Foundations of quantum mechanics

Foundations of Quantum Mechanics
Dr. H. Osborn1
Michælmas 1997

1 A
LT

EXed by Paul Metcalfe – comments and corrections to


Revision: 2.5
Date: 1999-06-06 14:10:19+01

The following people have maintained these notes.
– date

Paul Metcalfe


Contents
Introduction
1

v

Basics
1.1 Review of earlier work . . . . . . . . . . . .
1.2 The Dirac Formalism . . . . . . . . . . . . .
1.2.1 Continuum basis . . . . . . . . . . .
1.2.2 Action of operators on wavefunctions
1.2.3 Momentum space . . . . . . . . . . .

1.2.4 Commuting operators . . . . . . . .
1.2.5 Unitary Operators . . . . . . . . . . .
1.2.6 Time dependence . . . . . . . . . . .

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3
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2

The Harmonic Oscillator
2.1 Relation to wavefunctions . . . . . . . . . . . . . . . . . . . . . . .
2.2 More comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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3

Multiparticle Systems
3.1 Combination of physical systems . . .
3.2 Multiparticle Systems . . . . . . . . .
3.2.1 Identical particles . . . . . . .
3.2.2 Spinless bosons . . . . . . . .
3.2.3 Spin 12 fermions . . . . . . .
3.3 Two particle states and centre of mass
3.4 Observation . . . . . . . . . . . . . .

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4

Perturbation Expansions
4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2 Non-degenerate perturbation theory . . . . . . . . . . . . . . . . . .
4.3 Degeneracy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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5

General theory of angular momentum
5.1 Introduction . . . . . . . . . . . . .
5.1.1 Spin 12 particles . . . . . . .
5.1.2 Spin 1 particles . . . . . . .
5.1.3 Electrons . . . . . . . . . .
5.2 Addition of angular momentum . . .
5.3 The meaning of quantum mechanics

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iv

CONTENTS


Introduction
These notes are based on the course “Foundations of Quantum Mechanics” given by
Dr. H. Osborn in Cambridge in the Michælmas Term 1997. Recommended books are
discussed in the bibliography at the back.
Other sets of notes are available for different courses. At the time of typing these
courses were:
Probability
Analysis
Methods
Fluid Dynamics 1
Geometry
Foundations of QM
Methods of Math. Phys
Waves (etc.)
General Relativity
Physiological Fluid Dynamics
Slow Viscous Flows
Acoustics
Seismic Waves

Discrete Mathematics
Further Analysis
Quantum Mechanics
Quadratic Mathematics
Dynamics of D.E.’s
Electrodynamics
Fluid Dynamics 2
Statistical Physics
Dynamical Systems
Bifurcations in Nonlinear Convection
Turbulence and Self-Similarity
Non-Newtonian Fluids

They may be downloaded from
or
/>or you can email to get a copy of the
sets you require.

v


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Chapter 1

The Basics of Quantum
Mechanics
Quantum mechanics is viewed as the most remarkable development in 20th century
physics. Its point of view is completely different from classical physics. Its predictions
are often probabilistic.
We will develop the mathematical formalism and some applications. We will emphasize vector spaces (to which wavefunctions belong). These vector spaces are sometimes finite-dimensional, but more often infinite dimensional. The pure mathematical
basis for these is in Hilbert Spaces but (fortunately!) no knowledge of this area is
required for this course.

1.1 Review of earlier work
This is a brief review of the salient points of the 1B Quantum Mechanics course. If
you anything here is unfamiliar it is as well to read up on the 1B Quantum Mechanics

course. This section can be omitted by the brave.
A wavefunction ψ(x) : R3 → C is associated with a single particle in three dimensions. ψ represents the state of a physical system for a single particle. If ψ is
normalised, that is
ψ

2

≡

2

d3 x |ψ| = 1

2

then we say that d3 x |ψ| is the probability of finding the particle in the infinitesimal
region d3 x (at x).
Superposition Principle
If ψ1 and ψ2 are two wavefunctions representing states of a particle, then so is the
linear combination a1 ψ1 + a2 ψ2 (a1 , a2 ∈ C). This is obviously the statement that
wavefunctions live in a vector space. If ψ = aψ (with a = 0) then ψ and ψ represent
the same physical state. If ψ and ψ are both normalised then a = eıα . We write
ψ ∼ eıα ψ to show that they represent the same physical state.
1


CHAPTER 1. BASICS

2

For two wavefunctions φ and ψ we can define a scalar product
(φ, ψ) ≡

d3 x φ∗ ψ ∈ C.

This has various properties which you can investigate at your leisure.
Interpretative Postulate
Given a particle in a state represented by a wavefunction ψ (henceforth “in a state
2
ψ”) then the probability of finding the particle in state φ is P = |(φ, ψ)| and if the
wavefunctions are normalised then 0 ≤ P ≤ 1. P = 1 if ψ ∼ φ.
We wish to define (linear) operators on our vector space — do the obvious thing.
In finite dimensions we can choose a basis and replace an operator with a matrix.
For a complex vector space we can define the Hermitian conjugate of the operator A
to be the operator A† satisfying (φ, Aψ) = (A† φ, ψ). If A = A† then A is Hermitian.
Note that if A is linear then so is A† .
In quantum mechanics dynamical variables (such as energy, momentum or angular
momentum) are represented by (linear) Hermitian operators, the values of the dynamical variables being given by the eigenvalues. For wavefunctions ψ(x), A is usually
a differential operator. For a single particle moving in a potential V (x) we get the
2
Hamiltonian H = − 2m ∇2 + V (x). Operators may have either a continuous or discrete spectrum.
If A is Hermitian then the eigenfunctions corresponding to different eigenvalues
are orthogonal. We assume completeness — that any wavefunction can be expanded
as a linear combination of eigenfunctions.
The expectation value for A in a state with wavefunction ψ is A ψ , defined to be
2
2
2
i λi |ai | = (ψ, Aψ). We define the square deviation ∆A to be (A − A ψ ) ψ
which is in general nonzero.

Time dependence
This is governed by the Schr¨odinger equation
ı

∂ψ
= Hψ,
∂t

where H is the Hamiltonian. H must be Hermitian for the consistency of quantum
mechanics:
ı

∂
(ψ, ψ) = (ψ, Hψ) − (Hψ, ψ) = 0
∂t

if H is Hermitian. Thus we can impose the condition (ψ, ψ) = 1 for all time (if ψ is
normalisable).
If we consider eigenfunctions ψi of H with eigenvalues Ei we can expand a general
wavefunction as
ψ(x, t) =

ai e −

ıEi

t

ψi (x).

If ψ is normalised then the probability of finding the system with energy Ei is |ai |2 .


1.2. THE DIRAC FORMALISM

3

1.2 The Dirac Formalism
This is where we take off into the wild blue yonder, or at least a more abstract form of
quantum mechanics than that previously discussed. The essential structure of quantum
mechanics is based on operators acting on vectors in some vector space. A wavefunction ψ corresponds to some abstract vector |ψ , a ket vector. |ψ represents the state of
some physical system described by the vector space.
If |ψ1 and |ψ2 are ket vectors then |ψ = a1 |ψ1 + a2 |ψ2 is a possible ket vector
describing a state — this is the superposition principle again.
We define a dual space of bra vectors φ| and a scalar product φ|ψ , a complex
number.1 For any |ψ there corresponds a unique ψ| and we require φ|ψ = ψ|φ ∗ .
We require the scalar product to be linear such that |ψ = a1 |ψ1 + a2 |ψ2 implies
φ|ψ = a1 φ|ψ1 + a2 φ|ψ2 . We see that ψ|φ = a∗1 ψ1 |φ + a∗2 ψ2 |φ and so
ψ| = a∗1 ψ1 | + a∗2 ψ2 |.
ˆ
We introduce linear operators A|ψ
= |ψ and we define operators acting on bra
ˆ
vectors to the left φ|Aˆ = φ | by requiring φ |ψ = φ|A|ψ
for all ψ. In general, in
ˆ , Aˆ can act either to the right or the left. We define the adjoint Aˆ† of Aˆ such
φ|A|ψ
ˆ
that if A|ψ
= |ψ then ψ|Aˆ† = ψ |. Aˆ is said to be Hermitian if Aˆ = Aˆ† .

If Aˆ = a1 Aˆ1 + a2 Aˆ2 then Aˆ† = a∗1 Aˆ†1 + a∗2 Aˆ†2 , which can be seen by appealing to
ˆ Aˆ as follows:
the definitions. We also find the adjoint of B
ˆ † and the result
ˆ † = ψ|Aˆ† B
ˆ
ˆ
ˆ
Let B A|ψ = B|ψ = |ψ . Then ψ | = ψ |B
†
ˆ
ˆ
follows. Also, if ψ|A = φ | then |φ = A |φ .
ˆ
We have eigenvectors A|ψ
= λ|ψ and it can be seen in the usual manner that the
eigenvalues of a Hermitian operator are real and the eigenvectors corresponding to two
different eigenvalues are orthogonal.
We assume completeness — that is any |φ can be expanded in terms of the basis ket
ˆ i = λi |ψi and ai = ψi |φ . If |ψ is normalised
vectors, |φ =
ai |ψi where A|ψ
ˆ , which is real if Aˆ
— ψ|ψ = 1 — then the expected value of Aˆ is Aˆ ψ = ψ|A|ψ
is Hermitian.
The completeness relation for eigenvectors of Aˆ can be written as ˆ1 = i |ψi ψi |,
which gives (as before)
|ψi ψi |ψ .

|ψ = ˆ

1|ψ =
i

We can also rewrite Aˆ =
i |ψi λi ψi | and if λj = 0 ∀j then we can define
−1
−1
ˆ
A = i |ψi λi ψi |.
We now choose an orthonormal basis {|n } with n|m = δnm and the completeness relation ˆ
1 = n |n n|. We can thus expand |ψ = n an |n with an = n|ψ .
ˆ
ˆ and then A|ψ
ˆ
= m am |m ,
We now consider a linear operator A,
= n an A|n
ˆ
ˆ
ˆ
with am = m|A|ψ = n an m|A|n . Further, putting Amn = m|A|n
we get
ˆ
am =
A
a
and
therefore
solving
A|ψ

=
λ|ψ
is
equivalent
to
solving
the
n mn n
ˆ
matrix equation Aa = λa. Amn is called the matrix representation of A. We also have
ψ| = n a∗n n|, with an ∗ = m a∗m A†mn , where A†mn = A∗nm gives the Hermitian
conjugate matrix. This is the matrix representation of Aˆ† .
1 bra

ket. Who said that mathematicians have no sense of humour?


CHAPTER 1. BASICS

4

1.2.1 Continuum basis
In the above we have assumed discrete eigenvalues λi and normalisable eigenvectors
|ψi . However, in general, in quantum mechanics operators often have continuous
ˆ in 3 dimensions. x
ˆ must have eigenspectrum — for instance the position operator x
ˆ |x = x|x for
values x for any point x ∈ R3 . There exist eigenvectors |x such that x
any x ∈ R3 .
ˆ must be Hermitian we have x|ˆ

As x
x = x x|. We define the vector space required
in the Dirac formalism as that spanned by |x .
For any state |ψ we can define a wavefunction ψ(x) = x|ψ .
We also need to find some normalisation criterion, which uses the 3 dimensional
Dirac delta function to get x|x = δ 3 (x − x ). Completeness gives
d3 x|x x| = 1.
We can also recover the ket vector from the wavefunction by
|ψ = ˆ1|ψ =

d3 x|x ψ(x).

ˆ on a wavefunction is multipliAlso x|ˆ
x|ψ = xψ(x); the action of the operator x
cation by x.
Something else reassuring is
ψ|ψ = ψ|ˆ1|ψ =
=

d3 x ψ|x x|ψ
2

d3 x |ψ(x)| .

ˆ is also expected to have continuum eigenvalues. We
The momentum operator p
ˆ |p = p|p . We can relate x
ˆ and p
ˆ using
can similarly define states |p which satisfy p

ˆ is defined by
the commutator, which for two operators Aˆ and B
ˆ B
ˆ = AˆB
ˆ −B
ˆ A.
ˆ
A,
ˆ and p
ˆ is [ˆ
x, pˆ] = ı .
The relationship between x
xi , pˆj ] = ı δij . In one dimension [ˆ
We have a useful rule for calculating commutators, that is:
ˆ Cˆ .
ˆ A,
ˆ B
ˆ Cˆ + B
ˆ B
ˆ Cˆ = A,
A,
This can be easily proved simply by expanding the right hand side out. We can use
this to calculate xˆ, pˆ2 .
xˆ, pˆ2 = [ˆ
x, pˆ] pˆ + pˆ [ˆ
x, pˆ]
= 2ı pˆ.
It is easy to show by induction that [ˆ
x, pˆn ] = nı pˆn−1 .
We can define an exponential by

e−

ıap
ˆ

∞

=

1
n!
n=0

−

ıaˆ
p

n

.


1.2. THE DIRAC FORMALISM
We can evaluate x
ˆ , e−

ıap
ˆ

x
ˆ , e−

ıap
ˆ

5

by
∞

ıaˆ
p

n

1
ıaˆ
p
x
ˆ, −
n!
n=0

n

= x
ˆ,
∞

=

1
n!
n=0

−

∞

=

1
ıa
−
n!
n=0

n

[ˆ
x, pˆn ]

∞

=

ıa
1
−

(n
−
1)!
n=1
∞

=a

−

ıa

n−1

n

ı pˆn−1

pˆn−1

n=1

= ae−

ıap
ˆ

and by rearranging this we get that
xˆe−

ıap
ˆ

= e−

ıap
ˆ

(ˆ
x + a)

ıap
ˆ

and it follows that e− |x is an eigenvalue of xˆ with eigenvalue x + a. Thus we
ıap
ˆ
see e− |x = |x + a . We can do the same to the bra vectors with the Hermiıap
ˆ
ıap
ˆ
to get x + a| = x|e . Then we also have the normalisation
tian conjugate e
x + a|x + a = x |x .
ıap
ˆ
ıap
x|p . Setting x = 0 gives
We now wish to consider x + a|p = x|e |p = e
ıap

a|p = e N , where N = 0|p is independent of x. We can determine N from the
normalisation of |p .
δ(p − p) = p |p =

da p |a a|p
2

= |N |

da e

ıa(p−p )

2

= |N | 2π δ(p − p)
So, because we are free to choose the phase of N , we can set N =
thus x|p =

1
2π

1
2

e

ıxp

|p =

1
2π

1
2

and

. We could define |p by
dx |x x|p =

1
2π

1
2

dx |x e

ıxp

,

but we then have to check things like completeness.

1.2.2 Action of operators on wavefunctions
We recall the definition of the wavefunction ψ as ψ(x) = x|ψ . We wish to see what
operators (the position and momentum operators discussed) do to wavefunctions.

CHAPTER 1. BASICS

6

Now x|ˆ
x|ψ = x x|ψ = xψ(x), so the position operator acts on wavefunctions
by multiplication. As for the momentum operator,

x|ˆ
p|ψ =

dp x|ˆ
p|p p|ψ

=

dp p x|p p|ψ

=

1
2

1
2π

dp pe

ıxp

p|ψ

d
dp x|p p|ψ
dx
d
d
x|ψ = −ı
ψ(x).
= −ı
dx
dx
= −ı

The commutation relation [ˆ
x, pˆ] = ı corresponds to x, −ı
ψ(x)).

d
dx

= ı (acting on

1.2.3 Momentum space
|x → ψ(x) = x|ψ defines a particular representation of the vector space. It is
˜
sometimes useful to use a momentum representation, ψ(p)
= p|ψ . We observe that
˜

ψ(p)
=
=

dx p|x x|ψ
1
2π

1
2

dx e−

ıxp

ψ(x).

In momentum space, the operators act differently on wavefunctions. It is easy to
d ˜
˜
see that p|ˆ
p|ψ = pψ(p)
and p|ˆ
x|ψ = ı dp
ψ(p).
We convert the Schr¨odinger equation into momentum space. We have the operator
ˆ = pˆ2 + V (ˆ
x) and we just need to calculate how the potential operates on
equation H
2m

the wavefunction.

p|V (ˆ
x)|ψ =

dx p|V (ˆ
x)|x x|ψ

1
=
2π
1
=
2π

1
2π

dx e−

dx e−

ıxp

V (x) x|ψ

˜ )e
dx dp V (x)ψ(p

ıx(p −p)

˜ ),
dp V˜ (p − p )ψ(p

=
where V˜ (p) =

1
2

ıxp

V (x). Thus in momentum space,

p2 ˜
˜
ψ(p) +
Hp ψ(p)
=
2m

˜ ).
dp V˜ (p − p )ψ(p


1.2. THE DIRAC FORMALISM

7

1.2.4 Commuting operators

ˆ are Hermitian and A,
ˆ B
ˆ = 0. Then Aˆ and B
ˆ have simultaneous
Suppose Aˆ and B
eigenvectors.
ˆ
Proof. Suppose A|ψ
= λ|ψ and the vector subspace Vλ is the span of the eigenvectors of Aˆ with eigenvalue λ. (If dim Vλ > 1 then λ is said to be degenerate.)
ˆ commute we know that λB|ψ
ˆ
ˆ
ˆ
As Aˆ and B
= AˆB|ψ
and so B|ψ
∈ Vλ . If λ is
ˆ
ˆ : Vλ → Vλ
non-degenerate then B|ψ
= µ|ψ for some µ. Otherwise we have that B
ˆ which lie entirely inside Vλ . We can label
and we can therefore find eigenvectors of B
these as |λ, µ , and we know that
ˆ µ = λ|λ, µ
A|λ,
ˆ µ = µ|λ, µ .
B|λ,

These may still be degenerate. However we can in principle remove this degeneracy by adding more commuting operators until each state is uniquely labeled by the

eigenvalues of each common eigenvector. This set of operators is called a complete
commuting set.
ˆ2
This isn’t so odd: for a single particle in 3 dimensions we have the operators xˆ1 , x
and xˆ3 . These all commute, so for a single particle with no other degrees of freedom
we can label states uniquely by |x . We also note from this example that a complete
commuting set is not unique, we might just as easily have taken the momentum operators and labeled states by |p . To ram the point in more, we could also have taken some
ˆ2 and pˆ3 .
weird combination like x
ˆ1 , x
For our single particle in 3 dimensions, a natural set of commuting operators inˆ =x
ˆ i = ijk x
ˆ∧p
ˆ , or L
ˆj pˆk .
volves the angular momentum operator, L
ˆ
We can find commutation relations between Li and the other operators we know.
These are summarised here, proof is straightforward.
ˆ i , xˆl = ı
• L

ˆj
ilj x

ˆ i, x
ˆ2 = 0
• L
ˆ i , pˆm = ı
• L

ˆk
imk p

ˆ i, p
ˆ2 = 0
• L
ˆ i, L
ˆj = ı
• L

ˆ

ijk Lk

ˆ i, L
ˆ2 = 0
• L
ˆ = pˆ 2 + V (|ˆ
ˆ H
ˆ = 0.
If we have a Hamiltonian H
x|) then we can also see that L,
2m
ˆ L
ˆ 2 and L
ˆ 3 and label states |E, l, m , where the
We choose as a commuting set H,
2
ˆ 3 is m.

ˆ is l(l + 1) and the eigenvalue of L
eigenvalue of L


CHAPTER 1. BASICS

8

1.2.5 Unitary Operators
ˆ = ˆ1, or equivalently U
ˆ −1 = U
ˆ is said to be unitary if U
ˆ †U
ˆ †.
An operator U
ˆ is unitary and U
ˆ |ψ = |ψ , U
ˆ |φ = |φ . Then φ | = φ|Uˆ † and
Suppose U
φ |ψ = φ|ψ . Thus the scalar product, which is the probability amplitude of finding
the state |φ given the state |ψ , is invariant under unitary transformations of states.
ˆ AˆU
ˆ † . Then φ |Aˆ |ψ = φ|A|ψ
ˆ
and
For any operator Aˆ we can define Aˆ = U
matrix elements are unchanged under unitary transformations. We also note that if
ˆ.
ˆ then Cˆ = Aˆ B
Cˆ = AˆB

ˆ B
ˆ etc. is the same as for |ψ , |φ , Aˆ ,
The quantum mechanics for the |ψ , |φ , A,
ˆ and so on. A unitary transform in quantum mechanics is analogous to a canonical
B
transformation in dynamics.
ˆ is Hermitian then U
ˆ = eıOˆ is unitary, as U
ˆ † = e−ıOˆ † = e−ıOˆ .
Note that if O

1.2.6 Time dependence
This is governed by the Schr¨odinger equation,
ı

∂
ˆ
|ψ(t) = H|ψ(t)
.
∂t

ˆ is the Hamiltonian and we require it to be Hermitian. We can get an explicit
H
ˆ does not depend explicitly on t. We set |ψ(t) = U(t)|ψ(0)
ˆ
solution of this if H
,
ˆ
ıHt
−

ˆ
ˆ
where U (t) = e
. As U (t) is unitary, φ(t)|ψ(t) = φ(0)|ψ(0) .
ˆ
If we measure the expectation of Aˆ at time t we get ψ(t)|A|ψ(t)
= a(t). This
description is called the Schr¨odinger picture. Alternatively we can absorb the time deˆ
ˆ † (t)AˆU(t)|ψ
.
pendence into the operator Aˆ to get the Heisenberg picture, a(t) = ψ|U
†
ˆ
ˆ
ˆ
ˆ
We write AH (t) = U (t)AU(t). In this description the operators are time dependent
(as opposed to the states). AˆH (t) is the Heisenberg picture time dependent operator.
Its evolution is governed by
ı
which is easily proven.
ˆ =
For a Hamiltonian H
for the operators xˆH and pˆH

∂ ˆ
ˆ ,
AH (t) = AˆH (t), H
∂t

1
ˆ(t)2
2m p

+ V (ˆ
x(t)) we can get the Heisenberg equations

d
1
x
ˆH (t) = pˆH (t)
dt
m
d
pˆH (t) = −V (ˆ
xH (t)).
dt
These ought to remind you of something.


Chapter 2

The Harmonic Oscillator
In quantum mechanics there are two basic solvable systems, the harmonic oscillator
and the hydrogen atom. We will examine the quantum harmonic oscillator using algebraic methods. In quantum mechanics the harmonic oscillator is governed by the
Hamiltonian
ˆ = 1 pˆ2 + 1 mω 2 xˆ2 ,
H
2
2m

ˆ
with the condition that [ˆ
x, pˆ] = ı . We wish to solve H|ψ
= E|ψ to find the energy
eigenvalues.
We define a new operator a
ˆ.
a
ˆ=
a
ˆ† =

mω
2
mω
2

1
2

1
2

ıˆ
p
mω
ıˆ
p
x
ˆ−

mω
x
ˆ+

.

a
ˆ and a
ˆ† are respectively called the annihilation and creation operators. We can
easily obtain the commutation relation a
ˆ, a
ˆ† = ˆ1. It is easy to show that, in terms
ˆ = 1 ω a
of the annihilation and creation operators, the Hamiltonian H
ˆa
ˆ† + a
ˆ† a
ˆ ,
2
1
†
†
† ˆ
ˆ
ˆ
ˆ + . Let N = a
ˆ. Then a
ˆ, N = a
ˆ and a
ˆ , N = −ˆ

a† .
which reduces to ω a
ˆ a
ˆ a
2

ˆ +1 .
ˆa
ˆ − 1 and N
ˆa
ˆ† N
Therefore N
ˆ=a
ˆ N
ˆ† = a
ˆ with eigenvalue λ. Then the commutation relaSuppose |ψ is an eigenvector of N
ˆa
tions give that N
ˆ|ψ = (λ − 1) a
ˆ|ψ and therefore unless a
ˆ|ψ = 0 it is an eigenvalue
ˆ with eigenvalue λ − 1. Similarly N
ˆa
of N
ˆ† |ψ = (λ + 1) a
ˆ† |ψ .
ˆ |ψ ≥ 0 and equals 0 iff a
But for any |ψ , ψ|N
ˆ|ψ = 0. Now suppose we have an
ˆ

ˆ
eigenvalue of H, λ ∈
/ {0, 1, 2, . . . }. Then ∃n such that a
ˆn |ψ is an eigenvector of N
with eigenvalue λ − n < 0 and so we must have λ ∈ {0, 1, 2, . . . }. Returning to the
Hamiltonian we get energy eigenvalues En = ω n + 12 , the same result as using the
Schr¨odinger equation for wavefunctions, but with much less effort.
We define |n = Cn a
ˆ†n |0 , where Cn is such as to make n|n = 1. We can take
an a
ˆ†n |0 to find Cn .
Cn ∈ R, and evaluate 0|ˆ
9


CHAPTER 2. THE HARMONIC OSCILLATOR

10

1 = n|n
an a
ˆ†n |0
= Cn2 0|ˆ
= Cn2 0|ˆ
an−1 a
ˆa
ˆ† a
ˆ†n−1 |0
=

Cn2
n − 1|ˆ
aa
ˆ† |n − 1
2
Cn−1

=

Cn2
ˆ + 1|n − 1
n − 1|N
2
Cn−1

=

Cn2
(n − 1 + 1) n − 1|n − 1
2
Cn−1

=n
We thus require Cn =

Cn2
.
2
Cn−1

Cn−1
√
n

and as C0 = 1 we get Cn = (n!)

− 12

and so we have
1
†n
ˆ
the normalised eigenstate (of N ) |n = √n! a
ˆ |0 (with eigenvalue n). |n is also an
ˆ
eigenvector of H with eigenvalue ω n + 12 . The space of states for the harmonic
oscillator is spanned by {|n }.
We also need to ask if there exists a non-zero state |ψ such that a
ˆ† |ψ = 0. Then
a† a
ˆ|ψ ≥ ψ|ψ > 0.
0 = ψ|ˆ
aa
ˆ† |ψ = ψ|ψ + ψ|ˆ
So there exist no non-zero states |ψ such that a
ˆ† |ψ = 0.

2.1 Relation to wavefunctions
We evaluate
0 = x|ˆ

a|0 =

1
2

mω
2

x+

d
mω dx

x|0

and we see that ψ0 (x) = x|0 satisfies the differential equation
d
mω
+
x ψ0 (x) = 0.
dx
1 mω

2

This (obviously) has solution ψ0 (x) = N e− 2 x for some normalisation constant N . This is the ground state wavefunction which has energy 12 ω.
For ψ1 (x) = x|1 = x|ˆ
a† |0 we find
ψ1 (x) =
=

=

mω
2
mω
2
2mω

1
2

x|ˆ
x−

1
2

x−
1
2

ı
ˆ|0
mω p

d
mω dx

xψ0 (x).

ψ0 (x)


2.2. MORE COMMENTS

11

2.2 More comments
Many harmonic oscillator problems are simplified using the creation and annihilation
operators.1 It is useful to summarise the action of the annihilation and creation operators on the basis states:
a
ˆ† |n =

√
n + 1|n + 1

and a
ˆ|n =

√
n|n − 1 .

For example

m|ˆ
x|n =
=
=

1

2

2mω
1
2

√
√
n m|n − 1 + n + 1 m|n + 1

1
2

√
√
n δm,n−1 + n + 1 δm,n+1 .

2mω
2mω

m|ˆ
a+a
ˆ† |n

This is non-zero only if m = n ± 1. We note that x
ˆr contains terms a
ˆs a
ˆ†r−s , where
r
0 ≤ s ≤ r and so m|ˆ

x |n can be non-zero only if n − r ≤ m ≤ n + r.
ˆ
ˆ
Ht
Ht
It is easy to see that in the Heisenberg picture a
ˆH (t) = eı a
ˆe−ı
= e−ıωt a
ˆ.
Then using the equations for xˆH (t) and pˆH (t), we see that
x
ˆH (t) = xˆ cos ωt +

1
ˆ sin ωt.
mω p

ˆ a† (t) = a
ˆ + ω), so if |ψ is an energy eigenstate with eigenvalue
Also, Hˆ
ˆ†H (t)(H
H
†
E then a
ˆH (t)|ψ is an energy eigenstate with eigenvalue E + ω.

1 And

such problems always occur in Tripos papers. You have been warned.

12

CHAPTER 2. THE HARMONIC OSCILLATOR


Chapter 3

Multiparticle Systems
3.1 Combination of physical systems
In quantum mechanics each physical system has its own vector space of physical states
and operators, which if Hermitian represent observed quantities.
If we consider two vector spaces V1 and V2 with bases {|r 1 } and {|s 2 } with
r = 1 . . . dim V1 and s = 1 . . . dim V2 . We define the tensor product V1 ⊗ V2 as the
vector space spanned by pairs of vectors
{|r 1 |s

2

: r = 1 . . . dim V1 , s = 1 . . . dim V2 }.

We see that dim(V1 ⊗ V2 ) = dim V1 dim V2 . We also write the basis vectors of
V1 ⊗V2 as |r, s . We can define a scalar product on V1 ⊗V2 in terms of the basis vectors:
r , s |r, s = r |r 1 s |s 2 . We can see that if {|r 1 } and {|s 2 } are orthonormal
bases for their respective vector spaces then {|r, s } is an orthonormal basis for V1 ⊗V2 .
ˆ2 is an operator on V2 we can define an
Suppose Aˆ1 is an operator on V1 and B
ˆ
ˆ

operator A1 × B2 on V1 ⊗ V2 by its operation on the basis vectors:
ˆ2 |r 1 |s
Aˆ1 × B

2

= Aˆ1 |r

1

ˆ2 |s
B

2

.

ˆ2 .
ˆ2 as Aˆ1 B
We write Aˆ1 × B
Two harmonic oscillators
We illustrate these comments by example. Suppose
2
ˆ i = pˆi + 1 mω x
H
ˆ2i
2m 2

i = 1, 2.

We have two independent vector spaces Vi with bases |n i where n = 0, 1, . . . and
ˆ†i are creation and annihilation operators on Vi , and
a
ˆi and a
ˆ i |n
H

i

= ω n+

1
2

|n i .

For the combined system we form the tensor product V1 ⊗ V2 with basis |n1 , n2
ˆ
ˆ =
ˆ
and Hamiltonian H
i Hi , so H|n1 , n2 = ω (n1 + n2 + 1) |n1 , n2 . There are
N + 1 ket vectors in the N th excited state.
13


CHAPTER 3. MULTIPARTICLE SYSTEMS

14

ˆ 1 and
The three dimensional harmonic oscillator follows similarly. In general if H
ˆ
H2 are two independent Hamiltonians which act on V1 and V2 respectively then the
ˆ =H
ˆ1 + H
ˆ 2 acting on V1 ⊗ V2 . If {|ψr }
Hamiltonian for the combined system is H
and {|ψs } are eigenbases for V1 and V2 with energy eigenvalues {Er1 } and {Es2 }
respectively then the basis vectors {|Ψ r,s } for V1 ⊗ V2 have energies Er,s = Er1 + Es2 .

3.2 Multiparticle Systems
We have considered single particle systems with states |ψ and wavefunctions ψ(x) =
x|ψ . The states belong to a space H.
Consider an N particle system. We say the states belong to Hn = H1 ⊗ · · · ⊗ HN
and define a basis of states |ψr1 1 |ψr2 2 . . . |ψrN N where {|ψri i } is a basis for Hi .
A general state |Ψ is a linear combination of basis vectors and we can define the
N particle wavefunction as Ψ(x1 , x2 , . . . , xN ) = x1 , x2 , . . . , xN |Ψ .
The normalisation condition is
Ψ|Ψ =

2

d3 x1 . . . d3 xN |Ψ(x1 , x2 , . . . , xN )| = 1

if normalised.

2

We can interpret d3 x1 . . . d3 xN |Ψ(x1 , x2 , . . . , xN )| as the probability density

that particle i is in the volume element d3 xi at xi . We can obtain the probability
density for one particle by integrating out all the other xj ’s.
∂
ˆ
ˆ is an operator
For time evolution we get the equation ı ∂t
|Ψ = H|Ψ
, where H
N
on H .
If the particles do not interact then
N

ˆi
H

ˆ =
H
i=1

ˆ i acts on Hi but leaves Hj alone for j = i. We have energy eigenstates in each
where H
ˆ i |ψr i = Er |ψr i and so |Ψ = |ψr1 1 |ψr2 2 . . . |ψrN N is an energy
Hi such that H
eigenstate with energy Er1 + · · · + ErN .

3.2.1 Identical particles
There are many such cases, for instance multielectron atoms. We will concentrate on
two identical particles.
“Identical” means that physical quantities are be invariant under interchange of

ˆ = H(ˆ
ˆ 1, x
ˆ2, p
ˆ 2 ) then this must equal the
particles. For instance if we have H
x1 , p
ˆ2, x
ˆ1, p
ˆ 1 ) if we have identical particles. We introduce
permuted Hamiltonian H(ˆ
x2 , p
ˆ such that
U
ˆ −1 = x
ˆx
ˆ1 U
ˆ2
U
−1
ˆ
ˆ
ˆ1U = p
ˆ2
Up

ˆx
ˆ −1 = x
ˆ2U
ˆ1
U

−1
ˆ
ˆ
ˆ 2U = p
ˆ 1.
Up

ˆH
ˆU
ˆ −1 = H
ˆ and more generally if Aˆ1 is an operator on
We should also have U
−1
ˆ is the corresponding operator on particle 2 (and vice versa).
ˆ Aˆ1 U
particle 1 then U


3.2. MULTIPARTICLE SYSTEMS

15

ˆ then so is U
ˆ |Ψ . Clearly U
ˆ 2 = ˆ1 and we
Note that if |Ψ is an energy eigenstate of H
ˆ
ˆ
require U to be unitary, which implies that U is Hermitian.
ˆ |Ψ to be the same states (for identical

In quantum mechanics we require |Ψ and U
ˆ
ˆ 2 = ˆ1 gives that
particles). This implies that U |Ψ = λ|Ψ and the requirement U
ˆ 2 ) = ±Ψ(ˆ
ˆ 1 ). If we
λ = ±1. In terms of wavefunctions this means that Ψ(ˆ
x1 , x
x2 , x
have a plus sign then the particles are bosons (which have integral spin) and if a minus
sign then the particles are fermions (which have spin 12 , 32 , . . . ).1
ˆij interThe generalisation to N identical particles is reasonably obvious. Let U
−1
ˆ
ˆ
ˆ
ˆ
change particles i and j. Then Uij H Uij = H for all pairs (i, j).
ˆij |Ψ = ±|Ψ for all pairs
The same physical requirement as before gives us that U
(i, j).
If we have bosons (plus sign) then in terms of wavefunctions we must have
ˆ N ) = Ψ(ˆ
ˆ pN ),
Ψ(ˆ
x1 , . . . , x
xp1 , . . . , x
where (p1 , . . . , pN ) is a permutation of (1, . . . , N ). If we have fermions then
ˆ N ) = λΨ(ˆ
ˆ pN ),

xp1 , . . . , x
Ψ(ˆ
x1 , . . . , x
where λ = +1 if we have an even permutation of (1, . . . , N ) and −1 if we have an odd
permutation.
Remark for pure mathematicians. 1 and {±1} are the two possible representations
of the permutation group in one dimension.

3.2.2 Spinless bosons
ˆ and p
ˆ .) Suppose
(Which means that the only variables for a single particle are x
ˆ 2 and we have
ˆ = H
ˆi + H
we have two identical non-interacting bosons. Then H
ˆ 1 |ψr i = Er |ψr i . The general space with two particles is H1 ⊗ H2 which has
H
a basis {|ψr 1 |ψs 2 }, but as the particles are identical the two particle state space is
(H1 ⊗ H2 )S where we restrict to symmetric combinations of the basis vectors. That
is, a basis for this in terms of the bases of H1 and H2 is
|ψr 1 |ψr 2 ; √12 (|ψr 1 |ψs

2

+ |ψs 1 |ψr 2 ) , r = s .

The corresponding wavefunctions are
ψr (x1 )ψr (x2 ) and

√1
2

(ψr (x1 )ψs (x2 ) + ψs (x1 )ψr (x2 ))
1

and the corresponding eigenvalues are 2Er and Er +Es . The factor of 2− 2 just ensures
normalisation and
√1 (1
2

ψr |2 ψs | + 1 ψs |2 ψr |) √12 (|ψr 1 |ψs

evaluates to δrr δss + δrs δr s .
ˆ =
For N spinless bosons with H
√1
N!
1 Spin

(|ψr1

1

. . . |ψrN

will be studied later in the course.

N

2

+ |ψs 1 |ψr 2 )

ˆ i we get
H
+ permutations thereof) if ri = rj


CHAPTER 3. MULTIPARTICLE SYSTEMS

16

3.2.3 Spin

1
2

fermions

In this case (which covers electrons, for example) a single particle state (or wavefunction) depends on an additional discrete variable s. The wavefunctions are ψ(x, s) or
ψs (x). The space of states for a single electron H = L2 (R3 ) ⊗ C2 has a basis of the
form |x |s ≡ |x, s and the wavefunctions can be written ψs (x) = x, s|ψ . A basis
of wavefunctions is {ψrλ (x, s) = ψr (x)χλ (s)}, where r and λ are labels for the basis.
λ takes two values and it will later be seen to be natural to take λ = ± 21 .
χλ (1)
We can also think of the vector χλ =
, in which case two possible basis
χλ (2)
1

0
vectors are
and
. Note that χ†λ χλ = δλλ .
0
1
The scalar product is defined in the obvious way: φr λ |φrλ = ψr |ψr χλ |χλ ,
which equals δrr δλλ if the initial basis states are orthonormal.
The two electron wavefunction is Ψ(x1 , s1 ; x2 , s2 ) and under the particle exchange
ˆ we must have Ψ(x1 , s1 ; x2 , s2 ) → −Ψ(x2 , s2 ; x1 , s1 ). The two particle
operator U
states belong to the antisymmetric combination (H1 ⊗ H2 )A .
For N electrons the obvious thing can be done.
Basis for symmetric or antisymmetric 2 particle spin states
There is only one antisymmetric basis state
1
χA (s1 , s2 ) = √ χ 1 (s1 )χ− 1 (s2 ) − χ− 1 (s1 )χ 1 (s2 ). ,
2
2
2
2
2
and three symmetric possibilities:

χ 1 (s )χ 1 (s )


 2 1 2 2
χS (s1 , s2 ) = √12 χ 1 (s1 )χ− 1 (s2 ) + χ− 1 (s1 )χ 1 (s2 ).
2

2
2
2


χ (s )χ (s ).
− 12

1

− 12

s1 = s2

2

ˆ = Hˆ1 + Hˆ2 and take
We can now examine two non-interacting electrons, with H
Hi independent of spin. The single particle states are |ψi |χs .
The two electron states live in (H1 ⊗ H2 )A , which has a basis
|ψr 1 |ψr 2 |χA ;
1
√ (|ψr 1 |ψs 2 + |ψs 1 |ψr 2 ) |χA ; r = s
2
1
√ (|ψr 1 |ψs 2 − |ψs 1 |ψr 2 ) |χS ; r = s,
2
with energy levels 2Er (one spin state) and Er + Es (one antisymmetric spin state
and three symmetric spin states).
We thus obtain the Pauli exclusion principle: no two electrons can occupy the same

state (taking account of spin).
As an example we can take the helium atom with Hamiltonian
ˆ 21
ˆ2
p
2e2
2e2
ˆ = p
H
+ 2 −
−
+
2m 2m 4π 0 |ˆ
x1 | 4π 0 |ˆ
x2 | 4π

e2
.
ˆ2 |
x1 − x
0 |ˆ


3.3. TWO PARTICLE STATES AND CENTRE OF MASS

17

If we neglect the interaction term we can analyse this as two hydrogen atoms and
glue the results back together as above. The hydrogen atom (with a nuclear charge 2e)
−2e2

has En = 8π
2 , so we get a ground state for the helium atom with energy 2E1 with
0n
no degeneracy and a first excited state with energy E1 + E2 with a degeneracy of four.
Hopefully these bear some relation to the results obtained by taking the interaction into
account.

3.3 Two particle states and centre of mass
2

2

ˆ = pˆ 1 + pˆ 2 + V (ˆ
ˆ 2 ) defined on H2 . We can
Suppose we have a Hamiltonian H
x1 − x
2m
2m
separate out the centre of mass motion by letting
1
ˆ 2)
(ˆ
p1 − p
2

ˆ =p
ˆ1 + p
ˆ2
P

ˆ=
p

ˆ = 1 (ˆ
ˆ2 )
x1 + x
X
2

ˆ=x
ˆ1 − x
ˆ2.
x

ˆ P
ˆ and x
ˆ i , Pˆj = ı δij , [ˆ
ˆ, p
ˆ commute respectively.
xi , pˆj ] = ı δij and X,
Then X
ˆ2
ˆ2
p
ˆ ˆ
ˆ = P
We can rewrite the Hamiltonian as H
x), where M = 2m and
2M + h, h = m + V (ˆ
2

ˆ and P
ˆ and has wavewe can decompose H into HCM ⊗ Hint . HCM is acted on by X
ˆ, p
ˆ and any spin operators. It has wavefunctions
functions φ(X). Hint is acted on by x
ψ(x, s1 , s2 ). We take wavefunctions Ψ(x1 , s1 ; x2 , s2 ) = Φ(X)ψ(x, s1 , s2 ) in H2 .
ıP.X
This simplifies the Schr¨odinger equation, we can just have φ(X) = e
and then
P2
ˆ = Eint ψ.
E = 2M
+ Eint . We thus need only to solve the one particle equation hψ
ˆ we have
Under the particle exchange operator U

ψ(x, s1 , s2 ) → ψ(−x, s2 , s1 ) = ±ψ(x, s1 , s2 ),
with a plus sign for bosons and a minus sign for fermions. In the spinless case then
ψ(x) = ψ(−x).
If we have a potential V (|ˆ
x|) then we may separate variables to get
ψ(x, s1 , s2 ) = Yl
x
= (−1)l Yl
with Yl − |x|

x
|x|

x

|x|

R(|x|)χ(s1 , s2 )

. For spinless bosons we therefore require l to be even.

3.4 Observation
Consider the tensor product of two systems H1 and H2 . A general state |Ψ in H1 ⊗H2
can be written as
aij |ψi 1 |φj

|Ψ =

2

i,j

with |ψi
spaces.

1

∈ H1 and |φj

2

∈ H2 assumed orthonormal bases for their respective vector


18

CHAPTER 3. MULTIPARTICLE SYSTEMS

Suppose we make a measurement on the first system leaving the second system
unchanged, and find the first system in a state |ψi 1 . Then 1 ψi |Ψ = j aij |φj 2 ,
which we write as Ai |φ 2 , where |φ 2 is a normalised state of the second system. We
2
interpret |Ai | as the probability of finding system 1 in state |ψi 1 . After measurement
system 2 is in a state |φ 2 .
If aij = λi δij (no summation) then Ai = λi and measurement of system 1 as |ψi 1
determines system 2 to be in state |φi 2 .


Chapter 4

Perturbation Expansions
4.1 Introduction
Most problems in quantum mechanics are not exactly solvable and it it necessary to
find approximate answers to them. The simplest method is a perturbation expansion.
ˆ where H
ˆ 0 describes a solvable system with known eigenvalues
ˆ =H
ˆ0 +H
We write H
ˆ is in some sense small.
and eigenvectors, and H
ˆ and expand the eigenvalues and eigenvectors in
ˆ
ˆ 0 + λH
We write H(λ)

= H
powers of λ. Finally we set λ = 1 to get the result. Note that we do not necessarily
have to introduce λ; the problem may have some small parameter which we can use.
This theory can be applied to the time dependent problem but here we will only discuss
the time independent Schr¨odinger equation.

4.2 Non-degenerate perturbation theory
ˆ 0 |n = n |n for n = 0, 1, . . . . We thus assume discrete energy levels
Suppose that H
ˆ
and we assume further that the energy levels are non-degenerate. We also require H
to be sufficiently non-singular to make a power series expansion possible.
ˆ
We have the equation H(λ)|ψ
n (λ) = En (λ)|ψn (λ) . We suppose that En (λ)
tends to n as λ → 0 and |ψn (λ) → |n as λ → 0. We pose the power series
expansions
En (λ) =

n

+ λEn(1) + λ2 En(2) + . . .

|ψn (λ) = N |n + λ|ψn(1) + . . . ,
substitute into the Schr¨odinger equation and require it to be satisfied at each power
of λ. The normalisation constant N is easily seen to be 1 + O(λ2 ). The O(1) equation
is automatically satisfied and the O(λ) equation is
ˆ 0 |ψn(1) + H
ˆ |n = En(1) |n +
H

(1)

(1)

(1)
n |ψn

.

Note that we can always replace |ψn with |ψn + α|n and leave this equation
(1)
unchanged. We can therefore impose the condition n|ψn = 0. If we apply n| to
19