PHAS3226: Quantum Mechanics
Jonathan Tennyson
(based on notes by Davids Bowler and Tovee)
2016
PHAS3226: Quantum Mechanics
Administration
Office hours will be Mondays at 1pm in my office (A4 on the top floor of the Physics Building). Attendance sheets must
be filled in. They’ll be given out at the start of a lecture and collated over the weeks.
Problem sheets will be given out through the term, roughly every two weeks. The best three problem sheets will count
for 10% of the final course mark. N.B. There will be four sheets during term.
Full sets of lecture notes will be made available approximately one week after the lecture. A complete PDF file will
be available at the end of the course.
Prerequisites
To have attended and passed the department’s introductory quantum mechanics courses, PHAS2222 or equivalent courses.
PHAS2224, the Atomic and Molecular Physics course is desirable. Frequent reference is made to the material in
PHAS2222.
The following topics should have been covered previously :
• The time-independent Schrödinger wave equation and its solution for:
1. quantum wells and quantum barriers/steps
2. The harmonic oscillator (classical and quantum)
3. The hydrogen atom including the radial equation as well as the angular equation and its solution with spherical
harmonics.
An understanding of atomic spectroscopic notation (n, l, m quantum numbers) and their physical basis is assumed.
• The expansion postulate; the Born interpretation of the wave function; simple calculations of probabilities and
expectation values.
• For a time-independent Hamiltonian, an understanding of the separability of the full Schrödinger equation into a
time-independent wave equation in position space and a time-dependent component is assumed. Familiarity with
applications to eigenstates and/or superpositions thereof is assumed.
• A basic understanding of the postulates of quantum theory is assumed.
Aims & Objectives
Aims This course aims to
• Introduce the basic concepts of Heisenberg’s matrix mechanics. The second year course (PHAS2222) dealt primarily with the Schrödinger’s matter wave dynamics. In PHAS3226, matrix mechanics is introduced as an alternative
approach to quantum dynamics. It is also shown to provide a complementary approach, enabling the treatment
of systems (such as spin systems) where solutions of the non-relativistic Schrodinger wave equation in position
coordinates is not possible.
• Apply matrix mechanics and operator algebra to the Quantum Harmonic Oscillator and its relation to the 2nd year
wave dynamics solutions using Hermite polynomials.
• Apply matrix mechanics and operator algebra to quantum angular momentum.
• Demonstrate that matrix mechanics predicts and permits solution of spin-1/2 systems using Pauli matrices.
• Develop understanding of fundamental concepts using these new methods. The Heisenberg uncertainty principle is
shown to be just one among a family of generalized uncertainty relations, which arise from the basic mathematical
structure of quantum theory, complementing arguments (e.g. Heisenberg microscope) introduced in the second
year.
ii
PHAS3226: Quantum Mechanics
• Explore some concepts of two-particle systems. The addition of two-spins is analysed including fundamental
implications, exemplified by the Einstein-Podolsky-Rosen paradox and Bell inequalities.
• Introduce approximate methods (time-independent perturbation theory, variational principle) to extend the PHAS2222
analytical solution of the hydrogen atom to encompass atoms in weak external electric and magnetic fields and twoelectron systems like helium atoms.
• Introduce symmetry requirements and the Pauli Principle .
Objectives After completing the module the student should be able to:
• Formulate most quantum expressions using abstract Dirac notation and understand that it is not simply shorthand
notation for second-year expressions.
• Understand how to formulate and solve simple quantum problems expressing quantum states as vectors and quantum operators as matrices.
• Use commutator algebra and creation/annihilation operators to solve for the Quantum Harmonic Oscillator.
• Derive generalized uncertainty relations; calculate variances and uncertainties for arbitrary observables. In this
context, have a clear understanding of the relation and difference between operators and observables.
• Use commutator algebra and raising/lowering operators to calculate angular momentum observables.
• Calculate the states and observables of spin-1/2 systems using Pauli matrices.
• Understand and estimate corrections to the energy and properties of low-lying states of helium atoms using approximate methods and symmetry.
• Understand time-independent perturbation theory.
• Understand the variational principle.
• Understand the Pauli Principle and symmetrisation requirements on quantum states including for combinations of
space/spin states.
Textbooks
Those which are closest to the material and level of the course are (in alphabetical order)
• B. H. Bransden and C. J. Joachain, Quantum Mechanics, Second Edition (Pearson, 2000)
• S. Gasiorowicz, Quantum Physics, Third Edition (Wiley, 2003).
• A.Rae, Quantum Mechanics, Fifth edition, (Taylor & Francis 2007).
You may also find some interest and useful perspective from the following books:
• P. A. M. Dirac, The Principles of Quantum Mechanics
• Feynman Lectures on Physics, Volume 3
Syllabus
The approximate allocation of lectures to topics is shown in brackets below.
iii
PHAS3226: Quantum Mechanics
Formal quantum mechanics [7]
Revision of year 2 concepts using Dirac bracket notation: Introduction to Dirac notation and application to PHAS2222
material including orthonormality of quantum states, scalar products, expansion postulate, Linear Hermitian Operators
and derivations of their properties; simple eigenvalue equations for energy, linear and angular momentum.
Representations of general operators as matrices and states as vectors using a basis of orthonormal states. Basis set
transformations; proof that eigenvalue spectrum is representation independent. Similarity transformations and applications to calculations of observables such as eigenvalues and expectation values.
The Quantum Harmonic Oscillator; Generalised Uncertainty Relations [7]
Introduction of creation and annihilation operators: main commutator relations, the number operator. Solution of the
QHO eigenstates and spectrum using this method. Relation to wave dynamics and Hermite polynomial solutions obtained
previously. The zero-point energy as a consequence of both commutator algebra as well as wave solutions. Uncertainty
relations for general operators. Solutions of simple examples including Heisenberg Uncertainty Principle.
Generalised Angular Momentum [4]
Commutator algebra and raising/lowering operators. Obtaining angular momentum eigenvalue and eigenstates using
raising and lowering operators. Matrix representation and solution of simple problems.
Spin-1/2 systems [8]
Introduction to spin-1/2 systems. Matrix representations of spin using eigenstates of z-component of spin: spinors and
Pauli matrices. Matrix representations of eigentates of spin operators along arbitrary directions. Basis set and similarity
transformations between different basis sets. Addition of two spins. The EPR paradox and derivation of Bell inequalities.
Approximate methods and many-body systems [7]
Derivation of time independent Perturbation theory. First and second order theory and examples: helium and atoms in
external fields. The variational principle; example with helium.
Symmetry, fermions, bosons and the Pauli principle. Two-particle space-spin symmetry. Slater determinants for many
body systems. Exchange corrections in helium spectrum.
iv
CONTENTS
Contents
1
Formal Quantum Mechanics
1.1 Introduction - Review of basic quantum mechanics . . . . . . . . . .
1.1.1 The Expansion Postulate . . . . . . . . . . . . . . . . . . . .
1.2 Dirac Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2.1 Dirac notation for state vectors . . . . . . . . . . . . . . . . .
1.3 The position (coordinate) representation . . . . . . . . . . . . . . . .
1.4 Basis states . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.4.1 Change of basis - change of representation . . . . . . . . . .
1.5 Matrix representation . . . . . . . . . . . . . . . . . . . . . . . . . .
1.5.1 Eigenvalue equation and eigenvalues in matrix representation
1.5.2 Change of basis - an example . . . . . . . . . . . . . . . . .
1.5.3 Solving for energy eigenvalues . . . . . . . . . . . . . . . . .
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3
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6
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19
23
2
QHO
2.1 Classical mechanics . . . . . . . . . . . . . . . . . . . .
2.2 Schrödinger equation for harmonic oscillator . . . . . .
2.3 Algebraic operator approach . . . . . . . . . . . . . . .
2.3.1 Eigenvectors of N̂ . . . . . . . . . . . . . . . .
2.3.2 Eigenvalues of the Hamiltonian . . . . . . . . .
2.3.3 Actions of operators â+ and â . . . . . . . . .
2.3.4 Spatial wavefunctions . . . . . . . . . . . . . .
2.3.5 Matrix representations . . . . . . . . . . . . . .
2.3.6 Minimum uncertainty for the harmonic oscillator
2.3.7 Eigenstates of â . . . . . . . . . . . . . . . . .
2.4 Compatible observables and commuting operators . . . .
2.5 Heisenberg Uncertainty Relations . . . . . . . . . . . .
2.5.1 Uncertainty relation for the harmonic oscillator .
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3
Generalised Angular Momentum
3.1 Orbital angular momentum . . . . . . . . . . .
3.2 Eigenvalues and eigenfunctions of L̂z and L̂2 .
3.3 Central potentials . . . . . . . . . . . . . . . .
3.3.1 Review of hydrogen atom . . . . . . .
3.4 Generalized angular momentum . . . . . . . .
3.4.1 Raising and lowering operators . . . .
3.5 General angular momentum eigenvalue problem
3.5.1 Actions of the Jˆ+ and Jˆ . . . . . . .
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41
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4
Spin 1/2 Systems
4.1 Useful relations for spin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2 Representation of spin 1/2 operators and eigenfunctions - Pauli matrices . . . . . . . . . . . . . . . . . .
4.2.1 Matrix representations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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PHAS3226: Quantum Mechanics
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64
Approximate methods & Many-body systems
5.1 Time-independent perturbation theory for a non-degenerate energy level
5.1.1 First-order . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.1.2 Second-order . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.1.3 Observations on the energy corrections . . . . . . . . . . . . .
5.2 The Degenerate case . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.3 Applications of perturbation theory . . . . . . . . . . . . . . . . . . . .
5.3.1 First-order examples . . . . . . . . . . . . . . . . . . . . . . .
5.3.2 Second-order example . . . . . . . . . . . . . . . . . . . . . .
5.3.3 Degenerate example - first order . . . . . . . . . . . . . . . . .
5.4 Variational method . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.4.1 Proof of the variational principle . . . . . . . . . . . . . . . . .
5.4.2 Excited States . . . . . . . . . . . . . . . . . . . . . . . . . . .
5.4.3 Variational examples . . . . . . . . . . . . . . . . . . . . . . .
5.5 Systems of identical particles . . . . . . . . . . . . . . . . . . . . . . .
5.5.1 Identical particles . . . . . . . . . . . . . . . . . . . . . . . . .
5.5.2 Exclusion principle . . . . . . . . . . . . . . . . . . . . . . . .
5.5.3 N-particle states . . . . . . . . . . . . . . . . . . . . . . . . .
5.5.4 Helium atom and the exchange force . . . . . . . . . . . . . . .
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4.3
4.4
5
CONTENTS
4.2.2 Commutators for the Pauli matrices . . . . . . . . .
4.2.3 Basis states . . . . . . . . . . . . . . . . . . . . . .
4.2.4 Determination of eigenvalues and eigenvectors . . .
4.2.5 Spin along an arbitrary direction . . . . . . . . . . .
Space-spin wavefunctions . . . . . . . . . . . . . . . . . . .
4.3.1 Stern-Gerlach experiment . . . . . . . . . . . . . .
Addition of angular momentum . . . . . . . . . . . . . . . .
4.4.1 Addition of two spin-1/2 angular momenta . . . . .
4.4.2 Coupling of spin-1/2 and orbital angular momentum
4.4.3 Addition of orbital and spin-1/2 angular momenta . .
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CHAPTER 1. FORMAL QUANTUM MECHANICS
Chapter 1
Formal Quantum Mechanics
1.1
Introduction - Review of basic quantum mechanics
Quantum mechanics is probabilistic and deterministic. The system of N particles has the same set of dynamical
variables (positions, momenta, kinetic energies, etc.) as in classical mechanics. However it is a fundamental feature of
quantum mechanics that it is impossible to have a precise knowledge of all the dynamical variables at the same time
t. The Heisenberg Uncertainty Principle states that it is impossible to know precisely the position and momentum of a
particle at any given time. In particular, the uncertainty in the x-component of position, x, and momentum, px , of a
particle are related by
1
x p
h̄ .
(1.1)
2
In the wave mechanics formulation of quantum mechanics due to Schrödinger, a basic postulate is that for every
dynamical system there exists a wavefunction that is a (single-valued) function of the parameters of the system and
time from which all possible information of the physical system can be obtained. In this formulation the wavefunction for
a single particle is denoted by (r, t), and by (r1 , r2 , . . . rN , t) for N particles.
The usual interpretation given to is that its square modulus | |2 | gives the probability density for finding the particle
at specified positions; | (r)|2 d3 r is the probability of finding the particle inside a volume element d3 r located at vector
position r . (Recall that the square modulus | (r)|2 = (r)? (r), where ⇤ (r) is the complex conjugate of (r).)
On the other hand, quantum mechanics is deterministic in that given the wavefunction at some time t its value at a
later time t0 can be found as (r, t) evolves according to the time-dependent Schrödinger equation
ih̄
@ (r, t)
= Ĥ (r, t) ,
@t
(1.2)
where Ĥ is the Hamiltonian operator for the system. For a particle of mass m moving in a potential V (x, t) in onedimension the Hamiltonian is
h̄2 d2
Ĥ =
+ V (x, t) .
(1.3)
2m dx2
If the potential is independent of time, the time-dependent Schrödinger equation separates with solutions of the form
(x) e iEt/h̄ and the general solution (by the expansion postulate) is
X
(x, t) =
cn n (x) e iEn t/h̄
(1.4)
n
with
n
(x) satisfying the time-independent Schrödinger equation
h̄2 d2 n (x)
+ V (x)
2m dx2
with energy eigenvalue En .
3
n
(x) = En
n
(x)
(1.5)
PHAS3226: Quantum Mechanics
CHAPTER 1. FORMAL QUANTUM MECHANICS
It is a basic principle of quantum mechanics that if 1 and 2 are two possible states of a given system then the linear
combination (superposition) = c1 1 + c2 2 is also a possible state of the system where c1 , c2 are constants. The
probability density associated with (r) becomes
2
| (r)|
2
1 (r) + c2 2 (r)|
(c⇤1 1⇤ (r) + c⇤2 2⇤ (r)) (c1 1 (r) + c2 2 (r))
2
2
2
2
|c1 | | 1 (r)| + |c2 | | 2 (r)| + c⇤1 c2 1⇤ (r) 2 (r) + c1 c⇤2
2
2
2
2
|c1 | | 1 (r)| + |c2 | | 2 (r)| + 2< [c⇤1 c2 1⇤ (r) 2 (r)]
2
2
|c1 | P1 (r) + |c2 | P2 (r) + 2< [c⇤1 c2 1⇤ (r) 2 (r)]
= |c1
=
P (r) =
=
=
2
(1.6)
1
(r)
⇤
2
(r)
(1.7)
(1.8)
(1.9)
2
where P1 (r) = | 1 (r)| and P2 (r) = | 2 (r)| are the probability densities associated with 1 (r) and 2 (r). However
the total probability density is not just the weighted sums of P1 (r) and P2 (r) as there is an additional interference term
2< [c⇤1 c2 1⇤ (r) 2 (r)].
The Hamiltonian operator was mentioned above. It is a postulate of quantum mechanics that every dynamical (variable) observable is represented by a linear Hermitian operator. A linear operator  is such that  (c1 1 + c2 2 ) =
c1 Â 1 + c2 Â 2 .
Another postulate is that the only possible results of a measurement of an observable A are the eigenvalues of the
corresponding operator Â. That is
 n = n n ,
(1.10)
where n is the eigenvalues for eigenfunction n . Immediately after such a measurement the wavefunction of the system
will be the same as the eigenfunction corresponding to the eigenvalue. Examples encountered in the previous course are
p̂
L̂z
@
@
; p̂ = p̂eikx = ih̄
= h̄keikx ,
@x
@x
@
= ih̄ ; L̂z = L̂z eim = mh̄eim ,
@
=
(1.11)
ih̄
L̂2 Y`m (✓, ) = ` (` + 1) h̄2 Y`m (✓, )
For an operator Â, the Hermitian conjugate † is defined by
Z
Z
Z ⇣
⇣
⌘
⌘⇤
⇤
†
Â
d⌧
=
Â
d⌧
=
2
1
2
1
for any functions
1
and
2.
⇤
2
⇣
†
If the operator is Hermitian then  = † . and so
Z
Z
Z ⇣
⇣
⌘
⌘⇤
⇣
⇤
⇤
 1
2 d⌧ =
1 Â 2 d⌧ =
2 Â
The definition of an Hermitian operator ensures that
1
⌘
⌘
d⌧
=
a⇤n
1
⇤
d⌧
(1.12)
⇤
(1.13)
1. the eigenvalues are real quantities (a necessity for a measurement).
Since Â
n
= an
n
then by Hermiticity
Z
Z
⇤
an
=
n n d⌧
an
⇤
n  n d⌧
= a⇤n
=
Z ⇣
Â
n
⌘⇤
n d⌧
Z
and so an is real.
2. the eigenfunctions
n,
m
corresponding to different eigenvalues are orthogonal, i.e.
Z
⇤
3
n (r) m (r) d r = 0.
4
⇤
n n d⌧
PHAS3226: Quantum Mechanics
If
ak
k
k
CHAPTER 1. FORMAL QUANTUM MECHANICS
and n are two eigenfunctions corresponding to different eigenvalues ak and an respectively then Â
and  n = an n .
Hence
Z ⇣
Z
Â
a⇤n
(an
⇤
n  k d⌧
ak )
As an 6= ak then
n
Z
⌘⇤
Z
k d⌧
= ak
⇤
n k d⌧
= ak
⇤
n k d⌧
= 0.
Z
i.e. the eigenfunctions are orthogonal.
= ak
⇤
n k d⌧
Z
k
=
⇤
n k d⌧
Z
⇤
n k d⌧
Z
⇤
n k d⌧
=0
(1.14)
{Note: This proof is not satisfying as it stands as sometimes an operator has more than one eigenfunction for one
eigenvalue (differing by more than a multiplicative factor). The eigenvalue is said to be degenerate. Suppose n and m
are two such degenerate eigenfunctions with an = am . Then = m + c n is also an eigenfunction of  with eigenvalue
an (=am );
 =  m + c n = am m + can n = am ( m + c n ) = am .
As c is arbitrary it can be chosen to make
Z
Z
⇤
d⌧
=
n
orthogonal to
as
Z
⇤
(
+
c
)
d⌧
=
m
n
n
if
Z
c=
n,
⇤
n m d⌧
⇤
n m d⌧
Z
/
|
2
n|
+c
Z
|
2
n|
d⌧ = 0
d⌧.
Thus if the eigenvalue is degenerate, the eigenfunctions can always be chosen to be orthogonal.}
Note on Hermitian operators.
If Â, B̂ are two Hermitian operators then ÂB̂ + B̂ Â and i(ÂB̂ B̂ Â) are also Hermitian. From the definition of an
Hermitian operator
Z
Z ⇣ ⌘⇤
⇣ ⌘
⇤
 d⌧ =
Â
d⌧
then
Similarly
Adding gives
so ÂB̂ + B̂ Â is Hermitian.
Likewise
Z
⇤
iÂB̂ d⌧
Z
⇤
iB̂ Â d⌧
Z
⇤
ÂB̂ d⌧ =
Z
Z
=
=
⇤
Z ⇣
⇤
Â
⌘⇤ ⇣
Z ⇣
Â
⌘
Z ⇣
B̂ Â d⌧ =
(ÂB̂ + B̂ Â) d⌧ =
Z ⇣
B̂
d⌧ =
ÂB̂
Z h
Z ⇣
⌘⇤
B̂ Â
⌘⇤
d⌧.
d⌧.
(ÂB̂ + B̂ Â)
i⇤
d⌧
Z ⇣
Z ⇣
⌘⇤ ⇣ ⌘
⌘⇤
⌘⇤
i B̂ d⌧ =
B̂ Â
i d⌧ =
iB̂ Â
d⌧
iÂB̂
⌘⇤
d⌧
5
PHAS3226: Quantum Mechanics
so subtracting gives
and so i(ÂB̂
1.1.1
Z
CHAPTER 1. FORMAL QUANTUM MECHANICS
Z ⇣
i(ÂB̂ B̂ Â) d⌧ =
i(ÂB̂
h
i
B̂ Â) is Hermitian, and i Â, B̂ is Hermitian.
⇤
B̂ Â)
⌘⇤
d⌧
The Expansion Postulate
The expansion postulate is an essential postulate of quantum mechanics. It says that for every dynamical observable that
can be measured its eigenfunctions form a complete set, { n } i.e. that any wavefunction describing a possible state of the
system can be expressed as a linear combination of the eigenstates (in same domain and dimensionality). Hence
X
=
cn n
(1.15)
n
where the constants, cn , are called the expansion coefficients. Taking the scalar product of equ(1.15) with
Z
X Z
X
⇤
⇤
d⌧
=
cn
cn kn = ck .
k
k n d⌧ =
n
k
gives
n
Thus the expansion coefficients are obtained from
cn =
Z
d⌧.
⇤
n
(1.16)
The number of eigenfunctions may be finite or infinite. In some cases the eigenfunctions of an observable form a
continuum, in which case the sum in equ(1.15) is replaced by an integral.
As an example, the functions cos (n⇡x/2a) and sin (n⇡x/2a) are eigenfunctions of the Hamiltonian for a particle in
a one-dimensional infinite potential well (V (x) = 0, |x|  a, V(x) = 0, |x| > a) and give rise to the Fourier expansion
of a function,
1 h
⇣ n⇡x ⌘
⇣ n⇡x ⌘i
X
f (x) = a0 +
an cos
+ bn sin
.
(1.17)
2a
2a
n=1
A further postulate is that if a measurement of observable A is repeated many times, always with the system in the
same initial state , the average value of A, the expectation value of A is given by
Z
⇤
hAi =
 d⌧.
(1.18)
Using the expansion postulate the expectation becomes
hAi =
=
Z X
(c⇤k
Z XX
XX
k
hAi =
X
n
X
(cn
n ) d⌧
n
k
k
=
⇤
k ) Â
(c⇤k
n
c⇤k cn an
n
2
|cn | an =
⇤
k ) (cn an n ) d⌧
Z
⇤
k n d⌧
X
=
pn an ,
=
k
n
Z
kn
(1.19)
n
2
c⇤k cn
c⇤k cn an
n
k
where pn is the probability of value an , i.e. pn = |cn | . Furthermore if
Z
Z X
X
⇤
1 =
d⌧ =
(c⇤k ⇤k )
(cn
XX
XX
is normalized
n ) d⌧
n
k
⇤
k n d⌧
=
XX
k
6
n
c⇤k cn
kn
=
X
n
2
|cn |
PHAS3226: Quantum Mechanics
CHAPTER 1. FORMAL QUANTUM MECHANICS
and
X
n
2
2
|cn | = 1,
(1.20)
as is to be expected if |pn | is interpreted as the probability of obtaining value an as a result of a measurement.
Many properties of a state are completely independent of how it is represented, e.g.probabilities of obtaining different
measured values of an observable; the expectation value of an observable cannot depend on whether the state was given
as a function of position or some other way. Thus it is desirable to have a way of discussing states which is independent of
the representation. When the concept of intrinsic angular momentum (spin) of a particle was introduced for electrons to
explain the results of a Stern-Gerlach experiment and the fine structure of spectral lines it was noted that the spin operator
cannot be represented in term of spatial coordinates, nor can the associated eigenfunction.
1.2
Dirac Notation
Dirac introduced a compact general notation for quantum states. The idea behind it is based on the observation that the
wavefunctions behave like vectors in many circumstances. The superposition of two wavefunctions
= c1
+ c2
1
2
(1.21)
,
where c1 and c2 are constants is similar to the addition of two vectors B and C:
D = c1 B + c2 C .
(1.22)
Furthermore if 1 and 2 are eigenfunctions of the same operator  they are orthonormal, so can be specified by its
two components (c1 , c2 ) just as the position vector r = xi + yj can be specified by its two components (x, y). For three
eigenfunctions
= c1
1
+ c2
2
+ c3
3
is analogous to
B = Bx i + By j + Bz k ,
(1.23)
where Bx , By and Bz are the Cartesian components of B. The full expansion
X
=
cn n
(1.24)
n
just expands the dimensionality of the space.
The wavefunction can be represented in terms of different complete orthonormal sets, {
X
=
c0n 0n .
0
n}
as
(1.25)
n
Similarly, the vector B can be expressed in an alternative set of Cartesian axes, having unit vectors i0 , j0 , k0 as
B = Bx0 i0 + By0 j0 + Bz0 k0 .
(1.26)
For two vectors B = Bx i + By j + Bz k and C = Cx i + Cy j + Cz k the scalar product is
B · C = Bx Cx + By Cy + Bz Cz .
If two wavefunctions
and
are expanded in terms of the orthonormal set of functions{
X
=
bm m
m
=
X
n
7
cn
n
n}
as
PHAS3226: Quantum Mechanics
CHAPTER 1. FORMAL QUANTUM MECHANICS
the overlap integral
Z
⇤
dx =
XX
m
b⇤m cn
n
Z
⇤
m n dx
=
XX
m
b⇤m cn
n
mn
=
X
b⇤n cn
(1.27)
n
is analogous to the scalar product B · C = Bx Cx +By Cy +Bz Cz , but with the difference that (1) it involved the complex
conjugate for one of the components, (2) it can involve more dimensions than three (even an infinite number).
Wavefunctions are acted on by linear operators, which map the wavefunctions onto each other e.g.
= Â .
(1.28)
Vectors can also be acted on by operators; a rotation by some chosen angle about a chosen axis; or a reflection in a chosen
plane. Under such operations vectors are mapped onto each other.
1.2.1
Dirac notation for state vectors
The wavefunction and its complex conjugate ? are denoted by the symbols | i and h |, which are called a “ket” and
a “bra” respectively so ! | i, and ⇤ ! h |. (The reason for these names is that a bra and a ket together make a
“bracket”!) The kets are members of a linear vector space whose dimensionality depends on the quantum system, and is
often infinite. The same is true of the bras. The kets are vectors in the sense that
(i) they can be multiplied by (generally complex) numbers c so if | i is a member of the space then so is c| i.
(ii) they can be added so if | i and | i are members of the vector space then so is | i + | i.
To each ket, there is a corresponding bra. With h | the bra corresponding to | i, the bra corresponding to c| i is
c? h |. The bra corresponding to c1 | i + c2 | i is c⇤1 h | + c⇤2 h |. The overlap integral of two wavefunctions and has
the properties of a scalar product,
Z
?
d⌧
!h | i
in the Dirac notation. The complex conjugate of such a quantity is
Z
Z
?
?
d⌧
=
?
d⌧
= h | i
h | i?
(1.29)
(1.30)
and is a key property of the scalar product of a bra and a ket. The normalization condition becomes h | i = 1 and the
orthogonality
Z
h
⇤
m n d⌧
=
mn
(1.31)
m| ni
=
mn .
(1.32)
The eigenfunctions of an operator  are usually characterised by some quantum number or numbers. For example,
the eigenfunctions of a particle in an infinite one-dimensional well are characterised by integers, n. The eigenfunctions
for a hydrogen atom are specified by knowing the principal quantum number, n, (related to the energy level), the orbital
angular momentum quantum number, `, and the magnetic quantum number, m, so that they are written n`m (r). In
Dirac notation these are denoted by n ! | n i ! |ni; n`m ! | n`m i ! |n, `, mi. The expansion
X
=
cn n
n
becomes
| i=
X
cn |
ni
8
=
X
n
cn |ni.
PHAS3226: Quantum Mechanics
CHAPTER 1. FORMAL QUANTUM MECHANICS
Taking the inner product (= scalar product) with a bra h
X
h m | i = hm| i =
cn h
m|
(= hm|) gives
X
X
cn hm|ni =
cn
m| ni =
cn = h
n|
n
Thus
The basis states |
mi
n
mn .
n
i = hn| i.
(1.33)
define a linear vector space. The terminology is adopted that in one representation
wavefunction {
state function
eigenfunction
()
state vector
eigenvector
vector space
}
A closure relation can be developed for a complete set of states since
X
| i =
cn |ni,
n
= hn| i
cn
and then
| i=
so that formally the closure relation
X
n
hn| i|ni =
X
n
X
n
|nihn| i
|nihn| = 1
(1.34)
behaves as an identity operator when the summation is over all eigenstates |ni. Know as the resolution of the identity.
The result of acting with operator  on ket | i is another ket written as Â| i. The scalar product of the ket Â| i with
the bra h | is written as h |Â| i. The Hermitian conjugate † of an operator  was defined by
Z
Z
Z ⇣
?
⇣ ⌘
⌘?
?
? †
Â
d⌧ =
†
d⌧ =
 d⌧
.
(1.35)
In Dirac notation, this relationship reads:
h |Â| i = h† | i = h |† | i? .
(1.36)
Thus the bra corresponding to the ket Â| i can be written as h |† . This means that the quantity h |Â| i can be
interpreted in two equivalent ways: (i) it is the scalar product of bra h | with ket Â| i; (ii) it is the scalar product of bra
h |Â with ket | i. This equivalence is an elegant feature of the Dirac notation.
If  is Hermitian,  = † and
h |Â| i = h |Â| i⇤ .
(1.37)
Hence in Dirac notation the proof that the eigenvalues are real becomes, if |ni is an eigenstate of operator  then
hn|Â|ni = hn|Â|ni⇤ = hÂn|ni⇤ ,
an hn|ni = a⇤n hn|ni
so an is real.
The expectation value of observable A is
hAi = h |Â| i =
The normalization is
1
1
= h | i=
=
X
n
2
XX
m
XX
m
n
n
c⇤m cn an hm|ni =
hm|c⇤m cn |ni =
|cn | .
9
X
n
XX
m
n
2
|cn | an .
c⇤m cn hm|ni,
(1.38)
PHAS3226: Quantum Mechanics
CHAPTER 1. FORMAL QUANTUM MECHANICS
2
Hence from the expression for the expectation value it follows that |cn | is probability that eigenvalue an is obtained
as a measurement of the observable A. After measurement the eigenfunction is the eigenstate |ni, i.e. the wavefunction
"collapses".
In Dirac notation an operator  transforms a ket |mi into another ket |ni as
|ni = Â|mi.
Taking the scalar product with the bra h | gives
hk|ni = hk|Â|ni = Akn .
The quantity Akn is called the matrix element of the operator  between the states |ki and |ni.
1.3
The position (coordinate) representation
Since a particle’s position along the x-axis can be measured there is an Hermitian operator x̂ for this observable. The
result of the measurement can be any real number, 1 < x < 1. If the eigenvalues of operator x̂ are x0 they form a
continuum. The corresponding eigenvector |x0 i satisfies
The expansion in discrete states |ni, | i =
P
x̂|x0 i = x0 |x0 i.
(1.39)
i|ni has to become an integral
Z
| i = hx0| i|x0 idx0 .
n hn|
(1.40)
The components hx0 | i constitute a complex-valued function of the real variable x0 . This function is identified with the
wavefunction as
(x0 ) = hx0 | i
so establishing the connection between the (abstract) state vector | i and the wavefunction (x0 ). Thus (x0 ) is a
component of the state vector in the infinitely dimensional abstract vector space. Thus (x0 ) is merely just one of many
possible ways of representing the state vector. This representation is called the coordinate or position representation.
Equally the momentum eigenvectors of the momentum operator p̂|p0 i = p0 |p0 i and the momentum wavefunction (p0 ) =
hp0 | i would be obtained in the momentum representation.
Note, by taking the scalar product of | i with |xi gives
Z
hx| i = hx|x0 ihx0 | idx0
so the integral version of the closure relation is
Z
|xihx|dx = 1.
(1.41)
This implies that normalization of the states |x0 i has to be
hx|x0 i = (x x0 )
(1.42)
R
where is the Dirac -function: (y) = 0 unless y = 0, and
(y)dy = 1 if the integral contains y = 0 or zero othewise.
As an advanced application, begin with the basic commutator
[x̂, p̂x ] = ih̄.
(1.43)
Its matrix element is
hx00 |x̂p̂x
(x
00
p̂x x̂| x0 i = ih̄ hx00 |x0 i = ih̄ (x00
x ) hx |p̂x |x i = ih̄ (x
x)
00
ih̄ (x
x0 )
.
hx00 |p̂x |x0 i =
00
0
(x
x)
0
00
0
00
10
x0 )
0
(1.44)
PHAS3226: Quantum Mechanics
CHAPTER 1. FORMAL QUANTUM MECHANICS
Making use of the property of the derivative of the Dirac delta-function, x 0 (x) =
R
Consider p̂x operating on a state | i =
hx00 |p̂x |x0 i =
ih̄
@
(x00
@x00
(x) then
x0 ) .
(1.45)
|x0 ihx0 | idx0 as
Z
p̂x | i = p̂x |x0 ihx0 | idx0
and so its expectation value is
h |p̂x | i =
Z Z
h |x00 idx00 hx00 |p̂x |x0 idx0 hx0 | i,

Z Z
@
00
00
=
h |x idx
ih̄ 00 (x00 x0 ) dx0 hx0 | i,
@x

Z Z
@
⇤
00
00
=
(x ) dx
ih̄ 00 (x00 x0 )
(x0 ) dx0 .
@x
Evaluating the (x00
x0 ) dx00 -function over x00 , (i.e. at x0 = x00 ) gives

Z
@
⇤
h |p̂x | i =
(x0 ) ih̄ 0
@x
(x0 ) dx0
(1.46)
(1.47)
as the expectation value of p̂x in the state | i. Hence the representation of the momentum operator p̂x in the coordinate
representation is
@
p̂x = ih̄
(1.48)
@x
as plausibly suggested in an earlier quantum mechanics course!
1.4
Basis states
The state of a quantum system can be specified by writing the wavefunction as a function of position, (r); (r1 , r2 . . . rN ).
However there are other ways of specifying the state. For example,
P can be expressed as a linear combination of the
complete set of eigenfunctions { n } of an operator Â, as
=
n cn n . In this representation the wavefunction is
specified by the list of coefficients cn instead of being given as an explicit function of position. {If is given as a function
of position, essentially this is a representation in terms of the eigenstates of the position operator x̂.} For example,
⇣ n⇡x ⌘ X
X
1
(x) =
An p sin
=
An n (x)
(1.49)
2a
a
n
n
as
(x) = p1a sin n⇡x
are eigenfunctions of the Hamiltonian Ĥ for the infinite potential well of width 2a.
2a
The wave function could be specified in terms of the eigenfunctions of the momentum operator
n
@
(x) = p (x) ,
@x
(p) = eipx/h̄ .
p̂ (x) =
Then
(x) =
ih̄
Z
1
˜ (p) eipx/h̄ dp
(1.50)
1
(a Fourier transform) and the state of the system is specified by the function ˜ (p). For discrete momentum values
(x) =
X
m
X
1
Bm p eikm x =
Bm
2a
m
11
m
(x) .
PHAS3226: Quantum Mechanics
CHAPTER 1. FORMAL QUANTUM MECHANICS
2
If the energy is measured (x) will reduce to an energy eigenstate n (x) with probability |A| ; if momentum is
2
measured the value p = h̄km is obtained with a probability |Bm | . In the above case, conversion from one representation
1
ikm x
ikm x
is straight forward since sin km x = 2i e
e
. Other cases are less straight forward and require formal
methods.
Many properties of a state are completely independent of how it is represented, e.g.probabilities of obtaining different
measured values of an observable cannot depend on whether the state was given as a function of position or some other
way. Thus it is desirable to have a way of discussing states which is independent of the representation. When the concept
of intrinsic angular momentum (spin) of a particle was introduced for electrons to explain the results of the Stern-Gerlach
experiment and fine structure of spectral lines it was noted that the spin operator cannot be represented in terms of spatial
coordinates, or their derivatives, and nor can the associated eigenfunctions.
1.4.1
Change of basis - change of representation
Transformation of states
Cold atom physicists can easily adjust the shape of the trapping potential which confines the atoms. Typically this is an
harmonic potential VT (x). The Hamiltonian is
Ĥ (x, p) =
p̂2
+ VT (x) ,
2m
(1.51)
with VT (x) = 12 m! 2 x2 . Solutions of the time-independent Schrödinger equation are
(x) = Nn Hn (x) e
n
↵2 x2 /2
(1.52)
1/2
where Hn (x) is a Hermite polynomial of degree n and ↵ = (m!/h̄) . The physicist can increase ! (and thus ↵) by
increasing the strength of the magnetic fields which confine the atom. A tight trap has large ! and ↵, and hence widely
2
spaced energy levels En = n + 12 h̄! and rapid decay of the tail of the probability distribution | (x)| showing the
atoms are tightly confined; conversely for a weak potential - small ! and ↵.
Suppose experimenter A prepares atoms in a well defined state by using dissipative laser cooling processes until all
the atoms are in the ground state. The cooling lasers are turned off and the atoms are in state 0 (x) of a potential
VT!0 (x) = 12 m!02 x2 . The confining fields are abruptly weakened and the trap centre is displaced to x = x1 . The state
function is still 0 (x) but it is no longer an eigenfunction of the new Hamiltonian of the system Ĥ =
A theorist B represents the state in the new basis as
X
| 0i =
cn | n i
p̂2
2m
+ VT! (x
x1 ).
(1.53)
n
with
1/2
with ↵ = (m!/h̄)
hx|
ni
⌘
n
(x) = Nn Hn (x
x1 ) e
↵(x x1 )2 /2
(1.54)
. The coefficients cn are calculated by
cn = h
n|
(1.55)
0i
2
to get probabilities |cn | and expectation values. However suppose the settings in the apparatus are wrong and the trap
1/2
frequency is really ! 0 , = (m! 0 /h̄)
with eigenstates | m i,
hx|
mi
⌘
m
(x) = Nm Hm (x
The theorist’s expansion should have been
|
0i
=
X
m
dm |
x1 ) e
m i.
2
(x x1 )2 /2
.
(1.56)
(1.57)
The eigenstates | n i and | m i both form a complete set of basis functions and mathematically both expansions are
equally valid. But only for eq(1.57) can the time evolution be written as
X
| 0 (t)i =
dm | m ie iEm t/h̄
(1.58)
m
12
PHAS3226: Quantum Mechanics
CHAPTER 1. FORMAL QUANTUM MECHANICS
as these states (and energies) correspond to the real apparatus. It is often advantageous to change representations, even in
mid-calculation. In the experiment different potentials VT (x) may exist for different times.
The basis states can be expanded in terms of each other, as each forms a complete set. Explicitly
X
X
| ni =
h m| ni | mi =
Smn | m i
(1.59)
m
m
and conversely
|
mi
X
=
h
(1.60)
n | m i | n i.
Hence from eq(1.57) and (1.53) the expansion coefficients are given by
X
X
dm = h m | 0 i = h m |
cn | n =
h
X
=
dm
n
n
(1.61)
m | n icn
(1.62)
Smn cn .
n
This is an example of matrix multiplication of column vectors d and c and matrix S with Smn = h
m | n i,
d = Sc,
0
B
B
B
@
d1
d2
..
.
dN
1
0
h
C B h
C B
C=B
A @
h
1| 1i
h
h
2| 1i
..
.
N|
1| 2i
···
···
2| 2i
..
.
i
h
1
N|
h
h
B
B
B
@
c1
c2
..
.
cN
1
C
C
C
A
c
0
h
B h
B
= B
@
h
= S d
†
1| 1i
h
h
···
i
···
2
2| 1i
..
.
N|
1| 2i
···
···
2| 2i
..
.
1i h N |
2| N i
..
.
h N|
S is the matrix which
P transforms between the two representations of |
expansion | 0 i = m dm | m i gives the inverse transformation
0
1| N i
Ni
0 i.
h
h
CB
CB
CB
A@
1
c1
c2
..
.
cN
C
C
C.
A
Noting that cn = h
1| N i
2| N i
..
.
h N|
···
2i · · ·
10
Ni
10
CB
CB
CB
A@
d1
d2
..
.
dN
1
C
C
C
A
(1.63)
n|
0i
and using the
(1.64)
(1.65)
⇤
where S † has elements Snm
(transpose and complex conjugation of S).
From
d = Sc
1
S
d = S
S
1
1
Sc = c = S † d
= S†
and so S is unitary, i.e.
SS † = In .
(1.66)
P The same results can easily be found using the closure property of a complete set of states. Recall that | i =
n cn | n i and cn = h n | 0 i so that
|
so that formally
0i
=
X
n
h
n|
0i | ni
X
|
n
=
n ih n |
13
X
n
= 1.
|
ni h n|
0i
(1.67)
(1.68)
PHAS3226: Quantum Mechanics
CHAPTER 1. FORMAL QUANTUM MECHANICS
Then since
dm = h
using the closure relation
dm =
X
n
with
h
m | n ih n |
Smn = h
and in matrix form
1.5
m|
(1.69)
i,
0i
=
X
(1.70)
Smn cn
n
(1.71)
m| ni
d = Sc.
(1.72)
Matrix representation
A state | i can be expanded in terms of a complete set of basis states {
X
| i=
cn | n i.
n}
as
(1.73)
n
The constants cn can be considered as the elements of a columns vector
0
1
c1
B c2 C
B
C
c=B . C
@ .. A
(1.74)
cn
with cn = h n | i.
The action of an operator  on the wavefunction is to generate another wavefunction | i = Â| i. The state | i can
also be expanded in terms of the basis states { n } as
X
| i=
dm | n i
(1.75)
m
with dm = h
m|
i. Thus
dm = h
The quantities Amn = h
becomes
m|
m |Â| n i
i=h
m |Â|
i=h
m |Â|
X
n
cn |
ni
=
X
n
h
(1.76)
m |Â| n icn .
are called the matrix elements of the operator  in the basis {
X
dm =
Amn cn
n }.
The above equation
n
and dm are elements of a column vector
0
B
B
d=B
@
with the matrix equation
d1
d2
..
.
dn
1
C
C
C
A
(1.77)
d = Ac
0
B
B
B
@
d1
d2
..
.
dn
1
0
C B
C B
C=B
A @
(1.78)
A11
A21
..
.
A12
A22
..
.
...
...
..
.
An1
An2
. . . Ann
14
A1n
A2n
..
.
10
CB
CB
CB
A@
c1
c2
..
.
cn
1
C
C
C
A
(1.79)
PHAS3226: Quantum Mechanics
CHAPTER 1. FORMAL QUANTUM MECHANICS
In the coordinate representation
Amn = h
m |Â|
h | i=
XX
The scalar product
as h
m| ni
=
mn
so
h | i=
X
n
m
ni =
n
Z
d⇤m cn h
⇤
m
(x) Â
m| ni
=
n
(x) d⌧.
X
(1.80)
d⇤n cn
n
0
B
B
d⇤n cn = d† c = (d⇤1 , d⇤2 , . . . d⇤n ) B
@
c1
c2
..
.
cn
1
C
C
C
A
(1.81)
These ideas lead to a matrix representation of quantum systems - a formulation put forward by Heisenberg.
1.5.1
Eigenvalue equation and eigenvalues in matrix representation
Suppose Â| i = a| i. In matrix representation this is written Âc = ac. Considering the i-th component,
X
X
Ain cn1 = aci1 = a
in cn1
This is equivalent to
0
B
B
B
@
X
n
(Ain
n
= 0.
in a) cn1
10
1
A11 a
A12
A13
...
c1
B
C
A21
A22 a
A23
... C
C B c2 C
= 0.
C
B
A31
A32
A33 a . . . A @ c3 C
A
..
..
..
.
..
..
.
.
.
.
(1.82)
This equation has a non-trivial solution for the c’s if
det |Ain
in a|
= 0 = det |A
aI| .
(1.83)
This is a good way of finding the eigenvalues for operators represented by finite matrices - not so simple for infinite
matrices!
Transformation of operators
Recall that the matrix representation of an operator equation such as
is
where
| i=
X
n
= Â
(1.84)
b = Ac
(1.85)
| i=
bn |ni;
in the same basis {|ni} and the operator is represented by the matrix
D
E
Akj = k|Â|j
whence
bk =
X
j
15
Akj cj .
X
n
cn |ni
(1.86)
(1.87)
(1.88)
PHAS3226: Quantum Mechanics
CHAPTER 1. FORMAL QUANTUM MECHANICS
D
E
D
The operator  in the basis {| n i}is represented by the matrix
i |Â| n and by
` |Â|
Using the closure relation twice
D
E
E
XXD
=
` |Â| m
` | i ih i Â| j ih j | m
⇣
Â
⌘
i
`m
i
and the converse
in the basis {|
⇤
Sli Âij Smj
n i}.
(1.89)
(1.90)
j
= S Â S † = S Â S
Â
E
j
XX
=
m
 = S
1
1
(1.91)
,
(1.92)
 S.
Eq(1.91) and (1.92) are called similarity transformations of the matrices representing the operator Â.
Let
D
E
A`m =
` |Â| m
and
But
|
and
so that
D
A0in =
D
` |Â| m
mi
|
⇤
`i
E
=
X
=
`|
X
i
h
i
n
E
X
h
`| ii h i|
h
(1.95)
` | i i h i |Â| n i h n | m i
(1.96)
n
|
⇤
S`i A0in Snm
n
0 †
= SA S = SA0 S
= SA S
X
1
(1.94)
ni h n| mi
` | i i h i |Â
n
(1.93)
n| mi | ni
i
XX
=
A
=
XX
=
i
A
h
n
=h
i |Â|
1
(1.97)
(1.98)
,
(1.99)
.
Conversely
AS
A
0
= SA0 S
= S
1
1
S = SA0
(1.100)
(1.101)
AS.
Eq(1.98) and (1.101) are called similarity transformations of the matrices representing the operator Â.
Transforming operators, especially the Hamiltonian between bases can be an efficient way to solve the Schrödinger
equation. As an example: represent Ĥ with the basis {| n i} (of harmonic oscillator states) with
✓
◆
1
Ĥ| n i = En | n i = n +
h̄!| n i,
(1.102)
2
so that
and H is a diagonal matrix
Hnn0 =
0
B
B
H=B
B
@
D
E1
0
..
.
..
.
n |Ĥ|
n0
E
= En
0
E2
···
0
0
..
.
E3
..
.
16
nn0
(1.103)
1
···
··· C
C
C.
··· C
A
..
.
(1.104)
PHAS3226: Quantum Mechanics
CHAPTER 1. FORMAL QUANTUM MECHANICS
{The matrix of an operator in a basis of its own eigenstates is diagonal, i.e.if Â|ni =
n |ni,
D
E
then n|Â|n0 =
n nn0 .}
However the matrix H 0 of Ĥ in the basis | n i is not diagonal in this example as | n i are not the eigenfunctions;
the | m i are. However the eigenvalues of H 0 are still n + 12 h̄!, i.e. the (energy) spectrum is independent of the
representation.
Proof that H and H 0 have the same spectrum To find the eigenvalues of H 0 ; H 0 |ni =
equation
det (H 0
I) = 0
must be solved. But H 0 = S
1
HS and I = S
I=S
H0
and
1
1
det (H 0
1
S
S=S
⇥
I) = det S
But det (AB) = det A det B so
the characteristic
S so and
HS
det (H 0
n |ni,
1
1
(H
I) S
⇤
I) S .
(H
I) = det S
1
det (H
= det S
1
det S det (H
= det (H
I) det S
I)
I) ,
since det S 1 det S = det S 1 S = det (I). The eigenvalues of H are known and hence so are those of H 0 . Also
both the eigenvalues as well as the trace of the matrix of any operator  are independent of the (basis) representation.
Demonstration that the expectation value is independent of the (basis) representation
The expectation value of an operator  can be evaluated from
D E D
E
 =
|Â|
.
If  is represented in a basis {|
diagonal, with elements
n i}
which is not formed from its eigenfunctions, the matrix representing  is nonA0nn0 =
Expanding | i in the same basis gives
and
D
|Â|
(1.105)
E
=
|Â|
E
X
n
=
cT
D
n |Â| n0
| i=
X
cn |
⇤
n |cn Â
X
cn0 |
h
⇤
A0 c
n
n0
E
(1.106)
.
(1.107)
ni
n0 i
=
XX
n
c⇤n A0nn0 cn0
(1.108)
n0
(1.109)
Using the closure relation
D
=
=
XXD
n
n0
n
†
n0
XX
|
n ih n |Â| n0 ih n0 |
c⇤n Ânn0 cn0 ,
E
,
(1.110)
(1.111)
= c A0 c.
If {|
j i}
is an eigenbasis of  then
Â|
ji
=
17
j| ji
(1.112)
PHAS3226: Quantum Mechanics
CHAPTER 1. FORMAL QUANTUM MECHANICS
and
h
=
j |A| i i
i.e. is diagonal. It could also have been used to expand | i as
X
| i=
dj |
D
|Â|
D
E
|Â|
X
=
j
=
⇤
j |dj Â
h
⇤
dT
X
j
XX
di | i i =
j
A d = d† A d.
0
E
(1.114)
ji
j
then
(1.113)
j ji
B
= (d⇤1 , d⇤2 , . . .) @ 0
..
.
X
=
d2j j .
j |Â| i idi ,
10
1
···
d1
B
C
··· C
A @ d2 A .
..
..
.
.
0
1
j
d⇤j h
2
(1.115)
(1.116)
(1.117)
(1.118)
j
Eq(1.109) and (1.118) give the same result. This follows from SS
(where S is defined in eq. 1.71):
= 1; d = Sc; d† = c† S † and A = SA0 S
1
†
d† A d = c† S† A Sc = c S
1
and thus
D
hAi =
0
E
|Â|
,
A Sc,
= c A c,
†
1
(1.119)
= c† A0 c = d† A d.
(1.120)
Transforming one state vector to another
Here we ask the question, “what operator  will transform a given state | i into | i ?”
Suppose operator  transforms state | i to state | i so that
| i = Â| i.
In the basis {|ni}
| i=
and
X
X
n
cn |ni
d = Ac
X
m
dm =
dm |mi = Â
X
n
But
| i=
dm |mi;
m
(1.121)
P
n
2
|cn | = 1 so
= dm
dm
X
=
n
(1.123)
X
n
cn hm|Â|ni =
dm
X
n
2
|cn | =
cn |ni,
(1.124)
X
(1.125)
Amn cn
n
X
n
hm| ih |nicn
18
(1.122)
hm| ic⇤n cn
(1.126)
PHAS3226: Quantum Mechanics
CHAPTER 1. FORMAL QUANTUM MECHANICS
Comparing the expressions eq(1.125) and (1.126) for dm gives
Amn = hm| ih |ni
(1.127)
 = | ih |.
(1.128)
or
as the required projection operator.
Alternatively, since
| i = Â| i
then
| ih |
= Â| ih |
X
X
XX
XX
= Â
cm |mi
c⇤n hn| = Â
cm c⇤n hn|mi = Â
cm c⇤n
m
= Â
X
n
n
m
n
m
(1.129)
mn
(1.130)
n
2
|cn | = Â
P
2
since for the expansion coefficients, normalization of | i requires n |cn | = 1.
Note that for any basis {|ni} which is a complete set of states, a state | i has the expansion
X
| i=
cn |ni
(1.131)
(1.132)
n
and
cn = hn| i.
Thus
| i=
Thus there is the identity
X
n
hn| i|ni =
1=
X
n
for a complete set of states.
1.5.2
X
n
(1.133)
|nihn| i.
|nihn|
(1.134)
(1.135)
Change of basis - an example
N.B. The example below uses ideas from later in the course, but is included for completeness Spin is associated with
a magnetic dipole moment µ carried by a spin- 12 particle, e.g. an electron where
µ=
with µB =
energy is
eh̄
2me
gs
µB
s
h̄
(1.136)
the Bohr magneton, gs the Landé g-factor for the electron (⇡ 2). In an external magnetic field B the
µB
s · B =µe s · B
(1.137)
h̄
with µe = gs µB /h̄. In MRI/NMR it is the nuclear proton’s spin that is involved with µN = eh̄/2mp instead of the Bohr
magneton.
Consider a spin- 12 particle (or system) in a magnetic field B = Bx ı̂ + Bz k̂. The Hamiltonian is
Ĥ = E =
Ĥ
µ · B =gs
= µe s · B
⇣
⌘
= µe Bx Ŝx + Bz Ŝz
h̄
(Bx ˆx + Bz ˆz )
2
= aˆz + bˆx .
= µe
Ĥ
19
(1.138)
PHAS3226: Quantum Mechanics
CHAPTER 1. FORMAL QUANTUM MECHANICS
Suppose a/h̄ = 4MHz and b/h̄ = 3MHz . The matrix representation of Ĥ in the basis functions of ˆz is
✓
◆
✓
◆ ✓
◆
1 0
0 1
a b
(z)
Ĥ = a
+b
=
.
0
1
1 0
b
a
The matrix representation of Ĥ in the basis functions of ˆx is
✓
x h+|Ĥ|+ix
Ĥ (x) =
x h |Ĥ|+ix
But
x h+|Ĥ|
ix
x h |Ĥ| ix
1
1
|+ix = p (|↵i + | i) = p
2
2
and
1
| ix = p (|↵i
2
and hence
z |↵i
= |↵i;
z|
i=
✓
1
| i) = p
2
1
1
1
1
.
(1.140)
◆
◆
1
(h↵| + h |) (aˆz + bˆx ) (|↵i + | i) .
2
| i; x |↵i = | i; x | i = |↵i so
x h+|Ĥ|+ix
But
✓
◆
(1.139)
x h+|Ĥ|+ix
=
=
=
1
(h↵| + h |) (a|↵i a| i + b| i + b|↵i)
2
1
(a + b a + b) = b.
2
(1.141)
Using the Pauli matrices
x h+|Ĥ|+ix
=
xh
|Ĥ|+ix
=
xh
|Ĥ| ix
=
✓
◆✓ ◆
✓
◆
1
1
a b
1
a+b
(1, 1)
= (1, 1)
= b,
b
a
1
b a
2
2
✓
◆✓ ◆
✓
◆
1
1
a b
1
a+b
(1, 1)
= (1, 1)
= a,
b
a
1
b a
2
2
✓
◆✓
◆
✓
◆
1
1
a b
1
a b
(1, 1)
= (1, 1)
= b.
b
a
1
b+a
2
2
Hence
Ĥ (x) =
✓
b
a
a
b
◆
(1.142)
Ĥ (x) and Ĥ (z) are the same Hamiltonian but represented in different bases. It is easy to verify that Ĥ (x) and Ĥ (z) have
the same spectrum (of eigenvalues). For Ĥ (z) one has
a
b
b
giving
(a
) (a + )
b2 = 0,
2
a2
=0
a
b2 = 0 and
p
= ± a2 + b2 = ±5. For Ĥ (x) one has
b
a
a
=0
b
p
(b
) (b + ) a2 = 0, 2 a2 b2 = 0 and = ± a2 + b2 = ±5. p
Suppose that the system is in the ground state | 0 i of Ĥ with eigenvalue
a2 + b2 = 5. What is this state in
terms of the basis of ˆz ? (If a measurement of a sample of these spins along the z-axis is measured the magnetization is
Mz = µe N hSz i.) If the eigenvalue is and | 0 i = p|↵i + q| i then
so
Ĥ|
0i
=
(aˆz + bˆx ) (p|↵i + q| i) =
20
|
0i
(p|↵i + q| i)
PHAS3226: Quantum Mechanics
CHAPTER 1. FORMAL QUANTUM MECHANICS
or in terms of matrices
✓
✓
so q =
For
2
a
b
b
a
4
3
3
4
◆✓
◆✓
◆
p
q
◆
p
q
✓
=
=
5
p
q
✓
4p + 3q
=
5p
3p
=
5q
p1
10
and q =
4q
2
3p. Normalization requires |p| + |q| = 1 so p =
◆
p
q
◆
p3
10
and
|
0i
1
= p |↵i
10
3
p | i.
10
(1.143)
|
1i
3
1
= p |↵i + p | i.
10
10
(1.144)
= +5 the eigenstate is
The expectation value
D
Ŝz
D
Ŝz
E
=
E
=
h̄
h 0 |ˆz | 0 i
2
✓
◆✓ ◆
h̄ 1
1
1 0
1
p (1, 3)⇤
p
0
1
3
2 10
10
✓ ◆
h̄ 1
4 h̄
( 8) =
,
2 10
5 2
=
1
h̄
9
h̄
which is just the probability of "up" 10
and "down" 10
.
2
2
What is the probability of the spin pointing along the +x axis? The ground state |
c1 |↵i + c2 | i needs to be expanded in terms of the eigenstates of Ŝx as
|
0i
0i
=
p1 |↵i
10
p3 |
10
i ⌘
= d1 |+ix + d2 | ix = c1 |↵i + c2 | i
Taking the scalar product with |+ix gives
d1x h+|+ix + d2x h+| ix = c1x h+|↵i + c2x h+| i
But |+ix and | ix are orthogonal so
d1x = x h+|↵ic1 + x h+| ic2 .
Similarly using | ix
d2x = x h |↵ic1 + x h | ic2 .
These equations can be represented by the matrix equation
✓
But |+ix =
p1
2
d1
d2
(|↵i + | i) and | ix =
d = Sc
◆
✓
x h+|↵i
=
x h |↵i
p1
2
S=
(|↵i
p1
2
p1
2
| i) so x h+|↵i =
p1
2
p1
2
!
21
◆✓
x h+| i
xh | i
1
=p
2
p1
2
✓
1
1
c1
c2
◆
(1.145)
.
etc. so that
1
1
◆
(1.146)
PHAS3226: Quantum Mechanics
CHAPTER 1. FORMAL QUANTUM MECHANICS
and
✓
1
S =p
2
†
1
1
◆
1
1
(1.147)
.
Thus S is the matrix that transforms representations in the z-basis to those of the x-basis. S † does the reverse. Note that
SS † =
1
2
✓
1
1
◆✓
1
1
1
1
◆
1
1
=
1
2
✓
2
0
0
2
◆
= I2
as it should!
{If working out S for the transformation between the z-basis and the y-basis care must be taken as there are i’s when
taking transposes and complex conjugations.}
In this case
✓
◆
✓
◆
✓
◆
✓
◆
1
1
1
d1
1 1
1
1
p
=p
=p
d2
1
1
3
2
2
10
5
and so
|
0i
=
1
2
p |+ix + p | ix .
5
5
2
The probability of measuring spin along the +x-axis is |d1 | =
2
is |d2 | = 45 and the magnetization
P+
P =h
1
5
(1.148)
and the probability of measuring spin along the x-axis
0 |ˆx | 0 i
=
3
.
5
(1.149)
The expectation (and magnetization) are measurable quantities and should be independent of the representation. This is
easily verified as in the z-representation
1
⇤
h 0 |ˆx | 0 i = p (1, 3)
10
✓
0
1
1
0
◆
1
p
10
✓
1
3
◆
3
.
5
=
(1.150)
Transformation of the operators between representations
The Hamiltonian operators Ĥ (x) and Ĥ (z) are related by a similarity transform
Ĥ (x) = S Ĥ (z) S
Explicitly
✓
b
a
a
b
◆
1
=p
2
✓
1
1
1
◆✓
1
1
= S Ĥ (z) S †
a
b
b
a
◆
1
p
2
(1.151)
✓
1
1
1
1
◆
(1.152)
For the simple 2 ⇥ 2 matrix and spin- 12 problems most quantities can be calculated using the transformation of the basis,
e.g.
1
1
|+ix = p (|↵i + | i) ; | ix = p (|↵i | i)
(1.153)
2
2
and their inverses
1
|↵i = p (|+ix + | ix ) ;
2
1
|↵i = p (|+ix
2
| ix )
(1.154)
But the similarity transforms are best for complex systems of many coupled spins where the matrices can be quite large.
Also if the magnet field is constantly changing as in Ĥ (t) = µe Ŝz Bz (t) + µe Ŝx Bx (t). Note that the matrix S is
independent of Bx (t) and Bz (t) so one can switch back and forth thousands of times to calculate the magnetizations
Mx , My , Mz at different times.
22
PHAS3226: Quantum Mechanics
1.5.3
CHAPTER 1. FORMAL QUANTUM MECHANICS
Solving for energy eigenvalues
Nowadays most solutions of the time-independent Schrödinger equation are achieved by diagonalizing a matrix such as
H 0 rather than solving second-order partial differential equations.
Suppose that the true eigenstates | i are not known, but that {| i} is a known basis. Then
X (n)
Ĥ n i = En | n i = Ĥ
cj | j i
(1.155)
j
(n)
but also the coefficients cj
Procedure:
are unknown (if they were known then so would the | i be known).
1. The effect of an operator on a state is given by the effect of its matrix on the state’s column vector
2. Represent Ĥ in the basis {| i} by H 0
3. Matrix H 0 has the correct energy spectrum, thus
H 0 c(n)
4. Solve det (H 0
0
0
H11
B 0
= @ H21
..
.
0
H12
0
H22
..
.
1 0 (n)
···
c
B 1(n)
··· C
B
A @ c2
..
..
.
.
1
0
1
(n)
c1
C
B
C
C = En B c(n)
C
A
@ 2 A
..
.
(1.156)
I) = 0 to get the eigenvalues {En }
5. Each eigenvector c(n) contains the coefficients for eq(1.155) so solve
H 0 c(n) = En c(n)
Hence now all the eigenvalues and eigenstates are known without solving the wave equation.
23
(1.157)
PHAS3226: Quantum Mechanics
CHAPTER 1. FORMAL QUANTUM MECHANICS
24
CHAPTER 2. QHO
Chapter 2
Quantum Harmonic Oscillator; Generalised
Uncertainty Relations
2.1
Classical mechanics
The harmonic oscillator is an important system to study as many physical systems are approximated by it. Many systems
have an interaction potential of this general form, with a quadratic potential.
Expanding the potential function about the minimum at x = x0 as a Taylor series gives
✓
◆
✓ 2 ◆
✓ 3 ◆
dV
1
d V
1
d V
2
3
V (x) = V (x0 ) + (x x0 )
+ (x x0 )
+ (x x0 )
+ ···
(2.1)
dx x0 2
dx2 x0 6
dx3 x0
At the minimum
dV
dx x0
= 0. The force of interaction and equation of motion of a particle of mass m are
dV
d2 x
x̂ = m 2 x̂
dx
dt
✓ 2 ◆
✓ 3 ◆
d V
1
d V
2
x0 )
(x
x
)
0
2
dx x0 2
dx3 x0
F=
m
d2 x
=
dt2
(x
··· .
For small displacements from the equilibrium position x = x0 the leading term is
✓
◆
d2 x
1 d2 V
=
(x x0 )
dt2
m dx2 x0
(2.2)
which is the equation of simple harmonic motion about x = x0 .
Defining the potentials with respect to the minimum V (x0 ) and displacements relative to x0 gives
V (x) =
with
k=
and
with angular frequency ! =
± 2E/m!
2 1/2
.
p
d2 x
=
dt2
✓
1 2
kx
2
(2.3)
◆
(2.4)
d2 V
dx2
k
x=
m
x0
! 2 x,
(2.5)
k/m and solution x (t) = A cos (!t) + sin (!t). The turning points are at x =
25
PHAS3226: Quantum Mechanics
2.2
CHAPTER 2. QHO
Schrödinger equation for harmonic oscillator
Expressing the potential as V (x) = 12 m! 2 x2 , the time-independent Schrödinger equation is
h2 d2 (x) 1
+ m! 2 x2 (x) = E (x) .
2m dx2
2
By introducing the dimensionless
⇣ variable⌘
d
d
Schrödinger equation becomes dx
= ↵ dy
(2.6)
1/2
= 2E/h̄! and y = ↵x where ↵ = (m!/h̄)
d2 (y)
+
dy 2
=
mk/h̄2
(y) = 0.
y2
1/4
the
(2.7)
This is a second-order differential equation which can be solved by a power series approach. The quantum harmonic
oscillator eigenfunctions are
n
(y) = Nn Hn (y) e
y 2 /2
(2.8)
,
n
where Hn (y) is Hermite polynomial of degree n, parity ( 1) and Nn is a normalisation constant
Nn =
The eigenvalues
n
✓
↵
p n
⇡2 n!
◆1/2
(2.9)
.
= 2n + 1 give energy eigenvalues
En =
✓
1
n+
2
◆
(2.10)
h̄!,
with n = 1, 2, . . ..
In unscaled variables
↵2 x2 /2
(x) = Nn (↵) Hn (↵x) e
(2.11)
.
Thus large ! implies a narrow potential well, large ↵ and a short tail (rapid fall off) of the wavefunction (and vice versa).
The eigenfunctions are orthonormal i.e.
Z
1
1
⇤
m
(x)
and
n (x) dx =
Z
1
e
y2
Z
Nm Nn
↵
1
e
y2
Hm (y) Hn (y) dy =
(2.12)
mn ,
1
Hm (y) Hn (y) dy =
p
⇡2n n!
(2.13)
mn .
1
Recurrence relations exist for the Hermite polynomials
Hn+1 (x)
2xHn (x) + 2nHn 1 (x) = 0
dHn (x)
= 2nHn
dx
1
(x) .
The recurrence relations enable evaluation of the matrix elements, e.g.
xkn =
Z
1
1
⇤
k
(x) x
n
(x) dx = h
k
|x|
ni
=
26
✓
h̄
2m!
◆1/2 ⇣
p
k+1
k+1,n
+
p
k
k 1,n
⌘
.
(2.14)
PHAS3226: Quantum Mechanics
2.3
CHAPTER 2. QHO
Algebraic operator approach
In many physical systems involving harmonic oscillators the interest is in the energy eigenvalues or the matrix elements
and the explicit form of the eigenfunctions is not needed (or is even hard to find). This can be achieved by an elegant
operator approach.
The Hamiltonian can be written as
p̂2
1
Ĥ =
+ m! 2 x2
(2.15)
2m 2
with the momentum and position operators satisfying
[x̂, p̂x ] = ih̄.
(2.16)
Two new operators â+ and â are defined by
â+
â+
â
✓
◆1/2
1
=
(m!x̂ ip̂) =
2h̄m!

1
i
= p ↵x̂
p̂ ,
h̄↵
2
✓
◆1/2
1
=
(m!x̂ + ip̂) =
2h̄m!
1
p
2
"
⇣ m! ⌘1/2
h̄
ip̂
x̂
1/2
(mh̄!)
!#
(2.17)
(2.18)

1
i
p ↵x̂ +
p̂ .
h̄↵
2
(2.19)
The operators â+ and â are called ladder operators and variously described as: (step-up and step-down), (raising and
lowering operators) or (creation and annihilation) operators. Since x̂ and p̂ are Hermitian operators, â+ and â are
Hermitian conjugates of each other,
â†+ = â ;
↠= â+
(2.20)
These operators satisfy the commutator
[â , â+ ] = 1.
(2.21)
Explicitly
[â , â+ ] =
1
([m!x̂ + ip̂, m!x̂ ip̂])
2h̄!m
1
im! ([x̂, p̂] + [p̂, x̂])
2h̄!m
since [x̂, x̂] = [p̂, p̂] = 0. But [x̂, p̂x ] = ih̄ so
[â , â+ ] =
i
( ih̄
2h̄
ih̄) = 1.
(2.22)
Introducing the Hermitian operator
N̂ = â+ â
(2.23)
then
â+ â
=
=
=
=
=
1
(m!x̂ ip̂) (m!x̂ + ip̂)
2h̄!m
1
m2 ! 2 x̂2 + p̂2 + im! (x̂p̂
2h̄!m
1
m2 ! 2 x̂2 + p̂2 + im!ih̄
2h̄!m
✓
◆
1
1
p̂2
1
m! 2 x̂2 +
h̄! 2
2m
2
1
1
Ĥ
.
h̄!
2
27
p̂x̂)
PHAS3226: Quantum Mechanics
CHAPTER 2. QHO
Hence the Hamiltonian can be written as
✓
◆
✓
◆
1
1
Ĥ = h̄! â+ â +
= h̄! N̂ +
.
2
2
(2.24)
The operator N̂ = â+ â is called the number operator for reasons that will be more obvious later! It follows that
N̂ = â+ â commutes with the⇥ Hamiltonian
⇤ Ĥ and thus they have a common set of eigenvectors.
i
{Using the form â+ = p12 ↵x̂ h̄↵
p̂
✓
◆ ✓
◆
1
i
i
[â , â+ ] =
↵x̂ +
p̂ , ↵x̂
p̂
2
h̄↵
h̄↵
Expanding using commutator algebra and noting that x̂ and p̂ commute with themselves, so


1
i
i
[â , â+ ] =
↵x̂,
p̂ +
p̂, ↵x̂
2
h̄↵
h̄↵
⇢✓
◆
1
i
i
=
[x̂, p̂] + [p̂, x̂]
2
h̄
h̄
⇢✓
◆
1
i
i
=
ih̄ + ( ih̄) = 1.
2
h̄
h̄
Also
â+ â
=
=
=
=
=

1
i
p ↵x̂
p̂
h̄↵
2
✓
1
p̂2
↵2 x̂2 + 2 2
2
h̄ ↵
✓
1
p̂2
↵2 x̂2 + 2 2
2
h̄ ↵
✓
1
p̂2
↵2 x̂2 + 2 2
2
h̄ ↵
✓
1
1
m! 2 x̂2 +
h̄! 2
and so
Ĥ =
as before.}
Useful commutators are
Proof:
h
h
2.3.1
Ĥ, â+
i
Ĥ, â
i
= h̄!
h
✓

1
i
p ↵x̂ +
p̂
h̄↵
2
◆
i
i
+ ↵x̂ p̂
p̂↵x̂
h̄↵
h̄↵
◆
✓
◆
i
1
p̂2
i
+ [x̂, p̂] =
↵2 x̂2 + 2 2 + ih̄
h̄
2
h̄
h̄ ↵
◆
⇣
⌘
2
1 m! 2
p̂
1
1 =
x̂ +
.
2 h̄
2h̄!m 2
◆
p̂2
1
1
1
=
Ĥ
,
2m
2
h̄!
2
✓
1
â+ â +
2
i
Ĥ, â+ = h̄!â+ ;
â+ â +
1
2
◆
h
◆
Ĥ, â
i
=
h̄!â .
= h̄! [â+ â , â ] = h̄! {â+ [â , â ] + [â+ , â ] â }
Eigenvectors of N̂
Suppose | i is an eigenvector of N̂ with eigenvalue , then
N̂ | i = | i
28
(2.26)
h̄!.
, â+ = h̄! [â+ â , â+ ] = h̄! {â+ [â , â+ ] + [â+ , â+ ] â }
= h̄!â+
✓
◆
1
= h̄! â+ â +
, â
2
=
h̄!â .
(2.25)
(2.27)
PHAS3226: Quantum Mechanics
CHAPTER 2. QHO
and
h
N̂
hâ
|â
i =
h | i=
h |â+ â | i =
i =
since â†+ = â . Consider the state | ˜i = â | i, then from above
D E
˜| ˜ = .
But by necessity h ˜| ˜i
Consider
0, and hence
0.
N̂ | ˜i = (â+ â ) â | i = (â â+
= â (â+ â ) | i
= â
N̂ (â | i) = (
| i
â | i
1) â | i
â | i = â N̂ | i
â | i
1) (â | i)
(2.28)
Thus â | i is also an eigenvector of N̂ but with eigenvalue
1. Hence the reason that â is called a lowering operator.
Note that â+ has the opposite effect, i.e. N̂ (â+ | i) = ( + 1) (â+ | i) and hence is a raising operator and obviously
â+ | ˜i = â+ â | i = | i.
2
Thus from the state | i a sequence of eigenstates can be formed by repeated application of â i.e. â| i, (â ) | i,
3
n
(â ) | i, . . . (â ) | i having eigenvalues
1,
2,
3, . . .
n. Hence can only have integer values (positive
or zero) since
0. If positive non-integer values of existed, the sequence ,
1,
2,
3, . . .
n would be
unending with some values which are negative contrary to the requirement that they are all greater or equal to zero. If
is an integer, then
n can be zero and its associated eigenstate | 0 i satisfies
â |
0i
= 0|
N̂ |
0i
= 0|
â+ â |
0i
= 0|
0i
= 0,
0i
(2.29)
0 i,
so the eigenvalues of N̂ are positive integers and zero. The corresponding eigenstate is denoted by | i = |ni, such that
N̂ |ni = n|ni
(2.30)
and
N̂ (â |ni) = (n
1) (â |ni) ,
N̂ (â+ |ni) = (n + 1) (â+ |ni) .
Since N̂ |ni = n|ni and En = n + 12 h̄! the state |ni can be interpreted as containing n quanta each of energy h̄!.
The state â+ |ni contains n + 1 quanta and so â+ creates a quantum of energy and is referred to as a creation operator.
Likewise â absorbs a quantum of energy and is referred to as the annihilation operator. The operator N̂ is called the
occupation number operator.
2.3.2
Eigenvalues of the Hamiltonian
⇣
⌘
Since Ĥ = h̄! N̂ + 12 then
✓
1
Ĥ| i = h̄! N̂ +
2
◆
| i=
✓
1
n+
2
◆
h̄!| i = En | i,
(2.31)
so the energy eigenvalues are En = n + 12 h̄! with n = 0, 1, 2, . . . Note that the lowest energy is E0 = 12 h̄!, the
so-called "zero-point" energy. In this approach it arises as a consequence of the non-commutability of the position and
momentim operators, and their assocated uncertainty x p 12 h̄.
29
PHAS3226: Quantum Mechanics
CHAPTER 2. QHO
Note also that the deternination of the energy eigenvalues has been done without solving the second-order differential
equation for the harmonic oscillator in the coordinate representation.
If | E i is a state of energy E then
Ĥ| E i = E| E i
h
i
⇣
⌘
and since Ĥ, â± = ±h̄!â± then Ĥâ± â± Ĥ = ±h̄!â± and
Ĥâ± |
Ei
⇣
=
⌘
â± Ĥ ± h̄!â± |
= (â± E ± h̄!â± ) |
= (E ± h̄!) â± |
Thus â± |
2.3.3
Ei
Ei
Ei
E i.
is an eigenstate of Ĥ with eigenvalue E ± h̄!, so â+ raise the value of E by h̄! and â lowers E by h̄!.
Actions of operators â+ and â
Assume that the eigenstates |ni of N̂ , are normalised so hn|ni = 1. Then
N̂ |ni = n|ni;
N̂ |n + 1i = (n + 1) |n + 1i
(2.32)
But â+ |ni is also an eigenstate of N̂ with eigenvalue n + 1, hence
but
N̂ (â+ |ni) = (n + 1) (â+ |ni)
(2.33)
N̂ |n + 1i = (n + 1) |n + 1i
(2.34)
â+ |ni = cn |n + 1i.
(2.35)
hence (â+ |ni) must be some multiple of |n + 1i, such as
Normalisation gives, since â†+ = â ,
2
2
hn|â â+ |ni = |cn | hn + 1|n + 1i = |cn |
(2.36)
But [â , â+ ] = 1, so â â+ = â+ â + 1 and
2
hn|â+ â + 1|ni = |cn |
D
E
2
n|N̂ + 1|n
= |cn |
n+1 =
so
(2.37)
n + 1|n + 1i.
(2.38)
p
cn =
2
|cn |
n+1
and
â+ |ni =
p
Similarly for â where â |ni is an eigenstate of N̂ with eigenvalue n
â |ni = cn |n
1. Hence â |ni is a multiple of |n
1i.
Then
p
â+ â+ |ni = cn â+ |n 1i = cn n|ni
p
p
hn|â+ â+ |ni = cn nhn|ni = cn n
p
hn|N̂ |ni = n = cn n
30
1i, as
PHAS3226: Quantum Mechanics
so cn =
p
CHAPTER 2. QHO
n and
â |ni =
If |0i is the lowest energy eigenstate then
p
1i.
n|n
(2.39)
â+ |0i = |1i,
p
2
(â+ ) |0i = â+ |1i = 2|2i,
p
p p
3
(â+ ) |0i = â+ 2|2i = 2 3|3i,
p p
p p p
4
(â+ ) |0i = â+ 2 3|3i = 2 3 4|4i
and by repeated action of â+
n
(â+ ) |0i =
and hence the state
p
n!|ni
(2.40)
1
n
|ni = p (â+ ) |0i
n!
(2.41)
generates all the eigenstates of N̂ .
2.3.4
Spatial wavefunctions
The eigenfunctions in the coordinate representation can be generated in a similar manner. The lowest eigenstate satisfies
â |0i = 0,
or equivalently
✓
In the coordinate representation this is
✓
This equation has solution
1
2h̄m!
0
(m!x̂ + ip̂) |0i = 0.
✓
◆◆
d
m!x̂ + i
ih̄
dx
d 0 (x) ⇣ m! ⌘
+
x
dx
h̄
0
satisfying the requirements that
◆1/2
(x) = Ae
0
(x) = 0
(2.42)
0
(x) = 0.
(2.43)
(m!/2h̄)x2
(2.44)
(x) ! 0 as x ! ±1. Normalisation requires
Z 1
2
2
|A| e (m!/h̄)x dx = 1
1
1/4
and gives A = (m!/h̄⇡)
and
and similarly for
2
etc. Hence from |ni =
(x) =
⇣ m! ⌘1/4
2
e (m!/2h̄)x = |0i.
h̄⇡
Then 1 (x) is obtained from 0 (x) using |1i = â+ |0i, i.e.
✓
◆1/2
1
(x)
=
(m!x̂ ip̂) 0 (x)
1
2h̄m!
✓
◆1/2 ✓
◆
1
d
=
m!x̂ h̄
0 (x)
2h̄m!
dx
0
p1
n!
1
n (x) = p
n!
(2.45)
(2.46)
n
(â+ ) |0i comes
✓
1
2h̄m!
◆1/2 ✓
31
m!x̂
d
h̄
dx
◆!n
0
(x) .
(2.47)
PHAS3226: Quantum Mechanics
2.3.5
CHAPTER 2. QHO
Matrix representations
Since â+ |ni =
p
n + 1|n + 1i then taking the scalar product with another state (ket) |ki gives
p
hk|â+ |ni =
and from â |ni =
p
n|n
n + 1hk|n + 1i =
p
n+1
(2.48)
k,n+1
1i comes
hk|â |ni =
p
n
(2.49)
k,n 1 .
Thus the matrix representation of â+ is
0
B
B
B
â+ = B
B
@
0
1
0
0
..
.
0
p0
2
0
..
.
✓
1
2h̄m!
1
0 ···
0 ··· C
C
0 ··· C
C.
0 ··· C
A
p
.
4 ..
0
0
p0
3
0
From the definitions of â+ and â ,
â+
=
â
=
✓
1
2h̄m!
◆1/2
◆1/2
(m!x̂
ip̂) ,
(2.50)
(m!x̂ + ip̂) .
(2.51)
adding gives the position operator in terms of â+ and â as
x̂ =
✓
h̄
2m!
◆1/2
1
(â+ + â ) = p (â+ + â ) ,
2↵
(2.52)
and by subtracting, the momentum operator is
p̂ = i
✓
m!h̄
2
◆1/2
ih̄↵
â ) = p (â+
2
(â+
â ) .
(2.53)
k,n 1
(2.54)
The matrix elements of x̂ are
hk|x̂|ni =
✓
h̄
2m!
◆1/2
p
n+1
k,n+1
+
p
for k, n = 1, 2, . . .. Note that there are no diagonal elements, whereas in
diagonal with
E D
E
0 D
1
0
0|Ĥ|0
0|Ĥ|1
···
B D
C
E D
E
B
B
C
1|Ĥ|1
Ĥ = B 1|Ĥ|0
C = h̄! @
@
A
..
..
..
.
.
.
n
the basis states {|ni}the Hamiltonian Ĥ is
1
2
0
..
.
0
3
2
..
.
1
···
··· C
A.
..
.
(2.55)
To obtain matrices for say x̂2 there are several ways to proceed; by repeated action of x̂; by expanding x̂2 in terms of â+ ,
and â , i.e.
32
PHAS3226: Quantum Mechanics
x̂|ni =
x̂x̂|ni =
x̂2 |ni =
=
x̂2 |ni =
hk|x̂2 |ni =
✓
✓
h̄
2m!
◆1/2
CHAPTER 2. QHO
(â+ + â ) |ni =
✓
h̄
2m!
◆1/2
◆1/2
p
n + 1|n + 1i +
p
n|n
1i ,
◆1/2
✓
p
p
h̄
h̄
(â+ + â )
n + 1|n + 1i + n|n 1i ,
2m!
2m!
✓
◆
p
p
⇥
⇤
p
p
h̄
â+
n + 1|n + 1i + n|n 1i + â
n + 1|n + 1i + n|n 1i ,
2m!
✓
◆
p
p
p
⇥p
p p
p p
h̄
n + 1 n + 2|n + 2i + n n|ni + n + 1 n + 1|ni + n n 1|n
2m!
✓
◆h
i
p
p
h̄
(n + 2) (n + 1)|n + 2i + (2n + 1) |ni + n (n 1)|n 2i ,
2m!
✓
◆h
i
p
p
h̄
(n + 2) (n + 1) k,n+2 + (2n + 1) kn + n (n 1) k,n 2 .
2m!
or using [â , â+ ] = 1 = â â+ â+ â ,
✓
◆1/2
h̄
x̂ =
(â+ + â ) ,
2m!
✓
◆1/2
✓
◆1/2
h̄
h̄
2
x̂ =
(â+ + â )
(â+ + â ) ,
2m!
2m!
✓
◆
✓
◆
h̄
h̄
=
(â+ + â ) (â+ + â ) =
â2+ + â+ â + â â+ + â2 ,
2m!
2m!
✓
◆
✓
◆⇣
⌘
h̄
h̄
2
2
2
x̂ =
â+ + â+ â + â+ â + 1 + â =
â2+ + 2N̂ + 1 + â2 ,
2m!
2m!
✓
◆h
i
p
p
h̄
hk|x̂2 |ni =
(n + 2) (n + 1) k,n+2 + (2n + 1) kn + n (n 1) k,n 2 .
2m!
2.3.6
⇤
2i ,
(2.56)
(2.57)
(2.58)
(2.59)
Minimum uncertainty for the harmonic oscillator
The uncertainty in the x-position
x is given by
D
2
( x) = (x̂
2
hx̂i)
E
⌦ ↵
= x̂2
2
hx̂i .
In terms of the raising and lowering operators â+ , â , the x̂ operator is
✓
◆1/2
h̄
x̂ =
(â+ + â )
2m!
so that
x̂|ni =
=
and hence
â†+
✓
✓
h̄
2m!
h̄
2m!
◆1/2
◆1/2
(2.60)
(â+ + â ) |ni
p
n + 1|n + 1i +
p
n|n
1i
hx̂i = hn|x̂|ni = 0.
(2.61)
†
Since x̂ is an Hermitian operator and
= â and â = â+ then
✓
◆
p
p
h̄
hn|x̂ x̂|ni =
hn + 1| n + 1 + hn 1| n
2m!
✓
◆
h̄
=
{(n + 1) hn + 1|n + 1i + nhn
2m!
33
p
p
n + 1|n + 1i + n|n 1i
✓
◆
(2n + 1) h̄
1|n 1i} =
,
2m!
PHAS3226: Quantum Mechanics
CHAPTER 2. QHO
and
✓
2
( x) =
In terms of â+ , â , the momentum operator p̂ is
p̂ = i
✓
mh̄!
2
◆
n+
1
2
◆1/2
(â+
h̄
.
m!
(2.62)
â )
(2.63)
and hence
p̂|ni = i
= i
✓
✓
mh̄!
2
mh̄!
2
and hence
◆1/2
(â+
◆1/2
p
â ) |ni
n + 1|n + 1i
p
n|n
hp̂i = hn|p̂|ni = 0.
Also
⌦ 2↵
p̂
✓
◆
p
= hn|p̂ p̂|ni =
hn + 1| n + 1 hn
✓
◆
✓
◆
mh̄!
1
=
(n + 1 + n) = mh̄! n +
,
2
2
mh̄!
2
and
2
( p) =
Thus the product
2
2
( x) ( p)
=
=
✓
✓
✓
◆
n+
1
2
1
2
◆
n+
1
n+
2
and
( x) ( p) =
✓
p
p
1| n
n + 1|n + 1i
✓
n+
1
2
◆
1
2
n|n
1i
mh̄!
h̄2
n+
p
(2.65)
mh̄!.
h̄
m!
◆2
(2.64)
(2.66)
◆
(2.67)
h̄.
Thus in the lowest state n = 0, the product of uncertainties is ( x) ( p) =
lower bound of the generalised uncertainty relation.
2.3.7
1i
1
2 h̄.
The quantum oscillator satisfies the
Eigenstates of â
Suppose that |↵i is an eigenstate of â with eigenvalue ↵, then
â |↵i = ↵|↵i.
(2.68)
Expanding |↵i in terms of the complete set of states {|ni} as
X
|↵i =
cn |ni.
n
Then
â |↵i =
X
n
cn â |ni =
X
p
cn n|n
n
34
1i = ↵
X
n
cn |ni.
PHAS3226: Quantum Mechanics
CHAPTER 2. QHO
Taking the scalar product with the state |ki gives
X p
hk|â |↵i =
cn nhk|n
n
X
=
p
cn n
1i = ↵
=
k,n 1
X
X
n
cn hk|ni
p
cn+1 n + 1
k+1,n
=↵
X
cn
(2.69)
kn .
p
Using the property of the Kronecker functions and equating terms in n gives ck+1 k + 1 = ↵ck . Thus for k =
0, 1, 2, . . . in turn,
p
p
c1 = ↵c0 ;
2c2 = ↵c1 ;
3c3 = ↵c2 ; . . .
(2.70)
Hence c1 = ↵c0 , c2 =
↵2
p
c ;c
2 0 3
=
3
p↵p c0 ,
3 2
etc. and in general
↵n
cn = p c0
n!
so that
|↵i = c0
Normalisation requires h↵|↵i = 1 so
2
1
= |c0 |
2
= |c0 |
Hence
1
X
↵n
p |ni.
n!
n=0
(2.72)
1
⇤ 1
X
(↵n ) X ↵m
p
p hn|mi
n! m=0 m!
n=0
1
2n
X
2
|↵|
2
= |c0 | e|↵| .
n!
n=0
|↵|2 /2
|c0 | = e
giving
(2.71)
1
X
↵n
p |ni.
n!
n=0
|↵|2 /2
|↵i = e
(2.73)
The state |ni contains n quanta each of energy h̄!. Thus the probability that state |↵i contains k quanta is (= |ck |2 )
2
= |hk|↵i| = e
Pk
=
Pk
↵k
p
k!
|↵|2 /2
e
1
X
↵n
p hk|ni
n!
n=0
⇣
⌘k
2
|↵|
2
=
e |↵| .
k!
2
|↵|2 /2
2
(2.74)
2
This is a Poisson distribution with a mean value |↵| . The average number of quanta in state |↵i can be found from the
expectation value of the occupation number operator N̂ in the state |↵i, i.e.
2
h↵|N̂ |↵i = h↵ |â+ â | ↵i = |↵| .
(2.75)
Alternatively using the probability Pk above, the average number of quanta is
n̄
=
1
X
kPk =
k=0
2
= |↵| e
1
X
k=0
|↵|2
k
⇣
⇣
2
|↵|
k!
⌘k
e
|↵|
2
=
1
X
e
|↵|
k=1
2
⇣
2
|↵|
(k
(k)!
2
= |↵| e
|↵|2 |↵|2
e
35
2
⌘k
1)!
2
1
|↵|
X
k=0
⌘k
= |↵| .
=
1
X
k=1
2
|↵| e
|↵|
2
⇣
2
|↵|
(k
⌘k
1)!
1
PHAS3226: Quantum Mechanics
2.4
CHAPTER 2. QHO
Compatible observables and commuting operators
Two physical observables A, B are said to be compatible if a measurement of A, giving a value a, followed immediately
by a measurement of B, giving value b, leaves the system in a state where an immediate remeasurement of A or B gives
the same values a or b again. {Statement requires slight modification for a degenerate system.} The requirement that the
two measurements and remeasurement be carried out immediately ensures that between measurements the values of the
observables have not altered owing to the state of the system evolving according to the equation of motion.
Since a definite value an can be assigned to a system only if it is in an eigenstate n of the operator Â, and likewise a
value bn for operator B̂ then n is an eigenstate of both  and B̂, i.e.  n = an n and B̂ n = bn n . The observables
A and B have simultaneous eigenfunctions. The two observables are compatible, otherwise they are incompatible. It
follows that
B̂ Â n = B̂an n = bn an n = an bn n = ÂB̂ n ,
Since any function
⇣
ÂB̂
B̂ Â
⌘
n
= 0.
can be expanded in terms of the complete set {
h
i
Â, B̂ = 0,
n}
(2.76)
⇣
then ÂB̂
B̂ Â
⌘
= 0 and the commutator
(2.77)
i.e.the two operators commute with each other. Thus compatibility implies commuting operators.
The converse is true. If A and B are two commuting observables then a common set of eigenfunctions exits.
⇣ If ⌘n is
an eigenfunction of  with eigenvalue an then  n = an n . Thus B̂  n = an B̂ n But B̂  = ÂB̂ so  B̂ n =
⇣
⌘
an B̂ n and hence B̂ n is an eigenfunction of  with eigenvalue an . If an is a non-degenerate value there is only one
eigenfunction having eigenvalue an , hence B̂ n must be a multiple of n , i.e. B̂ n = bn n and n is an eigenfunction
of B̂ with eigenvalue bn . Thus every eigenfunction of  is an eigenfunction of B̂.
{The argument requires modification if the eigenvalue an is degenerate. Following a measurement of observable A
the system will have an wavefunction corresponding to one of the set of ↵-degenerate eigenfunctions ns , s = 1, 2, . . . , ↵
for eigenvalue an . In general this need not be an eigenfunction of observable B and the exact result of a measurement
of B is not predictable even though A and B commute. Nevertheless a subsequent measurement of B will result in a
wavefunction being equivalent to one of the set of common eigenfunctions so that the results of further measurements of
A or B are completely predictable. If an is ↵-fold degenerate,
nk , k = 1, 2, . . . , ↵ are ↵ orthogonal eigenfunctions of
P↵
 with eigenvalue an . Any linear combination n = k=1 ck nk is also an eigenfunction. If this is an eigenfunction of
B̂ then B̂ n = bn n and
↵
↵
X
X
B̂
ck nk = bn
ck nk .
k=1
Taking the scalar product with
k=1
nm , m = 1, 2, . . . , ↵ gives
Z
Z
↵
↵
X
X
⇤
ck
B̂
d⌧
=
b
c
nk
n
k
nm
k=1
Writing
R
⇤
nm B̂ nk
= Bmk gives
⇤
nm nk d⌧.
k=1
↵
X
(Bmk
bn
mk ) ck
= 0.
(2.78)
k=1
This set of ↵ homogeneous linear equations in the unknowns c1 , c2 , . . . , c↵ possess a non-zero (non-trivial) solution if
det |Bmk bn mk | = 0. This is a polynomial equation of degree ↵ in bn having ↵ roots. Corresponding to each
(p)
root bn where p = 1, 2, . . . , ↵ there is a solution for ckp giving the corresponding eigenfunction of B̂ which is also an
eigenfunction of Â. }
The theorem can be extended to more than two observables. If A1 , A2 ,. . . An are commuting observables whose
operators satisfy
h
i
Âr , Âs = 0 r, s = 1, 2, . . . , n,
(2.79)
36
PHAS3226: Quantum Mechanics
CHAPTER 2. QHO
they possess a complete set of simultaneous eigenfunctions (and vice-versa). This is a powerful result as it is often easier
to evaluate commutators, even in complicated situations, when there is no hope of obtaining explicit eigenfunctions.
To exploit this result some properties of commutators are be needed,
[A, B] =
[B, A]
[A, B + C] = [A, B] + [A, C]
(2.80)
(2.81)
[A, BC] = [A, B] C + B [A, C]
(2.82)
[AB, C] = A [B, C] + [A, C] B
(2.83)
[A, [B, C]] + [B, [C, A]] + [C, [A, B]] = 0.
(2.84)
2
The order of the factors must be preserved.
⇣ ⌘FromP[A, BC] = [A, B] C + B [A, C] it follows that  commutes with  ,
and more generally with Ân and so if fn  = n cn Ân then
h
2.5
⇣ ⌘i
Â, fn  = 0.
(2.85)
Heisenberg Uncertainty Relations
The most famous pair of non-commuting observables are position and momentum in the same dimension, i.e.
[x̂, p̂x ] = ih̄.
Recall that in the coordinate representation p̂x =
[x̂, p̂x ]
(2.86)
@
i @x
, so
(x) =
ih̄x
@
@ (x )
+ ih̄
= ih̄.
@x
@x
For each component
[x̂, p̂x ] = [ŷ, p̂y ] = [ẑ, p̂z ] = ih̄
(2.87)
but different Cartesian components commute, e.g.
[x̂, p̂y ] = [x̂, p̂z ] = 0.
(2.88)
The product of the uncertainties ( x) ( px ) forms the Heisenberg Uncertainty Principle
( x) ( px )
1
h̄.
2
(2.89)
This is an example of a general principle that the product of the uncertainties of two observables is related to the expectation value of their commutator.
Suppose the state of the quantum system is . In this state measurement
of the observable A results in obtaining one
P
2
of its eigenvalues of operator  with probability |cn | where = n cn n . The average value is hAi and the observed
values are scattered about this value. The observable A hAi will be scattered about zero, so a measure of the uncertainty
is
⌧⇣
D E⌘2
2
( A) =
Â
Â
,
(2.90)
D E R
where  =
⇤
 d⌧ . (This is just the standard deviation of A.) Similarly for another observable B,
2
( B) =
⌧⇣
B̂
37
D E⌘2
B̂
.
(2.91)
PHAS3226: Quantum Mechanics
CHAPTER 2. QHO
Note that
2
( A)
⌧⇣
=
⌧
=
D
=
2
( A)
D
=
D E⌘2
Â
Â
Â2
Â2
Â2
E
=
D⇣
D E⌘ ⇣
Â
Â
Â
D E D E2
2Â Â + Â
E
D E⌘E
Â
D E D E D E2
2 Â Â + Â
D E2
 .
D E
Define Hermitian operators ↵
ˆ = Â
 , ˆ = B̂
Ô = ↵
ˆ + i ˆ where is a real parameter. Clearly
Z
(2.92)
D E
B̂ and their linear combination (which is non-Hermitian)
2
Ô
d⌧
0
and is real and non-negative. Then
From the Hermiticity of ↵
ˆ and ˆ,
But from definitions of ↵
ˆ and ˆ,
⇣
↵
ˆ
Z h⇣
Z ⇣
⌘ i⇤ h⇣
⌘ i
↵
ˆ+i ˆ
↵
ˆ+i ˆ
d⌧
0.
⇤
⇣
i h
↵
ˆ , ˆ = Â
Thus the function
and implies that
Dh
Â, B̂
iE
⇤
⇣
↵
ˆ2 +
⌦ 2↵
f( )= ↵
ˆ +
2 ˆ2
+i
= ↵
ˆ2 +
2 ˆ2
+i
2 ˆ2
2
+i
D
0.
⇣
h
↵
ˆˆ
ˆ↵
ˆ
i
↵
ˆ, ˆ .
⌘
D Ei h
i
B̂ = Â, B̂
h
Â, B̂
i⌘
d⌧
E
Dh
iE
ˆ2 + i
Â, B̂
0.
0,
is purely imaginary. This function has a minimum value for
0
The value of f ( ) at this minimum is
f(
= ↵
ˆ2 +
D E
 , B̂
D E
D E
since  and B̂ are just numbers. Hence
Z
Ô
⌘⇣
⌘
i ˆ ↵
ˆ + i ˆ d⌧
↵
ˆ
⌘⇣
⌘
i ˆ ↵
ˆ+i ˆ
h
⌘
0,
Z
Also
⌘⇤ ⇣
d⌧
Ô
0)
⌦ 2↵
= ↵
ˆ
=
i
Dh
iE
Â, B̂
D E .
2 ˆ2
Dh
iE2
Dh
iE2
Dh
iE2
Â, B̂
Â, B̂
Â,
B̂
⌦ 2↵
D E +
D E = ↵
D E
ˆ +
2
2
4 ˆ
2 ˆ
4 ˆ2
38
0.
PHAS3226: Quantum Mechanics
CHAPTER 2. QHO
Thus
But
Dh
Â, B̂
iE
is purely imaginary, say
⌦
↵
ˆ2
Â
ˆ2
E
Dh
where C is real. Then
⌧⇣
↵D
D E⌘2 ⌧⇣
Â
B̂
iE2
1 Dh
Â, B̂
.
4
Â, B̂
iE
= iC
D E⌘2
B̂
2
1
1
2
(iC) = C 2 ,
4
4
2
( A) ( B)
( A) ( B)
i.e.
( A) ( B)
This is the Generalised Uncertainty Principle.
1 2
C
4
1
C
2
iE
1 Dh
Â, B̂
.
2
For the position-momentum commutator [x̂, p̂x ] = ih̄ so that ( x) ( px )
39
(2.93)
1
2 h̄.
PHAS3226: Quantum Mechanics
2.5.1
CHAPTER 2. QHO
Uncertainty relation for the harmonic oscillator
2
The uncertainties are ( x) =
eigenfunction is
D
(x̂
2
hx̂i)
E
⌦ ↵
= x̂2
✓
0 (x) =
1/2
with ↵ = (m!/h̄)
. Then clearly
hx̂i =
✓
↵
p
⇡
as the integrand is an odd function of x. Then
✓
◆Z 1
⌦ 2↵
↵
x̂ = p
x2 e
⇡
1
R1 2
using the standard integral 1 x exp ax2 dx =
2
↵
p
⇡
◆Z
◆1/2
1
(p̂
2
hp̂i)
E
. In the ground state the
↵2 x2 /2
e
↵2 x2
xe
D
(2.94)
dx = 0
1
↵2 x2
1
2a
2
hx̂i and ( p) =
dx =
p⇡
a,
✓
↵
p
⇡
◆
1
2↵2
r
⇡
1
=
,
↵2
2↵2
hence
1
( x) = p .
2↵
(2.95)
2
Note comparing the probability distribution P (x) = | 0 | = p↵⇡ exp ↵2 x2 with the normal (Gaussian) distribution
p1
exp( x2 /2 2 ) the variance is 2 1 2 = ↵ or = p12↵ .
2⇡
For the momentum operator,
✓ ◆⇣
Z 1
Z 1
⌘
2 2
↵
d
↵
↵2 x2 /2
↵2 x2 /2
hp̂x i = ih̄ p
e
e
dx = ih̄ p
↵2 xe ↵ x dx = 0,
dx
⇡
⇡
1
1
⌦ 2↵
as integrand is an odd function of x. To evaluate p̂x use integration by parts as
⌦
p̂2x
↵
=
=
=
and
✓ 2 ◆⇣
Z 1
⌘
2 2
2 2
↵
d
h̄2 p
e ↵ x /2
e ↵ x /2 dx
2
dx
⇡
1
(
)
✓ ◆⇣
Z 1 ✓ ◆ ⇣
2
⌘ 1
⌘ 2
↵h̄
d
d
↵2 x2 /2
↵2 x2 /2
↵2 x2 /2
p
e
e
e
dx
dx
dx
⇡
1
1
r
Z
Z
i2
2 2
↵h̄2 1 h 2
↵h̄2 1 4 2 ↵2 x2
↵5 h̄2 1
⇡
h̄2 ↵2
p
↵ xe ↵ x /2 dx = p
↵ x e
dx = p
=
.
2
⇡
⇡
⇡ 2↵2 ↵2
1
1
h̄↵
( px ) = p .
2
Two useful sandard integrals are
Z 1
1
e
ax2
dx =
r
Z
⇡
;
a
1
1
The product of the uncertainties
(2.96)
x2 e
ax2
dx =
1
2a
r
⇡
.
a
(2.97)
h̄
.
(2.98)
2
Thus the ground state of the harmonic oscillator minimises the uncertainty principle. In general for a harmonic oscillator
in energy state En = n + 12 h̄!, the product of the uncertainties is
✓
◆
1
( x) ( px ) = n +
h̄.
(2.99)
2
( x) ( px ) =
40
CHAPTER 3. GENERALISED ANGULAR MOMENTUM
Chapter 3
Generalised Angular Momentum
3.1
Orbital angular momentum
One set of non-commuting (and incompatible) operators which have already been met are the components of orbital
momentum. In general the components of orbital angular momentum cannot be assigned definite values simultaneously
(except for Lx = Ly = Lz = 0). A review of some of the properties of orbital angular momentum is given before
moving on to its generalizations.
In classical mechanics the principle of conservation of angular momentum is a powerful tool in the solution of
rotational problems, e.g. motion of satellites, planets, gyroscopes, etc. The role of angular momentum in quantum
mechanics is probably even more important. This is due to the fact that the angular momentum properties of a system are
largely independent of the details of the forces acting. (The angular momentum properties are related to the rotational
symmetry properties.) For example, for a particle moving in a central potential V (r) a part of the wavefunction is fully
determined by its angular momentum properties, independent of the exact form of the potential. As a consequence, the
angular momentum and hence qualitative features of the energy level schemes and spectroscopy of many-electron atoms
can be understood without formally solving the Schrödinger equation.
Classically the orbital angular momentum L with respect to the origin of a particle at position r moving with momentum p is
L = r ⇥ p.
(3.1)
This has Cartesian components
Lx
= ypz
zpy
(3.2)
Ly
= zpx
xpz
(3.3)
Lz
= xpy
ypx .
(3.4)
L2 = L2x + L2y + L2z .
(3.5)
and square of the magnitude of the orbital angular momentum
Using the fundamental position-momentum commutator relations
[x̂, p̂x ] = [ŷ, p̂y ] = [ẑ, p̂z ] = ih̄
and commutator algebra leads to
h
i
L̂x , L̂y = ih̄L̂z ;
Proof:
h
L̂x , L̂y
i
= [ŷ p̂z
h
i
h
i
L̂y , L̂z = ih̄L̂x ; L̂z , L̂x = ih̄L̂y .
ẑ p̂y , ẑ p̂x
x̂p̂z ]
= [ŷ p̂z , ẑ p̂x ]
[ẑ p̂y , ẑ p̂x ]
= ŷ p̂x [p̂z , ẑ]
0
= ih̄ (x̂p̂y
[ŷ p̂z , x̂p̂z ] + [ẑ p̂y , x̂p̂z ]
0 + p̂y x̂ [ẑ, p̂z ] = ŷ p̂x ( ih̄) + p̂y x̂ (ih̄)
ŷ p̂x ) = ih̄L̂z .
41
(3.6)
PHAS3226: Quantum Mechanics
CHAPTER 3. GENERALISED ANGULAR MOMENTUM
Relations for the other components may be obtained by cyclic permutation of the x, y, z as there can be nothing special
about the choice of x, y, z. These commutation relations mean that two components of the angular momentum can not be
measure simultaneously. This is very different from classical mechanics. The commutator relations can be expressed as
L̂ ⇥ L̂ = ih̄L̂.
The generalized uncertainty relation
( A) ( B)
gives
⇣
⇣
L̂x
L̂x
⌘⇣
⌘⇣
L̂y
L̂y
⌘
(3.7)
iE
1 Dh
| Â, B̂ |
2
iE 1 D
E
1 Dh
| L̂x , L̂y =
ih̄L̂z
2
2
1 D E
h̄ L̂z .
2
⌘
(3.8)
So in general all three components can not be known except for Lx = Ly = Lz = 0.
The components of L may not commute among themselves but each commutes with L̂2 , as for example
h
i
h
i h
i h
i
L̂x , L̂2
= L̂x , L̂2x + L̂x , L̂2y + L̂x , L̂2z
h
i
h
i h
i
h
i
= 0 + L̂x , L̂y L̂y + L̂y L̂x , L̂y + L̂x , L̂z L̂z + L̂z L̂x , L̂z
h
L̂x , L̂2
i
= ih̄L̂z L̂y + ih̄L̂y L̂z
ih̄L̂y L̂z
ih̄L̂z L̂y
= 0.
(3.9)
Similarly for the other components so that
h
i h
i h
i
L̂x , L̂2 = L̂y , L̂2 = L̂z , L̂2 = 0.
(3.10)
Thus it is possible to construct simultaneous eigenfunctions of L̂2 and one component of L. This is by convention chosen
to be the z-component L̂z (owing to its simple representation in spherical polar coordinates). Thus
⌦ ↵
Since L2 = L2x + L2y + L2z then L2
D
L̂2z
E
L̂2
= a ,
(3.11)
L̂z
= b .
(3.12)
which implies that a
b2 .
The differential operators for the components of L̂ are obtained from L̂ = r̂ ⇥ p̂ using p̂ =
In Cartesian form they are
✓
◆
@
@
L̂x =
ih̄ y
z
,
@z
@y
✓
◆
@
@
L̂y =
ih̄ z
x
,
@x
@z
✓
◆
@
@
.
L̂y =
ih̄ z
x
@x
@z
ih̄r as L̂ =
ih̄r̂ ⇥ r.
(3.13)
(3.14)
(3.15)
However as orbital angular momentum is intimately related to rotations it is more convenient to express the operators in
spherical polar coordinates (r, ✓, ) related to the Cartesian coordinates (x, y, z) by
x = r sin ✓ cos ,
(3.16)
y
= r sin ✓ sin ,
(3.17)
z
= r cos ✓
(3.18)
42
PHAS3226: Quantum Mechanics
CHAPTER 3. GENERALISED ANGULAR MOMENTUM
with 0  r  1, 0  ✓  ⇡, 0   2⇡. A tedious bit of algebra relating the partial derivatives with respect to
(x, y, z) to those with respect to (r, ✓, ) gives
✓
◆
@
@
L̂x = ih̄ sin
+ cot ✓ cos
,
(3.19)
@✓
@
✓
◆
@
@
L̂y = ih̄
cos
+ cot ✓ sin
,
(3.20)
@✓
@
@
L̂z =
ih̄ ,
(3.21)
@
and
h̄2
L̂2 =

1 @
sin ✓ @✓
✓
sin ✓
@
@✓
◆
+
1 @2
sin2 ✓ @ 2
(3.22)
.
{Note as the interest is only in L̂z and L̂2 the following procedure can be used. In spherical polar coordinates
r = r̂
then L̂ =
ˆ @
@
✓ˆ @
+
+
@r r @✓ r sin ✓ @
ih̄r̂ ⇥ r becomes
L̂ =
Noting that ẑ = cos ✓r̂
ih̄
sin ✓✓ˆ then Lz = L · ẑ =
=
L̂2
=
=
✓ˆ @
@
+ˆ
sin ✓ @
@✓
!
.
ih̄ @@ . For
h̄2 (r ⇥ r) · (r ⇥ r)
h̄2 r · [r⇥ (r ⇥ r)]
!
✓ˆ @
@
2
h̄ r · r⇥
+ˆ
sin ✓ @
@✓
using the general relation a · (c ⇥ d) = c · (d ⇥ a). Evaluating the curl
✓ˆ @
@
+ˆ
r sin ✓ @
@✓
r⇥
!
1
= 2
r sin ✓
r̂
r✓ˆ
r sin ✓ ˆ
@
@r
@
@✓
r
@
sin ✓ @
@
@
0
.
r sin ✓ @@
The interest is only in the r̂ component so
3.2
2
L̂
=
L̂2
=
 ✓
◆
✓
◆
1
@
@
@
r @
h̄ r 2
r sin ✓
r sin ✓ @✓
@✓
@
sin ✓ @

✓
◆
2
1 @
@
1 @
h̄2
sin ✓
+
.
sin ✓ @✓
@✓
sin2 ✓ @ 2
2
Eigenvalues and eigenfunctions of L̂z and L̂2
The operators L̂z and L̂2 have simultaneous eigenfunctions
(✓, ) satisfying
L̂2 (✓, ) = a (✓, ) ,
L̂2z (✓, ) = b (✓, ) .
In polar coordinates
ih̄
@ (✓, )
= b (✓, ) .
@
43
(3.23)
PHAS3226: Quantum Mechanics
CHAPTER 3. GENERALISED ANGULAR MOMENTUM
This clearly has solution (✓, ) = ⇥ (✓) ( ) = ⇥ (✓) eib /h̄ . It is usual to impose the condition that must be a
single-valued function. In this case that ( + 2⇡) = ( ). This requires eib( +2⇡)/h̄ = eib /h̄ i.e. ei2⇡b/h̄ = ei2m⇡
and so b = mh̄ where m = 0, ±1, ±2, . . ..
2
{The single-valued requirement looks reasonable in view of the probability interpretation of | (✓, )| . However this
2
would only seem to require that | (✓, )| is single valued. This would admit both integer and half-integer values of m.
However in the present context the half-integer values of m can be ruled out because they lead to solutions for ⇥ (✓) for
which the probability flux has a physically impossible behaviour (infinite at angles ✓ = 0 and ⇡). Half-integer values of
m do arise in connection with eigenstates used to describe "spin" of elementary particles.}
Normalization is achieved by
Z 2⇡
⇤
m( ) m( )d = 1
0
so that
1
( ) = p eim .
2⇡
The full wavefunction (✓, ) = ⇥ (✓) ( ) is such that it satisfies

✓
◆
1 @
@
1 @2
1
1
L̂2 (✓, ) = h̄2
sin ✓
+
⇥ (✓) p eim = a⇥ (✓) p eim
sin ✓ @✓
@✓
sin2 ✓ @ 2
2⇡
2⇡
m
and carrying out the differentiation with respect to gives
✓
◆ ✓
1 @
d⇥ (✓)
a
sin ✓
+
sin ✓ @✓
d✓
h̄2
Putting x = cos ✓,
d
d✓
=
d
sin ✓ dx
=
"
d
1
dx
1
2
x
x2
1/2 d
dx ,
m2
sin2 ✓
◆
⇥ (✓) = 0.
(3.24)
(3.25)
P (cos ✓) = ⇥ (✓) gives
#

|m|
dP` (x)
a
+ 2
dx
h̄
m2
|m|
P
(x) = 0
(1 x2 ) `
(3.26)
which is the associated Legendre differential equation. Physically acceptable solutions which remain finite over the range
0  ✓  ⇡ exist only if
a
= ` (` + 1)
h̄2
where ` = 0, 1, 2, . . ., and m = `, ` + 1, . . . ` 1, `. That is ` is a positive integer or zero. The solutions are called
|m|
the associated Legendre polynomials denoted by P` (x). The functions ⇥ (✓) are normalized so that
Z ⇡
⇥⇤`0 m (✓) ⇥`m (✓) sin ✓d✓ = `0 `
0
|m|
and are given in terms of the Legendre polynomials P`
m
⇥`m (✓) = ( 1)

|m|
= ( 1)
The simultaneous eigenfunctions
(2` + 1) (` m)!
2 (` + m)!
|m|
P`
(x) ,
0,
m
(3.28)
( ) are called the spherical harmonics Y`m (✓, ) given by

1/2
|m|
P`
(x) eim
m
0
44
0,
m < 0.
The spherical harmonics satisfy the orthonormality relation
Z
Z 2⇡
Z ⇡
⇤
Y`0 m0 (✓, ) Y`m (✓, ) d⌦ =
d
sin ✓Y`⇤0 m0 (✓, ) Y`m (✓, ) d✓ =
0
(3.27)
m < 0.
(2` + 1) (` m)!
4⇡ (` + m)!
m ⇤
m (✓, ) = ( 1) Y`m (✓, )
m
1/2
⇥`|m| (✓)
(✓, ) = ⇥ (✓)
Y`m (✓, ) = ( 1)
Y`,
(x) as
`0 ` m0 m .
(3.29)
(3.30)
(3.31)
PHAS3226: Quantum Mechanics
CHAPTER 3. GENERALISED ANGULAR MOMENTUM
The first few spherical harmonics are
Y00
Y1,±1
Y10
Y11 + Y1
1
1
4⇡,
r
3
= ⌥
sin ✓e±im ,
8⇡
r
r
3
3 ⇣z ⌘
=
cos ✓ =
,
4⇡
4⇡ r
r
r
3
3 ⇣x⌘
= 2
sin ✓ cos = 2
.
8⇡
8⇡ r
=
p
(3.32)
(3.33)
(3.34)
(3.35)
The basic eigenvalue equations for L̂2 and L̂z are
L̂2 Y`m
L̂z
= ` (` + 1) h̄2 Y`m ,
(3.36)
= mh̄Y`m ,
(3.37)
with ` = 0, 1, 2, . . ., and m = `, ` + 1, . . . ` 1, ` . Thus the ` (` + 1) h̄2 eigenvalue is (2` + 1)-fold degenerate. The
quantities `, and m` are called the orbital angular momentum quantum number and the magnetic quantum number
p respectively. A particle is referred to to as being in a state with angular momentum ` - not as one of magnitude h̄ ` (` + 1)!
For historical reasons the states with ` = 0, 1, 2, 3 are called s, p, d, and f states. The higher values of ` (= 4, 5, . . .) are
referred to in alphabetical order as g, h, etc. states.
3.3
Central potentials
The Hamiltonian for a particle moving in a potential V (r) is
Ĥ =
h̄2 2
r + V (r) .
2m
For a central potential V depends only on the magnitude of r and is spherically symmetric giving
Ĥ =
h̄2 2
r + V (r) .
2m
In polar coordinates
Ĥ
=
=
=

✓
◆
✓
◆
h̄2 1 @
1
@
@
1
@2
2 @
r
+
sin
✓
+ 2 2
+ V (r) ,
2
2
2m r @r
@r
r sin ✓ @✓
@✓
r sin ✓ @ 2
"
#
✓
◆
h̄2 1 @
L̂2
2 @
r
+ V (r) ,
2m r2 @r
@r
h̄2 r2
✓
◆
h̄2 1 @
L̂2
2 @
r
+ V (r) +
.
2
2m r @r
@r
2mr2
Since L̂2 and L̂z operate only on the angular coordinates (✓, ), then
h
i h
i
Ĥ, L̂2 = Ĥ, L̂z = 0,
(3.38)
(3.39)
(3.40)
and hence there are simultaneous eigenfunctions of the total energy E, orbital angular momentum and one component,
E`m
(r, ✓, ) = RE` (r) Y`m (✓, ) .
The energy level corresponding to the same value of ` are (2` + 1)-fold degenerate.
45
(3.41)
PHAS3226: Quantum Mechanics
3.3.1
CHAPTER 3. GENERALISED ANGULAR MOMENTUM
Review of hydrogen atom
The time-independent Schrödinger equation is
✓
h̄2 2
r
2me
e2
4⇡✏0 r
◆
(r, ✓, ) = E (r, ✓, ) .
In atomic units where e = me = h̄ = 1/4⇡✏0 = 1 so
Ĥ
and thus
Ĥ
=
"
In general since V (r) is a central potential
1 d
2r2 dr
h
1 2
r
2
1
,
r
L̂2
= f (r) + 2
2r
=
✓
d
r
dr
2
◆
1
r
L̂2
+ 2
2r
#
=E .
i h
i
Ĥ, L̂2 = Ĥ, L̂z = 0
and hence there are simultaneous eigenfunctions of Ĥ, the orbital angular momentum L̂2 and one component.
wavefunction can be written as
E`m (r, ✓, ) = RE` (r) Y`m (✓, )
and the energy eigenvalues
En`m ⌘ En =
The radial wavefunctions have the form
1
,
2n2
The
n = 1, 2, 3, . . . .
Rn` (r) = Pn` (r) e
r/n
where Pn` (r) is a polynomial of degree n ` 1. Note that 0  `  n 1, and m = `, ` + 1, . . . ` 1, `. For a
given ` there are 2` + 1 values of m all giving the same energy. Without a spin-orbit interaction all states of given n have
the same energy regardless of the ` value, i.e.are degenerate. The spin-orbit interaction removes some of this degeneracy.
The spectroscopic notation is
`
= 0 1 2 3 4 5 ···
.
notation
s p d f g h ···
3.4
Generalized angular momentum
An orbiting electron constitutes a current loop which by electromagnetic theory (Biot-Savart law for example) has associated with it a magnetic moment µL proportional to the electron’s angular momentum L. Thus a hydrogen atom in its
ground state (` = 0) would not be expected to have a magnetic moment. However when a beam of such atoms is passed
through an inhomogeneous magnetic field, the beam splits into two components (Stern-Gerlach type experiment). This
suggests that the H-atom has indeed a magnetic moment which can have two quantized orientations to the direction of
the magnetic field. Furthermore this magnetic moment µS is ascribed to an intrinsic property of the electron. Then by
analogy with orbital magnetic moment µL one deduces that the electron possesses an intrinsic angular momentum s in
its own rest frame proportional to its magnetic moment µS . This angular momentum is called spin. This spin has two
projections on the z-axis and this m = 12 or 12 as 2` + 1 = 2 gives ` = 12 . {The fine structure of spectral lines is also
explained in terms of the interaction between the magnetic moments associated with spin and the orbital angular momenta
of the electron, via the spin-orbit coupling s · `.}
Clearly spin is not an angular momentum of the familiar r ⇥ p kind. It has no classical analogue. To discuss it a
generalized definition of angular momentum is needed. Also clearly the eigenfunctions associated with the spin variable
are not ordinary functions of position, x̂, and momentum, p̂, or time of the system. Thus one has to generalize the idea of
angular momentum in quantum mechanics. The Dirac notation offers a good way to do this.
46
PHAS3226: Quantum Mechanics
CHAPTER 3. GENERALISED ANGULAR MOMENTUM
Orbital angular momentum satisfies the commutator relations
h
i
L̂x , L̂y = ih̄L̂z
and with cyclic permutations of the components. The will be the starting position for a general definition of angular momentum. As a definition J is an angular momentum if its components are Hermitian operators satisfying the commutation
relations
h
i
Jˆx , Jˆy = ih̄Jˆz
(3.42)
and with its cyclic permutations., i.e. Ĵ ⇥ Ĵ = ih̄Ĵ.
3.4.1
Raising and lowering operators
Define two operators Jˆ+ and Jˆ by
Jˆ+
Jˆ
= Jˆx + iJˆy ,
= Jˆx iJˆy .
(3.43)
(3.44)
These operators are not Hermitian (but they are self-adjoint, or Hermitian conjugates) nor do they represent any physically
measurable quantity, but they are very useful in the mathematical developments. Note that
D
Commutator relations for Jˆ+ and Jˆ
E
|Jˆ+
D
= Jˆ
|
Jˆ+ Jˆ
h
D
|Jˆ
E⇤
.
⇣
=
i
Jˆ2 , Jˆ± = 0
(3.45)
Jˆ+ Jˆ
Similarly
= Jˆ2
Jˆ+ Jˆ
Finally
h
Jˆz , Jˆ±
i
=
h
Jˆz , Jˆ±
i
= ±h̄Jˆ± .
(3.47)
Jˆz2
(3.48)
Jˆz2 + h̄Jˆz
Jˆ Jˆ+ = Jˆ2
⇣
(3.46)
⌘⇣
⌘
Jˆx iJˆy
h
i
i Jˆx Jˆy Jˆy Jˆx
Jˆx + iJˆy
= Jˆx2 + Jˆy2
and in
=
Clearly
as Jˆ2 commutes with Jˆx and Jˆy . The product
Subtracting gives
E
h̄Jˆz .
⌘ h
i
Jˆ Jˆ+ = Jˆ+ , Jˆ = 2h̄Jˆz .
h
i
Jˆ± , Jˆ⌥ = ±2h̄Jˆz .
i h
i
h
i
Jˆz , Jˆx ± iJˆy = Jˆz , Jˆx ± i Jˆz , Jˆy
⇣
⌘
⇣
⌘
= ih̄Jˆy ± i
ih̄Jˆx = ih̄Jˆy ± h̄Jˆx = ±h̄ Jˆx ± iJˆy
(3.49)
h
47
(3.50)
PHAS3226: Quantum Mechanics
3.5
CHAPTER 3. GENERALISED ANGULAR MOMENTUM
General angular momentum eigenvalue problem
Let the eigenvalues of Jˆ2 and Jˆz be ↵ and
respectively and the corresponding simultaneous eigenfunction be . Thus
Jˆ2 | i = ↵| i,
Jˆz | i =
| i.
(3.51)
(3.52)
The eigenvalues of Jˆ2 and Jˆz are real as Jˆ2 and Jˆz are Hermitian operators. Moreover the eigenvalues of Jˆ2 must be
positive or D
zero because
E the expectation value of the square
D of an Hermitian
E
D operator
E⇤must be positive or zero.
⇤
2
{Note
Jˆx
0. Choose | i = Jˆx | i then
Jˆx
=
Jˆx
= h | i = h | i
0, as Jˆx is
D E D E D E
D E D E D E D E
D E D E
Hermitian. Therefore Jˆx2 , Jˆy2 , Jˆz2 are all 0. But Jˆ2 = Jˆx2 + Jˆy2 + Jˆz2
0 and so Jˆ2
Jˆz2
2
and hence ↵
.
Operating with Jˆ+
h
i
Using Jˆz , Jˆ+ = h̄Jˆ+ i.e. Jˆz Jˆ+
Jˆ+ Jˆz = h̄Jˆ+
Similarly operating with Jˆ gives
⇣
Jˆz Jˆ+
Jˆ+ Jˆz | i = Jˆ+ | i.
⌘
h̄Jˆ+ | i =
Jˆ+ | i
⇣
⌘
⇣
⌘
Jˆz Jˆ+ | i
= ( + h̄) Jˆ+ | i .
(3.53)
⇣
⌘
Jˆz Jˆ | i = (
⇣
⌘
h̄) Jˆ | i .
(3.54)
⇣
⌘
⇣
⌘
Thus Jˆ+ | i is an eigenfunction of Jˆz with eigenvalue ( + h̄) and Jˆ | i is an eigenfunction of Jˆz with eigenvalue
(
h̄). Thus Jˆ+ and Jˆ raise or lower respectively the eigenvalues of Jˆz (They are called step-up, step-down or raising,
lowering, or ladder operators.)
h
i
Also from Jˆ2 | i = ↵| and using Jˆ2 , Jˆ± = 0
Jˆ± Jˆ2 | i = ↵Jˆ± |
⇣
⌘
⇣
⌘
Jˆ2 Jˆ± | i
= ↵ Jˆ± |
(3.55)
so Jˆ± | i are also eigenfunctions of Jˆ2 with eigenvalue ↵.
Hence by repeatedly operating with Jˆ+ a sequence of eigenstates of Jˆz can be constructed with eigenvalues + h̄,
+ 2h̄, . . . + ph̄, each eigenstate being an eigenfunction of Jˆ2 with eigenvalue ↵. Similarly repeated application of
2
ˆ
J produces a sequence of eigenfunctions of Jˆz with eigenvalues
h̄,
2h̄, . . .
qh̄. Since ↵ ( + ph̄) , and
2
↵ (
qh̄) these sequences must terminate. There will be a maximum value of + ph̄, say mT with eigenfunction
| T i such that
Jˆ+ | T i = 0
(3.56)
and
Similarly there is a minimum value of
Jˆz |
Ti
= mT |
qh̄ = mB with eigenfunction |
Jˆ |
Jˆz |
Bi
Bi
(3.57)
T i.
Bi
= 0,
= mB |
such that
(3.58)
Bi
(3.59)
Furthermore as Jˆ+ and Jˆ raise and lower the eigenvalues in steps of h̄, then
mT
mB = nh̄
where n = 0, 1, 2, . . ..
48
(3.60)
PHAS3226: Quantum Mechanics
Applying Jˆ to Jˆ+ |
Ti
CHAPTER 3. GENERALISED ANGULAR MOMENTUM
= 0 then Jˆ Jˆ+ |
Thus evaluating using equ(3.51, 3.59)
= 0. But Jˆ Jˆ+ = Jˆ2 Jˆz2
⇣
⌘
Jˆ2 Jˆz2 h̄Jˆz | T i = 0.
Ti
↵
and
Similarly from Jˆ+ Jˆ |
Bi
ˆ2
= 0, but Jˆ+ Jˆ = J
↵
and
m2T
mT h̄ |
Ti
(3.61)
=0
↵ = mT (mT + h̄)
Jˆz2 + h̄Jˆz and so
m2B + mB h̄ |
↵ = mB (mB
Equations (3.62, 3.63) give
h̄Jˆz so
Ti
(3.62)
=0
mT (mT + h̄) = mB (mB
m2T + (mT + mB ) h̄
(3.63)
h̄)
h̄) ,
m2B = 0.
This equation has two solutions
The second one violates mT
mT
=
mT
= mB
h̄.
mB thus the only valid solution is mT =
h̄
mT = n ,
2
Putting n/2 = j means mT =
mB ,
mB and so mT
n = 0, 1, 2, . . . .
0. But mT
mB = nh̄ so
(3.64)
mB = jh̄ and
↵ = j (j + 1) h̄2
(3.65)
with j = 0, 12 , 1, 32 , . . . , i.e. j is an integer or half-integer and
= mh̄
with m ranging between j and j in integer steps., i.e. m has 2j + 1 values for a given j value.
The general definition of angular momentum based on the commutator relations has led to a positive integer (or zero)
and half-integer values for j (and for m). The half odd-integer values for j are excluded for orbital angular momentum for
which j = 0, 1, 2, . . .. There is overwhelming evidence that electrons, protons, neutrons and many more particles possess
an intrinsic angular momentum, called spin, of 12 . For pions s = 0, for photons s = 1, for the ⌦ particle s = 32 . Note
properties of electron spin can be derived from Dirac’s relativistic formulation of quantum mechanics.
3.5.1
Actions of the Jˆ+ and Jˆ
The state |j, mi is an eigenstate of Jˆz with eigenvalue mh̄ so
Jˆz |j, mi = mh̄|j, mi
so
⇣
Jˆ+ Jˆz |j, mi = mh̄Jˆ+ |j, mi
⌘
⇣
⌘
Jˆz Jˆ+ h̄Jˆ+ |j, mi = mh̄ Jˆ+ |j, mi
⇣
⌘
⇣
⌘
Jˆz Jˆ+ |j, mi
= (m + 1) h̄ Jˆ+ |j, mi ,
49
(3.66)
PHAS3226: Quantum Mechanics
CHAPTER 3. GENERALISED ANGULAR MOMENTUM
⇣
⌘
so the state Jˆ+ |j, mi is an eigenstate of Jˆz with eigenvalue (m + 1) h̄. But also
Jˆz |j, m + 1i = (m + 1) h̄|j, m + 1i
(3.67)
and so the state Jˆ+ |j, mi must be a multiple of state |j, m + 1i. Hence
Similarly for the lowering operator
Jˆ+ |j, mi = C+ (j, m) |j, m + 1i.
(3.68)
Jˆ |j, mi = C (j, m) |j, m
(3.69)
1i.
Applying Jˆ to equ(3.68) and taking the scalar product with state |j, mi,
hj, m Jˆ Jˆ+ |j, mi = C+ (j, m) hj, m|Jˆ |j, m + 1i.
(3.70)
But Jˆ+ and Jˆ are Hermitian adjoint operators such that
h |Jˆ+ i = hJˆ
| i = h |Jˆ
i⇤
and so
hj, m|Jˆ |j, m + 1i = hj, m + 1|Jˆ+ |j, mi⇤
= hj, m + 1|C+ (j, m) |j, m + 1i⇤
⇤
= C+
(j, m) .
Equ(3.70) reads
2
hj, m|Jˆ Jˆ+ |j, mi = |C+ (j, m)| .
(3.71)
Alternatively noting that since Jˆ+ and Jˆ are Hermitian adjoints of each other then as
Jˆ+ |j, mi = C+ (j, m) |j, m + 1i
then
† ˆ
⇤
hj, m|Jˆ+
J+ |j, mi = hj, m + 1|C+
(j, m) C+ (j, m) |j, m + 1i
2
hj, m|Jˆ Jˆ+ |j, mi = |C+ (j, m)|
But Jˆ Jˆ+ = Jˆ2
and
Jˆz2
h̄Jˆz so
2
hj, m|Jˆ2 Jˆz2 h̄Jˆz |j, mi = |C+ (j, m)|
⇥
⇤
2
j (j + 1) m2 m h̄2 = |C+ (j, m)|
C+ (j, m) = [j (j + 1)
m (m + 1)]
1/2
h̄,
the usual convention being to take C+ (j, m) as positive so that
and similarly
Jˆ+ |j, mi = [j (j + 1)
m (m + 1)]
1/2
h̄|j, m + 1i,
(3.72)
Jˆ |j, mi = [j (j + 1)
m (m
1/2
h̄|j, m
1i.
(3.73)
50
1)]
CHAPTER 4. SPIN 1/2 SYSTEMS
Chapter 4
Spin 1/2 Systems
The theory of the last section provides the formalism for a discussion of the spin angular momentum of a particle. The
usual convention is that s denotes the spin angular momentum number, Ŝ the spin angular momentum operator. The
corresponding quantities for orbital angular momentum are ` and L̂. The discussion will be confined to spin- 12 particles an important case as electrons and protons have spin- 12 . For such particles the only possible values of m (denoted now by
ms to distinguish it from the orbital magnetic quantum number) are 12 h̄ and 12 h̄. These are often referred to as "spin-up"
q
p
and "spin-down" states (with respect to quantization axis). Since |s| = s (s + 1)h̄ = 34 h̄ and sz = 12 h̄ the classical
vector for s makes an angle of 54.7 to the z-axis!
The eigenstates of Sz are denoted by | 12 , 12 i ⌘ |↵i for ms = 12 h̄, and | 12 , 12 i ⌘ | i for ms = 12 h̄. Then
Ŝ 2 |↵i = s (s + 1) h̄2 |↵i =
Ŝ 2 | i =
Ŝz |↵i =
Ŝz | i =
3 2
h̄ | i,
4
1
h̄|↵i,
2
1
h̄| i.
2
3 2
h̄ |↵i,
4
(4.1)
(4.2)
(4.3)
(4.4)
Since |↵i and | i belong to different eigenvalues of Sz they are orthogonal, h↵| i = h |↵i = 0 and are assumed to be
normalized, h↵|↵i = h | i = 1.
The angular momentum commutation relations are
h
i
Ŝx , Ŝy = ih̄Ŝz
(4.5)
with cyclic permutations. The action of the raising and lowering operators Ŝ+ = Ŝx + iŜy and Ŝ
general,
1/2
Ŝ± |s, ms i = [s (s + 1) ms (ms ± 1)] h̄|s, ms ± 1i,

1/2
1
3
1
Ŝ± | , ms i =
ms (ms ± 1)
h̄| , ms ± 1i.
2
4
2
= Ŝx
iŜy are, in
(4.6)
(4.7)
In particular
Ŝ+ |↵i = 0;
Ŝ+ | i = h̄|↵i,
⌘
1⇣
Ŝ+ + Ŝ ;
2
Ŝy =
Ŝ |↵i = h̄| i;
Since
Ŝx =
51
(4.8)
Ŝ | i = 0.
1 ⇣
Ŝ+
2i
(4.9)
Ŝ
⌘
(4.10)
PHAS3226: Quantum Mechanics
this leads to the set
CHAPTER 4. SPIN 1/2 SYSTEMS
Ŝx |↵i = h̄2 | i
Ŝy |↵i = i h̄2 | i
Ŝz |↵i = h̄2 |↵i
Ŝx | i = h̄2 |↵i
Ŝy | i = i h̄2 |↵i .
Ŝz | i = h̄2 | i
(4.11)
An arbitrary spin- 12 state | i can be written as a linear superposition of states |↵i and | i as
| i = a|↵i + b| i.
(4.12)
Taking the scalar product with |↵i and | i in turn gives
a = h↵| i;
so
b = h | i,
| i = |↵ih↵| i + | ih | i
(4.13)
which is just an example of the general expansion in eigenstates. If | i is normalized as h | i = 1 then
1
= h |↵ih↵| i + h | ih | i
2
1
2
= |a| + |b| .
(4.14)
The expectation value of Ŝz in the state | i from
Ŝz | i = Ŝz |↵ih↵| i + Ŝz | ih | i
h̄
h̄
=
|↵ih↵| i
| ih | i
2
2
h̄
h̄
h̄ 2
2
2
h |Ŝz | i =
|h↵| i|
|h | i| ⌘ |a|
2
2
2
2
h̄ 2
|b| ,
2
(4.15)
2
as would be expected since |a| is the probability of finding the particle in state | i with spin-up and |b| the probability
of spin-down.
4.1
Useful relations for spin
The spin- 12 operators have some other useful properties. One is that Ŝ2 is a purely numerical operator since
Ŝ 2 | i = s (s + 1) h̄2 | i =
for any state | i, so
Ŝ 2 ⌘
3 2
h̄ | i
4
(4.16)
3 2
h̄ .
4
For spin- 12 only:
Ŝx2 |↵i =
Ŝx2 | i =
h̄
Ŝx | i =
2
h̄
Ŝx |↵i =
2
so
Ŝx2 | i =
h̄2
|↵i,
4
2
h̄
| i
4
h̄2
| i.
4
(4.17)
Similar results hold for Ŝy2 and Ŝz2 , hence
Ŝx2 = Ŝy2 = Ŝz2 =
52
h̄2
.
4
(4.18)
PHAS3226: Quantum Mechanics
CHAPTER 4. SPIN 1/2 SYSTEMS
Since
Ŝ+ |↵i = 0;
Ŝ+ | i = h̄|↵i,
2
Ŝ+
|↵i = 0;
then for an arbitrary state | i
Similarly for Ŝ
and thus
⇣
0
=
0
= Ŝx2
It follows that
2
Ŝ+
| i=0
2
Ŝ+
| i = 0.
(4.19)
Ŝ 2 | i = 0
(4.20)
⌘⇣
⌘
Ŝx ± iŜy ,
⇣
⌘
Ŝy2 ± i Ŝx Ŝy + Ŝy Ŝx .
Ŝx ± iŜy
(4.21)
Ŝx Ŝy + Ŝy Ŝx = 0
(4.22)
n
o
Ŝx , Ŝy = Ŝx Ŝy + Ŝy Ŝ = 0.
(4.23)
and that Ŝx and Ŝy are said to anticommute,
Similar anticommutation relations exist between the other components so,
n
o n
o n
o
Ŝx , Ŝy = Ŝy , Ŝz = Ŝz , Ŝx = 0.
(4.24)
Combining the anticommutation relations above with the commutation relations, e.g.
Ŝx Ŝy + Ŝy Ŝx = 0
and
Ŝx Ŝy
Ŝy Ŝx = ih̄Ŝz
gives
h̄
Ŝx Ŝy = i Ŝz .
2
A similar relation holds for a cyclic permutation of the indices x, y, z so that in general
h̄
Ŝx Ŝy = i Ŝz
2
(4.25)
(4.26)
with cyclic permutation of (x, y, z).
2
This relation is useful because when combined with Ŝx2 = Ŝy2 = Ŝz2 = h̄4 it shows that any arbitrary product of spin- 12
operators can be reduced to a spin independent term or a term linear in Ŝx , Ŝy , Ŝz , i.e. the most general spin- 12 operator
is
 = Â0 + B̂ · Ŝ =Â0 + B̂x Ŝx + B̂y Ŝy + B̂z Ŝz .
(4.27)
4.2
Representation of spin 1/2 operators and eigenfunctions - Pauli matrices
So far the discussion of spin has made use solely of the spin operators and their eigenstates |↵i and | i. For some
calculations it is useful to have an explicit representation for the operators and eigenfunctions. It should be obvious from
the earlier discussion that they cannot be represented as functions of the particle’s position and momentum as in the case
of orbital angular momentum. They can, however, be very conveniently handled by representing the spin operators by
2 ⇥ 2 matrices and the spin states by column vectors.
53
PHAS3226: Quantum Mechanics
4.2.1
CHAPTER 4. SPIN 1/2 SYSTEMS
Matrix representations
Since Ŝ 2 |s, ms i = s (s + 1) h̄2 |s, ms i then hs0 , m0s |Ŝ 2 |s, ms i = s (s + 1) h̄2
Ŝ 2 is diagonal with
✓
◆
3 2 1 0
2
Ŝ = h̄
.
0 1
4
s0 s m0s ms
and the matrix representation of
(4.28)
Since Ŝz |s, ms i = ms h̄|s, ms i then hs, m0s |Ŝz |s, ms i = ms h̄ m0s ms and
✓
◆
1
1
1 0
Ŝz = h̄
= h̄ˆz
0
1
2
2
(4.29)
where z is a Pauli matrix.
The matrix representations of Ŝx and Ŝy can be found using
Ŝx =
and
⌘
1⇣
Ŝ+ + Ŝ ;
2
Ŝ± |s, ms i = [s (s + 1)
Hence
Ŝx |s, ms i =
1
{[s (s + 1)
2
Explicity Ŝx | 12 , 12 i = h̄2 | 12 ,
Ŝx ⌘
✓
ms (ms + 1)]
1
2i
and Ŝx | 12 ,
1/2
1
2i
1/2
ms (ms ± 1)]
= h̄2 | 12 , 12 i and so
h̄
Ŝx =
2
By similar steps
✓
0
1
1
0
0
1
h̄
Ŝy =
2
✓
0
i
◆
=
✓
I=
✓
The three Pauli matrices
=
✓
;
y
and the identity
◆
1
2i
1
2i
1
0
◆
=
=
h̄
2
◆
=
h̄
2
0
i
i
0
◆
;
1
0
0
1
i
0
◆
Ŝ
⌘
h̄|s, ms ± 1i.
h̄|s, ms + 1i + [s (s + 1)
h 12 , 12 |Ŝx | 12 , 12 i
h 12 , 12 |Ŝx | 12 ,
1
1
1 1
h 2 , 2 |Ŝx | 2 , 2 i h 12 , 12 |Ŝx | 12 ,
x
1 ⇣
Ŝ+
2i
Ŝy =
✓
1/2
ms (ms
1)]
h̄|s, ms
h↵|Ŝx |↵i h↵|Ŝx | i
h |Ŝx |↵i h |Ŝx | i
◆
1i}.
,
(4.30)
(4.31)
(4.32)
x.
(4.33)
y.
z
=
✓
1
0
0
1
◆
;
(4.34)
(4.35)
form a complete set in terms of which any 2 ⇥ 2 matrix may be expanded, possibly with complex coefficients.
4.2.2
Commutators for the Pauli matrices
Since Pauli matrices are defined by
and
then
h̄
Ŝ = ˆ
2
h
i
Ŝx , Ŝy = ih̄Ŝz
[ˆx , ˆy ] = 2iˆz
54
(4.36)
(4.37)
PHAS3226: Quantum Mechanics
CHAPTER 4. SPIN 1/2 SYSTEMS
together with the cyclic permutations of the indices.
Also since Ŝx Ŝy = i h̄2 Ŝz then
ˆx ˆy = iˆz .
(4.38)
Note also that the traces of all the Pauli matrices vanish,
Tr
and
4.2.3
det
x
x
= Tr
= det
= Tr
y
y
z
= det
=0
z
=
(4.39)
1.
(4.40)
Basis states
The eigenfunctions |↵i and | i are represented by the column vectors
✓
◆
✓
◆
1
0
|↵i =
; | i=
.
0
1
Thus the operator equation
1
h̄|↵i
2
Ŝz |↵i =
becomes
1
h̄
2
✓
1
0
0
1
◆✓
(4.41)
1
0
◆
=
1
h̄
2
✓
1
0
◆
(4.42)
.
The arbitrary spin state | i = a|↵i + b| i becomes
✓
◆
✓
◆ ✓
◆
1
0
a
| i=a
+b
=
.
0
1
b
If |⌘i = c|↵i + d| i then
(4.43)
h⌘| i = c⇤ a + d⇤ b.
(4.44)
h⌘| = (c⇤ , d⇤ )
(4.45)
This expression can be obtained if the bra h⌘| is represented by a two-component row vector
and so
h⌘| i = (c⇤ , d⇤ )
✓
a
b
◆
= c⇤ a + d⇤ b
by the rules of matrix multiplication.
As an example consider the expectation value of Ŝz in the state | i;
✓
◆✓
◆
h̄ ⇤ ⇤
h̄
1 0
a
h |Ŝs | i = (a , b )
= (a⇤ a
0
1
b
2
2
b⇤ b) =
as before.
4.2.4
2
|b|
⌘
(4.46)
Determination of eigenvalues and eigenvectors
To solve, say, Ŝx | i = a| i in the matrix representation. Write a = h̄2 and Ŝx =
✓
◆
u
| i=
,
v
so that
h̄ ⇣ 2
|a|
2
h̄
2
✓
0
1
1
0
◆✓
u
v
◆
55
=
h̄
2
✓
u
v
◆
,
h̄
2 ˆx
and express
(4.47)
PHAS3226: Quantum Mechanics
CHAPTER 4. SPIN 1/2 SYSTEMS
then
✓
◆✓
1
1
◆
u
v
= 0.
(4.48)
This equation has a non-trivial solution if
det
1
1
=0
(4.49)
i.e.
1=0
2
and hence = ±1.
For = +1,
h̄
2
✓
0
1
1
0
◆✓
◆
u
v
h̄
=
2
✓
u
v
◆
gives
v
= u
u
= v.
2
2
Normalization of the eigenvector | i as 1 = h | i = |u| + |v| gives u = v =
eigenvector is
✓
◆
1
1
1
p
| +i =
= p (|↵i + | i) .
1
2
2
For
=
1 one gets u =
q
1
2.
The eigenvalue is thus
h̄
2
and the
(4.50)
v and eigenvector
|
1
i= p
2
✓
1
1
◆
1
= p (|↵i
2
| i) .
(4.51)
In a similar way the eigenvalues and eigenvectors of Ŝy and Ŝz can be found. These are summarized below:
eigenvalue
Ŝx
1
2 h̄
1
2 h̄
Ŝy
1
2 h̄
1
2 h̄
Ŝz
1
2 h̄
1
2 h̄
eigenstate
|
|
= |+ix =
p1
2
(|↵i + | i)
i = | ix =
p1
2
(|↵i
+i
| i)
|+iy =
p1
2
(|↵i + i| i)
| iy =
p1
2
(|↵i
i| i)
|+iz = |↵i
| iz = | i
matrix
✓ form
◆
1
p1
2
✓ 1 ◆
1
p1
2
✓ 1◆
1
p1
2
✓ i ◆
1
p1
2
i
✓
◆
1
✓ 0 ◆
0
1
(4.52)
The eigenvectors can also be found via the basis states |↵i and | i Express | i = a|↵i + b| i then
Ŝx | i = a| i =
⌘
1⇣
Ŝ+ + Ŝ (a|↵i + b| ) =
2
h̄
h̄
b |↵i + a | i =
2
2
56
1
h̄ | i
2
1
h̄ | i
2
1
h̄ (a|↵i + b| i)
2
(4.53)
PHAS3226: Quantum Mechanics
CHAPTER 4. SPIN 1/2 SYSTEMS
Taking the scalar product with |↵i and | i, or noting that |↵i and | i are linearly independent so that coefficients on each
side may be equated, gives
h̄
2
h̄
a
2
Thus
4.2.5
2
= 1, a2 = b2 ,
= ±1. For
1
h̄ a
2
1
h̄ b.
2
=
b
=
= +1, a = b and normalization gives a = b =
p1
2
as before.
Spin along an arbitrary direction
The component of spin S along a direction n̂ is Ŝn = Ŝ · n̂. Taking the direction
n̂ = (cos sin ✓, sin sin ✓, cos ✓)
then
Ŝn
=
=
=
Ŝn
=
h̄
(ˆx cos sin ✓ + ˆy sin sin ✓ + ˆz cos ✓)
2 ⇢✓
◆
✓
◆
✓
h̄
0 1
0
i
1
cos sin ✓ +
sin sin ✓ +
1 0
i 0
0
2
✓
◆
h̄
cos ✓
cos sin ✓ i sin sin ✓
cos sin ✓ + i sin sin ✓
cos ✓
2
✓
◆
i
h̄
cos ✓
sin ✓e
.
sin ✓ei
cos ✓
2
The eigenvalue equation is
Ŝn | i =
with | i = a|↵i + b| i, or in matrix form
✓
cos ✓
sin ✓ei
Hence for a non-trivial solution
sin ✓e i
cos ✓
cos ✓
sin ✓ei
(cos ✓
0
1
◆
cos ✓
h̄
| i
2
◆✓
a
b
sin ✓e
cos ✓
◆
=
i
) (cos ✓ + )
✓
a
b
◆
(4.55)
.
=0
sin2 ✓ = 0
= 1,
2
= ±1.
(4.56)
Thus the eigenvalues are ±h̄/2 for a spin- 12 system regardless of the axis chosen. To find the eigenvectors, for
a cos ✓ + b sin ✓e
✓
✓
b2 sin cos e
2
2
so
Choose for
i
= a
i
= a (1
cos ✓) = 2a sin2
sin ✓2 ei
b
=
.
a
cos ✓2
= +1,
|+in̂ =
(4.54)
✓
cos ✓2
sin ✓2 ei
57
= +1
✓
2
(4.57)
◆
,
PHAS3226: Quantum Mechanics
=
and for
1,
| in̂ =
CHAPTER 4. SPIN 1/2 SYSTEMS
✓
sin ✓2
cos ✓2 ei
◆
;
or
| in̂ =
✓
sin ✓2
cos ✓2 ei
◆
.
(4.58)
In state |+in̂ the probability of "spin-up" (along +ẑ) is P+ = cos2 ✓2 ; probability of "spin-down" (along -ẑ) is P
sin2 ✓2 .
4.3
=
Space-spin wavefunctions
The wavefunction of a spin- 12 particle also has a spatial dependence. The complete wavefunction is
ms =1/2
X
=
cms
ms
(r)
(4.59)
ms ,
ms = 1/2
which for a spin- 12 particle can be written a
(r, s) = c 12 12 (r) |↵i + c 12
✓
◆
c 12 12 (r)
(r, s) =
.
1 (r)
c 12
2
1
2
(r) | i,
(4.60)
(4.61)
Thus a spin- 12 particle is described by a two-component wavefunction (a spinor). The probability that the particle will be
2
found in a volume element d⌧ with "spin up" c 12 12 (r) , and the probability that it is found anywhere with "spin-up" is
2
2
R
1 (r)
c 12 provided 12 (r) is normalized, as
d⌧ = 1. The expectation value
2
D
|Ŝz |
E
h̄
=
2
✓
c 12
2
c
2
1
2
◆
(4.62)
Suppose the electron has spin fully aligned along the z-axis, i.e. it’s spin state is |↵i and it is moving along the x-axis in
h̄2 d2
a particle free region. Then Ĥ = 2m
dx2 and the wavefunction is
(x) = Aeikx |↵i = Aeikx
with
4.3.1
+
(x) = Aeikx and
✓
1
0
◆
=
✓
+
(x)
(x)
◆
(4.63)
(x) = 0.
Stern-Gerlach experiment
This is one of the famous experiments that demonstrated the existence of electron spin (1922, Nobel prize 1943). Originally performed with Ag atoms which have an unpaired electron in the outer shell. The beam of Ag atoms is directed
along the x-axis and passes between the poles of a magnet with a very non-uniform field along the z-axis. The potential
energy is
⇣
⌘
µB
µB
V = µ·B=
gs
s · B = gs
s · B.
(4.64)
h̄
h̄
and the force is
dV
µB dB
µB dB
Fz =
= gs
s·
k̂ = gs
Sz
.
(4.65)
dz
h̄
dz
h̄
dz
Depending on the sign of Sz (+h̄/2 or h̄/2) the force is upwards or downwards. Hence two "spots" are seen on a screen.
Classically s would be orientated at random, so all points on the screen between the two spot limits would be filled in.
Suppose that the atoms are initially polarized with ms = +h̄/2, i.e. ˆz = +1. The incoming wavefunction is
"
in
(r, t) = f (r
58
vt) eikx |↵i
(4.66)
PHAS3226: Quantum Mechanics
CHAPTER 4. SPIN 1/2 SYSTEMS
where f is an "envelope" function describing the shape of the wave packet - the spatial part of the wave packet at some
time t. Particles are located within the envelope whose maximum advances at speed v = h̄k/m. When the atoms emerge
from the magnet they will have been deflected by an amount r = z k̂ and the outgoing wavefunction is
⇣
⌘
"
(r,
t)
=
f
r
vt
+
z
k̂
eikx |↵i.
(4.67)
out
Similarly the "spin-down" atoms have
z
=
1 and
#
in
(r, t) = f (r
⇣
#
(r,
t)
=
f
r
out
vt) eikx | i,
⌘
vt
z k̂ eikx | i.
(4.68)
(4.69)
Suppose that initially the atoms were fully polarized along the +x-axis, so then
1
|+ix = p (|↵i + | i) .
2
Then before the magnet
After the magnet induce deflections
✓
✓
"
out
#
out
"
in
#
in
(r, t)
(r, t)
(r, t)
(r, t)
◆
◆
1
= p eikx
2
✓
f (r
f (r
0 ⇣
1 ikx @ f ⇣r
=p e
2
f r
vt)
vt)
vt +
vt
◆
.
⌘ 1
z k̂
⌘ A.
z k̂
(4.70)
(4.71)
The intensity of each beam will be equal. If the lower beam (state | i) is taken through a second magnet with B = Bx (x)
the atoms will be found with a 50:50 mix along the +x-axis and the x-axis since the lower z-beam was in a state
| i = p12 (|+ix + | ix ).
4.4
Addition of angular momentum
For an atom in free space, its total angular momentum is conserved (i.e. it commutes with the Hamiltonian). This means
that the stationary states (energy eigenstates) of the atom can be chosen so that the square of total angular momentum and
the z-component of angular momentum are precisely defined. But the total angular momentum of an atom is generally
the sum of a number of individual angular momenta. Each electron in the atom has an orbital angular momentum as well
as an intrinsic spin angular momentum. The vector sum of all these angular momenta is the total angular momentum of
the electrons. In addition, the atomic nucleus generally has an intrinsic angular momentum, which must also be included
to obtain the total angular momentum of the atom. So in order to understand atoms, we need to know how to add angular
momenta. The question is: if two (quantised) angular momenta are add them together, what are the possible values of
the j and m quantum numbers of the resulting total angular momentum, and how are the eigenvectors of total angular
momentum expressed in terms of the eigenvectors of the individual angular momenta? The general theory of addition
of angular momenta is rather complicated, and is presented in the 4th-year quantum mechanics course. Here, only two
important special cases are studied: the addition of two spin- 12 angular momenta, and the addition of a spin- 12 angular
momentum and an orbital angular momentum. The first case is relevant to the addition of the spin of two electrons in
an atom (e.g. the He atom), and the second case is relevant to the addition of the orbital and spin angular momenta of an
electron.
4.4.1
Addition of two spin-1/2 angular momenta
The simplest, but important example, is the addition of the spins of two spin- 12 particles, e.g. two electrons. The vector
operators representing the spin angular momenta of these two particles are denoted by Ŝ1 and Ŝ2 . Then the total spin
angular momentum Ŝ is the vector sum of Ŝ1 and Ŝ2 :
Ŝ = Ŝ1 + Ŝ2 .
59
(4.72)
PHAS3226: Quantum Mechanics
CHAPTER 4. SPIN 1/2 SYSTEMS
Since Ŝ is an angular momentum, it is quantized: its square can only have values S(S + 1)h̄2 , with S an integer or a
half-odd-integer; its z-component Ŝz can only have the values ms h̄, with ms = S, S + 1, . . . S 1, S.
Let the two eigenvectors of ŝ1z be called ↵1 and 1 then
Ŝ1z ↵1 =
1
h̄ ↵1 ,
2
Ŝ1z
1
=
Ŝ12 ↵1 =
3 2
h̄ ↵1 ,
4
Ŝ12
1
=
These are also eigenstates of Ŝ12 :
Similarly, for particle 2, denote the eigenvectors of Ŝ2z by ↵2 ,
Ŝ2z ↵2
=
Ŝ22 ↵2
=
1
h̄
2
3 2
h̄
4
1
(4.73)
.
1
(4.74)
.
2:
1
h̄ ↵2 , Ŝ2z
2
3 2
h̄ ↵2 , Ŝ22
4
2
=
2
=
1
h̄
2
3 2
h̄
4
2
2
.
(4.75)
Note that the state vectors ↵1 and 1 on the one hand, and ↵2 and 2 on the other hand, inhabit completely different
spaces. Linear combinations of ↵1 and 1 , e.g. p12 (↵1 + 1 ), can be constructed, describing different quantum states of
particle 1. Similarly, linear combinations of ↵2 and 2 describe states of particle 2. But ↵1 + ↵2 has no meaning.
All possible states of the system of two spin- 12 particles can be represented as linear combinations of the following
four basic states:
↵1 ↵2 , ↵1 2 ,
(4.76)
1 ↵2 ,
1 2.
These are eigenstates of both Ŝ1z and Ŝ2z , for example,
Ŝ1z ↵1 ↵2 =
1
h̄ ↵1 ↵2 ,
2
Ŝ2z ↵1 ↵2 =
1
h̄ ↵1 ↵2 ,
2
(4.77)
so that in the state ↵1 ↵2 , S1z has the precisely defined value 12 h̄, and S2z also has the precisely defined value 12 h̄. Similarly
in the state ↵1 2 , S1z and S2z have the values 12 h̄ and 21 h̄ respectively; in state 1 ↵2 they have values 12 h̄ and 12 h̄; and
in state 1 2 , they have values 12 h̄ and 12 h̄. (Note that ↵1 ↵2 does not mean ↵1 multiplied by ↵2 in the normal sense
of the word ‘multiplied’. It is just a notation for the state of the system in which S1z and S2z have the values 12 h̄ and 12 h̄.)
The total spin is
Ŝ = Ŝ1 + Ŝ2
(4.78)
and
Ŝz = Ŝ1z + Ŝ2z .
(4.79)
Denote the general eigenstate by |S, M i. There are four possible product combinations ↵1 ↵2 , ↵1 2 , 1 ↵2 and 1 2
which are simultaneous eigenvectors of Ŝz with eigenvalues (M = ms1 + ms2 ) of 1, 0, 0, 1 respectively. The state
↵1 ↵2 with M = 1 must belong to the state S = 1 as
⇣
⌘
h̄
h̄
Ŝz |1, 1i = Ŝ1z + Ŝ2z ↵1 ↵2 = ↵1 ↵2 + ↵1 ↵2 = h̄↵1 ↵2 .
2
2
(4.80)
There are two M = 0 states, ↵1 2 , 1 ↵2 . From these can be formed two linearly independent combinations. One
corresponds to the state |1, 0i and the other to state |0, 0i. These can be found by using the lowering operators Ŝ =
Ŝ1 + S2 and
⇣
⌘
Ŝ |1, 1i = Ŝ1 + S2 ↵1 ↵2 = h̄ 1 ↵2 + h̄↵1 2 = h̄ ( 1 ↵2 + ↵1 2 ) .
(4.81)
But in general Ŝ |s, ms i = [s (s + 1)
and hence
ms (ms
1/2
1)]
h̄|s, ms 1i so
p
Ŝ |1, 1i = 2h̄|1, 0i
1
|1, 0i = p (↵1
2
60
2
+
1 ↵2 )
(4.82)
PHAS3226: Quantum Mechanics
CHAPTER 4. SPIN 1/2 SYSTEMS
Note that Ŝ is symmetric in the label 1, 2 of the two electrons. Hence operating on ↵1 ↵2 which is also symmetric under
the interchange of labels 1 and 2 must produce a symmetric state. Note also that |1, 0i is already normalized. The state
with S = 1 and M = 1 is clearly 1 2 . It could be found by operating with Ŝ on |1, 0i.
The state |0, 0i with S = 0, M = 0 must also be a linear combination of ↵1 2 and 1 ↵2 . It is orthogonal to |1, 0i. As
|1, 0i is symmetric then |0, 0i is antisymmetric, so
1
|0, 0i = p (↵1
2
Formally one can write |0, 0i = a↵1
+b
1 ↵2 ) .
2
(4.83)
then Ŝ+ or Ŝ applied to it must give zero, i.e.
⇣
⌘
Ŝ1+ + Ŝ2+ (a↵1 2 + b 1 ↵2 ) = 0
2 + b 1 ↵2 ) =
2
Ŝ+ (a↵1
1 ↵2
= bh̄↵1 ↵2 + ah̄↵1 ↵2 = 0
2
2
so a = b. Normalization requires |a| + |b| = 1 so a =
The triplet states with S = 1 are
(4.85)
p1 .
2
1
|1, 0i = p (↵1
2
|1, 1i = ↵1 ↵2 ;
(4.84)
2
+
1 ↵2 ) ;
|1, 1i =
(4.86)
1 2,
and the singlet S = 0 is
1
|0, 0i = p (↵1 2
1 ↵2 ) .
2
An alternative approach can be followed that shows that these eigenstates are also eigenstates of
Since Ŝx =
1
2
⇣
Ŝ+ + Ŝ
⌘
⇣
=
Ŝ 2
⌘ ⇣
⌘
Ŝ1 + Ŝ2 · Ŝ1 + Ŝ2
1
2i
⇣
Ŝ+
Ŝ
⌘
Express |1, 0i = a↵1
2
+ b↵2
1
h
(4.90)
⌘
1⇣
Ŝ1+ Ŝ2 + Ŝ1 Ŝ2+
2
(4.91)
⇣
⌘
Ŝ 2 = Ŝ12 + Ŝ22 + 2Ŝ1z Ŝ2z + Ŝ1+ Ŝ2 + Ŝ1 Ŝ2+ .
(4.92)
then
Ŝ 2 |1, 0i = 1 (1 + 1) h̄2 |1, 0i = 2a↵1
and
(4.89)
then
Ŝ1x Ŝ2x + Ŝ1y Ŝ2y =
and
(4.88)
= Ŝ12 + Ŝ22 + 2Ŝ1 · Ŝ2
⇣
⌘
= Ŝ12 + Ŝ22 + 2 Ŝ1x Ŝ2x + Ŝ1y Ŝ2y + Ŝ1z Ŝ2z
and Ŝy =
(4.87)
2
+ 2b↵2
⇣
⌘i
Ŝ12 + Ŝ22 + 2Ŝ1z Ŝ2z + Ŝ1+ Ŝ2 + Ŝ1 Ŝ2+ (a↵1

3 2
h̄ ↵1
4

3
+b h̄2 ↵2
4
+a
= ↵1
2
= ↵1
2

3 2
h̄ a
2
3 2
2 + h̄ ↵1
4
3 2
1 + h̄ ↵2
4
1 ↵2
(a + b) h̄2 .
61
+ b↵2
✓ ◆✓
◆
h̄
h̄
↵1
2
2
✓
◆✓ ◆
h̄
h̄
+
2
↵2
1
2
2
2+2
1 2
h̄ a + h̄2 b + ↵2
2
(a + b) h̄2 +
2
1

3 2
h̄ b
2
(4.93)
1
1)
=
2
+ h̄2
1 ↵2
1
+ h̄2
2 ↵1
1 2
h̄ b + h̄2 a
2
(4.94)
PHAS3226: Quantum Mechanics
CHAPTER 4. SPIN 1/2 SYSTEMS
Thus from eq(4.93,4.94) comparing coefficients of ↵1
2
and
2a = a + b;
1 ↵2
gives
2b = a + b
a = b.
2
2
Normalization requires |a| + |b| = 1 so a = b =
p1
2
and
1
|1, 0i = p (↵1
2
If a = b then the eigenvalue is 0 and p12 (↵1
Applying the same method to ↵1 ↵2 :
2
↵2
1)
2
+ ↵2
1) .
(4.95)
corresponds to the state |0, 0i.
⇣
⌘i
Ŝ12 + Ŝ22 + 2Ŝ1z Ŝ2z + Ŝ1+ Ŝ2 + Ŝ1 Ŝ2+ ↵1 ↵2

3 2 3 2
h̄ h̄
=
h̄ + h̄ + 2
+) + 0 + 0 ↵1 ↵2 = 2h̄2 ↵1 ↵2
4
4
22
=
Ŝ 2 ↵1 ↵2
h
= S (S + 1) h̄2 ↵1 ↵2
and so S = 1. For the z-component
⇣
⌘
h̄
h̄
Ŝz ↵1 ↵2 = Ŝ1z + Ŝ2z ↵1 ↵2 = ↵1 ↵2 + ↵1 ↵2
2
2
so M = 1 and the state ↵1 ↵2 = |1, 1i.
4.4.2
Coupling of spin-1/2 and orbital angular momentum
If a spin- 12 particle is in a state such that its orbital angular momentum quantum number is `, what are the possible values
of total angular momentum quantum number j, and how are eigenstates of Jˆ2 and Jˆz , where Ĵ is total angular momentum
constructed?
All possible states of the system having orbital angular momentum quantum number ` can be represented as linear
combinations of the state vectors Y`m (✓, ) ↵ and Y`m (✓, ) , where ↵ and represent spin-up and spin-down, as before.
(For a complete specification of the state, a radial function R(r) is needed, but this is of no interest here, since it does
not affect the angular momentum.) Denote the magnetic quantum number for orbital angular momentum by m` , and
the magnetic quantum number for total angular momentum by mj . Recalling the restriction |m` |  `, there are 2` + 1
possible values of m, so that the total number of state vectors Ylm ↵ and Ylm is 2(2` + 1).
The total angular momentum Ĵ is:
Ĵ = L̂ + Ŝ ,
(4.96)
where L̂ is the orbital angular momentum and Ŝ is the spin angular momentum. The states Y`m (✓, ) ↵ and Y`m (✓, )
are already eigenstates of Jˆz :
✓
◆
1
Y`m (✓, ) ↵
(4.97)
Jˆz Y`m (✓, ) ↵ = L̂z Y`m (✓, ) ↵ + Ŝz Y`m (✓, ) ↵ = h̄ m +
2
✓
◆
1
Jˆz Y`m (✓, )
= L̂z Y`m (✓, ) + Ŝz Y`m (✓, ) = h̄ m
Y`m (✓, ) .
(4.98)
2
From this, it is clear that the highest value of total magnetic quantum number mj is ` + 12 , and the lowest possible value
is ` 12 . This means that the highest value of J is ` + 12 , and the values of mj are ` 12 , ` + 12 . . ., ` + 12 . There are
therefore 2 ` + 12 + 1 = 2` + 2 states having j = ` + 12 . Since the total number of states is 2(2` + 1), the other 2` states
must be associated with J = ` 12 . This is conirmed by the fact that there is only a single state for which mj = ` + 12 ,
and a single state for which mj = ` 12 , but there are two states for all the other values of mj .
Note that most of the states Y`m ↵ and Y`m , although they are eigenvectors of Jˆz , are not eigenvectors of Jˆ2 . The
only exceptions to this statement are the states Y`` ↵ and Y` ` . One way to construct all the eigenstates of Jˆ2 having
62
PHAS3226: Quantum Mechanics
CHAPTER 4. SPIN 1/2 SYSTEMS
J = ` + 12 is to start with Y`` ↵ and act repeatedly with the total lowering operator Jˆ = L̂ + Ŝ . The eigenstates of
Jˆ2 for which J = ` 12 can then be obtained by finding the states that are eigenvectors of Jˆz but are orthogonal to the
eigenvectors of Jˆz having J = ` + 12 .
As an explicit example, take the case of j = ` + 12 and mJ = ` + 12 i.e. the maximal projection state Y`` ↵ ⌘
|j, ji. Applying Jˆ to it generates the state of j = ` + 12 with mJ = ` 12 . For any angular momenta, Jˆ |j, mi =
p
j (j + 1) m (m 1)h̄|j, m 1i, so
s ✓
◆
p
1
1
1
1
1
Jˆ |j, ji =
2jh̄|j, j 1i;
Jˆ |` + , ` + i = 2 ` +
h̄|` + , `
i.
2
2
2
2
2
⇣
⌘
⇣
⌘
p
L̂ + Ŝ Y`` ↵ =
L̂ + Ŝ |`, `i|↵i = 2`h̄|`, ` 1i|↵i + |`, `ih̄| i
so
s ✓
◆
1
1
2 `+
h̄|` + , `
2
2
and the state
The state |`
1
|` + , `
2
1
2, `
1
2i
= a|`, `
1
i=
2
r
p
1
i = 2`h̄|`, `
2
2`
|`, `
2` + 1
1i|↵i +
1i|↵i + |`, `ih̄| i,
r
1
|`, `ih̄| i.
2` + 1
1i|↵i + b|`, `i| i is orthogonal to |` + 12 , `
1
h` + , `
2
1
|`
2
1
,`
2
1
2i
1
i = 0,
2
r
2`
h`, `
2` + 1
|`
1
,`
2
1
i=
2
r
1
|`, `
2` + 1
r
1i|↵i
(4.100)
so
!
1
1|h↵| +
h`, `|h | (a|`, ` 1i|↵i + b|`, `i| i) = 0,
2` + 1
r
r
2`
1
a
+b
= 0,
2` + 1
2` + 1
p
b = a 2`.
q
q
2
2
1
2`
Since normalization requires |a| + |b| = 1, then a = 2`+1
and b =
2`+1 , and
r
(4.99)
2`
|`, `i| i.
2` + 1
(4.101)
(4.102)
(4.103)
(4.104)
In summary the total angular momentum Ĵ is:
Ĵ = L̂ + Ŝ ,
(4.105)
where L̂ is the orbital angular momentum and Ŝ is the spin angular momentum and
= ` (` + 1) h̄2 Y`m ,
L̂2 Y`m
= s (s + 1) h̄
2
Ŝ
Jˆ2
also
2
,
= J (J + 1) h̄2 ,
Jˆz = L̂z + Ŝz .
(4.106)
(4.107)
(4.108)
(4.109)
Four good quantum numbers are J, MJ , `, s where
MJ
Jˆz
= m` + ms
(4.110)
= MJ h̄ ,
(4.111)
63
PHAS3226: Quantum Mechanics
CHAPTER 4. SPIN 1/2 SYSTEMS
and
with |`
4.4.3
Jˆ2 |J, MJ , `, si = J (J + 1) h̄2 |J, MJ , `, si
s|  J  ` + s and
Jˆz |J, MJ , `, si = MJ |J, MJ , `, si
✓
◆
1
=
m` ±
h̄|J, MJ , `, si.
2
Addition of orbital and spin-1/2 angular momenta
{This section can be ignored} If only orbital and one spin angular momentum are to be added the method outlined below
can be followed:
The linear combination
1
|j, m + i = a|l, mi↵ + b|l, m + 1i
(4.112)
2
is constructed which is an eigenfunction of Jˆz with eigenvalue m + 12 h̄. The constants a and b have to be found such
that this state is also an eigenstate of Jˆ2 . Since
Ĵ
Ĵ
2
Ĵ2
= L̂ + Ŝ,
= L̂2 + Ŝ2 + 2L̂ · Ŝ
= L̂2 + Ŝ2 + 2Lz Sz + L̂+ Ŝ + L̂ Ŝ+
(4.113)
Then using
Jˆ± |j, mi = [j (j + 1)
1/2
m (m ± 1)]
= [(j ⌥ m) (j ± m + 1)]
1/2
h̄|j, m ± 1i,
h̄|j, m ± 1i
⇣
This will be equivalent to
⌘
L̂2 + Ŝ2 + 2Lz Sz + L̂+ Ŝ + L̂ Ŝ+ (a|l, mi↵ + b|l, m + 1i )
✓
✓ ◆◆
3
1
2
= ah̄ ` (` + 1) + + 2m
|`, mi↵
4
2
✓
✓
◆◆
3
1
2
+bh̄ ` (` + 1) + 2 (m + 1)
|`, m + 1i
4
2
⇣
⌘
1
+ah̄2 [(` m) (` + m + 1)] 2 |`, m + 1i
⇣
⌘
1
+bh̄2 [(` + m + 1) (` (m + 1) + 1)] 2 |`, mi↵.
1
Ĵ2 |j, m + i = j (j + 1) h̄2 (a|l, mi↵ + b|l, m + 1i )
2
provided (by comparing coefficients of |l, mi↵ and |l, m + 1i ) that
✓
✓ ◆◆
⇣
⌘
1
3
1
ah̄2 ` (` + 1) + + 2m
+ bh̄2 [(` + m + 1) (` (m + 1) + 1)] 2
4
2
✓
✓
◆◆
⇣
⌘
1
3
1
2
2
2
ah̄ [(` m) (` + m + 1)] + bh̄ ` (` + 1) + 2 (m + 1)
4
2
i.e.

3
` (` + 1) + + m
4
[(`
j (j + 1) a + [(`

3
1/2
m) (` + m + 1)] a + ` (` + 1) +
4
64
m) (` + m + 1)]
m
1
1/2
= j (j + 1) h̄2 a
= j (j + 1) h̄2 b
b
= 0
j (j + 1) b
= 0
PHAS3226: Quantum Mechanics
CHAPTER 4. SPIN 1/2 SYSTEMS
These are two simultaneous equations for a and b which have a non-trivial solution if the determinant is zero,


3
3
` (` + 1) + + m j (j + 1) ` (` + 1) +
m 1 j (j + 1)
(` m) (` + m + 1) = 0
4
4
Putting A = j (j + 1)
` (` + 1)
3
4,
(A
gives
m) (A + m + 1)
[` (` + 1)
and
1
±
2
A=
so A = ` or A =
`
1, and j = ` +
or j = `
1
2
` (` + 1) + 34 + m j (j + 1)
1/2
[(` m) (` + m + 1)]
gives for j = ` +
1
2
(m
2
s✓
`+
m (m + 1)] = 0,
1
2
◆2
1
2.
To solve for a and b the matrix equation
!✓
◆
1/2
[(` m) (` + m + 1)]
a
=0
b
` (` + 1) + 34 m 1 j (j + 1)
`) a + [(` m) (` + m + 1)]
✓
◆1/2
a
`+m+1
=
.
b
` m
1/2
b=0
2
Normalization requires |a| + |b| = 1, yielding
r
b=
` m
;
2` + 1
a=
r
`+m+1
.
2` + 1
Hence the state
1
1
1
|j, m + i ⌘ |` + , m + i =
2
2
2
Choosing j = `
1
2
r
`+m+1
|l, mi↵ +
2` + 1
r
` m
|l, mi↵
2` + 1
r
` m
|l, m + 1i .
2` + 1
(4.114)
`+m+1
|l, m + 1i .
2` + 1
(4.115)
gives
1
|j, m + i ⌘ |`
2
1
1
,m + i =
2
2
65
r
PHAS3226: Quantum Mechanics
CHAPTER 4. SPIN 1/2 SYSTEMS
66
CHAPTER 5. APPROXIMATE METHODS & MANY-BODY SYSTEMS
Chapter 5
Approximate methods & Many-body systems
As in classical mechanics there are relatively few interesting problems in quantum mechanics that can be solved exactly. Approximate methods are necessary to obtain eigenvalues and eigenfunctions for potentials in the Schrödinger
equation. Such methods conveniently divide into two groups according to whether the Hamiltonian of the system is
time-independent or time-dependent.
In the first instance the approximate determination of the discrete eigenvalues and eigenfunctions for stationary states
of a time-indpendent Hamiltonian will be discussed.
5.1
Time-independent perturbation theory for a non-degenerate energy level
Suppose that the time-independent Hamiltonian Ĥ of a system can be expressed as
Ĥ = Ĥ0 + H 0
(5.1)
where Ĥ0 is the unperturbed Hamiltonian whose corresponding time-independent Schrödinger equation can be solved
exactly, i.e.
Ĥ0 | n(0) i = En(0) | n(0) i
(5.2)
(0)
(0)
where En is the energy eigenvalue (known) with eigenfunction | n i (also known). These eigenfunctions form a com(0) (0)
plete orthonormal set, with h k | n i = kn . The additional term H 0 is in some sense a "small" change to Ĥ0 - a
perturbation. In perturbation theory the solution is expanded as a power series in the perturbation. For the theory to be
valid the series must be convergent in a mathematical sense. For the theory to be useful, however, the series must be
rapidly convergent such that only the first few terms are important. (The convergence of the series will not be discussed.
In some cases it can even be shown that the series cannot converge and yet the first few terms do satisfactorily describe
the physical system.) To keep track of the order of the perturbation a dimensionless parameter is introduced into the
Hamiltonian via
Ĥ = Ĥ0 + H 0
(5.3)
such that = 0 gives the unperturbed system and = 1 gives the actual system to be solved. It is assumed that Ĥ has a
discrete set of eigenvalues which are non-degenerate and satisfy
⇣
Ĥ|
⌘
Ĥ0 + H 0 |
ni
ni
= En |
= En |
(5.4)
n i,
(5.5)
n i.
(0)
(0)
Considering a particular unperturbed energy En it is assumed that H 0 is small enough that En is closer to En than
any other unperturbed level. Thus
lim En ! En(0)
!0
and
67
lim |
!0
ni
!|
(0)
n i.
(5.6)
PHAS3226: Quantum Mechanics
CHAPTER 5. APPROXIMATE METHODS & MANY-BODY SYSTEMS
The basis idea of perturbation theory is to assume that the eigenvalues and eigenfunctions can be expanded in a power
series of the perturbation parameter as
En =
1
X
j
En(j) = En(0) + En(1) +
2
En(2) + . . .
(5.7)
j=0
and
|
ni
=
1
X
j
j=0
|
(j)
n i
=|
(0)
n i
+ |
+
(1)
n i
2
(2)
n i
|
+ ...
(5.8)
where j, the power of , is the order of the perturbation. Substituting these expressions into equ(5.5) gives
⇣
⌘⇣
⌘
Ĥ0 + H 0 | n(0) i + | n(1) i + 2 | n(2) i + . . .
⇣
⌘⇣
⌘
= En(0) + En(1) + 2 En(2) + . . . | n(0) i + | n(1) i + 2 | n(2) i + . . .
1
X
j=0
j
Ĥ0 |
(j)
n
+
1
X
j
Ĥ|
0
1
X
= @
(j 1)
n
j=1
j=0
X
=
j
1
j+k
j,k
1
X
En(j) A
En(j) |
k=0
(k)
n i
=
As equ(5.9) is to be true for all values of , coefficients of equal powers of
various orders give
0
1
;
2
;
etc.
Ĥ0 |
Ĥ0 |
;
(1)
n i
(2)
n i
Ĥ0 |
+ Ĥ |
0
+ Ĥ 0 |
Ĥ0 |
5.1.1
(k)
n i
(0)
n i
(0)
n i
(1)
n i
= En(0) |
=
En(0) |
En(0) |
=
+ Ĥ 0 |
(k 1)
i
n
=
(0)
n i,
(1)
n i+
(2)
n i+
k
X
r=0
k
|
(5.9)
!
(k)
n i
j
1 X
X
j=0 k=0
j
En(k) |
(j k)
.
n
must be the same on both sides. Thus the
(5.10)
En(1) |
En(1) |
En(r) |
(0)
n i,
(1)
n i+
(5.11)
En(2) |
(0)
n i.
(5.12)
(k r)
i.
n
First-order
(1)
To obtain the first-order correction to the energy, En then equ(5.11) is
⇣
⌘
⇣
⌘
Ĥ0 En(0) | n(1) i = En(1) Ĥ 0 |
(0)
n i
(5.13)
(0)
and taking the scalar product with | n i gives
⇣
⌘
h n(0) | Ĥ0 En(0) | n(1) i = h
as |
(0)
n i
(0
(0)
n |
⇣
En(1)
⌘
Ĥ 0 |
(0)
n i
= En(1)
h
(0)
0 (0)
n |Ĥ | n i
are normalized. The L.H.S. is zero as Ĥ0 is Hermitian. To see this explicitly,
h
(0)
(1)
n |Ĥ0 | n i
=h
as En is real. Thus first-order correction to
(1)
(0) ⇤
(1)
(0 (0) ⇤
n |Ĥ0 | n i = h n |En | n i
(1)
the energy, En is given by
En(1) = h
= En(0 h
(0) (1)
n | n i
(0)
0 (0)
n |Ĥ | n i.
(5.14)
(1)
This is one of the most useful equations in quantum mechanics. The first-order correction to the energy, En is just the
(0)
expectation value of the perturbation in the state | n i. Hence to first-order
En ' En(0) + En(1) .
68
(5.15)
PHAS3226: Quantum Mechanics
5.1.2
CHAPTER 5. APPROXIMATE METHODS & MANY-BODY SYSTEMS
Second-order
Taking the scalar product of equ(5.12) with |
h
(0)
(2)
(0)
0
n |Ĥ0 | n i + h n |Ĥ |
h n(0) |Ĥ0 En(0) |
(0)
n i
gives
= En(0) h
(1)
n i
(2)
n i
(0) (2)
n | n i+
En(2) + h n(0) |En(1)
=
En(1) h
(0) (1)
n | n i
(1)
n i.
Ĥ 0 |
+ En(2) h
(0) (0)
n | n i
The L.H.S. is zero for the same reason as in the first-order case, so that
En(2) = h
(0)
0
n |Ĥ
En(1) |
(1)
n i.
(5.16)
(2)
(1)
Hence to find the second-order correction En it is necessary to find the first-order correction, | n i, to the eigenfunction
(0)
(0)
(0)
(1)
(1)
(2)
(3)
(1)
| n i . In fact in general | nn i yields
o En , En ; | n i yields En , En etc. To find | n i it can be expanded in terms
of the complete set of states |
(0)
n i
as
|
Substituting this into eq(5.13)
leads to
⇣
⇣
k
X
k
But h
(0) (0)
` | k i
=
`k
(1)
ank h
(0)
` |
⇣
(1)
(0)
ank Ek
⌘
En(0) |
⇣
X
⌘
En(0) |
(0)
k i
(1)
(0)
k i.
⇣
= En(1)
(1)
n i
(0)
k i
(5.17)
⌘
Ĥ 0 |
⇣
= En(1)
(0)
n i
⌘
Ĥ 0 |
(0)
n i
gives
Ĥ0
⌘
En(0) h
= h
(0) (0)
` | k i
(0)
` |
= En(1) h
⇣
⌘
Ĥ 0 |
En(1)
(0) (0)
` | n i
h
(0)
n i
(0)
0 (0)
` |Ĥ | n i
(5.18)
(5.19)
and by introducing the matrix element of the perturbation
(0)
0 (0)
` |Ĥ | n i,
0
Ĥ`n
=h
then
(1)
ank |
k
ank |
k
(0)
` i
=
⌘X
En(0)
Ĥ0
and taking the scalar product with |
X
Ĥ0
(1)
n i
X
k
⇣
(1)
(0)
ank Ek
En(0)
⌘
`k
= En(1)
(5.20)
0
Ĥ`n
.
`n
(5.21)
(1)
0
For ` = n the first-oder result is obtained En = Ĥnn
. For ` 6= n
⇣
⌘
(1)
(0)
0
an` E`
En(0)
=
Ĥ`n
,
(1)
=
an`
and so
|
(1)
n i
=
X
k6=n
Note that a sufficient condition for this method is that
⇣
⇣
(0)
0
Ĥ`n
(0)
E`
⌘|
(0)
k i.
En
0
Ĥkn
(0)
En
0
Ĥkn
En
⇣
(0)
Ek
69
(0)
Ek
⌘ ⌧ 1.
(0)
⌘
(5.22)
PHAS3226: Quantum Mechanics
CHAPTER 5. APPROXIMATE METHODS & MANY-BODY SYSTEMS
(1)
(1)
n i
Note also that the coefficient ann , the component of |
En(2)
= h
= h
En(2)
but h
(0)
0
n |Ĥ
(1)
En |
(0)
n i
= h
(0)
0
n |Ĥ
(0)
0
n |Ĥ
En(1) |
En(1) |
(1)
n i
X
k
(1)
En(1) | @a(1)
nn |
cannot be found. Returning to
(0)
k i
ank |
0
(0)
0
n |Ĥ
(0)
n i
along |
(0)
n i
X
+
k6=n
= 0 (the first-order correction formula), so
X (1)
En(2) = h n(0) |Ĥ 0 En(1) |
ank |
X
=
k6=n
X
k6=n
X
=
(1)
ank |
(1)
(0)
0
n |Ĥ
⇣
(1)
0
ank Ĥnk
En(1) |
En(1)
nk
(1)
(0) A
k i ,
(5.23)
(0)
k i
k6=n
ank h
1
(5.24)
(0)
k i
⌘
0
ank Ĥnk
k6=n
X
=
En(2)
k6=n
En(2)
=
X
k6=n
and
|
ni
⇣
0
Ĥkn
(0)
En
=|
(0)
n i
to first-order in the eigenfunction.
Note that if the term with k = n is included then
|
ni
⇣
= (1 + a(1)
nn |
0
Ĥkn
(0)
En
(0)
Ek
2
0
⌘ Ĥnk
,
(5.25)
(0)
(0)
2
X h k |Ĥ 0 | n i
⌘=
⇣
⌘ .
(0)
(0)
(0)
Ek
En
Ek
k6=n
+
X
k6=n
(0)
n i
+
⇣
0
Ĥkn
(0)
En
X
k6=n
(1)
⇣
(0)
Ek
0
Ĥkn
(0)
En
⌘|
(5.26)
(0)
k i
(5.27)
2
(0)
Ek
⌘|
(0)
k i.
(5.28)
The coefficients ann are not determined by these methods. It can be shown that coefficients of this type can not be found
in any order by the method above. Thus one in tempted to conclude that the choice of these quantities has no physical
(1)
significance, thus could set ann = 0!
(1)
(0) (1)
{One could attempt to find ann = h n | n i by using the normalization condition
⇣
⌘⇣
⌘
1 = h n | n i = h n(0) | + h n(1) | + 2 h n(2) | + . . . | n(0) i + | n(1) i + 2 | n(2) i + . . . .
This could be done to any order. For example, to first-order in
⇣
⌘⇣
⌘
1 =
h n(0) | + h n(0) | | n(0) i + | n(1) i = 1 + h
1
=
(0) (1)
n | n i
+ h
(0) (1)
n | n i
(1)⇤
1 + a(1)
nn + ann
(1)
(1)
shows that the real parts of ann vanish, but gives no information on the imaginary parts, so ann = i say, with
normalization to second order is carried out one gets
X (1) 2
(2)⇤
a(2)
ank = 0
nn + ann +
k
70
real. If
PHAS3226: Quantum Mechanics
CHAPTER 5. APPROXIMATE METHODS & MANY-BODY SYSTEMS
(2)
which again is only an equation for the real parts of ann . This feature is found for any order. The imaginary parts of
(j)
ann cannot be found from the normalization condition. This remaining arbitrariness corresponds to the fact that one can
(j)
i↵
multiply a wave function
| ⌘
(↵ real) without changing the normalization. Thus if ann
n i by an arbitrary phase factor e
⇣
(j)
is imaginary then 1 + ann |
of the coefficients
5.1.3
(j)
ann
(0)
n i
to zero. }
⌘ Aei↵ |
(0)
n i.
Therefore without loss of generality one can set the imaginary parts
Observations on the energy corrections
(1)
(0)
1. Unlike the first-order energy shift En which only involves the state | n i, the second-order correction depends
on the complete set of unperturbed states and, in general, leads to formidable calculations.
(0)
(0)
(2)
Ek < 0 (always) and thus En < 0.
2. If n is the ground state, En
(1)
3. The first-order energy shift En often vanishes exactly on symmetry grounds thus necessitating evaluation of the
second-order correction.
(0)
(2)
4. The states | k i over which the summation is performed are often called intermediate states. Each term in En
is viewed as a succession of two first-order "transitions" n ! k; k ! n weighted by the energy denominator
(0)
(0)
(0)
(0)
En
Ek . System appears to leave state | n i, propagate to intermediate state | k i then fall back again to
(0)
state | n i.
5. If all matrix elements of Ĥ 0 are of the same order of magnitude - a reasonable first guess - then the "nearby" levels
(2)
have a bigger effect on En than "distant" ones.
0
6. If n is not the ground state and if an important level k - lying nearby or having Ĥkn
large - lies above level n then
(2)
En is <0; if level k lies below the shift is upwards. One speaks of a tendency of levels to repel each other!
7. For perturbation theory to be useful it must give small corrections so that calculations to low order suffice. For
(0)
(0)
(0)
0
first-order changes to | n i to be small one needs Ĥnk
⌧ En
Ek for k 6= n . In general the diagonal
0
0
matrix elements Ĥnn
will be of the same order of magnitude as the non-diagonal elements Ĥnk
. Thus the condition
(0)
0
implies Ĥnn
⌧ En
(0)
Ek
(0)
0
and therefore Ĥnn
⌧ min En
(0)
must be small compared to the level spacing min En
(0)
Ek
(0)
Ek
(1)
. That is the first-order shift En
(0)
between En and the nearest lying energy level.
(0)
These conditions break down if the level is degenerate. In this case there are some states | k i, k 6= n with
(0)
(0)
(0)
(0)
(1)
the same energy Ek = En . Thus some denominators En
Ek vanish making the expressions for ank
meaningless. One can, however, relax the condition that all the unperturbed energy levels are non-degenerate as
one only needs that the level whose energy shift is being calculated is non-degenerate. The difficulties only arise
(0)
(0)
from the degeneracy of level n, the other levels with Ek 6= En can be degenerate.
5.2
The Degenerate case
(0)
(0)
(0)
(0)
As well as the difficulty referred to above, i.e. En
Ek being zero for | k i =
6 | n i for a degenerate level
(0)
(0)
there is another difficulty. Suppose level En is ↵-fold degenerate. There are ↵ unperturbed wavefunctions | nµ i,
(µ = 1, 2, . . . ↵), corresponding to the level and it is not known a priori to which of these functions, or linear combinations,
the perturbed eigenfunction tends as ! 0.
(0)
(0)
(0)
The ↵-unperturbed eigenfunctions | nµ i, (µ = 1, 2, . . . ↵), for level En are orthogonal to those | k i corresponding
(0)
(0)
to any other level Ek 6= En . Although not necessarily
D orthogonal
E amongst themselves it is always possible to construct
linear combinations of them that are orthonormal, i.e.
n
o
(0)
done so that the set
nµ are orthonormal.
(0) (0)
nµ | n⌫
71
=
µ⌫ .
It will be assumed that this has already been
PHAS3226: Quantum Mechanics
CHAPTER 5. APPROXIMATE METHODS & MANY-BODY SYSTEMS
The "correct" unperturbed eigenfunctions |
(0)
| nµ i, so that
lim |
!0
(0)
nµ i
nµ i
that |
=|
(0)
nµ i
! 0 are some linear combination of the
tend to as
nµ i
↵
X
=
⌫=1
cµ⌫ |
(0)
nv i.
(5.29)
To avoid too many summations, consideration will be given explicitly to a doubly degenerate level (↵ = 2) as the final
(0)
result is easily generalized to an ↵-fold degenerate level. In this case the degenerate unperturbed eigenfunctions are | n1 i,
(0)
| n2 i. The correct zeroth-order eigenfunctions are
(0)
n1 i
(0)
n2 i
|
|
(0)
n1 i
(0)
n1 i
= c11 |
= c21 |
+ c12 |
+ c22 |
(0)
n2 i,
(0)
n2 i
(5.30)
(5.31)
Corresponding to eqs(5.7, 5.8) of the non-degenerate case is
|
nµ i
=|
(0)
nµ i
+ |
(1)
nµ i
+
(1)
Enµ = En(0) + Enµ
+
for µ = 1, 2. Thus for first-order eq(5.13) becomes
⇣
⌘
Ĥ0 En(0) |
For a doubly degenerate case this is
⇣
⌘
Ĥ0 En(0) |
and
⇣
Ĥ0
⌘
En(0) |
Taking the scalar product of eq(5.36)with |
⇣
⌘
(0)
h n1 | Ĥ0 En(0) |
⇣
⌘
(0)
h n2 | Ĥ0 En(0) |
(1)
n1 i
=
=
(1)
n2 i
=
=
(0)
n1 i
and |
(1)
nµ i
En1
⇣
En2
⇣
2
(1)
(1)
En1
(1)
(1)
En2
(0)
n2 i
(2)
nµ i
|
+ ...,
(5.32)
(2)
Enµ
+ ...
(5.33)
⌘
Ĥ 0 |
(5.34)
⇣
(1)
= Enµ
⇣
⇣
2
(0)
nµ i.
⌘
(0)
Ĥ 0 | n1 i
⌘⇣
Ĥ 0 c11 |
⌘
(0)
(0)
n1 i + c12 | n2 i
⌘
(0)
Ĥ 0 | n2 i
⌘⇣
Ĥ 0 c21 |
⌘
(0)
(0)
n1 i + c22 | n2 i
(5.35)
(5.37)
in turn gives
⇣
⌘⇣
(1)
(0)
(1)
0
i
=
h
|
E
Ĥ
c11 |
n1
n1
n1
⇣
⌘⇣
(1)
(0)
(1)
Ĥ 0 c11 |
n1 i = h n2 | En1
(0)
n1 i
+ c12 |
(0)
0 (0)
ni |Ĥ | nj i
and have a matrix form
these equations are
⇣
⌘
(1)
0
0
Ĥ11
En1 c11 + Ĥ12
c12
⇣
⌘
(1)
0
0
Ĥ21
c11 + Ĥ22
En1 c12
(1)
0
Ĥ11
En1
0
Ĥ21
0
Ĥ12
(1)
0
Ĥ22
En1
72
!✓
(5.38)
⌘
(0)
n2 i
⌘
(0)
(0)
n1 i + c12 | n2 i
(0)
The L.H.S is zero just as in the non-degenerate case. Using the orthonormality of | n1 i and |
⇣
⌘
(1)
(0)
(0)
(0)
(0)
c11 En1 h n1 |Ĥ 0 | n1 i
c12 h n1 |Ĥ 0 | n2 i = 0,
⇣
⌘
(0)
(0)
(1)
(0)
(0)
c11 h n2 |Ĥ 0 | n1 i + c12 En1 h n2 |Ĥ 0 | n2 i
= 0.
0
Writing Ĥij
=h
(5.36)
(0)
n2 i
(5.39)
(5.40)
gives
(5.41)
(5.42)
= 0,
= 0
c11
c12
◆
= 0.
(5.43)
PHAS3226: Quantum Mechanics
CHAPTER 5. APPROXIMATE METHODS & MANY-BODY SYSTEMS
These simultaneous equations for c11 and c12 have a non-trivial solution if and only if the determinant is zero,
(1)
0
Ĥ11
En1
0
Ĥ21
0
Ĥ12
(1)
0
Ĥ22
En1
=0
(1)
(5.44)
(1)
(1)
This is a quadratic equation in En1 leading to two possible values of En1 . One solution gives En1 , the other gives
(1)
En2 . Substituting these back into eq(5.43) leads to two sets of values for the coefficients, One set for c11 and c12 and
(0)
(0)
the other set corresponds to c21 and c22 . Thus two zero-order eigenfunctions | n1 i and | n2 i are found that are linear
(0)
(0)
combinations of | n1 i and | n2 i, and which, in general, correspond to different first-order corrections to the energy
(1)
(1)
En1 and En2 . These two linear combinations are the correct zeroth-order eigenfunctions and can thus be used along
with the unperturbed eigenfunctions of the other states to evaluate higher order corrections in a similar way as in the
non-degenerate case.
(0)
nµ i
The generalization to the ↵-fold degenerate case is straightforward. Since |
nantal equation becomes
(1)
0
Ĥ11
En1
0
Ĥ21
0
Ĥ31
..
.
0
Ĥ12
(1)
En1
0
Ĥ32
..
.
0
Ĥ22
0
Ĥ↵1
0
Ĥ13
0
Ĥ23
(1)
0
Ĥ33
En1
..
.
...
...
...
..
.
0
Ĥ↵3
...
0
Ĥ↵2
=
P↵
(0)
⌫=1 cµ⌫ | nv i
0
Ĥ1↵
0
Ĥ2↵
0
Ĥ3↵
..
.
then the determi-
= 0.
(5.45)
(1)
0
Ĥ↵↵
En1
(1)
which has ↵-roots for En1 . If all the roots are different the degeneracy is completely removed by the perturbation. If
some of them are the same the degeneracy is only partially removed. The residual degeneracy may be removed in a higher
order of perturbation. In general the determination of the coefficients cµ⌫ is a lengthy process.
(0)
The determination of the eigenfunctions | nµ i amounts to finding the correct
combination of the
D orthogonal linear
E
original zero-order degenerate eigenfunctions |
to the indices µ, ⌫.
A few special cases are worth noting:
(0)
0
nv i such that the matrix Ĥµ⌫
(0)
nµ
=
Ĥ 0
(0)
n⌫
is diagonal with respect
1. The matrix of Ĥ 0 is already diagonal, as in
(1)
0
Ĥ11
0
(1)
0
En1
0
Ĥ22
(1)
(1)
En1
=0
(0)
(5.46)
(0)
0
0
then En1 = Ĥ11
and En2 = Ĥ22
. Thus the two states | n1 i and | n2 i are not connected in first-order. The
(0)
(0)
0
0
degeneracy is completely removed and the energies are En1 = En + Ĥ11
, and En2 = En + Ĥ22
. This
0
happens when the perturbation Ĥ commutes with the operator
whose
eigenvalues
the
label
µ
represents.
Suppose
h
i
h
i
an Hermitian operator  commutes with Ĥ0 and Ĥ 0 , i.e. Ĥ0 ,  = 0, and Ĥ 0 ,  = 0. Thus Ĥ0 and  possess
(0)
(0)
a complete set of mutual eigenstates. The eigenstates | nµ i, µ = 1, . . . ↵, belonging to eigenvalue En of Ĥ0 are
also eigenfunctions of Â, with
(0)
(0)
Â| nµ
i = anµ | nµ
i.
(5.47)
Then
D
(0)
nµ
h
Ĥ 0 , Â
i
(0)
n⌫
E
=
=
D
(0)
nµ
an⌫
Ĥ 0 Â
a⇤nµ
ÂĤ 0
D
(0)
nµ
(0)
n⌫
Ĥ 0
E
= 0,
E
(0)
= 0.
n⌫
(5.48)
(5.49)
0
0
and Ĥµ⌫
= 0 if an⌫ 6= anµ (note: the eigenvalues are real). Hence if all the ↵-eigenvalues anµ are different Ĥµ⌫
is
already diagonal and the |
(0)
nµ i
are already the correct zero-order states.
73
PHAS3226: Quantum Mechanics
CHAPTER 5. APPROXIMATE METHODS & MANY-BODY SYSTEMS
In the hydrogen atom a degeneracy is associated with the eigenvalues of L̂z in that all 2` + 1 m values have the
same energy. Suppose that
h
i
Ĥ 0 , L̂z = 0,
then choose |
(0)
nµ i
to be eigenfunctions of L̂z i.e. µ ⌘ m, so that
(0)
nm i
L̂z |
Then
D
(0)
nm0
h
Ĥ 0 , L̂z
i
(0)
nm
E
D
=
= mh̄|
(0)
nm i.
(0)
0
(0)
nm0 |Ĥ L̂z | nm
D
= mh̄
= (m
(5.50)
E
(0)
0 (0)
nm0 |Ĥ | nm
0
m0⇤ ) h̄Ĥm
0n.
D
E
(0)
0 (0)
nm0 |L̂z Ĥ | nm
m0⇤ h̄
D
E
(5.51)
(0)
0 (0)
nm0 |Ĥ | nm
E
(5.52)
0
0
But the m values are real and thus if m0 6= m then Ĥm
0 n = 0, i.e. Ĥm0 n is a diagonal matrix.
This example illustrates a general result: a degeneracy due to a symmetry of Ĥ0 is reduced or removed altogether
by a perturbation of a lower symmetry. In this example, Ĥ0 is spherically symmetric but Ĥ 0 possesses only axial
symmetry.
The coefficients c11 , c12 are easily found from
(1)
0
Ĥ11
0
0
En1
(1)
0
Ĥ22
(1)
0
For En1 = Ĥ11
, c12 = 0 so can choose c11 = 1 and |
(0)
(0)
c22 = 1 and | n2 i = | n2 i.
En1
(0)
n1 i
!✓
=|
c11
c12
(0)
n1 i.
◆
= 0.
(5.53)
(1)
0
For En2 = Ĥ22
, c21 = 0 so can choose
0
0
2. The other interesting case occurs for Ĥ11
= Ĥ22
= 0. Then the determinant is
(1)
En1
0
Ĥ21
and
⇣
since
(1)
En1
⌘2
Hence the first-order energy correction is
=0
0
0
0
= Ĥ12
Ĥ21
= Ĥ12
(0)
0 (0)
n1 |Ĥ | n2 i
0
Ĥ12
=h
0
Ĥ12
(1)
En1
=h
(5.54)
2
(0)
0 (0) ⇤
n2 |Ĥ | n1 i
(5.55)
⇣
⌘⇤
0
= Ĥ21
.
(5.56)
(1)
0
En1 = ± Ĥ12
.
The coefficients satisfy
(5.57)
(1)
0
c11 En1 + c12 Ĥ12
=0
so that
0
c11
Ĥ12
= (1)
.
c12
En1
(1)
0
For En1 = + Ĥ12
,
eigenstates |
(0)
n1 i
and
(5.58)
c11
0
c12 = ±1 depending on the sign of Ĥ12 . For
(0)
| n2 iare said to be fully mixed with the correct
|
(0)
n1 i
=
|
(0)
n2 i
=
1 ⇣
p |
2
1 ⇣
p |
2
74
(1)
En1 =
0
Ĥ12
,
c11
c12
= ⌥1. The original
zeroth-order eigenfunctions being
⌘
(0)
(0)
i
±
|
i
,
n1
n2
(0)
n1 i
⌥|
⌘
(0)
n2 i
.
(5.59)
(5.60)
PHAS3226: Quantum Mechanics
5.3
5.3.1
CHAPTER 5. APPROXIMATE METHODS & MANY-BODY SYSTEMS
Applications of perturbation theory
First-order examples
The anharmonic oscillator To calculate the energy eigenvalues of an anharmonic oscillator represented by the Hamiltonian
p̂2
1
Ĥ =
+ kx̂2 + x̂4 .
(5.61)
2m 2
The presence of the anharmonic term x̂4 makes this problem difficult to solve exactly, but if the term x̂4 is small (in a
sense to be clarified below), first-order perturbation theory can be used. The unperturbed Hamiltonian Ĥ0 is the harmonic
oscillator,
p̂2
1
Ĥ0 =
+ kx̂2 ,
(5.62)
2m 2
and the perturbation Hamiltonian is:
Ĥ 0 = x̂4 .
(5.63)
(0)
Consider the change of energy of the ground state. The unperturbed ground-state energy is E0 = 12 h̄!, where !
is the frequency of the harmonic oscillator ! = (k/m)1/2 . According to eqn (5.14), the change of ground-state energy
caused by the term x4 is:
(1)
E0 = h 0 |Ĥ 0 | 0 i ,
(5.64)
where | 0 i is the ground-state eigenvector. This quantity could be evaluated using the operator methods for the harmonic
oscillator but here the normal wavefunction methods are used .
The normalized ground-state wave-function of the harmonic oscillator is
0 (x)
Hence
(1)
E0
=
In order to calculate this integral note that
=
⇣ m! ⌘1/4
⇡h̄
⇣ m! ⌘1/2 Z
1
e
(5.65)
x4 exp( m!x2 /h̄)dx .
(5.66)
1
⇡h̄
Z
exp( m!x2 /2h̄) .
1
↵x2
dx = ⇡ 1/2 ↵
1/2
(5.67)
.
1
Differentiating this with respect to ↵ once gives
Z 1
Z
2
d
e ↵x dx =
d↵ 1
1
x2 e
↵x2
dx =
1
1 1/2
⇡ ↵
2
3/2
(5.68)
and differentiating again
d
d↵
It follows that
and hence that
Z
Z
(1)
E0
1
2
x e
1
1
↵x2
dx =
Z
1
1
x4 exp( m!x2 /h̄)dx =
1
=
x4 e
↵x2
dx =
3 1/2
⇡ ↵
4
3 1/2
⇡ (m!/h̄)
4
5/2
5/2
✓
◆1/2 ✓
◆2
3 ⇣ m! ⌘1/2 ⇡h̄
h̄
3h̄2
=
.
4
⇡h̄
m!
m!
4m2 ! 2
.
(5.69)
(5.70)
(5.71)
Thus the ground-state energy is raised by an amount proportional to , as should be expected.
(0)
(1)
It is useful to consider how the total ground-state energy E0 ' E0 +E0 deviates from the unperturbed ground-state
(0)
energy E0 ;
⇣
⌘
1
3h̄
(0)
(0)
(1)
(0)
(1)
E0 /E0 ' E0 + E0 /E0 = 1 + E0 /( h̄!) = 1 +
.
(5.72)
2
2m2 ! 3
75
PHAS3226: Quantum Mechanics
CHAPTER 5. APPROXIMATE METHODS & MANY-BODY SYSTEMS
The strength of the perturbation can therefore be characterized by the dimensionless quantity g = 2h̄ /m2 ! 3 . The table
(0)
shows how the approximation for E0 /E0 from first-order perturbation theory compares with essentially exact numerical
predictions. This shows that the energy shift is given very accurately for small values of g, but becomes less accurate as g
increases.
g = 2h̄ /m2 ! 3
0.01
0.10
0.2
(0)
E0 /E0 (pert. th.)
1.00750
1.075
1.15
(0)
E0 /E0 (exact)
1.00735
1.065
1.12
Table 5.1: The ground-state energy E0 of the anharmonic oscillator divided by the ground-state energy of the unperturbed
(0)
harmonic oscillator E0 for different values of . The size of is specified by the dimensionless quantity g = 2h̄ /m! 3 .
(0)
Second and third columns give E0 /E0 from first-order perturbation theory and from almost exact numerical solution of
the Schrödinger equation.
Harmonic oscillator As an example to demonstrate the validity of the perturbation approach consider a harmonic
oscillator where the perturbation is itself harmonic. The unperturbed Hamiltonian
Ĥ0 =
has energy eigenvalues
En(0) =
p̂2
1
+ kx̂2
2m 2
✓
n+
1
2
◆
(5.73)
h̄!0 ,
p
where !0 = k/m. Suppose that the force constant k is changed slightly, k ! k 0 such that k 0
The Hamiltonian is now
p̂2
1
1
1
+ kx̂2 + ✏x̂2 = Ĥ0 + ✏x̂2
2m 2
2
2
p̂2
1
2
=
+ (k + ✏) x̂
2m 2
and the new force constant is k + ✏ so the new frequency of oscillation
✓
◆1/2 ✓ ◆1/2 ⇣
⇣
k+✏
k
✏ ⌘1/2
✏ ⌘1/2
!=
=
1+
= !0 1 +
m
m
k
k
Ĥ
and the exact energy levels are
En
En
=
◆
⇣
1
✏ ⌘1/2
h̄!0 1 +
2
k
✓
◆
✓
◆
1
1✏
1 ⇣ ✏ ⌘2
=
n+
h̄!0 1 +
+ ... .
2
2k 8 k
=
✓
k = ✏ with 0 < ✏ ⌧ k.
(5.74)
(5.75)
n+
(5.76)
The same problem will now be solved using time-independent perturbation theory. The perturbation is Ĥ 0 = 12 ✏x̂2 .
(0)
The energy levels are non-degenerate, so the first-order correction to En = n + 12 h̄!0 is
En(1) = h
(0)
0 (0)
n |Ĥ | n i
=
1
✏h
2
(0) 2 (0)
n |x̂ | n i
=
1
✏ x̂2
2
nn
(0)
.
The matrix elements may be evaluated several ways; using the eigenfunctions for n ; or using the operator approach.
The latter is easier. The matrix elements x̂2 nn were calculated earlier when considering the generalized uncertainty
relation for the harmonic oscillator, i.e.
✓
◆
2n + 1
1
h̄
hn|x̂2 |ni =
h̄ = n +
.
(5.77)
2m!0
2 m!
76
PHAS3226: Quantum Mechanics
CHAPTER 5. APPROXIMATE METHODS & MANY-BODY SYSTEMS
For the
correction matrix elements between different states will be needed. In anticipation of this recall that
⇣ second-order
⌘
h̄
x̂ = 2m!0 (â+ + â ) and
x̂|ni =
so
✓
✓
h̄
2m!0
◆
◆1/2
(â+ + â ) |ni =
✓
h̄
2m!0
◆1/2
p
n + 1|n + 1i +
p
n|n
1i
⇣p
⌘
p
p
n + 1hn + 1| + nhn 1|
k + 1|k + 1i + k|k 1i ,
p
✓
◆✓ p
◆
h̄
(n + p
1) (k + 1)hn + 1|k + 1i + p(n + 1) khn + 1|k 1i
=
,
+ n (k + 1)hn 1|k + 1i + nkhn 1|k 1i
2m!0
✓
◆⇣
p
p
p
p
h̄
=
(n + 1) (k + 1) nk + (n + 1) k n,k 2 + n (k + 1) n,k+2 + nk
2m!0
hn|x̂x̂|ki =
h̄
2m!0
hn|x̂2 |ki =
✓
p
h̄
2m!0
The first-order correction is
◆⇣
(2n + 1)
En(1) =
nk
+
1
✏ x̂2
2
p
(n + 1) (n + 2)
nn
=
n,k 2
+
✓
◆
1
1
h̄
✏ n+
.
2
2 m!0
p
n (n
1)
n,k+2
⌘
.
nk
⌘
,
(5.78)
(5.79)
This is the same as the first term in the expansion equ(5.76) since
✓
◆
✓
◆
✓
◆
1
1✏
1
1 ✏
1
h̄✏
n+
h̄!0
= n+
h̄!0
= n+
.
2
2k
2
2 m!02
2 2m!0
(5.80)
The second-order correction is obtained from
En(2)
=
X
2
⌘=
X
2
hk| 12 ✏x̂2 |ni
h̄!0
k+
1
(0)
(0)
n + 12
En
Ek
2 h̄!0
k6=n
2
X ✓ 1 ◆2 hk|x̂2 |ni 2
1 ✏2 X hk|x̂2 |ni
✏
=
.
2
(n k) h̄!0
4 h̄!0
(n k)
k6=n
=
⇣
hk|Ĥ 0 |ni
k6=n
(5.81)
k6=n
But from equ(5.78) the only non-zero matrix elements are for k = n ± 2; k = n already being excluded from the
summation. Hence
"
#
2
2
2
2
2
hn
+
2|x̂
|ni
hn
2|x̂
|ni
1
✏
En(2) =
+
,
(5.82)
4 h̄!0 (n (n + 2))
(n (n 2))
2
3
2
2
✓
◆2 p(n + 1) (n + 2)
✓
◆2 pn (n 1)
2
1 ✏ 6
h̄
h̄
7
=
+
4
5,
4 h̄!0
2m!0
2
2m!0
2
✓
◆2 
1 ✏2
h̄
|(n + 1) (n + 2)| |n (n
+
4 h̄!0 2m!0
2
2
✓
◆
✓
◆✓
2
2
h̄✏
h̄✏
=
( 4n 2) =
n+
32m2 !03
8m2 !03
=
En(2)
1)|
1
2
◆
.
,
(5.83)
The perturbation result is in agreement with the exact result equ(5.76), since from the expansion the second-order correction is
✓
◆
✓
◆
✓
◆
1
1 ⇣ ✏ ⌘2
1
1
h̄✏2
n+
h̄!0
=
n+
.
2
8 k
8
2 m2 !03
77
PHAS3226: Quantum Mechanics
CHAPTER 5. APPROXIMATE METHODS & MANY-BODY SYSTEMS
Ground state of helium atom The neutral He atom consists of two electrons bound to a nucleus having charge Z|e|,
with Z = 2. Because of the electrostatic repulsion between the electrons, the Schrödinger equation for this system cannot
be solved exactly. One way of tackling it is to treat the inter-electron repulsion by perturbation theory.
The Hamiltonian for the two-electron system is:
✓
◆
h̄2 2
h̄2 2
Ze2
1
1
e2 1
Ĥ (r1 , r2 ) =
r1
r2
+
+
.
(5.84)
2m
2m
4⇡✏0 r1
r2
4⇡✏0 r12
Here, r1 and r2 are the positions of the two electrons, and r12 = |r1 r2 | is the distance between the electrons; r21 and
r22 are the Laplacian operators for the two electron positions. Since the electrons have spin- 12 the full space-spin wavefunction should describe both the spin state and the space state. The electrons are also fermions so the total wave-function
must be antisymmetric with respect to interchange of electrons.
p However, the ground state is a spin singlet so that the spin
part of the two-electron wave-function is (↵1 2
↵
)/
2. This is already antisymmetric, so that the spatial part of
1 2
the wave-function (r1 , r2 ) is symmetric. Since the Hamiltonian does not depend on spin, Schrödinger’s equation is:
✓
✓
◆
◆
h̄2 2
h̄2 2
Ze2
1
1
e2 1
r1
r2
+
+
(r1 , r2 ) = E (r1 , r2 ) .
(5.85)
2m
2m
4⇡✏0 r1
r2
4⇡✏0 r12
The inter-electron repulsion e2 /(4⇡✏0 r12 ) makes it impossible to solve Schrödinger’s equation exactly, and perturbation theory offers one way forward. The unperturbed Hamiltonian Ĥ0 omits the repulsion
✓
◆
h̄2 2
h̄2 2
Ze2
1
1
Ĥ0 =
r
r
+
,
(5.86)
2m 1 2m 2 4⇡✏0 r1
r2
and is a separable Hamiltonian with
Ĥ0 (r1 , r2 ) = Ĥ0 (r1 ) + Ĥ0 (r1 ) .
In atomic units
1 2
r
2
Ĥ0 (r) =
(5.87)
Z
.
r
(5.88)
The unperturbed Schrödinger equation,
Ĥ0 (r1 , r2 )
(0)
0
(r1 , r2 ) = E0
(0)
0
(r1 , r2 ) ,
(5.89)
is easy to solve, because it represents two independent electrons. The ground state wavefunction of Ĥ0 is just the product
of the ground-state wavefunctions of the individual electrons,
(0)
0 (r1 , r2 )
=
100 (r1 ) 100 (r2 )
(5.90)
,
where 100 (r) is the ground-state wavefunction for a single electron bound to a nucleus of charge Z|e|. This can be
2
obtained from the usual hydrogenic wave-function by replacing e2 by (Ze) and the energy for each electron is
En(0) =
with
100 (r)
Z2
,
2n2
= Ae
Zr/a0
(5.91)
(5.92)
,
(0)
and with a0 = 4⇡✏0 h̄2 /me2 ' 0.529 Å the usual Bohr radius. The ground-state energy E0 of the He atom in this approximation is twice the ground-state energy of a single electron bound to a nucleus of charge Z|e|, which is Z 2 /2 a.u, so
(0)
that E0 = Z 2 = 4 a.u. ("Hartrees"). Not surprisingly, this is a poor approximation compared with the experimental
value of the He ground-state energy, which is 2.905 a.u. .
The electron-electron mutual repulsion is taken as the perturbation part
Ĥ 0 =
e2
1
⌘
,
4⇡✏0 r12
r12
78
(5.93)
PHAS3226: Quantum Mechanics
CHAPTER 5. APPROXIMATE METHODS & MANY-BODY SYSTEMS
so that by first-order perturbation theory, the total energy is shifted by:
⌧
⌧
e2
(1)
(0)
(0)
E0 =
⌘
0
4⇡✏0 r12 0
This can be evaluated exactly, but it is long and tedious as
1
r12
=
(1)
E0
1
|r1 r2 |
(0)
0
1
r12
(0)
0
.
(5.94)
depends on the angular coordinates. The result is
= 1.25 a.u., so that the corrected ground-state energy is 4+1.25 = 2.75 a.u. This is still not in perfect agreement
with the experimental value of 2.905 a.u, but it is much better than the value obtained by omitting the repulsion between
electrons.
5.3.2
Second-order example
Stark effect in hydrogen To illustrate the application of perturbation theory to a real physical problem consider the
effect of a time-independent, spatially uniform external electric field on the energy levels of an atom - the so-called
Stark effect. The Stark effect is the shift of the optical adsorption and emission frequencies of atoms induced by the
application of an external electric field. This shift arises from the shift of the atomic energy levels. To keep the discussion
as manageable as possible it will be restricted to hydrogen-like atoms, i.e.to one electron atoms.
The unperturbed Hamiltonian Ĥ0 is, as usual,
h̄2 2
r
2m
Ĥ0 =
e2
,
4⇡✏0 r
(5.95)
whose zero-order eigenfunctions are the hydrogenic wavefunctions. It is useful to recall some of their properties:
(1) The eigenfunctions are n`m (r, ✓, ) = Rn` (r) Y`m (✓, ), with Rn` (r) the radial wavefunction and Y`m (✓, )
the spherical harmonic,
`
(2) the parity of n`m (r, ✓, ) is ( 1) ,
R 1 R ⇡ R 2⇡ ⇤
2
(3) the eigenfunctions are orthonormal, 0 0 0
n0 `0 m0 n`m r dr sin ✓d✓d = n0 n `0 ` m0 m
2
2
V (r) = eEz =
p·E
2
2
m Ze 1
mc
Z↵
(4) the unperturbed energy levels are En = 2h̄
where n = 1, 2, 3, . . . and ↵ = 4⇡✏e0 h̄c '
2 4⇡✏ n2 =
2
n
0
1
137 is the fine structure constant
(5) for a given `, m has 2` + 1 values, `, ` + 1, . . . ` 1, `
(6) for a given n, 0  `  n 1.
Pn 1
(7) each level is (without considering spin) `=0 (2` + 1) = n2 degenerate; including spin doubles this value.
The applied electric field E is uniform over the atomic dimensions and assumed to be directed along the the positive
z-axis, so that the additional potential energy of the electron is
(5.96)
where p = ez is the electric dipole moment of the electron. This interaction energy must be added to the Coulomb
potential energy of the electron in the electric field of the nucleus.
(1)
Consider first the ground state (n = 1, ` = 0, m = 0). The first-order shift E100 is
Z
(1)
?
E100 = h 100 | V | 100 i = dr 100
(r) V (r) 100 (r) .
The ground-state wave-function
100 (r)
is:
100 (r)
=
1
3/2
⇡ 1/2 a0
so that
(1)
E100 = eE
2
(5.97)
Z
|
e
r/a0
100
1
⌘p e
⇡
2
(r)| zdr .
r
,
(5.98)
(5.99)
This is exactly zero, since | 100 (r)| is an even function under parity (r ! r) whereas the perturbation is an odd
function as z ! z and hence overall the integrand is odd. {Positive values of the integrand cancel negative values. The
79
PHAS3226: Quantum Mechanics
CHAPTER 5. APPROXIMATE METHODS & MANY-BODY SYSTEMS
reason for this is that the value of the integrand at any point r is exactly equal and opposite to its value at r}. This is
the result of symmetry; the charge distribution in the unperturbed atom is spherically symmetric, whereas the perturbation
Hamiltonian V changes sign when r ! r. Thus the first-order shift is zero. Hence for the ground state there is no
energy shift linear in the electric field strength E. Classically a system with electric dipole moment p will experience an
energy shift p · E hence it is concluded that the atom has, in the ground state, no permanent dipole moment.
Since the first-order shift is zero a second-order calculation must be done. For the ground state the second-order
correction is given by
(0)
(2)
E100
1 n
X
X1 m=`
X h
⇣
= e2 E 2
ṅ>1 `=0 m= `
(0)
(0)
(0)
2
1 n
X
X1 m=`
X h n`m |Ĥ 0 | 100 i
⇣
⌘ ,
=
(0)
(0)
E100 En`m
ṅ>1 `=0 m= `
(2)
(0)
(0)
n`m |z| 100 i
(0)
(0)
E100
En`m
2
⌘.
Since E100 En`m < 0, then En < 0 i.e. the ground state energy is lowered by the interaction. To evaluate the matrix
elements put z = r cos ✓ then
Z Z
(0)
(0)
⇤
h n`m |z| 100 i =
Rn` (r) Y`m
(✓, ) r cos ✓R10 Y00 r2 dr d⌦.
(5.100)
q
p
But Y00 = 1/ 4⇡ and cos ✓ = 4⇡
3 Y10 so that
h
(0)
(0)
n`m |z| 100 i
=
Z
Z
Rn` (r) R10 r3 dr
1
⇤
Y`m
(✓, ) p Y10 d⌦.
3
From the orthonormality of the spherical harmonics this integral vanishes unless ` = 1 and m = 0. Thus only the states
(0)
| n10 i contribute to the integral and
(2)
E100
1 h
X
⇣
= e2 E 2
ṅ=2
(0)
Explicit evaluation of h n10 |z|
on n. Thus one can write
(0)
100 i
2
(2)
E100
=
2 2
e E
(0)
E100
(0)
En10
2
(5.101)
⌘.
is very difficult but clearly it is proportional to some length squared and depends
h
with a0 the Bohr radius, {Actually h
(0)
(0)
n10 |z| 100 i
(0)
(0)
n10 |z| 100 i
(0)
(0)
n10 |z| 100 i
1
⇣ a ⌘2 X
2
ṅ=2
=
⇣ a ⌘2
0
Z
8 7
2n
1 2 n (n 1)
3
(n+1)2n+5
f (n)
0
Z
=
2
mc2
2
2
(Z↵)
1
1
n2
f (n) ,
5
(5.102)
a20 !} so that
=
1
2e2 E 2 a20 X n2 f (n)
.
↵2 Z 4 mc2
(n2 1)
(5.103)
ṅ=2
(2)
Hence E100 / e2 E 2 as is to be expected from electromagnetic theory as it is due to an induced dipole moment p = e E
which has an energy 12 pE / E 2 .
5.3.3
Degenerate example - first order
Stark effect in hydrogen - excited states The first excited state has n = 2. If fine-structure (i.e. spin-orbit) effects are
neglected this level is 4-fold degenerate. The degenerate unperturbed eigenfunctions are for n = 2 , ` = 0, | 200 i, and
n = 2, ` = 1, m = 1, 0, 1, | 211 i , | 210 i , | 21 1 i . Consequently to find even the first-order correction for the excited
states, degenerate perturbation theory needs to be used. At first sight it appears that one needs to calculate the 16 matrix
0
elements `0 m0 Ĥ`m
= h 2`0 m0 |eEz| 2`m i (in a slight deviation from the usual notation to emphasise the states involved)
in the table below
80
PHAS3226: Quantum Mechanics
CHAPTER 5. APPROXIMATE METHODS & MANY-BODY SYSTEMS
0, 0
`0 m0 \ `, m
0, 0
1, 0
1, 1
1, 1
(1)
0
00 Ĥ00
E2µ
0
10 Ĥ00
0
11 Ĥ00
0
1 1 Ĥ00
1, 0
1, 1
1, 1
0
00 Ĥ10
(1)
0
E2µ
10 Ĥ10
0
11 Ĥ10
0
1 1 Ĥ10
0
00 Ĥ11
0
10 Ĥ11
(1)
0
E2µ
11 Ĥ11
0
1 1 Ĥ11
0
00 Ĥ1 1
0
10 Ĥ1 1
0
11 Ĥ1 1
(1)
0
E2µ
1 1 Ĥ1 1
and evaluate a 4⇥4 determinant. However an examination of the form of these elements can result in a dramatic reduction
in the number (and effort) of them to evaluate explicitly.
`
1. The perturbation H 0 = eEz has odd parity and the eigenstates | 2`m i have parity ( 1) . Thus
D
E
0
`0 = `.
2`0 m0 Ĥ
2`m = 0 if
(5.104)
0
0
0
Thus all the diagonal elements of Ĥ`00 m0 `m vanish, i.e. 00 Ĥ00
= 10 Ĥ10
= 11 Ĥ11
0 = 1 1 Ĥ10 1 = 0. In addition
0
0
0
0
the elements 10 Ĥ11
, 10 Ĥ10 1 , 11 Ĥ10
, 11 Ĥ10 1 , 1 1 Ĥ10
, and 1 1 Ĥ11
are also zero - resulting in a reduction of 10
elements that do no need to be evaluated explicitly, leaving
0, 0
(1)
E2µ
0
10 Ĥ00
0
11 Ĥ00
0
1 1 Ĥ00
`0 m0 \ `, m
0, 0
1, 0
1, 1
1, 1
1, 0
0
00 Ĥ10
(1)
E2µ
0
0
1, 1
0
00 Ĥ11
0
(1)
E2µ
0
1, 1
0
00 Ĥ1 1
0
.
0
(1)
E2µ
h
i
2. The perturbation H 0 = eEz commutes with L̂z whose eigenvalues result in the degeneracy index m, i.e. L̂z , Ĥ 0 =
h
i
eE L̂z , z = 0. Hence
D
0
=
0
= h̄ (m0
Thus if m0 6= m then
D
2`0 m0
2`0 m0
Ĥ 0
0
0
0
0
11 Ĥ1 1 , 1 1 Ĥ00 , 1 1 Ĥ10 , 1 1 Ĥ11
h
i
E D
h
L̂z , Ĥ 0
L̂z Ĥ 0
2`m =
2`0 m0
D
E
m) 2`0 m0 Ĥ 0 2`m .
2`m
E
Ĥ 0 L̂z
0
= 0. Hence elements 00 Ĥ11
, 00 Ĥ10
i
2`m
E
,
(5.105)
0
0
0
0
1 , 10 Ĥ11 , 10 Ĥ1 1 , 11 Ĥ00 , 11 Ĥ10 ,
are all zero, giving
`0 m0 \ `, m
0, 0
1, 0
1, 1
1, 1
0, 0
(1)
E2µ
0
10 Ĥ00
0
0
0
Hence only two non-zero matrix elements exist, 00 Ĥ10
0⇤
=10 Ĥ00
1, 0
1, 1
1, 1
0
0
0
00 Ĥ10
(1)
E2µ
0
0
.
(1)
0
E2µ
0
(1)
0
0
E2µ
D
E
D
0
0
=
200 Ĥ
210 and 10 Ĥ00 =
210
Ĥ 0
200
E
. But
as Ĥ is Hermitian then
so only one matrix element need be calculated! {As a general policy it is
always advisable to examine the symmetries and origin of the degeneracy before using brute force to evaluate all
the matrix elements}
0
0
00 Ĥ10
The 4 ⇥ 4 matrix
0
B
B
B
@
0
00 Ĥ00
(1)
E2µ
0
10 Ĥ00
0
11 Ĥ00
0
1 1 Ĥ00
0
00 Ĥ10
(1)
0
E2µ
10 Ĥ10
0
11 Ĥ10
0
1 1 Ĥ10
0
00 Ĥ11
0
10 Ĥ11
(1)
0
E2µ
11 Ĥ11
0
1 1 Ĥ11
81
0
00 Ĥ1 1
0
10 Ĥ1 1
0
11 Ĥ1 1
(1)
0
E2µ
1 1 Ĥ1 1
10
1
c1
CB
C B c2 C
C=0
C@
c3 A
A
c4
(5.106)
PHAS3226: Quantum Mechanics
reduces to
CHAPTER 5. APPROXIMATE METHODS & MANY-BODY SYSTEMS
0
and only the secular equation
(1)
E2µ
0
10 Ĥ00
0
0
B
B
B
@
0
0
(1)
E2µ
0
0
00 Ĥ10
(1)
E2µ
0
0
(1)
0
00 Ĥ10
(1)
E2µ
E2µ
0
10 Ĥ00
(1)
needs to be solved. This has two roots E2µ = ±
0
00 Ĥ10
=h
200
|eEz|
210 i
= eE h
200
10
1
c1
CB
C B c2 C
C=0
C@
c3 A
A
c4
(5.107)
=0
(5.108)
. Explicit evaluation
0
00 Ĥ10
|z|
0
0
0
(1)
E2µ
210 i
= eE
Z Z
⇤
⇤
R20
Y00
(r cos ✓) R21 Y10 r2 dr d⌦.
(5.109)
0
of 00 Ĥ10
is now needed. Using expressions for the hydrogenic wavefunctions,
R20 (r) = 2
✓
1
p
3
R21 (r) =
Z
2a0
✓
◆3/2 ✓
Z
2a0
Zr
2a0
1
◆3/2
Zr
e
a0
◆
e
Zr/2a0
(5.110)
,
Zr/2a0
with Z = 1, and z = r cos ✓
0
00 Ĥ10
= eE
Z
2
✓
Z
2a0
◆3/2 ✓
Zr
2a0
1
◆
e
1
r p
3
Zr/2a0 3
✓
Z
2a0
◆3/2
Zr
e
a0
Zr/2a0
dr
Z
1/2
1
Evaluating the angular integral first (always a wise move!), and noting that Y00 = 4⇡
r
Z
Z
1
4⇡
1
⇤
p
Y00 cos ✓Y10 d⌦ =
Y10 Y10 d⌦ = p
3
4⇡
3
⇤
Y00
(cos ✓) Y10 d⌦.
and Y10 =
3 1/2
4⇡
(5.111)
cos ✓,
(5.112)
using the orthonormality of the spherical harmonics. Hence
=
0
00 Ĥ10
=
2
eE
3
2
eE
3
✓
✓
Z
2a0
Z
2a0
setting ⇢ = Zr/a0 . Then
0
00 Ĥ10
=
=
using the standard integral,
R1
0
xn e
ax
(1)
E2µ
◆3 Z
1
r3
0
◆3 ⇣
a0 ⌘4
Z
✓
Z
Zr
a0
1
0
◆✓
1
⇣
⇢4 1
◆
Zr
e Zr/a0 dr
2a0
⇢⌘ ⇢
e d⇢
2
Z
⇣ a ⌘ ⇢Z 1
1
1 1 5
0
eE
⇢4 e ⇢ d⇢
⇢ e
12
Z
2 0
0
⇣a ⌘ ⇢
1
1
a0
0
eE
4!
5! = 3eE ,
12
Z
2
Z
⇢
d⇢
(5.113)
dx = n!/an+1 . Hence the first-order energy correction is
= ±
= 0
0
00 Ĥ10
= ±3eE
for µ = 3, 4.
a0
,
Z
for µ = 1, 2,
(5.114)
(5.115)
Thus the energy shift is proportional to E, i.e. a linear Stark effect. The linear Stark effect is only non-zero because the
n = 2 state is degenerate.
82
PHAS3226: Quantum Mechanics
c,
CHAPTER 5. APPROXIMATE METHODS & MANY-BODY SYSTEMS
The correct linear combination of the unperturbed eigenfunctions |
(1)
0
00 Ĥ10
(1)
E2µ
E2µ
0
10 Ĥ00
so
!✓
2`m i
◆
c1
c2
can be found by solving for the coefficients
=0
0
c1
00 Ĥ10
=
(1)
c2
E2µ
0
Since 00 Ĥ10
=
(1)
3eE aZ0 then for E21 = +
|
(1)
For E22 =
0
00 Ĥ10
=
3eE aZ0 ,
|
(1)
(0)
21 i
= c1 |
0
00 Ĥ10
200 i
= 3eE aZ0 ,
c1
c2
(5.116)
=
1 and
+ c2 |
210 i
1
= p (|
2
200 i
|
210 i) .
+ c2 |
210 i
1
= p (|
2
200 i
+|
210 i) .
= +1 and
c1
c2
(0)
22 i
= c1 |
200 i
(0)
(1)
(0)
For completeness, E23 = 0 and | 23 i = | 211 i and E24 = 0 and | 24 i = | 21 1 i.
(0)
(0)
The states in the presence of the electric field are no longer eigenstates of L̂2 or parity as | 21 i and | 22 i mix these
states and so neither ` nor parity are "good" quantum numbers. The magnetic quantum number m is still a good quantum
number (as Ĥ 0 commutes with L̂z ) and the system is invariant under rotations about the z-axis. Physically the external
electric field defines a preferred direction in space so the system cannot be invariant under under arbitrary rotations (i.e.
L̂2 ) but still invariant for rotations about this preferred direction (i.e. L̂z ).
The degeneracy of the states | 211 i and | 21 1 i is not removed in first-order. The 4-fold degenerate n = 2 level splits
symmetrically into three sub-levels.
This means that the hydrogen atom (in its unperturbed first-excited state) behaves as if it had a permanent electric
dipole moment of magnitude 3ea0 which can be orientated in three different ways,
1. antiparallel to electric field E for |
W = µ · E.)
2. parallel to electric field E for |
(0)
22 i
(0)
21 i
=
=
p1
2
(|
p1
2
(|
200 i
200 i
+|
|
(Note energy in electric field E of a dipole µ is
210 i)
3. two states with zero component along electric field E for |
5.4
210 i)
(0)
23 i
=|
211 i
and |
(0)
24 i
=|
21 1 i.
Variational method
There are many stationary state problems which cannot be solved exactly and for which the perturbation approach is unsatisfactory because the first-order is not sufficiently accurate and higher orders involve enormous (complex) calculations.
Basically the perturbation expansion converges too slowly. The variational method - also known as the Rayleigh-Ritz
method - does not presuppose a knowledge of the simpler problem and is, therefore, very versatile. The method is particularly suitable for calculating the ground state energies of a system.
The quantum ground state of a system has a special importance. For systems of electrons, in particular, one is often
interested only in the ground state. For example, when quantum mechanics is used to calculate the energies of chemical
reactions, or the properties of materials, excited states are usually irrelevant. Because of this, methods for calculating the
ground-state energy of quantum systems are important. In developing such methods, the variational principle of quantum
mechanics usually plays a key role. This principle states that
“The expectation value of the Hamiltonian, evaluated in an arbitrary state, is always greater than or equal
to the ground-state energy.”
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PHAS3226: Quantum Mechanics
CHAPTER 5. APPROXIMATE METHODS & MANY-BODY SYSTEMS
This principle forms the basis for practical approximate methods of calculating the ground-state energy of a complex
system. The idea is to guess an approximate ground-state wave-function, called the “trial wave-function”, which is
used to calculate the expectation value of the Hamiltonian. To improve the estimate of the ground-state energy, the
trial wave-function contains variable parameters. The expectation value of the Hamiltonian is minimized with respect to
these parameters, and the minimum energy provides the best estimate of the true ground-state energy, within the given
parameterized ‘family’ of wavefunctions.
5.4.1
Proof of the variational principle
The Hamiltonian Ĥ possesses a complete set of orthonormal eigenstates { n } , 0 , 1 , 2 , . . . , h
corresponding energies E0 , E1 , E2 , . . . with the ground state energy E0  E1  E2 . . . .and
ni
Ĥ|
= En |
n| mi
=
nm
and
(5.117)
n i.
Let | i be an approximate wavefunction which satisfies the correct boundary conditions. Expand | i in terms of the
complete set { n } as
X
| i=
cn | n i
(5.118)
n
then the expectation value of the energy in the state | i is
E [ ] = hEi =
=
hEi
and
hEi
E0 =
=
P
h
i
Ĥ
,
h | i
P P ⇤
c c h |Ĥ|
P n P m ⇤m n m
n
m cm cn En h m |
P
2
n |cn | En
P
2 .
n |cn |
2
|cn | En
P
2
n |cn |
E0 =
n
P
n
ni
ni
,
2
|cn | (En E0 )
.
P
2
n |cn |
(5.119)
Since every term on the R.H.S. of equ(5.119) is positive then it follows that
hEi
h
E0 ,
Ĥ
h | i
i
E0 .
(5.120)
and gives an upper bound on the ground state energy. Equality will occur if the trial function | i should be the same as
h |Ĥ | i
the correct ground state eigenfunction | 0 i in which case |c0 | = 1 and cn = 0 for all n 6= 0. If h | i = E0 but cn 6= 0
for more than one n then the ground state is degenerate.
In practice a trial function is chosen which depends on some parameters ↵1 ↵2 , . . . ↵s , i.e. | T i = | T (↵1 , ↵2 , . . . ↵s )i
and one calculates
h T Ĥ T i
hEi =
= E (↵1 , ↵2 , . . . ↵s ) .
(5.121)
h T| Ti
This energy is minimized with respect to the variational parameters ↵1 , ↵2 , . . . ↵s by solving the equations
@E (↵1 , ↵2 , . . . ↵s )
= 0,
@↵i
i = 1, 2, . . . , s.
(5.122)
The resulting minimum value of E (↵1 , ↵2 , . . . ↵s ) represents the best estimate of the ground state energy with that
particular trial function.
84
PHAS3226: Quantum Mechanics
CHAPTER 5. APPROXIMATE METHODS & MANY-BODY SYSTEMS
Clearly the success of the variational method depends on choosing a trial function that incorporates the correct features
of the ground state. Symmetry and other physical properties of the system are useful guides. The use of powerful computers allows trial functions of great complexity and hence great flexibility leading to very accurate results. One should
bear in mind that the variational method optimizes the energy. The trial function is not necessarily a good approximation
to the true wavefunction and may give poor results when used in calculations of quantities other than energy.
If | i is almost the true eigenfunction for the ground state | 0 i but with a bit of | 1 i mixed in so that | i = | 0 i +
a1 | 1 i then the approximation to E0 is as good as
!
2
2
E0 + |a| E1
1 + |a| E1 /E0
hEi
=
= E0
(5.123)
2
2
1 + |a|
1 + |a|
hEi
E0
2
= |a|
(E1
E0 )
(5.124)
.
2
1 + |a|
and the error E is quadratic in the coefficients. This means that accurate values of the ground-state energy can often be
obtained with rather simple guesses for the approximate ground-state wavefunction.
5.4.2
Excited States
The variational method can be used to obtain an upper bound for the energy of an excited state provided the trial function
is made orthogonal to all the energy eigenfunctions corresponding to states having an energy lower than the energy level
being considered. This can be a difficult in function space but is autoamtically satisfied in matrix formaulation which are
widely used for actual variational calculations.
5.4.3
Variational examples
Ground state of hydrogen - actually an unrealistic example as it can be solved exactly! However it is useful to see
how to choose a trial function and perform the integrals and the variation.
The ground state is an s-state with ` = 0. Thus | i depends only on r with | (0)i 6= 0. At large distances a bound
state wavefunction must vanish. Thus take the trial function as (r) = Ce ↵r . Normalization requires
Z 1
h | i = 4⇡C 2
e 2↵r r2 dr = 1.
(5.125)
0
Using the standard integral
R1
0
rn e
ar
dr = n!/an+1 ,
1 = 4⇡C 2
and
C=
and
(r) =
The Hamiltonian is Ĥ =
h̄2
2
2m r
h |
e2
4⇡✏0 r .
✓
✓
2!
↵3
⇡
↵3
⇡
3
(2↵)
◆1/2
◆1/2
e
↵r
(5.126)
.
The expectation value of the potential energy is
e2
| i =
4⇡✏0 r
=
=
✓
◆ 2
Z 1
↵3
e
4⇡
e
⇡ 4⇡✏0
0
✓ 3◆ 2
↵
e
1
4⇡
,
⇡ 4⇡✏0 (2↵)2
e2
↵.
4⇡✏0
85
2↵r
rdr,
(5.127)
(5.128)
PHAS3226: Quantum Mechanics
CHAPTER 5. APPROXIMATE METHODS & MANY-BODY SYSTEMS
The following "trick" frequently simplifies the kinetic energy calculation. For any function
Z
Z
Z
⇤
r· ( ⇤ r ) d⌧ =
r2 d⌧ + r ⇤ · (r ) d⌧,
(r) a vector identity is
V
and using Gauss’ theorem on the L.H.S.
Z
Z
r· ( ⇤ r ) d⌧ =
(
V
S
⇤
r ) · n̂dS =
Z
⇤
d⌧ +
r2
Z
2
|(r )| d⌧.
As ! 0 sufficiently rapidly as r ! 1 the surface integral vanishes (certainly the case for the exponential function) so
that
Z
Z
⇤
2
d⌧ =
r2
|(r )| d⌧.
(5.129)
Using this result on the expectation value of the kinetic energy term
h |
h̄2 2
h̄2
r | i=
2m
2m
Z
2
|(r )| d⌧ =
But for the trial function
2
|(r )| =
Hence
h |
Hence the expectation value of the energy is
E (↵) =
and
↵=
2
|(r )| 4⇡r2 dr.
2
= ↵2 | | .
Z
h̄2 ↵2
2m
@E (↵)
h̄2 ↵
=
@↵
m
and
Z
2
@
@r
h̄2 2
h̄2 ↵2
r | i=
2m
2m
h̄2
2m
2
| | 4⇡r2 dr =
e2
↵
4⇡✏0
h̄2 ↵2
2m
(5.130)
(5.131)
e2
4⇡✏0
e2 m
1
= ,
4⇡✏0 h̄
a0
where a0 is the Bohr radius. Consequently the minimum energy is
✓ ◆
1
1 e2
E
=
=
a0
2 4⇡✏0 a0
(5.132)
1
a.u.
2
(5.133)
Harmonic oscillator As a simple illustration of how the variational principle works consider the harmonic oscillator,
with the usual Hamiltonian:
h̄2 d2
1
Ĥ =
+ kx2 .
(5.134)
2m dx2
2
For this system the exact ground-state energy E0 = 12 h̄! is known, where ! is the frequency ! = (k/m)1/2 ; also the
1
exact ground-state wavefunction which is 0 (x) = (m!/⇡h̄) 4 exp( m!x2 /2h̄). However, to show that the variational
principle really works suppose that this wave function is not known and a trial function is constructed. Because the
potential is symmetric about x = 0 and the wavefunction must satisfy (x) ! 0 as x ! 1 then take as a trial
wavefunction as
2
1/4
e ↵x ,
(5.135)
T (x) = (2↵/⇡)
with ↵ (positive) variational parameter. To demonstrate that the expectation value of Ĥ evaluated with this wavefunction
is always greater than or equal to the exact ground-state energy.
86
PHAS3226: Quantum Mechanics
T (x)
First, verify that the
For any positive number ,
Z 1
e
CHAPTER 5. APPROXIMATE METHODS & MANY-BODY SYSTEMS
as given by equ (5.135) is already correctly normalized,
✓ ◆1/2 Z 1
Z 1
2
2↵
| T (x)|2 dx =
e 2↵x dx .
⇡
1
1
x2
dx =
1
In the present case,
= 2↵, so that
R1
✓ ◆1/2
⇡
1
;
Z
and
1
x2 e
x2
dx =
1
1
2
(5.136)
✓ ◆1/2
⇡
.
exp( 2↵x2 ) dx = (⇡/2↵)1/2 . This confirms that
Z 1
| T (x)|2 dx = 1 .
(5.137)
(5.138)
1
Now to evaluate the expectation value of H, work out separately the expectation values of the kinetic energy T and
the potential energy V ,
◆
✓
Z 1
h̄2 d2
?
hT i = h T |T̂ | T i =
(5.139)
T (x)
T (x) dx
2m dx2
1
In general by integrating by parts
Z 1
? 2
T (x) r
T (x)dx
= [
T (x)
=
Z
1
Thus
?
r
|r
1
T (x)] 1
2
T (x)|
Z
⇤
T (x)r T (x)dx
dx
"✓ ◆
Z 1(
1/4
h̄2
d
2↵
hT i = h T |T̂ | T i =
e
2m 1 dx
⇡
✓ ◆1/2 Z 11/4
2
h̄2 2↵
=
4↵2 x2 e ↵x dx,
2m ⇡
1
◆1/2
2 ✓
h̄
2↵
1 ⇣ ⇡ ⌘1/2
=
4↵2
,
2m ⇡
2 (2↵) 2↵
hT i =
r
(5.140)
2
↵x
#)2
dx,
(5.141)
(5.142)
h̄2 ↵
.
2m
(5.143)
The expectation value of the potential energy is
✓ ◆1/2 Z 1
1
2↵
hV i = h T |V̂ | T i = k
x2 e
2
⇡
1
✓ ◆1/2
1
2↵
1 ⇣ ⇡ ⌘1/2
=
k
,
2
⇡
2 (2↵) 2↵
k
hV i =
.
8↵
2↵x2
dx,
(5.144)
(5.145)
(5.146)
Adding hT i and hV i the expectation value of the Hamiltonian is
hHi =
h̄2 ↵
k
+
.
2m
8↵
(5.147)
Now note that as ↵ ! 0, hHi ! k/8↵ ! +1, and as ↵ ! 1, hHi ! h̄2 ↵/2m ! 1. Clearly, hHi must have a
minimum for some value of ↵. To find this minimum, differentiate with respect to ↵,
d
h̄2
hHi =
d↵
2m
87
k
= 0,
8↵2
(5.148)
PHAS3226: Quantum Mechanics
CHAPTER 5. APPROXIMATE METHODS & MANY-BODY SYSTEMS
so that
mk
m2 ! 2
.
(5.149)
2 =
4h̄
4h̄2
This means that for ↵ > 0 there is only a single stationary point, which must be a minimum. The value of ↵ that gives
this minimum is
↵min = m!/2h̄.
(5.150)
↵2 =
Hence the trial wave-function that gives the minimum value of hHi is:
min (x)
= (m!/⇡h̄)1/4 exp( m!x2 /2h̄) .
(5.151)
This is exactly the correct ground-state wave-function.
From equ (5.147), the value of hHi given by min (x) is:
h̄2
m! k
2h̄
1
1
1
·
+ ·
= h̄! + h̄! = h̄!,
(5.152)
2m
2h̄
8 m!
4
4
2
which (not surprisingly) is exactly the correct result.
In this case, the variational principle gives exactly the correct result, because the trial wavefunction used is capable of
giving the correct result. If some other kind of trial wavefunction had been taken, this would not have happened.
hHimin =
Ground state of helium atom The Schrödinger equation for the helium atom is impossible to solve exactly because of
the electrostatic repulsion between the two electrons. The Hamiltonian is:
✓
◆
h̄2 2
h̄2 2
Ze2
1
1
e2 1
Ĥ =
r1
r2
+
+
,
(5.153)
2m
2m
4⇡✏0 r1
r2
4⇡✏0 r12
1 2 1 2 Z
Z
1
=
r1
r2
+
.
(5.154)
2
2
r1
r2
r12
Here, Z is the charge on the nucleus in units of e, which for the helium atom is Z = 2.
This was treated earlier as an example of perturbation theory, where Ĥ was separated into an unperturbed part Ĥ0 ,
in which electron-electron repulsion is ignored, and Ĥ 0 = e2 /4⇡✏0 r12 , which is the electron-electron repulsion. A
reasonably good estimate of the ground-state energy can be obtained by using first-order perturbation theory to calculate
the energy shift due to Ĥ 0 . The zero-th order wavefunctions used in the perturbation calculation assume that each electron
moves in the Coulomb field of the nucleus, with Z = 2. In fact the effect of the mutual repulsion of the electrons is to
reduce the effect of the nuclear filed experienced by each electron. This screening can be accounted for by assuming an
effective nuclear charge (1  < 2). The ground-state wavefunction of a single electron bound to a nucleus of charge
Z is
✓ ◆3/2
1
Z
1
p
(r)
=
e Zr/a0 ⌘ p Z 3/2 e Zr ,
(5.155)
100
⇡ a0
⇡
and the single-electron ground-state energy is
E1 =
Z 2 h̄2
=
2ma20
Z 2 e2
=
8⇡✏0 a0
Z2
a.u.
2
(5.156)
The two-electron ground-state wave-function is
0 (r1 , r2 ) =
100 (r1 )
100 (r2 ) =
1
⇡
✓
Z
a0
◆3
e
Z(r1 +r2 )/a0
⌘
1 3
Z e
⇡
Z(r1 +r2 )
(5.157)
i.e.the product of the two single-electron wave-functions.
This form of the approximate wave-function immediately suggests how to do a variational calculation. This suggests
that Z is replaced in 0 (r1 , r2 ) by a variable parameter , an effective charge, so that the trial wave-function is:
✓ ◆3
1
e (r1 +r2 )/a0 ,
(5.158)
T (r1 , r2 ) =
⇡ a0
3
⌘
⇡
e
88
(r1 +r2 )
.
(5.159)
PHAS3226: Quantum Mechanics
CHAPTER 5. APPROXIMATE METHODS & MANY-BODY SYSTEMS
100 (r1 )
This trial function is already normalized since
100 (r2 )
and
were normalized.
The Hamiltonian can be written as
1 2 1 2 Z
r
r
2 1 2 2 r1
1 2 Z
1 2
=
r1
r
2
r1
2 2
= Ĥ1 + Ĥ2 + Vee ,
Z
1
+
r2
r12
Z
1
+
r2
r12
=
Ĥ
where
1 2
r
2 1
Ĥ1 =
(5.160)
(5.161)
Z
r1
(5.162)
and similarly for Ĥ2 . Ĥ1 can be re-written as
1 2
(Z
)
r
,
2 1 r1
r1
(Z
)
= ĥ
,
r1
=
Ĥ1
(5.163)
where ĥ is the Hamiltonian for a hydrogenic atom of nuclear charge . Thus by analogy with the hydrogen wavefunctions
⇣ 3 ⌘1/2
2
e r1 is an eigenfunction of ĥ with eigenvalue E1 = 2n2 . Hence
100 (r1 ) =
⇡
D
hĤ1 i =
hĤ1 i =
⌧
E
(Z
)
+
r1
⌧
2
(Z
)
+
2
r1
ĥ
,
(5.164)
since
D
ĥ
E
=
=
D
100
2
2
(r1 )
100
(r2 ) |ĥ |
100
(r1 )
100
⇥ 1.
E D
(r2 ) =
100
(r1 ) |ĥ |
100
(r1 )ih
100
(r2 )
100
E
(r2 )
Now
⌧
(Z
)
r1
= (Z
)
= (Z
noting that h
100
(r2 )
100
3
⇡
)4
(r2 )i = 1 and
R1
0
Z
0
3
4⇡
1
2
(2 )
re
hĤ1 i =
By the interchange r1 $ r2 it follows that
1
2 r
2 r1 2
r1 dr1
=
(Z
= (Z
)
),
3
⇡
Z
1
4⇡e
2 r1
r1 dr1
0
(5.165)
2
dr = 1/ (2 ) and hence
2
2
1
e
r1
(Z
hĤ2 i =
)=
2
2
Z.
(5.166)
2
Z
2
and
hĤ1 i + hĤ2 i =
89
2
2 Z.
(5.167)
PHAS3226: Quantum Mechanics
CHAPTER 5. APPROXIMATE METHODS & MANY-BODY SYSTEMS
The electron-electron interaction expectation value is much more difficult to evaluate as it involves the relative angles
between the position vectors r1 , r2 as r12 = |r1 r2 | = r12 + r22 2r1 r2 cos ✓12 . The result is
hVee i =
=
✓
3
⇡
5
.
8
◆2 Z
1
dr1
0
Z
1
0
dr2
1
e
r12
2 (r1 +r2 )
(5.168)
Adding the three terms, we have the expectation value of the total Hamiltonian
hHi =
2 Z+
2
5
.
8
(5.169)
Since this is a quadratic form that goes to 1 for large , it has a single minimum. To find this, differentiating with respect
to gives
dhHi
5
=2
2Z + = 0,
(5.170)
d
8
so that
5
27
=Z
=
.
(5.171)
16
16
The term 5/16 reduces from the value of = Z = 2 that it would have in the absence of the electron-electron
repulsion. The value of the total energy for this value of is:
ET =
2
2
✓
+
5
16
◆
+
5
=
8
2
=
✓
27
16
◆2
,
(5.172)
and
ET =
(27/16)2 '
2.85 au.
The experimental value of the total energy of the He atom is 2.905 au. The variational estimate of
slightly above the correct value (as it must be!), but the accuracy is good; the error is less than 2 %.
An example of a more elaborate trial function with two screening constants is
h
i
(r1 , r2 ) = C e (↵r1 + r2 )/a0 + e (↵r2 + r1 )/a0
(5.173)
2.85 au which is
and gives E = 2.875 au. With improved variational wavefunctions, which include explicitly the correlation between
the electrons, essentially exact values of the ground-state energy can be obtained.
5.5
Systems of identical particles
A wide range of physical systems, e.g. electrons in atoms or molecules; protons and neutrons in nuclei; cold atoms in a
Bose-Einstein condensate (BEC) consist of many identical particles. If they interact strongly solutions of the Hamiltonian
are very difficult due to the complexity. Generally the difficulty increases "exponentially" with the number of particles.
Even for "non-interacting" particles attention must be paid to the fact that the particles are indistinguishable.
In classical dynamics particles move in well defined trajectories and their positions can be tracked with arbitrary precision. Hence they can be distinguished. In quantum mechanics if their wavefunctions overlap they cannot be distinguished,
especially if the Hamiltonian does not "label" the particles in some way.
5.5.1
Identical particles
Identical particles are often called indistinguishable particles to emphasize that they cannot be distinguished by any
physical measurement. For example, two electrons may be at different positions, have different momenta and even have
different z-component of spin as these are dynamical parameters of the system but they have the same charge, mass and
intrinsic magnetic moment.
90
PHAS3226: Quantum Mechanics
CHAPTER 5. APPROXIMATE METHODS & MANY-BODY SYSTEMS
Any operator representing a physical measurement on the system must remain unchanged if the labels assigned to the
particles are interchanged. For example, if one writes the Hamiltonian of two identical particles as Ĥ (1, 2) then one must
have
Ĥ (1, 2) = Ĥ (2, 1) .
(5.174)
Such a Hamiltonian is that for the helium atom
Ĥ (r1 , r2 ) =
h̄2 2
r
2m 1
Ze2
4⇡✏0 r1
h̄2 2
r
2m 2
Ze2
e2
+
4⇡✏0 r2
4⇡✏0 |r1
r2 |
,
(5.175)
which is symmetric in the electron coordinates, i.e. invariant under the interchange of the labels 1 , 2. For such systems
it is meaningless to ask for the probability that helium in its ground state has electron 1 in a volume element dVa at
position a and electron 2 in a volume element dVb at position b. One can only ask for the probability that one electron is
in dVa and the other in dVb .
An immediate consequence of the symmetry in equ(5.174) is that the eigenvalues of Ĥ are degenerate. Suppose that
Interchanging 1 $ 2 gives
but Ĥ (1, 2) = Ĥ (2, 1) so
Ĥ (1, 2)
(1, 2) = E (1, 2) .
Ĥ (2, 1)
(2, 1) = E (2, 1)
Ĥ (1, 2)
(2, 1) = E (2, 1) .
(5.176)
(5.177)
Thus if (1, 2) is an eigenfunction with energy eigenvalue E, then so also is (2, 1). Hence the eigenvalue is two-fold
degenerate. This is known as exchange degeneracy. Thus (1, 2), (2, 1) or any linear combination of them is also an
eigenfunction. Two particular combinations are
S
1
= p [ (1, 2) +
2
(2, 1)]
(5.178)
and
1
= p [ (1, 2)
(2, 1)] ,
(5.179)
2
which are, respectively, symmetric and antisymmetric under the interchange 1 $ 2.
It follows from the Schrödinger equation that the symmetry of the wavefunction is a constant of the motion. In fact
physically acceptable wavefunctions representing identical particles must be either symmetric or antisymmetric. Define
a particle exchange operator P12 such that P12 acting on any function of the variables of the two particles interchanges
their labels. Then
P12 Ĥ (1, 2) = Ĥ (2, 1)
A
also
so
But
h
P12 Ĥ (1, 2)
h
i
(1, 2)
= H (2, 1)
(2, 1) ,
= Ĥ (1, 2) P12 (1, 2) ,
i
P12 , Ĥ (1, 2)
(1, 2) = 0.
(5.180)
(1, 2) is any two-particle wavefunction so
h
i
P12 , Ĥ (1, 2) = 0.
Thus P12 and Ĥ (1, 2) are compatible and have a common set of eigenfunctions. If
then to be an eigenfunction of P12 it satisfies
P12 (1, 2) = p (1, 2) ,
(2, 1) = p (1, 2) .
91
(5.181)
(1, 2) is such an eigenfunction,
PHAS3226: Quantum Mechanics
Operate with P12 again
CHAPTER 5. APPROXIMATE METHODS & MANY-BODY SYSTEMS
P12 (2, 1) =
(1, 2) = pP12 (1, 2) = p2 (1, 2)
(5.182)
so
p2
p
Hence
= 1,
= ±1.
(5.183)
(1, 2) = ± (2, 1)
and is either symmetric or antisymmetric under the interchange 1 $ 2. Not only the Hamiltonian but any operator
representing a physical property of the system must be symmetric under 1 $ 2 and must therefore commute with P12 .
Thus whatever measurement is made in the system the resulting wavefunction will be an eigenfunction of P12 . No loss of
generality occurs if one assumes that the wavefunction always has this property.
This property can be extended to a many-body wavefunction for identical particles. The wavefunction must be either
symmetric or antisymmetric with respect to interchange of any pair of particles. It follows that every particle belongs
to one of two classes depending on whether a number of them is symmetric or antisymmetric under particle exchange.
Experimental evidence is that for a given kind of particle the symmetry is always of one kind only. For example, electrons,
positrons, protons, neutrons, neutrinos are described by antisymmetric wavefunctions; photons, pions, K-mesons, 4 He
(alpha particles) have symmetric wavefunctions. Particles with antisymmetric wavefunctions are called fermions, while
those with symmetric wavefunctions are bosons. The choice of symmetry is closely related with the value of the total spin
of the particle. It an empirical fact that fermions have spin s = 21 , 32 , 52 , . . ., i.e. odd half-integer values, and bosons have
s = 0, 1, 2, 3, . . ., i.e. integral values. This one-to-one correspondence between spin and the interchange symmetry can
be shown to be a necessary consequence of a relativistic quantum field theory of identical particles (and the spin-statistics
theorem).
5.5.2
Exclusion principle
Consider two identical particles which interact very weakly with one another. To a first approximation one may neglect
their mutual interaction and hence write the Hamiltonian Ĥ (1, 2) as the sum of the two single-particle Hamiltonians as
Ĥ (1, 2) = Ĥ1 (1) + Ĥ2 (2) .
(5.184)
If Ĥ1 (1) has a complete set of eigenfunctions a (1) with eigenvalues Ea , such that Ĥ1 (1)
similarly for Ĥ2 (2) b (2) = Eb b (2) then
⇣
⌘
Ĥ (1, 2) a (1) b (2) = Ĥ1 (1) + Ĥ2 (2) a (1) b (2) = (Ea + Eb ) a (1)
a
b
(1) = Ea
(2) .
a
(1) and
(5.185)
The symmetric and antisymmetric eigenfunctions of equ(5.184) are
S
A
1
(1, 2) = p [
2
a
(1)
b
(2) ±
a
(2)
b
(1)] .
(5.186)
For fermions, the antisymmetric wavefunction vanishes identically if states a = b, i.e.two identical fermions cannot exits
in the same single-particle state. This is known as the Pauli Exclusion Principle.
{The Exclusion Principle actually states that each single-particle eigenfunction can only be used once in constructing
products, linear combinations of which form the total wavefunction.}
This obviously generalizes to more than two fermions; such that in a quantum system at most one particle can occupy
any one single-particle state. In this form it is restricted to non-interacting particles. The requirement of antisymmetric
wavefunctions for identical particles with odd half-integer spin always holds and represents a general statement of the
Pauli Exclusion Principle.
No such restriction applies to bosons. In this case states a = b are allowed and
(1, 2) =
a
(1)
a
(2)
(5.187)
and any number of bosons can occupy the same single-particle state. The helium atom is a good example to illustrate the
ideas of symmetric and antisymmetric wavefunctions.
92
PHAS3226: Quantum Mechanics
5.5.3
CHAPTER 5. APPROXIMATE METHODS & MANY-BODY SYSTEMS
N-particle states
The discussion can be extended to N -particle Hamiltonians for non-interacting particles,
Ĥ =
N
N
N
X
h̄2 X 2 X
ri +
Vi (ri ) =
ĥi
2m i=1
i=1
i=1
(5.188)
2
h̄
where ĥi = 2m
r2i + Vi (ri ) is a single particle
P Hamiltonian. The physics becomes more interesting - genuine manybody - if there are particle-particle interactions i6=j V (rij ) e.g. the Coulomb interaction
V (rij ) =
Z2
(5.189)
,
rj |
|ri
or a short-range spin exchange as in a spin chain,
Vij = KSi · Sj
i,j±1 .
(5.190)
(ri )
(5.191)
For the i-th particle
ĥi
with h
n
(ri ) |
m
(ri )i =
nm .
ni
(ri ) = ✏i
ni
If the total Hamiltonian is separable the solutions are the products
n1 n2 ...nN
(r1 , r2 , . . . rN ) =
n1
(r1 )
n2
(r2 ) . . .
nN
(rN ) ,
n1 n2 ...nN
(r1 , r2 , . . . rN )
(5.192)
and
N
X
i=1
ĥi
!
Ĥ
n1
n1 n2 ...nN
(r1 )
n2
(r1 , r2 , . . . rN ) = Ej
(r2 ) . . .
nN
(rN ) =
N
X
(j)
✏i
i=1
!
n2
(r2 ) . . .
nN
(rN )
(5.193)
(5.194)
The j-th energy eigenvalue of the system is
Ej =
N
X
(j)
(5.195)
✏i .
i=1
Owing to the separability of Ĥ then if
Ĥ
n1 n2 ...nN
(r1 , . . . rk , . . . r` , . . . rN ) = Ej
Ej
n1 n2 ...nN
(r1 , . . . rk , . . . r` , . . . rN ) ;
= ✏1 + . . . ✏k . . . ✏` . . . ✏N
(5.196)
(5.197)
the interchange of the position of two particles (e.g.rk $ r` ) leaves the energy unchanged,
Ĥ
n1 n2 ...nN
(r1 , . . . r` , . . . rk , . . . rN ) = Ej
Ej
n1 n2 ...nN
(r1 , . . . r` , . . . rk , . . . rN ) ;
= ✏1 + . . . ✏` . . . ✏k . . . ✏N .
(5.198)
(5.199)
This will hold for Hamiltonians which are symmetric with respect to exchange whether with or without interaction, e.g.
Ĥ (r1 , r2 ) = ĥ1 (r1 ) + ĥ2 (r2 ) + V (r1 , r2 )
if V (r1 , r2 ) = V (r2 , r1 ).
Let P̂k` be the operator which interchanges particle k with particle `. Then applied to equ(5.196) gives
P̂k` Ĥ
n1 n2 ...nN
(r1 , . . . rk , . . . r` , . . . rN ) = Ej P̂k`
P̂k` Ĥ
n1 n2 ...nN
(r1 , . . . rk , . . . r` , . . . rN ) = Ej
93
n1 n2 ...nN
n1 n2 ...nN
(r1 , . . . rk , . . . r` , . . . rN )
(r1 , . . . r` , . . . rk , . . . rN ) ,
(5.200)
PHAS3226: Quantum Mechanics
CHAPTER 5. APPROXIMATE METHODS & MANY-BODY SYSTEMS
but equ(5.198) can be re-written as
Ĥ P̂k`
n1 n2 ...nN
(r1 , . . . rk , . . . r` , . . . rN ) = Ej
n1 n2 ...nN
(r1 , . . . r` , . . . rk , . . . rN ) .
But the R.H.S. of equ(5.200, 5.201) are the same so from the L.H.S. one has
⇣
⌘
Ĥ P̂k` P̂k` Ĥ
n1 n2 ...nN (r1 , . . . rk , . . . r` , . . . rN ) = 0,
⇣
⌘
Ĥ P̂k` P̂k` Ĥ
h
i
Ĥ, P̂k`
(5.201)
= 0,
= 0,
(5.202)
so P̂k` and Ĥ commute and there are simultaneous eigenstates of P̂k` and Ĥ for all k, ` combinations.
Also two operations of exchange of particles k and `
2
P̂k`
n1 n2 ...nN
(r1 , . . . rk , . . . r` , . . . rN ) =
n1 n2 ...nN
(r1 , . . . rk , . . . r` , . . . rN )
(5.203)
2
leaves the wavefunction unchanged. Hence the eigenvalue of P̂k`
is 1 and that of P̂k` are ±1. Hence the states are either
symmetric
(S)
k`
P̂k`
=+
(S)
k`
(5.204)
=
(A)
`k .
(5.205)
or antisymmetric
P̂k`
(A)
k`
Explicitly for two non-interacting indistinguishable particles the symmetric wavefunction is
(S)
1
(r1 , r2 ) = p [
2
1
(r1 )
2
(r2 ) +
1
(r2 )
2
(r1 )]
(5.206)
(A)
1
(r1 , r2 ) = p [
2
1
(r1 )
2
(r2 )
1
(r2 )
2
(r1 )] .
(5.207)
and the antisymmetric one is
With interactions the individual wavefunctions are more complex (not necessarily products) but the same symmetry
applies, e.g.
1
(A)
(r1 , r2 ) = p [ (r1 , r2 )
(r2 , r1 )]
(5.208)
2
and often
X
(r1 , r2 ) =
cnm m (r1 ) n (r2 ) .
(5.209)
m,n
Symmetrizing for N > 2 particles
Anti-symmetrizing the wavefunction when there are more than two particles is important when doing molecular or atomic
structure calculations, e.g. "Hartree-Fock", ab-initio calculations in quantum chemistry. The generalization for N > 2 is
the Slater Determinant,
(r1 )
n2 (r1 )
..
.
n1
(A)
n1 n2 ...nN
1
(r1 , r2 , . . . rN ) = p
N!
nN
(r1 )
(r2 )
n2 (r2 )
..
.
n1
nN
(rN )
n2 (rN )
..
.
···
···
..
.
n1
(r2 ) · · ·
nN
(5.210)
(rN )
which makes use of the rules for evaluating determinants. For example if N = 3 the antisymmetric state is
(A)
n 1 n 2 n3
(r1 , r2 , r3 ) =
1
p [ n1 (r1 ) n2 (r2 ) n3 (r3 )
6
+ n1 (r2 ) n2 (r3 ) n3 (r1 )
n1
(r3 )
n2
(r2 )
n3
(r1 )
+
n1
(r1 )
n2
(r3 )
n3
(r2 )]
n1
(r3 )
n2
(r1 )
94
n3
(r2 )
n1
(r2 )
n2
(r1 )
n3
(r3 )
(5.211)
PHAS3226: Quantum Mechanics
or equivalently
(A)
n1 n2 n3
CHAPTER 5. APPROXIMATE METHODS & MANY-BODY SYSTEMS
1 X
P
(r1 , r2 , r3 ) = p
( 1) P
6 perm
n1
(r1 )
n2
(r2 )
n3
(r3 )
(5.212)
where the summation is over all permutations of 1, 2, 3.
The symmetric wavefunction is
(S)
n1 n2 n3
1 X
(r1 , r2 , r3 ) = p
P
6 perm
n1
(r1 )
n2
(r2 )
n3
(r3 )
(5.213)
which replaces all the signs in equ(5.212) with + signs.
Consider a two-electron state . Its wavefunction must be antisymmetric with respect to interchange of both position
and spin coordinates, i.e.exchange of the particles. The wavefunction is
n 1 n2
1
(r1 , s1 , r2 , s2 ) = p [
2
n1
(r1 , s1 )
n2
(r2 , s2 )
n1
(r2 , s2 )
n2
(r1 , s1 )]
(5.214)
where s1 , s2 denote "spin-up" or "spin-down" states. If the space-spin states are the same, i.e. n1 = n2 , s1 = s2 ,
= 0 for all r. If s1 6= s2 there is no problem, nor if n1 6= n2 . The same applies to the
n1 (r, s1 ) = n2 (r, s2 ) then
Slater determinant. If two rows of a determinant are identical it evaluates to zero. To include spin, replace n by n, s.
5.5.4
Helium atom and the exchange force
The Hamiltonian for the helium atom is
Ĥ =
h̄2 2
r
2m 1
h̄2 2
r
2m 2
Ze2
4⇡✏0 r1
Ze2
e2
+
4⇡✏0 r2
4⇡✏0 |r1
r2 |
if small corrections due to spin-spin and spin-orbit interactions are neglected. Since Ĥ is spin-independent, spin operators
2
S2 = (s1 + s2 ) and Sz commute with Ĥ and so are constants of the motion. Operators L2 , Lz also commute with Ĥ,
so the simultaneous eigenfunctions of Ĥ , L2 , Lz , S2 , Sz can be written as products of functions. The two electrons can
be in spin state S = 0, a singlet,
1
|0, 0i = p (↵1 2
(5.215)
1 ↵2 ) ,
2
or S = 1, a triplet,
|1, 1i = ↵1 ↵2 ,
1
|1, 0i = p (↵1
2
|1, 1i =
1 2.
(5.216)
2
+
1 ↵2 ) ,
(5.217)
(5.218)
The singlet is antisymmetric with respect to particle exchange, 1 $ 2; the triplet is symmetric. Since the overall
wavefunction must be antisymmetric with respect to particle exchange, 1 $ 2 the products are spatially symmetric wavefunctions (S) (r1 , r2 ) with antisymmetric spin wavefunctions, (A) (s1 , s2 ), i.e. of the form (S) (r1 , r2 ) (A) (s1 , s2 )
or spatially antisymmetric wavefunctions (A) (r1 , r2 ) with antisymmetric spin wavefunctions, (S) (s1 , s2 ), i.e. of the
form (A) (r1 , r2 ) (S) (s1 , s2 ). The lowest energy single particle state would have n1 = 1, n2 = 1 and be the spatially
symmetric state
(S)
(5.219)
11 (r1 , r2 ) = 100 (r1 ) 100 (r2 )
which is combined with an antisymmetric singlet spin state |0, 0i to give
(Sing)
(r1 , r2 ) =
100
(r1 )
100
1
(r2 ) p (↵1
2
2
1 ↵2 )
as the ground state eigenfunction.
For the first excited state n1 = 1, (` = 0, m = 0) and n2 = 2, (` = 0, 1) so there are two possibilities,
95
(5.220)
PHAS3226: Quantum Mechanics
CHAPTER 5. APPROXIMATE METHODS & MANY-BODY SYSTEMS
(S)
(r1 , r2 ) =
(A)
(r1 , r2 ) =
The symmetric spatial wavefunction
(S)
1
p [
2
1
p [
2
(r1 )
2`m
(r2 ) +
2`m
(r1 )
100
(r2 )] ,
(5.221)
100
(r1 )
2`m
(r2 )
2`m
(r1 )
100
(r2 )] .
(5.222)
must combine with the antisymmetric spin function |0, 0i to give
(Sing)
and the antisymmetric spatial function
give
100
(A)
(Trip)
(r1 , r2 ) =
(S)
(r1 , r2 ) |0, 0i
(5.223)
combines with the symmetric triplet spin functions |1, 1i, |1, 0i, |1, 1i to
(r1 , r2 ) =
8
<
(r1 , r2 ) |1, 1i
.
(r1 , r2 ) |1, 0i
(A)
(r1 , r2 ) |1, 1i
(A)
(5.224)
(A)
:
A perturbation calculation of the first excited state should use degenerate perturbation theory. However
2
(a) the perturbation 4⇡✏0 |re 1 r2 | does not involve spin so all perturbation matrix elements between different spin states
D
E
vanish as s, ms | r112 |s0 , ms0 = ss0 ms ms0 , i.e.S is a constant.
(b) the total angular momentum cannot change as there are no external torques, so matrix elements between different
`-values are also zero.
(c) the perturbation is symmetric in exchange 1 $ 2 so matrix elements between states of different symmetry also
vanish.
The expectation value of the inter-electron repulsion for the singlet and triplet states are
⌧
1
(Sing,Trip)
E (S,T ) =
(r1 , r2 ) |
| (Sing,Trip) (r1 , r2 )
(5.225)
r12
=
h p12 [
(r1 ) 2`m (r2 ) ± 2`m (r1 ) 100 (r2 )] r112
,
[ 100 (r1 ) 2`m (r2 ) ± 2`m (r1 ) 100 (r2 )]i
D
E
= 12 { 100 (r1 ) 2`m (r2 ) r112 100 (r1 ) 2`m (r2 )
D
E
+ 2`m (r1 ) 100 (r2 ) r112 2`m (r1 ) 100 (r2 )
.
±h 2`m (r1 ) 100 (r2 ) r112 100 (r1 ) 2`m (r2 )i
D
E
± 100 (r1 ) 2`m (r2 ) r112 2`m (r1 ) 100 (r2 ) }
100
p1
2
(5.226)
(5.227)
The first two terms are symmetric in 1 $ 2 and so equal, similarly the last two terms are symmetric in 1 $ 2 and so
equal so that
⌧
1
(S,T )
E
=
100 (r1 ) 2`m (r2 )
100 (r1 ) 2`m (r2 )
r12
⌧
1
±
(5.228)
100 (r1 ) 2`m (r2 )
2`m (r1 ) 100 (r2 )
r12
E (S,T )
=
Z
±
|
Z
The first integral
J=
Z
100
2
(r1 )|
⇤
100
|
(r1 )
100
1
|
r12
2`m
⇤
2`m
(r2 )
2
(r1 )|
1
|
r12
96
2
(r2 )| dr1 dr2
1
r12
2`m
2`m
(r1 )
2
100
(r2 )| dr1 dr2
(r2 ) dr1 dr2 .
(5.229)
(5.230)
PHAS3226: Quantum Mechanics
CHAPTER 5. APPROXIMATE METHODS & MANY-BODY SYSTEMS
is called the direct integral. It is clearly positive and represents the mutual electrostatic energy of two charge densities
2
2
| 100 (r1 )| and | 2`m (r2 )| separated by a distance r12 , i.e. the Coulomb repulsive energy and thus has a classical
analogue.
The second term
Z
1
⇤
⇤
(5.231)
K=
2`m (r1 ) 100 (r2 ) dr1 dr2
100 (r1 ) 2`m (r2 )
r12
has no classical analogue. Its origin lies in the Pauli Principle and is referred to as the exchange integral. Because of this
exchange contribution the singlet and triplet states are no longer degenerate, with
E (S)
E
(T )
= J + K.
(5.232)
= J
(5.233)
K.
The J term is obviously positive. For the K term, it may be evaluated for the hydrogenic wavefunctions and is also
positive. For ` = n 1 this is (obvious) because the wavefunctions appearing in K have no nodes. That the triplet state
has lower energy than the singlet state can be seen on qualitative grounds. For the triplet state the spatial wavefunction
is antisymmetric and so the electrons tend to stay away from each other ( (A) ⇠ 0; r ' 0; hr12 iTrip > hr12 iSing ); the
probability of finding the two electrons at the same place is low. This makes the repulsion between the electrons in the
triplet state less than in the singlet., hV12 iTrip < hV12 iSing . An interesting feature of the singlet-triplet energy splitting
2
is that although the perturbing interaction 4⇡✏0 |re 1 r2 | is spin-independent the symmetry of the wavefunction make the
potential behave as if it were spin-dependent.
Note that S2 = (s1 + s2 ) = s21 + s22 + 2s1 · s2 so that
h2s1 · s2 i =
⌦
S2
↵
⌦ 2↵
s1
= S (S + 1) h̄2
3 2
h̄
4
⌦ 2↵
s2 = S (S + 1) h̄2
3 2
h̄ .
2
3 2
h̄ ,
4
(5.234)
Thus as S = 0 for the singlet and S = 1 for the triplet,
2
1
2 hs1 · s2 i = 2 h
h̄
1
·
2i
= S (S + 1)
Hence the energy shift is
3
=
2
⇢
3
2
1
2
sin glet
,
triplet
(5.235)
1
(1 + 1 · 2 ) K.
(5.236)
2
An effective "exchange interaction" term appears having a similar magnitude to the electrostatic Coulomb repulsion.
Note that the magnetic dipole interaction which gives rise to the spin-orbit coupling is not responsible for the spin-spin
coupling - it is too weak. The exchange
force arises owing to symmetry effects alone. It gives rise to ferromagnetism via
PN
an interaction of the form V̂ = i=1 Aˆi · ˆi+1 between neighbouring atoms in a ferromagnet. (In ferromagnetism spins
tend to align because K 1 · 2 lowers the energy if the spins are aligned and K > 0).
For helium the direct integral depends on the average separation of the electrons. Since the probability density of the
electrons in the 100 state is spherically symmetric, the average distance between the electrons depends only on the radial
2
probability of the two particles, i.e. / r2 Rn`
(r). Thus the average separation is greater for ` = 0 , ` = 0 than ` = 0,
` = 1 (see radial density plots, e.g. Eisberg p 306), so that hr12 i00 > hr12 i01 , and hence hV12 i00 < hV12 i01 and the
p-state energy is increased more than the s-state. Singlets lie above triplets in a multiplet - an example of Hund’s rule;
other things being equal, state of highest spin has lowest energy.
E=J
97