D18L.
Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission.
Foundations of Quantum Mechanics
Dr. H. Osborn1
Michælmas 1997
1 A
LT
EXed by Paul Metcalfe – comments and corrections to [email protected]
Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission.
T O C LAIRE
T HANKS FOR YOUR PATIENCE
Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission.
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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1
1
2
3
5
6
7
8
8
2
The Harmonic Oscillator
2.1 Relation to wavefunctions . . . . . . . . . . . . . . . . . . . . . . .
2.2 More comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9
10
11
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 21 fermions . . . . . . .
3.3 Two particle states and centre of mass
3.4 Observation . . . . . . . . . . . . . .
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13
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17
17
4
Perturbation Expansions
4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.2 Non-degenerate perturbation theory . . . . . . . . . . . . . . . . . .
4.3 Degeneracy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
19
19
19
21
5
General theory of angular momentum
5.1 Introduction . . . . . . . . . . . . .
5.1.1 Spin 21 particles . . . . . . .
5.1.2 Spin 1 particles . . . . . . .
5.1.3 Electrons . . . . . . . . . .
5.2 Addition of angular momentum . . .
5.3 The meaning of quantum mechanics
23
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25
26
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Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission.
iv
CONTENTS
Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission.
Introduction
These notes are based on the course “Foundations of Quantum Mechanics” given by
Dr. H. Osborn in Cambridge in the Michælmas Term 1997. These typeset notes have
been produced mainly for my own benefit but seem to be officially supported. Recommended books are discussed in the bibliography at the back.
A word or two about the philosophy of these notes seem in order. They are based in
content on the lectures given, but I have felt free to expand and contract various details,
as well as to clarify explanations and improve the narrative flow. Errors in content are
(hopefully) mine and mine alone but I accept no responsibility for your use of these
notes.
Other sets of notes are available for different courses. At the time of typing, these
courses were:
Discrete Mathematics
Further Analysis
Fluid Dynamics 1
Geometry
Foundations of QM
Methods of Math. Phys
Waves (etc.)
Dynamical Systems
Probability
Analysis
Quantum Mechanics
Quadratic Mathematics
Dynamics of D.E.’s
Electrodynamics
Fluid Dynamics 2
Applications of QM
Statistical Physics
They may be downloaded from
http://pdm23.trin.cam.ac.uk/˜pdm23/maths/
http://www.damtp.cam.ac.uk/
or
or you can email me on [email protected] to get a copy of the sets you require.
Even if you download them please email me to let me know, so that I can keep you up
to date with the errata and new note sets. The other people who have contributed time
and effort to these note sets are:
Richard Cameron
Claire Gough
Kate Metcalfe
Hugh Osborn
Malcolm Perry
David Sanders
Analysis
Proofreading
Probability
Proofreading
Accomodation
Proofreading
Although these notes are free of charge anyone who wishes to express their thanks
could send a couple of bottles of interesting beer to Y1 Burrell’s Field, Grange Road.
Paul Metcalfe
15th December 1997
v
Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission.
vi
INTRODUCTION
Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission.
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 7→ 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
Z
2
2
kψk ≡ 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 6= 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.
For two wavefunctions φ and ψ we can define a scalar product
Z
(φ, ψ) ≡ d3 x φ∗ ψ ∈ C.
1
CHAPTER 1. BASICS
2
This has various properties which you can investigate at your leisure.
Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission.
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
∇2 + V (x). Operators may have either a continuous or disHamiltonian H = − 2m
crete 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 hAiψ , defined to be
P
2
2
2
i λi |ai | = (ψ, Aψ). We define the square deviation ∆A to be h(A − hAiψ ) iψ
which is in general nonzero.
Time dependence
This is governed by the Schrödinger 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
X
ıEi
ψ(x, t) =
ai e− ~ t ψi (x).
i
2
If ψ is normalised then the probability of finding the system with energy Ei is |ai | .
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
Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission.
1.2. THE DIRAC FORMALISM
3
mechanics is based on operators acting on vectors in some vector space. A wavefunction ψ corresponds to some abstract vector |ψi, a ket vector. |ψi represents the state of
some physical system described by the vector space.
If |ψ1 i and |ψ2 i are ket vectors then |ψi = a1 |ψ1 i + a2 |ψ2 i is a possible ket vector
describing a state — this is the superposition principle again.
We define a dual space of bra vectors hφ| and a scalar product hφ|ψi, a complex
number.1 For any |ψi there corresponds a unique hψ| and we require hφ|ψi = hψ|φi∗ .
We require the scalar product to be linear such that |ψi = a1 |ψ1 i + a2 |ψ2 i implies
hφ|ψi = a1 hφ|ψ1 i + a2 hφ|ψ2 i. We see that hψ|φi = a∗1 hψ1 |φi + a∗2 hψ2 |φi and so
hψ| = a∗1 hψ1 | + a∗2 hψ2 |.
We introduce linear operators Â|ψi = |ψ ′ i and we define operators acting on bra
vectors to the left hφ|Â = hφ′ | by requiring hφ′ |ψi = hφ|Â|ψi for all ψ. In general, in
hφ|Â|ψi,  can act either to the right or the left. We define the adjoint † of  such
that if Â|ψi = |ψ ′ i then hψ|† = hψ ′ |.  is said to be Hermitian if  = † .
If  = a1 Â1 + a2 Â2 then † = a∗1 †1 + a∗2 †2 , which can be seen by appealing to
the definitions. We also find the adjoint of B̂ Â as follows:
Let B̂ Â|ψi = B̂|ψ ′ i = |ψ ′′ i. Then hψ ′′ | = hψ ′ |B̂ † = hψ|† B̂ † and the result
follows. Also, if hψ| = hφ′ | then |φ′ i = † |φi.
We have eigenvectors Â|ψi = λ|ψi 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 |φi can be expanded in terms of the basis
P
a
|ψ
i
where Â|ψi i = λi |ψi i and ai = hψi |ψi. If |ψi is
ket vectors, |φi =
i
i
i
normalised — hψ|ψi = 1 — then the expected value of  is hÂiψ = hψ|Â|ψi, which
is real if  is Hermitian.
P
The completeness relation for eigenvectors of  can be written as 1̂ = i |ψi ihψi |,
which gives (as before)
|ψi = 1̂|ψi =
X
i
|ψi ihψi |ψi.
P
We can also rewrite  =
i |ψi iλi hψi | and if λj 6= 0 ∀j then we can define
P
hψ
|.
Â−1 = i |ψi iλ−1
i
i
We now choose
an
P orthonormal basis {|ni} with hn|mi
P = δnm and the completeness relation 1̂ = n |nihn|. We can thus expand |ψi = n an |ni with an = hn|ψi.
P
P
We now consider a linear operator Â, and then Â|ψi = n an Â|ni = m a′m |mi,
P
with a′m = hm|Â|ψi = n an hm|Â|ni. Further, putting Amn = hm|Â|ni we get
P
a′m =
n Amn an and therefore solving Â|ψi = λ|ψi is equivalent to solving the
matrix P
equation Aa = λa. AmnPis called the matrix representation of Â. We also have
hψ| = n a∗n hn|, with a′n ∗ = m a∗m A†mn , where A†mn = A∗nm gives the Hermitian
conjugate matrix. This is the matrix representation of † .
1.2.1
Continuum basis
In the above we have assumed discrete eigenvalues λi and normalisable eigenvectors
|ψi i. However, in general, in quantum mechanics operators often have continuous
1 bra
ket. Who said that mathematicians have no sense of humour?
CHAPTER 1. BASICS
Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission.
4
spectrum — for instance the position operator x̂ in 3 dimensions. x̂ must have eigenvalues x for any point x ∈ R3 . There exist eigenvectors |xi such that x̂|xi = x|xi for
any x ∈ R3 .
As x̂ must be Hermitian we have hx|x̂ = xhx|. We define the vector space required
in the Dirac formalism as that spanned by |xi.
For any state |ψi we can define a wavefunction ψ(x) = hx|ψi.
We also need to find some normalisation criterion, which uses the 3 dimensional
Dirac delta function to get hx|x′ i = δ 3 (x − x′ ). Completeness gives
Z
d3 x|xihx| = 1.
We can also recover the ket vector from the wavefunction by
Z
|ψi = 1̂|ψi = d3 x|xiψ(x).
Also hx|x̂|ψi = xψ(x); the action of the operator x̂ on a wavefunction is multiplication by x.
Something else reassuring is
Z
hψ|ψi = hψ|1̂|ψi = d3 xhψ|xihx|ψi
Z
2
= d3 x |ψ(x)| .
The momentum operator p̂ is also expected to have continuum eigenvalues. We
can similarly define states |pi which satisfy p̂|pi = p|pi. We can relate x̂ and p̂ using
the commutator, which for two operators  and B̂ is defined by
h
i
Â, B̂ = ÂB̂ − B̂ Â.
The relationship between x̂ and p̂ is [x̂i , p̂j ] = ı~δij . In one dimension [x̂, p̂] = ı~.
We have a useful rule for calculating commutators, that is:
h
i h
i
h
i
Â, B̂ Ĉ = Â, B̂ Ĉ + B̂ Â, Ĉ .
This can be easily
£
¤proved simply by expanding the right hand side out. We can use this
to calculate x̂, p̂2 .
£
¤
x̂, p̂2 = [x̂, p̂] p̂ + p̂ [x̂, p̂]
= 2ı~p̂.
It is easy to show by induction that [x̂, p̂n ] = nı~p̂n−1 .
We can define an exponential by
− ıa~p̂
e
µ
¶n
∞
X
ıap̂
1
−
=
.
n!
~
n=0
1.2. THE DIRAC FORMALISM
h
i
ıap̂
We can evaluate x̂, e− ~ by
Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission.
h
− ıa~p̂
x̂, e
i
5
µ
¶n #
∞
X
ıap̂
1
−
= x̂,
n!
~
n=0
·
µ
¶n ¸
∞
X 1
ıap̂
=
x̂, −
n!
~
n=0
=
=
"
∞
X
1 ³ ıa ´n
−
[x̂, p̂n ]
n!
~
n=0
∞
X
³ ıa ´n
1
−
ı~p̂n−1
(n
−
1)!
~
n=1
=a
∞ ³
X
ıa ´n−1 n−1
p̂
−
~
n=1
= ae−
ıap̂
~
and by rearranging this we get that
x̂e−
ıap̂
~
= e−
ıap̂
~
(x̂ + a)
ıap̂
and it follows that e− ~ |xi is an eigenvalue of x̂ with eigenvalue x + a. Thus we
ıap̂
see e− ~ |xi = |x + ai. We can do the same to the bra vectors with the Hermiıap̂
ıap̂
tian conjugate e ~ to get hx + a| = hx|e ~ . Then we also have the normalisation
hx′ + a|x + ai = hx′ |xi.
ıap̂
ıap
We now wish to consider hx + a|pi = hx|e ~ |pi = e ~ hx|pi. Setting x = 0 gives
ıap
ha|pi = e ~ N , where N = h0|pi is independent of x. We can determine N from the
normalisation of |pi.
′
′
Z
da hp′ |aiha|pi
Z
ıa(p−p′ )
2
= |N |
da e ~
δ(p − p) = hp |pi =
2
= |N | 2π~ δ(p′ − p)
So, because we are free to choose the phase of N , we can set N =
¡ 1 ¢ 12 ıxp
e ~ . We could define |pi by
thus hx|pi = 2π~
|pi =
Z
dx |xihx|pi =
µ
1
2π~
¶ 12 Z
dx |xie
ıxp
~
¡
1
2π~
¢ 12
and
,
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) = hx|ψi. We wish to see what
operators (the position and momentum operators discussed) do to wavefunctions.
CHAPTER 1. BASICS
6
Now hx|x̂|ψi = xhx|ψi = xψ(x), so the position operator acts on wavefunctions
by multiplication. As for the momentum operator,
hx|p̂|ψi =
Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission.
=
Z
Z
dp hx|p̂|pihp|ψi
dp phx|pihp|ψi
¶ 12 Z
ıxp
1
dp pe ~ hp|ψi
=
2π~
Z
d
= −ı~
dp hx|pihp|ψi
dx
d
d
= −ı~ hx|ψi = −ı~ ψ(x).
dx
dx
£
¤
d
The commutation relation [x̂, p̂] = ı~ corresponds to x, −ı~ dx
= ı~ (acting on
ψ(x)).
µ
1.2.3
Momentum space
|xi 7→ ψ(x) = hx|ψi defines a particular representation of the vector space. It is
sometimes useful to use a momentum representation, ψ̃(p) = hp|ψi. We observe that
Z
ψ̃(p) = dx hp|xihx|ψi
=
µ
1
2π~
¶ 21 Z
dx e−
ıxp
~
ψ(x).
In momentum space, the operators act differently on wavefunctions. It is easy to
d
see that hp|p̂|ψi = pψ̃(p) and hp|x̂|ψi = ı~ dp
ψ̃(p).
We convert the Schrödinger equation into momentum space. We have the operator
p̂2
equation Ĥ = 2m
+ V (x̂) and we just need to calculate how the potential operates on
the wavefunction.
hp|V (x̂)|ψi =
Z
dx hp|V (x̂)|xihx|ψi
¶ 12 Z
ıxp
1
dx e− ~ V (x)hx|ψi
2π~
ZZ
ıx(p′ −p)
1
=
dx dp′ V (x)ψ̃(p′ )e ~
2π~
Z
= dp′ Ṽ (p − p′ )ψ̃(p′ ),
=
where Ṽ (p) =
1
2π~
R
dp e−
µ
ıxp
~
V (x). Thus in momentum space,
p2
ψ̃(p) +
Hp ψ̃(p) =
2m
Z
dp′ Ṽ (p − p′ )ψ̃(p′ ).
1.2. THE DIRAC FORMALISM
1.2.4
7
Commuting operators
Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission.
h
i
Suppose  and B̂ are Hermitian and Â, B̂ = 0. Then  and B̂ have simultaneous
eigenvectors.
Proof. Suppose Â|ψi = λ|ψi and the vector subspace Vλ is the span of the eigenvectors of  with eigenvalue λ. (If dim Vλ > 1 then λ is said to be degenerate.)
As  and B̂ commute we know that λB̂|ψi = ÂB̂|ψi and so B̂|ψi ∈ Vλ . If λ is
non-degenerate then B̂|ψi = µ|ψi for some µ. Otherwise we have that B̂ : Vλ 7→ Vλ
and we can therefore find eigenvectors of B̂ which lie entirely inside Vλ . We can label
these as |λ, µi, and we know that
Â|λ, µi = λ|λ, µi
B̂|λ, µi = µ|λ, µi.
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.
This isn’t so odd: for a single particle in 3 dimensions we have the operators x̂1 , x̂2
and x̂3 . These all commute, so for a single particle with no other degrees of freedom
we can label states uniquely by |xi. 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 |pi. To ram the point in more, we could also have taken some
weird combination like x̂1 , x̂2 and p̂3 .
For our single particle in 3 dimensions, a natural set of commuting operators involves the angular momentum operator, L̂ = x̂ ∧ p̂, or L̂i = ǫijk x̂j p̂k .
We can find commutation relations between L̂i and the other operators we know.
These are summarised here, proof is straightforward.
i
h
• L̂i , x̂l = ı~ǫilj x̂j
h
i
• L̂i , x̂2 = 0
i
h
• L̂i , p̂m = ı~ǫimk p̂k
h
i
• L̂i , p̂2 = 0
i
h
• L̂i , L̂j = ı~ǫijk L̂k
h
i
• L̂i , L̂2 = 0
If we have a Hamiltonian Ĥ =
p̂2
2m
h
i
+ V (|x̂|) then we can also see that L̂, Ĥ = 0.
We choose as a commuting set Ĥ, L̂2 and L̂3 and label states |E, l, mi, where the
eigenvalue of L̂2 is l(l + 1) and the eigenvalue of L̂3 is m.
CHAPTER 1. BASICS
8
Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission.
1.2.5
Unitary Operators
An operator Û is said to be unitary if Û † Û = 1̂, or equivalently Û −1 = Û † .
Suppose Û is unitary and Û |ψi = |ψ ′ i, Û |φi = |φ′ i. Then hφ′ | = hφ|Û † and
′ ′
hφ |ψ i = hφ|ψi. Thus the scalar product, which is the probability amplitude of finding
the state |φi given the state |ψi, is invariant under unitary transformations of states.
For any operator  we can define Â′ = Û ÂÛ † . Then hφ′ |Â′ |ψ ′ i = hφ|Â|ψi and
matrix elements are unchanged under unitary transformations. We also note that if
Ĉ = ÂB̂ then Ĉ ′ = Â′ B̂ ′ .
The quantum mechanics for the |ψi, |φi, Â, B̂ etc. is the same as for |ψ ′ i, |φ′ i, Â′ ,
′
B̂ and so on. A unitary transform in quantum mechanics is analogous to a canonical
transformation in dynamics.
†
Note that if Ô is Hermitian then Û = eıÔ is unitary, as Û † = e−ıÔ = e−ıÔ .
1.2.6
Time dependence
This is governed by the Schrödinger equation,
ı~
∂
|ψ(t)i = Ĥ|ψ(t)i.
∂t
Ĥ is the Hamiltonian and we require it to be Hermitian. We can get an explicit
solution of this if Ĥ does not depend explicitly on t. We set |ψ(t)i = Û (t)|ψ(0)i,
ıĤt
where Û (t) = e− ~ . As Û (t) is unitary, hφ(t)|ψ(t)i = hφ(0)|ψ(0)i.
If we measure the expectation of  at time t we get hψ(t)|Â|ψ(t)i = a(t). This
description is called the Schrödinger picture. Alternatively we can absorb the time dependence into the operator  to get the Heisenberg picture, a(t) = hψ|Û † (t)ÂÛ (t)|ψi.
We write ÂH (t) = Û † (t)ÂÛ (t). In this description the operators are time dependent
(as opposed to the states). ÂH (t) is the Heisenberg picture time dependent operator.
Its evolution is governed by
ı~
which is easily proven.
For a Hamiltonian Ĥ =
for the operators x̂H and p̂H
h
i
∂
ÂH (t) = ÂH (t), Ĥ ,
∂t
1
2
2m p̂(t)
+ V (x̂(t)) we can get the Heisenberg equations
d
1
x̂H (t) = p̂H (t)
dt
m
d
p̂H (t) = −V ′ (x̂H (t)).
dt
These ought to remind you of something.
Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission.
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
1 2 1
Ĥ =
p̂ + 2 mω 2 x̂2 ,
2m
with the condition that [x̂, p̂] = ı~. We wish to solve Ĥ|ψi = E|ψi to find the energy
eigenvalues.
We define a new operator â.
³ mω ´ 12 µ
¶
ıp̂
â =
x̂ +
2~
mω
¶
µ
³ mω ´ 12
ıp̂
†
.
â =
x̂ −
2~
mω
â and ↠are respectively called the£ annihilation
and creation operators. We can
¤
easily obtain the commutation relation â, ↠= 1̂. It is easy to show that, in terms
¡ †
¢
of the annihilation and creation operators, the Hamiltonian
Ĥ = 21 ~ω
ââ i + ↠â ,
h
i
h
¢
¡
which reduces to ~ω ↠â + 12 . Let N̂ = ↠â. Then â, N̂ = â and ↠, N̂ = −↠.
³
³
´
´
Therefore N̂ â = â N̂ − 1 and N̂ ↠= ↠N̂ + 1 .
Suppose |ψi is an eigenvector of N̂ with eigenvalue λ. Then the commutation relations give that N̂ â|ψi = (λ − 1) â|ψi and therefore unless â|ψi = 0 it is an eigenvalue
of N̂ with eigenvalue λ − 1. Similarly N̂ ↠|ψi = (λ + 1) ↠|ψi.
But for any |ψi, hψ|N̂ |ψi ≥ 0 and equals 0 iff â|ψi = 0. Now suppose we have
an eigenvalue λ ∈
/ {0, 1, 2, . . . }. Then ∃n such that ân |ψi is an eigenvector of N̂
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
Schrödinger equation for wavefunctions, but with much less effort.
We define |ni = Cn â†n |0i, where Cn is such as to make hn|ni = 1. We can take
Cn ∈ R, and evaluate h0|ân â†n |0i to find Cn .
9
CHAPTER 2. THE HARMONIC OSCILLATOR
10
1 = hn|ni
= Cn2 h0|ân â†n |0i
Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission.
= Cn2 h0|ân−1 â↠â†n−1 |0i
=
Cn2
hn − 1|â↠|n − 1i
2
Cn−1
=
Cn2
hn − 1|N̂ + 1|n − 1i
2
Cn−1
=
Cn2
(n − 1 + 1)hn − 1|n − 1i
2
Cn−1
=n
Cn2
.
2
Cn−1
√
−1
We thus require Cn = Cn−1 / n and as C0 = 1 we get Cn = (n!) 2 and so we
have the normalised eigenstate (of N̂ ) |ni = √1n! ↠|0i (with eigenvalue n). |ni is also
¢
¡
an eigenvector of Ĥ with eigenvalue ~ω n + 21 . The space of states for the harmonic
oscillator is spanned by {|ni}.
We also need to ask if there exists a non-zero state |ψi such that ↠|ψi = 0. Then
0 = hψ|â↠|ψi = hψ|ψi + hψ|↠â|ψi ≥ hψ|ψi > 0.
So there exist no non-zero states |ψi such that ↠|ψi = 0.
2.1 Relation to wavefunctions
We evaluate
0 = hx|â|0i =
³ mω ´ 12 µ
2~
~ d
x+
mω dx
¶
hx|0i
and we see that ψ0 (x) = hx|0i satisfies the differential equation
µ
¶
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) = hx|1i = hx|↠|0i we find
³ mω ´ 12
³
ı ´
p̂ |0i
hx| x̂ −
2~
mω ¶
µ
³ mω ´ 12
~ d
=
ψ0 (x)
x−
2~
mω dx
µ
¶1
2mω 2
=
xψ0 (x).
~
ψ1 (x) =
2.2. MORE COMMENTS
11
2.2 More comments
Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission.
Many harmonic oscillator problems are simplified using the creation and annihilation
operators.1 For example
hm|x̂|ni =
µ
~
2mω
=
µ
~
2mω
=
µ
~
2mω
¶ 12
¶ 12
¶ 12
hm|â + ↠|ni
√
¢
¡√
n hm|n − 1i + n + 1 hm|n + 1i
√
¢
¡√
n δm,n−1 + n + 1 δm,n+1 .
This is non-zero only if m = n ± 1. We note that x̂r contains terms âs â†r−s , where
0 ≤ s ≤ r and so hm|x̂r |ni can be non-zero only if n − r ≤ m ≤ n + r.
Ĥt
Ĥt
It is easy to see that in the Heisenberg picture âH (t) = eı ~ âe−ı ~ = e−ıωt â.
Then using the equations for x̂H (t) and p̂H (t), we see that
x̂H (t) = x̂ cos ωt +
1
mω p̂ sin ωt.
Also, Ĥâ†H (t) = â†H (t)(Ĥ + ~ω), so if |ψi is an energy eigenstate with eigenvalue
E then â†H (t)|ψi is an energy eigenstate with eigenvalue E + ~ω.
1 And
such problems always occur in Tripos papers. You have been warned.
Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission.
12
CHAPTER 2. THE HARMONIC OSCILLATOR
Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission.
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 {|ri1 } and {|si2 } 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
{|ri1 |si2 : 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, si. We can define a scalar product on V1 ⊗V2 in terms of the basis vectors:
hr′ , s′ |r, si = hr′ |ri1 hs′ |si2 . We can see that if {|ri1 } and {|si2 } are orthonormal
bases for their respective vector spaces then {|r, si} is an orthonormal basis for V1 ⊗V2 .
Suppose Â1 is an operator on V1 and B̂2 is an operator on V2 we can define an
operator Â1 × B̂2 on V1 ⊗ V2 by its operation on the basis vectors:
´
´³
´
³
³
Â1 × B̂2 |ri1 |si2 = Â1 |ri1 B̂2 |si2 .
We write Â1 × B̂2 as Â1 B̂2 .
Two harmonic oscillators
We illustrate these comments by example. Suppose
Ĥi =
p̂2i
+ 1 mωx̂2i
2m 2
i = 1, 2.
We have two independent vector spaces Vi with bases |nii where n = 0, 1, . . . and
âi and â†i are creation and annihilation operators on Vi , and
¢
¡
Ĥi |nii = ~ω n + 12 |nii .
For the combined system
we form the tensor product V1 ⊗ V2 with basis |n1 , n2 i
P
and Hamiltonian Ĥ = i Ĥi , so Ĥ|n1 , n2 i = ~ω (n1 + n2 + 1) |n1 , n2 i. There are
N + 1 ket vectors in the N th excited state.
13
CHAPTER 3. MULTIPARTICLE SYSTEMS
14
The three dimensional harmonic oscillator follows similarly. In general if Ĥ1 and
Ĥ2 are two independent Hamiltonians which act on V1 and V2 respectively then the
Hamiltonian for the combined system is Ĥ = Ĥ1 + Ĥ2 acting on V1 ⊗ V2 . If {|ψr i}
and {|ψs i} are eigenbases for V1 and V2 with energy eigenvalues {Er1 } and {Es2 }
respectively then the basis vectors {|Ψir,s } for V1 ⊗ V2 have energies Er,s = Er1 + Es2 .
Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission.
3.2 Multiparticle Systems
We have considered single particle systems with states |ψi and wavefunctions ψ(x) =
hx|ψi. 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 i1 |ψr2 i2 . . . |ψrN iN where {|ψri ii } is a basis for Hi .
A general state |Ψi is a linear combination of basis vectors and we can define the
N particle wavefunction as Ψ(x1 , x2 , . . . , xN ) = hx1 , x2 , . . . , xN |Ψi.
The normalisation condition is
Z
2
hΨ|Ψi = 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.
∂
|Ψi = Ĥ|Ψi, where Ĥ is an operator
For time evolution we get the equation ı~ ∂t
N
on H .
If the particles do not interact then
Ĥ =
N
X
Ĥi
i=1
where Ĥi acts on Hi but leaves Hj alone for j 6= i. We have energy eigenstates in each
Hi such that Ĥi |ψr ii = Er |ψr ii and so |Ψi = |ψr1 i1 |ψr2 i2 . . . |ψrN iN is an energy
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
particles. For instance if we have Ĥ = H(x̂1 , p̂1 , x̂2 , p̂2 ) then this must equal the
permuted Hamiltonian H(x̂2 , p̂2 , x̂1 , p̂1 ) if we have identical particles. We introduce
Û such that
Û x̂1 Û −1 = x̂2
Û x̂2 Û −1 = x̂1
Û p̂1 Û −1 = p̂2
Û p̂2 Û −1 = p̂1 .
We should also have Û Ĥ Û −1 = Ĥ and more generally if Â1 is an operator on
particle 1 then Û Â1 Û −1 is the corresponding operator on particle 2 (and vice versa).
Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission.
3.2. MULTIPARTICLE SYSTEMS
15
Note that if |Ψi is an energy eigenstate of Ĥ then so is Û |Ψi. Clearly Û 2 = 1̂ and we
require Û to be unitary, which implies that Û is Hermitian.
In quantum mechanics we require |Ψi and Û |Ψi to be the same states (for identical
particles). This implies that Û |Ψi = λ|Ψi and the requirement Û 2 = 1̂ gives that
λ = ±1. In terms of wavefunctions this means that Ψ(x̂1 , x̂2 ) = ±Ψ(x̂2 , x̂1 ). If we
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 , 23 , . . . ).1
The generalisation to N identical particles is reasonably obvious. Let Ûij inter−1
change particles i and j. Then Ûij Ĥ Ûij
= Ĥ for all pairs (i, j).
The same physical requirement as before gives us that Ûij |Ψi = ±|Ψi for all pairs
(i, j).
If we have bosons (plus sign) then in terms of wavefunctions we must have
Ψ(x̂1 , . . . , x̂N ) = Ψ(x̂p1 , . . . , x̂pN ),
where (p1 , . . . , pN ) is a permutation of (1, . . . , N ). If we have fermions then
Ψ(x̂1 , . . . , x̂N ) = λΨ(x̂p1 , . . . , x̂pN ),
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 one dimensional
representations of the permutation group.
3.2.2
Spinless bosons
(Which means that the only variables for a single particle are x̂ and p̂.) Suppose
we have two identical non-interacting bosons. Then Ĥ = Ĥi + Ĥ2 and we have
Ĥ1 |ψr ii = Er |ψr ii . The general space with two particles is H1 ⊗ H2 which has
a basis {|ψr i1 |ψs i2 }, 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
o
n
|ψr i1 |ψr i2 ; √12 (|ψr i1 |ψs i2 + |ψs i1 |ψr i2 ) , r 6= 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 (1hψr ′ |2hψs′ |
2
+ 1hψs′ |2hψr′ |) √12 (|ψr i1 |ψs i2 + |ψs i1 |ψr i2 )
evaluates to δrr′ δss′ + δrs′ δr′ s .
P
For N spinless bosons with Ĥ =
Ĥi the appropriate completely symmetric
states are
´
³
√1
i
+
permutations
thereof
if ri 6= rj
i
.
.
.
|ψ
|ψ
rN N
r1 1
N!
1 Spin
will be studied later in the course.
CHAPTER 3. MULTIPARTICLE SYSTEMS
16
Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission.
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 |xi|si ≡ |x, si and the wavefunctions can be written ψs (x) = hx, s|ψi. 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 µ
to be natural
to take λ = ± 12 .
¶
χλ (1)
, in which case two possible basis
We can also think of the vector χλ =
χλ (2)
µ ¶
µ ¶
1
0
vectors are
and
. Note that χ†λ′ χλ = δλλ′ .
0
1
The scalar product is defined in the obvious way: hφr′ λ′ |φrλ i = hψr′ |ψr ihχλ′ |χλ i,
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
operator Û we must have Ψ(x1 , s1 ; x2 , s2 ) 7→ −Ψ(x2 , s2 ; x1 , s1 ). The two particle
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 ³
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 ³1 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 ).
− 21
1
− 12
s1 6= s2
2
We can now examine two non-interacting electrons, with Ĥ = Ĥ1 + Ĥ2 and take
Hi independent of spin. The single particle states are |ψi i|χs i.
The two electron states live in (H1 ⊗ H2 )A , which has a basis
|ψr i1 |ψr i2 |χA i;
1
√ (|ψr i1 |ψs i2 + |ψs i1 |ψr i2 ) |χA i; r 6= s
2
1
√ (|ψr i1 |ψs i2 − |ψs i1 |ψr i2 ) |χS i; r 6= 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
Ĥ =
2e2
2e2
e2
p̂2
p̂21
+ 2 −
−
+
.
2m 2m 4πǫ0 |x̂1 | 4πǫ0 |x̂2 | 4πǫ0 |x̂1 − x̂2 |
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) has
2e2
En = − 8πǫ
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.
Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission.
3.3 Two particle states and centre of mass
p̂2
p̂2
Suppose we have a Hamiltonian Ĥ = 2m1 + 2m2 + V (x̂1 − x̂2 ) defined on H2 . We can
separate out the centre of mass motion by letting
p̂ =
P̂ = p̂1 + p̂2
1
2
1
2
(p̂1 − p̂2 )
x̂ = x̂1 − x̂2 .
X̂ = (x̂1 + x̂2 )
i
Then X̂i , P̂j = ı~δij , [x̂i , p̂j ] = ı~δij and X̂, P̂ and x̂, p̂ commute respectively.
h
2
2
P̂
We can rewrite the Hamiltonian as Ĥ = 2M
+ ĥ, ĥ = p̂m + V (x̂), where M = 2m and
2
we can decompose H into HCM ⊗ Hint . HCM is acted on by X̂ and P̂ and has wavefunctions φ(X). Hint is acted on by x̂, p̂ and any spin operators. It has wavefunctions
ψ(x, s1 , s2 ). We take wavefunctions Ψ(x1 , s1 ; x2 , s2 ) = Φ(X)ψ(x, s1 , s2 ) in H2 .
P.X
This simplifies the Schrödinger equation, we can just have φ(X) = eı ~ and then
2
P
+ Eint . We thus need only to solve the one particle equation ĥψ = Eint ψ.
E = 2M
Under the particle exchange operator Û we have
ψ(x, s1 , s2 ) 7→ 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 (|x̂|) then we may separate variables to get
µ ¶
x
R(|x|)χ(s1 , s2 )
ψ(x, s1 , s2 ) = Yl
|x|
³
´
³ ´
x
x
with Yl − |x|
= (−1)l Yl |x|
. 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 |Ψi in H1 ⊗H2
can be written as
X
|Ψi =
aij |ψi i1 |φj i2
i,j
with |ψi i1 ∈ H1 and |φj i2 ∈ H2 assumed orthonormal bases for their respective vector
spaces.
Suppose we make a measurement on the first system leaving the second
P system
unchanged, and find the first system in a state |ψi i1 . Then 1hψi |Ψi = j aij |φj i2 ,
which we write as Ai |φi2 , where |φi2 is a normalised state of the second system. We
CHAPTER 3. MULTIPARTICLE SYSTEMS
18
2
Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission.
interpret |Ai | as the probability of finding system 1 in state |ψi i1 . After measurement
system 2 is in a state |φi2 .
If aij = λi δij (no summation) then Ai = λi and measurement of system 1 as |ψi i1
determines system 2 to be in state |φi i2 .
Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission.
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.
We write Ĥ = Ĥ0 + Ĥ ′ where Ĥ0 describes a solvable system with known eigenvalues
and eigenvectors, and Ĥ ′ is in some sense small.
We write Ĥ(λ) = Ĥ0 + λĤ ′ and expand the eigenvalues and eigenvectors in
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 Schrödinger equation.
4.2 Non-degenerate perturbation theory
Suppose that Ĥ0 |ni = ǫn |ni for n = 0, 1, . . . . We thus assume discrete energy levels
and we assume further that the energy levels are non-degenerate. We also require Ĥ ′
to be sufficiently non-singular to make a power series expansion possible.
We have the equation Ĥ(λ)|ψn (λ)i = En (λ)|ψn (λ)i. We suppose that En (λ)
tends to ǫn as λ → 0 and |ψn (λ)i → |ni as λ → 0. We pose the power series
expansions
En (λ) = ǫn + λEn(1) + λ2 En(2) + . . .
|ψn (λ)i = N |ni + λ|ψn(1) i + . . . ,
substitute into the Schrödinger 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
Ĥ0 |ψn(1) i + Ĥ ′ |ni = En(1) |ni + ǫn |ψn(1) i.
(1)
(1)
Note that we can always replace |ψn i with |ψn i + α|ni and leave this equation
(1)
unchanged. We can therefore impose the condition hn|ψn i = 0. If we apply hn| to
19
CHAPTER 4. PERTURBATION EXPANSIONS
20
(1)
this equation we get En = hn|Ĥ ′ |ni — the first order perturbation in energy. If we
apply hr| where r 6= n we see that
hr|ψn(1) i = −
hr|Ĥ ′ |ni
ǫr − ǫn
Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission.
and therefore
|ψn(1) i = −
X |rihr|Ĥ ′ |ni
.
ǫr − ǫn
r6=n
Note that we are justified in these divisions as we have assumed that the eigenvalues
are non-degenerate. On doing the same thing to the O(λ2 ) equation we see that
En(2) = hn|Ĥ ′ |ψn(1) i
¯2
¯
¯
¯
X ¯hr|Ĥ ′ |ni¯
=−
.
ǫr − ǫn
r6=n
This procedure is valid if ǫr − ǫn is not very small when hr|Ĥ ′ |ni =
6 0.
d
Using these results we can see that dλ
En (λ) = hψn (λ)|Ĥ ′ |ψn (λ)i and
X
∂
1
|ψn (λ)i = −
|ψr (λ)ihψr (λ)|Ĥ ′ |ψn (λ)i.
∂λ
Er (λ) − En (λ)
r6=n
Also
∂ Ĥ
∂λ
= Ĥ ′ and so
∂
∂2
En (λ) = 2hψn (λ)|Ĥ ′ |ψn (λ)i.
∂λ2
∂λ
Example: harmonic oscillator
2
p̂
Consider Ĥ = 2m
+ 21 mω 2 x̂2 + λmω 2 x̂2 , which can be viewed as Ĥ0 + Ĥ ′ , where
Ĥ0 is the plain vanilla quantum harmonic oscillator Hamiltonian.
Calculating the matrix elements hr|x̂2 |ni required is an extended exercise in manipulations of the annihilation and creation operators and is omitted. The results are
¡
¢
En(1) = ~ω n + 12
¡
¢
En(2) = − 21 ~ω n + 21 .
We thus get the perturbation expansion for En′
¢
¢¡
¡
En′ = ~ω n + 21 1 + λ − 12 λ2 + O(λ3 ) .
¡
¢√
1 + 2λ which
This system can also be solved exactly to give En′ = ~ω n + 12
agrees with the perturbation expansion.
4.3. DEGENERACY
21
Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission.
4.3 Degeneracy
The method given here breaks down if ǫr = ǫn for r 6= n. Perturbation theory can be
extended to the degenerate case, but we will consider only the first order shift in ǫr .
We suppose that the states |n, si, s = 1 . . . Nn have the same energy ǫn . Ns is the
degeneracy of this energy level.
As before we pose a Hamiltonian Ĥ = Ĥ0 + λĤ ′ such that Ĥ0 |n, si = ǫn |n, si
and look for states |ψ(λ)i with energy E(λ) → ǫn as λ → 0.
The difference with the previous method is that we expand |ψ(λ)i as a power series
in λ in the basis of eigenvectors of Ĥ0 . That is
X
|ψ(λ)i =
|n, sias + λ|ψ (1) i.
s
As the as are arbitrary we can impose the conditions hn, s|ψ (1) i = 0 for each s and
n. We thus have to solve Ĥ|ψ(λ)i = E(λ)|ψ(λ)i with E(λ) = ǫn + λE (1) . If we take
the O(λ) equation and apply hn, r| to it we get
X
as hn, r|Ĥ ′ |n, si = ar Er(1)
s
which is a matrix eigenvalue problem. Thus the first order perturbations in ǫn are
the eigenvalues of the matrix hn, r|Ĥ ′ |n, si. If all the eigenvalues are distinct then
the perturbation “lifts the degeneracy”. It is convenient for the purpose of calculation
to choose a basis for the space spanned by the degenerate eigenvectors in which this
matrix is “as diagonal as possible”.1
1 Don’t
ask...
Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission.
22
CHAPTER 4. PERTURBATION EXPANSIONS
Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission.
Chapter 5
General theory of angular
momentum
For a particle with position and momentum operators x̂ and p̂ with the commutation
relations [x̂i , pˆj ] =h ı~δij iwe define L̂ = x̂ ∧ p̂. It can be seen that L̂ is Hermitian and
it is easy to show L̂i , L̂j = ı~ǫijk L̂k .
5.1 Introduction
More generally we define Hermitian angular momentum operators J with the commutation relation [Ji , Jj ] = ı~ǫijk Jk .1 We ask on what space of states can this algebra of
operators be realised, or alternatively, what are the representations?
We choose one component of J whose eigenvalues label the states; in accordance
with convention we choose J3 . Then we have J3 |mi = m~|mi for m ∈ R (discrete)
and we also have hm′ |mi = δm′ m .
¤
£
By applying the commutation relation we get (easily) that J3 , J2 = 0, so as J2
is Hermitian we can choose a simultaneous eigenbasis. That is, J2 |mi = λ~2 |mi. We
know that λ ≥ 0 since J2 is the sum of the squares of Hermitian operators.
At this stage we choose an alternate basis; that is we split J into J+ , J− and J3 ,
†
with J± = J1 ± ıJ2 . Note that J±
= J∓ . The commutation relations for Ji give that
[J3 , J± ] = ±ı~J± and [J+ , J− ] = 2~J3 . It will be useful later to note that
(
J− J+ + J32 + ~J3
J2 = 12 (J+ J− + J− J+ ) + J32 =
J+ J− + J32 − ~J3
.
Proof of this is immediate. We can now rewrite the [J3 , J± ] commutation relation as J3 J± = J± (J3 ± ~) and so we see that J± |mi is an eigenvector of J3 with
±
eigenvalue (m + 1)~ and so J± |mi = ~Nm
|m ± 1i, with some normalisation con±
stant Nm . By evaluating the norm of J± |mi, and noting that hm|mi = 1 we see that
± 2
) = λ − m2 ∓ m.
(Nm
We can now define states |m ± ni for n = 0, 1, 2, . . . . We can pin them down more
± 2
by noting that (Nm
) ≥ 0 for positive norms. However the formulae we have are,
1 Because
we are now grown up we will omit the hats if they do not add to clarity.
23
CHAPTER 5. GENERAL THEORY OF ANGULAR MOMENTUM
24
given λ, negative for sufficiently large |m| and so to avoid this we must have mmax = j
¡
¢2
such that J+ |ji = 0 and so Nj+ = λ − j 2 − j = 0 and so λ = j(j + 1).
We can perform a similar trick with J− ; there must exist mmin = −j ′ such that
J− | − j ′ i = 0; thus λ = j ′ (j ′ + 1). So j ′ = j and as −j ′ = −j = j − n for some
n ∈ {0, 1, 2, . . . } we have j = 0, 12 , 1, 23 , . . . .
In summary the states can be labelled by |j mi such that
Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission.
J2 |j mi = ~2 j(j + 1)|j mi
J3 |j mi = ~m|j mi
1
J± |j mi = ~ ((j ∓ m) (j ± m + 1)) 2 |j m + 1i
with m ∈ {−j, −j + 1, . . . , j − 1, j} and j ∈ {0, 12 , 1, 23 , . . . }. There are 2j + 1 states
with different m for the same j. |j mi is the standard basis of the angular momentum
states.
We have obtained a representation of the algebra labelled by j. If J = L = x̂ ∧ p̂
we must have j an integer.
P
′
′
′
Recall that if we have
P Â we can define a matrix Aλ λ by Â|λi = λ′ |λ iAλ λ .
Note that (BA)λ′ λ =
Bλ′ µ Aµλ . Given j, we have (J3 )m′ m = ~m δm′ m and
µ
p
(J± )m′ m = ~ (j ∓ m) (j ± m + 1) δm′ ,m±1 , giving us (2j + 1)×(2j + 1) matrices
satisfying the three commutation relations [J3 , J± ] = ±~J± and [J+ , J− ] = 2~J3 .
If J are angular momentum operators which act on a vector space V and we have
|ψi ∈ V such that J3 |ψi = ~k|ψi and J+ |ψi = 0 then ψ is a state with angular
n
momentum j = k. The other states are given by J−
|ψi, 1 ≤ n ≤ 2k. The conditions
2
2
also give J |ψi = ~ k (k + 1) |ψi
5.1.1
Spin
1
2
particles
This is the simplest non-trivial case. We have j = 12 and a two dimensional state space
with a basis | 21 12 i and | 12 − 21 i. We have the relations J3 | 12 ± 21 i = ± 21 ~| 12 ± 21 i and
J+ | 12
1
2i
1
J− | 2 − 21 i
J− | 21
1
2i
1
J+ | 2 − 21 i
=0
=0
= ~| 21 − 12 i
= ~| 21
1
2 i.
It is convenient to introduce explicit matrices σ such that
X
J| 12 mi =
| 12 m′ i 21 ~ (σ)m′ m .
m′
are
The matrices σ are 2 × 2 matrices (called the Pauli spin matrices). Explicitly, they
σ+ =
σ1 =
1
(σ+ + σ− ) =
2
µ
µ
0 2
0 0
0 1
1 0
¶
¶
¶
µ
1 0
σ3 =
0 −1
µ
¶
ı
0 −ı
σ2 = − (σ+ − σ− ) =
.
ı 0
2
σ− =
µ
0
2
¶
0
0
Note that σ12 = σ22 = σ32 = 1 and σ † = σ. These satisfy the commutation relations
[σi , σj ] = 2ıǫijk σk (a slightly modified angular momentum commutation relation)
and we also have σ2 σ3 = ıσ1 (and the relations obtained by cyclic permutation), so
5.2. ADDITION OF ANGULAR MOMENTUM
25
2
σi σj + σj σi = 2δij 1. Thus if n̂ is a unit vector we have (σ.n̂) = 1 and we see that
σ.n̂ has eigenvalues ±1.
1
3 2
2
We define the angular momentum
2 ~σ and so s = 4 ~ 1.
µ ¶ matrices s µ= ¶
1
0
and χ− 12 =
.
The basis states are χ 12 =
0
1
Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission.
5.1.2
Spin 1 particles
We apply the theory as above to get


1 0 0
S3 = 0 0 0 
0 0 −1
and S− = S+† .
5.1.3

0
S+ = 0
0
√

2 √0
0
2
0
0
Electrons
Electrons are particles with intrinsic spin 21 . The angular momentum J = x̂ ∧ p̂ + s,
where s are the spin operators for spin 12 .
The basic operators for an electron are x̂, p̂ and s. We can represent these operators
by their action on two component wavefunctions:
X
ψ(x) =
ψλ (x)χλ .
λ=± 21
In this basis x̂ 7→ x, p̂ 7→ −ı~∇ and s 7→ 21 ~σ. All other operators are constructed
in terms of these, for instance we may have a Hamiltonian
H=
p̂2
+ V (x) + U (x)σ.L
2m
where L = x̂ ∧ p̂.
If V and U depend only on |x| then [J, H] = 0.
5.2 Addition of angular momentum
Consider two independent angular momentum operators J(1) and J(2) with J(r) acting
on some space V (r) and V (r) having spin jr for r = 1, 2.
We now define an angular momentum J acting on V (1) ⊗ V (2) by J = J(1) + J(2) .
Using the commutation relations for J(r) we can get [Ji , Jj ] = ı~ǫijk Jk .
We want to construct states |J M i forming a standard angular momentum basis,
that is such that:
J3 |J M i = ~M |J M i
±
J± |J M i = ~NJ,M
|J M ±1i
p
±
with NJ,M
= (J ∓ M ) (J ± M + 1). We look first for states in V which satisfy
J+ |J Ji = 0 and J3 |J Ji = ~J|J Ji. The maximum value of J we can get is j1 + j2 ;
and |j1 +j2 j1 +j2 i = |j1 j1 i1 |j2 j2 i2 . Then J+ |j1 +j2 j1 +j2 i = 0. Similarly this is
CHAPTER 5. GENERAL THEORY OF ANGULAR MOMENTUM
26
an eigenvector of J3 with eigenvalue ~ (j1 + j2 ). We can now apply J− repeatedly to
form all the |J M i states. Applying J− we get
|J M −1i = α|j1 j1 −1i1 |j2 j2 i2 + β|j1 j1 i1 |j2 j2 −1i2 .
−
, and we must
The coefficents α and β can be determined from the coefficents Na,b
2
2
have α + β = 1. If we choose |ψi a state orthogonal to this
Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission.
|ψi = −β|j1 j1 −1i1 |j2 j2 i2 + α|j2 j2 i1 |j2 j2 −1i2 .
J3 |ψi can be computed and it shows that |ψi is an eigenvector of J3 with eigenvalue
~ (j1 + j2 − 1). Now
0 = hψ|j1 +j2 j1 +j2 −1i ∝ hψ|J− |j1 +j2 j1 +j2 i
and so hψ|J− |φi = 0 for all states |φi in V . Thus J+ |ψi = 0 and hence we have |ψi =
|j1 +j2 −1 j1 +j2 −1i. We can then construct the states |j1 +j2 −1 M i by repeatedly
applying J− .
For each J such that |j1 − j2 | ≤ J ≤ j1 + j2 we can construct a state |J Ji. We
define the Clebsch-Gordan coefficients hj1 m1 j2 m2 |J M i, and so
X
hj1 m1 j2 m2 |J M i|j1 m1 i|j2 m2 i.
|J M i =
m1 ,m2
The Clebsch-Gordan coefficients are nonzero only when M = m1 + m2 .
We can check the number of states;
jX
1 +j2
J=|j1 −j2 |
(2J + 1) =
jX
1 +j2
J=|j1 −j2 |
o
n
2
(J + 1) − J 2 = (2j1 + 1) (2j2 + 1) .
Electrons
Electrons have spin 21 and we can represent their spin states with χ± 12 (s). Using this
notation we see that two electrons can form a symmetric spin 1 triplet


1
)χ 1 (s )

χ 2 (s
³1 2 2
´
1
√
1 (s1 )χ 1 (s2 ) + χ 1 (s1 )χ 1 (s2 )
χm (s1 , s2 ) =
χ
−
−
2
2
2
2
2


χ 1 (s )χ 1 (s )
−2
1
−2
2
and an antisymmetric spin 0 singlet;
´
1 ³
χ0 (s1 , s2 ) = √ χ 12 (s1 )χ− 21 (s2 ) − χ− 12 (s1 )χ 21 (s2 ) .
2
5.3 The meaning of quantum mechanics
Quantum mechanics deals in probabilities, whereas classical mechanics is deterministic if we have complete information. If we have incomplete information classical
mechanics is also probabilistic.
Inspired by this we ask if there can be “hidden variables” in quantum mechanics
such that the theory is deterministic. Assuming that local effects have local causes, this
is not possible.
5.3. THE MEANING OF QUANTUM MECHANICS
27
Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission.
We will take a spin example to show this. Consider a spin 12 particle, with two
spin states |↑i and |↓i which are eigenvectors of S3 = 21 ~σ3 . If we choose to use two
component vectors we have
µ ¶
µ ¶
0
1
.
χ↓ =
χ↑ =
1
0
Suppose n = (sin θ, 0, cos θ) (a unit vector) and let us find the eigenvectors of
µ
¶
cos θ
sin θ
σ.n =
.
sin θ − cos θ
2
As (σ.n) = 1 we must have eigenvalues ±1 and an inspired guess gives χ↑,n and
χ↓,n as
χ↑,n = cos θ2 χ↑ + sin θ2 χ↓
and
χ↓,n = − sin θ2 χ↑ + cos θ2 χ↓ .
Thus (reverting to ket vector notation) if an electron is in a state |↑i then the probability of finding it in a state |↑, ni is cos2 θ2 and the probability of finding it in a state
|↓, ni is sin2 θ2 .
Now, suppose we have two electrons in a spin 0 singlet state;
1
|Φi = √ {|↑i1 |↓i2 − |↓i1 |↑i2 } .
2
Then the probability of finding electron 1 with spin up is 12 , and after making this
measurement electron 2 must be spin down. Similarly, if we find electron 1 with spin
down (probability 12 again) then electron 2 must have spin up. More generally, suppose
we measure electron 1’s spin along direction n. Then we see that the probability for
electron 1 to have spin up in direction n (aligned) is 21 and then electron 2 must be in
the state |↓, ni2 .
If we have two electrons (say electron 1 and electron 2) in a spin 0 state we may
physically separate them and consider independent experiments on them.
We will consider three directions as sketched. For electron 1 there are three vari(1)
(1)
ables which we may measure (in separate experiments); Sz = ±1, Sn = ±1 and
(1)
Sm = ±1. We can also do this for electron 2.
(1)
(2)
We see that if we find electron 1 has Sz = 1 then electron 2 has Sz = −1 (etc.).
If there exists an underlying deterministic theory then we could expect some probability distribution p for this set of experiments;
³
´
(1)
(2)
0 ≤ p Sz(1) , Sn(1) , Sm
, Sz(2) , Sn(2) , Sm
≤1
(1)
(2)
which is nonzero only if Sdirn = −Sdirn and
X
p({s}) = 1.
{s}
Bell inequality
Suppose we have a probability P
distribution p(a, b, c) with a, b, c = ±1. We define
partial probabilities pbc (b, c) = a p(a, b, c) and similarly for pac (a, c) and pab (a, b).
Then
28
CHAPTER 5. GENERAL THEORY OF ANGULAR MOMENTUM
pbc (1, −1) = p(1, 1, −1) + p(−1, 1, −1)
≤ p(1, 1, −1) + p(1, 1, 1) + p(−1, 1, −1) + p(−1, −1, −1)
≤ pab (1, 1) + pac (−1, −1).
Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission.
Applying this to the two electron system we get
³
´
³
´
³
´
(2)
(2)
P Sn(1) = 1, Sm
= 1 ≤ P Sz(1) = 1, Sn(2) = −1 + P Sz(1) = −1, Sm
=1 .
We can calculate these probabilities from quantum mechanics
³
´
³
´
P Sz(1) = 1, Sn(2) = −1 = P Sz(1) = 1, Sn(1) = 1 = cos2
³
´
(2)
P Sz(1) = −1, Sm
= 1 = cos2 θ+φ
and
2
³
´
(2)
P Sn(1) = 1, Sm
= 1 = sin2 θ2 .
The Bell inequality gives sin2
φ
2
≤ cos2
θ
2
+ cos2
θ+φ
2
θ
2
which is not in general true.
Copyright © 2004 University of Cambridge. Not to be quoted or reproduced without permission.
References
◦ P.A.M. Dirac, The Principles of Quantum Mechanics, Fourth ed., OUP, 1958.
I enjoyed this book as a good read but it is also an eminently suitable textbook.
◦ Feynman, Leighton, Sands, The Feynman Lectures on Physics Vol. 3, AddisonWesley, 1964.
Excellent reading but not a very appropriate textbook for this course. I read it as a
companion to the course and enjoyed every chapter.
◦ E. Merzbacher, Quantum Mechanics, Wiley, 1970.
This is a recommended textbook for this course (according to the Schedules). I wasn’t
particularly impressed but you may like it.
There must be a good, modern textbook for this course. If you know of one please
send me a brief review and I will include it if I think it is suitable. In any case, with
these marvellous notes you don’t need a textbook, do you?
Related courses
There are courses on Statistical Physics and Applications of Quantum Mechanics in
Part 2B and courses on Quantum Physics and Symmetries and Groups in Quantum
Physics in Part 2A.
29
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