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A3: Quantum Mechanics
Toby Adkins

June 20, 2016

Contents
1 Introduction

3

1
...
1
...
1
...
2 Operators and Observables
1
...
1 The Hamiltonian Operator
1
...
2 Shared Eigenstates
1
...
3 The Uncertainty Principle
1
...
4 The Momentum Representation

4
6
6
7
8
8
9
10
11

¨
2 The Schrodinger Equations

12

3 Quantum Mechanics and Waves

20

4 The Quantum Harmonic Oscillator

32

2
...
1
...
2 The Time-Dependent Schr¨dinger Equation
o
2
...
1 Time Evolution
2
...
4 Ehrenfest's Theorem
3
...
1
...
2 The Potential Step
3
...
3
...
3
...
4 The Dirac-Delta Well

13
13
15
15
16
17
21
21
23
24
25
28
30

4
...
1
...
2 Stationary States
4
...
1 Normalisation
4
...
2 The Number Operator
4
...
3 Matrix Representations
4
...
3
...
4 Dynamics of Oscillators
4
...
1 Heisenberg's Uncertainty Principle

1

33
33
35
36
36
37
38
39
40
42

Toby Adkins

A3

5 Angular Momentum

43

6 Perturbation Theory

67

7 Multiple Particle Systems

88

5
...
1
...
1
...
1
...
1
...
2 Orbital Angular Momentum
5
...
1 Eigenfunctions of Lz and L2
5
...
2 Angular Momentum and Orbits
5
...
3
...
3
...
3
...
4 Composite Systems
5
...
1 Product States
5
...
2 Correlated States
5
...
3 Combining Angular Momentum
5
...
5
...
5
...
1 Time-Independent Perturbation Theory
6
...
1 Some Examples
6
...
2 Degenerate Perturbation Theory
6
...
3 Time-Dependent Perturbation Theory
6
...
1 The Sudden Approximation
6
...
2 The Adiabatic Approximation
6
...
3 A Time-Dependent Perturbation
6
...
4
...
4
...
5 Atoms in a Weak Magnetic Field
6
...
1 Gauge Transformations
6
...
2 The Classical Limit
6
...
3 The Zeeman Eect
6
...
4 The Spin-Orbit Interaction
7
...
1
...
1
...
1
...
2 The Helium Atom
7
...
1 The Electron Interaction
7
...
2 Higher Excited States
7
...
3
...


Introduction

This chapter will serve as an introduction to the fundamental concepts that form the basis
of Quantum Mechanics, including:
• Quantum States
• Operators and Observables
• The Position Representation
• The Momentum Representation

This material will assume familiarity with Dirac notation and the mathematics of operators
found in Chapter 1 of the Mathematical Methods notes
...
Many texts will make use of the hat symbol to denote an operator
...
It is the piece of Physics the least
understood
...
" - Professor James Binney

3

Toby Adkins
1
...
Here, we shall dene it more rigorously
...

We will see what exactly a quantum amplitude represents shortly
...
This can take on
two values; ±1/2
...

The knowledge of any quantum-mechanical system can be encoded in some quantum state
that is written as |ψ
...
1)

ai |i
i

where ai are the quantum amplitudes of each state
...
Consider ψ|ψ
...
Then we have
the sum of the moduli of the quantum amplitudes are equal to one
...
Thus, we can think of the modulus-square of
the quantum amplitudes ai as being the probability of nding the system in the state |i
...
We can then nd ak by
ai |i

k|ψ = k|
i

=

ai k|i
i

=

ai δik
i

= ak

In a very similar way to writing a vector as a linear combination of some basis, we can nd
the quantum-amplitudes as
(1
...
A particle can 'tunnel' between potential wells that form a
chain, and so it's state can be written as
∞

|ψ =

an |n
n=−∞

where
1
an = √
2

−i
3

|n|/2

einπ

Find the probability of the particle being in the centre well, or anywhere to the 'right' (towards positive n) of it
...
Evidently, this is just
square of the real part of an
...

P≥0 = |ao |2 + |a1 |2 + |a2 |2 +
...

2 2 3 2 3
∞
1
1
=
2
3n
n=0

→ P≥0

1
1
= ·
2 1 − 1/3
3
=
4

We could have also performed this calculation very quickly by recognising the symmetry
of the problem; the probability of it being on both sides of the centre must be the same
...


5

Toby Adkins

A3

1
...
1 Energy Representation
Suppose that our system is has a set (which may be innite) of allowable energy states Ei ,
each with quantum amplitudes ai
...
We clearly then
have that ai = 0 for i = k and ak = 1
...
The state |k = |Ek is called a
state of well dened energy
...
This is a logical necessity of the way in which we want
to interpret the mathematics in Quantum Mechanics
...
1
...
We then measure the energy and nd that |ψ is in
the energy state Ek
...
This is what is
known as the 'collapse of the wave-function' that is probably the most common thing that
popular science points to when talking about the weird nature of Quantum Mechanics
...


6

Toby Adkins
1
...
In general, these operators correspond to some observable, such
as momentum, position or spin
...
Instead, we shall focus on examining some of the consequences
of these properties of operators for the 'physics' of the situation
...

Q |qk =

qi |qi qi | |qk
i

qi |qi qi |qk

=
i

qi |qi δik

=
i

→ Q |qk = qk |qk

This reduces to an eigenvalue problem
...
By denition, the product
qi |qj = aij = δij

gives the amplitude to nd the system in state qi given that it is already in qj
...
This means that most, but not all, operators
of observables turn out to be hermitian as a result of this logical necessity
...
Let us postulate that the expectation value Q for a hermitian operator Q for a
system in a state |ψ is given by ψ|Q|ψ
...
3)

P i qi
i

In general, the quantum analogue to the determination of an observable in classical physics
is nding the expectation value of the operator that corresponds to that observable
...
2
...
It is dened as
(1
...


1
...
2 Shared Eigenstates
Suppose that we have two operators A and B
...
Thus, we obtain the condition that
if [A, B] = 0, then the operators A and B share a complete set of mutual eigenstates
...
This is a very powerful statement
...
This is in fact too weak of a condition
...
Can it have well dened momentum for a particular V (x)? If we consider
the energy to be the sum of the kinetic (p2 /2m) and potential (V (x)) energies, then energy
is only well dened given that the kinetic energy (and thus momentum) is well dened
...
This means that under certain
conditions, the particle can have well-dened momentum
...

Another trap that students can fall into is assuming that if [A, B] = 0, and the system is
in an eigenstate of A, then it is also in an eigenstate of B
...
Essentially, just because there is a complete set of eigenstates which
spans both operators, this does not mean that both operators contain all of said eigenstates
...
But we might also have B |u = b1 |u and B |v = b2 |v
...

8

Toby Adkins

A3

1
...
3 The Uncertainty Principle
Consider two hermitian operators A and B
...
For
some real number s, and properly normalised state |ψ :
|φ = (F + isG) |ψ
φ|φ = φ| |(F ∗ − isG∗ )(F + isG)| |φ
= F 2 + s2 G2 + is [F, G]

By the denition of the product φ|φ , this must be real and positive
...
5)

As we shall see later, this gives rise to some of the more commonly known uncertainty
relations
...
3

A3

The Position Representation

The position operator unsurprisingly takes the form x
...
This means that instead
of working in a discrete basis, we instead have to work over a continuous interval
...
6)

By analogy to (1
...
This is what is known as the wave-function in Quantum Mechanics,
as is essentially the extension of the discrete probability amplitudes ai that we saw before
to continuous space; a probability density function
...
7)

dx |x x|

I=
−∞

Let us check that this works:
∞

∞

I |ψ =

dx |x x|
−∞
∞

dx ψ(x) |x
−∞

dx ψ(x) |x x|x

=
−∞
∞

dx ψ(x) |x

=
−∞

= |ψ

The condition for a wave-function to be properly normalised is then give by
ψ|ψ = ψ |I| ψ
∞

=
!

dx ψ|x x|ψ
−∞
∞

1=

dx |ψ(x)|2

−∞

This is why the function space L2 is so important; it is the space in which wave-functions,
w
amongst other important functions, exist
...
Another thing we might want to
consider is the expectation value:
Q = ψ|Q|ψ
= ψ|I|Q|I|ψ
∞

=

dx ψ|x x|Q|x x|ψ
−∞

Thus, the expectation value of an operator in position space is given by
∞

Q =

dx ψ|x Qx x|ψ
−∞

(1
...
For the position
operator, we nd simply that x = x · ψ(x)
...

10

Toby Adkins
1
...
Suppose that we have some wave-function ψ(x) in
x-space, and perform a Fourier Transform to k -space
...
Next, let us calculate
the expectation value of momentum:
∞

p =

dk ψ ∗ (x) ( k) ψ(x)

−∞

1
=
2π
1
=
2π
∞

=
−∞
∞

=

∞

∞

dk
dx
−∞

dx ( k) ψ(x ) e−ikx

−∞

−∞
∞

−∞
∞

∞

dx ψ ∗ (x) eikx
∞

dx

dk eik(x−x ) ψ ∗ (x ) −i

−∞

−∞

dx δ(x − x ) ψ ∗ (x ) −i
dx ψ ∗ (x) −i

−∞

∂
∂x

∂
∂x

∂
∂x

ψ(x )

ψ(x )

ψ(x)

Thus, the momentum operator is given by
p = −i

= −i

∂
∂x

(1
...
That last phrase is important, as the momentum operator
is not always a slave to the x-operator
...


11

∂φ
∂x

∗

∂φ∗
∂x

2
...
All of Quantum
o
Mechanics results from the manipulation of these two equations, which is quite a fascinating
thing to come to grips with
...
1

A3

¨

The Time-Independent Schrodinger Equation

Suppose that a system is in a state of well dened energy, such that |ψ can be written as
a linear combination of energy states En
...
1)

This is known as the The Time-Independent Schr¨dinger Equation (TISE)
...
Not all wave-functions will satisfy the TISE, as this requires that they are in
a state of well dened energy
...
2
...

The form of the Hamiltonian is entirely dependent on the system being examined
...
2)

Evidently, V (x) may take multiple forms
...


2
...
1 The Nodal Theorem
A small disclaimer to begin with: to our knowledge, there is no actual theorem with this
name
...

We want to prove that An nth order wave-function that satises the TISE has n-nodes for
any time independent Hamiltonian
...
The proof makes use of the results of
(3
...

Consider a particle of energy E(x) that obeys the TISE under some one-dimensional potential V (x) over all space
...
For such a particle in an innite well, we know the
solutions to be
(n + 1)πx
ε
(2n + 1)πx
ψ(x) even ∝ cos
2ε
ψ(x) odd ∝ sin

for n = 0, 1, 2, 3,
...
e
to increase the sides of the well by dε, where dε is small
...
We assume that the solutions are well-behaved, meaning that they must be
zero outside the well
...

This can occur in one of two ways:
13

Toby Adkins

A3

1
...
However,
in this case, ψ will be discontinuous at the boundary in this case as ψ = 0
...

2
...
There must be a value for x at which it touches the
zero-axis, becoming tangent at this point, which we will call xo
...
Then:
V (k) ψ + · · · +

k
k
V ψ (k) = E (k) ψ + · · · +
Eψ (k)
k
k

This implies that all derivatives from j = 0 to j = k will be zero, as V (x) = E(x) by
the denition of H
...

This means that the wave-function must vanish if it is to become tangent at xo
...

Continually deforming ε in small changes dε such that ε → ∞ means that it cannot develop
a node over all space
...


14

Toby Adkins
2
...
3)

This governs the time evolution of the state of the system |ψ
...
As the equation is rst order, the initial
information required to solve for t > 0 is the initial state |ψ, 0 that consists of a complete
set of amplitudes
...
However, this would mean that |ψ, 0 was in fact not a complete
set of amplitudes, which would break the 'predicting power' of Physics
...
2
...
Then according to the TDSE, the time-evolution of this state is given
by
i

∂ |En
= H |En = En |En
∂t

which implies that
|En , t = |En , 0 e−iEn t/

We can use this to nd the time evolution of some arbitrary state |ψ
...

˙
This means that the time evolution of |ψ is simply given by
an e−iEn t/ |En , 0

|ψ, t =

(2
...
This means that the probability that the
system will be in any energy state En is time-independent as Pn = |an |2 and an = 0
...


15

Toby Adkins
2
...
3), we saw the idea that ψ(x, t) is the probability density function for a particle
to be at a particular point in space
...
Let us dene ρ(x, t) = |ψ(x, t)|2
...
Then we are left with
i

ψ∗

2
∂ψ
∂ψ ∗
+ψ
=−
ψ
∂t
∂t
2m
∂ρ
i
=−
ψ
∂t
2m

2

ψ∗ − ψ∗

2

ψ

2

ψ∗ − ψ∗

2

ψ

We can now dene the probability current as
J=

i
(ψ ψ ∗ − ψ ∗ ψ)
2m

(2
...
This demonstrates the conversation of probability
...
5)
ψ = ( |ψ| + i|ψ| θ) eiθ
i
J=
[|ψ| ( |ψ| − i|ψ| θ) − |ψ| ( |ψ| + i|ψ| θ)]
2m
i
=
(−2i θ)|ψ|2
2m

Hence we obtain the useful expression of
J = |ψ|2

m

θ

(2
...
An
interesting case to be considered is that of a decaying or evanescent wave of the form
ψ(x) = |ψ|e−qx

If this is substituted into (2
...
This means that a decaying wave
does not actually carry any probability with it; we can actually have decaying waves in
classically forbidden regions without breaking probability conservation
...
4

A3

Ehrenfest's Theorem

Given the results of Section (2
...

∂ ψ|
∂Q
∂ |ψ
Q |ψ + ψ|
|ψ + ψ| Q
∂t
∂t
∂t
∂ ψ|
∂ |ψ
∂Q
= i
Q |ψ + ψ| Q i
+ i ψ|
|ψ
∂t
∂t
∂t
∂Q
|ψ
= − ψ| HQ |ψ + ψ| QH |ψ + i ψ|
∂t
∂
∂Q
i
ψ|Q|ψ = ψ| [Q, H] |ψ + i ψ|
|ψ
∂t
∂t
i

∂
ψ|Q|ψ = i
∂t

This expression is known as Ehrenfrest's Theorem
...
7)

If Q commutes with H , then Q is a constant of motion
...
Suppose that at t = 0, |ψ = qi
...

Let us look at some specic, and interesting, cases of Ehrenfest's Theorem:
• The Hamiltonian (H ) - Evidently, [H, H] = 0, and so for a time-independent Hamilto-

nian, it is a constant of motion, meaning that energy is conserved
...
This makes sense,
as it takes energy to modify the kinetic and potential energies
...
2)
...


x| [x, p] |ψ = x| xp |ψ − x| px |ψ
= −i x

∂ψ
+i
∂x

x

∂ψ
+ψ
∂x

=i ψ
=i

x|ψ

Hence,
[x, p] = i

17

(2
...

i

∂
ψ| x |ψ = ψ| [x, H] |ψ
∂t
p2
= ψ| x,
+ V (x) |ψ
2m
1
=
ψ| [x, p2 ] |ψ
2m
1
ψ| ([x, p]p + p[x, p]) |ψ
=
2m
i
=
ψ| p |ψ
m

We thus obtain

p
∂
ψ| x |ψ = ψ| |ψ
∂t
m

(2
...

• The Momentum Operator (p) - As the rate of change of momentum (in Classical

physics) is equal to the applied force, we expect to nd the quantum analogue of
Newton's Second Law
...
10)

This is indeed the analogue of NII, with the force of the form F = − V
...

∂
E| xp |E
∂t
p2
= E| xp,
+ V (x)
2m

0=

|E

Calculating the commutators:
[xp, p2 ] = [x, p2 ]p + x[p, p2 ]
= 2i p2
[xp, V ] = [x, V ]p + x[p, V ]
= −i x ·

18

V

Toby Adkins

A3

Then,
0 = 2i

E|

p2
|E − i
2m

E| x ·

V |E

Re-arranging, this becomes
2 T = E| x ·

V |E

Suppose that the potential is of the form
V (x) = A|x|α

This could be, for example, a Coulomb potential with a = −1
...
11)

This is known as the Virial Theorem
...
Astute students will have noticed this relationship cropping
up a lot when dealing with gravitational forces and orbits
...


19

3
...
However, it can lead to some interesting results that
illustrate some of the weird behaviours that can result from the consideration of systems
quantum-mechanically
...
1

A3

A General Framework

This entire chapter focusses on nding solutions to the equation
−

h2 d2 ψ
+ V (x)ψ = Eψ
2m dx2

for some wave-function ψ in the position representation
...
2) for a particle with some initial energy E
...
We then impose the continuity conditions of ψ and ∂ψ/∂x
at the boundary, and solve the resulting equations for some sort of interpretable solution
...


3
...
1 Parity and Phase
The idea of 'parity' can be used to help signicantly in this context, as well as many others
...
For
now, we shall simply state that the Parity Operator P acts in the following way:
P Q(x) = Q(−x)

It can be shown that if
[P, H] = 0

the solutions outside of the nite, symmetric potential V (x) form a complete set of stationary states of H
...

Consider a particle travelling along the x-axis towards positive x that reaches a large square
potential step of the form
V =

Vo
0

for |x| < a
otherwise

Graphically, we can represent this as

Figure 3
...
This imposes the
condition that
B = −Bei(φ −φ)

Taking a linear combination of the odd and even parity solutions for x > a,
ψ(x) = ψe (x) + ψo (x)
= B sin(kx + φ) + B sin kx + φ
eikx
e−ikx
Beiφ + B eiφ −
Be−iφ + B e−iφ
2i
2i
eikx iφ
1 − 2ei(φ −φ)
Be
=
2i

=

Similarly for x < −a,
ψ(x) = ψe (x) + ψo (x)
= B sin(−kx + φ) − B sin −kx + φ
= eikx iBe−iφ +

e−ikx iφ
Be
1 + 2e2i(φ −φ)
2i

Thus, we can conclude that the amplitudes for the incoming, reected and transmitted
waves are
Ai = iBe−iφ
B
Ar = eiφ 1 + e2i∆φ
2i
B
At = eiφ 1 − e2i∆φ
2i

where ∆φ = φ − φ is the phase dierence between the odd and even parity solutions
...
1)
(3
...
Notice that these formulae for the transmission and reection
probabilities have been obtained without reference to the form of the wave-function within
the barrier; this means that they hold for any potential V (x) that has odd and even parity,
and vanishes outside some nite region
...

22

Toby Adkins
3
...
2: A potential step
Let us assume that E > Vo (the other case is left as an exercise to the reader)
...
The solutions for x < 0
will be of the form
ψ(x) = Ai/r e±ikx for k =

2mE
2

where Ai and Ar are the amplitudes for the incident and reected waves respectively
...
Imposing continuity at the boundary:
Ai + Ar = At
ikAi − ikAr = iKAt

It follows quickly that
k−K
k−K
Ar
=
−→ Pr =
Ai
k+K
k+K
At
2k
4kK
=
−→ Pt =
Ai
k+K
(k + K)2

2

Unlike in normal wave mechanics, transmission and reection coecients give the probability of transmission
...

Recalling (2
...
3)

This gives rise to the probabilities shown above
...

We can re-obtain a 'classical' result by looking at the limiting case for initial energies
E
Vo , k K , meaning that t = 1 as we would expect
...

23

Toby Adkins
3
...
3: A square potential well
Consider the bounded solutions for E < Vo
...
Outside the well:
ψ(x) = Ae±kx for k =

2m(Vo − E)
2

The odd parity solutions inside the well are of the form
2mE

ψ(x) = B sin Kx for K =

2

We need continuity at |x| = a in order for the solutions to be stable, and so the conditions
we obtain are
B sin Ka = Ae−ka
KB cos Ka = −kAe−ka

Dividing the second equation by the rst
cos Ka
ke−ka
= − −ka
sin Ka
e
K cot Ka = −k

K

2mVo

=−

2

2mVo

=−

2mE
2

− K2

W2
−1
(Ka)2

cot Ka = −

for W =

−

(2mVo a2 )/ 2
...


24

Toby Adkins

A3

Figure 3
...
Evidently, for an odd parity solution
to exist, the minimum value that Ka/π can take is 1/2
...

We can repeat the same calculation for the even parity solution, and we obtain the relationship that
W2
−1
(Ka)2

tan Ka =

This always has a solution, regardless of the values of W and a
...
Essentially, if we are able to 't' a square well within
the larger potential, then the larger potential must also have a bound state
...
3
...
Considering the solutions to the TISE in each of the regions, we
have the following wave-functions
• x < −a: ψ1 (x) = eikx + Re−ikx
• |x| < a: ψ2 (x) = AeiKx + B −iKx
• x > a: ψ3 (x) = T eikx

25

Toby Adkins

A3

where k and K have the same values are before
...

T eika = AeiKa + Be−iKa
ikT eika = iKAeiKa − iKBe−iKa

Eliminating A and B , as we are not interested in the solutions inside the well
...
4)
(3
...
4) and (3
...

1
e−ika + Reika = T eika
2
1
e−ika − Reika = T eika
2

1+

k
K

e−2iKa + 1 −

K
+ 1 e−2iKa −
k

k
K

e2iKa

K
− 1 e2iKa
k

Adding these two equations to eliminate R
...
Interestingly, this
also gives us the reection and transmission probabilities for a square potential step as in
Section (3
...
1) for the case where E < Vo by making the substitution that
K=i

2m(Vo − E)
= iκ
2m

26

Toby Adkins

A3

This will give a transmission coecient of
T =

2kκe−2ika
2kκ cosh(2κa) − i(k 2 − κ2 ) sinh(2κa)

and so a transmission probability of
Pt =

1
2

cosh (2κa) +

(k 2

−

κ2 )2 sinh2 (2κa)/(4k 2 κ2 )

Consider the limiting case of a very think potential barrier such that κa

1
...
This is actually the principle for the tunnelling electron
microscope
...
The distance between the tip
and the surface is varied, and due to the sensitive dependence of the expression on a, we
can obtain a very detailed picture of the surface of the substance
...
Instead, let
us nd the same results by considering parity
...
As [P, H] = 0, we know that the waves outside the well must be
stationary states
...

tan(ka + φ) = 0
tan ka + φ = ∞

−→
−→

27

ka + φ = nπ
ka + φ = (2n + 1)

π
2

Toby Adkins

A3

Thus,
∆φ = φ − φ =
rπ
2

Pr = cos2

rπ
2

=0

Thus, we were able to obtain the same result through much simpler algebra, though we do
not have explicit forms for the reection and transmission coecients
...
3
...
As Vo → ∞ with a xed, W → ∞, meaning
that the values of k that solve the dening equations for the odd and even solutions tend
to k = nπ/2a and k = (2n + 1)π/2a respectively
...
6)

for n = 0, 1, 2, 3,
...
We can infer from this that wave-functions vanish at the edges of a region with
innite potential energy
...
7)

n2

for n = 1, 2, 3,
...
4)
...
The Correspondence
Principle dictates that for high energies, and thus high n, that the quantum and classical
results must agree
...
For this, the odd parity solutions will remain
the same
...

2a

1
nπx
√ sin
2a
a

dx x2

x2 =
0

2

1
a3
32π 3 n3 − 12πn
·
a 24π 3 n3
4
1
= a2
−
3 2(πn)2

=

Thus
4
1
−
3 2(πn)2
1
1
−
3 2(πn)2

σ 2 = a2
= a2

− a2

Classically, the probability density function must also be normalised, and equal to a constant
...

2a

x

c

dx x

=
0

1
√
2a

2

=a
2a

x2

c

dx x2

=
0

1 x3
2a 3
4
= a3
3
1
2
σc = a2
3

1
√
2a

2

2a

=

0

2
Thus, it is clear that in the limit as n → ∞ that σ 2 = σc , conrming our expectations
...
4

A3

The Dirac-Delta Well

Consider a very narrow well with a potential of the form V (x) = −Vδ δ(x)
...

ε

−

ε

d2 ψ
2m
= 2
2
dx
−ε
dψ ε
2m
−
= 2
dx −ε
dx

Vδ ψ(0) + E

dx ψ
−ε
ε

dx ψ

Vδ ψ(0) + E
−ε

The last integral will vanish as ε → 0, and so we obtain the condition
dψ
dx

ε

=−

2mVδ

−ε

2

ψ(0)

(3
...
Again,
as [P, H] = 0, we can consider stationary states outside the well
...
Imposing the boundary conditions at x = 0 for the even parity
solutions
...

2mVδ
2k cos φ = − 2 sin φ
tan φ =

2k

mVδ

Doing the same for the odd parity solutions
...
It follows that
2k

∆φ = φ − φ = − tan−1

mVδ

Thus, it follows that the transmission probability is given by
Pt = cos2 (φ − φ)

2
mVδ

=
4 k 2 + m2 V 2
δ
→ Pt =

m2 Vδ2
1+ 4 2
k

30

−1

Toby Adkins

A3

Inside the well, we expect solutions to exponentially decay, and so they will be of the form
ψδ (x) = Ae

Kx

where the negative sign applies for x = 0+
...
8):
−KA − KA = −
−2K = −
K=

2mVδ
2

A

2mVδ
2

mVδ
2

Letting ψδ (x) = Ae−Kx into the TISE for x > 0 and re-arranging, we nd that
Eδ = −

mVδ2
2 2

Thus, the energy of the particle is dependent on the depth of the well
...
Using symmetry
arguments, nd the implicit equations for the energy eigenvalues in the odd and even cases
...
This means that we will have
solutions of the form
even
odd

x < −a
AeKx
AeKx

−a < x < a
B cosh(x)
C sinh(x)

x>a
Ae−Kx
Ae−Kx

where K = 2m|E|/ 2
...
As in (3
...
For the
even case, let us impose the boundary conditions at x = a:
Continuity in ψ : B cosh(Ka) = A−Ka
Discontinuity in ψ : kB sinh(Ka) + KAe−Ka =

2mVδ
2

B cosh(Ka)

Solving these equations, and their odd counterparts, simultaneously will yield:
tanh(Ka) =

W
−1
ka

±1

where the positive sign corresponds to the even solutions, and W = 2m|E|/ 2
...


31

4
...

Almost any system near equilibrium is at least approximately harmonic as one can expand
the potential energy as a Taylor series around this equilibrium point, and the linear term
is zero by construction
...


32

Toby Adkins
4
...
Let us
suppose that a system moves under a well-dened potential V (x) through a point of stable
equilibrium xo
...

V (x)

V (xo ) + V (1) (xo )(x − xo ) +

1 (2)
V (xo )(x − xo )2 +
...

V (1) (xo ) must vanish, meaning that the potential can be approximated by
V (x)

1 (2)
V (xo )(x − xo )2
2

In this case, let us assume that V (x) is a quadratic potential, and that is equal to zero at
the equilibrium point xo = 0
...
2), it becomes
H=

p2 + (mωx)2
2m

(4
...
This will allow us to nd the time evolution of the system assuming that
the system was initially in a linear combination of these energy eigenstates
...
1
...
2)

creation

(4
...
Consider the
operator product A† A
...

33

(4
...
5)

This is all just groundwork for the derivations in the next sections, so do not worry if it
feels a little disjointed
...
2

A3

Stationary States

Suppose that we have a stationary state |E such that
H |E = E |E

Apply (4
...
The
application of the creation operator thus raises the energy of the system (excitation) by
ω ; hence it's name
...
However, this seems to
imply that we can continue lowering the energy of the system arbitrarily
...

E = E| H |E
1
m2 ω 2
E| p† p |E +
E| x† x |E
2m
2m
|p |E |2 + m2 ω 2 |x |E |2
=
2m
≥0
=

This means that the annihilation operator A can only act assuming that A |E = 0
...
This means

that the energy eigenvalues are given by

En =

n+

1
2

ω

(4
...
This is why A† and A are
often called 'ladder operators'
...
2
...
We saw that when the creation operator is applied,
the states satisfy
A† |n = α |n + 1

for some constant α
...
Thus, we obtain the very useful relations that
1
A† |n
n+1
1
|n − 1 = √ A |n
n
|n + 1 = √

(4
...
8)

These equations can be remembered quite easily as dividing through by the square root of
the highest ladder number involved in the operation
...
2
...
9)

This is known as the number operator
...

N (H |n ) = N (En |n )
(A† A)H |n = En (A† A) |n
A† (A(H |n )) = En A† (A |n )
√
A† (H |n − 1 ) = En nA† |n
(H |n ) = n(En |n )

Thus, the number operator does not modify the state of the system; it simply returns the
number of the energy eigenstate that the system is currently occupying
...
2
...

2mωx
A + A† = √
2m ω
(A + A† )
2mω
= (A + A† )

x=

where we have dened
given by

/(2mω)
...
The rst term will contribute
entries on the upper diagonal, and the second will contribute entries to the lower diagonal
...

0
√
 1 0
2 √
0
0
...

0


=  0
3 √
0
4
...





xjk


...


...


...



...


...


...



...


...


Similarly, using the fact that
p=

i
(A† − A)
2

we arrive at the expression
Pjk =

i
2

√

nk+1 δj,k+1 −

√

nk δj,k−1

This gives us the matrix
√
0
0
0
0
√ − 1
√
 1
0
− 2
0
0

√
√
 0
2
0
− 3
0
i 
√
√
=
 0
0
3
0
− 4
2 
√
 0
0
0
4
0



pjk


...


...


...



...


...


...



...


...


...



...




...
3

A3

Wave-functions of Stationary States

We want the nd the position representation of these energy eigenstates
...
Now, remark that x|0 is of the form of a Gaussian
...
That is;
x|n = (A† )n x|0

In order to do this, it is helpful to write A† in a more useful form
...
10)

As an example, let us apply this to the ground-state to obtain the wave-function of the
rst excited state
...
11)

2

where Hn (x) are Hermite polynomials, as we saw in the Mathematical Methods notes
...
3
...
Looking back at the results
of Section (4
...
3), we have already shown that both x = p = 0 as the diagonal elements
of the matrix are equal to zero
...

Let us also examine the second moments of position and momentum
...

1
V = mω 2 x2
2
1
En
= mω 2
2
mω 2
1
= En
2

1
p2
2m
1
=
mEn
2m
1
= En
2

T =

We re-obtain the classical result as expected
...
4

A3

Dynamics of Oscillators

If a system is dynamic (that is, it is moving in a classical sense) then it cannot be in a
state of well-dened energy by denition
...
Suppose that the system is initially in some state
an |n

|ψ, 0 =
n

Then the time-evolution of the system is given simply by
1

an e−i(n+ 2 )ωt |n

|ψ, t =
n

Let us now nd the expectation value of position
...

a∗ −1 an
n

x =

√

n e−iωt +

a∗ an−1
n

√

n eiωt

n

n

Doing another arbitrary relabelling, we arrive at
√

x =

(4
...
If we let 2 n a∗ an−1 = Xn eiφn
n
where both Xn and φn are real, then it follows that
x =

Xn cos(ωt + φn )
n

Thus, we have obtained sinusoidal oscillations, with each of the initial stationary states
oscillating at the same frequency ω regardless of their amplitudes an
...

By analogy, the expectation value of momentum in the same general state is given by
p =

i
2

√
n a∗ an−1 eiωt − a∗ an e−iωt
n
n−1
n

As an example, let the initial state of the system be
|ψ, 0 =

1
1
1
|N − 1 + √ |N + |N + 1
2
2
2

40

(4
...
Using (4
...
Let X be the maximum
displacement of the oscillator
...
In this case, let us assume that the initial state of the system is
1
|ψ, 0 = √
K

for N

K

N +K−1

|k
k=N

1
...
12):
N +K−1 √

x =

k a∗ ak e−iωt + a∗ ak−1 eiωt
k−1
k

k=N
N +K−1 √

=

=

K
2
K

k eiωt + e−iωt

k=N
N +K−1 √

k cos(ωt)
k=N

We can use the fact that K N to say that all K of the
all close enough to this large number N
...
If the correspondence principle does not appear to
initially hold, it is usually because the problems being considered are in fact not equivalent,
rather than a fault in Quantum Theory!
41

Toby Adkins

A3

4
...
1 Heisenberg's Uncertainty Principle
Another thing to be considered with an oscillator is it's uncertainty relation, as this gives
us constraints on x and p for a dynamical system
...
3
...
14)

This is what most people will commonly recognise as Heisenberg's Uncertainty Principle
...


Angular Momentum

This chapter aims to cover the basics of Angular Momentum in Quantum Mechanics,
including:
• Symmetries and Conservation Laws
• Orbital Angular Momentum
• Spin Angular Momentum
• Composite Systems
• The Hydrogen Atom

Hopefully, the previous chapters will have built in the reader a solid understanding of the
basic concepts and manipulations that are involved in Quantum Mechanics
...
This is a particularly important area, as it will allow us to better understand
the behaviour of particles, as well as more complex systems such as the Hydrogen atom
...


43

Toby Adkins
5
...
Let C be
some operator that performs such an invariant transformation
...

If a system in invariant under C , we say that the system posses the symmetry corresponding
to C
...
This means that for the two states |ψ = C |ψ and|φ = C |φ , we require
that
ψ φ

2

= | ψ|φ |2 −→

ψ |φ = ψ|φ
ψ |φ = φ|ψ

This means that C is either unitary or anti-unitary
...

Now let us consider the action of C on the TISE
...

Applying the operator to both sides of (2
...


5
...
1 Parity
The parity operator P is dened by the equation
P x|ψ = −x|ψ

(5
...
Finding it's eigenvalues:
P |ψ = λ |ψ
P 2 |ψ =

P

|ψ = λ2 |ψ

Unitary

This means that it's eigenvalues are λ = ±1
...
We met this concept in Section (3
...
A property that
results from this is that concerning vector operators
...

ψ| v |ψ = − ψ| P vP |ψ = −(±1) ψ| vP |ψ = −(±1)2 ψ| v |ψ

This means that the expectation value of any vector quantity in a state of well-dened parity
is zero
...
For example, consider a particle trapped in an
innite square well, and it's classical analogue of a ball bouncing back and forth between
two walls
...

44

Toby Adkins

A3

5
...
2 Continuous Symmetries
Suppose that our transformation operator C now depends on some parameter θ
...
If
δθ is small, we can expand our operator as
C(δθ) = I − iδθτ + O(δθ2 )

We require that T is hermitian, and so
I = C † (δθ)C(δθ) = I + iδθ(τ † − τ ) + O(δθ2 )

This means that we require our complex coecient τ to be Hermitian, and so it might
correspond to an observable
...
Suppose that θ = N δθ
...
2)

= e−iθτ |ψ

Dierentiating both sides of this equation with respect to the parameter θ:
i

∂
ψ
∂θ

=τ ψ

Thus, τ gives the rate of change of the original state of the system as we vary our transformation parameter θ
...
With this in mind, consider again the commutator of our transformation C with
the Hamiltonian
...
This means that we usually do not work with the transformation itself, but
rather the associated generator
...
1
...
How does the state change when we make the shift x → x − a?
x − a|ψ = x|ψ − a ·

∂ψ 1 2 ∂ 2 ψ
∂
+ a
+ · · · = exp −a ·
2
∂x 2 ∂x
∂x

i
x|ψ = exp − a · p

x|ψ

T (a)

Our transformation operator for our parameter a is thus T (a) as shown, with the generator
of the transformation being the momentum operator p
...

ψ p ψ

= ψ| T † pT |ψ = ψ| p |ψ

45

Toby Adkins

A3

as the generator must commute with the operation by denition
...

ψ x ψ

= ψ| T † xT |ψ = ψ| T † T x + T † [T, x] |ψ = ψ| x − a |ψ

As one could have guessed, the expected value of position is changed simply by the original
translation that was made
...
1
...
Then, by direct analogy, the
transformation operator will be given by
i
U (α) = exp − α · J

(5
...
This can be thought of as an operator associated
with angular momentum, though we can only substantiate this claim by saying that in
Classical Mechanics, dynamical symmetry about some axis implies that the component of
angular momentum about that axis is conserved
...
Consider the state |ψ = U |ψ ,
and the expectation value in this state of:
• Some scalar operator a
...
Then, we can describe the action of U on the expectation
value of v by a single rotational operator R:
ψ v ψ

= R(α) ψ| v |ψ −→ U † vU = Rv

where the second expression follows from the fact that this must hold for any state
|ψ
...
Then, using a small angle
expansion of U to rst order:
i
i
I + δα · J v I − δα · J
i

= v + δα × v

[δα · J, v] + O(δα2 ) = δα × v
[vi , Jj ] = i

ijk vk

as this must hold for arbitrary δα
...
Since δα is an arbitrary vector, the
invariance of δα · J under rotations implies that under rotations the components of
J transform like those of a vector, meaning that we can simply substitute vi = Ji
into the last relationship
...
A summary of these results is shown
in the box below
...
4)
(5
...
We can use the
above commutation relations to nd the associated eigenvalues, though we will need more
specic information about the form of J to be able to nd the eigenfunctions
...
We are going to choose |j, m to
be our ket that is simultaneously an eigenket of Jz and J 2 , such that
Jz |j, m = m |j, m

and J 2 |j, m = β

2

|j, m

Note that our labelling of the eigenvalue of J 2 is arbitrary; we have simply chosen to call
it β
...
6)

J± = Jx ± iJy

These evidently commute with J 2 , meaning that they have no eect on the total angular
momentum
...
7)

[Jz , L± ] = [Jz , Jx ] ± i[Jz , Jy ] = i Jy ± i(−i Jx ) = ± (Jx ± iJy ) = ± J±

The kets J± |j, m are eigenkets of J 2 with eigenvalue j(j + 1)
...
We then write
J± |j, m = α± |j, m ± 1

where α± are some (possibly complex) coecients
...
2
...
All that can stop us doing this is the vanishing of α+ when we reach
some maximum eigenvalue mmax , and similarly for J− and α−
...
This also means that −j < m < j ; there
is thus a 2j + 1 degeneracy in m for each value of j
...

Jz |j, m = m |j, m
2

2

J |j, m = j(j + 1)
J± |j, m =

|j, m

j(j + 1) − m(m ± 1) |j, m ± 1

(5
...
9)
(5
...
Suppose that we can model a
diatomic molecule as two hard spheres connected by a light spring that is aligned along
the z -axis
...
1: A diatomic molecule with it's axis aligned with the z -axis
We can then write the Hamiltonian in the form
H=

2
2
Jy
Jx
J2
J2
+
+ z =
+ Jz
2Ix 2Iy
2Iz
2Ix

1
1
−
2Iz
2Ix

as Ix = Iy
...

|j, m

This means that it is very hard to create rotation around the axis of the molecule (aligned
with the z -axis), and so we can eectively set m = 0
...
11)

j(j + 1)

It is quite easy to show using the centre of mass that Ix = µs2 , where µ is the reduced
mass of the molecule, and s is the typical separation of the two molecules
...
It's energy is then given by
2

ω = Ej − Ej−1 =

Ix

j

If we are given two dierent frequencies of photons that are emitted, and their corresponding transitions, we can approximate the spring constant for the molecule by nding the
change in the force F = µsω 2 and the change in the distance s, as k = ∆F/∆s
...
We know from (1
...
As Jx = 0, σJx = Jx
...
This means that angular momentum does in fact
satisfy the uncertainty relation
...
As we shall see in following sections, L corresponds to orbital angular momentum (that is analogous to the classical angular momentum that we are used to), while S
corresponds to the intrinsically quantum-mechanical eect that is spin angular momentum
...
12)

This means that our rotation can be decomposed into a rotation associated with the orbital angular momentum, and a rotation associated with the spin angular momentum
...

The last important thing to note about this decomposition is that due to linearity, both
L and S obey the same commutation relations, and thus eigenvalue equations as J
...
1
...
We will make the
changes j → and j → s respectively
...
2

A3

Orbital Angular Momentum

We are rst going to take a look orbital angular momentum, as it has an already quite
familiar classical analogue
...
13)

L = r × p = −i r ×

The forms of each of the components of L are thus very easy to work out in Cartesian
coordinates by evaluating the above cross product
...
It can be shown that the polar coordinate representations of
these operators are
∂
∂φ

Lz = −i
L2 = −

2

(5
...
15)

Note that the former of these can be proven by observing that
∂
∂x ∂
∂y ∂
∂z ∂
=
+
+
∂φ
∂φ ∂x ∂φ ∂φ ∂φ ∂z

and using the polar coordinate representations of x, y, z in comparison to the Cartesian
denition of Lz
...
2
...
However, this has to be a single valued function, due to rotational symmetry,
meaning that m (and thus ) must be an integer
...

We now want to nd the eigenfunctions of the operator L2
...
1
...
We are going to consider the eigenfunctions
of the state | , , and work backwards from here
...
Then:

Y =0
Y
...
16)

Successive applications of the lowering operator L− to this expression should yield all
other eigenfunctions
...
It can be shown that the
eigenfunctions of L2 are given by the spherical harmonics
(5
...
By the denition of m, we have
2 + 1 possible eigenfunctions for a given value of , and so we have to specify both and
m when denoting a particular eigenfunction
...
However, we often nd that normalisation is
irrelevant, as we are more interested in determining the values of and m based on the
form of the eigenfunctions
...
What are the possible measurements of
Lz and L2 ? Give the probabilities of each outcome
...
This can be done by observing
that
1
2
sin2 θ = 1 − cos2 θ = − (3 cos2 θ − 1) +
3
3

This means that
θ, φ|ψ ∝ −

1
3

√
16π 0 2 √
Y2 +
4π Y00 ∝ −Y20 + 5 Y00
5
3

Thus, clearly a measurement of Lz will always yield zero, though we could have read this
o immediately from the fact that the angular part of the wavefunction is independent of
5
φ
...

6
We are now going to consider the parity of the spherical harmonics, as this can often
become very useful in order to simplify integrals, and in other such calculations
...
Recalling Equation (5
...
This means that the parity of a general state is given by
P | , m = (−1) | , m

(5
...
2
...
Our rst instinct would be to dene the radial momentum operator as p · r; however, this clearly does not work as this would make pr manifestly not
Hermitian
...
19)

Squaring up this expression:
p2 = −
r

2

∂
1
+
∂r r

∂
1
+
∂r r

=−

2

∂2
2 ∂
1
1
+
− 2+ 2
2
∂r
r ∂r r
r

2

=−

r2

∂
∂r

r2

∂
∂r

We note how this is the radial part of the Laplacian in spherical polar coordinates
...
20)

Angular

This means that we re-obtain the classical result, but instead dealing with Hermitian
operators
...
3

A3

Spin Angular Momentum

Spin is an intrinsic form of angular momentum carried by elementary particles who's existence was inferred from experiments, such as the Stern-Gerlach experiment (discussed
below), in which particles are observed to possess angular momentum that cannot be accounted for by orbital angular momentum alone
...

As previously stated, S shares all the commutation and eigenvalue relations with J
...
3
...
One particularly common way of doing this is by nding their matrix representation
...
Recall that for some state |s, m , m can
take values −s < m < s
...


...



...


...


...



...


...


...



...


This matrix will always have an odd number of diagonal entries (2s+1), where the 'middle'
entry will always be zero
...
However, we will quote the results for the
s = 1 matrices here for reference:



0 1 0
Sx = √ 1 0 1
2 0 1 0




0 −1 0
i
Sy = √ 1 0 −1
2 0 1
0

53




1 0 0
Sz = 0 0 0 
0 0 −1

Toby Adkins

A3

Pauli Spin Matrices
In this section, we will consider the case of the above results for s = 1 ; as most of matter
2
is made up of spin- 1 particles, we will often be working in this regime
...
21)
(5
...
Suppose that the eigenstates of Sz are |+ and |− , often referred to as spin-up and
spin-down states
...
This state vector is known
as a spinor
...
Then:
n·σ =

cos θ
sin θe−iφ
iφ
sin θe
− cos θ

This has eigenvectors corresponding to eigenvalues ±1 of
v+ =

cos θ/2 e−iφ/2
sin θ/2 eiφ/2

and v − =

− sin θ/2 e−iφ/2
cos θ/2 eiφ/2

Then, in the basis (|+ , |− ), we can write the states of a spin-half particle in which the
measurement of a component of spin along n is certain to yield ± 1 as
2
|+, n = sin θ/2 eiφ/2 |− + cos θ/2 e−iφ/2 |+
|−, n = cos θ/2 eiφ/2 |− − sin θ/2 e−iφ/2 |+

Let us look at some cases of these results
...
For θ = π , there is complete certainty to measure the spin
in the negative z direction, as n(π) = −z
...
23)

Toby Adkins

A3

5
...
2 Spin and Magnetic Fields
As we have seen in the A2 course, magnetic dipoles are inuenced by magnetic elds
...

When this is placed in an external magnetic eld B , it experiences a torque µ × B
...
24)

The Hamiltonian lacks any kinetic energy term, as if the particle is initially at rest (we
can move to a frame in which it is), there is no change in kinetic energy as the external
magnetic eld does no work on the particle
...
It is found that at t = 0, the particle is in an eigenstate of |+, x of Sx
...

The rst step is to evaluate our expression for the Hamiltonian
...
The initial state of the system in the basis (|+ , |− ) is
1
|ψ, 0 = |+, x = √ (|+ + |− )
2
˙
As H = 0, the time evolution of the system is given by (2
...

0 1
|ψ, t =
e2iE0 t/ + e−2iE0 t/ = cos(γB0 t)
1 0
4
2
i
0 −i
=
ψ, t| σy |ψ, t =
ψ, t|
|ψ, t =
e2iE0 t/ − e−2iE0 t/ = − sin(γB0 t)
i 0
2
2
4
2

Sx =
Sy

2

ψ, t| σx |ψ, t =

2

ψ, t|

From the eigenvectors of Sx , it is clear that in the basis (|+ , |− ) that
1
|−, x = √ (|+ − |− )
2

Thus, the probability of being along −x is
P−x = | −, x|ψ, t |2 = sin2

1
γB0 t
2

All of these results describe the procession of the spin of the particle in the magnetic eld
...
3
...
It involved the use of an inhomogeneous magnetic eld (in
the z direction) that results in a force being exerted on anything with a magnetic dipole
moment, given by
F = µz

∂Bz
∂z

If we had a continuous spectrum of values for the magnetic dipole moment, we would
expect to observe a continuous spread of beams coming out of the apparatus
...
This was the rst experimental
demonstration of the eect of spin
...
2: A schematic diagram of the Stern-Gerlach apparatus
The Stern-Gerlach experiment can be best understood by realizing that it eectively counts
as a measurement, and so the wavefunction is collapsed into whatever eigenstate we measure it to be in
...
We can
nd the transmission amplitudes for subsequent lters with inner products - for example,
say we pass a beam of spin-up particles through a lter aligned along n that transmits
plus along the positive n direction
...
3
...


A beam of spin-one particles emerges from an oven and enters a Stern-Gerlach lter that
passes only particles with Jz =
...
What is the probability
that a particle makes it through all three lters?
Before the lter, all three spins are equally likely, and so the probability of transmission for
a given particle through the rst lter is P1 = 1
...

 
1
1 √ 
Sx |+, x = |+, x −→ |+, x =
2
2
1

Then, the probability of transmission through the second lter is
  
1
1
1 √   
0
P2 = | +, x|+ |2 =
2
2
0
1

2

=

1
4

=

1
4

Similarly, for the third lter
  
1
0
1 √   
0
P3 = | +, x|− |2 =
2
2
1
1

2

Thus, the probability that any one particle passes through the whole system is given by
Pt = P1 P2 P3 =

57

1
48

Toby Adkins
5
...


5
...
1 Product States
In some cases, we can write the state of the composite system as a product of the two
separate systems
|ψ = |A ⊗ |B
(5
...

Pai bj = | ai bj |ψ |2 = | ai |A |2 | bj |B |2 = Pai Pbj

In such states, each system behaves independently of the other
...
We can write their separate states as
|A = a1 |+
|B = b1 |+

A

+ a2 |−

A

B

+ b2 |−

B

The product state is then given by
|ψ = a1 b1 |+, + + a1 b2 |+, − + a2 b1 |−, + + a2 b2 |−, −

Suppose that we wanted to nd the probability that A is in |+ given that B is in |−
...
This is why
product states are said to be uncorrelated or separable states
...
Let
|a, b = |a |b describe the state of the combined system
...
For two independent
systems, we must have that [HA , HB ] = 0 as they cannot modify one another's equations
of motion, and thus their energy
...
4
...
In this case, the systems must be interacting
...


where the last term is the interaction Hamiltonian that describes how the two sub-systems
inuence one-another
...
|a, b
∂t

(5
...
, and so governs how the motion of each particle is modied in time
...
, HA ] = 0; neither are constants of motion, as energy is continuously
transferred between systems A and B
...
Their
interaction Hamiltonian is given by
Hint
...
Calculating the matrix elements with respect to this basis, we nd that


1 0
0 0
2 

0 −1 2 0
S1 · S2 =
0 2 −1 0
4
0 0
0 1

This allows us to nd the eigenvalues and eigenvectors of the interaction Hamiltonian as
State

Energy Eigenvalue

|+, +
1
√ (|+, − + |−, + )
2
|−, −
1
√ (|+, − − |−, + )
2

A/4
A/4
A/4
−3A/4

Total z component of spin
0
−

0

Thus, the energy eigenstates consist of a triplet of levels at E = A/4 and a singlet level at

E = −3A/4
...
4
...
Suppose that we have two angular momenta denoted by quantum
numbers j1 and j2 that lead to a combined angular momenta denoted by the quantum
2
2
number J with z -component M
...
This means that we can
know
• j1 , m1 , j2 , m2 and M but not J

OR

• j1 , j2 , M and J but not m1 or m2

We know that M = m1 + m2 ≤ j1 + j2
...
the number of states will increase until we
reach M = |j1 − j2 |
...
, j1 + j2

(5
...
27), we multiplicity of the combined angular momentum is then given by
j1 +j2

j1 +j2

J

g(J) =

2j2

M=
J=|j1 −j2 | M =−J

2(j1 − j2 + n) + 1

2J + 1 =
n=0

J=|j1 −j2 |
2j2

= (2j2 + 1)(2j1 − 2j2 + 1) + 2

n = (2j1 + 1)(2j2 + 1)
n=0

We hence nd that
g(J) = g(j1 + j2 ) = g(j1 )g(j2 )

(5
...


Clebsh-Gordan Coecients
We can write the total state of the system as
j1

j2

|J, M =

CJM |j1 , m1 , j2 , m2
m1 =−j1 m2 =−j2

where CJM = j1 , m1 , j2 , m2 |J, M are known as the Clebsch-Gordan coecients
...


A box containing two spin-1 objects A and B is found to have angular momentum quantum
numbers J = 2 and M = 1
...


60

Toby Adkins

A3

Figure 5
...
We can
nd this state by using
J− = J1− + J2− = (Jx1 − iJy1 )(Jx2 − iJy2 )

on the state |J, J
...
10)
...
Then:
1
|2, 1 = √ (|1, 0 |1, 1 + |1, 1 |1, 0 )
2

This means that the probabilities that we require are
P (m = − ) = 0
P (m = ) = 1/2
P (m = 0) = 1/2

where we have simply read o the coecients from the state vector above
...
5

A3

The Hydrogen Atom

We have spent the vast majority of this chapter building up our mathematical and physical
apparatus concerning angular momentum in Quantum Mechanics; it is now to time to put
this into practise by looking at the hydrogen atom
...

We are going to ignore the eects of relativity, spin and magnetism
...
5
...
This
classical model is of course wrong, but gives a lot of correct answer to some simple questions
...

µ=

me mn
me + mn

In most circumstances, we can write that µ me as mn
me even for just a single
proton
...
Classically, we can write the energy of
this system as
E=

L2
Ze2
−
2µr2 4π 0 r

If we minimise this energy curve, with the assumption that L = n , we nd that the most
probable radius is
r = aµ

Note that aµ =

me
µ a0 ,

n2
Z

(5
...
30)

This is known as the Bohr radius, and is the standard unit for length in atomic physics
...
31)

Putting these results together, we nd that that the energy levels of the system are given
by
Z2
1
En = − µ(αc)2 2
2
n

(5
...
6 eV
...

Interestingly, as we shall see, this turns out to be the correct answer when you treat the
same problem quantum mechanically
...
5
...
Then, we can write the Hamiltonian of the system as
H=

p2
p2
e
+ n + V (re − rn ) =
2me 2mn

P2
2M

p2
+ V (r)
2µ

+

centre of mass motion

relative mass motion

We can arbitrarily move to a coordinate system in which there is no motion of the centre
of mass
...
2
...
Therefore, there exists a complete set of mutual
eigenkets of H , L2 and Lz , which we use to denote the states of hydrogenic atoms
...
This
means that we can automatically write that
r, θ, φ|n, , m = r|n,
radial

θ, φ| , m = R (r) Y m (θ, φ)
angular

Substituting this trial solution into the TISE:
H r, θ, φ|ψ = E r, θ, φ|ψ
Hr +

L2
RY = ERY
2µr2
1 2
1
L Y
= 2µr2 E − HR
Y
R

function of θ, φ only

function of r only

Evidently, as we know that the angular eigenfunctions are the spherical harmonics, we
have an easy choice of separation constant
...
Letting y = 2λρ, we arrive at
yL (y) + L (y) [2( + 1) − y] −

+1−

1
L(y) = 0
λ

This dierential equation can be solved by the methods covered in the Mathematical
Methods course, but the solutions turn out to be associated Laguerre polynomials
L2
k

+1

(y) for λ =

1
k+ +1

where k sums over integers
...
We call n the principle quantum number
...
When normalising
these functions, remember to integrate over r, θ and φ
...

Let us quickly consider an interesting property of the spherical harmonics in the context of
an 'orbiting' electron
...
This means that the only fair
assumption we can make is that the electron as an equal probability of being in any of the
states for a given , θ and φ
...


Radial Wavefunctions
Evidently, one does not have to be able to recall all of the radial wave-functions, but the
one worth remembering is that of the ground-state, given by
1
r|1, 0, 0 = √
π

Z
aµ

3/2

e−Zr/aµ

(5
...
Note that this had been normalised over all space
...
Some graphs of the lower order radial wavefunctions
are shown in the gure overleaf
...
There are
n − 1 nodes for each value of n, and so as you increase n, you will obtain more frequent,
and sharper peaks
...


Size of Orbit
Let us nish by calculating the expectation value of r to give us an idea of the typical
2
size of a 'circular' orbit
...
This means that L0 +1 = 1,
and we have a much simpler radial wavefunction of the form
R(r) ∝ rn−1 e−Zr/(naµ )

64

Toby Adkins

A3

Figure 5
...
34)

aµ

Similarly, it can be shown that
r2 = n2 (n + 1) n +

1
2

a2
µ

This means that for large n, r ∝ n2 while σr ∝ n3/2
...
In this limit, we recover the Bohr model
...


|n, , m

for


1 ≤ n

0≤ ≤n−1


− ≤m≤

(5
...
36)
(5
...
38)

It is, to varying degrees of accuracy, often a good approximation to use hydrogenic formulae
for dierent atoms - however relativistic eects start to become important for helium and
anything more complex
...
= µ(αc)2 2
2
n
2
E rel
...
The fractional error is given by
δE
E rel
...

=
E rel
...


1 v
2 c

2

α2

10−4

So the fractional error isn't particularly large when neglecting relativistic eects in the
hydrogen atom
...


Perturbation Theory

This chapter aims to extend the knowledge gained in the preceding chapters by introducing
the various aspects of Perturbation Theory, including:
• Time-Independent Perturbation Theory
• The Variational Principle
• Time-Dependent Perturbation Theory
• Selection Rules and Transitions
• Atoms in a Weak Magnetic Field

Perturbation Theory is used to calculate the changes that a system experiences when the
Hamiltonian is modied in some way
...
These systems are relatively 'boring'; perturbation
theory is where the interesting Quantum Mechanics actually lies
...
1

A3

Time-Independent Perturbation Theory

Let us begin by considering time-independent changes to our Hamiltonian, for which we will
have to consider solutions to Equation (2
...
Suppose that we can write our Hamiltonian
in the form
H=

+

H0
original Hamiltonian

δH
small perturbation

This will give rise to some changes in energy that we will denote by
E=

+

En

δEn

+
...


perturbed eigenfunction

In order for these approximations to be valid, they must be a solution to the TISE
...
) = (En + δEn + δ 2 En +
...
)

We now equate the various orders of terms that appear in this equation, as we know that
each order must satisfy separate equalities as we can arbitrarily get rid of certain orders
by scaling the size of the perturbation term
...
1)
(6
...
Let us nd the rst order correction in the energy
...
2):
H0 En |δEn + En | δH |En = En En |δEn + δEn En |En

This gives the rst order change in the energy as the matrix element
δEn = En | δH |En

(6
...
4)

Toby Adkins

A3

The second order change in the energy is then given by the expectation value of the
perturbation given the rst order change in energy:
δ 2 En = En | δH |δEn =
n=m

En | δH |Em Em | δH |En
En − Em

This gives the simple expression of
δ 2 En =
n=m

| En | δH |Em |2
En − Em

(6
...
In some cases, one can actually nd
the exact change in the energy or the eigenfunctions; these can then be Taylor expanded
to show agreement with the predictions of Perturbation Theory
...
1
...

•

When treating the hydrogen atom, we have so far assumed that the nucleus is pointlike, when it fact it has nite size
...

We need to nd our perturbing Hamiltonian in order to calculate these energy
changes
...
3) with n = 1 for the ground-state:
δE = E1 | δH |E1

Ze2
=
2π 0 ap

Z
a0

3

ap

dr r2

0

2ap
r2
−3+
a2
r
p

e−2Zr/a0

We know that ap /a0 104 1
...

δE

Ze2
2π 0 ap

Z
a0

3

ap

dr r2

0

69

2ap
r2
−3+
2
ap
r

ze2 Z 3
2π 0 ap 5

ap
a0

2

Toby Adkins

A3

The ground-state energy for hydrogen is given by
|E1 | =

Z 2 e2
8π 0 a0

This means that we can write our value for the rst order change in the energy as
4
δE = |E1 |
5

Zap
a0

2

Evidently, as ap /a0 10−4 , this correction in the energy is vanishingly small
...

•

Suppose that a Quantum Harmonic Oscillator is subject to a perturbing potential
δH = λx3
...

We can consider x3 to be a vector operator
...
2); the terms in the
expansion of x3 all have un-even powers of A† and A, meaning that the resultant
states are orthogonal to |En
...
As A |0 = 0, the only terms that remain in the expansion
2
of x3 are those with A† or higher
...
5), we nd the second order correction to the energy as
δ 2 En = λ2

6

6
9
+
2
(E0 − E3 )
(E0 − E1 )2

= 11

λ2 6
ω

As expected, the correction to the energy scales with both the constant λ, as well as
the typical scale of the system
...
1
...
We can see that the sums in Equation (6
...
5) have a denominator that will
clearly diverge for m = n
...
This means that small perturbations can give large changes in the state,
and we have to abandon the assumption that the change in the wave-function is small in
comparison to the original wave-function; that is, our Taylor expansion is no longer valid
...

In order to be able to use the techniques of perturbation theory on degenerate states, we
need to move to a basis in which the perturbed state is diagonal, as this means that we
will not have to worry about degeneracy
...
Express the perturbation Hamiltonian δH in the basis of all new degenerate vectors
...
Evaluate all the matrix elements to give a Hermitian matrix
...
Diagonalise this matrix to nd the eigenvalues and eigenvectors
...


The Quadratic Stark Eect
This is a very common example of a problem involving degenerate perturbation theory
...
The perturbation to the system will then be
δH = eE z = eEr cos θ

(6
...

The perturbing matrix is thus


2, 0, 0| δH |2, 0, 0
 2, 1, 0| δH |2, 0, 0
δHij = 
 2, 1, 1| δH |2, 0, 0
2, 1, −1| δH |2, 0, 0

2, 0, 0| δH |2, 1, 0
2, 1, 0| δH |2, 1, 0
2, 1, 1| δH |2, 1, 0
2, 1, −1| δH |2, 1, 0

2, 0, 0| δH |2, 1, 1
2, 1, 0| δH |2, 1, 1
2, 1, 1| δH |2, 1, 1
2, 1, −1| δH |2, 1, 1


2, 0, 0| δH |2, 1, −1
2, 1, 0| δH |2, 1, −1 

2, 1, 1| δH |2, 1, −1 
2, 1, −1| δH |2, 1, −1

This looks like a horrible matrix to evaluate, but in fact we have some tricks up our sleeve
...
Recalling
Equation (5
...
Also, remarking that [Lz , z] = 0:
Lz z |n, , m = zLz |n, , m = m z |n, , m

This means that z |n, , m is an eigenket of Lz , meaning that only terms where m = 0 are
able to contribute
...
It is easy to show that the new eigenvalues are
3eEa0 , 0, 0, with corresponding eigenstates
1
√ (|2, 0, 0 ± |2, 1, 0 ), |2, 1, 1 , |2, 1, −1
2

Note how two of the eigenstates have remained the same; these are unaected by the
perturbing Hamiltonian, and so we only observe the 'mixing' of the two states eected
...
2

A3

The Variational Principle

This is alternative method to perturbation theory that allows us to obtain some upper
bound estimate for the ground-state energy given that we do not have explicit solutions
for the form of the wave-functions for the new system
...
7)

E0 ≤ ψ| H |ψ

We can use this in conjunction with Rayleigh's Theorem :

The stationary points of the expectation value of an Hermitian operator
occur at the eigenstates of the operator
...
Thus, if we take |ψ to be a reasonable guess of the wave-function with some
adjustable parameter, we can then minimise ψ| H |ψ (assuming that |ψ is normalised)
in order to obtain an upper bound on the ground-state energy of the system
...

As this is a spherically symmetric wave-function, the angular components will cancel in
the normalisation
...
We now need to consider the kinetic energy term
...
19):
|pr |ψ |2 =

2

dr r2

∂
1
+
∂r r

2

ψ

=

2

dr r2

72

∂ψ
∂r

2

+

∂
r|ψ|2 =
∂r

2

dr r2

∂ψ
∂r

2

Toby Adkins

A3

where the last equality follows from evaluating the second term in the integrand, assuming
that the wave-function is well dened in space
...

Now for ψ = e−br
...
By Rayleigh's Theorem, we want to minimise
this expression:
∂ H
:
∂b

2

m

b−

me2
e2
= 0 −→ b =
= a−1
0
4π 0
4π 2 0

We thus re-obtain the un-normalised form of the ground-state wave-function with the Bohr
radius
...
32)
...
7) can also be used to prove the useful result that any potential well has at
least one bound state
...
Let Vsq > Vw be the potential describing a square well that 'ts'
inside the potential Vw
...


73

Toby Adkins
6
...
4) only holds for the case where the Hamiltonian in time-independent, meaning that we need another method to nd the time evolution
of states if H = H(t)
...
8)

where En and Em are the energy eigenvalues of two energy levels n and m respectively
...
We shall
detail the main methods used in the following sections
...
3
...
Suppose that
the system is initially in a state that satises some time-independent Hamiltonian Hi , which
is suddenly (almost instantaneously) changed to another time-independent Hamiltonian
Hf
...
Suppose that the eigenstates
of Hf are labelled by |En with corresponding eigenvalues En
...
9)

This means that calculations using the sudden approximation are typically very easy, as
they simply involve calculating matrix elements
...
Assuming that the decay occurs over a short time interval, calculate the
probability that this helium ion is in the 1s state
...
Let at be the amplitude for the transition
...
702

As we will see later, the probability of the atom remaining in the same state is much greater
than the probability that it undergoes a transition to any other state over a short time
interval
...
3
...
In
this case, the change in the system occurs on a time-scale t τH
...

Consider an interval [t, t + δt] for small δt
...
By analogy
to Equation (6
...
Make the substitution that
s = t/τ so that s = 0 initially, and s = 1 nally
...
This means that am is constant, and hence that the system remains
in the eigenstate that it was in initially
...
For t > 0, the frequency slowly increases until a time t1 where it
has reached a value 2ω0
...

Recalling Equation (4
...

According to the adiabatic principle, the oscillator is still in the n = 1 state, except the
frequency is now 2ω0
...


6
...
3 A Time-Dependent Perturbation
In this case, we assume that the original time-independent Hamiltonian H0 is modied by
some small time-dependent variation δH(t)
...
Substitute
this into the TDSE:

e−iEn t/ (i an + n an ) |n =
˙
E
n


e−iEn t/ an ( + δH |n )
H0 |n
n

i

an e
˙

−iEn t/

n

Bra through by m| and

e−iEn t/ an δH |n

|n =
n

1 iEm t/
i e

am = −
˙

:
i

ei(Em −En )t/ m| δH |n an
n

Suppose that the system is initially in the state n
...
This means that the
amplitude of the transition to some state m is given by
am
˙

i
− ei(Em −En )t/ m| δH |n

(6
...
At long times, however, all bets are o
...

With an explicit expression for δH , one can then nd expressions for the probability of a
transition n → m, which we shall denote by Pnm
...
In this case, we use the following steps
...
Find the matrix corresponding to the perturbed Hamiltonian
...
Diagonalise this matrix to nd the eigenvectors and eigenvalues
...

3
...

4
...

5
...


A spin-half particle of magnetic moment µ is travelling at a speed v under the inuence of
a magnetic eld of magnitude B that is orientated along z
...
Given that the Hamiltonian satises
H(t) =

−µ(Bσz + bσx ) for 0 < t < /v
−µB
otherwise

and that the initial spin-state of the particle was |+ , nd the probability that it transitions
to |−
...
Consider the perturbed Hamiltonian:
0 1
1 0
− µb
1 0
0 −1

H = −µB

= −µ

B
b
b −B

As above, let us diagonalise this to nd the eigenvalues and eigenstates:
B−λ
b
!
= 0 −→ λ = ±µ
b
−B − λ

b2 + B 2

By the denition of some eigenvector (α, β):
B−

√

b2 + B 2
b
√
b
−B − b2 + B 2

α
β

=0

This means that
r=

α
B
=
±
β
b

1+

B
b

2

2B
b

where we have used the fact that b << B
...

|+ = c1 + + c2 − =

c1 −

2b
b
c2 |+ + c2 +
c1 |−
B
2B

77

Toby Adkins

A3

This means that
|ψ, 0 = + −

b
−
2B

Note that this state is already properly normalised as we can neglect the contribution from
√
the second term
...

b
−|ψ, 0 =
e−iEτ / − eiEτ /
2B

b
= sin
B

√
µ b2 + B 2 τ

b
sin
B

µB

τ

Thus, the required probability for the transition is
Pt =

b
B

2

sin2 (

µB

τ

Let us now turn to perturbation theory
...
For ωnm = (Em − En )/ , we
can write Equation (6
...
11)

In this case, it is clear that Λnm = µB , and that ωnm = 2µB
...
Notice how using perturbation theory is
signicantly faster; approximation is (nearly) always faster than not!

78

Toby Adkins
6
...
Let
δH(t) = V0 e−iωt

Let ωnm = (Em − En )/ , meaning that our amplitude for the transition can be written as
am = −

i

dt eiωnm t m| δH |n = −

1

m| V0 |n

ei(ωnm −ω)t − 1
ωnm − ω

with associated probability
Pnm =

ωnm −ω
t
2
2
ωnm −ω
2

sin2

| m| V0 |n |2
2

transition cross section σ(t)

For a given t, σ(t) is dominated by a bump around the origin that is of height t2 and width
2π/t
...
Thus, the rate at which the transition n → m occurs is given by
νmn =

2π
2

| m| V0 |n |2 δ(ωnm − ω)

(6
...

It is very easy to see that for δH = V0 eiωt , there will be a transition to a new state that
is lower in energy by ω
...
There
are two possible cases:
1
...
However, there may be a range of ω in the incoming energy, with
associated density of states g(ω)
...
The energies may vary continuously (meaning a continuous range in Em ), but ω of the
incoming energy remains constant
...

νnm =

dEm g(Em )νnm =

Note that in this last expression, a factor of
energy
...
4
...
As λ a0 for most incident radiation, we can
consider the strength of the electric and magnetic elds to be roughly constant over the
transition cross-section
...
This is known as
the electric dipole approximation
...

Suppose that the system is orientated such that the direction of the electric eld is along
the z -axis
...
12), the corresponding transition rates are given by:
νnm =

2π
2

e2 E 2 | m| z |n |2 g(ωnm )

The relative magnitude of these expressions essentially translates to how strong an absorption or emission line is in a particular spectrum
...


6
...
2 Selection Rules
As we had already seen in the case of the transitions considered in Section (6
...
2), not all
transitions are possible due to parity and symmetry arguments
...
Note that we will be working in the regime of the electric dipole approximation
...
18) that the eigenstates of hydrogen are of well-dened parity
satisfying
P | , m = (−1) | , m

We also know that the expectation value of any vector operator is zero in a state of welldened parity
...
This automatically forces us to
conclude that
| − |=1

in order for the state to be one of mixed parity
...
In a similar argument used before, as [Lz , z] = 0, we can write
Lz (z |n, , m ) = zLz |n, , m = m (z |n, , m )

This means that z |n, , m is an eigenket of Lz , and is therefore orthogonal to all other
eigenkets of Lz
...
It follows quickly from the commutation relations
of angular momentum with vector operators that [Lz , x± ] = ± x±
...
Given that x and y can both be
written in terms of x± , we conclude that the matrix elements for x and y are zero unless
|m − m | = 1

for orthogonality reasons
...
1
...

In any case, as summary of the important results is shown in the box below
...

∆ = ±1
∆m = 0 for z
∆m = ±1 for x, y

(6
...
14)
(6
...

Suppose that the photon is emitted in the same direction as the imposed electric eld
...
In fact, ∆m = 1 corresponds to left-hand
circularly polarised light, while ∆m = −1 corresponds to right-hand circular
...
This means that linear polarisation can only be observed in
the x-y plane, as linear polarisation requires no angular momentum
...
This argument is bogus
...
left-handed has higher energy)
...


81

Toby Adkins
6
...
We will now bring
this back into consideration
...
This leads
to a problem, as it means that we cannot simply write it as a potential that we can then
include in our Hamiltonian
...

H=

1
(p − qA)2 + qφ
2m

(6
...
Note that as we have change the momentum term in
the Hamiltonian, our expression for probability current must be modied
...

J=

Using the same simplication as in Section (2
...
17)

A particle moves in the x-y plane with a uniform magnetic eld in the z -direction, represented by
A=

B
(−y, x, 0)
2

Show that the function
φ(x, y) = exp −[x2 + y 2 ]/4

2
B

is an eigenfunction for a suitably chosen value for B , which you should nd
...

This problem has to be treated component-wise
...
Then:



px − γy
p − qA = py + γx
pz

Evidently, φ(x, y) is an eigenfunction of pz with eigenvalue zero
...

H=

1
1 2
(p − qA)2 =
(p + p2 + γ 2 (x2 + y 2 ) − γpx y − γypx + γpy x + γxpy )
y
2m
2m x

Then, consider derivatives of our wave-function φ(x, y)
...
17)
...
This has a maximum around r = B , and goes to zero for r → ∞
...


6
...
1 Gauge Transformations
We have a choice of the potentials φ and A that we can use to represent E and B but
introducing some scalar gauge f (t)
...
Consider the
gauge transformation:
1
(p − qA )2 + qφ
2m
= eif q/ |ψ

H→H =
|ψ → ψ

We can now substitute these into the TDSE
...

83

Toby Adkins

A3

6
...
2 The Classical Limit
Let us examine whether our form of the Hamiltonian given by Equation (6
...
By Ehrenfest's
Theorem, for some operator Q:
∂2 Q
1
=
[H, [H, Q]]
2
∂t
(i )2

If we take Q = xk , where xk is a particular component of the position of the particle, we
can nd some sort of analogue of Newton's Second Law
...
Now for the second commutator:
i
1
(pj − qAj )2 , pk − qAk + [qφ, pk − qAk ]
m 2m
∂Aj
∂Ak
∂φ
(−i )2 q
(pj − qAj )
−
−q
=
m
m
∂xk
∂xj
∂xk

[H, [H, xk ]] = −

Putting this all together:
m

∂ 2 xk
=q
∂t2

ij

(vj × (

× A) ) − q

∂φ
= (q(v × B) − qE) · k
∂xk

We have thus re-produced the familiar Lorentz Force result for the k component of position,
meaning that we should have at least a little condence that our Hamiltonian is along the
right lines
...
5
...

We dene the Bohr magneton as
µB =

e
2m

(6
...
This will have Hamiltonian
H=

p2
+ V (r)
2m

1
(p − qA)2 + V (r) =
2m

−

1
(2qp · A − q 2 A · A)
2m

Original HamiltonianH0

Now assume that the magnetic eld is weak
...

H

H0 −

H0 , and so

q
p·A
m

We shall consider the magnetic eld to be uniform, as it is very dicult to create a eld that
has signicant variation in space over the size of an atom
...
This means that the total dipole moment for an electron in an
external magnetic eld is given by
µ=−

µB

(6
...
We will also assume that
the eect of this magnetic eld is small in comparison to the spin-orbit interaction (see
Section (6
...
4))
...
Consequently, when
we use perturbation theory to calculate the smaller eect of an imposed magnetic eld,
the degenerate eigenspace in which we have to work is that spanned by the states that
have given values of j , and s but dier in their eigenvalues mj of Jz
...
The rst order change in the
energy is thus given by
δE =

µB

B j, mj , , s| (Lz + 2Sz ) |j, mj , , s =

µB

B (mj + j, mj , , s| Sz |j, mj , , s )

where we have used the fact that Jz = Lz +Sz
...
This means that be obtain:
j, mj , , s| Sz |j, mj , , s =

We can then write that
J ·S =

J · S mj
j(j + 1)

1 2
J − L2 + S 2
2

(6
...
21)

It is clear from this equation that the Zeeman eect splits the degenerate energy levels
corresponding to a given j into 2j + 1 levels, proportionally to the applied magnetic eld
...
1: Zeeman splitting for = 1 and s = 1/2
...
This also gives rise to pairs of lines in emission spectra; they are split by the
weak magnetic eld of the Earth
...
Thus, we obtain the familiar energy shift of δE = ±µB B
...
However, suppose that we neglect the
spin-orbit interaction; this means that we can no longer treat it as being small, and so
cannot ignore the torque that it applies to the system
...
This means that J 2 is no longer a constant of the motion,
and so we characterise our states by L2 , Lz , S 2 and Sz
...
In general, this will be the case that we are dealing with,
as the spin-orbit interaction is not on syllabus
...
5
...
Suppose that the potential is of the form φ(r)
...

δHSO = −µ · B = −

µB
S · (v ×
c2

φ) = −

e
r ∂φ
S· p×
2 c2
2m
r ∂r

86

=−

e
1 ∂φ
S·L
2 c2
2m
r ∂r

Toby Adkins

A3

where we have observed that L = r × p
...
This means that we can use the J 2
and Jz basis, labelled by the kets |j, , m
...
For hydrogen, the rst of these is
1
r3

=

2
1
( + 1)(2 + 1) (na0 )3

The second can be calculated by recognising that
1
S · L = (J 2 − L2 − S 2 )
2

(6
...
23)

In a similar way to the Zeeman eects, the spin orbit interaction splits the energy levels
labelled by j into the 2j + 1 energy levels
...
1)
...
The
'spin-up' states are shifted by an amount
δE+ ∝ j+ (j+ + 1) − ( + 1) −

3
∝
4

+

1
2

+

3
2

− ( + 1) −

3
∝
4

Likewise, the 'spin-down' states are shifted by an amount
δE− ∝ j− (j− + 1) − ( + 1) −

3
∝
4

−

1
2

+

1
2

− ( + 1) −

3
∝ −( + 1)
4

For each value of j , the spin-orbit interaction creates another 2j + 1 energy levels (due to
degeneracy), and so the mean energy shift for the system is proportional to
¯
δ E ∝ (2j+ + 1)δE+ + (2j− + 1)δE− ∝ 0

Thus, the mean energy shift for a system under the spin-orbit interaction is zero
...

87

7
...
4), but here will extend
this fully to systems involving multiple particles, where we take account of all aspects of the
wave-function
...


88

Toby Adkins
7
...
Let us now 'swap' the two
particles
...

| |a, b |2 = | |b, a |2
|b, a = eiφ |a, b

If we now swap the particles in the second ket, we obtain
|b, a = e2iφ |b, a
eiφ =±1

This means that there are two possible exchange symmetries :
1
...
|a, b = − |b, a for fermions that have half odd-integer spin
Suppose now that that the states "a" and "b" are in fact the same state
...
This is known, quite famously, as the Pauli Exclusion Principle
...
1
...
In general, for a system of N identical particles, the Hamiltonian will be of
the form
N

H=

Hi

+

Hint
...
= 0 (the particles are non-interacting), then the TISE can simply be solved via
separation of variables, as the states are uncorrelated
...
When solving a partial dierential equation like the above, the general
solution is usually a superposition of factored solutions
...
Then we can write the spatial wavefunction as
1
ψ(r1 , r2 ) = √ [φ1 (r1 )φ2 (r2 ) ± φ1 (r2 )φ2 (r1 )]
(7
...
Notice
how the wave-function disappears in the case where φ1 = φ2 ; this is the Pauli Exclusion
Principle in action
...
How many distinct states of the system are possible? The best way to answer
this equation is combinatorially
...
Then for fermions, we have to
89

Toby Adkins

A3

nd the number of ways of putting p particles into n states, where as for bosons we instead
need to nd a way to divide up p particles into n states using n − 1 'divisions'
...
In the case of
bosons, we need to create a wave-function that is completely symmetric under the exchange
of particles
...
For
fermions, however, there is a much more systematic way of computing the wave-function,
given orbitals φ1 ,
...


1
φ1 (r2 )
ψ (r1 ,
...


...



...



...


...


...


...
2)


...
We know from linear algebra that a determinant
is antisymmetric under row swapping, and this is analogous to the swapping of particle
coordinates, so the correct symmetry is preserved
...
1) for the N = 2 case
...

Let us now bring our attention to the issue of spin
...

Consider two spin-half particles of angular momentum quantum numbers j1 and j2
...
Let us denote
the total angular momentum by J with z -component M , and the total spin by S
...

Symmetric:


|+, +

|ψ = |−, −
 1
√
(|+, − + |−, + )
2
1
2

Antisymmetric: |ψ = √ (|+, − − |−, + )

M = 1, S = 1
M = −1, S = 1
M = 0, S = 1
M = 0, S = 0

These are known as the triplet and singlet states respectively, which we have already
encountered in Section (5
...
2)
...
Looking at the values of S for each of the states listed above, we
can conclude that S = 1 implies a symmetric spin wave-function, and an anti-symmetric
spatial wave-function, while S = 0 implies an antisymmetric spin wave-function, and a
symmetric spatial wave-function
...
1
...
This itself is determined by the
symmetry of the spin wave-function
...
We can calculate this using rst order perturbation theory
...

δH = V (|r1 − r2 |)

Equation (6
...
1)
...
3)

d3 r1 d3 r2 φ∗ (r1 )φ∗ (r2 )φ2 (r1 )φ1 (r2 ) V (|r1 − r2 |)
1
2

(7
...

Assuming that V (|r1 − r2 |) is large and positive for a small separation (such as a Coulomb
potential), then δEexchange is positive
...


7
...
3 Systems of Fermions and Bosons
Suppose that we have two identical macroscopic particles that are each made up of NF
fermions and NB bosons
...

Let the spins of the particles be labelled by quantum numbers s1 and s2 , while the total
spin of the system be S
...
This means that depending on whether s1 and s2 are integers or halfintegers, the spin of the composite system with either be an integer or half integer, as
shown in the table below
...
2

A3

The Helium Atom

The Helium atom is the simplest multi-particle system, consisting of two electrons and a
nucleus that provides a xed Coulomb potential
...

2

H=−

2m

(

2
1

+

2
2)

kinetic energy term

−

Ze2
4π 0

1
1
+
r1 r2

e2
1
4π 0 |r1 − r2 |

+

coulomb interaction with nucleus

(7
...
This is usually to ignore the interaction term between the two electrons,
as the equation then simply becomes separable in r1 and r2
...
In this case,
Equation (5
...

1
He
En = − µ(αc)2 Z 2
2

1
1
2 + n2
n1
2

(7
...
In this case, we have that:
n1 = n2 = 1
1

=

2

=0

m1 = m2 = 0
1
1
s1 = , s2 = −
2
2

The spatial wave-function is thus symmetric, as all of the quantum numbers are the same
apart from the spin part
...
Conventionally, this state is notated as 1s2 1 S 0
...
The signicance
of each of the terms used is detailed in the gure below
...
1: An explanation of spectroscopic notation
For L = 0, we use S , and then P , D and F for the successive values of L
...

92

Toby Adkins

A3

7
...
1 The Electron Interaction
We will now estimate the eect of the electron-electron interaction term in Equation (7
...

We shall assume that the atom is in the ground state, meaning that each electron has a
wave-function given by Equation (5
...
From rst order perturbation theory:
δE1 =

d3 r1 d3 r2 |ψ(r1 , r2 )|2

e2
e2 Z 6
= 3 6
4π 0 |r1 − r2 |
4π 0 a0

d3 r1 d3 r2

e−2Z(r1 +r2 )/a0
|r1 − r2 |

We can evaluate this integral using the result that
d3 r1 d3 r2

e−(r1 +r2 )
= 20π 2
|r1 − r2 |

Then:
δE1 =

a5
e2 Z 6
0
4π 3 0 a6 32Z 5
0

d3 r1 d3 r2

e−(r1 +r2 )
5
= Zµ(αc)2
|r1 − r2 |
8

This means that the energy shift (from perturbation theory) due to the Coulomb interaction
between the two electrons in the ground-state is
5
δE1 = Zµ(αc)2
8

5
RZ
4

(7
...
8 eV
4
2
The actual value is −79 eV, meaning that our rst-order estimate is not too inaccurate
...
2 eV (from calculation) higher in energy than

the triplet state due to the exchange energy dierence
...
Let the individual particle Hamiltonians be H1 (Z) and H2 (Z), with the
variational parameter in our wave-function being Z
...
The expectation value of the rst two terms is simply
given by Equation (5
...
7) above
...
5 eV,
which is a more accurate result that that obtained previously
...

min

93

Z
a0

Toby Adkins

A3

7
...
2 Higher Excited States
For a single particle in a Coulomb potential, the energy only depends on the principle
quantum number
...
Let us consider the potential seen by an electron for
dierent values of r
...
At large radius (r a0 ), the potential is dominated by the interaction
between electrons, as the other screen the nucleic charge and thus reduce the strength of
the potential
...
This is due to the conversation of angular momentum; small radii
imply large momenta, which prevents the electron from getting close to the nucleus if is
large: L = r × p ∝ r

...

Consider the rst excited state of helium
...
This means that the conguration state of
the system is 1s1 2s1
...
This means that
we can write the two possible states as
symmetric :
antisymmetric :

1s1 2s1 3 S 1
1s1 2s1 1 S 0

However, we can also consider cases of non-zero angular momentum
...
This state, notated as 1s1 2p1 , is higher in energy than 1s1 2s1
...


94

Toby Adkins
7
...
Initially, it will be assumed that the electrons only experience a Coulomb
interaction with the nucleus; we are neglecting the electron-electron interaction
...
Let
us consider the number of states associated with each value of n
...

• n=2:

= 0, m = 0, ms = ± 1
2
= 1, m = −1, 0, 1, ms = ± 1
2

This gives rise to 8 states, accounting for Li to Ne
...
We would thus expect 18 elements across the third row
...
For example,
it turned out that n = 4, = 0 is lower in energy than n = 3, = 1 because the electrons
are on average found closer to the nucleus, and are more eected by nucleic screening
...
3
...
In general, we want to take the highest spin state allowed by the Pauli
Exclusion Principle as this minimises the exchange energy, and then take the highest L
consistent with this
...
= 1, m = −1, 0, 1,
ms = ±1/2
...
This means that we have
a symmetric spin state, and an antisymmetric spatial state, giving the lowest Coulomb
energy: M = 0, L = 0, J = 3/2

---