψ 00 (x) = −k 2 ψ(x)
√
(1.55)
Thus, k 2 = 2mE/h̄2 , or h̄k = 2mE. Let’s work in units in which h̄ = 1 and me = 1. Energy in
these units is the Hartree (≈ 27.eV.) Posted on the web-page is a file (c-header file) which has a
number of useful conversion factors.
20
Since ψ(x) must vanish at x = 0 and x = `
A+B = 0
A exp(ik`) + B exp(−ik`) = 0
(1.56)
(1.57)
We can see immediately that A = −B and that the solutions must correspond to a family of sine
functions:
ψ(x) = A sin(nπ/`x)
(1.58)
ψ(`) = A sin(nπ/``) = A sin(nπ) = 0.
(1.59)
Just a check,
To obtain the coefficient, we simply require that the wavefunctions be normalized over the range
x = [0, `].
Z
`
sin(nπx/`)2 dx =
0
`
2
(1.60)
Thus, the normalized solutions are
s
2
sin(nπ/`x)
(1.61)
`
The eigenenergies are obtained by applying the Hamiltonian to the wavefunction solution
ψn (x) =
h̄2 2
∂ ψn (x)
En ψn (x) = −
2m x
h̄2 n2 π 2
ψn (x)
=
2a2 m
Thus we can write En as a function of n
(1.62)
(1.63)
h̄2 π 2 2
n
(1.64)
2a2 m
for n = 0, 1, 2, .... What about the case where n = 0? Clearly it’s an allowed solution of
the Schrödinger Equation. However, we also required that the probability to find the particle
anywhere must be 1. Thus, the n = 0 solution cannot be permitted.
Note also that the cosine functions are also allowed solutions. However, the restriction of
ψ(0) = 0 and ψ(`) = 0 discounts these solutions.
In Fig. 1.5 we show the first few eigenstates for an electron trapped in a well of length a = π.
The potential is shown in gray. Notice that the number of nodes increases as the energy increases.
In fact, one can determine the state of the system by simply counting nodes.
What about orthonormality. We stated that the solution of the eigenvalue problem form an
orthonormal basis. In Dirac notation we can write
En =
hψn |ψm i =
=
Z
Z
0
dxhψn |xihx|ψm i
`
dxψn∗ (x)ψm (x)
2Z `
=
dx sin(nπx/`) sin(mπx/`)
` 0
= δnm .
21
(1.65)
(1.66)
(1.67)
(1.68)
14
12
10
8
6
4
2
-1
1
2
3
4
Figure 1.5: Particle in a box states
Thus, we can see in fact that these solutions do form a complete set of orthogonal states on
the range x = [0, `]. Note that it’s important to specify “on the range...” since clearly the sin
functions are not a set of orthogonal functions over the entire x axis.
1.3.2
Particle in a finite Box
Now, suppose our box is finite. That is
(
V (x) =
−Vo if −a < x < a
0
otherwise
(1.69)
Let’s consider the case for E < 0. The case E > 0 will correspond to scattering solutions. In
side the well, the wavefunction oscillates, much like in the previous case.
ψW (x) = A sin(ki x) + B cos(ki x)
(1.70)
where ki comes from the equation for the momentum inside the well
h̄ki =
q
2m(En + Vo )
(1.71)
We actually have two classes of solution, a symmetric solution when A = 0 and an antisymmetric solution when B = 0.
Outside the well the potential is 0 and we have the solutions
ψO (x) = c1 eρx andc2 e−ρx
22
(1.72)
We will choose the coefficients c1 and c2 as to create two cases, ψL and ψR on the left and right
hand sides of the well. Also,
√
h̄ρ = −2mE
(1.73)
Thus, we have three pieces of the full solution which we must hook together.
ψL (x) = Ceρx for x < −a
(1.74)
ψR (x) = De−ρx for x > −a
(1.75)
(1.76)
ψW (x) = A sin(ki x) + B cos(ki x)for inside the well
(1.77)
To find the coefficients, we need to set up a series of simultaneous equations by applying the
conditions that a.) the wavefunction be a continuous function of x and that b.) it have continuous
first derivatives with respect to x. Thus, applying the two conditions at the boundaries:
ψL (−a) − ψW (−a) = 0
(1.78)
(1.79)
ψR (a) − ψW (a) = 0
(1.80)
(1.81)
0
ψL0 (−a) − ψW
(−a) = 0
(1.82)
(1.83)
0
ψR0 (a) − ψW
(a) = 0
(1.84)
The matching conditions at x = a
The final results are (after the chalk dust settles):
1. For A = 0. B = D sec(aki )e−aρ and C = D. (Symmetric Solution)
2. For B = 0, A = C csc(aki )e−aρ and C = −D. (Antisymmetric Solution)
So, now we have all the coefficients expressed in terms of D, which we can determine by normalization (if so inclined). We’ll not do that integral, as it is pretty straightforward.
For the energies, we substitute the symmetric and antisymmetric solutions into the Eigenvalue
equation and obtain:
ρ cos(aki ) = ki sin(ki )
23
(1.85)
or
ρ
= tan(aki )
ki
s
q
E
= tan(a 2m(Vo − E)/h̄)
Vo − E
(1.86)
(1.87)
for the symmetric case and
ρ sin(aki ) = −ki cos(aki )
(1.88)
ρ
= cot(aki )
ki
(1.89)
for the anti-symmetric case, or
(1.90)
s
q
E
= cot(a 2m(Vo − E)/h̄)
Vo − E
(1.91)
Substituting the expressions for ki and ρ into final results for each case we find a set of
matching conditions: For the symmetric case, eigenvalues occur when ever the two curves
q
q
(1.92)
q
(1.93)
1 − Vo /E = tan(a 2m(E − Vo )/h̄)
and for the anti-symmetric case,
q
1 − Vo /E = cot(a 2m(E − Vo )/h̄)
These are called “transcendental” equations and closed form solutions are generally impossible
to obtain. Graphical solutions are helpful. In Fig. ?? we show the graphical solution to the
transendental equations for an electron in a Vo = −10 well of width a = 2. The black dots
indicate the presence of two bound states, one symmetric and one anti-symmetric at E = 2.03
and 3.78 repectively.
1.3.3
Scattering states and resonances.
Now let’s take the same example as above, except look at states for which E > 0. In this case, we
have to consider where the particles are coming from and where they are going. We will assume
that the particles are emitted with precise energy E towards the well from −∞ and travel from
left to right. As in the case above we have three distinct regions,
1. x > −a where ψ(x) = eik1 x + Re−ik1 x = ψL (x)
2. −a ≤ x ≤ +a where ψ(x) = Ae−ik2 x + Be+ik2 x = ψW (x)
24
symêasym
4
3
2
1
E
2
4
6
8
10
-1
-2
-3
-4
Figure 1.6: Graphical solution to transendental equations for an electron in a truncated hard
well of depth Vo = 10 and width a = 2. The short-dashed blue curve corresponds to the
symmetric
case and the long-dashed blue curve corresponds to the asymetric case. The red line
q
is 1 − V o/E. Bound state solution are such that the red and blue curves cross.
3. x > +a where ψ(x) = T e+ik1 x = ψR (x)
q
√
where k1 = 2mE/h̄ is the momentum outside the well, k2 = 2m(E − V )/h̄ is the momentum
inside the well, and A, B, T , and R are coefficients we need to determine. We also have the
matching conditions:
ψL (−a) − ψW (−a) = 0
0
ψL0 (−a) − ψW
(−a) = 0
ψR (a) − ψW (a) = 0
0
ψR0 (a) − ψW
(a) = 0
This can be solved by hand, however, Mathematica make it easy. The results are a series of rules
which we can use to determine the transmission and reflection coefficients.
−4e−2iak1 +2iak2 k1 k2
,
−k1 2 + e4iak2 k1 2 − 2k1 k2 − 2e4iak2 k1 k2 − k2 2 + e4iak2 k2 2
2e−iak1 +3iak2 k1 (k1 − k2 )
A →
,
−k1 2 + e4iak2 k1 2 − 2k1 k2 − 2e4iak2 k1 k2 − k2 2 + e4iak2 k2 2
−2e−iak1 +iak2 k1 (k1 + k2 )
B →
,
−k1 2 + e4iak2 k1 2 − 2k1 k2 − 2e4iak2 k1 k2 − k2 2 + e4iak2 k2 2
−1 + e4iak2 k1 2 − k2 2
R →
e2iak1 −k1 2 + e4iak2 k1 2 − 2k1 k2 − 2e4iak2 k1 k2 − k2 2 + e4iak2 k2 2
T →
25
R,T
1
0.8
0.6
0.4
0.2
10
20
30
40
En HhartreeL
Figure 1.7: Transmission (blue) and Reflection (red) coefficients for an electron scattering over
a square well (V = −40 and a = 1 ).
The R and T coefficients are related to the rations of the reflected and transimitted flux to
the incoming flux. The current operator is given by
j(x) =
h̄
(ψ ∗ ∇ψ − ψ∇ψ ∗ )
2mi
(1.94)
Inserting the wavefunctions above yields:
jin =
h̄k1
m
h̄k1 R2
m
= f rach̄k1 T 2 m
jref = −
jtrans
Thus, R2 = −jref /jin and T 2 = jtrans /jin . In Fig. 1.7 we show the transmitted and reflection
coefficients for an electron passing over a well of depth V = −40, a = 1 as a function of incident
energy, E.
Notice that the transmission and reflection coefficients under go a series oscillations as the
incident energy is increased. These are due to resonance states which lie in the continuum. The
condition for these states is such that an integer number of de Broglie wavelength of the wave in
the well matches the total length of the well.
λ/2 = na
Fig. 1.8,show the transmission coefficient as a function of both incident energy and the well
depth and (or height) over a wide range indicating that resonances can occur for both wells and
bumps. Figures 1.9 show various scattering wavefunctions for on an off-resonance cases. Lastly,
Fig. ?? shows an Argand plot of both complex components of ψ.
26
1
0.95
T
10
0.9
5
0.85
0
V
10
-5
20
En
30
40
-10
Figure 1.8: Transmission Coefficient for particle passing over a bump. Here we have plotted T as
a function of V and incident energy En . The oscillations correspond to resonance states which
occur as the particle passes over the well (for V < 0) or bump V > 0.
1.4
Summary
In three lectures we’ve covered a lot of ground. We now have enough tools at hand to begin
to study some physical systems. The traditional approach to studying quantum mechanics is
to progressively a series of differential equations related to physical systems (harmonic oscillators, angular momentum, hydrogen atom, etc...). We will return to those models in a week or
so. Next week, we’re going to look at 2 and 3 level systems using both time dependent and
time-independent methods. We’ll develop a perturbative approach for computing the transition
amplitude between states. We will also look at the decay of a state when its coupled to a continuum. These are useful models for a wide variety of phenomena. After this, we will move on
to the harmonic oscillator.
1.5
Problems and Exercises
Exercise 1.1
1. Derive the expression for
Go (x, x0 ) = hx| exp(−iho t/h̄)|x0 i
(1.95)
where ho is the free particle Hamiltonian,
h̄2 ∂ 2
ho = −
2m ∂x2
27
(1.96)
2. Show that Go is a solution of the free particle Schrödinger Equation
ih̄∂t Go (t) = ho Go (t).
(1.97)
Exercise 1.2 Show that the normalization of a wavefunction is independent of time.
Solution:
i∂t hψ(t)|ψ(t)i = (ihψ̇(t)|)(|ψ(t)i) + (hψ(t)|)(i|ψ̇(t)i)
= −hψ(t)|Ĥ † |ψ(t)i + hψ(t)|Ĥ|ψ(t)i
= −hψ(t)|Ĥ|ψ(t)i + hψ(t)|Ĥ|ψ(t)i = 0
(1.98)
(1.99)
(1.100)
Exercise 1.3 Compute the bound state solutions (E < 0) for a square well of depth Vo where
(
V (x) =
−Vo −a/2 ≤ x ≤ a/2
0
otherwise
(1.101)
1. How many energy levels are supported by a well of width a.
2. Show that a very narrow well can support only 1 bound state, and that this state is an even
function of x.
3. Show that the energy of the lowest bound state is
E≈
mVo2 a2
2h̄2
(1.102)
2mE
→0
h̄2
(1.103)
4. Show that as
s
−
ρ=
the probability of finding the particle inside the well vanishes.
Exercise 1.4 Consider a particle with the potential
V (x) =


 0


−Vo
∞
for x > a
for 0 ≤ x ≤ a
for x < 0
(1.104)
1. Let φ(x) be a stationary state. Show that φ(x) can be extended to give an odd wavefunction
corresponding to a stationary state of the symmetric well of width 2a (i.e the one studied
above) and depth Vo .
2. Discuss with respect to a and Vo the number of bound states and argue that there is always
at least one such state.
3. Now turn your attention toward the E > 0 states of the well. Show that the transmission
of the particle into the well region vanishes as E → 0 and that the wavefunction is perfectly
reflected off the sudden change in potential at x = a.
28
Exercise 1.5 Which of the following are eigenfunctions of the kinetic energy operator
T̂ = −
h̄2 ∂ 2
2m ∂x2
(1.105)
ex , x2 , xn ,3 cos(2x), sin(x) + cos(x), e−ikx ,
f (x − x0 ) =
Z
∞
0
dke−ik(x−x ) e−ik
2 /(2m)
(1.106)
−∞
.
Solution Going in order:
1. ex
2. x2
3. xn
4. 3 cos(2x)
5. sin(x) + cos(x)
6. e−ikx
Exercise 1.6 Which of the following would be acceptable one dimensional wavefunctions for a
2
2
bound particle (upon normalization): f (x) = e−x , f (x) = e−x , f (x) = xe−x , and
(
f (x) =
2
e−x
x≥0
−x2
2e
x<0
(1.107)
Solution In order:
1. f (x) = e−x
2. f (x) = e−x
2
2
3. f (x) = xe−x
4.
(
f (x) =
2
e−x
x≥0
−x2
2e
x<0
(1.108)
Exercise 1.7 For a one dimensional problem, consider a particle with wavefunction
exp(ipo x/h̄)
ψ(x) = N √ 2
x + a2
where a and po are real constants and N the normalization.
29
(1.109)
1. Determine N so that ψ(x) is normalized.
Z
∞
2
dx|ψ(x)|
= N
2
−∞
Z
∞
−∞
= N2
dx
x2
1
+ a2
(1.110)
pi
a
(1.111)
Thus ψ(x) is normalized when
r
N=
a
π
(1.112)
2. The √
position of the
√ particle is measured. What is the probability of finding a result between
−a/ 3 and +a/ 3?
√
Z +a/√3
1
a Z +a/ 3
2
=
√ dx 2
√ dx|ψ(x)|
x + a2
π −a/ 3
−a/ 3
(1.113)
√
+a/ 3
1
=
tan−1 (x/a) √
π
−a/ 3
1
=
3
(1.114)
(1.115)
3. Compute the mean value of a particle which has ψ(x) as its wavefunction.
aZ∞
x
hxi =
dx 2
π −∞ x + a2
= 0
(1.116)
(1.117)
Exercise 1.8 Consider the Hamiltonian of a particle in a 1 dimensional well given by
H=
1 2
p̂ + x̂2
2m
(1.118)
where x̂ and p̂ are position and momentum operators. Let |φn i be a solution of
H|φn i = En |φn i
(1.119)
hφn |p̂|φm i = αnm hφn |x̂|φm i
(1.120)
for n = 0, 1, 2, · · ·. Show that
where αnm is a coefficient depending upon En − Em . Compute αnm . (Hint: you will need to use
the commutation relations of [x̂, H] and [p̂, H] to get this). Finally, from all this, deduce that
X
(En − Em )2 |φn |x̂|φm i|2 =
m
30
h̄2
hφn |p̂2 |φn i
2m
(1.121)
Exercise 1.9 The state space of a certain physical system is three-dimensional. Let |u1 i, |u2 i,
and |u3 i be an orthonormal basis of the space in which the kets |ψ1 i and |ψ2 i are defined by
1
i
1
|ψ1 i = √ |u1 i + |u2 i + |u3 i
2
2
2
(1.122)
1
i
|ψ2 i = √ |u1 i + √ |u3 i
3
3
(1.123)
1. Are the states normalized?
2. Calculate the matrices, ρ1 and ρ2 representing in the {|ui ii basis, the projection operators
onto |ψ1 i and |ψ2 i. Verify that these matrices are Hermitian.
Exercise 1.10 Let ψ(r) = ψ(x, y, z) be the normalized wavefunction of a particle. Express in
terms of ψ(r):
1. A measurement along the x-axis to yield a result between x1 and x2 .
2. A measurement of momentum component px to yield a result between p1 and p2 .
3. Simultaneous measurements of x and pz to yield x1 ≤ x ≤ x2 and pz > 0.
4. Simultaneous measurements of px , py , and pz , to yield
p1 ≤ px ≤ p2
(1.124)
(1.125)
p3 ≤ py ≤ p4
(1.126)
(1.127)
p5 ≤ pz ≤ p6
(1.128)
Show that this result is equal to the result of part 2 when p3 , p5 → −∞ and p4 , p6 → +∞.
Exercise 1.11 Consider a particle of mass m whose potential energy is
V (x) = −α(δ(x + l/2) + δ(x − l/2))
1. Calculate the bound states of the particle, setting
h̄2 rho2
E=−
.
2m
Show that the possible energies are given by
−pl
e
2p
=± 1−
µ
!
where µ = 2mα/h̄2 . Give a graphic solution of this equation.
31
(a) The Ground State. Show that the ground state is even about the origin and that it’s
energy, Es is less than the bound state of a particle in a single δ-function potential,
−EL (c.f Sec 2.5 in Griffiths). Interpret this physically. Plot the corresponding wavefunction.
(b) Excited State. Show that when l is greater than some value (which you need to determine), there exists an odd excited state of energy EA with energy greater than −EL .
Determine and plot the corresponding wavefunction.
(c) Explain how the preceeding calculations enable us to construct a model for an ionized
diatomic molecule, eg. H2+ , whose nuclei are separated by l. Plot the energies of the
two states as functions of l, what happens as l → ∞ and l → 0?
(d) If we take Coulombic repulsion of the nuclei into account, what is the total energy
of the system? Show that a curve which gives the variation with respect to l of the
energies thus obtained enables us to predict in certain cases the existence of bound
states of H2+ and to determine the equilibrium bond length.
2. Calculate the reflection and transmission coefficients for this system. Plot R and T as
functions of l. Show that resonances occur when l is an integer multiple of the de Broglie
wavelength of the particle. Why?
32
1.5
1
0.5
-10
-5
5
10
5
10
-0.5
-1
-1.5
1
0.5
-10
-5
-0.5
-1
Figure 1.9: Scattering waves for particle passing over a well. In the top graphic, the particle is
partially reflected from the well (V < 0) and in the bottom graphic, the particle passes over the
well with a slightly different energy than above, this time with little reflection.
33
4
2
Im@yD
0
-2
-4
5
-10
0
Re@yD
0
x
-5
10
Figure 1.10: Argand plot of a scattering wavefunction passing over a well. (Same parameters as
in the top figure in Fig. 1.9).
34
Chapter 2
Postulates of Quantum Mechanics
When I hear the words “Schrödinger’s cat”, I wish I were able to reach for my gun.
Stephen Hawkings.
The dynamics of physical processes at a microscopic level is very much beyond the realm of
our macroscopic comprehension. In fact, it is difficult to imagine what it is like to move about on
the length and timescales for whcih quantum mechanics is important. However, for molecules,
quantum mechanics is an everyday reality. Thus, in order to understand how molecules move and
behave, we must develop a model of that dynamics in terms which we can comprehend. Making
a model means developing a consistent mathematical framework in which the mathematical
operations and constructs mimic the physical processes being studied.
Before moving on to develop the mathematical framework required for quantum mechanics,
let us consider a simple thought experiment. WE could do the experiment, however, we would
have to deal with some additional technical terms, like funding. The experiment I want to
consider goes as follows: Take a machine gun which shoots bullets at a target. It’s not a very
accurate gun, in fact, it sprays bullets randomly in the general direction of the target.
The distribution of bullets or histogram of the amount of lead accumulated in the target is
roughly a Gaussian, C exp(−x2 /a). The probability of finding a bullet at x is given by
P (x) = Ce−x
2 /a
.
(2.1)
C here is a normalization factor such that the probability of finding a bullet anywhere is 1. i.e.
Z
∞
dxP (x) = 1
(2.2)
−∞
The probability of finding a bullet over a small interval is
Z
b
dxP (x)i0.
(2.3)
a
Now suppose we have a bunker with 2 windows between the machine gun and the target such
that the bunker is thick enough that the bullets coming through the windows rattle around a
few times before emerging in random directions. Also, let’s suppose we can “color” the bullets
with some magical (or mundane) means s.t. bullets going through 1 slit are colored “red” and
35
Figure 2.1: Gaussian distribution function
1
0.8
0.6
0.4
0.2
-10
-5
5
36
10
Figure 2.2: Combination of two distrubitions.
1
0.8
0.6
0.4
0.2
-10
-5
5
10
bullets going throught the other slit are colored “blue”. Thus the distribution of bullets at a
target behind the bunker is now
P12 (x) = P1 (x) + P2 (x)
(2.4)
where P1 is the distribution of bullets from window 1 (the blue bullets) and P2 the “red” bullets.
Thus, the probability of finding a bullet that passed through either 1 or 2 is the sum of the
probabilies of going through 1 and 2. This is shown in Fig. ??
Now, let’s make an “electron gun” by taking a tungsten filiment heated up so that electrons
boil off and can be accellerated toward a phosphor screen after passing through a metal foil with
a pinhole in the middle We start to see little pin points of light flicker on the screen–these are
the individual electron “bullets” crashing into the phosphor.
If we count the number of electrons which strike the screen over a period of time–just as in
the machine gun experiment, we get a histogram as before. The reason we get a histogram is
slightly different than before. If we make the pin hole smaller, the distribution gets wider. This
is a manifestation of the Heisenberg Uncertainty Principle which states:
∆x · δp ≥ h̄/2
(2.5)
In otherwords, the more I restrict where the electron can be (via the pin hole) the more uncertain
I am about which direction is is going (i.e. its momentum parallel to the foil.) Thus, I wind up
with a distribution of momenta leaving the foil.
Now, let’s poke another hole in the foil and consider the distribution of electrons on the foil.
Based upon our experience with bullets, we would expect:
P12 = P1 + P2
(2.6)
BUT electrons obey quantum mechanics! And in quantum mechanics we represent a particle
via an amplitude. And one of the rules of quantum mechanics is that we first add amplitudes
and that probabilities are akin to the intensity of the combinded amplitudes. I.e.
P = |ψ1 + ψ2 |2
37
(2.7)
Figure 2.3: Constructive and destructive interference from electron/two-slit experiment. The
superimposed red and blue curves are P1 and P2 from the classical probabilities
1
0.8
0.6
0.4
0.2
-10
-5
5
10
where ψ1 and ψ2 are the complex amplitudes associated with going through hole 1 and hole 2.
Since they are complex numbers,
ψ1 = a1 + ib1 = |psi1 |eiφ1
ψ2 = a2 + ib2 = |psi2 |eiφ2
(2.8)
(2.9)
ψ1 + ψ2 = |ψ1 |eiφ1 + |ψ2 |eiφ2
(2.10)
Thus,
|ψ1 + ψ2 |2 = (|ψ1 |eiφ1 + |ψ2 |eiφ2 )
× (|ψ1 |e−iφ1 + |ψ2 |e−iφ2 )
P12 = |ψ1 |2 + |ψ2 |2 + 2|ψ1 ||ψ2 | cos(φ1 − φ2 )
q
P12 = P1 + P2 + 2 P1 P2 cos(φ1 − φ2 )
(2.11)
(2.12)
(2.13)
In other words, I get the same envelope as before, but it’s modulated by the cos(φ1 − φ2 )
“interference” term. This is shown in Fig. 2.3. Here the actual experimental data is shown as a
dashed line and the red and blue curves are the P1 and P2 . Just as if a wave of electrons struck
the two slits and diffracted (or interfered) with itself. However, we know that electrons come in
definite chunks–we can observe individual specks on the screen–only whole lumps arrive. There
are no fractional electrons.
Conjecture 1 Electrons–being indivisible chunks of matter–either go through slit 1 or slit 2.
Assuming Preposition 1 is true, we can divide the electrons into two classes:
1. Those that go through slit 1
38
2. Those that go through slit 2.
We can check this preposition by plugging up hole 1 and we get P2 as the resulting distribution.
Plugging up hole 2, we get P1 . Perhaps our preposition is wrong and electrons can be split in
half and half of it went through slit 1 and half through slit 2. NO! Perhaps, the electron went
through 1 wound about and went through 2 and through some round-about way made its way
to the screen.
Notice that in the center region of P12 , P12 > P1 +P2 , as if closing 1 hole actually decreased the
number of electrons going through the other hole. It seems very hard to justify both observations
by proposing that the electrons travel in complicated pathways.
In fact, it is very mysterious. And the more you study quantum mechanics, the more mysterious it seems. Many ideas have been cooked up which try to get the P12 curve in terms of
electrons going in complicated paths–all have failed.
Surprisingly, the math is simple (in this case). It’s just adding complex valued amplitudes.
So we conclude the following:
Electrons always arrive in discrete, indivisible chunks–like particles. However, the
probability of finding a chunk at a given position is like the distribution of the intensity
of a wave.
We could conclude that our conjecture is false since P12 6= P1 + P2 . This we can test.
Let’s put a laser behind the slits so that an electron going through either slit scatters a bit
of light which we can detect. So, we can see flashes of light from electrons going through slit
1, flashes of light from electrons going through slit 2, but NEVER two flashes at the same
time. Conjecture 1 is true. But if we look at the resulting distribution: we get P12 = P1 + P2 .
Measuring which slit the electon passes through destroys the phase information. When we make
a measurement in quantum mechanics, we really disturb the system. There is always the same
amount of disturbance because electrons and photons always interact in the same way every time
and produce the same sized effects. These effects “rescatter” the electrons and the phase info is
smeared out.
It is totally impossible to devise an experiment to measure any quantum phenomina without
disturbing the system you’re trying to measure. This is one of the most fundimental and perhaps
most disturbing aspects of quantum mechanics.
So, once we have accepted the idea that matter comes in discrete bits but that its behavour
is much like that of waves, we have to adjust our way of thinking about matter and dynamics
away from the classical concepts we are used to dealing with in our ordinary life.
These are the basic building blocks of quantum mechanics. Needless to say they are stated in
a rather formal language. However, each postulate has a specific physical reason for its existance.
For any physical theory, we need to be able to say what the system is, how does it move, and
what are the possible outcomes of a measurement. These postulates provide a sufficient basis for
the development of a consistent theory of quantum mechanics.
39
2.0.1
The description of a physical state:
The state of a physical system at time t is defined by specifying a vector |ψ(t)i belonging to a
state space H. We shall assume that this state vector can be normalized to one:
hψ|ψi = 1
2.0.2
Description of Physical Quantities:
Every measurable physical quantity, A, is described by an operator acting in H; this operator is
an observable.
A consequence of this is that any operator related to a physical observable must be Hermitian.
This we can prove. Hermitian means that
hx|O|yi = hy|O|xi∗
(2.14)
Thus, if O is a Hermitian operator and hOi = hψ|O|ψi = λhψ|ψi,
hOi = hx|O|xi + hx|O|yi + hy|O|xi + hy|O|yi.
(2.15)
Likewise,
hOi∗ =
=
=
=
hx|O|xi∗ + hx|O|yi∗ + hy|O|xi∗ + hy|O|yi∗
hx|O|xi + hy|O|xi∗ + hx|O|yi + hy|O|yi
hOi
λ
(2.16)
(2.17)
If O is Hermitian, we can also write
hψ|O = λhψ|.
(2.18)
which shows that hψ| is an eigenbra of O with real eigenvalue λ. Therefore, for an arbitrary ket,
hψ|O|φi = λhψ|φi
(2.19)
Now, consider eigenvectors of a Hermitian operator, |ψi and |φi. Obviously we have:
O|ψi = λ|ψi
O|φi = µ|φi
(2.20)
(2.21)
hψ|O = λhψ|
hφ|O = µhφ|
(2.22)
(2.23)
hφ|O|ψi = λhφ|ψi
hφ|O|ψi = µhφ|ψi
(2.24)
(2.25)
Since O is Hermitian, we also have
Thus, we can write:
Subtracting the two: (λ − µ)hφ|ψi = 0. Thus, if λ 6= µ, |ψi and |φi must be orthogonal.
40
2.0.3
Quantum Measurement:
The only possible result of the measurement of a physical quantity is one of the eigenvalues of
the corresponding observable. To any physical observable we ascribe an operator, O. The result
of a physical measurement must be an eigenvalue, a. With each eigenvalue, there corresponds
an eigenstate of O, |φa i. This function is such that the if the state vector, |ψ(t◦ )i = |φa i where
t◦ corresponds to the time at which the measurement was preformed, O|ψi = a|ψi and the
measurement will yield a.
Suppose the state-function of our system is not an eigenfunction of the operator we are
interested in. Using the superposition principle, we can write an arbitrary state function as a
linear combination of eigenstates of O
|ψ(t◦ )i =
X
hφa |ψ(t◦ )i|φa i
a
=
X
ca |φa i.
(2.26)
a
where the sum is over all eigenstates of O. Thus, the probability of observing answer a is |ca |2 .
IF the measurement DOES INDEED YIELD ANSWER a, the wavefunction of the system
at an infinitesmimal bit after the measurement must be in an eigenstate of O.
|ψ(t+
◦ )i = |φa i.
2.0.4
(2.27)
The Principle of Spectral Decomposition:
For a non-discrete spectrum: When the physical quantity, A, is measured on a system in a
normalized state |ψi, the probability P(an ) of obtaining the non-degenerate eigenvalue an of the
corresponding observable is given by
P(an ) = |hun |ψi|2
(2.28)
where |un i is a normalized eigenvector of A associated with the eigenvalue an . i.e.
A|un i = an |un i
For a discrete spectrum: the sampe principle applies as in the non-discrete case, except we
sum over all possible degeneracies of an
P(an ) =
gn
X
|hun |ψi|2
i=1
Finally, for the case of a continuous spectrum: the probability of obtaining a result between
α and α + dα is
dPα = |hα|ψi|2 dα
41
2.0.5
The Superposition Principle
Let’s formalize the above discussion a bit and write the electron’s state |ψi = a|1i + b|2i where
|1i and |2i are “basis states” corresponding to the electron passing through slit 1 or 2. The
coefficients, a and b, are just the complex numbers ψ1 and ψ2 written above. This |ψi is a vector
in a 2-dimensional complex space with unit length since ψ12 + ψ12 = 1. 1
Let us define a Vector Space by defining a set of objects {|ψi}, an addition rule: |φi =
|ψi + |ψ 0 > which allows us to construct new vectors, and a scaler multiplication rule |φi = a|ψi
which scales the length of a vector. A non-trivial example of a vector space is the x, y plane.
Adding two vectors gives another vector also on the x, y plane and multiplying a vector by a
constant gives another vector pointed in the same direction but with a new length.
The inner product of two vectors is written as
ψx
hφ|ψi = (φx φy )
ψy
∗
∗
= φx ψx + φy ψy
= hψ|φi∗ .
!
(2.29)
(2.30)
(2.31)
The length of a vector is just the inner product of the vector with itself, i.e. ψ|ψi = 1 for the
state vector we defined above.
The basis vectors for the slits can be used as a basis for an arbitrary state |ψi by writing it
as a linear combination of the basis vectors.
|ψi = ψ1 |1i + ψ1 |1i
(2.32)
In fact, any vector in the vector space can always be written as a linear combination of basis
vectors. This is the superposition principle.
The different ways of writing the vector |ψi are termed representations. Often it is easier to
work in one representation than another knowing fully that one can always switch back in forth at
will. Each different basis defines a unique representation. An example of a representation are the
unit vectors on the x, y plane. We can also define another orthonormal representation of the x, y
plane by introducing the unit vectors |ri,
√|θi, which define−1a polar coordinate system. One can
write the vector v = a|xi + b|y > as v = a2 + b2 |ri + tan (b/a)|θi or v = r sin θ|xi + r cos θ|yi
and be perfectly correct. Usually experience and insight is the only way to determine a priori
which basis (or representation) best suits the problem at hand.
Transforming between representations is accomplished by first defining an object called an
operator which has the form:
I =
X
|iihi|.
(2.33)
i
The sum means “sum over all members of a given basis”. For the xy basis,
I = |xihx| + |yihy|
1
(2.34)
The notation we are introducing here is known as “bra-ket” notation and was invented by Paul Dirac. The
vector |ψi is called a “ket”. The corresponding “bra” is the vector hψ| = (ψx∗ ψy∗ ), where the ∗ means complex
conjugation. The notation is quite powerful and we shall use is extensively throughout this course.
42
This operator is called the “idempotent” operator and is similar to multiplying by 1. For example,
I|ψi = |1ih1|ψi + |2ih2|ψi
= ψ1 |1i + ψ2 |2i
= |ψi
(2.35)
(2.36)
(2.37)
|ψi = |1ih1|ψi + |2ih2|ψi
(2.38)
We can also write the following:
The state of a system is specified completely by the complex vector |ψi which can be written
as a linear superposition of basis vectors spanning a complex vector space (Hilbert space). Inner
products of vectors in the space are as defined above and the length of any vector in the space
must be finite.
Note, that for state vectors in continuous representations, the inner product relation can be
written as an integral:
hφ|ψi =
Z
dqφ∗ (q)φ(q)
(2.39)
dq|φ(q)|2 ≤ ∞.
(2.40)
and normalization is given by
hψ|ψi =
Z
The functions, ψ(q) are termed square integrable because of the requirement that the inner
product integral remain finite. The physical motivation for this will become apparent in a
moment when ascribe physical meaning to the mathematical objects we are defining. The class
of functions satisfying this requirement are also known as L2 functions. (L is for Lebesgue,
referring to the class of integral.)
The action of the laser can also be represented mathematically as an object of the form
P1 = |1ih1|.
(2.41)
P2 = |2ih2|
(2.42)
and
and note that P1 + P2 = I.
When P1 acts on |ψi it projects out only the |1i component of |ψi
P1 |ψi = ψ1 |1i.
(2.43)
The expectation value of an operator is formed by writing:
hP1 i = hψ|P1 |ψi
(2.44)
hPx i = hψ|1ih1|ψi
= ψ12
(2.45)
Let’s evaluate this:
43
Similarly for P2 .
Part of our job is to insure that the operators which we define have physical counterparts.
We defined the projection operator, P1 = |1ih1|, knowing that the physical polarization filter
removed all “non” |1i components of the wave. We could have also written it in another basis,
the math would have been slightly more complex, but the result the same. |ψ1 |2 is a real number
which we presumably could set off to measure in a laboratory.
2.0.6
Reduction of the wavepacket:
If a measurement of a physical quantity A on the system in the state |ψi yields the result, an ,
the state of the physical system immediately after the measurement is the normalized projection
Pn |ψi onto the eigen subspace associated with an .
In more plain language, if you observe the system at x, then it is at x. This is perhaps
the most controversial posulate since it implies that the act of observing the system somehow
changes the state of the system.
Suppose the state-function of our system is not an eigenfunction of the operator we are
interested in. Using the superposition principle, we can write an arbitrary state function as a
linear combination of eigenstates of O
|ψ(t◦ )i =
X
hφa |ψ(t◦ )i|φa i
a
=
X
ca |φa i.
(2.46)
a
where the sum is over all eigenstates of O. Thus, the probability of observing answer a is |ca |2 .
IF the measurement DOES INDEED YIELD ANSWER a, the wavefunction of the system
at an infinitesmimal bit after the measurement must be in an eigenstate of O.
|ψ(t+
◦ )i = |φa i.
(2.47)
This is the only postulate which is a bit touchy deals with the reduction of the wavepacket
as the result of a measurement. On one hand, you could simply accept this as the way one
goes about business and simply state that quantum mechanics is an algorithm for predicting
the outcome of experiments and that’s that. It says nothing about the inner workings of the
universe. This is what is known as the “Reductionist” view point. In essence, the Reductionist
view point simply wants to know the answer: “How many?”, “How wide?”, “How long?”.
On the other hand, in the Holistic view, quantum mechanics is the underlying physical theory
of the universe and say that the process of measurement does play an important role in how the
universe works. In otherwords, in the Holist wants the (w)hole picture.
The Reductionist vs. Holist argument has been the subject of numerous articles and books
in both the popular and scholarly arenas. We may return to the philosophical discussion, but
for now we will simply take a reductionist view point and first learn to use quantum mechanics
as a way to make physical predictions.
44
2.0.7
The temporal evolution of the system:
The time evolution of the state vector is given by the Schrödinger equation
ih̄
∂
|ψ(t)i = H(t)|ψ(t)i
∂t
where H(t) is an operator/observable associated withthe total energy of the system.
As we shall see, H is the Hamiltonian operator and can be obtained from the classical Hamiltonian of the system.
2.0.8
Dirac Quantum Condition
One of the crucial aspects of any theory is that we need to be able to construct physical observables. Moreover, we would like to be able to connect the operators and observables in quantum
mechanics to the observables in classical mechanics. At some point there must be a correspondence. This connection can be made formally by relating what is known as the Poisson bracket
in classical mechanics:
{f (p, q), g(p, q)} =
∂f ∂g ∂g ∂f
−
∂q ∂p ∂q ∂p
(2.48)
which looks a lot like the commutation relation between two linear operators:
[Â, B̂] = ÂB̂ − B̂ Â
(2.49)
Of course, f (p, q) and g(p, q) are functions over the classical position and momentum of the
physical system. For position and momentum, it is easy to show that the classical Poisson
bracket is
{q, p} = 1.
Moreover, the quantum commutation relation between the observable x and p is
[x̂, p̂] = ih̄.
Dirac proposed that the two are related and that this relation defines an acceptible set of
quantum operations.
The quantum mechanical operators fˆ and ĝ, which in classical theory replace the
classically defined functions f and g, must always be such that the commutator of fˆ
and ĝ corresponds to the Poisson bracket of f and g according to
ih̄{f, g} = [fˆ, ĝ]
(2.50)
To see how this works, we write the momentum operator as
p̂ =
h̄ ∂
i ∂x
45
(2.51)
Thus,
p̂ψ(x) =
h̄ ∂ψ(x)
i ∂x
(2.52)
Let’s see if x̂ and p̂ commute. First of all
∂
xf (x) = f (x) + xf 0 (x)
∂x
(2.53)
Thus,
h̄ ∂
∂
(x f (x) −
xf (x)
i ∂x
∂x
h̄
=
(xf 0 (x) − f (x) − xf 0 (x))
i
= ih̄f (x)
[x̂, p̂]f (x) =
(2.54)
The fact that x̂ and p̂ do not commute has a rather significant consequence:
In other words, if two operators do not commute, one cannot devise and experiment to
simultaneously measure the physical quantities associated with each operator. This in fact limits
the precision in which we can preform any physical measurement.
The principle result of the postulates is that the wavefunction or state vector of the system
carries all the physical information we can obtain regarding the system and allows us to make
predictions regarding the probable outcomes of any experiment. As you may well know, if one
make a series of experimental measurements on identically prepared systems, one obtains a
distribution of results–usually centered about some peak in the distribution.
When we report data, we usually don’t report the result of every single experiment. For
a spectrscopy experiment, we may have made upwards of a million individual measurement,
all distributed about some average value. From statistics, we know that the average of any
distribution is the expectation value of some quantity, in this case x:
E(x) =
Z
P(x)xdx
(2.55)
for the case of a discrete spectra, we would write
E[h] =
X
hn Pn
(2.56)
n
where hn is some value and Pn the number of times you got that value normalized so that
X
Pn = 1
n
. In the language above, the hn ’s are the possible eigenvalues of the h operator.
A similar relation holds in quantum mechanics:
Postulate 2.1 Observable quantities are computed as the expectation value of an operator hOi =
hψ|O|ψi. The expectation value of an operator related to a physical observable must be real.
46
For example, the expectation value of x̂ the position operator is computed by the integral
hxi =
Z
+∞
ψ ∗ (x)xψ(x)dx.
−∞
or for the discrete case:
hOi =
X
on |hn|ψi|2 .
n
Of course, simply reporting the average or expectation values of an experiment is not enough,
the data is usually distributed about either side of this value. If we assume the distribution is
Gaussian, then we have the position of the peak center xo = hxi as well as the width of the
Gaussian σ 2 .
The mean-squared width or uncertainty of any measurement can be computed by taking
σA2 = h(A − hAi)i.
In statistical mechanics, this the fluctuation about the average of some physical quantity, A. In
quantum mechanics, we can push this definition a bit further.
Writing the uncertainty relation as
σA2 = h(A − hAi)(A − hAi)i
= hψ|(A − hAi)(A − hAi)|ψi
= hf |f i
(2.57)
(2.58)
(2.59)
where the new vector |f i is simply short hand for |f i = (A − hAi)|ψi. Likewise for a different
operator B
σB2 = = hψ|(B − hBi)(B − hBi)|ψi
= hg|gi.
(2.60)
(2.61)
We now invoke what is called the Schwartz inequality
σA2 σB2 = hf |f ihg|gi ≥ |hf |gi|2
(2.62)
So if we write hf |gi as a complex number, then
|hf |gi|2 = |z|2
= <(z)2 + =(z)2
2
1
∗
2
≥ =(z) =
(z − z )
2i
2
1
≥
(hf |gi − hg|f i)
2i
(2.63)
So we conclude
σA2 σB2
2
1
(hf |gi − hg|f i)
≥
2i
47
(2.64)
Now, we reinsert the definitions of |f i and |gi.
hf |gi =
=
=
=
hψ|(A − hAi)(B − hBi)|ψi
hψ|(AB − hAiB − AhBi + hAihBi)|ψi
hψ|AB|ψi − hAihψ|B|ψi − hBihψ|A|ψi + hAihBi
hABi − hAihBi
(2.65)
Likewise
hg|f i = hBAi − hAihBi.
(2.66)
Combining these results, we obtain
hf |gi − hg|f i = hABi − hBAi = hAB − BAi = h[A, B]i.
(2.67)
So we finally can conclude that the general uncertainty product between any two operators is
given by
2
1
h[A, B]i
(2.68)
2i
This is commonly referred to as the Generalized Uncertainty Principle. What is means is that
for any pair of observables whose corresponding operators do not commute there will always be
some uncertainty in making simultaneous measurements. In essence, if you try to measure two
non-commuting properties simultaneously, you cannot have an infinitely precise determination of
both. A precise determination of one implies that you must give up some certainty in the other.
In the language of matrices and linear algebra this implies that if two matrices do not commute, then one can not bring both matrices into diagonal form using the same transformation
matrix. in other words, they do not share a common set of eigenvectors. Matrices which do
commute share a common set of eigenvectors and the transformation which diagonalizes one will
also diagonalize the other.
σA2 σB2 =
Theorem 2.1 If two operators A and B commute and if |ψi is an eigenvector of A, then B|ψi
is also an eigenvector of A with the same eigenvalue.
Proof: If |ψi is an eigenvector of A, then A|ψi = a|ψi. Thus,
BA|ψi = aB|ψi
(2.69)
Assuming A and B commute, i.e. [A, B] = AB − BA = 0,
AB|ψi = a(B|ψi)
(2.70)
Thus, (B|ψi) is an eigenvector of A with eigenvalue, a.
Exercise 2.1
1. Show that matrix multiplication is associative, i.e. A(BC) = (AB)C, but
not commutative (in general), i.e. BC 6= CB
2. Show that (A + B)(A − B) = A2 + B 2 only of A and B commute.
3. Show that if A and B are both Hermitian matrices, AB + BA and i(AB − BA) are also
Hermitian. Note that Hermitian matrices are defined such that Aij = A∗ji where ∗ denotes
complex conjugation.
48
2.1
Dirac Notation and Linear Algebra
Part of the difficulty in learning quantum mechanics comes from fact that one must also learn a
new mathematical language. It seems very complex from the start. However, the mathematical
objects which we manipulate actually make life easier. Let’s explore the Dirac notation and the
related mathematics.
We have stated all along that the physical state of the system is wholly specified by the
state-vector |ψi and that the probability of finding a particle at a given point x is obtained via
|ψ(x)|2 . Say at some initial time |ψi = |si where s is some point along the x axis. Now, the
amplitude to find the particle at some other point is hx|si. If something happens between the
two points we write
hx|operator describing process|si
(2.71)
The braket is always read from right to left and we interpret this as the amplitude for“starting
off at s, something happens, and winding up at i”. An example of this is the Go function in the
homework. Here, I ask “what is the amplitude for a particle to start off at x and to wind up at
x0 after some time interval t?”
Another Example: Electrons have an intrinsic angular momentum called “spin” . Accordingly,
they have an associated magnetic moment which causes electrons to align with or against an
imposed magnetic field (eg.this gives rise to ESR). Lets say we have an electron source which
produces spin up and spin down electrons with equal probability. Thus, my initial state is:
|ii = a|+i + b|−i
(2.72)
√
Since I’ve stated that P (a) = P (b), |a|2 = |b|2 . Also, since P (a) + P (b) = 1, a = b = 1/ 2.
Thus,
1
|ii = √ (|+i + |−i)
2
(2.73)
Let’s say that the spin ups can be separated from the spin down via a magnetic field, B and we
filter off the spin down states. Our new state is |i0 i and is related to the original state by
hi0 |ii = ah+|+i + bh+|−i = a.
2.1.1
(2.74)
Transformations and Representations
If I know the amplitudes for |ψi in a representation with a basis |ii , it is always possible to
find the amplitudes describing the same state in a different basis |µi. Note, that the amplitude
between two states will not change. For example:
|ai =
X
|iihi|ai
(2.75)
|µihµ|ai
(2.76)
i
also
|ai =
X
µ
49
Therefore,
X
hµ|ai =
hµ|iihi|ai
(2.77)
hi|µihµ|ai.
(2.78)
i
and
hi|ai =
X
µ
Thus, the coefficients in |µi are related to the coefficients in |ii by hµ|ii = hi|µi∗ . Thus, we can
define a transformation matrix Sµi as


hµ|ii hµ|ji hµ|ki


Sµi =  hν|ii hν|ji hν|ki 
hλ|ii hλ|ji hλ|ki
(2.79)
and a set of vectors


hi|ai


ai =  hj|ai 
hk|ai

(2.80)

hµ|ai

aµ =  hν|ai 

hλ|ai
(2.81)
X
(2.82)
Thus, we can see that
aµ =
Sµi ai
i
Now, we can also write
hµ|ii∗ hµ|ji∗ hµ|ki∗


∗
=  hν|ii hν|ji hν|ki∗  = Sµi
∗
∗
∗
hλ|ii hλ|ji hλ|ki

S iµ

(2.83)
Thus,
ai =
X
S iµ aµ
(2.84)
µ
Since hi|µi = hµ|ii∗ , S is the Hermitian conjugate of S. So we write
S = S†
50
(2.85)
(2.86)
S † = (S T )∗
(2.87)
(2.88)
∗
(S † )ij = Sji
(2.89)
a = Sa = SSa = S † Sa
(2.90)
S †S = 1
(2.91)
So in short;
thus
and S is called a unitary transformation matrix.
2.1.2
Operators
A linear operator  maps a vector in space H on to another vector in the same space. We can
write this in a number of ways:
|φi 7→ |χi
Â
(2.92)
|χi = Â|φi
(2.93)
Â(a|φi + b|χi) = aÂ|φi + bÂ|χi
(2.94)
or
Linear operators have the property that
Since superposition is rigidly enforced in quantum mechanics, all QM operators are linear operators.
The Matrix Representation of an operator is obtained by writing
Aij = hi|Â|ji
(2.95)
For example: Say we know the representation of A in the |ii basis, we can then write
|χi = Â|φi =
X
Â|iihi|φi =
i
X
|jihj|χi
(2.96)
j
Thus,
hj|χi =
X
hj|A|iihi|φi
i
We can keep going if we want by continuing to insert 1’s where ever we need.
51
(2.97)
The matrix A is Hermitian if A = A† . If it is Hermitian, then I can always find a basis |µi in
which it is diagonal, i.e.
Aµν = aµ δµν
(2.98)
X
(2.99)
So, what is Â|µi ?
Â|µi =
|iihi|A|jihj|µi
ij
(2.100)
=
X
|iiAij δjµ
(2.101)
ij
(2.102)
=
X
|iiAiµ
(2.103)
i
(2.104)
=
X
|iiaµ δiµ
(2.105)
i
(2.106)
= aµ |µi
(2.107)
An important example of this is the “time-independent” Schroedinger Equation:
Ĥ|ψi = E|ψi
(2.108)
which we spend some time in solving above.
Finally, if Â|φi = |χi then hφ|A† = hχ|.
2.1.3
Products of Operators
An operator product is defined as
(ÂB̂)|ψi = Â[B̂|ψi]
(2.109)
where we operate in order from right to left. We proved that in general the ordering of the
operations is important. In other words, we cannot in general write ÂB̂ = B̂ Â. An example of
this is the position and momentum operators. We have also defined the “commutator”
[Â, B̂] = ÂB̂ − B̂ Â.
(2.110)
Let’s now briefly go over how to perform algebraic manipulations using operators and commutators. These are straightforward to prove
52
1. [Â, B̂] = −[B̂, Â]
2. [Â, Â] = −[Â, Â] = 0
3. [Â, B̂ Ĉ] = [Â, B̂]Ĉ + B̂[Â, Ĉ]
4. [Â, B̂ + Ĉ] = [Â, B̂] + [Â, Ĉ]
5. [Â, [B̂, Ĉ]] + [B̂, [Ĉ, Â]] + [Ĉ, [Â, B̂]] = 0 (Jacobi Identity)
6. [Â, B̂]† = [† , B̂ † ]
2.1.4
Functions Involving Operators
Another property of linear operators is that the inverse operator always can be found. I.e. if
|χi = Â|φi then there exists another operator B̂ such that |φi = B̂|χi. In other words B̂ = Â−1 .
We also need to know how to evaluate functions of operators. Say we have a function, F (z)
which can be expanded as a series
F (z) =
∞
X
fn z n
(2.111)
fn Ân .
(2.112)
n=0
Thus, by analogy
F (Â) =
∞
X
n=0
For example, take exp(Â)
exp(x) =
∞
X
xn
n=0
n!
= 1 + x + x2 /2 + · · ·
(2.113)
thus
exp(Â) =
∞
X
Ân
n=0
n!
(2.114)
If  is Hermitian, then F (Â) is also Hermitian. Also, note that
[Â, F (Â)] = 0.
Likewise, if
Â|φa i = a|φa i
(2.115)
Ân |φa i = an |φa i.
(2.116)
then
53
Thus, we can show that
X
F (Â)|φa i =
fn Ân |φa i
(2.117)
n
(2.118)
=
X
fn an |φa i
(2.119)
n
(2.120)
= F (a)
(2.121)
Note, however, that care must be taken when we evaluate F (Â + B̂) if the two operators
do not commute. We ran into this briefly in breaking up the propagator for the Schroedinger
Equation in the last lecture (Trotter Product). For example,
exp(Â + B̂) 6= exp(Â) exp(B̂)
(2.122)
unless [Â, B̂] = 0. One can derive, however, a useful formula (Glauber)
exp(Â + B̂) = exp(Â) exp(B̂) exp(−[Â, B̂]/2)
(2.123)
Exercise 2.2 Let H be the Hamiltonian of a physical system and |φn i the solution of
H|φn i = En |φn i
(2.124)
1. For an arbitrary operator, Â, show that
hφn |[Â, H]|φn i = 0
(2.125)
1 2
p̂ + V (x̂)
2m
(2.126)
2. Let
H=
(a) Compute [H, p̂], [H, x̂], and [H, x̂p̂].
(b) Show hφn |p̂|φn i = 0.
(c) Establish a relationship between the average of the kinetic energy given by
Ekin = hφn |
p̂2
|φn i
2m
(2.127)
and the average force on a particle given by
F = hφn |x̂
∂V (x)
|φn i.
∂x
(2.128)
Finally, relate the average of the potential for a particle in state |φn i to the average
kinetic energy.
54
Exercise 2.3 Consider the following Hamiltonian for 1d motion with a potential obeying a simple power-law
H=
p2
+ αxn
2m
(2.129)
where α is a constant and n is an integer. Calculate
hAi = hψ|[xp, H]||ψi
(2.130)
and use the result to relate the average potential energy to the average kinetic energy of the
system.
2.2
Constants of the Motion
In a dynamical system (quantum, classical, or otherwise) a constant of the motion is any quantity
such that
∂t A = 0.
(2.131)
[A, H] = 0
(2.132)
For quantum systems, this means that
(What’s the equivalent relation for classical systems?) In other words, any quantity which
commutes with H is a constant of the motion. Furthermore, for any conservative system (in
which there is no net flow of energy to or from the system),
[H, H] = 0.
(2.133)
∂t hAi = ∂t hψ(t)|A|ψ(t)i
(2.134)
From Eq.2.131, we can write that
Since [A, H] = 0, we know that if the state |φn i is an eigenstate of H,
H|φn i = En |φn i
(2.135)
A|φn i = an |φn i
(2.136)
then
The an are often referred to as “good quantum numbers”. What are some constants of the
motion for systems that we have studied thus far? (Bonus: how are constants of motion related
to particular symmetries of the system?)
A state which is in an eigenstate of H it’s also in an eigenstate of A. Thus, I can simultaneously
measure quantities associated with H and A. Also, after I measure with A, the system remains
in a the original state.
55
2.3
Bohr Frequency and Selection Rules
What if I have another operator, B, which does not commute with H? What is hB(t)i? This we
can compute by first writing
|ψ(t)i =
X
cn e−iEn t/h̄ |φn i.
(2.137)
n
Then
hB(t)i = hψ|B|ψ(t)i
=
X
cn c∗m e−i(En −Em )t/h̄ hφm |B|φn i.
(2.138)
(2.139)
n
Let’s define the “Bohr Frequency” as ωnm = (En − Em )/h̄.
hB(t)i =
X
cn c∗m e−iωnm t hφm |B|φn i.
(2.140)
n
Now, the observed expectation value of B oscillates in time at number of frequencies corresponding to the energy differences between the stationary states. The matrix elements Bnm =
hφm |B|φn i do not change with time. Neither do the coefficients, {cn }. Thus, let’s write
B(ω) = cn c∗m hφm |B|φn iδ(ω − ωnm )
(2.141)
and transform the discrete sum into an continuous integral
hB(t)i =
1 Z ∞ −iωt
e
B(ω)
2π 0
(2.142)
where B(ω) is the power spectral of B. In other words, say I monitor < B(t) > with my
instrument for a long period of time, then take the Fourier Transform of the time-series. I get
the power-spectrum. What is the power spectrum for a set of discrete frequencies: If I observe
the time-sequence for an infinite amount of time, I will get a series of discretely spaced sticks
along the frequency axis at precisely the energy difference between the n and m states. The
intensity is related to the probability of making a transition from n to m under the influence
of B. Certainly, some transitions will not be allowed because hφn |B|φm i = 0. These are the
“selection rules”.
We now prove an important result regarding the integrated intensity of a series of transitions:
Exercise 2.4 Prove the Thomas-Reiche-Kuhn sum rule:2
X 2m|xn0 |2
n
h̄2
(En − Eo ) = 1
(2.143)
where the sum is over a compete set of states, |ψn i of energy En of a particle of mass m which
moves in a potential; |ψo i represents a bound state, and xn0 = hψn |x|ψo i. (Hint: use the commutator identity: [x, [x, H]] = h̄2 /m)
2
This is perhaps one of the most important results of quantum mechanics since it is gives the total spectral
intensity for a series of transitions. c.f Bethe and Jackiw for a great description of sum-rules.
56
Figure 2.4: The diffraction function sin(x)/x
1
0.8
0.6
0.4
0.2
-20
-10
10
20
-0.2
2.4
Example using the particle in a box states
What are the constants of motion for a particle in a box?
Recall that the energy levels and wavefunctions for this system are
n2 π 2 h̄2
En =
2ma2
s
2
nπ
φn (x) =
sin( x)
a
a
(2.144)
(2.145)
Say our system in in the nth state. What’s the probability of measuring the momentum and
obtaining a result between p and p + dp?
Pn (p)dp = |φn (p)|2 dp
(2.146)
where
s
1 Za
2
φn (p) = √
dx
sin(nπx/a)e−ipx/h̄
(2.147)
a
2πh̄ 0
#
"
1 1
ei(nπ/a−p/h̄)a − 1 e−i(nπ/a−p/h̄)a − 1
√
=
−
(2.148)
2i 2πh̄ i(nπ/a − p/h̄)
−i(nπ/a − p/h̄)
r
1
a
=
exp i(nπ/2 − pa/(2h̄))[F (p − nπh̄/2) + (−1)n+1 F (p + nπh̄/a)] (2.149)
2i πh̄
Where the F (p) are “diffraction functions”
F (p) =
sin(pa/(2h̄))
pa/(2h̄)
(2.150)
Note that the width 4πh̄/a does not change as I change n. Nor does the amplitude. However,
note that (F (x + n) ± FF (x − n))2 is always an even function of x. Thus, we can say
hpin =
Z
+∞
−∞
Pn (p)pdp = 0
57
(2.151)
We can also compute:
2
2
hp i = h̄
Z
∂φ (x) 2
n
dx ∂x (2.152)
cos(nπx/a)dx
(2.153)
= 2mEn
(2.154)
∞
0
2
= h̄
Z
a
0
=
2 nπ
a a
nπh̄
a
2
!2
Thus, the RMS deviation of the momentum:
∆pn =
q
hp2 in − hpi2n =
nπh̄
a
(2.155)
Thus, as n increases, the relative accuracy at which we can measure p increases due the fact
that we can resolve the wavefunction into two distinct peaks corresponding to the particle either
going to the left or to the right. ∆p increases due to the fact that the two possible choices for
the measurement are becoming farther and farther apart and hence reflects the distance between
the two most likely values.
2.5
Time Evolution of Wave and Observable
Now, suppose we put our system into a superposition of box-states:
1
|ψ(0)i = √ (|φ1 i + |φ2 i)
2
(2.156)
What is the time evolution of this state? We know the eigen-energies, so we can immediately
write:
1
|ψ(t)i = √ (exp(−iE1 t/h̄)|φ1 i + exp(−iE2 t/h̄)|φ2 i)
2
(2.157)
Let’s factor out a common phase factor of e−iE1 t/h̄ and write this as
1
|ψ(t)i ∝ √ (|φ1 i + exp(−i(E2 − E1 )t/h̄)|φ2 i)
2
(2.158)
and call (E2 − E1 )/h̄ = ω21 the Bohr frequency.
1
|ψ(t)i ∝ √ (|φ1 i + exp(−iω21 t)|φ2 i)
2
(2.159)
where
ω21
3π 2 h̄
=
.
2ma2
58
(2.160)
The phase factor is relatively unimportant and cancels out when I make a measurement. Eg.
the prob. density:
|ψ(x, t)|2 = |hx|ψ(t)i|2
1
1
= φ21 (x) + φ22 (x) + φ1 (x)φ2 (x) cos(ω21 t)
2
2
(2.161)
(2.162)
(2.163)
Now, let’s compute hx(t)i for the two state system. To do so, let’s first define x0 = x − a/2
as the center of the well to make the integrals easier. The first two are easy:
0
hφ1 |x |φ1 i ∝
0
hφ2 |x |φ2 i ∝
a
Z
dx(x − a/2) sin2 (πx/a) = 0
(2.164)
dx(x − a/2) sin2 (2πx/a) = 0
(2.165)
0
Z
a
0
which we can do by symmetry. Thus,
hx0 (t)i = Re{e−iω21 t hφ1 |x0 |φ2 i}
hφ1 |x0 |φ2 i = hφ1 |x|φ2 i − (a/2)hφ1 |φ2 i
2Z a
=
dxx sin(πx/a) sin(2πx/a)
a 0
16a
= − 2
9π
(2.166)
(2.167)
(2.168)
(2.169)
Thus,
hx(t)i =
a 16a
−
cos(ω21 t)
2 9π 2
(2.170)
Compare this to the classical trajectory. Also, what about hE(t)i?
2.6
“Unstable States”
So far in this course, we have been talking about systems which are totally isolated from the
rest of the universe. In these systems, there is no influx or efflux of energy and all our dynamics
are governed by the three principle postulates I mentioned a the start of the lecture. IN essence,
if at t = 0 I prepare the system in an eigenstate of H, then for all times later, it’s still in that
state (to within a phase factor). Thus, in a strictly conservative system, a system prepared in
an eigenstate of H will remain in an eigenstate forever.
However, this is not exactly what is observed in nature. We know from experience that atoms
and molecules, if prepared in an excited state (say via the absorption of a photon) can relax
59
back to the ground state or some lower state via the emission of a photon or a series of photons.
Thus, these eigenstates are “unstable”.
What’s wrong here? The problem is not so much to do with what is wrong with our description
of the isolated system, it has to do with full description is not included. A isolated atom or
molecule can still interact with the electro-magnetic field (unless we do some tricky confinement
experiments). Thus, there is always some interaction with an outside “environment”. Thus,
while it is totally correct to describe the evolution of the global system in terms of some global
“atom” + “environment” Hamiltonian, it it NOT totally rigorous to construct a Hamiltonian
which describes only part of the story. But, as the great Prof. Karl Freed (at U. Chicago) once
told me “Too much rigor makes rigor mortis”.
Thankfully, the coupling between an atom and the electromagnetic field is pretty weak. Each
photon emission probability is weighted by the fine-structure constant, α ≈ 1/137. Thus a 2
photon process is weighted by α2 . Thus, the isolated system approximation is pretty good. Also,
we can pretty much say that most photon emission processes occur as single photon events.
Let’s play a bit “fast and loose” with this idea. We know from experience that if we prepare
the system in an excited state at t = 0, the probability of finding it still in the excited state at
some time t later, is
P (t) = e−t/τ
(2.171)
where τ is some time constant which we’ll take as the lifetime of the state. One way to “prove”
this relation is to go back to Problem Set 0. Let’s say we have a large number of identical systems
N , each prepared in the excited state at t = 0. At time t, there are
N (t) = N e−t/τ
(2.172)
systems in the excited state. Between time t and t + dt a certain number, dn(t) will leave the
excited state via photon emission.
dn(t) = N (t) − N (t + dt) = −
dN (t)
dt
dt = N (t)
dt
τ
(2.173)
Thus,
dn(t)
dt
=
N (t)
τ
(2.174)
Thus, 1/τ is the probability per unit time for leaving the unstable state.
The Avg. time a system spends in the unstable state is given by:
1Z∞
dtte−t/τ = τ
τ 0
(2.175)
For a stable state P (t) = 1 thus, τ → ∞.
The time a system spends in the state is independent of its history. This is a characteristic of
an unstable state. (Also has to do with the fact that the various systems involved to not interact
with each other. )
60
Finally, according to the time-energy uncertainty relation:
∆Eτ ≈ h̄.
(2.176)
Thus, an unstable system has an intrinsic “energy width” associated with the finite time the
systems spends in the state.
For a stable state:
|ψ(t)i = e−iEn t/h̄ |φn i
(2.177)
Pn (t) = |e−iEn t/h̄ |2 = 1
(2.178)
and
for real energies.
What if I instead write: En0 = En − ih̄γn /h̄? Then
Pn (t) = |e−iEn t/h̄ e−γn /2t |2 = e−γn t
(2.179)
γn = 1/τn
(2.180)
Thus,
is the “Energy Width” of the unstable state.
The surprising part of all this is that in order to include dissipative effects (photon emission,
etc..) the Eigenvalues of H become complex. In other words, the system now evolves under a
non-hermitian Hamiltonian! Recall the evolution operator for an isolated system:
U (t) = e−iHt/h̄
(2.181)
(2.182)
U † (t) = eiHt/h̄
(2.183)
where the first is the forward evolution of the system and the second corresponds to the backwards evolution of the system. Thus, Unitarity is thus related to the time-reversal symmetry
of conservative systems. The inclusion of an “environment” breaks the intrinsic time-reversal
symmetry of an isolated system.
2.7
Problems and Exercises
Exercise 2.5 Find the eigenvalues and eigenvectors of the matrix:




M =
0
0
0
1
0
0
1
0
61
0
1
0
0
1
0
0
0





(2.184)
Solution: You can either do this the hard way by solving the secular determinant and then
finding the eigenvectors by Gramm-Schmidt orthogonalization, or realize that since M = M −1
and M = M † , M is a unitary matrix, this, its eigenvalues can only be ± 1. Furthermore, since
the trace of M is 0, the sum of the eigenvalues must be 0 as well. Thus, λ = (1, 1, −1, −1) are
the eigenvalues. To get the eigenvectors, consider the following. Let φm u be an eigenvector of
M , thus,




φµ = 
x1
x2
x3
x4



.

(2.185)
Since M φµ = λµ φm u, x1 = λµ x4 and x2 = λµ x3 Thus, 4 eigenvectors are





−1
0
0
1
0
1
−1
0
 
 
 
,
 
 
 
 
,
 
1
0
0
1
 
 
 
,
 
0
1
1
0





(2.186)
for λ = (−1, −1, 1, 1).
Exercise 2.6 Let λi be the eigenvalues of the matrix:


2 −1 −3

2 
H =  −1 1

−3 2
3
(2.187)
calculate the sums:
3
X
λi
(2.188)
λ2i
(2.189)
i
and
3
X
i
Hint: use the fact that the trace of a matrix is invariant to choice representation.
Solution: Using the hint,
trH =
X
λi =
i
X
Hii = 2 + 3 + 1 = 6
(2.190)
i
and
X
i
λ2i =
X
Hij Hji =
ij
X
ij
62
Hij2 = 42
(2.191)
Exercise 2.7
1. Let |φn i be the eigenstates of the Hamiltonian, H of some arbitrary system
which form a discrete, orthonormal basis.
H|φn i = En |φn i.
Define the operator, Unm as
Unm = |φn ihφm |.
†
(a) Calculate the adjoint Unm
of Unm .
(b) Calculate the commutator, [H, Unm ].
†
(c) Prove: Umn Upq
= δnq Ump
(d) For an arbitrary operator, A, prove that
hφn |[A, H]|φn i = 0.
(e) Now consider some arbitrary one dimensional problem for a particle of mass m and
potential V (x). For here on, let
H=
p2
+ V (x).
2m
i. In terms of p, x, and V (x), compute: [H, p], [H, x], and [H, xp].
ii. Show that hφn |p|φn i = 0.
iii. Establish a relation between the average value of the kinetic energy of a state
p2
hT i = hφn |
|φn i
2m
and
dV
|φn i.
dx
The average potential energy in the state φn is
hφn |x
hV i = hφn |V |φn i,
find a relation between hV i and hT i when V (x) = Vo xλ for λ = 2, 4, 6, . . ..
(f ) Show that
hφn |p|φm i = αhφn |x|φm i
where α is some constant which depends upon En − Em . Calculate α, (hint, consider
the commutator [x, H] which you computed above.
(g) Deduce the following sum-rule for the linear -response function.
hφ0 |[x, [H, x]]|φ0 i = 2
X
(En − E0 )|hφ0 |x|φn i|2
n>0
Here |φ0 i is the ground state of the system. Give a physical interpretation of this last
result.
63
Exercise 2.8 For this section, consider the following 5 × 5 matrix:




H=



0
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
0








(2.192)
1. Using Mathematica determine the eigenvalues, λj , and eigenvectors, φn , of H using the
Eigensystem[] command. Determine the eigenvalues only by solving the secular determinant.
|H − Iλ| = 0
Compare the computational effort required to perform both calculations. Note: in entering
H into Mathematica, enter the numbers as real numbers rather than as integers (i.e. 1.0
vs 1 ).
2. Show that the column matrix of the eigenvectors of H,
T = {φ1 , . . . , φ5 },
provides a unitary transformation of H between the original basis and the eigenvector basis.
T † HT = Λ
where Λ is the diagonal matrix of the eigenvalues λj . i.e. Λij = λi δij .
3. Show that the trace of a matrix is invarient to representation.
4. First, without using Mathematica, compute: T r(H 2 ). Now check your result with Mathematica.
64
Chapter 3
Bound States of The Schrödinger
Equation
A #2 pencil and a dream can take you anywhere.
– Joyce A. Myers
Thus far we have introduced a series of postulates and discussed some of their physical implications. We have introduced a powerful notation (Dirac Notation) and have been studying how
we describe dynamics at the atomic and molecular level where h̄ is not a small number, but is
of order unity. We now move to a topic which will serve as the bulk of our course, the study
of stationary systems for various physical systems. We shall start with some general principles,
(most of which we have seen already), and then tackle the following systems in roughly this
order:
1. Harmonic Oscillators: Molecular vibrational spectroscopy, phonons, photons, equilibrium
quantum dynamics.
2. Angular Momentum: Spin systems, molecular rotations.
3. Hydrogen Atom: Hydrogenic Systems, basis for atomic theory
3.1
Introduction to Bound States
Before moving on to these systems, let’s first consider what is meant by a “bound state”. Say
we have a potential well which has an arbitrary shape except that at x = ±a, V (x) = 0 and
remains so in either direction. Also, in the range of −a ≤ x ≤ a, V (x) < 0. The Schrodinger
Equation for the stationary states is:
"
−h̄2 ∂ 2
+ V (x) φn (x) = En φn (x)
2m ∂x2
#
(3.1)
Rather than solve this exactly (which we can not do since we haven’t specified more about
V ) let’s examine the topology of the allowed bound state solutions. As we have done with the
square well cases, let’s cut the x axis into three domains: Domain 1 for x < −a, Domain 2 for
−a ≤ x ≤ a, Domain 3 for x > a. What are the matching conditions that must be met?
65
For Domain 1 we have:
∂2
2mE
φI (x) = − 2 φI (x) ⇒ (+)φI (x)
2
∂x
h̄
(3.2)
∂2
2m(V (x) − E)
φ
(x)
=
φII (x) ⇒ (−)φII (x)
II
∂x2
h̄2
(3.3)
For Domain 2 we have:
For Domain 3 we have:
∂2
2m(E)
φ
(x)
=
−
φIII (x) ⇒ (+)φIII (x)
III
∂x2
h̄2
(3.4)
At the rightmost end of each equation, the (±) indicates the sign of the second derivative of
the wavefunction. (i.e. the curvature must have the same or opposite sign as the function itself.)
For the + curvature functions, the wavefunctions curve away from the x-axis. For − curvature,
the wavefunctions are curved towards the x-axis.
Therefore, we can conclude that for regions outside the well, the solutions behave much like
exponentials and within the well, the behave like superpositions of sine and cosine functions.
Thus, we adopt the asymptotic solution that
φ(x) ≈ exp(+αx)for x < a as x → −∞
(3.5)
φ(x) ≈ exp(−αx)for x > a as x → +∞
(3.6)
and
Finally, in the well region, φ(x) oscillates about the x-axis. We can try to obtain a more complete
solution by combining the solutions that we know. To do so, we must find solutions which are
both continuous functions of x and have continuous first derivatives of x.
Say we pick an arbitrary energy, E, and seek a solution at this energy. Define the righthand
part of the solution to within a multiplicative factor, then the left hand solution is then a
complicated function of the exact potential curve and can be written as
φIII (x) = B(E)e+ρx + B 0 (E)e−ρx
(3.7)
where B(E) and B 0 (E) are both real functions of E and depend upon the potential function.
Since the solutions must be L2 , the only appropriate bound states are those for which B(E) = 0.
Any other value of B(E) leads to diverging solutions.
Thus we make the following observations concerning bound states:
1. They have negative energy.
2. They vanish exponentially outside the potential well and oscillate within.
3. They form a discrete spectrum as the result of the boundary conditions imposed by the
potential.
66
3.2
The Variational Principle
Often the interaction potential is so complicated that an exact solution is not possible. This
is often the case in molecular problems in which the potential energy surface is a complicated
multidimensional function which we know only at a few points connected by some interpolation function. We can, however, make some approximations. The method, we shall use is the
“Variational method”.
3.2.1
Variational Calculus
The basic principle of the variational method lies at the heart of most physical principles. The
idea is that we represent a quantity as a stationary integral
J=
Z
x2
f (y, yx , x)dx
x1
(3.8)
where f (y, yx , x) is some known function which depened upon three variables, which are also
functions of x, y(x), yx = dy/dx, and x itself. The dependency of y on x uis generally unknown.
This means that while we have fixed the end-points of the integral, the path that we actually
take between the endpoints is not known.
Picking different paths leads to different values of J. However ever certain paths will minimize,
maximize, or find the saddle points of J. For most cases of physical interest, its the extrema
that we are interested. Lets say that there is one path, yo (t) which minimizes J (See Fig. 3.1).
If we distort that path slightly, we get another path y(x) which is not too unlike yo (x) and we
will write it as y(x) = yo (x) + η(x) where η(x1 ) = η(x2 ) = 0 so that the two paths meet at the
terminal points. If η(x) differs from 0 only over a small region, we can write the new path as
y(x, α) = yo (x) + αη(x)
and the variation from the minima as
δy = y(x, α) − yo (x, 0) = αη(x).
Since yo is the path which minimizes J, and y(x, α) is some other path, then J is also a
function of α.
Z x2
J(α) =
f (y(x, α), y 0 (x, α), x)dx
x1
and will be minimized when
∂J
∂α
!
=0
α=0
Because J depends upon α, we can examine the α dependence of the integral
∂J
∂α
Since
!
=
α=0
Z
x2
x1
∂f ∂y
∂f ∂y 0
+ 0
dx
∂y ∂α ∂y ∂α
!
∂y
= η(x)
∂α
67
and
∂y 0
∂η
=
∂α
∂x
we have
∂J
∂α
!
=
Z
x2
x1
α=0
!
∂f
∂f ∂η
η(x) + 0
dx.
∂y
∂y ∂x
Now, we need to integrate the second term by parts to get η as a common factor. Remember
integration by parts?
Z
Z
udv = vu − vdu
From this
Z
x2
x1
x
!
∂f ∂η
∂f 2 Z x2
d ∂f
dx
=
η(x)
−
η(x)
dx
0
∂y ∂x
∂x x1
dx ∂y 0
x1
The boundaty term vanishes since η vanishes at the end points. So putting it all together and
setting it equal to zero:
!
Z x2
∂f
d ∂f
η(x)
−
dx. = 0
∂y dx ∂y 0
x1
We’re not done yet, since we still have to evaluate this. Notice that α has disappeared from
the expression. In effect, we can take an arbitrary variation and still find the desired path tha
minimizes J. Since η(x) is arbitrary subject to the boundary conditions, we can make it have the
same sign as the remaining part of the integrand so that the integrand is always non-negative.
Thus, the only way for the integral to vanish is if the bracketed term is zero everywhere.
∂f
d ∂f
−
∂y dx ∂y 0
!
=0
(3.9)
This is known as the Euler equation and it has an enormous number of applications. Perhaps
the simplest is the proof that the shortest distance between two points is a straight line (or
on a curved space, a geodesic). The straightline distance between
q two points on the
q xy plane
√ 2
2
2
2
is s = x + y and the differential element of distance is ds = (dx) + (dy) = 1 + yx2 dx.
Thus, we can write a distance along some line in the xy plane as
J=
Z
x2 y2
x1 y1
ds =
Z
x2 y2
x1 y1
q
1 + yx2 dx.
If we knew y(x) then J would be the arclength or path-length along the function y(x) between
two points. Sort of like, how many steps you would take along a trail between two points. The
trail may be curvy or straight and there is certainly a single trail which is the shortest. So,
setting
q
f (y, yx , x) = 1 + yx2
and substituting it into the Euler equation one gets
d ∂f
d
1
= − q
dx ∂yx
dx 1 + yx2
= 0.
68
(3.10)
So, the only way for this to be true is if
1
q
1 + yx2
= constant.
(3.11)
Solving for yx produces a second constant: yx = a, which immediatly yields that y(x) = ax + b.
In other words, its a straight line! Not too surprising.
An important application of this principle is when the integrand f is the classical Lagrangian
for a mechanical system. The Lagrangian is related to Hamiltonian and is defined as the difference
between the kinetic and potential energy.
L=T −V
(3.12)
where as H is the sum of T + V . Rather than taking x as the independent variable, we take time,
t, and position and velocity oa a particle as the dependent variables. The statement of δJ = 0
is a mathematical statement of Hamilton’s principle of least action
δ
Z
t2
L(x, ẋ, t)dt = 0.
(3.13)
t1
In essence, Hamilton’s principle asserts that the motion of the system from one point to another
is along a path which minimizes the integral of the Lagrangian. The equations of motion for that
path come from the Euler-Lagrange equations,
d ∂L ∂L
−
= 0.
dt ∂ ẋ
∂x
(3.14)
1
L = mẋ2 − V (x)
2
(3.15)
So if we write the Lagrangian as
and substitute this into the Euler-Lagarange equation, we get
mẍ = −
∂V
∂x
(3.16)
which is Newton’s law of motion: F = ma.
3.2.2
Constraints and Lagrange Multipliers
Before we can apply this principle to a quantum mechanical problem, we need to ask our selves
what happens if there is a constraint on the system which exclues certain values or paths so that
not all the η’s may be varied arbitrarily? Typically, we can write the constraint as
φi (y, x) = 0
(3.17)
For example, for a bead on a wire we need to constrain the path to always lie on the wire or for
a pendulum, the path must lie on in a hemisphere defined by the length of the pendulum from
69
the pivot point. In any case, the general proceedure is to introduce another function, λi (x) and
integrate
x2
Z
x1
λi (x)φi (y, x)dx = 0
(3.18)
λi (x)φi (y, x)dx = 0
(3.19)
so that
δ
Z
x2
x1
as well. In fact it turns out that the λi (x) can be even be taken to be a constant, λi for this
whole proceedure to work.
Regardless of the case, we can always write the new stationary integral as
δ
Z
(f (y, yx , x) +
X
λi φi (y, x))dx = 0.
(3.20)
i
The multiplying constants are called Lagrange Mulipliers. In your statistical mechanics course,
these will occur when you minimize various thermodynamic functions subject to the various
extensive constraints, such as total number of particles in the system, the average energy or
temperature, and so on.
In a sence, we have redefined the original function or Lagrangian to incorporate the constraint
into the dynamics. So, in the presence of a constraint, the Euler-Lagrange equations become
d ∂L ∂L X ∂φi
−
=
λi
dt ∂ ẋ
∂x
i ∂x
(3.21)
where the term on the right hand side of the equation represents a force due to the constraint.
The next issue is that we still need to be able to determine the λi Lagrange multipliers.
70
Figure 3.1: Variational paths between endpoints. The thick line is the stationary path, yo (x)
and the dashed blue curves are variations y(x, α) = yo (x) + αη(x).
fHxL
2
1
-1.5
-1
-0.5
0.5
1
-1
-2
71
1.5
x
3.2.3
Variational method applied to Schrödinger equation
The goal of all this is to develop a procedure for computing the ground state of some quantum
mechanical system. What this means is that we want to minimize the energy of the system
with respect to arbitrary variations in the state function subject to the constraint that the state
function is normalized (i.e. number of particles remains fixed). This means we want to construct
the variation:
δhψ|H|ψi = 0
(3.22)
with the constraint hψ|ψi = 0.
In the coordinate representation, the integral involves taking the expectation value of the
kinetic energy operator...which is a second derivative operator. That form is not too convenient
for our purposes since it will allow us to write Eq.3.22 in a form suitable for the Euler-Lagrange
equations. But, we can integrate by parts!
Z
∗∂
2
Z
ψ
∗ ∂ψ ψ
dx = ψ
−
∂x2
∂x ∂ψ ∗
∂x
!
!
∂ψ
dx
∂x
(3.23)
Assuming that the wavefunction vanishes at the limits of the integration, the surface term vanishes leaving only the second term. We can now write the energy expectation value in terms
of two dependent variables, ∇ψ and ψ. OK, they’re functions, but we can still treat them as
dependent variables just like we treated the y(x)0 s above.
E=
Z
h̄2
(∇ψ ∗ )(∇ψ) + V ψ ∗ ψ dx
2m
!
(3.24)
Adding on the constraint and defining the Lagrangian as
h̄2
(∇ψ ∗ )(∇ψ) + V ψ ∗ ψ − λψ ∗ ψ,
2m
!
L=
(3.25)
we can substitute this into the Euler-Lagrange equations
∂L
∂L
− ∇∂x
= 0.
∗
∂ψ
∂(∇ψ ∗
(3.26)
h̄2 2
∇ ψ,
(V − λ)ψ =
2m
(3.27)
This produces the result
which we immediately recognize as the Schrödinger equation.
While this may be a rather academic result, it gives us the key to recognize that we can
make an expansion of ψ in an arbitrary basis and take variations with respect to the coeffients
of that basis to find the lowest energy state. This is the basis of a number of powerful numerical
methods used solve the Schrödinger equation for extremely complicated systems.
72
3.2.4
Variational theorems: Rayleigh-Ritz Technique
We now discuss two important theorems:
Theorem 3.1 The expectation value of the Hamiltonian is stationary in the neighborhood of its
eigenstates.
To demonstrate this, let |ψi be a state in which we compute the expectation value of H. Also,
let’s modify the state just a bit and write
|ψi → |ψi + |δψi.
(3.28)
Expectation values are computed as
hHi =
hψ|H|ψi
hψ|ψi
(3.29)
(where we assume arb. normalization). In other words
hψ|ψihHi = hψ|H|ψi
(3.30)
hψ|ψiδhHi + hδψ|ψihHi + hψ|δψihHi = hδψ|H|ψi + hψ|H|δψi
(3.31)
hψ|ψiδhHi = hδψ|H − hHi|ψi − hψ|H − hHi|δψi
(3.32)
Now, insert the variation
or
If the expectation value is to be stationary, then δhHi = 0. Thus the RHS must vanish for an
arbitrary variation in the wavefunction. Let’s pick
|δψi = |ψi.
(3.33)
(H − hHi)|ψi = 0
(3.34)
Thus,
That is to say that |ψi is an eigenstate of H. Thus proving the theorem.
The second theorem goes:
Theorem 3.2 The Expectation value of the Hamiltonian in an arbitrary state is greater than or
equal to the ground-state energy.
The proof goes as this: Assume that H has a discrete spectrum of states (which we demonstrated
that it must) such that
H|ni = En |ni
73
(3.35)
Thus, we can expand any state |ψi as
|ψi =
X
cn |ni.
(3.36)
n
Consequently
X
hψ|ψi =
|cn |2 ,
(3.37)
|cn |2 En .
(3.38)
n
and
X
hψ|H|ψi =
n
Thus, (assuming that |ψi is normalized)
hHi =
X
En |cn |2 ≥
X
Eo |cn |2 ≥ Eo
(3.39)
n
n
quid est demonstrato.
Using these two theorems, we can estimate the ground state energy and wavefunctions for a
variery of systems. Let’s first look at the Harmonic Oscillator.
Exercise 3.1 Use the variational principle to estimate the ground-state energy of a particle in
the potential
(
V (x) =
Cx x > 0
+∞ x ≤ 0
(3.40)
Use xe−ax as the trial function.
3.2.5
Variational solution of harmonic oscillator ground State
The Schrödinger Equation for the Harmonic Osc. (HO) is
"
h̄2 ∂ 2
k2 2
−
+
x φ(x) − Eφ(x) = 0
2m ∂x2
2
#
(3.41)
Take as a trial function,
φ(x) = exp(−λx2 )
(3.42)
where λ is a positive, non-zero constant to be determined. The variational principle states that
the energy reaches a minimum
∂hHi
= 0.
∂λ
(3.43)
when φ(x) is the ground state solution. Let us first derive hHi(λ).
hHi(λ) =
hφ|H|φi
hφ|φi
74
(3.44)
To evaluate this, we break the problem into a series of integrals:
hφ|φi =
∞
Z
dx|φ(x)|2 =
r
−∞
hφ|p2 |φi =
Z
∞
π
2λ
(3.45)
dxφ00 (x)φ(x) = −2λhφ|φi + 4λ2 hφ|x2 |φi
(3.46)
−∞
and
< φ|x2 |φi =
Z
∞
1
hφ|φi.
4λ
(3.47)
1
k 1
+
4λ
2 4λ
(3.48)
dxx2 |φ(x)|2 =
−∞
Putting it all together:
hφ|H|φi
h̄2
= −
hφ|φi
2m
!
−2λ + 4λ2
h̄2
k
λ+
2m
8λ
!
hφ|H|φi
=
hφ|φi
(3.49)
Taking the derivative with respect to λ:
∂hHi
h̄2
k
=
− 2 =0
∂λ
2m 8λ
(3.50)
Thus,
√
mk
2h̄
λ=±
(3.51)
Since only positive values of λ are allowed.
√
λ=
mk
2h̄
(3.52)
Using this we can calculate the ground state energy by substituting λ back into hHi(λ).
√
h̄2
k 4h̄2
+
2m 8 mk
mk
hHi(λ) =
2h̄
Now, define the angular frequency: ω =
!
h̄
=
2
s
k
m
(3.53)
q
k/m.
hHi(λ) =
h̄
ω
2
which ( as we can easily prove) is the ground state energy of the harmonic oscillator.
Furthermore, we can write the HO ground state wavefunction as
75
(3.54)
1
φo (x) = q
hφ|φi
φo (x) =
2λ
π
!1/4
√
φo (x) =
mk
h̄π
φ(x)
(3.55)
√
mk 2
exp −
x
2h̄
!1/4
√
!
mk 2
exp −
x
2h̄
(3.56)
!
(3.57)
To compute the “error” in our estimate, let’s substitute the variational solution back into the
Schrodinger equation:
"
h̄2 ∂ 2
k2 2
h̄2 00
k2
−
+
x
φ
(x)
=
−
φ
(x)
+
φo (x)
o
2m ∂x2
2
2m o
2
#
h̄2 00
k2
h̄2
−
φ (x) + φo (x) = −
2m o
2
2m
√ !
kmx2 − h̄ km
k
φo (x) + x2 φo (x)
2
2
h̄
h̄2 00
k2
h̄
−
φo (x) + φo (x) =
2m
2
2
s
k
φo (x)
m
(3.58)
(3.59)
(3.60)
Thus, φo (x) is in fact the correct ground state wavefunction for this system. If it were not
the correct function, we could re-introduce the solution as a new trial function, re-compute the
energy, etc... and iterate through until we either find a solution, or run out of patience! (Usually
it’s the latter than the former.)
3.3
The Harmonic Oscillator
Now that we have the HO ground state and the HO ground state energy, let us derive the whole
HO energy spectrum. To do so, we introduce “dimensionless” quantities: X and P related to
the physical position and momentum by
mω 1/2
) x
(3.61)
X=(
2h̄
P =(
1
)1/2 p
2h̄mω
(3.62)
This will save us from carrying around a bunch of coefficients. In these units, the HO Hamiltonian
is
H = h̄ω(P 2 + X 2 ).
76
(3.63)
The X and P obey the canonical comutational relation:
[X, P ] =
1
i
[x, p] =
2h̄
2
(3.64)
We can also write the following:
(X + iP )(X − iP ) = X 2 + P 2 + 1/2
(3.65)
(X − iP )(X + iP ) = X 2 + P 2 − 1/2.
(3.66)
Thus, I can construct the commutator:
[(X + iP ), (X − iP )] = (X + iP )(X − iP ) − (X − iP )(X + iP )
= 1/2 + 1/2
= 1
(3.67)
Let’s define the following two operators:
a = (X + iP )
a† = (X + iP )† = (X − iP ).
(3.68)
(3.69)
Therefore, the a and a† commute as
[a, a† ] = 1
(3.70)
Let’s write H in terms of the a and a† operators:
H = h̄ω(X 2 + P 2 ) = h̄ω(X − iP )(X + iP ) + h̄ω/2
(3.71)
or in terms of the a and a† operators:
H = h̄ω(a† a + 1/2)
(3.72)
Now, consider that |φn i is the nth eigenstate of H. Thus, we write:
h̄ω(a† a + 1/2)|φn i = En |φn i
(3.73)
What happens when I multiply the whole equation by a? Thus, we write:
ah̄ω(a† a + 1/2)|φn i = aEn |φn i
h̄ω(aa† + 1/2)(a|φn i) = En (a|φn i)
(3.74)
(3.75)
Now, since aa† − a† a = 1,
h̄ω(a† a + 1/2 − 1)(a|φn i) = En (a|φn i)
In other words, a|φn i is an eigenstate of H with energy E = En − h̄ω.
77
(3.76)
What happends if I do the same procedure, this time using a† ? Thus, we write:
a† h̄ω(a† a + 1/2)|φn i = a† En |φn i
(3.77)
[a, a† ] = aa† − a† a
(3.78)
a† a = aa† − 1
(3.79)
Since
we have
we can write
a† a† a = a† (aa† − 1)
= (a† a − 1)a† .
(3.80)
(3.81)
Thus,
a† h̄ω(a† a + 1/2)|φn i = h̄ω((a† a − 1 + 1/2)a† )|φn i
(3.82)
h̄ω(a† a − 1/2)(a† |φn i) = En (a† |φn i).
(3.83)
or,
Thus, a† |φn i is an eigenstate of H with energy E = En + h̄ω.
Since a† and a act on harmonic oscillator eigenstates to give eigenstates with one more or one
less h̄ω quanta of energy, these are termed “creation” and “annihilation” operators since they
act to create additional quata of excitation or decrease the number of quanta of excitation in
the system. Using these operators, we can effectively “ladder” our way up the energy scale and
determine any eigenstate one we know just one.
Well, we know the ground state solution. That we got via the variational calculation. What
happens when I apply a† to the φo (x) we derived above? In coordinate form:
(X − iP ) φo (x) =
mω
2h̄
1/2
mω
2h̄
1/2
1
x+
2mωh̄
1/2
∂
φo (x)
∂x
1
2mωh̄
1/2
!
!
(3.84)
=
x+
∂
∂x
mω
h̄π
1/4
2 mω
e(−x 2h̄ )
(3.85)
X acting on φo is:
mω
xφo (x)
2h̄
(3.86)
1
∂
φo (x)
mω2h̄ ∂x
(3.87)
r
Xφo (x) =
iP acting on φo is
s
iP φo (x) = −h̄
78
mω
φo (x)
2h̄
= −Xφo (x)
r
iP φo (x) = −x
(3.88)
(3.89)
After cleaning things up:
(X − iP ) φo (x) = 2Xφo (x)
r
mω
= 2
xφo (x)
2h̄
= 2Xφo (x)
r
mω
mω 1/4
2 mω
= 2
x
exp −x
2h̄
2h̄
2h̄
3.3.1
(3.90)
(3.91)
(3.92)
(3.93)
Harmonic Oscillators and Nuclear Vibrations
We introduced one of the most important applications of quantum mechanics...the solution of
the Schrödinger equation for harmonic systems. These are systems in which the amplitude of
motion is either small enough so that the physical potential energy operator can be expanded
about its minimum to second order in the displacement from the minima. When we do so, the
Hamiltonian can be written in the form
H = h̄ω(P 2 + X 2 )
(3.94)
where P and X are dimensionless operators related to the physical momentum and position
operators via
mω
x
2h̄
(3.95)
1
p.
2h̄mω
(3.96)
r
X=
and
s
P =
We also used the variational method to deduce the ground state wavefunction and demonstrated
that the spectrum of H is a series of levels separated by h̄ω and that the ground-state energy is
h̄ω/2 above the energy minimum of the potential.
We also defined a new set of operators by taking linear combinations of the X and P .
a = X + iP
(3.97)
a† = X − iP.
(3.98)
We also showed that the commutation relation for these operators is
[a, a† ] = 1.
(3.99)
These operators are non-hermitian operators, and hence, do not correspond to a physical observable. However, we demonstrated that when a acts on a eigenstate of H, it produces another
79
eigenstate withe energy En − h̄ω. Also, a† acting on an eigenstate of H produces another eigenstate with energy En + h̄ω. Thus,we called a the destruction or annihilation operator since it
removes a quanta of excitation from the system and a† the creation operator since it adds a
quanta of excitation to the system. We also wrote H using these operators as
1
H = h̄ω(a† a + )
(3.100)
2
Finally, ω is the angular frequency of the classical harmonic motion, as obtained via Hooke’s
law:
k
ẍ = − x.
(3.101)
m
Solving this produces
x(t) = xo sin(ωt + φ)
(3.102)
p(t) = po cos(ωt + φ).
(3.103)
and
Thus, the classical motion in the x, p phase space traces out the circumference of a circle every
1/ω regardless of the initial amplitude.
The great advantage of using the a, and a† operators is that they we can replace a differential
equation with an algebraic equation. Furthermore, since we can represent any Hermitian operator
acting on the HO states as a combination of the creation/annihilation operators, we can replace
a potentially complicated series of differentiations, integrations, etc... with simple algebraic
manipulations. We just have to remember a few simple rules regarding the commutation of the
two operators. Two operators which we may want to construct are:
• position operator:
2h̄
mω
• momentum operator: i
1/2
(a† + a)
h̄mω
2
1/2
(a† − a).
Another important operator is
N = a† a.
(3.104)
H = h̄ω(N + 1/2).
(3.105)
and
Since [H, N ] = 0, eigenvalues of N are “good quantum numbers” and N is a constant of the
motion. Also, since
H|φn i = En |φn i = h̄ω(N + 1/2)|φn i
(3.106)
N |φn i = n|φn i,
(3.107)
then if
then n must be an integer n = 0, 1, 2, · · · corresponding to the number of quanta of excitation in
the state. This gets the name “Number Operator”.
Some useful relations (that you should prove )
80
1. [N, a] = [a† a, a] = −a
2. [N, a† ] = [a† a, a† ] = a†
To summarize, we have the following relations using the a and a† operators:
√
1. a|φn i = n|φn−1 i
√
2. a† |φn i = n + 1|φn+1 i
√
3. hφn |a = n + 1hφn+1 = (a|φn i)†
√
4. hφn |a† = n + 1hφn−1 |
5. N |φn i = n|φn i
6. hφn |N = nhφn |
Using the second of these relations we can write
|φn+1 i = √
a†
|φn i
n+1
(3.108)
which can be iterated back to the ground state to produce
(a† )n
|φn i = √ |φo i
n!
(3.109)
This is the “generating relation” for the eigenstates.
Now, let’s look at x and p acting on |φn i.
s
x|φn i =
s
=
h̄
(a† + a)|φn i
2mω
√
h̄ √
( n + 1|φn+1 i + n|φn−1 i)
2mω
(3.110)
(3.111)
Also,
s
p|φn i = i
s
=i
mh̄ω †
(a − a)|φn i
2
√
mh̄ω √
( n + 1|φn+1 i − n|φn−1 i)
2
(3.112)
(3.113)
Thus,the matrix elements of x and p in the HO basis are:
s
hφm |x|φn i =
√
h̄ √
n + 1δm,n+1 + nδm,n−1
2mω
81
(3.114)
s
hφm |p|φn i = i
√
mωh̄ √
n + 1δm,n+1 − nδm,n−1
2
(3.115)
The harmonic oscillator wavefunctions can be obtained by solving the equation:
hx|a|φo i = (X + iP )φo (x) = 0
(3.116)
!
mω
∂
x+
φo (x) = 0
h̄
∂x
(3.117)
The solution of this first order differential equation is easy:
φo (x) = c exp(−
mω 2
x)
2
(3.118)
where c is a constant of integration which we can obtain via normalization:
Z
dx|φo (x)|2 = 1
(3.119)
Doing the integration produces:
φo (x) =
mω
h̄π
1/4
82
mω
2
e− 2h̄ x
(3.120)
Figure 3.2: Hermite Polynomials, Hn up to n = 3.
Hn HxL
10
7.5
5
2.5
-3
-2
-1
1
2
-2.5
-5
-7.5
-10
83
3
x
Since we know that a† acting on |φo i gives the next eigenstate, we can write
!
∂
mω
x−
φo (x)
h̄
∂x
φ1 (x) =
(3.121)
Finally, using the generating relation, we can write
1
φn (x) = √
n!
mω
∂
x−
h̄
∂x
!n
φo (x).
(3.122)
Lastly, we have the “recursion relations” which generates the next solution one step higher or
lower in energy given any other solution.
!
∂
mω
x−
φn (x)
h̄
∂x
1
φn+1 (x) = √
n+1
(3.123)
and
1
φn−1 (x) = √
n
!
mω
∂
x+
φn (x).
h̄
∂x
(3.124)
These are the recursion relationships for a class of polynomials called Hermite polynomials, after
the 19th French mathematician who studied such functions. These are also termed “GaussHermite” and form a set of orthogonal polynomials. The first few Hermite Polynomials, Hn (x)
are {1, 2 x, −2 + 4 x2 , −12 x + 8 x3 , 12 − 48 x2 + 16 x4 } for n = 0 to 4. Some of these are plotted
in Fig. 3.2
The functions themselves are defined by the generating function
2 +2tx
g(x, t) = e−t
=
∞
X
Hn (x)
n=0
tn
.
n!
(3.125)
Differentiating the generating function n times and setting t = 0 produces the nth Hermite
polynomial
n
dn
2 d
2
Hn (x) = n g(x, t) = (−1)n ex
e−x
n
dt
dx
(3.126)
Another useful relation is the Fourier transform relation:
1 Z ∞ itx −x2 /2
2
√
e e
Hn (x)dx = −in e−t /2 Hn (t)
2π −∞
(3.127)
which is useful in generating the momentum space representation of the harmonic oscillator
functions. Also, from the generating function, we can arrive at the recurrence relation:
Hn+1 = 2xHn − 2nHn−1
(3.128)
Hn0 (x) = 2nHn−1 (x).
(3.129)
and
84
Consequently, the hermite polynomials are solutions of the second-order differental equation:
Hn00 − 2xHn0 + 2nHn = 0
(3.130)
which is not self-adjoint! To put this into self-adjoint form, we multiply by the weighting function
2
w = e−x , which leads to the orthogonality integral
Z
∞
2
−∞
Hn (x)Hm (x)e−x dx = δnm .
(3.131)
For the harmonic oscillator functions, we absorb the weighting function into the wavefunction
itself
2
ψn (x) = e−x /2 Hn (x).
When we substitute this function into the differential equation for Hn we get
ψn00 + (2n + 1 − x2 )ψn = 0.
(3.132)
To normalize the functions, we first multipy g by itself and then multiply by w
2
e−x e−s
2 +2sx
2 +2tx
e−t
=
X
2
e−x Hn (x)Hm (x)
mn
sm tn
n!m!
(3.133)
When we integrate over −∞ to ∞ the cross terms drop out by orthogonality and we are left
with
X (st)n Z
n=0
n!n!
∞
−∞
2
e−x Hn2 (x)dx =
=
Z
∞
−∞
Z ∞
e−x
2 −s2 +2sx−t2 +2xt
dx
2
e−(x−s−t) dx
−∞
= π 1/2 e2st =
X 2n (st)n
n=0
n!
.
(3.134)
Equating like powers of st we obtain,
Z
∞
−∞
2
e−x Hn2 (x)dx = 2n π 1/2 n!.
(3.135)
When we apply this technology to the SHO, the solutions are
2
ψn (z) = 2−n/2 π −1/4 (n!)−1/2 e−z Hn (z)
where z = αx and
α2 =
(3.136)
mω
.
h̄
A few gratuitous solutions:
"
4
φ1 (x) =
π
mω
φ2 (x) =
4πh̄
mω
h̄
1/4 3 #1/4
1
x exp(− mωx2 )
2
mω 2
1
2
x − 1 exp(− mωx2 )
h̄
2
(3.137)
Fig. 3.3 shows the first 4 of these functions.
85
(3.138)
3.3.2
Classical interpretation.
In Fig. 3.3 are a few of the lowest energy states for the harmonic oscillator. Notice that as the
quantum number increases the amplitude of the wavefunction is pushed more and more towards
larger values of ±x. This becomes more pronounced when we look at the actual probability
distribution functions, |ψn (x)|2 | for the same 4 states as shown in Fig 3.4.
Here, in blue are the actual quantum distributions for the ground-state through n = 3. In
gray are the classical probability distrubutions for the corresponding energies. The gray curves
tell us the probabilty per unit length of finding the classical particle at some point x and any
point in time. This is inversely proportional to how long a particle spends at a given point...i.e.
Pc (x) ∝ 1/v(x). Since E = mv 2 /2 + V (x),
v(x) =
q
and
2(E − V (x))/m
s
P (x) ∝
m
2(E − V (x))
For the Harmonic Oscillator:
s
Pn (x) ∝
m
.
2(h̄ω(n + 1/2) − kx2 /2)
Notice that the denominator goes to zero at the classical turning points, in other words,
the particle comes to a dead-stop at the turning point and consequently we have the greatest
likelyhood of finding the particle in these regions. Likewise in the quantum case, as we increase
the quantum number, the quantum distrubution function becomes more and more like its classical
counterpart. This is shown in the last four frames of Fig. 3.4 where we have the same plots as in
Fig. 3.4, except we look at much higher quantum numbers. For the last case, where n = 19 the
classical and quantum distributions are nearly identical. This is an example of the correspondence
principle. As the quantum number increases, we expect the quantum system to look more and
more like its classical counter part.
86
Figure 3.3: Harmonic oscillator functions for n = 0 to 3
1
1
0.8
0.5
0.6
0.4
-4
-2
2
4
2
4
-0.5
0.2
-1
-4
-2
2
4
2
4
1
-4
-2
2
2
4
-4
-2
-2
-1
-4
-2
87
Figure 3.4: Quantum and Classical Probability Distribution Functions for Harmonic Oscillator
for n = 0, 1, 2, 3, 4, 5, 9, 14, 19
2.5
1.75
1.5
1.25
1
0.75
0.5
0.25
2
1.5
1
0.5
-4
-2
2
4
-4
-2
1
0.75
0.5
0.25
-2
2
4
0.6
0.4
0.2
2.5
5
7.5
-2
-7.5 -5 -2.5
2
4
2.5
5
7.5
2.5
5
7.5
0.6
0.5
0.4
0.3
0.2
0.1
0.6
0.5
0.4
0.3
0.2
0.1
-7.5 -5 -2.5
-4
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.8
-7.5 -5 -2.5
4
1.4
1.2
1
0.8
0.6
0.4
0.2
1.5
1.25
-4
2
2.5
5
7.5
-7.5 -5 -2.5
88
3.3.3
Molecular Vibrations
The fully quantum mechanical treatment of both the electronic and nuclear dynamics of even a
diatomic molecule is a complicated affair. The reason for this is that we are forced to find the
stationary states for a potentially large number of particles–all of which interact, constrained
by a number of symmetry relations (such as the fact that no two electrons can be in the same
state at the same time.) In general, the exact solution of a many-body problem (such as this)
is impossible. (In fact, believe it it is rigorously impossible for even three classically interacting
particles..although many have tried. ) However, the mass of the electron is on the order of 103 to
104 times smaller than the mass of a typical nuclei. Thus, the typical velocities of the electrons
is much larger than the typical nuclear velocities. We can then assume that the electronic cloud
surrounding the nuclei will respond instantaneously to small and slow changes to the nuclear
positions. Thus, to a very good approximation, we can separate the nuclear motion from the
electonic motion. This separation of the nuclear and electronic motion is called the BornOppenheimer Approximation or the Adiabatic Approximation. This approximation is
one of the MOST important concepts in chemical physics and is covered in more detail in Section
7.4.1.
Fundimental notion is that the nuclear motion of a molecule occurs in the average field of
the electrons. In other words, the electronic charge distribution acts as an extremely complex
multi-dimensional potential energy surface which governs the motion and dynamics of the atoms
in a molecule. Consequently, since chemistry is the science of chemical structure, changes, and
dynamics, nearly all chemical reactions can be described in terms of nuclear motion on one (or
more) potential energy surface. In Fig. ?? is the London-Eyring-Polanyi-Sato (LEPS) [1]surface
for the F + H2 → HF + H reaction using the Mukerman V set of parameters.[3] The LEPS
surface is an empirical potential energy surface based upon the London-Heitler valance bond
theory. Highly accurate potential functions are typically obtained by performing high level ab
initio electronic structure calculations sampling over numerous configurations of the molecule.[2]
For diatomic molecules, the nuclear stretching potential can be approximated as a Morse
potential curve
V (r) = De (1 − e−α(r−req )2 − De
(3.139)
where De is the dissociation energy, α sets the range of the potential, and req is the equilibrium
bond length. The Morse potential for HF is shown in Fig. 3.6 and is parameterized by De =
−1
591.1kcal/mol, α = 2.2189Å , and req = 0.917Å.
Close to the very bottom of the potential well, where r − re is small, the potential is nearly
harmonic and we can replace the nuclear SE with the HO equation by simply writing that the
angular frequancy is
s
V 00 (re )
(3.140)
m
So, measuring the vibrational spectrum of the well will give us the curvature of the well since
(En − Em )/h̄ is always an integer multiple of ω for harmonic systems. The red curve in Fig. 3.6
is a parabolic approximation for the bottom of the well.
ω=
V (r) = De (−1 + α2 (r − re )2 /2 + . . .)
89
(3.141)
Figure 3.5: London-Eyring-Polanyi-Sato (LEPS) empirical potential for the F + H2 → F H + H
chemical reaction
rHH
4
3.5
3
2.5
2
1.5
1
0.5
0.5
1
1.5
90
2
2.5
3
3.5
4
rFH
Figure 3.6: Morse well and harmonic approximation for HF
V HkCalêmolL
600
400
200
rFH
1
2
3
4
-200
-400
-600
q
Clearly, De α2 is the force constant. So the harmonic frequency for the well is ω = k/µ, where
µ is the reduced mass, µ = m1 m2 /(m1 + m2 ) and one would expect that the vibrational energy
levels would be evenly spaced according to a harmonic progression. Deviations from this are due
to anharmonic effects introduced by the inclusion of higher order terms in the Taylor expansion
of the well. As one might expect, the harmonic expansion provides a descent estimate of the
potential energy surface close to the equilibrium geometry.
3.4
3.4.1
Numerical Solution of the Schrödinger Equation
Numerov Method
Clearly, finding bound state soutions for the Schrödinger equation is an important task. Unfortunately, we can only solve a few systems exactly. For the vast majority of system which
we cannot handle exactly, we need to turn to approximate means to finde solutions. In later
chapters, we will examine variational methods and perturbative methods. Here, we will look at
a very simple scheme based upon propagating a trial solution at a given point. This methods is
called the ”Numerov” approach after the Russian astronomer who developed the method. It can
be used to solve any differential equation of the form:
f 00 (r) = f (r)u(r)
(3.142)
where u(r) is simply a function of r and f 00 is the second derivative of f (r) with respect to r and
which is the solution we are looking for. For the Schrödinger equation we would write:
ψ 00 =
2m
(V (r) − E)ψ
h̄2
91
(3.143)
Figure 3.7: Model potential for proton tunneling.
V Hcm-1 L
6000
4000
2000
-1
-0.5
0.5
1
x HbohrL
-2000
-4000
The basic proceedure is to expand the second derivative as a finite difference so that if we know
the solution at one point, xn , we can get the solution at some small point a distance h away,
xn+1 = xn + h.
h2 00
h3
h4
f [n] + f 000 [n] + f (4) [n] . . .
2!
3!
4!
2
3
h
h
h4
f [n − 1] = f [n] − hf 0 [n] + f 00 [n] − f 000 [n] + f (4) [n]
2!
3!
4!
f [n + 1] = f [n] + hf 0 [n] +
(3.144)
(3.145)
If we combine these two equations and solve for f [n + 1] we get a result:
f [n + 1] = −f [n − 1] + 2f [n] + f 00 [n]h2 +
h4 (4)
f [n] + O[h6 ]
12
(3.146)
Since f is a solution to the Schrödinger equation, f 00 = Gf where
G=
2m
(V − E)
h̄2
we can get the second derivative of f very easily. However, for the higher order terms, we have
to work a bit harder. So let’s expand f 00
f 00 [n + 1] = −2f 00 [n] − f 00 [n − 1] + s2 f (4) [n]
and truncate at order h6 . Now, solving for f (4) [n] and substituting f 00 = Gf we get
f [n + 1] =
h2
(G[n − 1]f [n −
12
2
1 − h12 G[n + 1]
2f [n] − f [n − 1] +
1] + 10G[n]f [n])
(3.147)
which is the working equation.
Here we take a case of proton tunneling in a double well potential. The potential in this case
is the V (x) = α(x4 − x2 ) function shown in Fig. 3.7. Here we have taken the parameter α = 0.1
92
3
1
2
0.5
1
-2
-1
1
2
-0.5
-2
-1
1
-1
2
-1
Figure 3.8: Double well tunneling states as determined by the Numerov approach. On the left
is the approximate lowest energy (symmetric) state with no nodes and on the rigth is the next
lowest (antisymmetric) state with a single node. The fact that the wavefunctions are heading off
towards infinity indicated the introduction of an additional node coming in from x = ∞.
and m = 1836 as the proton and use atomic units throughout. Also show in Fig. 3.7 are effective
harmonic oscillator wells for each side of the barrier. Notice that the harmonic approximation
is pretty crude since the harminic well tends to over estimate the steepness of the inner portion
and under estimate the steepness of the outer portions. Nonetheless, we can use the harminic
oscillator ground states in each well as starting points.
To use the Numerov method, one starts by guessing an initial energy, E, and then propagating
a trial solution to the Schrödinger equation. The Curve you obtain is in fact a solution to the
equation, but it will ususally not obey the correct boundary conditions. For bound states, the
boundary condition is that ψ must vanish exponentially outside the well. So, we initialize the
method by forcing ψ[1] to be exactly 0 and ψ[2] to be some small number. The exact values
really make no difference. If we are off by a bit, the Numerov wave will diverge towards ±∞ as x
increases. As we close in on a physically acceptible solution, the Numerov solution will begin to
exhibit the correct asymptotic behavour for a while before diverging. We know we have hit upon
an eigenstate when the divergence goes from +∞ to −∞ or vice versa, signeling the presence
of an additional node in the wavefunction. The proceedure then is to back up in energy a bit,
change the energy step and gradually narrow in on the exact energy. In Figs. 3.8a and 3.8b are
the results of a Numerov search for the lowest two states in the double well potential. One at
-3946.59cm−1 and the other at -3943.75cm−1 . Notice that the lowest energy state is symmetric
about the origin and the next state is anti-symmetric about the origin. Also in both cases, the
Numerov function diverges since we are not precisely at a stationary solution of the Schrödinger
equation...but we are within 0.01cm−1 of the true eigenvalue.
The advantage of the Numerov method is that it is really easy to code. In fact you can
even code it in Excell. Another advantage is that for radial scattering problems, the out going
boundary conditions occur naturally, making it a method of choice for simple scattering problems.
In the Mathematica notebooks, I show how one can use the Numerov method to compute the
scattering phase shifts and locate resonance for atomic collisions. The disadvantage is that you
have to search by hand for the eigenvalues which can be extremely tedious.
93
3.4.2
Numerical Diagonalization
A more general approach is based upon the variational principle (which we will discuss later)
and the use of matrix representations. If we express the Hamiltonian operator in matrix form in
some suitable basis, then the eigenfunctions of H can also be expressed as linear combinations
of those basis functions, subject to the constraint that the eigenfunctions be orthonormal. So,
what we do is write:
hφn |H|φm i = Hnm
and
|ψj i =
X
hφn |ψj i|φn i
n
The hφn |ψj i coefficients are also elements of a matrix, Tnj which transforms a vector in the φ
basis to the ψ basis. Conseqently, there is a one-to-one relation between the number of basis
functions in the ψ basis and the basis functions in the φ basis.
If |ψn i is an eigenstate of H, then
H|ψj i = Ej |ψj i.
Multiplying by hφm | and resolving the identity,
X
hφm |H|φn ihφn |ψj i = hφm |ψj iEj
n
X
Hmn Tnj = Ej Tmj
(3.148)
n
Thus,
X
Tmj Hmn Tnj = Ej
(3.149)
mn
or in more compact form
T † HT = E Iˆ
where Iˆ is the identity matrix. In otherwords, the T -matrix is simply the matrix which brings
H to diagonal form.
Diagonalizing a matrix by hand is very tedeous for anything beyond a 3×3 matrix. Since this
is an extremely common numerical task, there are some very powerful numerical diagonalization
routines available. Most of the common ones are in the Lapack package and are included as part
of the Mathematica kernel. So, all we need to do is to pick a basis, cast our Hamiltonian into
that basis, truncate the basis (usually determined by some energy cut-off) and diagonalize away.
Usually the diagonalization part is the most time consuming. Of course you have to be prudent
in choosing your basis.
A useful set of basis functions are the trigonmetric forms of the Tchebychev polynomials.1
These are a set of orthogonal functions which obey the following recurrence relation
Tn+1 (x) − 2xTn (x) + Tn−1 (x) = 0
94
(3.150)
1
0.75
0.5
0.25
-1
-0.5
0.5
1
-0.25
-0.5
-0.75
-1
Figure 3.9: Tchebyshev Polynomials for n = 1 − 5
Table 3.1: Tchebychev polynomials of the first type
To
T1
T2
T3
T4
T5
=
=
=
=
=
=
1
x
2x2 − 1
4x3 − 3x
8x4 − 8x2 − 1
16x5 − 20x3 + 5x
95
Table 3.1 lists a the first few of these polynomials as functions of x and a few of these are plotted
in Fig. 3.9
It is important to realize that these functions are orthogonal on a finite
range and that
√
integrals over these functions must include a weighting function w(x) = 1/ 1 − x2 . The orthogonality relation for the Tn polynomials is
Z
+1
−1


 0
m 6= m
m = n 6= 0
Tm (x)Tn (x)w(x)dx =


π m=n=0
π
2
(3.151)
Arfkin’s Mathematical Methods for Physicists has a pretty complete overview of these special
functions as well as many others. As usual, These are encorporated into the kernel of Mathematica and the Mathematica book and on-line help pages has some useful information regarding
these functions as well as a plethera of other functions.
From the recurrence relation it is easy to show that the Tn (x) polynomials satisfy the differential equation:
(1 − x2 )Tn00 − xTx0 + n2 Tn = 0
(3.152)
If we make a change of variables from x = cos(θ) and dx = − sin θdθ, then the differential
equation reads
dTn
+ n2 Tn = 0
dθ
(3.153)
This is a harmonic oscillator and has solutions sin nθ and cos nθ. From the boundary conditions
we have two linearly independent solutions
Tn = cos nθ = cos n(arccosx)
and
Vn = sin nθ.
The normalization condition then becomes:
Z
+1
−1
Tm (x)Tn (x)w(x)dx =
π
Z
cos(mθ) cos(nθ)dθ
(3.154)
0
and
Z
+1
−1
Vm (x)Vn (x)w(x)dx =
Z
π/2
sin(mθ) sin(nθ)dθ
(3.155)
−π/2
which is precisely the normalization integral we perform for the particle in a box state assuming
the width of the box was π. For more generic applications, we can scale θ and its range to any
range.
1
There are at least 10 ways to spell Tchebychev’s last name Tchebychev, Tchebyshev, Chebyshev are the
most common, as well as Tchebysheff, Tchebycheff, Chebysheff, Chevychef, . . .
96
The way we use this is to use the φn = N sin nx basis functions as a finite basis and truncate
any expansion in this basis at some point. For example, since we are usually interested in low
lying energy states, setting an energy cut-off to basis is exactly equivalent to keeping only the
lowest ncut states. The kinetic energy part of the Hamiltonian is diagonal in this basis, so we get
that part for free. However, the potential energy part is not diagonal in the φn = N sin nx basis,
so we have to compute its matrix elements:
Vnm =
Z
φn (x)V (x)φm (x)dx
(3.156)
To calculate this integral, let us first realize that [V, x] = 0, so the eigenstates of x are also
eigenstates of the potential. Taking matrix elements in the finite basis,
xnm = N
2
Z
φn (x)xφm (x)dx,
and diagonalizing it yields a finite set of ”position” eigenvalues, {xi } and a transformation for
converting between the ”position representation” and the ”basis representation”,
Tin = hxi |φn i,
which is simply a matrix of the basis functions evaluated at each of the eigenvalues. The special
set of points defined by the eigenvalues of the position operator are the Gaussian quadrature
points over some finite range.
This proceedure termed the ”discrete variable representation” was developed by Light and
coworkers in the 80’s and is a very powerful way to generate coordinate representations of Hamiltonian matrixes. Any matrix in the basis representation (termed the FBR for finite basis representation) can be transformed to the discrete variable representation (DVR) via the transformation
matrix T . Moreover, there is a 1-1 correspondency between the number of DVR points and the
number of FBR basis functions. Here we have used only the Tchebychev functions. One can
generate DVRs for any set of orthogonal polynomial function. The Mathematica code below generates the required transformations, the points, the eigenvalues of the second-derivative operator,
and a set of quadrature weights for the Tchebychev sine functions over a specified range:
dv2fb[DVR_, T_] := T.DVR.Transpose[T];
fb2dv[FBR_, T_] := Transpose[T].FBR.T;
tcheby[npts_, xmin_, xmax_] := Module[{pts, fb, del},
del = xmax - xmin;
pts = Table[i*del*(1/(npts + 1)) + xmin, {i, npts}] // N;
fbrke = Table[(i*(Pi/del))^2, {i, npts}] // N;
w = Table[del/(npts + 1), {i, npts}] // N;
T = Table[
Sqrt[2.0/(npts + 1)]*Sin[(i*j)*Pi/(npts + 1)],
{i, npts}, {j, npts}] // N;
Return[{pts, T, fbrke, w}]
]
To use this, we first define a potential surface, set up the Hamiltonian matrix, and simply
diagonalize. For this example, we will take the same double well system described above and
compare results and timings.
97
V[x_] := a*(x^4 - x^2);
cmm = 8064*27.3;
params = {a -> 0.1, m -> 1836};
{x, T, K, w} = tcheby[100, -1.3, 1.3];
Kdvr = (fb2dv[DiagonalMatrix[K], T]*m)/2 /. params;
Vdvr = DiagonalMatrix[V[x]] /. params;
Hdvr = Kdvr + Vdvr;
tt = Timing[{w, psi} = Transpose[
Sort[Transpose[Eigensystem[Hdvr]]]]];
Print[tt]
(Select[w*cmm , (# < 3000) &]) // TableForm
This code sets up the DVR points x, the transformation T and the FBR eigenvalues K using
the tcheby[n,xmin,xmax] Mathematica module defined above. We then generate the kinetic
energy matrix in the DVR using the transformation
KDV R = T † KF BR T
and form the DVR Hamiltonian
HDV R = KDV R + VDV R .
The eigenvalues and eigenvectors are computed via the Eigensystem[] routine. These are then
sorted according to their energy. Finally we print out only those states with energy less than
3000 cm−1 and check how long it took. On my 300 MHz G3 laptop, this took 0.3333 seconds to
complete. The first few of these are shown in Table 3.2 below. For comparison, each Numerov
iteration took roughly 1 second for each trial function. Even then, the eigenvalues we found are
probabily not as accurate as those computed here.
Table 3.2: Eigenvalues for double well potential computed via DVR and Numerov approaches
i ωi (cm−1 ) Numerov
1 -3946.574 -3946.59
2 -3943.7354 -3943.75
3 -1247.0974
4 -1093.5204
5
591.366
6 1617.424
98
3.5
Problems and Exercises
Exercise 3.2 Consider a harmonic oscillator of mass m and angular frequency ω. At time
t = 0, the state of this system is given by
|ψ(0)i =
X
cn |φn i
(3.157)
n
where the states |φn i are stationary states with energy En = (n + 1/2)h̄ω.
1. What is the probability, P , that at a measurement of the energy of the oscillator at some
later time will yield a result greater than 2h̄ω. When P = 0, what are the non-zero coefficients, cn ?
2. From now on, let only co and c1 be non zero. Write the normalization condition for |ψ(0)i
and the mean value hHi of the energy in terms of co and c1 . With the additional requirement
that hHi = h̄ω, calculate |co |2 and |c1 |2 .
3. As the normalized state vector |ψi is defined only to within an arbitrary global phase factor,
as can fix this factor by setting co to be real and positive. We set c1 = |c1 |eiφ . We assume
also that hHi = h̄ω and show that
1
hxi =
2
s
h̄
.
mω
(3.158)
Calculate φ.
4. With |ψi so determined, write |ψ(t)i for t > 0 and calculate the value of φ at time t.
Deduce the mean of hxi(t) of the position at time t.
Exercise 3.3 Find hxi, hpi, hx2 i and hp2 i for the ground state of a simple harmonic oscillator.
What is the uncertainty relation for the ground state.
Exercise 3.4 In this problem we consider the the interaction between molecule adsorbed on a
surface and the surface phonons. Represent the vibrational motion of the molecule (with reduced
mass µ) as harmonic with force constant K
Ho =
−h̄2 ∂ 2
K
+ x2
2
2µ ∂x
2
(3.159)
and the coupling to the phonons as
H 0 = −x
X
Vk cos(Ωk t)
(3.160)
k
where Vk is the coupling between the molecule and phonon of wavevector k and frequency Ωk .
99
1. Express the total Hamiltonian as a displaced harmonic well. What happens to the well as
a function of time?
2. What is the Golden-Rule transition rate between the ground state and the nth excited state
of the system due to phonon interactions? Are there any restrictions as to which final state
can be reached? Which phonons are responsible for this process?
3. From now on, let the perturbing force be constant in time
H0 = x
X
Vk
(3.161)
k
where Vk is the interaction with a phonon with wavevector k. Use the lowest order level
of perturbation theory necessary to construct the transition probability between the ground
state and the second-excited state.
Exercise 3.5 Let
mω 1/2
) x
2h̄
(3.162)
1
)1/2 p
2h̄mω
(3.163)
X=(
P =(
Show that the Harmonic Oscillator Hamiltonian is
H = h̄ω(P 2 + X 2 )
(3.164)
Now, define the operator: a† = X − iP . Show that a† acting on the harmonic oscillator ground
state is also an eigenstate of H. What is the energy of this state? Use a† to define a generating
relationship for all the eigenstates of H.
Exercise 3.6 Show that if one expands an arbitrary potential, V (x) about its minimum at xmin ,
and neglects terms of order x3 and above, one always obtains a harmonic well. Show that a
harmonic oscillator subject to a linear perturbation can be expressed as an unperturbed harmonic
oscillator shifted from the origin.
Exercise 3.7 Consider the one-dimensional Schrödinger equation with potential
(
V (x) =
m 2 2
ω x
2
+∞
x>0
x≤0
(3.165)
Find the energy eigenvalues and wavefunctions.
Exercise 3.8 An electron is contained inside a hard sphere of radius R. The radial components
of the lowest S and P state wavefunctions are approximately
sin(kr)
kr
cos(kr) sin(kr)
∂ψS (kr)
ψP (x) ≈
−
=
.
kr
(kr)2
∂(kr)
ψS (x) ≈
100
(3.166)
(3.167)
1. What boundary conditions must each state obey?
2. Using E = k 2 h̄2 /(2m) and the above boundary conditions, what are the energies of each
state?
3. What is the pressure exerted on the surface of the sphere if the electron is in the a.) S state,
b.) the P state. (Hint, recall from thermodynamics: dW = P dV = −(dE(R)/dR)dR.)
4. For a solvated e− in water, the S to P energy gap is about 1.7 eV. Estimate the size of
the the hard-sphere radius for the aqueous electron. If the ground state is fully solvated,
the pressure of the solvent on the electron must equal the pressure of the electron on the
solvent. What happens to the system when the electron is excited to the P -state from the
equilibrated S state? What happens to the energy gap between the S and P as a result of
this?
Exercise 3.9 A particle moves in a three dimensional potential well of the form:
(
V (x) =
mω 2
2
∞
z 2 > a2
(x2 + y 2 ), otherwise
(3.168)
Obtain an equation for the eigenvalues and the associated eigenfunctions.
Exercise 3.10 A particle moving in one-dimension has a ground state wavefunction (not-normalized)
of the form:
ψo (x) = e−α
4 x4 /4
(3.169)
where α is a real constant with eigenvalue Eo = h̄2 α2 /m. Determine the potential in which the
particle moves. (You do not have to determine the normalization.)
Exercise 3.11 A two dimensional oscillator has the Hamiltonian
1
1
H = (p2x + p2y ) + (1 + δxy)(x2 + y 2 )
2
2
(3.170)
where h̄ = 1 and δ << 1. Give the wavefunctions for the three lowest energy levels when δ = 0.
Evaluate using first order perturbation theory these energy levels for δ 6= 0.
Exercise 3.12 For the ground state of a harmonic oscillator, find the average potential and
kinetic energy. Verify the virial theorem, hT i = hV i, holds in this case.
Exercise 3.13 In this problem we will explore the use of two computational tools to calculate the
vibrational eigenstates of molecules. The the Spartan structure package is available on one of the
PCs in the computer lab in the basement of Fleming. First, we will generate a potential energy
surface using ab initio methods. We will then fit that surface to a functional form and try to
calculate the vibrational states. The system we will consider is the ammonia molecule undergoing
inversion from its degenerate C3v configuration through a D3h transition state.
101
Figure 3.10: Ammonia Inversion and Tunneling
1. Using the Spartan electronic structure package (or any other one you have access to), build
a model of N H3 , and determine its ground state geometry using various levels of ab initio
theory. Make a table of N − H bond lengths and θ = 6 H − N − H bond angles for the
equilibrium geometries as a function of at least 2 or 3 different basis sets. Looking in the
literature, find experimental values for the equilibrium configuration. Which method comes
closest to the experimental values? Which method has the lowest energy for its equilibrium
configuration.
2. Using the method which you deamed best in part 1, repeat the calculations you performed
above by systematically constraining the H − N − H bond angle to sample configurations
around the equilibrium configuration and up to the planar D3h configuration. Note, it may
be best to constrain two H − N − H angles and then optimize the bond lengths. Sample
enough points on either side of the minimum to get a descent potential curve. This is your
Born-Oppenheimer potential as a function of θ.
3. Defining the orgin of a coordinate system to be the θ = 120o D3h point on the surface, fit
your ab initio data to the ”W”-potential
V (x) = αx2 + βx4
(3.171)
What are the theoretical values of α and β?
4. We will now use perturbation theory to compute the tunneling dynamics.
(a) Show that the points of minimum potential energy are at
xmin
α
=±
2β
!1/2
and that the energy difference between the top of the barrier and the minimum energy
is given by
V
= V (0) − V (xmin )
α2
=
4β
102
(3.172)
(3.173)
(b) We first will consider the barrier to be infinitely high so that we can expand the potential function around each xmin . Show that by truncating the Taylor series expansion
above the (x − xmin )2 terms that the potential for the left and right hand sides are
given by
VL = 2α (x + xmin )2 − V
and
VR = 2α (x − xmin )2 − V.
What are the vibrational energy levels for each well?
(c) The wavefunctions for the lowest energy states in each well are given by
γ2
γ 1/2
ψ(x) = 1/4 exp[− (x ± xmin )2 ]
π
2
with
γ=
(4µα)1/2
h̄
!1/2
.
The energy levels for both sides are degenerate in the limit that the barrier height is
infinite. The total ground state wavefunction for this case is
Ψ(x) =
ψL (x)
ψR (x)
!
.
However, as the barrier height decreases, the degenerate states begin to mix causing
the energy levels to split. Define the ”high barrier” hamiltonian as
h̄2 ∂ 2
H=−
+ VL (x)
2µ ∂x
for x < 0 and
H=−
h̄2 ∂ 2
+ VR (x)
2µ ∂x
for x > 0. Calculate the matrix elements of H which mix the two degenerate left and
right hand ground state wavefunctions: i.e.
hΨ|H|Ψi =
HRR HLR
HRL HLL
!
where
HRR = hψR |H|ψR i
, with similar definitions for HRL , HLL and HLR . Obtain numerical values of each
matrix element using the values of α and β you determined above (in cm−1 ). Use the
mass of a H atom for the reduced mass µ.
103
(d) Since the ψL and ψR basis functions are non-orthogonal, you will need to consider the
overlap matrix, S, when computing the eigenvalues of H. The eigenvalues for this
system can be determined by solving the secular equation
α−λ
β − λS
β − λS
α−λ
=0
(3.174)
where α = HRR = HLL and β = HLR = HRL (not to be confused with the potential
parameters above). Using Eq. 3.174, solve for λ and determine the energy splitting
in the ground state as a function the unperturbed harmonic frequency and the barrier
height, V . Calculate this splitting using the parameters you computed above. What is
the tunneling frequency? The experimental results is ∆E = 0.794cm−12 .
Exercise 3.14 Consider a system in which the Lagrangian is given by
L(qi , q̇i ) = T (qi , q̇i ) − V (qi )
(3.175)
where we assume T is quadratic in the velocities. The potential is independent of the velocity
and neither T nor V carry any explicit time dependency. Show that


d X ∂L
q̇j
− L = 0
dt j
∂ q̇j
The constant quantity in the (. . .) defines a Hamiltonian, H. Show that under the assumed
conditions, H = T + V
Exercise 3.15 The Fermat principle in optics states that a light ray will follow the path, y(x)
which minimizes its optical length, S, through a media
S=
Z
x2 ,y2
n(y, x)ds
x1 ,y1
where n is the index of refraction. For y2 = y1 = 1 and −x1 = x2 = 1 find the ray-path for
1. n = exp(y)
2. n = a(y − yo ) for y > yo
Make plots of each of these trajectories.
Exercise 3.16 In a quantum mechanical system there are gi distinct quantum states between
energy Ei and Ei + dEi . In this problem we will use the variational principle and Lagrange
multipliers to determine how ni particles are distributed amongst these states subject to the constraints
1. The number of particles is fixed:
n=
X
ni
i
2
From Molecular Structure and Dynamics, by W. Flygare, (Prentice Hall, 1978)
104
2. the total energy is fixed
X
ni Ei = E
i
We consider two cases:
1. For identical particles obeying the Pauli exclusion principle, the probability of a given configuration is
WF D =
Y
i
gi
ni !(gi − ni )!
(3.176)
Show that maximizing WF D subject to the constraints above leads to
ni =
gi
λ
+λ
e 1 2 Ei
+1
with the Lagrange multipliers λ1 = −Eo /kT and λ2 = 1/kT . Hint: try working with the
log W and use Sterling’s approximation in the limit of a large number of particles .
2. In this case we still consider identical particles, but relax the restriction on the fixed number
of particles in a given state. The probability for a given distribution is then
WBE =
Y (ni + gi − 1)!
i
ni !(gi − 1)!
.
Show that by minimizing WBE subject to the constraints above leads to the occupation
numbers:
gi
ni = λ1 +λ2 Ei
e
−1
where again, the Lagrange multipliers are λ1 = −Eo /kT and λ2 = 1/kT . This yields the
Bose-Einstein statistics. Note: assume that gi 1
3. Photons satisfy the Bose-Einstein distribution and the constraint that the total energy is
constant. However, there is no constrain regarding the total number of photons. Show that
by eliminating the fixed number constraint leads to the foregoing result with λ1 = 0.
105
Bibliography
[1] C. A. Parr and D. G. Trular, J. Phys. Chem. 75, 1844 (1971).
[2] H. F. Schaefer III, J. Phys. Chem. 89, 5336 (1985).
[3] P. A. Whitlock and J. T. Muckermann, J. Chem. Phys. 61, 4624 (1974).
106
Chapter 4
Quantum Mechanics in 3D
In the next few lectures, we will focus upon one particular symmetry, the isotropy of free space.
As a collection of particles rotates about an arbitrary axis, the Hamiltonian does not change. If
the Hamoltonian does in fact depend explicitly upon the choice of axis, the system is “gauged”,
meaning all measurements will depend upon how we set up the coordinate frame. A Hamiltonian
with a potential function which depends only upon the coordinates, e.g. V = f (x, y, z), is gauge
invarient, meaning any measurement that I make will not depend upon my choice of reference
frame. On the other hand, if our Hamiltonian contains terms which couple one reference frame
to another (as in the case of non-rigid body rotations), we have to be careful in how we select
the “gauge”. While this sounds like a fairly specialized case, it turns out that many ordinary
phenimina depend upon this, eg. figure skaters, falling cats, floppy molecules. We focus upon
rigid body rotations first.
For further insight and information into the quantum mechanics of angular momentum, I
recommend the following texts and references:
1. Theory of Atomic Structure, E. Condon and G. Shortley. This is the classical book on
atomic physics and theory of atomic spectroscopy and has inspired generations since it
came out in 1935.
2. Angular Momentum–understanding spatial aspects in chemistry and physics, R. N. Zare.
This book is the text for the second-semester quantum mechanics at Stanford taught by
Zare (when he’s not out looking for Martians). It’s a great book with loads of examples in
spectroscopy.
3. Quantum Theory of Angular Momentum, D. A. Varshalovich, A. Moskalev, and V. Khersonskii. Not to much physics in this book, but if you need to know some relation between
Wigner-D functions and Racah coefficients, or how to derive 12j symbols, this book is for
you.
First, we need to look at what happens to a Hamiltonian under rotation. In order to show
that H is invarient to any rotations, we need only to show that it is invarient under an infintesimal
rotation.
107
4.1
Quantum Theory of Rotations
~ by a vector of a small rotation equal in magnitude to the angle δφ directed long an
Let δ φ
~ changes the direction vectors ~rα by δ~rα .
arbitrary axis. Rotating the system by δ φ
~ × ~rα
δ~rα = δ φ
(4.1)
Note that the × denotes the vector “cross” product. Since we will be using cross-products
through out these lectures, we pause to review the operation.
A cross product between two vectors is computed as
~c = ~a × ~b
î
ĵ k̂ = ai aj ak bi bj bk = î(aj bk − bj ak ) − ĵ(ai bk − bi ak ) + k̂(ai bj − bi aj )
= ijk aj bk
(4.2)
Where ijk is the Levi-Cevita symbol or the “anti-symmetric unit tensor” defined as
ijk


 0
if any of the indices are the same
=  1 for even permuations of the indices

−1 for odd permutations of the indices
(4.3)
(Note that we also have assumed a “summation convention” where by we sum over all repeated
indices. Some elementary properties are ikl ikm = δlm and ikl ikl = 6.)
So, an arbitrary function ψ(r1 , r2 , · · ·) is transformed by the rotation into:
ψ1 (r1 + δr1 , r2 + δr2 , · · ·) = ψ(r1 , r2 , · · ·) +
X
= ψ(r1 , r2 , · · ·) +
X
~ a
δra · ∇ψ
a
~ × ra · ∇
~ a ψa
δφ
a
!
~·
1 + δφ
=
X
~ a ψa
~ra × ∇
(4.4)
a
Thus, we conclude, that the operator
~·
1 + δφ
X
~a
~ra × ∇
(4.5)
a
is the operator for an infintesimal rotation of a system of particles. Since δφ is a constant, we
can show that this operator commutes with the Hamiltonian
!
[
X
~ a , H] = 0
~ra × ∇
(4.6)
a
This implies then a particular conservation law related to the isotropy of space. This is of course
angular momentum so that
!
X
~a
~ra × ∇
a
108
(4.7)
must be at least proportional to the angular momentum operator, L. The exact relation is
~
h̄L = ~r × p~ = −ih̄~r × ∇
(4.8)
which is much like its classical counterpart
L=
1
~r × ~v .
m
(4.9)
The operator is of course a vector quantity, meaning that is has direction. The components of
the angular momentum vector are:
h̄Lx = ypz − zpy
(4.10)
h̄Ly = zpx − zpz
(4.11)
h̄Lz = xpy − ypx
(4.12)
h̄Li = ijk xj pk
(4.13)
For a system in a external field, antgular momentum is in general not conserved. However, if
the field posesses spherical symmetry about a central point, all directions in space are equivalent
and angular momentum about this point is conserved. Likewise, in an axially symmetric field,
motion about the axis is conserved. In fact all the conservation laws which apply in classical
mechanics have quantum mechanical analogues.
We now move on to compute the commutation rules between the Li operators and the x and
p operators First we note:
[Lx , x] = [Ly , y] = [Lz , z] = 0
[Lx , y] =
1
z
((ypz − zpy )y − y(ypz − zpy )) = − [py , y] = iz
h̄
h̄
(4.14)
(4.15)
In short hand:
[Li , xk ] = iikl xl
(4.16)
We need also to know how the various components commute with one another:
h̄[Lx , Ly ] = Lx (zp x − xp z) − (zpx − xpz )Lx
(4.17)
= (Lx z − zLx )px − x(Lx pz − pz Lx )
(4.18)
= −iypx + ixpy
(4.19)
109
= ih̄Lz
(4.20)
[Ly , Lz ] = iLx
(4.21)
[Lz , Lx ] = iLy
(4.22)
[Lx , Ly ] = iLz
(4.23)
[Li , Lj ] = iijk Lk
(4.24)
Which we can summarize as
Now, denote the square of the modulus of the total angular momentum by L2 , where
L2 = L2x + L2y + L2z
(4.25)
Notice that this operator commutes with all the other Lj operators,
[L2 , Lx ] = [L2 , Ly ] = [L2 , Lz ] = 0
(4.26)
For example:
[L2x , Lz ] = Lx [Lx , Lz ] + [Lx , Lz ]Lx = −i(Lx Ly + Ly Lx )
(4.27)
also,
[L2y , Lz ] = i(Lx Ly + Ly Lx )
(4.28)
[L2 , Lz ] = 0
(4.29)
Thus,
Thus, I can measure L2 and Lz simultaneously. (Actually I can measure L2 and any one component Lk simultaneously. However, we ueually pick this one as the z axis to make the math
easier, as we shall soon see.)
A consequence of the fact that Lx , Ly , and Lz do not commute is that the angular momentum
~ can never lie exactly along the z axis (or exactly along any other axis for that matter).
vector L
q
We can interpret this in a classical context as a vector of length |L| = h̄ L(L + 1) with the Lz
component being h̄m. The vector is then constrained to lie in a cone as shown in Fig. ??. We
will take up this model at the end of this chapter in the semi-classical context.
It is also convienent to write Lx and Ly as a clinear combination
L+ = Lx + iLy L− = Lx − iLy
(4.30)
(Recall what we did for Harmonic oscillators?) It’s easy to see that
[L+ , L− ] = 2Lz
110
(4.31)
Figure 4.1: Vector Model of angular momentum
»Y4,0»2
»Y10,0»2
1
1.5
0.8
1.25
0.6
1
0.75
0.4
0.5
0.2
0.25
0.5
1
1.5
»Y4,2»2
2
2.5
3
0.5
1
1
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
0.5
1
1.5
»Y4,4»2
2
2.5
3
1
1
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
0.5
1
1.5
2
2.5
3
1
0.5
1
0.5
1
111
1.5
»Y10,2»2
2
2.5
3
1.5
2
2.5
3
1.5
2
2.5
3
»Y10,10 »2
[Lz , L+ ] = L+
(4.32)
[Lz , L− ] = −L−
(4.33)
L2 = L+ L− + L2z − Lz = L− L+ + L2z + Lz
(4.34)
Likewise:
We now give some frequently used expressions for the angular momentum operator for a
single particle in spherical polar coordinates (SP). In SP coordinates,
x = r sin θ cos φ
y = r sin θ sin φ
z = r cos θ
(4.35)
(4.36)
(4.37)
It’s easy and straightforward to demonstrate that
Lz = −i
∂
∂φ
(4.38)
and
±φ
L± = e
∂
∂
± + i cot θ
∂θ
∂φ
!
(4.39)
Thus,
1
1 ∂2
∂
∂
L =−
+
sin θ
2
sin θ sin θ ∂φ
∂θ
∂θ
"
#
2
(4.40)
which is the angular part of the Laplacian in SP coordinates.
1
∂
∂
∂
∂
1 ∂2
sin θ r2
+
sin θ
+
∇ = 2
r sin θ
∂r ∂r ∂θ
∂θ sin θ ∂φ2
"
2
=
1 ∂ 2∂
1
r
− 2 L2
2
r ∂r ∂r r
#
(4.41)
(4.42)
In other words, the kinetic energy operator in SP coordinates is
h̄2 2
h̄2
∇ =−
−
2m
2m
1 ∂ 2∂
1 2
r
−
L
r2 ∂r ∂r r2
112
!
(4.43)
4.2
Eigenvalues of the Angular Momentum Operator
Using the SP form
Lz ψ = i
∂ψ
= lz ψ
∂φ
(4.44)
Thus, we conclude that ψ = f (r, θ)eilz φ . This must be single valued and thus periodic in φ with
period 2π. Thus,
lz = m = 0, ±1, ±2, · · ·
(4.45)
Thus, we write the azimuthal solutions as
1
Φm (φ) = √ eimφ
2π
(4.46)
Φ∗m (φ)Φm0 (φ)dφ = δmm0
(4.47)
which are orthonormal functions:
Z
0
2π
In a centrally symmetric case, stationary states which differ only in their m quantum number
must have the same energy.
We now look for the eigenvalues and eigenfunctions of the L2 operator belonging to a set of
degenerate energy levels distinguished only by m. Since the +z−axis is physically equivalent to
the −z−axis, for every +m there must be a −m. Let L denote the greatest possible m for a
given L2 eigenstate. This upper limit must exist because of the fact that L2 − L2z = L2x + L2y is a
operator for an essentially positive quantity. Thus, its eigenvalues cannot be negative. We now
apply Lz L± to ψm .
Lz (L± ψm ) = (Lz ± 1)(L± ψm ) = (m ± 1)(L± ψm )
(4.48)
(note: we used [Lz , L± ] = ±L± ) Thus, L± ψm is an engenfunction of Lz with eigenvalue m ± 1.
ψm+1 ∝ L+ ψm
(4.49)
ψm−1 ∝ L− ψm
(4.50)
If m = l then, we must have L+ ψl = 0. Thus,
L− L+ ψl = (L2 − L2z − Lz )ψl = 0
(4.51)
L2 ψl = (L2z + Lz )ψl = l(l + 1)ψl
(4.52)
i.e.
Thus, the eigenvalues of L2 operator are l(l + 1) for l any positive integer (including 0). For a
given value of l, the component Lz can take values
l, l − 1, · · · , 0, −l
(4.53)
or 2l + 1 different values. Thus an energy level with angular momentum l has 2l + 1 degenerate
states.
113
4.3
Eigenstates of L2
Since l ansd m are the good quantum numbers, we’ll denote the eigenstates of L2 as
L2 |lmi = l(l + 1)|lmi.
(4.54)
This we will often write in short hand after specifying l as
L2 |mi = l(l + 1)|mi.
(4.55)
Since L2 = L+ L− + L2z − Lz , we have
hm|L2 |mi = m2 − m −
X
hm|L+ |m0 ihm0 |L− |mi = l(l + 1)
(4.56)
m0
Also, note that
hm − 1|L− |mi = hm|L+ |m − 1i∗ ,
(4.57)
|hm|L+ |m − 1i|2 = l(l + 1) − m(m − 1)
(4.58)
thus we have
Choosing the phase (Condon and Shortly phase convention) so that
hm − 1|L− |mi = hm|L+ |m − 1i
hm|L+ |m − 1i =
q
l(l + 1) − m(m − 1) =
q
(l + m)(l − m + 1)
(4.59)
(4.60)
Using this relation, we note that
1q
hm|Lx |m − 1i = hm − 1|Lx |mi =
(l + m)(l − m + 1)
2
(4.61)
−i q
hm|Ly |m − 1i = hm − 1|Lx |mi =
(l + m)(l − m + 1)
2
(4.62)
Thus, the diagonal elements of Lx and Ly are zero in states with definite values of hLz i = m.
4.4
Eigenfunctions of L2
The wavefunction of a particle is not entirely determined when l and m are presribed. We still
need to specify the radial component. Thus, all the angular momentum operators (in SP coords)
contain and explicit r dependency. For the time, we’ll take r to be fixed and denote the angular
momentum eigenfunctions in SP coordinates as Ylm (θ, φ) with normalization
Z
|Ylm (θ, φ)|2 dΩ
114
(4.63)
where dΩ = sin θdθdφ = d(cos θ)dφ and the integral is over all solid angles. Since we can
determine common eigenfunctions for L2 and Lz , there must be a separation of variables, θ and
φ, so we seek solutions of the form:
(4.64)
Ylm (θ, φ) = Φm (φ)Θlm (θ)
The normalization requirement is that
π
Z
0
|Θlm (θ)|2 sin θdθ = 1
(4.65)
and I require
Z
0
2π
Z
π
0
Yl∗0 m0 Ylm dΩ = δll0 δmm0 .
(4.66)
I thus seek solution of
1 ∂
∂
1 ∂2
sin θ
+
ψ + l(l + 1)ψ = 0
sin θ ∂θ
∂θ sin2 θ ∂φ2
(4.67)
1 ∂
∂
m2
sin θ
−
+ l(l + 1) Θlm (θ) = 0
sin θ ∂θ
∂θ sin2 θ
(4.68)
!
i.e.
!
which is well known from the theory of spherical harmonics.
v
u
u
m l t (2l + 1)(l − m)! m
Θlm (θ) = (−1) i
Pl (cos θ)
2(l − m)!
(4.69)
for m > 0. Where Plm are associated Legendre Polynomials. For m < 0 we get
Θl,−|m| = (−1)m Θl,|m|
(4.70)
Thus, the angular momentum eigenfunctions are the spherical harmonics, normalized so that
the matrix relations defined above hold true. The complete expression is
"
(m+|m|)/2 l
Ylm = (−1)
i
2l + 1 (l − |m|)!
4π (l + |m|)!
#1/2
|m|
Pl (cos θ)eimφ
(4.71)
For the case of m = 0,
Yl0 = i
l
2l + 1
4π
!1/2
Pl (cos θ)
(4.72)
Other useful relations are in cartesian form, obtained by using the relations
z
cos θ = ,
r
115
(4.73)
Table 4.1: Spherical Harmonics (Condon-Shortley Phase convention.
1
Y00 = √
4π
3 1/2
Y1,0 =
cos(θ)
4π
3 1/2
Y1,±1 = ∓
sin(θ)e±iφ
8π
s
5
Y2,±2 = 3
sin2 θe∓iφ
96π
s
Y2,±1 = ∓3
s
Y2,0 =
5
4π
5
sin θ cos θeiφ
24π
3
1
cos2 θ −
2
2
These can also be generated by the SphericalHarmonicY[l,m,θ,φ] function in Mathematica.
Figure 4.2: Spherical Harmonic Functions for up to l = 2. The color indicates the phase of the
function.
116
x
,
r
(4.74)
y
sin θ sin φ = .
r
(4.75)
sin θ cos φ =
and
3
=
4π
Y1,0
1/2
Y1,1
3
=
8π
Y1,−1
3
=
8π
1/2
1/2
z
r
(4.76)
x + iy
r
(4.77)
x − iy
r
(4.78)
The orthogonality integral of the Ylm functions is given by
Z
0
2π
Z
π
0
∗
Ylm
(θ, φ)Yl0 m0 (θ, φ) sin θdθdφ = δll0 δmm0 .
(4.79)
Another useful relation is that
∗
Yl,−m = (−1)m Ylm
.
(4.80)
This relation is useful in deriving real-valued combinations of the spherical harmonic functions.
Exercise 4.1 Demonstrate the following:
1. [L+ , L2 ] = 0
2. [L− , L2 ] = 0
Exercise 4.2 Derive the following relations
v
u
u
ψl,m (θ, φ) = t
and
v
u
u
ψl,m (θ, φ) = t
(l + m)!
(L− )l−m ψl,l (θ, φ)
(2l!)(l − m)!
(l − m)!
(L+ )l+m ψl,−l (θ, φ)
(2l!)(l + m)!
where ψl,m = Yl,m are eigenstates of the L2 operator.
117
4.5
Addition theorem and matrix elements
In the quantum mechanics of rotations, we will come across integrals of the general form
Z
or
Yl1∗m1 Yl2 m2 Yl3 m3 dΩ
Z
Yl1∗m1 Pl2 Yl3 m3 dΩ
in computing matrix elements between angular momentum states. For example, we may be
asked to compute the matrix elements for dipole induced transitions between rotational states of
a spherical molecule or between different orbital angular momentum states of an atom. In either
case, we need to evaluate an integral/matrix element of the form
hl1 m1 |z|l2 m2 i =
Z
Yl1∗m1 zYl2 m2 dΩ
(4.81)
q
Realizing that z = r cos θ = r 4π/3Y10 (θ, φ), Eq. 4.87 becomes
s
hl1 m1 |z|l2 m2 i =
4π Z ∗
r Yl1 m1 Y10 Yl2 m2 dΩ
3
(4.82)
Integrals of this form can be evaluated by group theoretical analysis and involves the introduc1
tion of Clebsch-Gordan coefficients, ClLM
which are tabulated in various places or can be
1 m1 l2 m2
computed using Mathematica. In short, some basic rules will always apply.
1. The integral will vanish unless the vector sum of the angular momenta sums to zero.
i.e.|l1 − l3 | ≤ l2 ≤ (l1 + l3 ). This is the “triangle” rule and basically means you have to be
able make a triangle with length of each side being l1 , l2 , and l3 .
2. The integral will vanish unless m2 + m3 = m1 . This reflects the conservation of the z
component of the angular momentum.
3. The integral vanishes unless l1 + l2 + l3 is an even integer. This is a parity conservation
law.
So the general proceedure for performing any calculation involving spherical harmonics is to first
check if the matrix element violates any of the three symmetry rules, if so, then the answer is 0
and you’re done. 2
To actually perform the integration, we first write the product of two of the Ylm ’s as a
Clebsch-Gordan expansion:
Yl1 m1 Yl2 m2
v
u
X u (2l1 + 1)(2l2 + 1)
t
=
ClL0
C LM
Y .
1 0l2 0 l1 m1 l2 m2 LM
LM
4π(2L + 1)
1
(4.83)
Our notation is based upon Varshalovich’s book. There at least 13 different notations that I know of for
expressing these coefficients which I list in a table at the end of this chapter.
2
In
Mathematica,
the
Clebsch-Gordan
coefficients
are
computed
using
the
function
ClebschGordan[j1,m1,j2,m2,j,m] for the decomposition of |jmi in to |j1 , m1 i and |j2 , m2 i.
118
We can use this to write
Z
∗
Ylm
Yl1 m2 Yl2 m2 dΩ =
v
u
X u (2l1 + 1)(2l2 + 1)
t
C L0
4π(2L + 1)
LM
=
LM
l1 0l2 0 Cl1 m1 l2 m2
Z
∗
Ylm
YLM dΩ
v
u
X u (2l1 + 1)(2l2 + 1)
t
C L0
4π(2L + 1)
LM
LM
l1 0l2 0 Cl1 m1 l2 m2 δlL δmM
v
u
u (2l1 + 1)(2l2 + 1)
= t
Cll01 0l2 0 Cllm
1 m1 l2 m2
4π(2l + 1)
(4.84)
In fact, the expansion we have done above for the product of two spherical harmonics can be
inverted to yield the decomposition of one angular momentum state into a pair of coupled angular
momentum states, such as would be the case for combining the orbital angular momentum of a
particle with, say, its spin angular momentum. In Dirac notation, this becomes pretty appearent
|LM i =
X
hl1 m1 l2 m2 |LM i|l1 m1 l2 m2 i
(4.85)
m1 m2
where the state |l1 m1 l2 m2 i is the product of two angular momentum states |l1 m1 i and |l2 m2 i.
The expansion coefficients are the Clebsch-Gordan coefficients
ClLM
= hl1 m1 l2 m2 |LM i
1 m1 l2 m2
(4.86)
Now, let’s go back the problem of computing the dipole transition matrix element between
two angular momentum states in Eq. 4.87. The integral we wish to evaluate is
hl1 m1 |z|l2 m2 i =
Z
Yl1∗m1 zYl2 m2 dΩ
(4.87)
and we noted that z was related to the Y10 spherical harmonic. So the integral over the angular
coordinates involves:
Z
Yl1∗m1 Y10 Yl2 m2 dΩ.
(4.88)
First, we evaluate which matrix elements are going to be permitted by symmetry.
1. Clearly, by the triangle inequality, |l1 − l2 | = 1. In other words, we change the angular
momentum quantum number by only ±1.
2. Also, by the second criteria, m1 = m2
3. Finally, by the third criteria: l1 + l2 + 1 must be even, which again implies that l1 and l2
differ by 1.
Thus the integral becomes
Z
∗
Yl+1,m
Y10 Ylm
v
u
u (2l + 1)(2 + 1) l+1,0
1m
= t
Cl010 Cl+1,ml0
4π(2l + 3)
119
(4.89)
From tables,
√
!
2 (1 + l)
√
=− √
2 + 2l 3 + 2l
√ √
√
!
2 1+l−m 1+l+m
√
√
=−
2 + 2l 3 + 2l
l+1,0
Cl010
1m
Cl+1,ml0
So
l+1,0 1m
Cl010
Cl+1,ml0
√
√
2 (1 + l) 1 + l − m 1 + l + m
=
(2 + 2 l) (3 + 2 l)
Thus,
Z
s
∗
Yl+1,m
Y10 Ylm dΩ =
v
u
3u
t (l + m + 1)(l − m + 1)
4π
(2l + 1)(2l + 3)
(4.90)
Finally, we can construct the matrix element for dipole-transitions as
v
u
u (l + m + 1)(l − m + 1)
hl1 m1 |z|l2 m2 i = rt
δl1 ±1,l2 δm1 ,m2 .
(2l + 1)(2l + 3)
(4.91)
Physically, this make sense because a photon carries a single quanta of angular momentum. So
in order for molecule or atom to emit or absorb a photon, its angular momentum can only change
by ±1.
Exercise 4.3 Verify the following relations
Z
Z
s
∗
Yl+1,m+1
Y11 Ylm dω =
s
(4.92)
v
u
3u
∗
t (l − m)(l − m − 1)
Yl−1,m−1
Y11 Ylm dω = −
8π
2l − 1)(2l + 1)
(4.93)
1
∗
Ylm
Y00 Ylm dΩ = √
4π
(4.94)
Z
4.6
v
u
3u
t (l + m + 1)(l + m + 2)
8π
2l + 1)(2l + 3)
Legendre Polynomials and Associated Legendre Polynomials
Ordinary Legendre polynomials are generated by
Pl (cos θ) =
1
dl
(cos2 θ − 1)l
2l l! (d cos θ)l
120
(4.95)
i.e. (x = cos θ)
Pl (x) =
1 ∂l 2
(x − 1)l
2l l! ∂xl
(4.96)
and satisfy
!
1 ∂
∂
sin θ
+ l(l + 1) Pl = 0
sin θ ∂θ
∂θ
(4.97)
The Associated Legendre Polynomials are derived from the Legendre Polynomials via
Plm (cos θ) = sinm θ
4.7
∂m
Pl (cos θ)
(∂ cos θ)m
(4.98)
Quantum rotations in a semi-classical context
Earlier we established the fact that the angular momentum vector can never exactly lie on a
single spatial axis. By convention we take the quantization axis to be the z axis, but this is
arbitrary and we can pick any axis as the quantization axis, it is just that picking the z axis
make the mathematics much simpler. Furthermore, we established that the maximum length
the angular momentum vector qcan have along the z axis is the eigenvalue of Lz when m = l,
so hlz i = l which is less than l(l + 1). Note, however, we can write the eigenvalue of L2 as
l2 (1+1/l2 ) . As l becomes very large the eigenvalue of Lz and the eigenvalue of L2 become nearly
identical. The 1/l term is in a sense a quantum mechanical effect resulting from the uncertainty
~
in determining the precise direction of L.
We can develop a more quantative model for this by examining both the uncertainty product
and the semi-classical limit of the angular momentum distribution function. First, recall, that if
we have an observable, A then the spread in the measurements of A is given by the varience.
∆A2 = h(A − hAi)2 i = hA2 i − hAi2 .
(4.99)
In any representation in which A is diagonal, ∆A2 = 0 and we can determine A to any level of
precision. But if we look at the sum of the variances of lx and ly we see
∆L2x + ∆L2y = l(l + 1) − m2 .
(4.100)
So for a fixed value of l and m, the sum of the two variences is constant and reaches its minimum
when |m| = l corresponding to the case when the vector points as close to the ±z axis as it
possible can. The conclusion we reach is that the angular momentum vector lies somewhere in a
cone in which the apex half-angle, θ satisfies the relation
m
cos θ = q
(4.101)
l(l + 1)
which we can varify geometrically. So as l becomes very large the denominator becomes for m = l
l
1
= q
→1
l(l + 1)
1 1 + 1/l2
q
121
(4.102)
and θ = 0,cooresponding to the case in which the angular momentum vector lies perfectly along
the z axis.
Exercise 4.4 Prove Eq. 4.100 by writing hL2 i = hL2x i + hL2y i + hL2z i.
To develop this further, let’s look at the asymptotic behavour of the Spherical Harmonics at
large values of angular momentum. The angular part of the Spherical Harmonic function satisfies
1 ∂
∂
m2
Θlm = 0
sin θ
+ l(l + 1) −
sin θ ∂θ
∂θ
sin2 θ
!
(4.103)
For m = 0 this reduces to the differential equation for the Legendre polynomials
∂2
∂
+ cot θ
+ l(l + 1) Pl (cos θ) = 0
2
∂θ
∂θ
!
(4.104)
If we make the substitution
Pl (cos θ) =
χl θ
(sin θ)1/2
(4.105)
then we wind up with a similar equation for χl (θ)
∂2
csc2 θ
2
+
(l
+
1/2)
+
χl = 0.
∂θ2
4
!
(4.106)
For very large l, the l + 1/2 term dominates and we can ignore the cscθ term everywhere except
for angles close to θ = 0 or θ = π. If we do so then our differential equation becomes
∂2
+ (l + 1/2)2 χl = 0,
∂θ2
!
(4.107)
which has the solution
χl (θ) = Al sin ((l + 1/2)θ + α)
(4.108)
where Al and α are constants we need to determine from the boundary conditions of the problem.
For large l and for θ l−1 and π − θ l−1 one obtains
Pl (cos θ) ≈ Al
sin((l + 1/2)θ + α)
.
(sin θ)1/2
(4.109)
Similarly,
Ylo (θ, φ) ≈
l + 1/2
2π
!1/2
Al
sin((l + 1/2)θ + α)
.
(sin θ)1/2
(4.110)
so that the angular probability distribution is
2
l + 1/2
2 sin ((l + 1/2)θ + α)
Al
.
2π
sin θ
!
2
|Ylo | =
122
(4.111)
When l is very large the sin2 ((l + 1/2)θ) factor is extremely oscillatory and we can replace it
by its average value of 1/2. Then, if we require the integral of our approximation for |Yl0 |2 to be
normalized, one obtains
|Yl0 |2 =
2π 2
1
sin(θ)
(4.112)
which holds for large values of l and all values of θ except for theta = 0 or θ = π.
We can also recover this result from a purely classical model. In classical mechanics, the
particle moves in a circular orbit in a plane perpendicular to the angular momentum vector. For
m = 0 this vector lies in the xy plane and we will define θ as the angle between the particle and
the z axis, φ as the azimuthal angle of the angular momentum vector in the xy plane. Since the
particles speed is uniform, its distribution in θ is uniform. Thus the probability of finding the
particle at any instant in time between θ and θ + dθ is dθ/π. Furthermore, we have not specified
the azimuthal angle, so we assume that the probability distribution is also uniform over φ and
the angular probability dθ/π must be smeared over some band on the uniform sphere defined by
the angles θ and θ + dθ. The area of this band is 2π sin θdθ. Thus, we can define the “classical
estimate” of the as a probability per unit area
dθ
1
1
= 2
π 2π sin θ
2π sin θ
which is in agreement with the estimate we made above.
For m 6= 0 we have to work a bit harder since the angular momentum vector is tilted out of
the plane. For this we define two new angles γ which is the azimuthal rotation of the particle’s
position about the L vector and α with is constrained by the length of the angular momentum
vector and its projection onto the z axis.
m
m
cos α = q
≈
l
l(l + 1
P (θ) =
The analysis is identical as before with the addition of the fact that the probability in γ (taken
to be uniform) is spread over a zone 2π sin θdθ. Thus the probability of finding the particle with
some angle θ is
dγ
1
P (θ) =
.
2
dθ 2π sin θ
Since γ is the dihedral angle between the plane containing z and l and the plane containing
l and r (the particle’s position vector), we can relate γ to θ and α by
π
π
cos θ = cos α cos + sin α sin cos γ = sin α cos γ
2
2
Thus,
sin θdθ = sin α sin γdγ.
This allows us to generalize our probability distribution to any value of m
1
sin α sin γ
1
=
2
2
2π (sin α − cos2 θ)1/2
|Ylm (θ, φ)|2 =
2π 2
123
(4.113)
(4.114)
Figure 4.3: Classical and Quantum Probability Distribution Functions for Angular Momentum.
»Y4,0»2
»Y10,0»2
1
1.5
0.8
1.25
0.6
1
0.75
0.4
0.5
0.2
0.25
0.5
1
1.5
»Y4,2»2
2
2.5
3
0.5
1
1
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
0.5
1
1.5
»Y4,4»2
2
2.5
3
1
1
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
0.5
1
1.5
2
2.5
3
1
0.5
1
0.5
1
1.5
»Y10,2»2
2
2.5
3
1.5
2
2.5
3
1.5
2
2.5
3
»Y10,10 »2
which holds so long as sin2 α > cos2 θ. This corresponds to the spatial region (π/2 − α) <
θ < π/2 + α. Outside this region, the distribution blows up and corresponds to the classically
forbidden region.
In Fig. 4.3 we compare the results of our semi-classical model with the exact results for l = 4
and l = 10. All in all we do pretty well with a semi-classical model, we do miss some of the
wiggles and the distribution is sharp close to the boundaries, but the generic features are all
there.
124
Table 4.2: Relation between various notations for Clebsch-Gordan Coefficients in the literature
Symbol
Cjjm
1 m1 j2 m2
j1 j2
Sjm
1 jm2
1 j2
Ajj
mm1 m2
Cm1 mj
2
(j1 j2 m1 m2 |j1 j2 jm)
j1 j2
(m1 m2 )
Cjm
X(j, m, j1 , j2 , m1 )
C(jm; m1 m2 )
j1 j2 j
Cm
1 m2 m
C(j
j
j,
m1 m2 )#
1
2
"
j1 j2 j
m1 m2 m
hj1 m1 j2 m2 |(j1 j2 )jmi
Author
Varshalovich a
Wignerb
Eckartc
Van der Weardend
Condon and Shortley
Fockf
Boysg
Blatt and Weisskopfh
Beidenharni
Rosej
e
Yutsis and Bandzaitisk
Fanol
a.) D. A. Varschalovich, et al. Quantum Theory of Angular Momentum, (World Scientific, 1988).
b.) E. Wigner, Group theory, (Academic Press, 1959).
c.) C. Eckart “The application of group theory to the quantum dynamics of monatomic systems”,
Rev. Mod. Phys. 2, 305 (1930).
d.) B. L. Van der Waerden, Die gruppentheorische methode in der quantenmechanik, (Springer,
1932).
e.)E. Condon and G. Shortley, Theory of Atomic Spectra, (Cambridge, 1932).
f.) V. A. Fock, “ New Deduction of the Vector Model”, JETP 10,383 (1940).
g.)S. F. Boys, “Electronic wave functions IV”, Proc. Roy. Soc., London, A207, 181 (1951).
h.) J. M. Blatt and V. F. Weisskopf, Theoretical Nuclear Physics, (McGraw-Hill, 1952).
i.) L. C. Beidenharn, ”Tables of Racah Coefficients”, ONRL-1098 (1952).
j.) M. E. Rose, Multipole Fields, (Wiley 1955).
k.) A. P. Yusis and A. A. Bandzaitit, The Theory of Angular Momentum in Quanutm Mechanics,
(Mintus, Vilinus, 1965).
l.) U. Fano, “Statistical matrix techniques and their application to the directional correlation of
radiation,” US Nat’l Bureau of Standards, Report 1214 (1951).
125
4.8
Motion in a central potential: The Hydrogen Atom
(under development)
The solution of the Schrödinger equation for the hydrogen atom was perhaps the most significant developments in quantum theory. Since it is one of the few problems in nature in which we
can derive an exact solution to the equations of motion, it deserves special attention and focus.
Perhaps more importantly, the hydrogen atomic orbitals form the basis of atomic physics and
quantum chemistry.
The potential energy function between the proton and the electron is the centrosymmetric
Coulombic potential
Ze2
V (r) = −
.
r
Since the potential is centrosymmetric and has no angular dependency the Hydrogen atom Hamiltonian separates in to radial and angular components.
h̄2
H=−
2µ
1 ∂ 2∂
L2
r
−
r2 ∂r ∂r h̄2 r2
!
e2
+
r
(4.115)
where L2 is the angular momentum operator we all know and love by now and µ is the reduced
mass of the electron/proton system
µ=
me mp
≈ me = 1
me + mp
Since [H, L] = 0, angular momentum and one component of the angular momentum must be
constants of the motion. Since there are three separable degrees of freedom, we have one other
constant of motion which must correspond to the radial motion. As a consequence, the hydrogen
wavefunction is separable into radial and angular components
ψnlm = Rnl (r)Ylm (θ, φ).
(4.116)
Using the Hamiltonian in Eq. 4.115 and this wavefunction, the radial Schrödinger equation reads
(in atomic units)
h̄2 1 ∂ 2 ∂
l(l + 1)
1
r
−
−
Rnl (R) = ERnl (r)
−
2
2
2 r ∂r ∂r
r
r
(
"
#
)
(4.117)
At this point, we introduce atomic units to make the notation more compact and drastically
simplify calculations. In atomic units, h̄ = 1 and e = 1. A list of conversions for energy, length,
etc. to SI units is listed in the appendix. The motivation is so that all of our numbers are of
order 1.
The kinetic energy term can be rearranged a bit
1 ∂ 2∂
∂2
2 ∂
r
=
+
r2 ∂r ∂r
∂r2 r ∂r
(4.118)
and the radial equation written as
h̄2 ∂ 2
2 ∂
l(l + 1)
1
−
+
−
−
Rnl (R) = ERnl (r)
2
2
2 ∂r
r ∂r
r
r
(
"
#
126
)
(4.119)
To solve this equation, we first have to figure out what approximate form the wavefunction must
have. For large values of r, the 1/r terms disappear and the asymptotic equation is
h̄2 ∂ 2
−
Rnl (R) = ERnl (r)
2 ∂r2
(4.120)
∂2R
= α2 R
∂r2
(4.121)
or
where α = −2mE/h̄2 . This differential equation we have seen before for the free particle, so the
solution must have the same form. Except in this case, the function is real. Furthermore, for
bound states with E < 0 the radial solution must go to zero as r → ∞, so of the two possible
asymptotic solutions, the exponentially damped term is the correct one.
R(r) ≡ e−αr
(4.122)
Now, we have to check if this is a solution everywhere. So, we take the asymptotic solution and
plug it into the complete equation:
2 −αr
α e
e2
+ E e−αr = 0.
r
!
2
2m
+ (−αe−αr ) + 2
r
h̄
(4.123)
Eliminating e−αr
2mE
1
α + 2
+
r
h̄
2
2me2
− 2α = 0
h̄2
!
(4.124)
For the solution to hold everywhere, it must also hold at r = 0, so two conditions must be met
α2 = −2mE/h̄2
(4.125)
2me2
− 2α = 0.
h̄2
(4.126)
which we defined above, and
!
If these conditions are met, then e−αr is a solution. This last equation also sets the length scale
of the system since
α = me2 /h̄2 = 1/ao
(4.127)
where ao is the Bohr radius. In atomic units, ao = 1. Likewise, the energy can be determined:
E=−
h̄2
h̄2 e2
e2
=
−
=
−
.
2ma2o
me2 2ao
2ao
In atomic units the ground states energy is E = −1/2hartree.
127
(4.128)
Finally, we have to normalize R
Z
−2αr
3
d re
= 4π
Z
∞
r2 e−2αr dr
(4.129)
0
The angular normalization
√ can be absorbed into the spherical harmonic term in the total wavefunction since Y00 = 1/ 4π. So, the ground state wavefunction is
ψn00 = N e−r/ao Y00
(4.130)
The radial integral can be evaluated using Leibnitz’ theorem for differentiation of a definite
integral
Z b
∂f (β, x)
∂ Zb
f (β, x)dx =
dx
∂β a
∂β
a
(4.131)
Thus,
Z
∞
r2 e−βr dr =
∞
Z
0
0
∂ 2 −βr
e dr
∂β 2
∂ 2 Z ∞ −βr
e dr
∂β 2 0
∂2 1
= − 2
∂β β
2
=
β3
=
(4.132)
Exercise 4.5 Generalize this result to show that
∞
Z
rn e−βr dr =
0
n!
β n+1
(4.133)
Thus, using this result and putting it all together, the normalized radial wavefunction is
R10 = 2
1
ao
3/2
e−r/ao .
(4.134)
For the higher energy states, we examine what happens at r → 0. Using a similar analysis
as above, one can show that close in, the radial solution must behave like a polynomial
R ≡ rl+1
which leads to a general solution
R = rl+1 e−αr
∞
X
s=0
128
as r s .
The proceedure is to substitute this back into the Schrodinger equation and evaluate term by
term. In the end one finds that the energies of the bound states are (in atomic units)
En = −
1
2n2
and the radial wavefunctions
2r
Rnl =
nao
l 2
nao
!
(n − l − 1)!
2r
e−r/nao L2l+1
n+1
3
2n((n + l)!)
nao
(4.135)
where the Lba are the associated Laguerre polynomials.
4.8.1
Radial Hydrogenic Functions
The radial wavefunctions for nuclei with atomic number Z are modified hydrogenic wavefunctions
with the Bohr radius scaled by Z. I.e a = ao /Z. The energy for 1 electron about a nucleus with
Z protons is
En = −
Z2 1
Z 2 Ry
=
−
n2 2ao
n2 2
(4.136)
Some radial wavefunctions are
Z
=2
ao
R1s
1 Z
=√
2 ao
R2s
3/2 1
Z
= √
2 6 ao
R2p
4.9
3/2
e−Zr/ao
Zr −Zr/2ao
1−
e
2ao
5/2
(4.137)
re−Zr/2ao
(4.138)
(4.139)
Spin 1/2 Systems
In this section we are going to illustrate the various postulates and concepts we have been
developing over the past few weeks. Rather than choosing as examples problems which are
pedagogic (such as the particle in a box and its variations) or or chosen for theor mathematical
simplicity, we are going to focus upon systems which are physically important. We are going to
examine, with out much theoretical introduction, the case in which the state space is limited to
two states. The quantum mechanical behaviour of these systems can be varified experimentally
and, in fact, were and still are used to test various assumptions regarding quantum behaviour.
Recall from undergraduate chemistry that particles, such as the electron, proton, and so forth,
~ called spin. This is a property which has no analogue
possess an intrinsic angular momentum, S,
in classical mechanics. Without going in to all the details of angular momentum and how it gets
quantized (don’t worry, it’s a coming event!) we are going to look at a spin 1/2 system, such as
129
a neutral paramagnetic Ag atom in its ground electronic state. We are going to dispense with
treating the other variables, the nuclear position and momentum,the motion of the electrons,
etc... and focus only upon the spin states of the system.
~ . This magnetic
The paramagnetic Ag atoms possess an electronic magnetic moment, M
~
moment can couple to an externally applied magnegit field, B, resulting on a net force being
applied to the atom. The potential energy in for this is
~ .B.
~
W = −M
(4.140)
We take this without further proof. We also take without proof that the magnetic moment and
the intrinsic angular momentum are proportional.
~ = γS
~
M
(4.141)
The proportionality constant is the gyromagnetic ratio of the level under consideration. When
the atoms traverse through the magnetic field, they are deflected according to how their angular
momentum vector is oriented with the applied field.
~ M
~ .B)
~
F~ = ∇(
(4.142)
Also, the total moment relative to the center of the atom is
~Γ = M
~ × B.
~
(4.143)
Thus, the time evolution of the angular momentum of the particle is
∂ ~ ~
S=Γ
∂t
(4.144)
∂ ~
~ × B.
~
S = γS
∂t
(4.145)
that it to say
~ and the angular momentum
Thus, the velocity of the angular momentum is perpendicular to S
vector acts like a gyroscope.
We can also show that for a homogeneous field the force acts parallel to z and is proportional
to Mz . Thus, the atoms are deflected according to how their angular momentum vector is oriented
with respect to the z axis. Experimentally, we get two distributions. Meaning that measurement
of Mz can give rise to two possible results.
4.9.1
Theoretical Description
We associate an observable, Sz , with the experimental observations. This has 2 eigenvalues, at
±h̄/2 We shall assume that the two are not degenerate. We also write the eigenvectors of Sz as
|±i corresponding to
h̄
Sz |+i = + |+i
2
130
(4.146)
h̄
Sz |−i = + |−i
2
(4.147)
h+|+i = h−|−i = 1
(4.148)
h+|−i = 0.
(4.149)
with
and
The closure, or idempotent relation is thus
|+ih+| + |−ih−| = 1.
(4.150)
The most general state vector is
|ψi = α|+i + β|−i
(4.151)
|α|2 + |β|2 = 1.
(4.152)
with
In the |±i basis, the matrix representation of Sz is diagonal and is written as
h̄
Sz =
2
4.9.2
!
1 0
0 −1
(4.153)
Other Spin Observables
We can also measure Sx and Sy . In the |±i basis these are written as
h̄
Sx =
2
0 1
1 0
!
(4.154)
and
h̄
Sy =
2
0 i
−i 0
!
(4.155)
You can verify that the eigenvalues of each of these are ±h̄/2.
4.9.3
Evolution of a state
The Hamiltonian for a spin 1/2 particle in a B-field is given by
H = −γ|B|Sz .
131
(4.156)
Where B is the magnitude of the field. This operator is time-independent, thus, we can solve the
Schrodinger Equation and see that the eigenvectors of H are also the eigenvectors of Sz . (This
the eigenvalues of Sz are “good quantum numbers”.) Let’s write ω = −γ|B| so that
H|+i = +
h̄ω
|+i
2
(4.157)
h̄ω
|−i
(4.158)
2
Therefore there are two energy levels, E± = ±h̄ω/2. The separation is proportional to the
magnetic field. They define a single “Bohr Frequency”.
H|−i = −
4.9.4
Larmor Precession
Using the |±i states, we can write any arb. angular momentum state as
θ
θ
|ψ(0)i = cos( )e−iφ/2 |+i + sin( )e+iφ/2 |−i
(4.159)
2
2
where θ and φ are polar coordinate angles specifing the directrion of the angular momentum
vector at a given time. The time evolution under H is
θ
θ
|ψ(0)i = cos( )e−iφ/2 e−iE+ t/h̄ |+i + sin( )e+iφ/2 e−iEm t/h̄ |−i,
2
2
or, using the values of E+ and E−
θ
θ
|ψ(0)i = cos( )e−i(φ+ωt)/2 |+i + sin( )e+i(φ+ωt)/2 |−i
2
2
In other words, I can write
(4.160)
(4.161)
θ(t) = θ
(4.162)
φ(t) = φ + ωt.
(4.163)
This corresponds to the precession of the angular momentum vector about the z axis at an angular
frequency of ω. More over, the expectation values of Sz , Sy , and Sx can also be computed:
hSz (t)i = h̄/2 cos(θ)
(4.164)
hSx (t)i = h̄/2 sin(θ/2) cos(φ + ωt)
(4.165)
hSy (t)i = h̄/2 sin(θ/2) sin(φ + ωt)
(4.166)
Finally, what are the “populations” of the |±i states as a function of time?
|h+|ψ(t)i|2 = cos2 (θ/2)
(4.167)
|h−|ψ(t)i|2 = sin2 (θ/2)
(4.168)
Thus, the populations do not change, neither does the normalization of the state.
132
4.10
Problems and Exercises
Exercise 4.6 A molecule (A) with orbital angular momentum S = 3/2 decomposes into two
products: product (B) with orbital angular momentum 1/2 and product (C) with orbital angular
momentum 0. We place ourselves in the rest frame of A) and angular momentum is conserved
throughout.
A3/2 → B 1/2 + C 0
(4.169)
1. What values can be taken on by the relative orbital angular momentum of the two final
products? Show that there is only one possible value of the parity of the relative orbital
state is fixed. Would this result remain the same if the spin of A was 3/2?
2. Assume that A is initially in the spin state characterized by the eigenvalue ma h̄ of its spin
component along the z-axis. We know that the final orbital state has a definite parity. Is it
possible to determine this parity by measuring the probabilities of finding B in either state
|+i or in state |−i?
Exercise 4.7 The quadrupole moment of a charge distribution, ρ(r), is given by
1Z
Qij =
3(xi xj − δij r2 )ρ(r)d3 r
e
(4.170)
where the total charge e = d3 rρ(r). The quantum mechanical equivalent of this can be written
in terms of the angular momentum operators as
R
1Z 2 3
Qij =
r
(Ji Jj + Jj Ji ) − δij J 2 ρ(r)d3 r
e
2
(4.171)
The quadrupole moment of a stationary state |n, ji, where n are other non-angular momentum
quantum numbers of the system, is given by the expectation value of Qzz in the state in which
m = j.
1. Evaluate
Qo = hQzz i = hnjm = j|Qzz |njm = ji
(4.172)
in terms of j and hr2 i = hnj|r2 |nji.
2. Can a proton (j = 1/2) have a quadrupole moment? What a bout a deuteron (j = 1)?
3. Evaluate the matrix element
hnjm|Qxy |nj 0 m0 i
(4.173)
What transitions are induced by this operator?
4. The quantum mechanical expression of the dipole moment is
r
po = hnjm = j| Jz |njm = ji
(4.174)
e
Can an eigenstate of a Hamiltonian with a centrally symmetric potential have an electric
dipole moment?
133
Exercise 4.8 The σx matrix is given by
0 1
1 0
σx =
!
,
(4.175)
prove that
exp(iασx ) = I cos(α) + iσx sin(α)
(4.176)
where α is a constant and I is the unit matrix.
Solution: To solve this you need to expand the exponential. To order α4 this is
eiασx = I + iασx −
α2 2 iα3 3 α4 4
σ −
σ + σx + · · ·
2 x
3!
4!
(4.177)
Also, note that σx .σx = I, thus, σx2n = I and σx2n+1 = σx . Collect all the real terms and all the
imaginary terms:
α2
α4
α3
= I +I
+I
+ · · · + iσx α − −i + · · ·
2
4!
3!
!
iασx
e
!
(4.178)
These are the series expansions for cos and sin.
eiασx = I cos(α) + iσx sin(α)
(4.179)
Exercise 4.9 Because of the interaction between the proton and the electron in the ground state
of the Hydrogen atom, the atom has hyperfine structure. The energy matrix is of the form:


H=


A 0
0 0
0 −A 2A 0
0 2A −A 0
0 0
0 A





(4.180)
in the basis defined by
|1i
|2i
|3i
|4i
=
=
=
=
|e+, p+i
|e+, p−i
|e−, p+i
|e−, p−i
(4.181)
(4.182)
(4.183)
(4.184)
where the notation e+ means that the electron’s spin is along the +Z-axis, and e− has the spin
pointed along the −Z axis. i.e. |e+, p+i is the state in which both the electron spin and proton
spin is along the +Z axis.
1. Find the energy of the stationary states and sketch an energy level diagram relating the
energies and the coupling.
134
2. Express the stationary states as linear combinations of the basis states.
3. A magnetic field of strength B applied in the +Z direction and couples the |e+, p+i and
|e−, p−i states. Write the new Hamiltonian matrix in the |e±, p±i basis. What happens to
the energy levels of the stationary states as a result of the coupling? Add this information
to the energy level diagram you sketched in part 1.
~ = γ S.
~ The spin space is
Exercise 4.10 Consider a spin 1/2 particle with magnetic moment M
spanned by the basis of |+i and |−i vectors, which are eigenvectors of Sz with eigenvalues ±h̄/2.
At time t = 0, the state of the system is given by
|ψ(0)i = |+i
1. If the observable Sx is measured at time t = 0, what results can be found and with what
probabilities?
2. Taking |ψ(0)i as the initial state, we apply a magnetic field parallel to the y axis with
strength Bo . Calculate the state of the system at some later time t in the {|±i} basis.
3. Plot as a function of time the expectation values fo the observables Sx , Sy , and Sz . What
are the values and probabilities? Is there a relation between Bo and t for the result of one
of the measurements to be certain? Give a physical interpretation of this condition.
4. Again, consider the same initial state, this time at t = 0, we measure Sy and find +h̄/2
What is the state vector |ψ(0+ )i immediately after this measurement?
5. Now we take |ψ(0+ )i and apply a uniform time-dependent field parallel to the z-axis. The
Hamiltonian operator of the spin is then given by
H(t) = ω(t)Sz
Assume that prior to t = 0, ω(t) = 0 and for t > 0 increases linearly from 0 to ωo at time
t = T . Show that for 0 ≤ t ≤ T , the state vector can be written as
1 |ψ(t)i = √ eiθ(t) |+i + ie−iθ(t) |−i
2
where θ(t) is a real function of t (which you need to determine).
6. Finally, at time t = τ > T , we measure Sy . What results can we find and with what
probabilities? Determine the relation which must exist between ωo and T in order for us to
be sure of the result. Give the physical interpretation.
135
Chapter 5
Perturbation theory
If you perturbate to much, you will go blind.
– T. A. Albright
In previous lectures, we discusses how , say through application of and external driving force,
the stationary states of a molecule or other quantum mechanical system can be come coupled
so that the system can make transitions from one state to another. We can write the transition
amplitude exactly as
G(i → j, t) = hj| exp(−iH(tj − ti ))/h̄)|ii
(5.1)
where H is the full Hamiltonian of the uncoupled system plus the applied perturbation. Thus, G
tells us the amplitude for the system prepared in state |ii at time ti and evolve under the applied
Hamiltonian for some time tj − ti and be found in state |ji. In general this is a complicated
quantity to calculate. Often, the coupling is very complex. In fact, we can only exactly determine
G for a few systems: linearly driven harmonic oscillators, coupled two level systems to name the
more important ones.
In today’s lecture and following lectures, we shall develop a series of well defined and systematic approximations which are widely used in all applications of quantum mechanics. We start
with a general solution of the time-independent Schrödinger equation in terms and eventually
expand the solution to infinite order. We will then look at what happens if we have a perturbation or coupling which depends explicitly upon time and derive perhaps the most important
rule in quantum mechanics which is called: “Fermi’s Golden Rule”. 1
5.1
Perturbation Theory
In most cases, it is simply impossible to obtain the exact solution to the Schrödinger equation.
In fact, the vast majority of problems which are of physical interest can not be resolved exactly
and one is forced to make a series of well posed approximations. The simplest approximation is
to say that the system we want to solve looks a lot like a much simpler system which we can
1
During a seminar, the speaker mentioned Fermi’s Golden Rule. Prof. Wenzel raised his arm and in German
spiked English chided the speaker that it was in fact HIS golden rule!
136
solve with some additional complexity (which hopefully is quite small). In other words we want
to be able to write our total Hamiltonian as
H = Ho + V
where Ho represents that part of the problem we can solve exactly and V some extra part which
we cannot. This we take as a correction or perturbation to the exact problem.
Perturbation theory can be formuated in a variery of ways, we begin with what is typically
termed Rayleigh-Schrödinger perturbation theory. This is the typical approach and used most
commonly. Let Ho |φn i = Wn |φn i and (Ho + λV )|ψi = En |ψi be the Schrödinger equations for
the uncoupled and perturbed systems. In what follows, we take λ as a small parameter and
expand the exact energy in terms of this parameter. Clearly, we write En as a function of λ and
write:
En (λ) = En(0) + λEn(1) + λ2 En(2) . . .
(5.2)
Likewise, we can expand the exact wavefunction in terms of λ
|ψn i = |ψn(0) i + λ|ψn(1) i + λ2 |ψn(2) i . . .
(5.3)
Since we require that |ψi be a solution of the exact Hamiltonian with energy En , then
H|ψi = (Ho + λV ) |ψn(0) i + λ|ψn(1) i + λ2 |ψn(2) i . . .
=
En(0) + λEn(1) + λ2 En(2) . . .
|ψn(0) i + λ|ψn(1) i + λ2 |ψn(2) i . . .
(5.4)
(5.5)
Now, we collect terms order by order in λ
• λ0 : Ho |ψn(0) i = En(0) |ψn(0) i
• λ1 : Ho |ψn(1) i + V |ψn(0) i = En(0) |ψ (1) i + En(1) |ψn(0) i
• λ2 : Ho |ψn(2) i + V |ψ (1) i = En(0) |ψn(2) i + En(1) |ψn(1) i + En(2) |ψn(0) i
and so on.
The λ0 problem is just the unperturbed problem we can solve. Taking the λ1 terms and
multiplying by hψn(0) | we obtain:
hψn(0) |Ho |ψn(0) i + hψn(0) |V |ψ (0) i = En(0) hψn(0) |ψn(1) i + En(1) hψn(0) |ψn(0) i
(5.6)
In other words, we obtain the 1st order correction for the nth eigenstate:
En(1) = hψn(0) |V |ψ (0) i.
Note to obtain this we assumed that hψn(1) |ψn(0) i = 0. This is easy to check by performing a
(0)
(0)
similar calculation, except by multiplying by hψm
| for m 6= n and noting that hψn(0) |ψm
i = 0 are
orthogonal state.
(0)
(0)
(0) (1)
hψm
|Ho |ψn(0) i + hψm
|V |ψ (0) i = En(0) hψm
|ψn i
137
(5.7)
Rearranging things a bit, one obtains an expression for the overlap between the unperturbed and
perturbed states:
(0) (1)
hψm
|ψn i =
(0)
hψm
|V |ψn(0) i
(5.8)
(0)
(0)
En − Em
Now, we use the resolution of the identity to project the perturbed state onto the unperturbed
states:
|ψn(1) i =
X
(0)
(0) (1)
|ψm
ihψm
|ψn i
m
(0)
X hψm
|V |ψn(0) i
=
(0)
m6=n En
−
(0)
Em
(0)
|ψm
i
(5.9)
where we explictly exclude the n = m term to avoid the singularity. Thus, the first-order
correction to the wavefunction is
|ψn i ≈ |ψn(0) i +
(0)
X hψm
|V |ψn(0) i
(0)
m6=n En
−
(0)
Em
(0)
|ψm
i.
(5.10)
This also justifies our assumption above.
5.2
Two level systems subject to a perturbation
Let’s say that in the |±i basis our total Hamiltonian is given by
H = ωSz + V Sx .
(5.11)
In matrix form:
H=
ω V
V −ω
!
Diagonalization of the matrix is easy, the eigenvalues are
√
E+ = ω 2 + V 2
√
E− = − ω 2 + V 2
(5.12)
(5.13)
(5.14)
We can also determine the eigenvectors:
|φ+ i = cos(θ/2)|+i + sin(θ/2)|−i
(5.15)
|φ− i = − sin(θ/2)|+i + cos(θ/2)|−i
(5.16)
where
|V |
(5.17)
ω
For constant coupling, the energy gap ω between the coupled states determines how the states
are mixed as the result of the coupling.
plot splitting as a function of unperturbed energy gap
tan θ =
138
5.2.1
Expansion of Energies in terms of the coupling
We can expand the exact equations for E± in terms of the coupling assuming that the coupling
is small compared to ω. To leading order in the coupling:
2
1 |V | E+ = ω(1 + · · ·)
2 ω (5.18)
2
1 |V | E− = ω(1 − · · ·)
2 ω (5.19)
On the otherhand, where the two unperturbed states are identical, we can not do this expansion and
E+ = |V |
(5.20)
E− = −|V |
(5.21)
and
We can do the same trick on the wavefunctions: When ω |V | (strong coupling) , θ ≈ π/2,
Thus,
1
|ψ+ i = √ (|+i + |−i)
2
(5.22)
1
|ψ− i = √ (−|+i + |−i).
2
(5.23)
In the weak coupling region, we have to first order in the coupling:
|ψ+ i = (|+i +
|V |
|−i)
ω
(5.24)
|ψ− i = (|−i +
|V |
|+i).
ω
(5.25)
In other words, in the weak coupling region, the perturbed states look a lot like the unperturbed
states. Where as in the regions of strong mixing they are a combination of the unperturbed
states.
139
5.2.2
Dipole molecule in homogenous electric field
Here we take the example of ammonia inversion in the presence of an electric field. From the
problem sets, we know that the N H3 molecule can tunnel between two equivalent C3v configurations and that as a result of the coupling between the two configurations, the unperturbed
energy levels Eo are split by an energy A. Defining the unperturbed states as |1i and |2i we can
define the tunneling Hamiltonian as:
H=
Eo −A
−A Eo
!
(5.26)
or in terms of Pauli matrices:
H = Eo σo − Aσx
Taking ψ to be the solution of the time-dependent Schrödinger equation
H|ψ(t)i = ih̄|ψ̇i
we can insert the identity |1ih1| + |2ih2| = 1 and re-write this as
ih̄ċ1 = Eo c1 − Ac2
ih̄ċ2 = Eo c2 − Ac1
(5.27)
(5.28)
where c1 = h1|ψi and c2 = h2|ψi. are the projections of the time-evolving wavefunction onto the
two basis states. Taking these last two equations and adding and subtracting them from each
other yields two new equations for the time-evolution:
ih̄ċ+ = (Eo − A)c+
ih̄ċ− = (Eo + A)c−
(5.29)
(5.30)
where c± = c1 ± c2 (we’ll normalize this later). These two new equations are easy to solve,
i
c± (t) = A± exp
(Eo ∓ A)t .
h̄
Thus,
1
c1 (t) = eiEo t/h̄ A+ e−iAt/h̄ + A− e+iAt/h̄
2
and
1
c2 (t) = eiEo t/h̄ A+ e−iAt/h̄ − A− e+iAt/h̄ .
2
Now we have to specify an initial condition. Let’s take c1 (0) = 1 and c2 (0) = 0 corresponding to
the system starting off in the |1i state. For this initial condition, A+ = A− = 1 and
c1 (t) = eiEo t/h̄ cos(At/h̄)
and
c2 (t) = eiEo t/h̄ sin(At/h̄).
140
So that the time evolution of the state vector is given by
|ψ(t)i = eiEo t/h̄ [|1i cos(At/h̄) + |2i sin(At/h̄)]
So, left alone, the molecule will oscillate between the two configurations at the tunneling frequency, A/h̄.
Now, we apply an electric field. When the dipole moment of the molecule is aligned parallel
with the field, the molecule is in a lower energy configuration, whereas for the anti-parrallel case,
the system is in a higher energy configuration. Denote the contribution to the Hamiltonian from
the electric field as:
H 0 = µe Eσz
The total Hamiltonian in the {|1i, |2i} basis is thus
H=
Eo + µe E
−A
−A
Eo − µ e E
!
(5.31)
Solving the eigenvalue problem:
|H − λI| = 0
we find two eigenvalues:
λ± = Eo ±
q
A2 + µ2e E 2 .
These are the exact eigenvalues.
In Fig. 5.1 we show the variation of the energy levels as a function of the field strength.
Figure 5.1: Variation of energy level splitting as a function of the applied field for an ammonia
molecule in an electric field
Weak field limit
If µe E/A 1, then we can use the binomial expansion
√
1 + x2 ≈ 1 + x2 /2 + . . .
to write
q
A2
+
µ2e E 2
!1
µe E 2
= A 1+
/2
A
!
1 µe E 2
≈ A 1+
2 A
(5.32)
Thus in the weak field limit, the system can still tunnel between configurations and the energy
splitting are given by
µ2 E 2
E± ≈ (Eo ∓ A) ∓ e
A
141
To understand this a bit further, let us use perturbation theory in which the tunneling
dominates and treat the external field as a perturbing force. The unperturbed hamiltonian can
be diagonalized by taking symmetric and anti-symmetric combinations of the |1i and |2i basis
functions. This is exactly what we did above with the time-dependent coefficients. Here the
stationary states are
1
|±i = √ (|1i ± |2i)
2
with energies E± = Eo ∓ A. So that in the |±i basis, the unperturbed Hamiltonian becomes:
Eo − A
0
0
Eo + A
H=
!
.
The first order correction to the ground state energy is given by
E (1) = E (0) + h+|H 0 |+i
To compute h+|H 0 |+i we need to transform H 0 from the {|1i, |2i} uncoupled basis to the new |±i
coupled basis. This is accomplished by inserting the identity on either side of H 0 and collecting
terms:
h+|H 0 |+i = h+|(|1i < 1| + |2ih2|)H 0 (|1i < 1| + |2ih2|)i
1
(h1| + h2|)H 0 (|1i + |2i)
=
2
= 0
(5.33)
(5.34)
(5.35)
Likewise for h−|H 0 |−i = 0. Thus, the first order correction vanish. However, since h+|H 0 |−i =
µe E does not vanish, we can use second order perturbation theory to find the energy correction.
(2)
W+
=
0
0
X Hmi
Him
m6=i
=
(5.36)
Ei − Em
h+|H 0 |−ih−|H 0 |+i
(0)
(0)
E+ − E−
(µe E)2
=
Eo − A − Eo − A
µ2 E 2
= − e
2A
(5.37)
(5.38)
(5.39)
(2)
Similarly for W− = +µ2e E 2 /A. So we get the same variation as we estimated above by expanding
the exact energy levels when the field was week.
Now let us examine the wavefunctions. Remember the first order correction to the eigenstates
is given by
h−|H 0 |−i
|−i
E+ − E−
µE
= − |−i
2A
|+(1) i =
142
(5.40)
(5.41)
Thus,
µE
|−i
2A
µE
|−i = |−(0) i +
|+i
2A
|+i = |+(0) i −
(5.42)
(5.43)
So we see that by turning on the field, we begin to mix the two tunneling states. However, since
we have assumed that µE/A 1, the final state is not too unlike our initial tunneling states.
Strong field limit
In the strong field limit, we expand the square-root term such that

q
A2 + µ2e E 2
A
= Eµe 
µe E
1
= Eµe 1 +
2
≈ Eµe +
A
µe E
2
1.
1/2
!2

+ 1
A
µe E
1 A2
2 µe E
!2 
...
(5.44)
For very strong fields, the first term dominates and the energy splitting becomes linear in the
field strength. In this limit, the tunneling has been effectively suppressed.
Let us analyze this limit using perturbation theory. Here we will work in the |1, 2i basis and
treat the tunneling as a perturbation. Since the electric field part of the Hamiltonian is diagonal
in the 1,2 basis, our unperturbed strong-field hamiltonian is simply
H=
Eo − µe E
0
0
Eo − µe E
!
(5.45)
and the perturbation is the tunneling component. As before, the first-order corrections to the
energy vanish and we are forced to resort to 2nd order perturbation theory to get the lowest
order energy correction. The results are
W (2) = ±
A2
2µE
which is exactly what we obtained by expanding the exact eigenenergies above. Likewise, the
lowest-order correction to the state-vectors are
A 0
|2 i
2µE
A 0
|2i = |20 i +
|1 i
2µE
|1i = |10 i −
(5.46)
(5.47)
So, for large E the second order correction to the energy vanishes, the correction to the wavefunction vanishes and we are left with the unperturbed (i.e. non-tunneling) states.
143
5.3
Dyson Expansion of the Schrödinger Equation
The Rayleigh-Schrödinger approach is useful for discrete spectra. However, it is not very useful
for scattering or systems with continuous spectra. On the otherhand, the Dyson expansion of the
wavefunction can be applied to both cases. Its development is similar to the Rayleigh-Schrödinger
case, We begin by writing the Schrödinger Equation as usual:
(Ho + V )|ψi = E|ψi
(5.48)
where we define |φi and W to be the eigenvectors and eigenvalues of part of the full problem.
We shall call this the “uncoupled” problem and assume it is something we can easily solve.
Ho |φi = W |φi
(5.49)
We want to write the solution of the fully coupled problem in terms of the solution of the
uncoupled problem. First we note that
(Ho − E)|ψi = V |ψi.
(5.50)
Using the “uncoupled problem” as a “homogeneous” solution and the coupling as an inhomogeneous term, we can solve the Schrödinger equation and obtain |ψi EXACTLY as
|ψi = |φi +
1
V |ψi
Ho − E
(5.51)
This may seem a bit circular. But we can iterate the solution:
|ψi = |φi +
1
1
1
V |φi +
V
V |ψi.
Ho − E
Ho − E Ho − W
(5.52)
Or, out to all orders:
|ψi = |φi +
∞ X
n=1
1
V
Ho − E
n
|φi
(5.53)
Assuming that the series converges rapidly (true for V << Ho weak coupling case), we can
truncate the series at various orders and write:
|ψ (0) i = |φi
(5.54)
1
|ψ (1) i = |φi +
V |φi
Ho − E
2
1
|ψ (2) i = |ψ (1) i +
V |φi
Ho − E
and so on. Let’s look at |ψ (1) i for a moment. We can insert 1 in the form of
|ψn(1) i
= |φn i +
X
n
1
|φm ihφn |V |φm i
Ho − W m
(5.55)
(5.56)
P
n
|φn ihφn |
144
(5.57)
i.e.
|ψn(1) i
= |φn i +
X
n
1
|φm ihφn |V |φm i
Wn − Wm
(5.58)
Likewise:
|ψn(2) i
=
|ψn(1) i
+
X
lm
1
(Wm − Wl )(Wn − Wm )
!2
Vlm Vmn |φn i
(5.59)
where
Vlm = hφl |V |φm i
(5.60)
is the matrix element of the coupling in the uncoupled basis. These last two expressions are the
first and second order corrections to the wavefunction.
Note two things. First that I can actually solve the perturbation series exactly by noting that
the series has the form of a geometric progression, for x < 1 converge uniformly to
∞
X
1
2
= 1 + x + x + ··· =
xn
1−x
n=0
(5.61)
Thus, I can write
∞ X
|ψi =
n=0
∞
X
=
1
V
Ho − E
n
|φi
(Go V )n |φi
(5.62)
(5.63)
n=0
1
|φi
1 − Go V
=
(5.64)
where Go = (Ho −E)−1 (This is the “time-independent” form of the propagator for the uncoupled
system). This particular analysis is particularly powerful in deriving the propagator for the fully
coupled problem.
We now calculate the first order and second order corrections to the energy of the system.
To do so, we make use of the wavefunctions we just derived and write
En(1) = hψn(0) |H|ψn(0) i = Wn + hφn |V |φn i = Wn + Vnn
(5.65)
So the lowest order correction to the energy is simply the matrix element of the perturbation
in the uncoupled or unperturbed basis. That was easy. What about the next order correction.
Same procedure as before: (assuming the states are normalized)
En(2) = hψn(1) |H|ψn(1) i
= hφn |H|φn i
1
hφm |V |φn i + O[V 3 ]
+
hφn |H|φm i
W
−
W
n
m
m6=n
X
= Wn + Vnn +
|Vnm |2
m6=n Wn − Wm
X
145
(5.66)
Notice that I am avoiding the case where m = n as that would cause the denominator to
be zero leading to an infinity. This must be avoided. The so called “degenerate case” must
be handled via explicit matrix diagonalization. Closed forms can be obtained for the doubly
degenerate case easily.
Also note that the successive approximations to the energy require one less level of approximation to the wavefunction. Thus, second-order energy corrections are obtained from first order
wavefunctions.
5.4
5.4.1
Van der Waals forces
Origin of long-ranged attractions between atoms and molecules
One of the underlyuing principles in chemistry is that molecules at long range are attractive
towards each other. This is clearly true for polar and oppositely charges species. It is also true
for non-polar and neutral species, such as methane, noble gases, and etc. These forces are due to
polarization forces or van der Waals forces, which is is attractive and decreases as 1/R7 , i.e. the
attractive part of the potential goes as −1/R6 . In this section we will use perturbation theory
to understand the origins of this force, restricting our attention to the interaction between two
hydrogen atoms separated by some distance R.
Let us take the two atoms to be motionless and separated by distance R with ~n being the
vector pointing from atom A to atom B. Now let ~ra be the vector connecting nuclei A to its
electron and likewise for ~rB . Thus each atom has an instantaneous electric dipole moment
~a
~µa = q R
~ b.
~µb = q R
(5.67)
(5.68)
(5.69)
We will assume that R ra & rb so that the electronic orbitals on each atom do not come into
contact.
Atom A creates an electrostatic potential, U , for atom B in which the charges in B can
interact. This creates an interaction energy W . Since both atoms are neutral, the most important
source for the interactions will come from the dipole-dipole interactions. This, the dipole of A
interacts with an electric field E = −∇U generated by the dipole field about B and vice versa. To
calculate the dipole-dipole interaction, we start with the expression for the electrostatic potential
created by µa at B.
1 µa · R
U (R) =
4πo R3
Thus,
~ = −∇U = − q 1 (~ra − 3(~ra · ~n)~n) .
E
4πo R3
Thus the dipole-dipole interaction energy is
~
W = −~µb · E
2
e
(~ra · ~rb − 3(~ra · ~n)(~rb · ~n))
=
R3
146
(5.70)
where e2 = q 2 /4πo . Now, let’s set the z axis to be along ~n so we can write
W =
e2
(xa xb + ya yb − 2za zb ).
R3
This will be our perturbing potential which we add to the total Hamiltonian:
H = Ha + Hb + W
where Ha are the unperturbed Hamiltonians for the atoms. Let’s take for example wo hydrogens
each in the 1s state. The unperturbed system has energy
H|1s1 ; 1s2 i = (E1 + E2 )|1s1 ; 1s2 i = −2EI |1s1 ; 1s2 i,
where EI is the ionization energy of the hydrogen 1s state (EI = 13.6eV). The first order vanishes
since it involves integrals over odd functions. This we can anticipate since the 1s orbitals are
spatially isotropic, so the time averaged valie of the dipole moments is zero. So, we have to look
towards second order corrections.
The second order energy correction is
E
(2)
=
XX
nlm
|hnlm; n0 l0 m0 |W |1sa ; 1sb i|2
nlm =
−2EI − En − En0
0 0
0
where we restrict the summation to avoid the |1sa ; 1ab i state. Since W ∝ 1/R3 and the deminator
is negative, we can write
C
E (2) = − 6
R
6
which explains the origin of the 1/R attraction.
Now we evaluate the proportionality constant C Written explicitly,
4
C=e
X X |hnlm0 n0 l0 m0 |(xa xb + ya yb − 2za zb )|1sa ; 1sb i|2
2EI + En + En0
nml n0 l0 m0
(5.71)
Since n and n0 ≥ 2 and |En | = EI /n2 < EI , we can replace En and En0 with 0 with out
appreciable error. Now, we can use the resolution of the identity
1=
X X
|nlm; n0 l0 m0 ihnlm; n0 l0 m0 |
nml n0 l0 m0
to remove the summation and we get
C=
e4
h1sa ; 1ab |(xa xb + ya yb − 2za zb )2 |1sa ; 1sb i
2EI
(5.72)
where EI is the ionization potential of the 1s state (EI = 1/2). Surprisingly, this is simple
to evaluate since we can use symmetry to our advantage. Since the 1s orbitals are spherically
symmetric, any term involving cross-terms of the sort
h1sa |xa ya |1si = 0
147
vanish. This leaves only terms of the sort
h1s|x2 |1si.
all of which are equal to 1/3 of the mean value of RA = x2a + ya2 + za2 . Thus,
e2
C=6
2EI
2
R
h1s| |1si = 6e2 ao
3
where ao is the Bohr radius. Thus,
ao
R6
What does all this mean. We stated at the beginning that the average dipole moment of a H
1s atom is zero. That does not mean that every single measurement of µa will yield zero. What is
means is that the probability of finding the atom with a dipole moment µa is the same for finding
the dipole vector pointed in the opposite direction. Adding the two together produces a net zero
dipole moment. So its the fluctuations about the mean which give the atom an instantaneous
dipole field. Moreover, the fluctuations in A are independent of the fluctuations in B, so first
order effects must be zero since the average interaction is zero.
Just because the fluctuations are independent does not mean they are not correlated. Consider
the field generated by A as felt by B. This field is due to the fluctuating dipole at A. This field
induces a dipole at B. This dipole field is in turn felt by A. As a result the fluctuations become
correlated and explains why this is a second order effect. In a sense, A interacts with is own
dipole field through “reflection” off B.
E (2) = −6e2
5.4.2
Attraction between an atom a conducting surface
The interaction between an atom or molecule and a surface is a fundimental physical process
in surface chemistry. In this example, we will use perturbation theory to understand the longranged attraction between an atom, again taking a H 1s atom as our species for simplicity, and
a conducting surface. We will take the z axis to be normal to the surface and assume that the
atom is high enough off the surface that its altitude is much larger than atomic dimensions.
Furthermore, we will assume that the surface is a metal conductor and we will ignore any atomic
level of detail in the surface. Consequently, the atom can only interact with its dipole image on
the opposite side of the surface.
We can use the same dipole-dipole interaction as before with the following substitutions
e2
R
xb
yb
zb
−→
−→
−→
−→
−→
−e2
2d
x0a = xa
ya0 = ya
za0 = −za
where the sign change reflects the sign difference in the image charges. So we get
e2 2
W = − 3 (xa + ya2 + 2za2 )
8d
148
(5.73)
(5.74)
(5.75)
(5.76)
(5.77)
as the interaction between a dipole and its image. Taking the atom to be in the 1s ground state,
the first order term is non-zero:
E (1) = h1s|W |1si.
Again, using spherical symmetry to our advantage:
E (1) = −
e2
e2 a2o
2
4h1s|r
|1si
=
−
.
8d3
2d3
Thus an atom is attracted to the wall with an interaction energy which varries as 1/d3 . This is
a first order effect since there is perfect correlation between the two dipoles.
5.5
Perturbations Acting over a Finite amount of Time
Perhaps the most important application of perturbation theory is in cases in which the coupling
acts for a finite amount of time. Such as the coupling of a molecule to a laser field. The laser field
impinges upon a molecule at some instant in time and some time later is turned off. Alternatively
we can consider cases in which the perturbation is slowly ramped up from 0 to a final value.
5.5.1
General form of time-dependent perturbation theory
In general, we can find the time-evolution of the coefficients by solving
ih̄ċn (t) = En cn (t) +
X
λWnk (t)ck (t)
(5.78)
k
where λWnk are the matrix elements of the perturbation. Now, let’s write the cn (t) as
cn (t) = bn (t)e−iEn t
and assume that bn (t) changes slowly in time. Thus, we can write a set of new equations for the
bn (t) as
ih̄b˙n =
X
eiωnk t λWnk bk (t)
(5.79)
k
Now we assume that the bn (t) can be written as a perturbation expansion
(1)
2 (2)
bn (t) = b(0)
n (t) + λbn (t) + λ bn (t) + · · ·
where as before λ is some dimensionless number of order unity. Taking its time derivative and
equating powers of λ one finds
ih̄ḃ(i)
n =
X
(i−1)
eiωnk t Wnk (t)bk
k
and that b(0)
n (t) = 0.
149
Now, we calculate the first order solution. For t < 0 the system is assumed to be in some
well defined initial state, |φi i. Thus, only one bn (t < 0) coefficient can be non-zero and must be
independent of time since the coupling has not been turned on. Thus,
bn (t = 0) = δni
At t = 0, we turn on the coupling and λW jumps from 0 to λW (0). This must hold for all order
in λ. So, we immediately get
b(0)
n (0) = δni
b(i)
n (0) = 0
(5.80)
(5.81)
Consequently, for all t > 0, b(0)
n (t) = δni which completely specifies the zeroth order result. This
also gives us the first order result.
ih̄ḃ(1)
n (t) =
X
eiωnk t Wnk δki
k
iωni t
= e
Wni (t)
(5.82)
which is simple to integrate
i Z t iωni s
=−
e
Wni (s)ds.
h̄ 0
Thus, our perturbation wavefunction is written as
b(1)
n (t)
|ψ(t)i = e−iEo t/h̄ |φo i +
X
−iEn t/h̄
λb(1)
|φn i
n (t)e
(5.83)
n6=0
5.5.2
Fermi’s Golden Rule
Let’s consider the time evolution of a state under a small perturbation which varies very slowly
in time. The expansion coefficients of the state in some basis of eigenstates of the unperturbed
Hamiltonian evolve according to:
ih̄ċs (t) =
X
Hsn cn (t)
(5.84)
n
where Hsn is the matrix element of the full Hamiltonian in the basis.
Hsn = Es δns + Vs n(t)
(5.85)
Assuming that V (t) is slowly varying and that Vsn << Es − En for all time. we can write the
approximate solution as
cs (t) = As (t) exp(−i/h̄Es t)
(5.86)
where As (t) is a function with explicit time dependence. Putting the approximate solution back
into the differential equation:
ih̄ċs (t) = ih̄Ȧs (t)cs (t) + Es cs (t)
150
(5.87)
(5.88)
= Es cs (t) +
X
Vsn An (t)e−i/h̄En t
(5.89)
n
We now proceed to solve this equation via series of well defined approximations. Our first
assumption is that Vsn << En − Es (weak coupling approximation). We can also write
Ȧs << 1
(5.90)
since we have assumed that A(t) varies slowly in time. For s 6= i we have the following initial
conditions:
Ai (0) = 1
(5.91)
As (0) = 0
(5.92)
Thus, at t = 0 we can set all the coefficients in zero except for Ai , which is 1. Thus,
Ȧs (t) = −i/h̄Vsi e−i/h̄(Es −Ei )t
(5.93)
which can be easily integrated
As (t) = −
iZt 0
0
dt Vsi e−i/h̄(Es −Ei )t
h̄ 0
(5.94)
where all the As (t) are assumed to be smaller than unity. Of course, to do the integral we need
to know how the perturbation depends upon time. Let’s assume that
V (t) = 2V̂ cos(ωt) = V̂ e+iωt + e−iωt
(5.95)
where V̂ is a time independent quantity. Thus we can determine A(t) as
h
i
iZt 0
0
0
dt hs|V̂ |ıi ei/h̄(Es −Ei +h̄ω)t ei/h̄(Es −Ei −h̄ω)t
As (t) = −
h̄ 0
(5.96)
(5.97)
1 − ei/h̄(Es −Ei +h̄ω)t 1 − ei/h̄(Es −Ei −h̄ω)t
= hs|V̂ |ıi
+
Es − En + h̄ω
Es − En − h̄ω
"
#
(5.98)
Since the coupling matrix element is presumed to be small. The only significant contribution
comes when the denominator is very close to zero. i.e. when
h̄ω ≈ |Es − En |
(5.99)
For the case when Es > En we get only one term making a significant contribution and thus the
transition probability as a function of time is
Psn (t) = |cs (t)|2 = |As (t)|2
151
(5.100)
(5.101)
"
2
= 4|hs|V̂ |ni|
sin2 ((Es − En − h̄ω)t/(2h̄))
.
(Es − En − h̄ω)2
#
(5.102)
This is the general form for a harmonic perturbation. The function
sin2 (ax)
x2
(5.103)
is called the sinc function or the “Mexican Hat Function”. At x = 0 the peak is sharply peaked
(corresponding to Es − En − h̄ω = 0). Thus, the transition is only significant when the energy
difference matches the frequency of the applied perturbation. As t → ∞ (a = t/(2h̄)), the peak
become very sharp and becomes a δ-function. The width of the peak is
∆x =
2πh̄
t
(5.104)
Thus, the longer we measure a transition, the more well resolved the measurement will become.
This has profound implications for making measurements of meta-stable systems.
We have an expression for the transition probability between two discrete states. We have
not taken into account the the fact that there may be more than one state close by nor have we
taken into account the finite “width” of the perturbation (instrument function). When there a
many states close the the transition, I must take into account the density of nearby states. Thus,
we define
ρ(E) =
∂N (E)
∂E
(5.105)
as the density of states close to energy E where N (E) is the number of states with energy E.
Thus, the transition probability to any state other than my original state is
Ps (t) =
X
Pns (t) =
X
s
|As (t)|2
(5.106)
s
to go from a discrete sum to an integral, we replace
X
→
Z
dN
dE = ρ(E)dE
dE
Z
n
(5.107)
Thus,
Ps (t) =
Z
dE|As (t)|2 ρ(E)
(5.108)
Since |As (t)|2 is peaked sharply at Es = Ei + h̄ω we can treat ρ(E) and V̂sn as constant and write
Ps (t) = 4|hs|V̂ |ni|2 ρ(Es )
Z
sin2 (ax)
dx
x2
(5.109)
Taking the limits of integration to ±∞
Z
∞
−∞
dx
sin2 (ax)
πt
= πa =
x
2h̄
152
(5.110)
In other words:
Ps (t) =
2πt
|hs|V̂ |ni|2 ρ(Es )
h̄
(5.111)
We can also define a transition rate as
R(t) =
P (t)
t
(5.112)
thus, the Golden Rule Transition Rate from state n to state s is
Rn→s (t) =
2π
|hs|V̂ |ni|2 ρ(Es )
h̄
(5.113)
This is perhaps one of the most important approximations in quantum mechanics in that it
has implications and applications in all areas, especially spectroscopy and other applications of
matter interacting with electro-magnetic radiation.
5.6
Interaction between an atom and light
What I am going to tell you about is what we teach our physics students in the third
or fourth year of graduate school... It is my task to convince you not to turn away
because you don’t understand it. You see my physics students don’t understand it...
That is because I don’t understand it. Nobody does.
Richard P. Feynman, QED, The Strange Theory of Light and Matter
Here we explore the basis of spectroscopy. We will consider how an atom interacts with a
photon field in the low intensity limit in which dipole interactions are important. We will then
examine non-resonant excitation and discuss the concept of oscillator strength. Finally we will
look at resonant emission and absorption concluding with a discussion of spontaneous emission.
In the next section, we will look at non-linear interactions.
5.6.1
Fields and potentials of a light wave
An electromagnetic wave consists of two oscillating vector field components which are perpendicular to each other and oscillate at at an angular frequency ω = ck where k is the magnitude
of the wavevector which points in the direction of propagation and c is the speed of light. For
such a wave, we can always set the scalar part of its potential to zero with a suitable choice in
~ given by
gauge and describe the fields associate with the wave in terms of a vector potential, A
~ t) = Ao ez eiky−iωt + A∗ ez e−iky+iωt
A(r,
o
Here, the wave-vector points in the +y direction, the electric field, E is polarized in the yz plane
and the magnetic field B is in the xy plane. Using Maxwell’s relations
~ t) = − ∂A = iωez (Ao ei(ky−ωt) − A∗ e−i(ky−ωt) )
E(r,
o
∂t
153
and
~ t) = ∇ × A
~ = ikex (Ao ei(ky−ωt) − A∗ e−i(ky−ωt) ).
B(r,
o
We are free to choose the time origin, so we will choos it as to make Ao purely imaginary and
set
iωAo = E/2
ikAo = B/2
(5.114)
(5.115)
where E and B are real quantities such that
E
ω
= = c.
B
k
Thus
E(r, t) = Eez cos(ky − ωt)
B(r, t) = Bez sin(ky − ωt)
(5.116)
(5.117)
where E and B are the magnitudes of the electric and magnetic field components of the plane
wave.
Lastly, we define what is known as the Poynting vector (yes, it’s pronounced pointing) which
is parallel to the direction of propagation:
~ = o c2 E
~ × B.
~
S
(5.118)
~ and B
~ above and averaging over several oscillation periods:
Using the expressions for E
~ = o c2 E ey
S
2
5.6.2
(5.119)
Interactions at Low Light Intensity
The electromagnetic wave we just discussed can interact with an atomic electron. The Hamiltonian of this electron can be given by
H=
q
1
(P − qA(r, t))2 + V (r) − S · B(r, t)
2m
m
where the first term represents the interaction between the electron and the electrical field of
the wave and the last term represents the interaction between the magnetic moment of the
electron and the magnetic moment of the wave. In expanding the kinetic energy term, we have
to remember that momentum and position do not commute. However, in the present case, A is
parallel to the z axis and Pz and y commute. So, we wind up with the following:
H = Ho + W
where
P2
Ho =
+ V (r)
2m
154
is the unperturbed (atomic) hamiltonian and
W =−
q
q
q2 2
P·A− S·B+
A.
m
m
2m
The first two depend linearly upon A and the second is quadratic in A. So, for low intensity we
can take
q
q
W = − P · A − S · B.
m
m
Before moving on, we will evaluate the relative importance of each term by orders of magnitude
for transitions between bound states. In the second term, the contribution of the spin operator
is on the order of h̄ and the contribution from B is on the order of kA. Thus,
WB
=
WE
q
S·B
m
q
P·A
m
≈
h̄k
p
h̄/p is on the order of an atomic radius, ao and k = 2π/λ where λ is the wavelength of the light,
typically on the order of 1000ao . Thus,
WB
ao
≈
1.
WE
λ
So, the magnetic coupling is not at all important and we focus only upon the coupling to the
electric field.
Using the expressions we derived previously, the coupling to the electric field component of
the light wave is given by:
WE = −
q
pz (Ao eiky e−iωt + A∗o e−iky e+iωt ).
m
Now, we expand the exponential in powers of y
1
e±iky = 1 ± iky − k 2 y 2 + . . .
2
since ky ≈ ao /λ 1, we can to a good approximation keep only the first term. Thus we get the
dipole operator
qE
WD =
pz sin(ωt).
mω
In the electric dipole approximation, W (t) = WD (t).
Note, that one might expect that WD should have been written as
WD = −qEz cos(ωt)
since we are, after all, talking about a dipole moment associated with the motion of the electron
about the nucleus. Actually, the two expressions are identical! The reason is that I can always
choose a differnent gauge to represent the physical problem without changing the physical result.
To get the present result, we used
E
A = ez sin(ωt)
ω
155
and
U (r) = 0
as the scalar potential. A gauge transformation is introduced by taking a function, f and defining
a new vector potential and a new scalar potential as
A0 = A + ∇f
∂f
∂t
We are free to choose f however we do so desire. Let’s take f = zE sin(ωt)/ω. Thus,
U0 = U −
E
A0 = ez (sin(ky − ωt) + sin(ωt))
ω
and
U 0 = −zE cos ωt
is the new scalar potential. In the electric dipole approximation, ky is small, so we set ky = 0
everywhere and obtain A0 = 0. Thus, the total Hamiltonian becomes
H = Ho + qU 0 (r, t)
with perturbation
WD0 = −qzE cos(ωt).
This is the usual form of the dipole coupling operator. However, when we do the gauge transformation, we have to transform the state vector as well.
Next, let us consider the matrix elements of the dipole operator between two stationary states
of Ho : |ψi i and |ψf i with eigenenergy Ei and Ef respectively. The matrix elements of WD are
given by
qE
sin(ωt)hψf |pz |ψi i
Wf i (t) =
mω
We can evaluate this by noting that
[z, Ho ] = ih̄
∂Ho
pz
= ih̄ .
∂pz
m
Thus,
hψf |pz |ψi i = imωf i hψf |z|ψi i.
Consequently,
sin(ωt)
zf i .
ω
Thus, the matrix elements of the dipole operator are those of the position operator. This determines the selection rules for the transition.
Before going through any specific details, let us consider what happens if the frequency ω
does not coincide with ωf i . Soecifically, we limit ourselves to transitions originating from the
ground state of the system, |ψo i. We will assume that the field is weak and that in the field the
Wf i (t) = iqEωf i
156
atom acquires a time-dependent dipole moment which oscillates at the same frequency as the
field via a forced oscillation. To simplify matters, let’s assume that the electron is harmonically
bound to the nucleus in a classical potential
1
V (r) = mωo r2
2
where ωo is the natural frequency of the electron.
The classical motion of the electron is given by the equations of motion (via the Ehrenfest
theorem)
qE
z̈ + ω 2 z =
cos(ωt).
m
This is the equation of motion for a harmonic oscillator subject to a periodic force. This inhomogeneous differential equation can be solved (using Fourier transform methods) and the result
is
qE
cos(ωt)
z(t) = A cos(ωo t − φ) +
2
m(ωo − ω 2 )
where the first term represents the harmonic motion of the electron in the absence of the driving
force. The two coefficients, A and φ are determined by the initial condition. If we have a very
slight damping of the natural motion, the first term dissappears after a while leaving only the
second, forced oscillation, so we write
qE
cos(ωt).
z=
2
m(ωo − ω 2 )
Thus, we can write the classical induced electric dipole moment of the atom in the field as
q2E
cos(ωt).
m(ωo2 − ω 2 )
Typically this is written in terms of a susceptibility, χ, where
D = qz =
q2
.
m(ωo2 − ω 2 )
Now we look at this from a quantum mechanical point of view. Again, take the initial state
to be the ground state and H = Ho + WD as the Hamiltonian. Since the time-evolved state can
be written as a superposition of eigenstates of Ho ,
χ=
|ψ(t)i =
X
cn (t)|φn i
n
To evaluate this we can us the results derived previously in our derivation of the golden rule,
X qE
|ψ(t)i = |φo i +
hn|pz |φo i
n6=0 2imh̄ω
(
×
e−iωno t − eiωt e−iωno t − e−iωt
−
|φn i
ωno + ω
ωno − ω
)
(5.120)
where we have removed a common phase factor. We can then calculate the dipole moment
expectation value, hD(t)i as
hD(t)i =
X ωon |hφn |z|φo i|2
2q 2
E cos(ωt)
2 − ω2
h̄
ωno
n
157
(5.121)
Oscillator Strength
We can now notice the similarity between a driven harmonic oscillator and the expectation value
of the dipole moment of an atom in an electric field. We can define the oscillator strength as a
dimensionless and real number characterizing the transition between |φo and |φn i
fno =
2mωno
|hφn |z|φo i|2
h̄
In Exercise 2.4, we proved the Thomas-Reiche-Kuhn sum rule, which we can write in terms of
the oscillator strengths,
X
fno = 1
n
This can be written in a very compact form:
m
hφo |[x, [H, x]]|φo i = 1.
h̄2
5.6.3
Photoionization of Hydrogen 1s
Up until now we have considered transitions between discrete states inrtoduced via some external
perturbation. Here we consider the single photon photoionization of the hydrogen 1s orbital to
illustrate how the golden rule formalism can be used to calculate photoionization cross-sections
as a function of the photon-frequency. We already have an expression for dipole coupling:
WD =
qE
pz sin(ωt)
mω
(5.122)
and we have derived the golden rule rate for transitions between states:
Rif =
2π
|hf |V |ii|2 δ(Ei − Ef + h̄ω).
h̄
(5.123)
For transitions to the continuum, the final states are the plane-waves.
ψ(k) =
1 ik·r
e .
Ω1/2
(5.124)
where Ω is the volume element. Thus the matrix element h1s|V |ki can be written as
h̄kz Z
h1s|pz |ki = 1/2 ψ1s (r)eik·r dr.
Ω
(5.125)
To evaluate the integral, we need to transform the plane-wave function in to spherical coordinates.
This can be done vie the expansion;
eik·r =
X
il (2l + 1)jl (kr)Pl (cos(θ))
(5.126)
l
where jl (kr) is the spherical Bessel function and Pl (x) is a Legendre polynomial, which we can
also write as a spherical harmonic function,
s
Pl (cos(θ)) =
4π
Yl0 (θ, φ).
2l + 1
158
(5.127)
Thus, the integral we need to perform is
1 X
h1s|ki = √
πΩ l
Z
q
Y00∗ Yl0 dΩ il 4π(2l + 1)
Z
0
∞
r2 e−r jl (kr)dr.
(5.128)
The angular integral we do by orthogonality and produces a delta-function which restricts the
sum to l = 0 only leaving,
1 Z ∞ 2 −r
h1s|ki = √
r e j0 (kr)dr.
Ω 0
(5.129)
The radial integral can be easily performed using
sin(kr)
kr
(5.130)
4 1
1
.
1/2
k Ω (1 + k 2 )2
(5.131)
qEh̄ 1
2
mω Ω1/2 (1 + k 2 )2
(5.132)
j0 (kr) =
leaving
h1s|ki =
Thus, the matrix element is given by
h1s|V |ki =
This we can indert directly in to the golden rule formula to get the photoionization rate to a
given k-state.
2πh̄ qE
Ω mω
R0k =
2
4
δ(Eo − Ek + h̄ω).
(1 + k 2 )4
(5.133)
which we can manipulate into reading as
R0k
16π
=
h̄Ω
qE
mω
2
δ(k 2 − K 2 )
m
(1 + k 2 )4
(5.134)
where we write K 2 = 2m(EI + h̄ω)/h̄2 to make our notation a bit more compact. Eventually,
we want to know the rate as a function of the photon frequency, so let’s put everything except
the frequency and the volume element into a single constant, I, which is related to the intensity
of the incident photon.
R0k =
I 1 δ(k 2 − K 2 )
.
Ω ω 2 (1 + k 2 )4
(5.135)
Now, we sum over all possible final states to get the total photoionization rate. To do this, we
need to turn the sum over final states into an integral, this is done by
X
k
=
Z ∞
Ω
4π
k 2 dk
(2π)3
0
159
(5.136)
Thus,
Z ∞
2
2
I 1 Ω
2 δ(k − K )
R =
4π
k
dk
Ω ω 2 (2π)3
(1 + k 2 )4
0
I 1 Z ∞ 2 δ(k 2 − K 2 )
=
k
dk
ω 2 2π 2 0
(1 + k 2 )2
Now we do a change of variables: y = k 2 and dy = 2kdk so that the integral become
Z
0
∞
− K 2)
1 Z ∞ y 1/2
k
dk =
δ(y − K 2 )dy
2
2
2
4
(1 + k )
2 0 (1 + y )
K
=
2(1 + K 2 )4
2 δ(k
2
(5.137)
Pulling everything together, we see that the total photoionization rate is given by
I 1
K
2
2
ω 2π (1 + K 2 )4
q √
I h̄m2 ω h̄ − o
= √
4
2 π 2 ω 2 1 + 2 m (ωh̄2h̄−o )
√
2ω − 1
= I
32 π 2 ω 6
R =
(5.138)
where in the last line we have converted to atomic units to clean things up a bit. This expression
is clearly valid only when h̄ω > EI = 1/2hartree (13.6 eV) and a plot of the photo-ionization
rate is given in Fig. 5.2
5.6.4
Spontaneous Emission of Light
The emission and absorption of light by an atom or molecule is perhaps the most spectacular
and important phenomena in the universe. It happens when an atom or molecule undergoes a
transition from one state to another due to its interaction with the electro-magnetic field. Because
the electron-magnetic field can not be entirely eliminated from any so called isolated system
(except for certain quantum confinement experiments), no atom or molecule is ever really isolated.
Thus, even in the absence of an explicitly applied field, an excited system can spontaneously emit
a photon and relax to a lower energy state. Since we have all done spectroscopy experiments
at one point in our education or another, we all know that the transitions are between discrete
energy levels. In fact, it was in the examination of light passing through glass and light emitted
from flames that people in the 19th century began to speculate that atoms can absorb and emit
light only at specific wavelengths.
We will use the GR to deduce the probability of a transition under the influence of an applied
light field (laser, or otherwise). We will argue that the system is in equilibrium with the electromagnetic field and that the laser drives the system out of equilibrium. From this we can deduce
the rate of spontaneous emission in the absence of the field.
160
R HarbL
0.8
0.6
0.4
0.2
0.5
1
1.5
2
Ñw HauL
Figure 5.2: Photo-ionization spectrum for hydrogen atom.
The electric field associated with a monochromatic light wave of average intensity I is
hIi = chρi

(5.139)

hE~o2 i
1 hB2o i 

= c o
+
2
µo 2
=
o
µo
!1/2
= co
Eo2
2
Eo2
2
(5.140)
(5.141)
(5.142)
~ and |Bo | = (1/c)|E|
~ are the maximum amplitudes
where ρ is the energy density of the field, |E|
of the E and B fields of the wave. Units are MKS units.
The em wave in reality contains a spread of frequencies, so we must also specify the intensity
density over a definite frequency interval:
dI
dω = cu(ω)dω
dω
(5.143)
where u(ω) is the energy density per unit frequency at ω.
Within the “semi-classical” dipole approximation, the coupling between a molecule and the
light wave is
~ = ~µ · ~ Eo cos(ωt)
~µ · E(t)
2
161
(5.144)
where ~µ is the dipole moment vector, and ~ is the polarization vector of the wave. Using this
result, we can go back to last week’s lecture and plug directly into the GR and deduce that
Pf i (ω, t) = 4|hf |~µ · ~|ii|2
Eo2 sin2 ((Ef − Ei − h̄ω)t/(2h̄))
4
(Ef − Ei − h̄ω)2
(5.145)
Now, we can take into account the spread of frequencies of the em wave around the resonant
value of ωo = (Ef − Ei )/h̄. To do this we note:
Eo2 = 2
hIi
co
(5.146)
and replace hIi with (dI/dω)dω.
2
=
co
dI
dω
0
!
2
|hf |~µ · ~|ii|
ωo
∞
Z
Pf i (t) =
Z
0
dωPf i (t, ω)
∞
(5.147)
sin2 ((h̄ωo − h̄ω)(t/(2h̄)))
dω
(h̄ωo − h̄ω)2
(5.148)
To get this we assume that dI/dω and the matrix element of the coupling vary slowly with
frequency as compared to the sin2 (x)/x2 term. Thus, as far as doing integrals are concerned,
they are both constants. With ωo so fixed, we can do the integral over dw and get πt/(2h̄2 ). and
we obtain the GR transition rate:
π
dI
µ · ~|ii|2
kf i =
2 |hf |~
dω
co h̄
!
(5.149)
ωo
Notice also that this equation predicts that the rate for excitation is identical to the rate for
de-excitation. This is because the radiation field contains both a +ω and −ω term (unless the
field is circularly polarized), this the transition rate to a state of lower energy to a higher energy
is the same as that of the transition from a higher energy state to a lower energy state.
However, we know that systems can emit spontaneously in which a state of higher energy
can go to a state of lower energy in the absence of an external field. This is difficult to explain
in the presence frame-work since we have assumed that |ii is stationary.
Let’s assume that we have an ensemble of atoms in a cavity containing em radiation and the
system is in thermodynamic equilibrium. (Thought you could escape thermodynamics, eh?) Let
E1 and E2 be the energies of two states of the atom with E2 > E1 . When equilibrium has been
established the number of atoms in the two states is determined by the Boltzmann equation:
N2
N e−E2 β
=
= e−β(E2 −E1 )
−E
β
1
N1
Ne
(5.150)
where β = 1/kT The number of atoms (per unit time) undergoing the transition from 1 to 20 is
proportional to k21 induced by the radiation and to the number of atoms in the initial state, N1 .
dN
(1 → 2) = N1 k21
dt
162
(5.151)
The number of atoms going from 2 to 1 is proportional to N2 and to k12 + A where A is the
spontaneous transition rate
dN
(2 → 1) = N2 (k21 + A)
dt
(5.152)
At equilibrium, these two rates must be equal. Thus,
k21 + A
N1
=
= eh̄ωβ
k21
N2
(5.153)
Now, let’s refer to the result for the induced rate k21 and express it in terms of the energy density
per unit frequency of the cavity, u(ω).
k21 =
π
|h2|~µ · ~|1i|2 u(ω) = B21 u(ω)
o h̄2
(5.154)
π
|h2|~µ · ~|1i|2 .
o h̄2
(5.155)
where
B21 =
For em radiation in equilibrium at temperature T the energy density per unit frequency is given
by Planck’s Law:
u(ω) =
1
h̄ω 3
π 2 c3 eh̄ωβ − 1
(5.156)
Combining the results we obtain
B12
A 1
+
= eh̄ωβ
B21 B21 u(ω)
(5.157)
B21
A π 2 c3 h̄ωβ
+
(e
− 1) = eh̄ωβ
B12 B21 h̄ω 3
(5.158)
(5.159)
which must hold for all temperatures. Since
B21
= 1.
B12
(5.160)
A π 2 c3
=1
B21 h̄ω 3
(5.161)
we get
and Thus, the spontaneous emission rate is
h̄ω 3
A = 2 3 B12
π c
163
(5.162)
ω3
|h2|~µ · ~|1i|2
o πh̄c3
=
(5.163)
This is a key result in that it determines the probability for the emission of light by atomic
and molecular systems. We can use it to compute the intensity of spectral lines in terms of the
electric dipole moment operator. The lifetime of the excited state is then inversely proportional
to the spontaneous decay rate.
τ=
1
A
(5.164)
To compute the matrix elements, we can make a rough approximation that hµi ∝ hxie where
e is the charge of an electron and hxi is on the order of atomic dimensions. We also must include
a factor of 1/3 for averaging over all orientations of (~µ · ~), since at any given time,the moments
are not all aligned.
4 ω 3 e2
1
=A=
|hxi|2
τ
3 h̄c3 4πo
(5.165)
e2
1
=α≈
4πo h̄c
137
(5.166)
The factor
is the fine structure constant. Also, ω/c = 2π/λ. So, setting hxi ≈ 1 Å
4 1
2π
A=
c
3 137
λ
3
(1Å)2 ≈
6 × 1018 −1
sec
[λ(Å)]3
(5.167)
So, for a typical wavelength, λ ≈ 4 × 103 Å.
τ = 10−8 sec
(5.168)
which is consistent with observed lifetimes.
We can also compare with classical radiation theory. The power radiated by an accelerated
particle of charge e is given by the Larmor formula (c.f Jackson).
P =
2 e2 (v̇)2
3 4πo c3
(5.169)
where v̇ is the acceleration of the charge. Assuming the particle moves in a circular orbit of
radius r with angular velocity ω, the acceleration is v̇ = ω 2 r. Thus, the time required to radiate
energy h̄ω/2 is equivalent to the lifetime τ .
1
τclass
=
=
2P
h̄ω
1 4 e2 ω 4 r2
h̄ω 3 4πo c3
164
(5.170)
(5.171)
=
4 ω 3 e2 2
r .
3 h̄c3 4πo
(5.172)
This qualitative agreement between the classical and quantum result is a manifestation of the
correspondence principle. However, it must be emphasized that the MECHANISM for radiation
is entirely different. The classical result will never predict a discrete spectrum. This was in fact
a very early indication that something was certainly amiss with the classical electro-magnetic
field theories of Maxwell and others.
5.7
Time-dependent golden rule
In the last lecture we derived the the Golden Rule (GR) transition rate as
k(t) =
2π
|hs|V̂ |ni|2 ρ(Es )
h̄
(5.173)
This is perhaps one of the most important approximations in quantum mechanics in that it
has implications and applications in all areas, especially spectroscopy and other applications of
matter interacting with electro-magnetic radiation. In today’s lecture, I want to show how we
can use the Golden Rule to simplify some very complex problems. Moreover, to show how we
used the GR to solve a real problem. The GR has been used by a number of people in chemistry
to look at a wide variery of problems. In fact, most of this lecture comes right out out of the
Journal of Chemical Physics. Some papers you may be interested in knowing about include:
1. B. J. Schwartz, E. R. Bittner, O. V. Prezhdo, and P. J. Rossky, J. Chem. Phys. 104, 5242
(1996).
2. E. Neria and A. Nitzan, J. Chem. Phys. 99, 1109 (1993).
3. A. Stiab and D. Borgis, J. Chem. Phys. 103, 2642 (1995)
4. E. J. Heller, J. Chem. Phys. 75, 2923 (1981).
5. W. Gelbart, K. Spears, K. F. Freed, J. Jortner, S. A. Rice, Chem. Phys. Lett. 6 345
(1970).
The focus of the lecture will be to use the GR to calculate the transition rate from one
adiabatic potential energy surface to another via non-radiative decay. Recall in a previous lecture,
we talked about potential energy curves of molecule and that these are obtained by solving the
Schrodinger equation for the electronic degrees of freedom assuming that the nuclei move very
slowly. We defined the adiabatic or Born-Oppenheimer potential energy curves for the nuclei in a
molecule by solving the Schrodinger equation for the electrons for fixed nuclear positions. These
potential curves are thus the electronic eigenvalues parameterized by the nuclear positions.
Vi (R) = Ei (R)
165
(5.174)
Under the BO approximation, the nuclei move about on a single energy surface and the electronic
wavefunction is simply |Ψi (R)i paramterized by the nuclear positions. However, when the nuclei
are moving fast, this assumption is no longer true since,
!
∂R ∂
d
+ Ei (R) |Ψi (R)i.
ih̄ |Ψi (R)i = ih̄
dt
∂t ∂R
(5.175)
That is to say, when the nuclear velocities are large in a direction that the wavefunction changes
a lot with varying nuclear position, the Born-Oppenheimer approximation is not so good and
the electronic states become coupled by the nuclear motion. This leads to a wide variety of
physical phenomina including non-radiative decay and intersystem crossing and is an important
mechanism in electron transfer dynamics.
THe picture I want to work with today is a hybred quantum/classical picture (or semiclassical). I want to treat the nuclear dynamics as being mostly “classical” with some “quantum
aspects”. IN this picture I will derive a semi-classical version of the Golden-Rule transition rate
which can be used in concert with a classical Molecular dynamics simulation.
We start with the GR transition rate we derived the last lecture. We shall for now assume
that the states we are coupling are the vibrational-electronic states of the system, written as
|ψi i = |αi (R)I(R)i
(5.176)
where R denotes the nuclear positions, |α(R)i is the adiabatic electronic eigenstate obtained at
position R and |I(R)i is the initial nuclear vibrational state on the α(R) potential energy surface.
Let this denote the initial quantum state and denote by
|ψf i = |αf (R)F (R)i
(5.177)
the final quantum state. The GR transition rate at nuclear position R is thus
kif =
2π X
|hψi |V̂ |ψf i|2 δ(Ei − Ef )
h̄ f
(5.178)
where the sum is over the final density of states (vibrational) and the energys in the δ-function
is the electronic energy gap measured with respect to a common origin. We can also define a
“thermal” rate constant by ensemble averaging over a collective set of initial states.
5.7.1
Non-radiative transitions between displaced Harmonic Wells
An important application of the GR comes in evaluating electron transfer rates between electronic
state of a molecule. Let’s approximate the diabatic electronic energy surfaces of a molecule as
harmonic wells off-set by by energy and with the well minimums displaced by some amount xo .
Let the curves cross at xs and assume there is a spatial dependent coupling V (x) which couples
the diabatic electronic states. Let T1 denote the upper surface and So denote the ground-state
surface. The diabatic coupling is maximized at the crossing point and decays rapidly as we move
away. Because these electronic states are coupled, the vibrational states on T1 become coupled
to the vibrational states on So and vibrational amplitude can tunnel from one surface to the
166
other. The tunneling rate can be estimated very well using the Golden Rule (assuming that the
amplitude only crosses from one surface to another once.)
kT S =
2π
|hΨT |V |ΨS i|2 ρ(Es )
h̄
(5.179)
The wavefunction on each surface is the “vibronic” function mentioned above. This we will write
as a product of an electronic term |ψT i (or |ψS i) and a vibrational term |nT i (or |nS i). For
shorthand, let’s write the electronic contribution as
VT S = hψT |V |ψS i
(5.180)
Say I want to know the probability of finding the system in some initial vibronic state after some
time. The rate of decay of this state is thus the sum over all possible decay channels. So, I must
sum over all final states that I can decay into. The decay rate is thus.
kT S =
2π X
|hnT |VT S |mS i|2 ρ(Es )
h̄ mS
(5.181)
This equation is completely exact (within the GR approximation) and can be used in this form.
However, let’s make a series of approximations and derive a set of approximate rates and compare
the various approximations.
Condon Approximation
First I note that the density of states can be rewritten as
ρ(Es ) = Tr(Ho − Es )−1
(5.182)
so that
kT S =
2π X |hnT |VT S |mS i|2
h̄ mS
En − Es
(5.183)
where En −Es is the energy gap between the initial and final states including the electronic energy
gap. What this means is that if the energy difference between the bottoms of the respective wells
is large, then the initial state will be coupled to the high-lying vibrational states in So . Next I
make the “Condon Approximation” that
hnT |VT S |mS i ≈ VT S hnT |mS i
(5.184)
where hnT |mS i is the overlap between vibrational state |nT i in the T1 well and state |mS i in the
So well. These are called Franck-Condon factors.
Evaluation of Franck Condon Factors
Define the Franck-Condon factor as
hnT |mS i =
Z
)∗
(S)
dxϕ(T
m (x)ϕn (x)
167
(5.185)
)
where ϕ(T
n (x) is the coordinate representation of a harmonic osc. state. We shall assume that
the two wells are off-set by xs and each has a freq. ω1 and ω2 . We can write the HO state for
each well as a Gauss-Hermite Polynomial (c.f. Compliment Bv in the text.)
β2
π
ϕn (x) =
where β =
!1/4
1
β2
√
exp − x2 Hn (βx)
2
2n n!
!
(5.186)
q
mω/h̄ and Hn (z) is a Hermite polynomial,
Hn (z) = (−1)n ez
2
∂ n −z2
e
∂z n
(5.187)
Thus the FC factor is
βT2
π
!1/4
!1/4
1
βS2
1
√
√
hnT |mS i =
π
2n n! 2m m!
!
!
Z
βT2 2
βS2
2
×
dx exp − x exp − (x − xs )
2
2
× Hni (βi x)Hmf (βf (x − xs ))
(5.188)
This integral is pretty difficult to solve analytically. In fact Mathematica even choked on this
one. Let’s try a simplification: We can expand the Hermite Polynomials as
n
Hn (z) = 2
√
1
nx3
2x
+ ···
π
−
−
Γ( 12 − n2 ) Γ(− n2 ) Γ( 12 − n2 )
#
"
(5.189)
Thus, any integral we want to do involves doing an integral of the form
β2
β2
dxx exp − 1 x2 − 2 (x − xs )2
2
2
!
Z
2
β1 2 β22 2
2
n
dxx exp − x − (x − 2xxs + xs )
2
2
!
!
Z
2
2
β1 + β2 2
β22
n
2
dxx exp −
x exp
(2xxs − xs )
2
2
!
!
β22 2 Z
β12 + β22 2
n
exp − xs
dxx exp −
x exp β22 (xxs )
2
2
!Z
2
β2 2
a 2
n
exp − xs
dxx exp − x exp (bx)
2
2
!
Z
In =
=
=
=
=
n
(5.190)
(5.191)
where I defined a = β12 + β22 and b = β22 xs .
Performing the integral:
In = 2
−
−1+n
2
√
1
a
2+n
2

s
n
(1 + (−1) )
1
1+n
1 + n 1 b2
a Γ(
) 1 F1 (
, ,
)
a
2
2
2 2a
2+n
2 + n 3 b2
2 (−1 + (−1) ) b Γ(
) 1 F1 (
, ,
)
2
2
2 2a
!
n
168
(5.192)
where 1 F1 (a, b, z) is the Hypergeometric function (c.f Landau and Lifshitz, QM) and Γ(z) is the
Gamma function. Not exactly the easist integral in the world to evaluate. (In other words, don’t
worry about having to solve this integral on an exam!) To make matters even worse, this is only
one term. In order to compute, say the FC factor between ni = 10 and mf = 12 I would need to
sum over 120 terms! However, Mathematica knows how to evaluate these functions, and we can
use it to compute FC factors very easily.
If the harmonic frequencies are the same in each well, life gets much easier. Furthermore, If I
make the initial vibrational state in the T1 well be the ground vibrational state, we can evaluate
the overlap exactly for this case. The answer is (see Mathematica hand out)
β n xns
√
M [n] =
2n n!
(5.193)
Note that this is different than the FCF calculated in Ref. 6 by Gelbart, et al. who do not have
the square-root factor (their denominator is my denominator squared.) 2
Finally, we can evaluate the matrix element as
β n xn
hnT |VT S |mS i ≈ VT S √ s
2n n!
(5.194)
Thus, the GR survival rate for the ground vibrational state of the T1 surface is
k=
X
2π
1
β n xn
√ s
VT S
h̄
2n n!
m h̄Ω − h̄ωm
(5.195)
where Ω is the energy difference between the T1 and So potential minimuma.
“Steep So Approximation”
In this approximation we assume that potential well of the So state is very steep and intesects
the T1 surface at xs . We also assume that the diabatic coupling is a function of x. Thus, the GR
survival rate is
k=
2π 2
hV iρS
h̄ T S
(5.196)
where
hVT S i =
Z
dxψT∗ (x)ψS (x)V (x)
(5.197)
When the So surface is steeply repulsive, the wavefunction on the So surface will be very oscillitory
at the classical turning point, which is nearly identical with xs for very steep potentials. Thus,
for purposes of doing integrations, we can assume that
ψS (x) = Cδ(x − xs ) + · · ·
2
I believe this is may be a mistake in their paper. I’ll have to call Karl Freed about this one.
169
(5.198)
where xs is the classical turning point at energy E on the So surface. The justification for this
is from the expansion of the “semi-classical” wavefunction on a linear potential, which are the
Airy functions, Ai(x).
1 Z
Ai (−ζ) =
+∞ds exp(is2 /3 − iζs)
2π −∞
(5.199)
Which can be expanded as
aAi(ax) = δ(x) + · · ·
(5.200)
Expansions of this form also let us estimate the coefficient C
Using the δ-function approximation
Z
dxψT∗ (x)ψS (x)V
(x) = C
Z
3
dxψT∗ (x)δ(x − xs )V (x) = CψT∗ (xs )V (xs )
(5.201)
Now, again assuming that we are in the ground vibrational state on the T1 surface,
ψT (x) =
Z
dxψT∗ (x)ψS (x)V
β2
π
!1/4
e−β
β2
(x) = C
π
2 x2 /2
(5.202)
!1/4
e−β
2 x2 /2
s
V (xs )
(5.203)
So, we get the approximation:
2π
β2
k=
CV (xs )
h̄
π
!1/4
e−β
2 x2 /2
s
1
h̄Ω
(5.204)
where C remains to be determined. For that, refer to the Heller-Brown paper.
Time-Dependent Semi-Classical Evaluation
We next do something tricky. There are a number of ways one can represent the δ-function. We
will use the Fourier representation of the function and write:
h̄ Z ∞
δ(Ei − Ef ) =
dtei/h̄(Ei −Ej )t
2π −∞
(5.205)
Thus, we can write
kif =
Z
∞
−∞
dt
X
hαi (R)I(R)|V |αf (R)F (R)i
f
× hαf (R)F (R)|e+iHf t/h̄ V e−iHi t/h̄ |αi (R)I(R)i
3
(see Heller and Brown, JCP 79 3336 (1983). )
170
(5.206)
Integrating over the electronic degrees of freedom, we can define
Vif (R) = hαi (R)|V |αf (R)i
(5.207)
and thus write
kif =
Z
∞
−∞
dt
X
hI(R)|Vif (R)|F (R)i
f
× hF (R)|e+iHf t/h̄ Vif (R)e−iHi t/h̄ |I(R)i
(5.208)
where Hi (R) is the nuclear Hamiltonian for the initial state and Hf (R) is the nuclear Hamiltonian
for the final state.
At this point I can remove the sum over f and obtain
kif =
Z
∞
−∞
dthI(R)|Vif (R)e+iHf t/h̄ Vif (R)e−iHi t/h̄ |I(R)i.
(5.209)
Vif (R) = Ṙ · hαf (R)|ih̄∇R |αi (R)i
= Ṙ(0) · Dif (R(0))
(5.210)
Next, we note that
is proportional to the nuclear velocity at the initial time. Likewise, the term in the middle
represents the non-adiabatic coupling at some later time. Thus, we can re-write the transition
rate as
kif =
Z
∞
−∞
dt(Ṙ(0) · Dif (R(0)))(Ṙ(t) · Dif (R(t)))
(5.211)
× hI(R)|e+iHf t/h̄ e−iHi t/h̄ |I(R)i.
(5.212)
Finally, we do an ensemble averate over the initial positions of the nuclei and obtain almost the
final result:
kif =
Z
∞
−∞
D
E
dt (Ṙ(0) · Dif (R(0)))(Ṙ(t) · Dif (R(t)))Jif (t) .
(5.213)
Jif (t) = hI(R)|e+iHf t/h̄ e−iHi t/h̄ |I(R)i
(5.214)
where I define
This term represents the evolution of the initial nuclear vibrational state moving forward in time
on the initial energy surface
|I(R(t))i = e−i/h̄Hi t |I(R(0))i
(5.215)
and backwards in time on the final energy surface.
hI(R(t))| = hI(R(0))|ei/h̄Hf t
171
(5.216)
So, J(t) represents the time-dependent overlap integral between nuclear wavefunctions evolving
on the different potential energy surfaces.
Let us assume that the potential wells are Harmonic with the centers off set by some amount
xs . We can define in each well a set of Harmonic Oscillator eigenstates, which we’ll write in short
hand as |ni i where the subscript i denotes the electronic state. At time t = 0, we can expand
the initial nuclear wavefunction as a superposition of these states:
|I(R(0))i =
X
γn |ni i
(5.217)
n
where γn = hni |I(R(0))i. The time evolution in the well is
|I(R(t))i =
X
γn exp(−i/2(n + 1)ωi t)|ni i.
(5.218)
n
We can also express the evolution of the ket as a superposition of states in the other well.
hI(R(t))| =
X
∗
ξm
exp(+i/2(m + 1)ωf t)hmf |
(5.219)
m
where ξm = hmf |I(R)i are the coefficients. Thus, J(t) is obtained by hooking the two results
together:
J(t) =
X
∗
ξm
γn e+i/2(m+1)ωf t e−i/2(n+1)ωi t hmf |ni i
(5.220)
mn
Now, we must compute the overlap between harmonic states in one well with harmonic states in
another well. This type of overlap is termed a Franck-Condon factor (FC). We will evaluate the
FC factor using two different approaches.
5.7.2
Semi-Classical Evaluation
I want to make a series of simplifying assumptions to the nuclear wavefunction. Many of these
assumptions follow from Heller’s paper referenced at the beginning. The assumptions are as
follows:
1. At the initial time, hx|I(R)i can be written as a Gaussian of width β centered about R.
hx|I(R)i =
β2
π
!1/4
β2
i
exp − (x − R(t))2 + p(t)(x − R(t))
2
h̄
!
(5.221)
where p(t) is the classical momentum. (c.f Heller)
2. We know that for a Gaussian wavepacket (especially one in a Harmonic Well), the center
of the Gaussian tracks the classical prediction. Thus, we can write that R(t) is the center
of the Gaussian evolves under Newton’s equation
mR̈(t) = Fi (R)
(5.222)
where Fi (R) is the force computed as the derivative of the ith energy surface w.r.t. R,
evaluated at the current nuclear position (i.e. the force we would get using the BornOppenheimer approximation.
Fi (R) = −
∂
E(R).
∂R
172
(5.223)
3. At t = 0, the initial classical velocities and positions of the nuclear waves evolving on the
i and f surface are the same.
4. For very short times, we assume that the wave does not spread appreciably. We can fix
this assuption very easily if need be.
Using these assumptions, we can approximate the time-dependent FC factor as4
β2
J(t) ≈ exp − (Rf (t) − Ri (t))2
4
"
#
1
2
× exp − 2 2 (pf (t) − pi (t))
4β h̄
i
× exp + (Rf (t) − Ri (t)) · (pj (t) − pi (t))
2h̄
"
#
(5.224)
Next, we expand Ri (t) and pi (t) as a Tayler series about t = 0.
Ri (t) = Ri (0) + tṘi (0) + t2
R̈i (0)
+ ···
2!
(5.225)
Using Newton’s Equation:
p(0)
Fi (0)
− t2
+ ···
m
m2!
1 ∂Fi 2
t ···
pi (t) = pi (0) + Fi (0)t +
2 ∂t
Ri (t) = Ri (0) + t
(5.226)
(5.227)
Thus the difference between the nuclear positions after a short amount of time will be
!
Ri (t) − Rf (t) ≈ −t2
Fi (0) Fj (0)
−
+ ···
2m
2m
(5.228)
Also, the momentum difference is
pf (t) − pi (t) ≈ (Ff (0) − Fi (0))t + · · ·
(5.229)
Thus,
β 2 t4
J(t) ≈ exp −
(Fi (0) − Fj (0))2
2
16m
#
"
t2
2
× exp − 2 2 ((Ff (0) − Fi (0))
4β h̄
i
(Ff (0) − Fi (0))t3
× exp +
4mh̄
"
#
(5.230)
If we include the oscillitory term, the integral does not converge (so much for a short time
approx.) However, when we do the ensemble average, each member of the ensemble contributes
4
B. J. Schwartz, E. R. Bittner, O. V. Prezhdo, and P. J. Rossky, J. Chem. Phys. 104, 5242 (1996).
173
a slightly different phase contribution, so, we can safely ignore it. Furthermore, for short times,
the decay of the overlap will be dominated by the term proportional to t2 . Thus, we defined the
approximate decay curve as
t2
J(t) = exp − 2 2 (Fj (0) − Fi (0))2
4β h̄
!
(5.231)
Now, pulling everything together, we write the GR rate constant as
∞
Z
kif =
−∞
D
dt (Ṙ(0) · Dif (R(0)))(Ṙ(t) · Dif (R(t)))
t2
× exp − 2 2 (Fj (0) − Fi (0))2
4β h̄
!+
.
(5.232)
The assumptions are that the overlap decays more rapidly than the oscillations in the autocorrelation of the matrix element. This actually bears itself out in reality.
Let’s assume that the overlap decay and the correlation function are un-correlated. (Condon
Approximation) Under this we can write:
∞
Z
kif =
−∞
E
D
dt (Ṙ(0) · Dif (R(0)))(Ṙ(t) · Dif (R(t)))
t2
exp − 2 2 (Fj (0) − Fi (0))2
4β h̄
*
×
!+
.
(5.233)
or defining
D
E
Cif (t) = (Ṙ(0) · Dif (R(0)))(Ṙ(t) · Dif (R(t)))
(5.234)
and using (c.f. Chandler, Stat. Mech.)
D
E
eA = ehAi ,
(5.235)
the desired result is
kif =
*
∞
Z
−∞
2
dtCif (t) exp −t
(Fj (0) − Fi (0))2
4β 2 h̄2
+!
.
(5.236)
Now, let’s assume that the correlation function is an oscillitory function of time.
Cif (t) = |Vif |2 cos(Ωt)
(5.237)
Then
kif =
Z
∞
−∞
r
=
*
2
2
dt|Vif | cos(Ωt) exp −t
π −Ω2 /(4b)
e
|Vif |2
b
(Fj (0) − Fi (0))2
4β 2 h̄2
+!
.
(5.238)
(5.239)
174
where
*
b=
(Fj (0) − Fi (0))2
4β 2 h̄2
+
(5.240)
In Ref 1. we used this equation (actually one a few lines up) to compute the non-radiative
relaxation rates between the p to s state of an aqueous electron in H2 0 and in D2 0 to estimate
the isotopic dependency of the transition rate. Briefly, the story goes as this. Paul Barbara’s
group at U. Minnisota and Yann Gauduel’s group in France measured the fluorescence decay of
an excited excess electron in H2 0 and in D2 0 and noted that there was no resolvable difference
between the two solvents. I.e. the non-radiative decay was not at all sensitive to isotopic changes
in the solvent. (The experimental life-times are roughly 310 fs for both sovents with resolution of
about 80 fs ) This is very surprising since, looking at the non-adiabatic coupling operator above,
you will notice that the matrix element coupling states is proportional to the nuclear velocities.
(The electronic matrix element
√ is between the s and p state of the electron.) Thus, since the
velocity of a proton is roughly 2 times faster than that of a deuteron of the same kinetic energy,
The non-adiabatic coupling matrix element between the s and p states in water should be twice
that as in heavy-water. Thus, the transition rate in water should be roughtly twice that as in
heavy-water. It turns out that since the D’s move slower than the H’s, the nuclear overlap decays
roughly twice as slowly. Thus we get competing factors of two which cancel out.
5.8
Problems and Exercises
Exercise 5.1 A one dimensional harmonic oscillator, with frequency ω, in its ground state is
subjected to a perturbation of the form
H 0 (t) = C p̂e−α|t| cos(Ωt)
(5.241)
where p̂ is the momentum operator and C, α, and Ω are constants. What is the probability that
as t → ∞ the oscillator will be found in its first excited state in first-order perturbation theory.
Discuss the result as a function of Ω, ω, and α.
Exercise 5.2 A particle is in a one-dimensional infinite well of width 2a. A time-dependent
perturbation of the form
H 0 (t) = To Vo sin(
πx
)δ(t)
a
(5.242)
acts on the system, where To and Vo are constants. What is the probability that the system will
be in the first excited state afterwards?
Exercise 5.3 Because of the finite size of the nucleus, the actual potential seen by the electron
is more like:
175
picture goes here
1. Calculate this effect on the ground state energy of the H atom using first order perturbation
theory with
(
0
H =
e2
r
−
e3
R
0
for r ≤ R
otherwise
(5.243)
2. Explain this choice for H 0 .
3. Expand your results in powers of R/ao 1. (Be careful!)
4. Evaluate numerically your result for R = 1 fm and R = 100 fm.
5. Give the fractional shift of the energy of the ground state.
Note that this effect is the “isotope shift” and can be observed in the spectral lines of the heavy
elements.
176
Chapter 6
Semi-Classical Quantum Mechanics
Good actions ennoble us, and we are the sons of our own deeds.
–Miguel de Cervantes
The use of classical mechanical analogs for quantum behavour holds a long and proud tradition in
the development and application of quantum theory. In Bohr’s original formulation of quantum
mechanics to explain the spectra of the hydrogen atom, Bohr used purely classical mechanical
notions of angular momentum and rotation for the basic theory and imposed a quantization
condition that the angular momentum should come in integer multiples of h̄. Bohr worked under
the assumption that at some point the laws of quantum mechanics which govern atoms and
molecules should correspond to the classical mechanical laws of ordinary objects like rocks and
stones. Bohr’s Principle of Correspondence states that quantum mechanics was not completely
separate from classical mechanics; rather, it incorporates classical theory.
From a computational viewpoint, this is an extremely powerful notion since performing a
classical trajectory calculation (even running 1000’s of them) is simpler than a single quantum
calculation of a similar dimension. Consequently, the development of semi-classical methods has
and remains an important part of the development and untilization of quantum theory. In fact
even in the most recent issues of the Journal of Chemical Physics, Phys. Rev. Lett, and other
leading physics and chemical physics journals, one finds new developments and applications of
this very old idea.
In this chapter we will explore this idea in some detail. The field of semi-classical mechanics
is vast and I would recommend the following for more information:
1. Chaos in Classical and Quantum Mechanics, Martin Gutzwiller (Springer-Verlag, 1990).
Chaos in quantum mechanics is a touchy subject and really has no clear-cut definition that
anyone seems to agree upon. Gutzwiller is one of the key figures in sorting all this out. This
is very nice and not too technical monograph on quantum and classical correspondence.
2. Semiclassical Physics, M. Brack and R. Bhaduri (Addison-Wesley, 1997). Very interesting
book, mostly focusing upon many-body applications and Thomas-Fermi approximations.
3. Computer Simulations of Liquids, M. P. Allen and D. J. Tildesley (Oxford, 1994). This
book mostly focus upon classical MD methods, but has a nice chapter on quantum methods
which were state of the art in 1994. Methods come and methods go.
There are many others, of course. These are just the ones on my bookshelf.
177
6.1
Bohr-Sommerfield quantization
Let’s first review Bohr’s original derivation of the hydrogen atom. We will go through this a bit
differently than Bohr since we already know part of the answer. In the chapter on the Hydrogen
atom we derived the energy levels in terms of the principle quantum number, n.
En = −
me4 1
2h̄2 n2
(6.1)
In Bohr’s correspondence principle, the quantum energy must equal the classical energy. So
for an electron moving about a proton, that energy is inversely proportional to the distance of
separation. So, we can write
−
me4 1
e2
=
−
2r
2h̄2 n2
(6.2)
Now we need to figure out how angular momentum gets pulled into this. For an orbiting body
the centrifugal force which pulls the body outward is counterbalenced by the inward tugs of the
centripetal force coming from the attractive Coulomb potential. Thus,
mrω 2 =
e2
,
r2
(6.3)
where ω is the angular frequency of the rotation. Rearranging this a bit, we can plug this into
the RHS of Eq. 6.2 and write
−
me4 1
mr3 ω 2
=
−
2r
2h̄2 n2
(6.4)
The numerator now looks amost like the classical definition of angular momentum: L = mr2 ω.
So we can write the last equation as
−
me4 1
L2
=
−
.
2mr2
2h̄2 n2
(6.5)
Solving for L2 :
L2 =
me4 2mr2
.
2h̄2 n2
(6.6)
Now, we need to pull in another one of Bohr’s results for the orbital radius of the H-atom:
h̄2 2
r=
n.
me2
(6.7)
Plug this into Eq.6.6 and after the dust settles, we find
L = h̄n.
(6.8)
But, why should electrons be confined to circular orbits? Eq. 6.8 should be applicable to
any closed path the electron should choose to take. If the quantization condition only holds
178
for circular orbits, then the theory itself is in deep trouble. At least that’s what Sommerfield
thought.
The numerical units of h̄ are energy times time. That is the unit of action in classical
mechanics. In classical mechanics, the action of a mechanical system is given by the integral of
the classical momentum along a classical path:
S=
Z
x2
pdx
(6.9)
x1
For an orbit, the initial point and the final point must coincide, x1 = x2 , so the action integral
must describe some the area circumscribed by a closed loop on the p−x plane called phase-space.
S=
I
pdx
(6.10)
So, Bohr and Sommerfield’s idea was that the circumscribed area in phase-space was quantized
as well.
As a check, let us consider the harmonic oscillator. The classical energy is given by
p2
k
+ q2.
2m 2
This is the equation for an ellipse in phase space since we can re-arrange this to read
E(p, q) =
p2
k 2
+
q
2mE 2E
p2 q 2
= 2+ 2
a
b
1 =
(6.11)
q
√
where a = 2mE and b = 2E/k describe the major and minor axes of the ellipse. The area of
an ellipse is A = πab, so the area circumscribed by a classical trajectory with energy E is
q
S(E) = 2Eπ m/k
(6.12)
q
Since k/m = ω, S = 2πE/ω = E/ν. Finally, since E/ν must be an integer multiple of h, the
Bohr-Sommerfield condition for quantization becomes
I
pdx = nh
(6.13)
q
where p is the classical momentum for a path of energy E, p = 2m(V (x) − E. Taking this a
bit farther, the de Broglie wavelength is p/h, so the Bohr-Sommerfield rule basically states that
stationary energies correspond to classical paths for which there are an integer number of de
Broglie wavelengths.
Now, perhaps you can see where the problem with quantum chaos. In classical chaos, chaotic
trajectories never return to their exact staring point in phase-space. They may come close, but
there are no closed orbits. For 1D systems, this is does not occur since the trajectories are the
contours of the energy function. For higher dimensions, the dimensionality of the system makes
it possible to have extremely complex trajectories which never return to their starting point.
Exercise 6.1 Apply the Bohr-Sommerfield proceedure to determine the stationary energies for
a particle in a box of length l.
179
6.2
The WKB Approximation
The original Bohr-Sommerfield idea can be imporoved upon considerably to produce an asymptotic (h̄ → 0) approximation to the Schrödinger wave function. The idea was put forward at about
the same time by three different theoreticians, Brillouin (in Belgium), Kramers (in Netherlands),
and Wentzel (in Germany). Depending upn your point of origin, this method is the WKB (US
& Germany), BWK (France, Belgium), JWKB (UK), you get the idea. The original references
are
1. “La mécanique odularatoire de Schrödinger; une méthode générale de résolution par approximations successives”, L. Brillouin, Comptes rendus (Paris). 183, 24 (1926).
2. “Wellenmechanik und halbzahlige Quantisierung”, H. A. Kramers, Zeitschrift für Physik
39, 828 (1926).
3. “Eine Verallgemeinerung der Quantenbedingungen für die Zwecke der Wellenmechanik”,
Zeitschrift für Physik 38, 518 (1926).
We will first go through how one can use the approach to determine the eigenvalues of the
Schrödinger equation via semi-classical methods, then show how one can approximate the actual
wavefunctions themselves.
6.2.1
Asymptotic expansion for eigenvalue spectrum
The WKB proceedure is initiated by writing the solution to the Schödinger equation
ψ 00 +
2m
(E − V (x))ψ = 0
h̄2
as
ψ(x) = exp
iZ
χdx .
h̄
(6.14)
We will soon discover that χ is the classical momentum of the system, but for now, let’s consider
it to be a function of the energy of the system. Substituting into the Schrodinger equation
produces a new differential equation for χ
h̄ dχ
= 2m(E − V ) − χ2 .
i dx
(6.15)
If we take h̄ → 0, it follows then that
χ = χo =
q
2m(E − V ) = |p|
(6.16)
which is the magnitude of the classical momentum of a particle. So, if we assume that this is
simply the leading order term in a series expansion in h̄ we would have
h̄
h̄
χ = χo + χ1 +
i
i
180
!2
χ2 . . .
(6.17)
Substituting Eq. 6.17 into
χ=
h̄ 1 ∂ψ
iψ x
(6.18)
and equating to zero coefficients with different powers of h̄, one obtains equations which determine
the χn corrections in succession:
n
X
d
χn−1 = −
χn−m χm
dx
m=0
(6.19)
for n = 1, 2, 3 . . .. For example,
χ1 = −
1 χ0o
1 V0
=
2 χo
4E −V
χ21 + χ01
2χo
(
)
1
V 02
V 02
V 00
+
+
.
= −
2χo 16(E − V )2 4(E − V )2 4(E − V )
5V 02
V 00
−
= −
32(2m)1/2 (E − V )5/2 8(2m)1/2 (E − V )3/2
(6.20)
χ2 = −
(6.21)
and so forth.
Exercise 6.2 Verify Eq. 6.19 and derive the first order correction in Eq.6.20.
Now, to use these equations to determine the spectrum, we replace x everywhere by a complex
coordinate z and suppose that V (z) is a regular and analytic function of z in any physically
relevant region. Consequently, we can then say that ψ(z) is an analytic function of z. So, we
can write the phase integral as
1Z
χ(z)dz
h C
1 Z ψn0 (z)
=
dz
2πi C ψn (z)
n =
(6.22)
where ψn is the nth discrete stationary solution to the Schrödinger equation and C is some
contour of integration on the z plane. If there is a discrete spectrum, we know that the number
of zeros, n, in the wavefunction is related to the quantum number corresponding to the n + 1
energy level. So if ψ has no real zeros, this is the ground state wavefunction with energy Eo , one
real zero corresponds to energy level E1 and so forth.
Suppose the contour of integration, C is taken such that it include only these zeros and no
others, then we can write
n=
Z
1Z
1 Z
χo dz +
−h̄ χ2 dz + . . .
h̄ C
2πi c
C
181
(6.23)
Each of these terms involves E − V in the denominator. At the classical turning points where
V (z) = E, we have poles and we can use the residue theorem to evaluate the integrals. For
example, χ1 has a pole at each turnining point with residue −1/4 at each point, hence,
1 Z
1
χ1 dz = − .
2πi C
2
The next term we evaluate by integration by parts
Z
C
V 00
3Z
V 02
dz
=
−
dz.
(E − V (z))3/2
2 C (E − V (z))5/2
(6.24)
(6.25)
Hence, we can write
Z
C
χ2 (z)dz =
Z
1
V 02
dz.
32(2m)1/2 C (E − V (z))5/2
(6.26)
Putting it all together
1Z q
2m(E − V (z))dz
h c
Z
h
V 02
−
dz + . . .
128π 2 (2m)1/2 c (E − V (z))5/2
n + 1/2 =
(6.27)
Granted, the above analysis is pretty formal! But, what we have is something new. Notice that
we have an extra 1/2 added here that we did not have in the original Bohr-Sommerfield (BS)
theory. What we have is something even more general. The original BS idea came from the
notion that energies and frequencies were related by integer multiples of h. But this is really
only valid for transitions between states. If we go back and ask what happens at n = 0 in
the Bohr-Sommerfield theory, this corresponds to a phase-space ellipse with major and minor
axes both of length 0–which violates the Heisenberg Uncertainly rule. This new quantization
condition forces the system to have some lowest energy state with phase-space area 1/2.
Where did this extra 1/2 come from? It originates from the classical turning points where
V (x) = E. Recall that for a 1D system bound by a potential, there are at least two such points.
Each contributes a π/4 to the phase. We will see this more explicitly in the next section.
6.2.2
WKB Wavefunction
Going back to our original wavefunction in Eq. 6.14 and writing
ψ = eiS/h̄
where S is the integral of χ, we can derive equations for S:
1
2m
∂S
∂x
!
−
ih̄ ∂ 2 S
+ V (x) = E.
2m ∂x2
(6.28)
Again, as above, one can seek a series expansion of S in powers of h̄. The result is simply the
integral of Eq. 6.17.
h̄
S = So + S 1 + . . .
i
182
(6.29)
If we make the approximation that h̄ = 0 we have the classical Hamilton-Jacobi equation for the
action, S. This, along with the definition of the momentum, p = dSo /dx = χo , allows us to make
a very firm contact between quantum mechanics and the motion of a classical particle.
Looking at Eq. 6.28, it is clear that the classical approximation is valid when the second term
is very small compared to the first. i.e.
S 00
1
S 02
!
!2
dS
dx
1
dx
dS
d 1
h̄
1
dx p
h̄
h̄
d
dx
(6.30)
where we equate dS/dx = p. Since p is related to the de Broglie wavelength of the particle
λ = h/p , the same condition implies that
1 dλ 1.
2π dx (6.31)
Thus the semi-classical approximation is only valid when the wavelength of the particle as determined by λ(x) = h/p(x) varies slightly over distances on the order of the wavelength itself.
Written another way by noting that the gradiant of the momentum is
dp
dq
m dV
=
2m(E − V (x)) = −
.
dx
dx
p dx
Thus, we can write the classical condition as
mh̄|F |/p3 1
(6.32)
Consequently, the semi-classical approximation can only be used in regions where the momentum
is not too small. This is especially important near the classical turning points where p → 0. In
classical mechanics, the particle rolls to a stop at the top of the potential hill. When this happens
the de Broglie wavelength heads off to infinity and is certainly not small!
Exercise 6.3 Verify the force condition given by Eq. 6.32.
Going back to the expansion for χ
χ1 = −
1 χ0o
1 V0
=
2 χo
4E −V
(6.33)
So00
p0
=
−
2S 0
2p
(6.34)
or equivalently for S1
S10 = −
So,
1
S1 (x) = − log p(x)
2
183
If we stick to regions where the semi-classical condition is met, then the wavefunction becomes
R
i
C1
C2 − i R p(x)dx
e h̄
ψ(x) ≈ q
e h̄ p(x)dx + q
p(x)
p(x)
(6.35)
√
The 1/ p prefactor has a remarkably simple interpretation. The probability of find the particle
in some region between x and x+dx is given by |ψ|2 so that the classical probability is essentially
proportional to 1/p. So, the fast the particle is moving, the less likely it is to be found in some
small region of space. Conversly, the slower a particle moves, the more likely it is to be found in
that region. So the time spend in a small dx is inversely proportional to the momentum of the
particle. We will return to this concept in a bit when we consider the idea of time in quantum
mechanics.
The C1 and C2 coefficients are yet to be determined. If we take x = a to be one classical
turning point so that x > a corresponds to the classically inaccessible region where E < V (x),
then the wavefunction in that region must be exponentially damped:
1Zx
exp −
|p(x)|dx
h̄ a
|p|
C
ψ(x) ≈ q
(6.36)
To the left of x = a, we have a combination of incoming and reflected components:
C1
iZa
C2
iZa
ψ(x) = √ exp
pdx + √ exp −
pdx
p
h̄ x
p
h̄ x
6.2.3
(6.37)
Semi-classical Tunneling and Barrier Penetration
Before solving the general problem of how to use this in an arbitrary well, let’s consider the case
for tunneling through a potential barrier that has some bumpy top or corresponds to some simple
potential. So, to the left of the barrier the wavefunction has incoming and reflected components:
ψL (x) = Aeikx + Be−ikx .
(6.38)
Inside we have
C
ψB (x) = q
i
|p(x)|
e+ h̄
R
|p|dx
R
i
D
+q
e− h̄ |p|dx
|p(x)|
(6.39)
and to the right of the barrier:
ψR (x) = F e+ikx .
(6.40)
If F is the transmitted amplitude, then the tunneling probability is the ratio of the transmitted
probability to the incident probability: T = |F |2 /|A|2 . If we assume that the barrier is high or
broad, then C = 0 and we obtain the semi-classical estimate for the tunneling probability:
2Zb
T ≈ exp −
|p(x)|dx
h̄ a
184
!
(6.41)
where a and b are the turning points on either side of the barrier.
Mathematically, we can “flip the potential upside down” and work in imaginary time. In this
case the action integral becomes
S=
Z
b
q
2m(V (x) − E)dx.
(6.42)
a
So we can think of tunneling as motion under the barrier in imaginary time.
There are a number of useful applications of this formula. Gamow’s theory of alpha-decay is
a common example (c.f. Griffiths). Another useful application is in the theory of reaction rates
where we want to determine tunneling corrections to the rate constant for a particular reaction.
Close to the top of the barrier, where tunneling may be important, we can expand the potential
and approximate the peak as an upside down parabola
k
V (x) ≈ Vo − x2
2
where +x represents the product side and −x represents the reactant side. See Fig. 6.1 Set the
zero in energy to be the barrier height, Vo so that any transmission for E < 0 corresponds to
tunneling. 1
e
0
-0.2
-0.4
-0.6
-0.8
-4
-2
2
4
x
Figure 6.1: Eckart Barrier and parabolic approximation of the transition state
At sufficiently large distances from the turning point, the motion is purely quasi-classical and
we can write the momentum as
q
q
√
p = 2m(E + kx2 /2) ≈ x mk + E m/k/x
(6.43)
and the asymptotic for of the Schrödinger wavefunction is
ψ = Ae+iξ
2 /2
ξ +i−1/2 + Be−iξ
2 /2
ξ −i−1/2
(6.44)
where A and B are coefficients we need to determine by the matching condition
q and ξ and are
1/4
dimensionless lengths and energies given by ξ = x(mk/h̄) , and = (E/h̄) m/k.
1
The analysis is from Kembel, 1935 as discussed in Landau and Lifshitz, QM
185
The particular case we are interested in is for a particle coming from the left and passing to
the right with the barrier in between. So, the wavefunctions in each of these regions must be
ψR = Be+iξ
2 /2
ξ i−1/2
(6.45)
and
ψL = e−iξ
2 /2
(−ξ)−i−1/2 + Ae+iξ
2 /2
(−ξ)i−1/2
(6.46)
where the first term is the incident wave and the second term is the reflected component. So,
|A|2 | is the reflection coefficient and |B|2 is the transmission coefficient normalized so that
|A|2 + |B|2 = 1.
Lets move to the complex plane and write a new coordinate, ξ = ρeiφ and consider what happens
as we rotate around in φ and take ρ to be large. Since iξ 2 = ρ2 (i cos 2φ − sin 2φ), we have
2
ψR (φ = 0) = Beiρ ρ+i−1/2
2
ψL (φ = 0) = Aeiρ (−ρ)+i−1/2
(6.47)
and at φ = π
2
ψR (φ = π) = Beiρ (−ρ)+i−1/2
2
ψL (φ = π) = Aeiρ ρ+i−1/2
(6.48)
So, in otherwords, ψR (φ = π) looks like ψL (φ = 0) when
A = B(eiπ )i−1/2
So, we have the relation A = −iBe−π . Finally, after we normalize this we get the transimission
coefficient:
1
T = |B|2 =
1 + e−2π
which must hold for any energy. If the energy is large and negative, then
T ≈ e−2π .
Also, we can compute the reflection coefficient for E > 0 as 1 − D,
R=
1
.
1 + e+2π
Exercise 6.4 Verify these last relationships by taking the ψR and ψL , performing the analytic
continuation.
186
This gives us the transmission probabilty as a function of incident energy. But, normal
chemical reactions are not done at constant energy, they are done at constant temperature.
To get the thermal transmission coefficient, we need to take a Boltzmann weighted average of
transmission coefficients
Tth (β) =
1Z
dEe−Eβ T (E)
Z
(6.49)
where β = 1/kT and Z is the partition function. If E represents a continuum of energy states
then
Tth (β) = −
βωh̄(ψ (0) ( βωh̄
) − ψ (0) ( 14 ( βωh̄
+ 2)))
4π
π
4π
(6.50)
where ψ (n) (z) is the Polygamma function which is the nth derivative of the digamma function,
ψ (0) (z), which is the logarithmic derivative of Eulers gamma function, ψ (0) (z) = Γ(z)/Γ(z).2
6.3
Connection Formulas
In what we have considered thus far, we have assumed that up until the turning point the
wavefunction was well behaved and smooth. We can think of the problem as having two domains:
an exterior and an interior. The exterior part we assumed to be simple and the boundary
conditions trivial to impose. The next task is to figure out the matching condition at the turning
point for an arbitrary system. So far what we have are two pieces, ψL and ψR , in the notation
above. What we need is a patch. To do so, we make a linearizing assumption for the force at
the classical turning point:
E − V (x) ≈ Fo (x − a)
(6.51)
where Fo = −dV /dx evaluated at x = a. Thus, the phase integral is easy:
2q
1Zx
pdx =
2mFo (x − a)3/2
h̄ a
3h̄
(6.52)
But, we can do better than that. We can actually solve the Schrodinger equation for the linear
potential and use the linearized solutions as our patch. The Mathematica Notebook AiryFunctions.nb goes through the solution of the linearized Schrodinger equation
−
h̄2 dψ
+ (E + V 0 )ψ = 0
2m dx2
(6.53)
which can be re-written as
ψ 00 = α3 xψ
with
2m 0
α=
V (0)
h̄2
2
See the Mathematica Book, sec 3.2.10.
187
(6.54)
1/3
.
Absorbing the coefficient into a new variable y, we get Airy’s equation
ψ 00 (y) = yψ.
The solutions of Airy’s equation are Airy functions, Ai(y) and Bi(y) for the regular and irregular
cases. The integral representation of the Ai and Bi are
!
s3
1Z∞
cos
Ai(y) =
+ sy ds
π 0
3
(6.55)
"
!#
s3
1 Z ∞ −s3 /3+sy
Bi(y) =
e
+ sin
+ sy ds
π 0
3
(6.56)
and
Plots of these functions are shown in Fig. 6.2.
Ai@yD, Bi@yD
1.5
1
0.5
y
-10
-8
-6
-4
-2
2
-0.5
Figure 6.2: Airy functions, Ai(y) (red) and Bi(y) (blue)
Since both Ai and Bi are acceptible solutions, we will take a linear combination of the two
as our patching function and figure out the coefficients later.
ψP = aAi(αx) + bBi(αx)
(6.57)
We now have to determine those coefficients. We need to make two assumptions. One,
that the overlap zones are sufficiently close to the turning point that a linearized potential is
reasonable. Second, the overlap zone is far enough from the turning point (at the origin) that
the WKB approximation is accurate and reliable. You can certainly cook up some potential
for which this will not work, but we will assume it’s reasonable. In the linearized region, the
momentum is
p(x) = h̄α3/2 (−x)3/2
(6.58)
|p(x)|dx = 2h̄(αx)3/2 /3
(6.59)
So for +x,
Z
x
0
188
and the WKB wavefunction becomes:
D
3/2
e−2(αx) /3 .
ψR (x) = √
3/4
1/4
h̄α x
(6.60)
In order to extend into this region, we will use the asymptotic form of the Ai and Bi functions
for y 0
3/2
e−2y /3
Ai(y) ≈ √ 1/4
2 πy
(6.61)
3/2
e+2y /3
Bi(y) ≈ √ 1/4 .
πy
(6.62)
Clearly, the Bi(y) term will not contribute, so b = 0 and
s
a=
4π
D.
αh̄
Now, for the other side, we do the same proceedure. Except this time x < 0 so the phase
integral is
Z
0
pdx = 2h̄(−αx)3/2 /3.
(6.63)
x
Thus the WKB wavefunction on the left hand side is
1 3/2
3/2
ψL (x) = √ Be2i(−αx) /3 + Ce−2i(−αx) /3
p
1
3/2
3/2
= √
Be2i(−αx) /3 + Ce−2i(−αx) /3
h̄α3/4 (−x)1/4
(6.64)
(6.65)
That’s the WKB part, to connect with the patching part, we again use the asymptotic forms for
y 0 and take only the regular solution,
1
3/2
sin
2(−y)
/3
+
π/4
π(−y)1/4
1
iπ/4 i2(−y)3/2 /3
−iπ/4 −i2(−y)3/2 /3
√
≈
e
e
−
e
e
2i π(−y)1/4
Ai(y) ≈ √
(6.66)
Comparing the WKB wave and the patching wave, we can match term-by-term
a
B
√ eiπ/4 = √
2i π
h̄α
−a −iπ/4
C
√ e
=√
2i π
h̄α
(6.67)
(6.68)
Since we know a in terms of the normalization constant D, B = ieiπ/4 D and C = ie−iπ/4 . This
is the connection! We can write the WKB function across the turning point as
h R
i

2D
1 0

 √p(x) sin h̄ x pdx + π/4
ψW KB (x) =  2D − 1 R 0 pdx
 √
e h̄ x
|p(x)|
189
x<0
x>0
(6.69)
Table 6.1: Location of nodes for Airy, Ai(x) function.
node
1
2
3
4
5
6
7
xn
-2.33811
-4.08795
-5.52056
-6.78671
-7.94413
-9.02265
-10.0402
Example: Bound states in the linear potential
Since we worked so hard, we have to use the results. So, consider a model problem for a particle
in a gravitational field. Actually, this problem is not so far fetched since one can prepare trapped
atoms above a parabolic reflector and make a quantum bouncing ball. Here the potential is
V (x) = mgx where m is the particle mass and g is the graviational constant (g = 9.80m/s).
We’ll take the case where the reflector is infinite so that the particle cannot penetrate into it.
The Schrödinger equation for this potential is
−
h̄2 00
ψ + (E − mgx)ψ = 0.
2m
(6.70)
The solutions are the Airy Ai(x) functions. Setting, β = mg and c = h̄2 /2m, the solutions are
β
ψ = CAi( −
c
!1/3
(x − E/β))
(6.71)
However, there is one caveat: ψ(0) = 0, thus the Airy functions must have their nodes at x = 0.
So we have to systematically shift the Ai(x) function in x until a node lines up at x = 0. The
nodes of the Ai(x) function can be determined and the first 7 of them are To find the energy
levels, we systematically solve the equation
β
−
c
!1/3
En
= xn
β
So the ground state is where the first node lands at x = 0,
2.33811β
(β/c)1/3
2.33811mg
=
(2m2 g/h̄2 )1/3
E1 =
(6.72)
and so on. Of course, we still have to normalize the wavefunction to get the correct energy.
We can make life a bit easier by using the quantization condition derived from the WKB
approximation. Since we require the wavefunction to vanish exactly at x = 0, we have:
1 Z xt
π
p(x)dx + = nπ.
h̄ 0
4
190
(6.73)
15
10
5
2
4
6
8
10
-5
-10
Figure 6.3: Bound states in a graviational well
This insures us that the wave vanishes at x = 0, xt in this case is the turning point E = mgxt .
(See Figure 6.3) As a consequence,
Z
xt
p(x)dx = (n − 1/4)π
0
q
Since p(x) =
Z
xt
2m(En − mgx), The integral can be evaluated
q
2m(E − mghdx =
√
0

2
√
2En En m
+
3gm
q
2 m (En − gmxt ) (−En + gmxt )
3gm

 (6.74)
Since xt = En /mg for the classical turning point, the phase intergral becomes
√
2 2En 2
{ √
} = (n − 1/4)πh̄.
3g En m
Solving for En yields the semi-classical approximation for the eigenvalues:
2
En =
1
g 3 m 3 (1 − 4 n)2
1
4 23
1
3
2
2
(3 π) 3 h̄ 3
(6.75)
In atomic units, the gravitional constant is g = 1.08563 × 10−22 bohr/au2 (Can you guess why
we rarely talk about gravitational effects on molecules?). For n = 0, we get for an electron
Eosc = 2.014 × 10−15 hartree or about 12.6 Hz. So, graviational effects on electrons are extremely
tiny compared to the electron’s total energy.
191
6.4
Problems and Exercises
Exercise 6.5 In this problem we will (again) consider the ammonia inversion problem, this time
we will proceed in a semi-classical context.
Recall that the ammonia inversion potential consists of two symmetrical potential wells separated by a barrier. If the barrier was impenetrable, one would find energy levels corresponding to
motion in one well or the other. Since the barrier is not infinite, there can be passage between
wells via tunneling. This causes the otherwise degenerate energy levels to split.
In this problem, we will make life a bit easier by taking
V (x) = α(x4 − x2 )
as in the examples in Chapter 3.
Let ψo be the semi-classical wavefunction describing the motion in one well with energy Eo .
Assume that ψo is exponentially damped on both sides of the well and that the wavefunction
is normalized so that the integral over ψo2 is unity. When tunning is taken into account, the
wavefunctions corresponding to the new energy levels, E1 and E2 are the symmetric and antisymmetric combinations of ψo (x) and ψo (−x)
√
ψ1 = (ψo (x) + ψo (−x)/ 2
√
ψ2 = (ψo (x) − ψo (−x)/ 2
where ψo (−x) can be thought of as the contribution from the zeroth order wavefunction in the
other well. In Well 1, ψo (−x) is very small and in well 2, ψo (+x) is very small and the product
ψo (x)ψo (−x) is vanishingly small everywhere. Also, by construction, ψ1 and ψ2 are normalized.
1. Assume that ψo and ψ1 are solutions of the Schrödinger equations
ψo00 +
2m
(Eo − V )ψo = 0
h̄2
and
2m
(E1 − V )ψ1 = 0,
h̄2
Multiply the former by ψ1 and the latter by ψo , combine and subtract equivalent terms, and
integrate over x from 0 to ∞ to show that
ψ100 +
h̄2
E1 − Eo = − ψo (0)ψo0 (0),
m
Perform a similar analysis to show that
E2 − Eo = +
h̄2
ψo (0)ψo0 (0),
m
2. Show that the unperturbed semiclassical wavefunction is
s
ψo (0) =
ω
1Za
exp −
|p|dx
2πvo
h̄ 0
192
and
ψo0 (0) =
where vo =
mvo
ψo (0)
h̄
q
2(Eo − V (0))/m and a is the classical turning point at Eo = V (a).
3. Combining your results, show that the tunneling splitting is
h̄ω
1 Z +a
exp −
|p|dx .
π
h̄ −a
∆E =
where the integral is taken between classical turning points on either side of the barrier.
4. Assuming that the potential in the barrier is an upside-down parabola
V (x) ≈ Vo − kx2 /2
what is the tunneling splitting.
5. Now, taking α = 0.1 expand the potential about the barrier and compute determine the
harmonic force constant for the upside-down parabola. Using the equations you derived and
compute the tunneling splitting for a proton in this well. How does this compare with the
calculations presented in Chapter 3.
193
Chapter 7
Many Body Quantum Mechanics
It is often stated that of all the theories proposed in this century, the silliest is
quantum theory. In fact, some say that the only thing that quantum theory has
going for it is that it is unquestionably correct.
–M. Kaku (Hyperspace, Oxford University Press, 1995)
7.1
Symmetry with respect to particle Exchange
Up to this point we have primarily dealt with quantum mechanical system for 1 particle or with
systems of distinguishable particles. By distinguishable we mean that one can assign a unique
label to the particle which distinguishes it from the other particles in the system. Electrons in
molecules and other systems, are identical and cannot be assigned a unique label. Thus, we must
concern our selves with the consequences of exchanging the labels we use. To establish a firm
formalism and notation, we shall write the many-particle wavefunction for a system as
hψN |ψN i =
=
Z
Z
···
Z
d3 r1 · · · d3 rN |ψN (r1 , r2 , · · · , rN )|2 < +∞
(7.1)
Z
d1d2 · · · dN |ψN (1, 2, · · · , N )|2 .
(7.2)
···
We will define the N -particle state space as the product of the individual single particle state
spaces thusly
|ψN i = |a1 a2 · · · aN ) = |a1 i ⊗ |a2 i ⊗ · · · ⊗ |aN i
(7.3)
For future reference, we will write the multiparticle state as with a curved bracket: | · · ·). These
states have wavefunctions
hr|ψN i = (r1 · · · rn |a1 a2 · · · aN ) = hr1 |a1 ihr2 |a2 i · · · hrN |aN i
(7.4)
= φa1 (r1 )φa1 (r2 ) · · · φaN (rN )
(7.5)
194
These states obey analogous rules for constructing overlap (projections) and idempotent relations.
They form a complete set of states (hence form a basis) and any multi-particle state in the state
space can be constructed as a linear combination of the basis states.
Thus far we have not taken into account the symmetry property of the wavefunction. There
are a multitude of possible states which one can construct using the states we defined above.
However, only symmetric and antisymmetric combinations of these state are actually observed in
nature. Particles occurring in symmetric or anti-symmetric states are called Bosons and Fermions
respectively.
Let’s define the permutation operator Pαβ which swaps the positions of particles α and β.
e.g.
P12 |1, 2) = |2, 1)
(7.6)
2
P12 P12 ψ(1, 2) = P12
ψ(1, 2) = ψ(1, 2)
(7.7)
Also,
thus ψ(1, 2) is an eigenstate of P12 with eigenvalue ±1. In other words, we can also write
P12 ψ(1, 2) = ζψ(1, 2)
(7.8)
where ζ = ±1.
A wavefunction of N bosons is totally symmetric and thus satisfies
ψ(P 1, P 2, · · · , P N ) = ψ(1, 2, · · · , N )
(7.9)
where (P 1, P 2, · · · , P N ) represents any permutation P of the set (1, 2, · · · , N ). A wavefunction
of N fermions is totally antisymmetric and thus satisfies
ψ(P 1, P 2, · · · , P N ) = (−1)P ψ(1, 2, · · · , N ).
(7.10)
Here, (−1)P denotes the sign or parity of the permutation and is defined as the number of binary
transpositions which brings the permutation (P 1, P 2, ...) to its original from (1, 2, 3...).
For example: what is the parity of the permutation (4,3,5,2,1)? A sequence of binary transpositions is
(4, 3, 5, 2, 1) → (2, 3, 5, 4, 1) → (3, 2, 5, 4, 1) → (5, 2, 3, 4, 1) → (1, 2, 3, 4, 5)
(7.11)
So P = 4. Thus, for a system of 5 fermions
ψ(4, 3, 5, 2, 1) = ψ(1, 2, 3, 4, 5)
(7.12)
In cases where we want to develop the many-body theory for both Fermions and Bosons
simultaneously, we will adopt the notation that ζ = ±1 and any wavefunction can be written as
ψ(P 1, P 2, · · · , P N ) = (ζ)P ψ(1, 2, · · · , N ).
where ζ = −1 for fermions and +1 for bosons.
195
(7.13)
While these symmetry requirement are observed in nature, they can also be derived in the
context of quantum field theory that given general assumptions of locality, causality and Lorentz
invariance, particles with half integer spin are fermions and those with integer spin are bosons.
Some examples of bosons are photon, photons, pions, mesons, gluons and the 4 He atom. Some
examples of fermions are protons, electrons, neutrons, muons, neutrinos, quarks, and the 3 He
atom. Composite particles composes of any number of bosons and even or odd numbers of
fermions behave as bosons or fermions respectively at temperatures low compared to their binding
energy. An example of this is super-conductivity where electron-phonon coupling induces the
pairing of electrons (Cooper pairs) which form a Bose-condensate.
Now, consider what happens if I place two fermion particles in the same state:
|ψ(1, 2)i = |α(1)α(2))
(7.14)
where α(1) is a state with the “spin up” quantum number. This state must be an eigenstate of
the permutation operator with eigenvalue ζ = −1.
P12 |ψ(1, 2)i = −|α(2)α(1))
(7.15)
However, |α(1)α(2)) = |α(2)α(1)), thus the wavefunction of the state must vanish everywhere.
For the general case of a system with N particles, the normalized wavefunction is
N1 !N2 !...
N!
ψ=
!1/2
X
ψp1 (1)ψp2 (2) · · · ψpN (N )
(7.16)
where the sum is over all permutations of different p1 , p2 ... and the numbers Ni indicate how
P
many of these have the same value (i.e. how many particles are in each state) with Ni = N .
For a system of 2 fermions, the wavefunction is
√
(7.17)
ψ(1, 2) = (ψp1 (1)ψp2 (2) − ψp1 (2)ψp2 (1))/ 2
Thus, in the example above:
√
ψ(1, 2) = (α(1)α(2) − α(2)α(1))/ 2 = 0
(7.18)
Likewise,
√
ψ(1, 2) = (β(1)α(2) − β(2)α(1))/ 2
√
= (β(1)α(2) − P12 β(1)α(2))/ 2
√
= (β(1)α(2))/ 2
(7.19)
We will write such symmetrized states using the curly brackets
|ψ} = |a1 a2 · · · aN } =
N1 !N2 !...
N!
!1/2
X
ψp1 (1)ψp2 (2) · · · ψpN (N )
(7.20)
For the general case of N particles, the fully anti-symmetrized form of the wavefunction takes
the form of a determinant
φ (1)
1 a
ψ = √ φb (1)
N ! φ (1)
c
φa (1) φa (1) φb (2) φb (2) φc (3) φc (3) 196
(7.21)
where the columns represent the particles and the rows are the different states. The interchange
of any two particles corresponds to the interchange of two columns, as a result, the determinant
changes sign. Consequently, if two rows are identical, corresponding to two particles occupying
the same state, the determinant vanishes.
Another example, let’s consider the possible states for the He atom ground state. Let’s
assume that the ground state wavefunction is the product of two single particle hydrogenic 1s
states with a spin wavefunction written thus
|ψi = |1s(1)α(1), 1s(2)β(2))
(7.22)
Let’s denote |αi as the spin up state and |βi as the spin down state. We have the following
possible spin combinations:
α(1)α(2)
α(1)β(2)
β(1)α(2)
β(1)β(2)
↑↑
↑↓
↓↑
↓↓
symmetric
nether
neither
symmetric
(7.23)
The ↑↑ and the ↓↓ states are clearly symmetric w.r.t particle exchanges. However, note that
the other two are neither symmetric nor anti-symmetric. Since we can construct linear combinations of these states, we can use the two allowed spin configurations to define the combined spin
state Thus, we get two possible total spin states:
1
√ (|α(1)β(2)) ± |β(1)α(2)))
2
(7.24)
Thus, the possible two particle spin states are
α(1)α(2)
↑↑
β(1)β(2)
↓↓
√1 (α(1)β(2) + β(1)α(2)) ↑↓ + ↑↓
2
√1 (α(1)β(2) − β(1)α(2)) ↑↓ − ↑↓
2
symmetric
symmetric
symmetric
anti-symmetric
(7.25)
These spin states multiply the spatial state and the full wavefunction must be anti-symmetric
w.r.t. exchange. For example, for the ground state of the He atom, the zero-th order spatial
state is |1s(1)1s(2)). This is symmetric w.r.t. exchange. Thus, the full ground-state wavefunction
must be the product
1
|ψi = |1s(1)1s(2)) √ (α(1)β(2) − β(1)α(2)
2
(7.26)
The full state is an eigenstate of P12 with eigenvalue -1, which is correct for a system of fermions.
What about the other states, where can we use them? What if we could construct a spatial
wavefunction that was anti-symmetric w.r.t particle exchange. Consider the first excited state
of He. The electron configuration for this state is
|1s(1)2s(2))
197
(7.27)
However, we could have also written
|1s(2)2s(1))
(7.28)
Taking the symmetric and anti-symmetric combinations
1
|ψ± i = √ (|1s(1)2s(2)) ± |1s(2)2s(1)))
2
(7.29)
The + state is symmetric w.r.t. particle exchange. Thus, the full state (including spin) must be
1
|ψ1 i = (|1s(1)2s(2)) + |1s(2)2s(1)))(α(1)β(2) − β(1)α(2)).
(7.30)
2
The other three states must be
1
|ψ2 i =
(|1s(1)2s(2)) − |1s(2)2s(1)))(α(1)β(2) + β(1)α(2))
(7.31)
2
1
(7.32)
|ψ3 i = √ (|1s(1)2s(2)) − |1s(2)2s(1)))(α(1)α(2))
2
1
|ψ4 i = √ (|1s(1)2s(2)) − |1s(2)2s(1)))(β(1)β(2)).
(7.33)
2
These states can also be constructed using the determinant wavefunction. For example, the
ground state configurations are generated using
1 1s(1)α(1) 1s(1)β(1)
|ψg } = √ 2 1s(2)α(2) 1s(2)β(2)
(7.34)
1
= √ |1s(1)1s(2))[α(1)β(2) − α(2)β(1)]
2
Likewise for the excited states, we have 4 possible determinant states
|ψ1 } =
|ψ2 } =
|ψ3 } =
|ψ4 } =
1 √ 2
1 √ 2
1 √ 2
1 √ 2
(7.35)
1s(1)α(1) 2s(1)α(1) 1s(2)α(2) 2s(2)α(2) 1s(1)α(1) 2s(1)β(1) 1s(2)α(2) 2s(2)β(2) 1s(1)β(1) 2s(1)α(1) 1s(2)β(2) 2s(2)α(2) 1s(1)β(1) 2s(1)β(1) 1s(2)β(2) 2s(2)β(2) (7.36)
The |ψm ) are related to the determinant states as follows
1
|ψ1 } = √ [1s(1)α(1)2s(2)α(2) − 1s(2)α(2)2s(1)α(1)]
2
1
= √ [1s(1)2s(2) − 1s(2)2s(1)] α(1)α(2)
2
= |ψ3 }
1
|ψ4 } = √ [1s(1)2s(2) − 1s(2)2s(1)] β(1)β(2) = |ψ4 }
2
198
(7.37)
The remaining two must be constructed from linear combinations of the determinant states:
1
|ψ2 } = √ [1s(1)α(1)2s(2)β(2) − 1s(2)α(2)2s(1)β(1)]
2
1
|ψ3 } = √ [1s(1)β(1)2s(2)α(2) − 1s(2)β(2)2s(1)α(1)]
2
1
|ψ2 i = √ [|ψ2 } + |ψ3 }]
2
1
|ψ1 i = √ [|ψ2 } − |ψ3 }]
2
(7.38)
(7.39)
(7.40)
(7.41)
When dealing with spin functions, a short hand notation is often used to reduced the notation
a bit. The notation
|1si ≡ |1sαi
(7.42)
|1si ≡ |1sβi
(7.43)
is used to denote a spin up state and
Using these, the above determinant functions can be written as
1 1s(1) 2s(1) √ 2 1s(2) 2s(2) i
1 h
√ |1s(1)2s(2)) − |1s(2)2s(1))
2
1 1s(1) 2s(1) √ 2 1s(2) 2s(2) i
1 h
√ |1s(1)2s(2)) − |1s(2)2s(1))
2
|ψ1 } =
=
|ψ2 } =
=
(7.44)
(7.45)
(7.46)
(7.47)
The symmetrization principal for fermions is often expressed as the Pauli exclusion principle
which states: no two fermions can occupy the same same state at the same time. This, as we all
well know gives rise to the periodic table and is the basis of all atomic structure.
7.2
Matrix Elements of Electronic Operators
We can write the Hamiltonian for a N -body problem as follows. Say our N -body Hamiltonian
consists of a sum of N single particle Hamiltonian, Ho , and two body interactions.
H=
N
X
n
Ho(n) +
X
V (ri − rj )
(7.48)
i6=j
Using the field operators, the expectation values of the Ho terms are
XZ
dxφ∗α (x)Ho φα (x) =
α
X
α
199
nα Wα
(7.49)
since φα (x) is an eigenstate of Ho with eigenvalue Eα . α should be regards as a collection of all
quantum numbers used to describe the Ho eigenstate.
For example, say we want a zeroth order approximation to the ground state of He and we
use hydrogenic functions,
|ψo i = |1s(2)1s(2)).
(7.50)
This state is symmetric w.r.t electron exchange, so the spin component must be anti-symmetric.
For now this will not contribute the the calculation. The zero-th order Schroedinger Equation is
(Ho (1) + Ho (2))|ψo i = Eo |ψo i
(7.51)
Where Ho (1) is the zero-th order Hamiltonian for particle 1. This is easy to solve
(Ho (1) + Ho (2))|ψo i = −Z 2 |ψo i
(7.52)
Z = 2, so the zero-th order guess to the He ground state energy is −4 (in Hartree units, recall 1
Hartree = 27.6 eV). The correct ground state energy is more like -2.90 Hartree. Let’s now evaluate
to first order in perturbation theory the direct Coulombic interaction between the electrons.
E (1) = (1s(1)1s(2)|
1
|1s(1)1s(2))
r12
(7.53)
The spatial wavefunction for the |1s(1)1s(2)) state is the product of two hydrogenic functions
Z 3 −Z(r1 +r2 )
e
π
(7.54)
1 Z 3 −2Z(r1 +r2 )
e
r12 π
(7.55)
(r1 r2 |1s(1)1s(2)) =
Therefore,
E
(1)
=
Z
dV1
Z
dV2
where dV = 4πr2 dr is the volume element. This integral is most readily solved if we instead
write it as the energy of a charge distribution, ρ(2) = |ψ(2)|2 , in the field of a charge distribution
ρ(1) = |ψ(1)|2 for r2 > r1 .
E (1) = 2
Z3 Z
1 Z r2
dV2 e−2Zr2
dV1 e−2Zr1
π
r2 0
(7.56)
The factor of 2 takes into account that we get the same result for when r1 > r2 . Doing the
integral, (See subsection 7.3.1.)
E (1) =
5Z
8
(7.57)
or 1.25 Hartee. Thus, for the He atom
E = Eo + E (1) = −Z 2 +
5Z
= −4 + 1.25 = −2.75
8
200
(7.58)
Not too bad, the actual result is −2.90 Hartee.
What we have not taken into consideration that there is an additional contribution to the
energy from the exchange interaction. In other words, we need to compute the Coulomb integral
exchanging electron 1 with electron 2. We really need to compute the perturbation energy with
respect the determinant wavefunction.
E (1) = {1s(1)1s(2)|v|1s(1)1s(2)}
1
=
(1s(1)1s(2)| − (1s(1)1s(2)|) v |1s(1)1s(2)) − |1s(1)1s(2))
2
1
=
(1s(1)1s(2)|v|1s(1)1s(2)) − (1s(1)1s(2)|v|1s(1)1s(2))
2
− (1s(1)1s(2)|v|1s(1)1s(2)) + (1s(1)1s(2)|v|1s(1)1s(2))
= 2(1s(1)1s(2)|v|1s(1)1s(2)) − (1s(1)1s(2)|v|1s(1)1s(2))
(7.59)
(7.60)
(7.61)
(7.62)
However, the potential does not depend upon spin. Thus any matrix element which exchanges
a spin from must vanish. This, we have no exchange contribution to the energy. We can in fact
move on to higher orders in perturbation theory and solve accordingly.
7.3
The Hartree-Fock Approximation
We just saw how to estimate the ground state energy of a system in the presence of interactions
using first order perturbation theory. To get this result, we assumed that the zeroth-order
wavefunctions were pretty good and calculated our results using these wavefunctions. Of course,
the true ground state energy is obtained by summing over all diagrams in the perturbation
expansion:
E − W = hψo |E − W |ψo i
(7.63)
The second order term contains explicit two-body correlation interactions. i.e. the motion of one
electron affects the motion of the other electron.
Let’s make a rather bold assumption that we can exclude connected two body interactions
and treat the electrons as independent but moving in an averaged field of the other electrons.
First we make some standard definitions:
Jαβ = (αβ|v|αβ)
Kαβ = (αβ|v|βα)
(7.64)
(7.65)
And write that
EHF − W =
X
(2Jαβ − Kαβ )
(7.66)
αβ
where sum over all occupied (n/2) spatial orbitals of a n−electron system. The J integral is the
direct interaction (Coulomb integral) and the K is the exchange interaction.
201
We now look for a a set of orbitals which minimize the variational integral EHF , subject to the
constrain that the wavefunction solutions be orthogonal. One can show (rather straightforwardly)
that if we write the Hamiltonian as a functional of the electron density, ρ,
H[ρ] = Ho [1, 2] +
Z Z
d3d4{12|v|34}ρ(3, 4)
= Ho [1] + 2J(1) − K(1)
(7.67)
(7.68)
where ρ(1, 2) = φ(1)∗ φ(2),
J(1) =
K(1) =
Z
d2|β(2)|2 v(12)
(7.69)
Z
d2α(2)∗ β(2)v(12)
(7.70)
The Hartree-Fock wavefunctions satisfy
H[ρ]ψ(1) = E(1)ψ(1)
(7.71)
In other words, we diagonalize the Fock-matrix H[ρ] given an initial guess of the electron density.
This gives a new set of electron orbitals, which we use to construct a new guess for the electron
densities. This procedure is iterated to convergence.
7.3.1
Two electron integrals
One of the difficulties encountered is in evaluating the Jαβ and Kαβ two electron integrals. Let’s
take the case that the φα and φβ orbitals are centred on the same atom. Two centred terms can
be evaluated, but the analysis is more difficult. Writing the J integral in Eq. 7.65 out explicitly
we have:
Jαβ = (φα (1)φβ (2)|v(12)|φα (1)φβ (1))
=
=
Z Z
Z
φ∗α (1)φ∗β (2)v(1 − 2)φα (1)φ2β d1d2
φ∗α (r1 )φα (r1 )
Z
φ∗β (r2 )φβ (r2 )v(r1 − r2 )dr2 dr1
(7.72)
If we can factor the single particle orbitals as φα (r, θ, φ) = Rnl (r)Ylm (θ, φ), then we can separate
the radial and angular integrals. Before we do that, we have to resolve the pair-interaction into
radial and angular components as well. For the Coulomb potential, we can use the expansion
+l
l
X X
1
4π r<
∗
=
l+1 Ylm (θ1 φ1 )Ylm (θ2 φ2 )
|r1 − r2 | l=0 m=−l 2l + 1 r<
(7.73)
where the notation r< denotes which is the smaller of r1 and r2 and r> the greater. For the
hydrogen 1s orbitals (normalized and in atomic units)
s
φ1s =
Z 3 −Zr
e ,
π
202
(7.74)
the J integral for the 1s1s configuration is
Z 6 Z 3 Z 3 −2Zr1 −2Zr2 1
d r1 d r2 e
e
π2
r12
√
Inserting the expansion and using Y00 = 1/ 4π,
J=
J = 16Z 6
l
X X
l
1 Z ∞ Z ∞ −2Zr1 −2Zr2 r<
2
2
e
e
l+1 r1 dr1 r2 dr2
2l
+
1
0
0
r
<
m=−l
l
×
Z
(7.75)
Z
∗
Ylm
(1)Y00 (1)dΩ1
Ylm (2)Y00∗ (2)dΩ2 .
(7.76)
The last two integrals are easy due to the orthogonality of the spherical harmonics. This leaves
the double integral,
J = 16Z 6
∞
Z
∞
Z
0
e−2Zr1 e−2Zr2
0
1 2
r1 dr1 r22 dr2
r>
(7.77)
which we evaluate by splitting into two parts,
J = 16Z
2
Z
∞
−2Zr1
e
0
+
Z
0
∞
e−2Zr1 r12
Z
Z
r1
∞
r1
0
r1
e−2Zr2 r22 dr2
e−2Zr2 r2 dr2 dr1
dr1
(7.78)
In this case the integrals are easy to evaluate and
J=
7.3.2
5Z
8
(7.79)
Koopman’s Theorem
Koopman’s theorem states that if the single particle energies are not affected by adding or
removing a single electron, then the ionization energy is energy of the highest occupied single
particle orbital (the HOMO) and the electron affinity is the energy of the lowest unoccupied
orbital (i.e. the LUMO). For the Hartree-Fock orbitals, this is theorem can be proven to be
exact since correlations cancel out at the HF level. For small molecules and atoms, the theorem
fails miserably since correlations play a significant role. On the other hand, for large polyatomic
molecules, Koopman’s theorem is extremely useful in predicting ionization energies and spectra.
From a physical point of view, the theorem is never exact since it discounts relaxation of both
the electrons and the nuclei.
7.4
Quantum Chemistry
Quantum chemical concepts play a crucial role in how we think about and describe chemical
processes. In particular, the term quantum chemistry usually denotes the field of electronic
structure theory. There is no possible way to cover this field to any depth in a single course and
this one section will certainly not prepare anyone for doing research in quantum chemistry. The
topic itself can be divided into two sub-fields:
203
• Method development: The development and implementation of new theories and computational strategies to take advantage of the increaseing power of computational hardware.
(Bigger, stronger, faster calculations)
• Application: The use of established methods for developing theoretical models of chemical
processes
Here we will go in to a brief bit of detail into various levels of theory and their implementation
in standard quantum chemical packages. For more in depth coverage, refer to
1. Quantum Chemistry, Ira Levine. The updated version of this text has a nice overview of
methods, basis sets, theories, and approaches for quantum chemistry.
2. Modern Quanutm Chemistry, A. Szabo and N. S. Ostlund.
3. Ab Initio Molecular Orbital Theory, W. J. Hehre, L. Radom, P. v. R. Schleyer, and J. A.
Pople.
4. Introduction to Quantum Mechanics in Chemistry, M. Ratner and G. Schatz.
7.4.1
The Born-Oppenheimer Approximation
The fundimental approximation in quantum chemistry is the Born Oppenheimer approximation we discussed earlier. The idea is that because the mass of an electron is at least 10−4 that of
a typical nuclei, the motion of the nuclei can be effectively ignored and we can write an electronic
Schrödinger equation in the field of fixed nuclei. If we write r for electronic coordinates and R
for the nuclear coordinates, the complete electronic/nuclear wavefunction becomes
Ψ(r, R) = ψ(r; R)χ(R)
(7.80)
where ψ(r; R) is the electronic part and χ(R) the nuclear part. The full Hamiltonian is
H = Tn + Te + Ven + Vnn + Vee
(7.81)
Tn is the nuclear kinetic energy, Te is the electronic kinetic energy, and the V ’s are the electronnuclear, nuclear-nuclear, and electron-electron Coulomb potential interactions. We want Ψ to be
a solution of the Schrödinger equation,
HΨ = (Tn + Te + Ven + Vnn + Vee )ψχ = Eψχ.
(7.82)
So, we divide through by ψχ and take advantage of the fact that Te does not depend upon the
nuclear component of the total wavefunction
1
1
Tn ψχ + Te ψ + Ven + Vnn + Vee = E.
ψχ
ψ
On the other hand, Tn operates on both components, and involves terms which look like
Tn ψχ =
X
n
−
1
(ψ∇2n χ + χ∇2n ψ + 2∇χ · ∇n ψ)
2Mn
204
where the sum is over the nuclei and ∇n is the gradient with respect to nuclear position n. The
crudest approximation we can make is to neglect the last two terms–those which involve the
derivatives of the electronic wave with respect to the nuclear coordinate. So, when we neglect
those terms, the Schrodinger equation is almost separable into nuclear and electronic terms
1
1
(Tn + Vn )χ + (Te + Ven + Vee )ψ = E.
χ
ψ
(7.83)
The equation is not really separtable since the second term depends upon the nuclear position.
So, what we do is say that the electronic part depends parametrically upon the nuclear position
giving a constant term, Eel (R), that is a function of R.
(Te + Ven (R) + Vee )ψ(r; R) = Eel (R)ψ(r; R).
(7.84)
The function, Eel , depends upon the particular electronic state. Since it an eigenvalue of Eq. 7.84,
there may be a manifold of these functions stacked upon each other.
Turning towards the nuclear part, we have the nuclear Schrödinger equation
(α)
(Tn + Vn (R) + Eel (R))χ = W χ.
(7.85)
(α)
Here, the potential governing the nuclear motion contains the electronic contribution, Eel (R),
which is the αth eigenvalue of Eq. 7.84 and the nuclear repulsion energy Vn . Taken together,
these form a potential energy surface
(α)
V (R) = Vn + Eel (R)
for the nuclear motion. Thus, the electronic energy serves as the interaction potential between
the nuclei and the motion of the nuclei occurs on an energy surface generated by the electronic
state.
Exercise 7.1 Derive the diagonal non-adiabatic correction term hψ|Tn |ψ > to produce a slightly
more accurate potential energy surface
(α)
V (R) = Vn + Eel + hψα |Tn |ψα i
.
The BO approximation breaks down when the nuclear motion becomes very fast and the
electronic states can become coupled via the nuclear kinetic energy operator. (One can see by
inspection that the electonic states are not eigenstates of the nuclear kinetic energy since Hele
does not commute with ∇2N . )
Let’s assume that the nuclear motion is a time dependent quantity, R(t). Now, take the time
derivative of |Ψen i
d e
∂R(t) ∂
∂
|Ψn (R(t))i =
|Ψen (R)i + |Ψen (R(t))i
dt
∂t ∂R
∂t
(7.86)
Now, multiply on the left by < Ψem (R(t))| where m 6= n
hΨem (R(t))|
d e
∂R(t) e
∂
|Ψn (R(t))i =
hΨm (R(t))|
|Ψen (R)i
dt
∂t
∂R
205
(7.87)
Cleaning things up,
hΨem (R(t))|
d e
|Ψ (R(t))i = Ṙ(t)· < Ψem (R(t))|∇N |Ψen (R)i
dt n
(7.88)
we see that the nuclear motion couples electronic states when the nuclear velocity vector Ṙ is
large in a direction in which the electronic wavefunction changes most rapidly with R.
For diatomic molecules, we can separate out the center of mass motion and write m as the
reduced mass
m1 m2
m=
(7.89)
m1 + m2
and write the nuclear Schrodinger equation (in one dimension)
−h̄2 ∂ 2
+ V (r) φ(r) = Eφ(r)
2m ∂x2
!
(7.90)
where V (r) is the adiabatic or Born-Oppenheimer energy surface discussed above. Since V (r) is
a polyomial of r, we can do a Taylor expansion of r about its minimum at re
1
1
V (r) = −Vo + V 00 (re )(r − re )2 + V 000 (re )(r − re )3 +
2
6
(7.91)
As an example of molecular bonding and how one computes the structure and dynamics of
a simple molecule, we turn towards the simplest molcular ion, H2+ . For a fixed H − H distance,
R, the electronic Schrödinger equation reads (in atomic units)
1
1
1
ψ(r1 , r2 ) = Eψ(r1 , r2 )
− ∇2 − −
2
r 1 r2
(7.92)
The problem can be solved exactly in elliptical coordinates, but the derivation of the result is
not terribly enlightining. What we will do, however, is use a variational approach by combining
hydrogenic 1s orbitals centered on each H nuclei. This proceedure is termed the Linear Combination of Atomic Orbitals and is the underlying idea behind most quantum chemical calculations.
The basis functions are the hydrogen 1s orbitals. The rationalle for this basis is that as R
becomes large, we have a H atom and a proton–a system we can handle pretty easily. Since the
electron can be on either nuclei, we take a linear combination of the two choices.
|ψi = c1 |φ1 i + c2 |φ2 i
(7.93)
We then use the variational proccedure to find the lowest energy subject to the constraint that
hψ|ψi = 1. This is an eigenvalue problem which we can write as
X
hφi |H|φj i = E
j=1,2
X
cj hφi |φj i.
(7.94)
j=1,2
or in Matrix form
H11 H12
H21 H22
!
c1
c2
!
=E
206
S11 S12
S21 S22
!
(7.95)
where Sij is the overlap between the two basis functions.
S12 = hφ1 |φ2 i =
Z
d3 rφ∗1 (r)φ2 (r)
and assuming the basis functions are normalized to begin with:
S11 = S22 = 1.
For the hydrogenic orbitals
e−(r1
ψ1 (r1 ) = √
π
and
e−(r2 )
ψ2 (r2 ) = √ .
π
A simple calculation yields1
R2
1+R+
.
3
!
−R
S=e
The matrix elements of the Hamiltonian need also to be computed. The diagonal terms are
easy and correspond to the hydrogen 1s energies plus the internuclear repulsion plus the Coulomb
interaction between nuclei 2 and the electron distribution about nuclei 1.
H11 = −EI +
1
− J11
R
(7.96)
where
1
|φ1 i
r2
Z
1
=
d3 r |φ1 (r)|2
r12
J11 = hφ1 |
(7.97)
(7.98)
This too, we evaluate in elliptic coordinates and the result reads
J11 =
1
2EI 1 − e−2R (1 + R) .
R
(7.99)
To derive this result, you need to first transform to elliptic coordinates u, v where
r1
=
r2
=
u+v
R
2
u−v
R
2
the volume element is then d3 r = R3 (u2 − v 2 )/8dudvdφ where φ is the azimuthal angle for rotation about the
H − H axis. The resulting integral reads
Z
Z +1 Z 2π
1 ∞
R3
S=
du
dv
dφ (u2 − v 2 )e−uR .
π 1
8
−1
0
207
Figure 7.1: Various contributions to the H2+ Hamiltonian.
S,J,A
1
0.8
0.6
0.4
0.2
2
4
6
8
RHbohrL
10
By symmetry, H11 = H22 and we have the diagonal elements.
We can think of J as being a modification of the nuclear repulsion due to the screening of the
electron about one of the atoms. |φ1 (r)|2 is the charge density of the hydrogen 1s orbital and is
spherically symmetric about nuclei 1. For large internuclear distances,
1
R
J=
(7.100)
and positive charge of nuclei 1 is completly counterballenced by the negative charge distribution
about it. At shorter ranges,
1
− J > 0.
R
(7.101)
However, screening along cannot explain a chemical bond since J does not go through a minimum
at some distance R. Figure shows the variation of J, H11 , and S as functions of R.
We now look at the off diagonal elements, H12 = H21 . Written explicitly
1
1
H12 = hφ1 |h(2)|φ2 > + S12 − hφ1 |
|φ2 i
R
r1 2
1
= (−EI + )S12 − A
R
(7.102)
where
A = hφ1 |
Z
1
1
|φ2 i = d3 rφ1 (r) φ2 (r),
r12
r12
(7.103)
which can also be evaluated using elliptical coordinates.
2
A = R EI
∞
Z
1
−R
= 2EI e
du2ue−uR
(1 + R)
208
(7.104)
(7.105)
Exercise 7.2 Verify the expressions for J, S, and A by performig the transformation to elliptic
coordinates and performing the integrations.
A is termed the resonance integral and gives the energy for moving an electron from one
nuclei to the other. When H12 6= 0, there is a finite probability for the electron to hop from one
site to the other and back. This oscillation results in the electron being delocalized between the
nuclei andis the primary contribution to the formation of a chemical bond.
To wrap this up, the terms in the Hamiltonian are
S11 = S22 = 1
S12 = S21 = S
1
H11 = H22 = −EI + − C
R
1
H12 = H21 = (−EI + )S − A
R
(7.106)
(7.107)
(7.108)
(7.109)
Since EI appears in each, we use that as our energy scale and set
E = EI
A = αEI
J = γEI
(7.110)
(7.111)
(7.112)
so that the secular equation now reads
−1 + R2 − γ − 2
−1 + R
S − α − S
2
R
−1 +
2
R
−1 +
S − α − S −γ−
=0
(7.113)
Solving the secular equation yields two eigenvalues:
± = −1
2
α−γ
±
.
R 1∓S
(7.114)
For large internuclear separations, ± ≈ −1, or −EI which is the ground state of an isolated H
atom EI = 1/2. Choosing this as the energy origin, and putting it all back together:
(
E± = EI
2
2e−R (1 + R) ∓ 2(1 − e−2R (1 + R))/R
±
R
1 ∓ e−R (1R + R2 /3
)
(7.115)
Plots of these two energy surfaces are shown in Fig. 7.2. The energy minimum for the lower state
(E− ) is at E− = −0.064831hartree when Req = 2.49283ao (or -0.5648 hartree if we don’t set our
zero to be the dissociation limit). These results are qualitatively correct, but are quantitatively
way off the mark. The experimental values are De = 0.1025 hartree and Re = 2.00ao . The
results can be improved upon by using improved basis function, using the charge as a variational
parameter and so forth. The important point is that even at this simple level of theory, we can
get chemical bonds and equilibrium geometries.
For the orbitals, we have a symmetric and anti-symmetric combination of the two 1s orbitals.
ψ± = N± (φ1 ± φ2 ).
209
(7.116)
Figure 7.2: Potential energy surface for H2+ molecular ion.
e+ ,e- HhartreeL
0.25
0.2
0.15
0.1
0.05
-0.05
2
4
6
8
10
R HbohrL
Figure 7.3: Three dimensional representations of ψ+ and ψ− for the H2+ molecular ion generated
using the Spartan ab initio quantum chemistry program.
In Fig. 7.3, we show the orbitals from an ab initio calculation using the 6-31∗ set of basis
functions. The first figure corresponds to the occupoled ground state orbital which forms a σ
bond between the two H atoms. The second shows the anti-bonding σ ∗ orbital formed by the
anti-symmetric combination of the 1s basis functions. The red and blue mesh indicates the phase
of the wavefunction.
210
Appendix: Creation and Annihilation Operators
Creation and annihilation operators are a convenient way to represent many-particle states and
many-particle operators. Recall, from the harmonic oscillator, a creation operator, a† , acts on
the ground state to produce 1 quanta of excitation in the system. We can generalize this to many
particles by saying that a†λ creates a particle in state λ. e.g.
√
a†λ |λ1 ...λN i = nλ + 1|λλ1 ...λN i
(7.117)
where nλ is the occupation of the |λi state. Physically, the operator αλ creates a particle in state
|λi and symmetries or antisymmetrizes the state as need be. For Bosons, the case is simple since
any number of particles can occupy a given state. For Fermions, the operation takes a simpler
form:
a†λ |λ1 ...λN i
(
=
|λλ1 ...λN i if the state |λi is not present in |λ1 ...λN i
0
otherwise
(7.118)
The anti-symmetrized basis vectors can be constructed using the a†j operators as
|λ1 ...λN } = a†λ1 a†λ2 · · · a†λN |0i
(7.119)
Note that when we write the |} states we do not need to keep track of the normalization factors.
We do need to keep track of them them when we use the |i or |) vectors.
|λ1 ...λN i = a†λ1 a†λ2 · · · a†λN |0i
s
=
Q
1
a†λ1 a†λ2 · · · a†λN |0i
λ nλ !
(7.120)
(7.121)
The symmetry requirement places certain requirement on the commutation of the creation
operators. For example,
a†µ a†ν |0i = |µν}
(7.122)
= ζ|νµ}
(7.123)
= ζa†ν a†µ |0i
(7.124)
a†µ a†ν − ζa†ν a†µ = 0
(7.125)
Thus,
In other words, for ζ = 1 (bosons) the two operators commute for ζ = −1 (fermions) the
operators anti-commute.
211
We can prove similar results for the adjoint of the a† operator. In sort, for bosons we have a
commutation relation between a†λ and aλ :
[aµ , a†ν ] = δµν
(7.126)
While for fermions, we have a anti-commutation relation:
{aµ , a†ν } = aµ a†ν + a†µ aν = δµν
(7.127)
Finally, we define a set of field operators which are related to the creation/annihilation operators as
ψ̂ † (x) =
X
hα|xia†α =
α
ψ̂(x) =
X
φ∗α (x)a†α
(7.128)
φα (x)aα
(7.129)
X
α
hα|xiaα =
α
X
α
These particular operators are useful in deriving various tight-binding approximations.
Say that a†α places a particle in state α and aα (1) deletes a particle from state α. The
occupation of state |αi is thus
n̂α |αi = a†α aα |αi = nα |αi
(7.130)
if the state is unoccupied, nα = 0 and if the state is occupied, the first operation removes the
particle and the second replaces, and nα = 1.
One-body operators can be evaluated as
U=
X
n α Uα =
α
X
hα|U |αia†α aα
(7.131)
α
Likewise,
Ûα = hα|U |αinα
(7.132)
Two body operators are written as
Z
1 XZ
V̂ =
dx dyφ∗α (x)φ∗β (y)V (x − y)φγ (x)φδ (y) = (αβ|v|γδ)a†α a†β aγ aδ
2 αβγδ
(7.133)
Occasionally, it is useful to write the symmetrized variant of this operator
V̂ =
1 X
(αβ|v[|γδ) − |δγ)]a†α a†β aγ aδ
4 αβγδ
=
1 X
{αβ|v|γδ}a†α a†β aγ aδ
4 αβγδ
212
(7.134)
(7.135)
7.5
Problems and Exercises
Exercise 7.3 Consider the case of two identical particles with positions r1 and r2 trapped in a
centeral harmonic well with a mutially repulsive harmonic interaction. The Hamiltonian for this
case can be written as
H=−
h̄2 2
1
λ
∇1 + ∇22 + mω 2 (r12 + r22 ) − mω 2 |r1 − r2 |2
2m
2
4
(7.136)
where λ is a dimensionalless scaling factor. This can be a model for two Bosons or Fermions
trapped in an optical trap and λmω 2 simply tunes the s-wave scattering cross-section for the two
atoms.
1. Show that upon an appropriate change of variables
√
u = (r1 + r2 )/ 2
√
v = (r1 − r2 )/ 2
(7.137)
(7.138)
the Hamiltonian simplifies to two separable three-dimensional harmonic oscillators:
h̄2 2 1
h̄2 2 1
2 2
H= −
∇u + mω u + −
∇v + (1 − λ)mω 2 v 2
2m
2
2m
2
"
#
"
#
(7.139)
2. What is the exact ground state of this system?
3. Assuming the particles are spin 1/2 fermions. What are the lowest energy triplet and singlet
states for this system?
4. What is the average distance of separation between the two particles in both the singlet and
triplet configurations.
5. Now, solve this problem via the variational approach by taking your trial wavefunction to
be a Slater determinant of the two lowest single particle states:
α(1)
ψ(r1 , r2 ) = φ(1)
β(2)
φ(2)
(7.140)
Where φ(1) and φ(2) are the lowest energy 3D harmonic oscillator states modified such
that we can take the width as a variational parameter:
φ(r) = N (ζ) exp(−r2 /(2ζ 2 )
where N (ζ) is the normalization factor. Construct the Hamiltonan and determine the
lowest energy state by taking the variation
δhψ|H|ψi = 0.
How does your variational estimate compare with the exact value for the energy?
213
Exercise 7.4 In this problem we consider an electron in a linear tri-atomic molecule formed by
three equidistant atoms. We will denote by |Ai, |Bi , and |Ci the three orthonormal state of
the electron, corresponding to three wavefunctions localized about the three nuclei, A, B, and C.
Whele there may be more than these states in the physical system, we will confine ourselves the
subspace spanned by these three vectors.
If we neglect the transfer of an electron from one site to an other site, its energy is described by
the Hamiltonian, Ho . The eigenstates of Ho are the three orthonormal states above with energies
EA , EB , and EC . For now, take EA = EB = EC = Eo . The coupling (i.e. electron hopping)
between the states is described by an additional term W defined by its action on the basis vectors.
W |Ai = −a|Ai
W |Bi = −a(|Ai + |Ci)
W |Ci = −a|Bi
where a is a real positive constant.
1. Write both Ho and W in matrix form in the orthonormal basis and determine the eigenvalues {E1 , E2 , E3 } and eigenvectors {|1i, |2i, |3i} for H = Ho + W . To do this numerically,
pick your energy scale in terms of Eo /a.
2. Using the eigenvectors and eigenvalues you just determined, calculate the unitary time
evolution operator in the original basis. Eg.
hA|U (t)|Bi = hA| exp (−iH/h̄t) |Bi
3. If at time t = 0 the electron is localized on site A (in state |Ai), calculate the probability
of finding the electron in any other state at some later time, t (i.e. PA , PB , and PC ). Plot
your results. Is there some later time at which the probability of finding the electron back
in the original state is exactly 1. Give a physical interpretation of this result.
4. Repeat your calculation in part 1 and 2, this time set EA = EC = Eo but set EB = 3Eo .
Again, Plot your results for PA , PB , and PC and give a physical interpretation of your
results.
214