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Part A: Mathematical Methods

Toby Adkins

June 20, 2016

Contents
1 Matrices and Operators

1
...
2 Linear Operators
1
...
1 The Identity Operator
1
...
2 Change of Basis
1
...
3 Hermitian Operators
1
...
4 Manipulations of Operators
1
...
3
...
3
...
3
...
4 Functions of Operators
1
...
1 The Exponential Operator

3

4
6
7
7
8
9
10
10
11
11
12
12

2 Function Spaces and Transforms

14

3 Sturm-Liouville Operators

28

2
...
1
...
1
...
2 Fourier Series
2
...
1 The Dirichlet Conditons
2
...
2 Symmetry and Periodicity
2
...
3
...
3
...
4 The Dirac Delta Function
2
...
1 Properties of the Dirac Delta
2
...
2 Fourier Series Representation
2
...
3 A basis for Fourier Space

3
...
2 Denition and Properties
3
...
1 Eigenvalue Properties
3
...
3
...
3
...
3
...
4 Known Dierential Equations
3
...
1 Legendre's Dierential Equation
3
...
2 Hermite's Dierential Equation
3
...
3 Bessel's Dierential Equation
1

15
16
16
18
19
20
22
23
24
25
26
26
26
29
29
31
32
32
32
32
33
33
34
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Toby Adkins

Mathematical Methods

3
...
1 Characteristics of the Solution
4
...
1 The Characteristic Equation
4
...
3 General Solutions to Common PDE's
4
...
1 The Wave Equation
4
...
2 The Heat Diusion Equation
4
...
3 Laplace's Equation
4
...
4
...
4
...
5 Lagrange Multipliers
4
...
1 Lagrange Multipliers in Classical Mechanics

5
...
1
...
1
...
1
...
1
...
2 Some Probability Distributions
5
...
1 Binomial
5
...
2 Poisson
5
...
3 Exponential
5
...
4 Normal
5
...
5 The Central Limit Theorem
5
...
6 Chi-Squared
5
...


Matrices and Operators

This chapter aims to re-familiarise the reader with the linear algebra of matrices, and
extend this to operators
...


3

Toby Adkins

Mathematical Methods

1
...
In order to familiarise the reader with it, we are going to introduce
it here, and use it throughout the remainder of these notes (where appropriate)
...
It can be thought of as a vector that contains all the
information of our system for some state ψ
...
)
α |ψ = (α a1 , α a2 ,
...
A basis set of kets |i is such that
any ket can be written as a linear combination of said basis kets:
|ψ =

ai |i
i

Now consider the adjoint space V of V
...
One can convert between the original space and the adjoint space
by taking the complex conjugate of the coecients of the ket, and changing the ket to a
bra
...
1)
We can thus dene a bra by a the following relationship:
(1
...
Consider ψ|φ :
a∗ i|
i

ψ| =
i

|φ =

bj |j
i


a∗ i| 
i

ψ|φ =


bj |j 

i

j

a∗ bj i|j
i

=
i,j

a∗ bj δij
i

=
i,j

a∗ bj
i

→ ψ|ψ =
i

Thus, it is clear that the operation ψ|φ is analogous to the inner product ψ, φ
...


Let us start by using the simplifying assumption that u|u = v|v = 1 (proper normalisation)
...
If we now let
|u =

|a
a|a

and |v =

|b
b|b

in this inequality, the result clearly follows
...


5

Toby Adkins

Mathematical Methods

1
...
For example, a linear operator could act on a ket to produce
another distinct ket
...
3)

In general, throughout the remainder of this chapter, we will not refer to operators being
linear explicitly; it shall be assumed from this point that all operators being dealt with are
linear unless something is stated to the contrary
...
Then, it is easy to show that any operator can
be written as a matrix
...

|φ = Q |ψ
bk |k = Q

aj |j
j

k

aj Q |j

‘=
i

bk =

i|Q|j aj
ij

We let Qij = i|Q|j be the matrix that corresponds to our operator Q
...
Then
Qij = qi |Q|qj
= qj qi |qj
= qj δij
→ Qij = qi

Thus, it becomes a diagonal matrix in this basis, as we have seen with matrices that are
written in their eigenbasis
...
4)

qi |qi qi |

Q=
i

We dene the adjoint operator Q† by the equation
φ|Q|ψ = ψ| Q† |φ

6

∗

Toby Adkins

Mathematical Methods

Using this denition, we can nd the matrix representation of Q†
...
5)

Hence, the matrix for Q† is the complex conjugate transpose of the matrix for Q
...
2
...
6)

|i i|

I=
i

To show that this is indeed the identity operator, apply it to some arbitrary state
...


1
...
2 Change of Basis
Suppose we want to make the basis transformation [i, j] → [i , j ]
...

j

Uij |i

=
i

7

Toby Adkins

Mathematical Methods

We want to restrict the transformation matrix such that the new basis it creates is orthonormal
...
Consider the transformation applied to kets:
q

= i q
= i |I|q
=

i j

j|q

j

=

i j qj
j

We know that Uij = i|j
...


1
...
3 Hermitian Operators
Hermitian operators obey the dening equation
[ φ|Q|ψ ]∗ = ψ|Q|φ

(1
...
Let Q be a Hermitian operator with eigenvalues
8

Toby Adkins

Mathematical Methods

qi and eigenvectors |qi
...
The left-hand side
will vanish as Q is hermitian, and so
(1
...
Thus, the eigenvalues are real

• For k = i, the eigenvalues are distinct
...


As we will see in Quantum Mechanics, the latter property is a logical necessity of the way
that we interpret the mathematics of quantum states, as the probability of being in one
state given another is by denition zero (see A3 notes)
...
2
...
Operators also obey the property that
(Q1 Q2 )† = Q† Q†
2 1

which can be shown by using the matrix representation for the operators
...

Assume that for the operators Q1 , Q2 , Q3 ,
...
Qk )† = Q†
...

(Q1 ,
...
Qk )(Qk+1 )]†
= (Qk+1 )† (Q1
...
Q†
1
k+1 k

using the inductive assumption
...
Qn )† = Q† Q†
...
9)
n n−1
1
is true for all n ≥ 0

9

Toby Adkins

Mathematical Methods

1
...
10)

[A, B] = AB − BA

Assuming that this is non-zero, this is also itself an operator, as linear operators form a
closed space, meaning that their linear combinations are also within said space
...
We never want to have to compute the
commutator if we can avoid it, so remembering and applying these relations is often the
key to the successful manipulation of operators
...
3
...
Then,
n
n

(A + B) =
k=0

n k n−k
A B
k

(1
...
The n = 0 case is trivial
...
For n = j + 1:
(A + B)j+1 = (A + B)(A + B)j
j

= (A + B)
k=0
j

=
k=0

j
Ak B j−k
k

j
Ak+1 B j−k +
k

j

k=0

j
Ak B j−k+1
k

We are able to make this last step because of the fact that the operators commute
...
This theorem can be used to prove a large variety of results involving
operators
...
3
...
Then, there exists a basis in which the matrices
representing A and B are both diagonal if and only if [A, B] = 0
...

ˆ
ˆ
• → If A and B are the diagonal forms of the matrices of the operators A and B

respectively, then by denition

ˆ
A = i|A|j = ai δij
ˆ
B = i|B|j = bi δij

Consequently,
ˆ ˆ
ˆˆ
ˆˆ
[A, B] = AB − B A
= ai bi δij δij − bi ai δij δij
=0

as all these terms commute
...

[A, B] = (AB − BA)ij
Aik Bkj − Bik Akj

=
k

ai δij Bkj − Bik ai δij

=
k

= (aj − ai )Bij
!

0 = (aj − ai )Bij

This means that for a non-zero matrix B , i = j and so B is diagonal
...
3
...
Let A be a linear operator in an n-dimensional vector space
...
As they are non-degenerate
by denition, we can construct an orthonormal basis
...
We will
not prove this result here as it is relatively tedious to do so
...
4 Functions of Operators
Now let us consider a complex valued function f (x), for x being some complex number
...
12)

f (qi ) |qi qi |

f (Q) =
i

It has the same eigenkets as Q, with the eigenvalues being the function evaluated on the
original eigenvalues
...

2!
3!

which follows from the Taylor series expansion of f (Q)
...
3, we can now nd an expression for
[f (B), A]
...
, A
2!

1
f2 [B 2 , A] +
...

2!
= f1 [B, A] +

At this stage, we have to make the assumption that [[B, A], B]
...

2

We thus arrive at the useful result that:
∂f
∂B

[f (B), A] =

(1
...
4
...

∞
exp(Q) = lim

n→∞

I+

1
Q
n

n

=
n=0

Qn
n!

(1
...
15)

This can be proven using (1
...

What does the exponential matrix represent? We can best demonstrate this by considering
an example
...
We observe
that G2 = −I , → G2m = (−1)m I and G2m+1 = (−1)m G
...

By considering
det(I + εQ)

show that
det(exp Q) = exp(tr Q)

We want to expand the determinant for this small parameter ε
...
in (e
i1 i2
...
(en + εQn )in

1 2

e
...
in (e e
...
en Q22 + · · · + e1 e2
...
14) for nite n
n
1
Q
n
n
1
= det 1 + Q
n
n
1
= 1 + tr Q +
...

n

n

det(exp Q) = exp(tr Q)

We have thus shown the desired result, not that it is particularly enlightening
...


Function Spaces and Transforms

This chapter aims to cover the basics of function spaces and transforms, including:
• Function Spaces
• Fourier Series
• Fourier Transforms
• The Dirac Delta Function
• Laplace Transforms

The techniques of this chapter are used extensively throughout most of Physics, and so it
is imperative that readers are very familiar and comfortable with the techniques in this
section
...
Students may nd that
many of the results shown in this Chapter are exactly analogous to many found in linear
algebra; this is because function spaces obey many of the rules and results of vectors and
matrices
...
1 Function Spaces
A function space is the set of functions that map a particular set X (domain) to a particular
set Y (co-domain), dened over some interval [a, b]
...
This is the space of
w
square integrable functions, namely that
b

dx |f (x)|2 w(x)

a

(2
...
Here, w(x) is known as the weight function for the function
w
space; essentially, it determines how "heavily" we sample a function within an interval
...
Note that in all of these
cases, w(x) → 0 at x = ±∞ in order for (2
...
In this function space,
the inner product is given by:
b

f |g =

dx f (x)∗ g(x)w(x)

a

(2
...

We know that for n orthogonal vectors, they are automatically linearly independent
...
Suppose that there is a non-trivial solution to the equation
∞

αi |ui = 0
i=0

Now, bra through by uj |
...
This means that the only term that remains is
αj uj |uj = 0

However, uj |uj ≥ 0 by denition of the inner product, and so we require that αj = 0
for this relationship to hold
...
We thus recover what we are used to with normal vectors
...
1
...
2
...
Using the denition of the
identity operator in function space:
∞

f |Q|g =

dx f |x
−∞
∞

x|Q|x x|g

d
+ x2 g(x)
dx
−∞
∞
∞
dg
=
dx f ∗ x2 g +
dx f ∗ −i
dx
−∞
−∞

=

∞

=
−∞
∞

dx f (x)∗ −i

dx f ∗ x2 g + [−if ∗ g]∞ + i
−∞

d
+ x2 f ∗
dx
∞
∞
d
=
dx g ∗ −i
+ x2 f
dx
∞
= [ g|Q|f ]∗
=

∞

dx g
∞

df ∗
dx

dx g i

∗

Thus, the operator Q is indeed hermitian
...
e integrating over x)
...
1
...

As long as these functions span L2 , then they form a basis of the space
...

∞

αi |ui

|f =
i=0

16

Toby Adkins

Mathematical Methods

What are the coecients αi ? We can exploit the fact that these functions are orthogonal
to nd them:
b

uj |f =

dx w(x) uj (x)f (x)
a
∞

b

dx w(x) uj (x)

=
a

αi ui (x)
i=0

∞

b

=

dx w(x) uj (x)ui (x)

αi
a

i=0
∞

=

αi δij
i=0

→ uj |f = aj

Once again, this is analogous to nding the coecients when writing a vector in a particular basis
...
We can use the Gram-Schmidt
procedure (see CP3 notes) to nd the orthonormal basis
...
Then:
1
1|1 = √
π

∞

2

dx e−x = 1

−∞

→ eo (x) = 1
e1 (x) = x − 1|x
1|x = 0
1
x|x = √
π
√
→ e1 (x) = 2x

∞

2

dx e−x x2 =

−∞

e2 (x) = x2 − 1 x2 −
1
1 x2 = √
π

∞

√
2

2x x2

dx e−x x2 =

−∞

1
2

√

2x

1
2

1
e2 (x) = x2 −
2
∞
1
1
2
e2 e2 = √
dx e−x (2x2 − 1)2 =
2
4 π −∞
1
→ e2 (x) = √ (2x2 − 1)
2

Evidently, this orthonormal basis is not unique, as it depends entirely on the weight function used
...
2 Fourier Series
A Fourier series of a function is simply the expansion of a function in sin and cos functions
...
1
...

2!
4!
x3 x5
sin x = x −
+
−
...
For the basis functions to be periodic over this interval,
we require that them to be of the form
cos

2πn
x
L

and

sin

2πn
x
L

Now, as these basis functions are evidently orthogonal, let us consider the results from
some of the possible inner products:
xo +L

dx cos

2πn
x sin
L

2πm
x
L

dx cos

2πn
x cos
L

2πm
x
L

2πn
x sin
L

2πm
x
L

xo
xo +L
xo

xo +L

dx sin
xo

= 0, for

L

= L/2


0

0

= L/2


0

all m and n

(2
...
4)

for n = m = 0
for n = m > 0
for n = m

(2
...

Let us put all this together, and write some function f (x) in terms of this basis:
f (x) =

ao
+
2

∞

an cos
n=1

2πn
x + bn sin
L

2πn
x
L

(2
...
We can nd the coecients ao , an and bn by taking the inner product of the
function with this basis:
ao =
an =
bn =

2
L
2
L
2
L

xo +L

dx f (x)
xo
xo +L

dx f (x) cos

2πn
x
L

dx f (x) sin

2πn
x
L

xo
xo +L
xo

Astute readers may have realised that this kind of expansion cannot be done for every
function; we can only do this for a restricted set of functions
...
2
...
It must be periodic
2
...
It must have only a nite number of maxima and minima within one period
4
...

This may make it seem like the Fourier series is a bit pointless, as apparently we can only
write already periodic functions, like the trigonometric functions, in terms of themselves
...

For example, one would not usually consider the function f (x) = |x| to be periodic
...


Figure 2
...


2
...
2 Symmetry and Periodicity
In the example above, we only had a single way of making the function periodic outside the
interval, as the function was even by denition
...
For example, consider f (x) = x sin x dened on [0, π]
...

This is shown in the Figure below
...


Figure 2
...
This is known as a sine series
• If f (−x) = f (x), then all bn = 0
...
That is, it converges to the midpoint of the interval at that point
...

It is often worth thinking about which extension of the function you wish to perform, as
it might be easier in one case or the other
...

Let us consider an example
...

We need to rst make the function odd; thus, let f (x) = −x2 for −2 < x < 0
...

bn =

2
L

2
=
4
1
=
2

xo +L

2πn
x
L

dx f (x) sin
xo
2

2πn
x
L

dx f (x) sin
−2
2

2πn
x −
L

dx x2 sin
0

=2 −

2x2
πn

0

dx sin
−2

2πn
x
L

πn
8x
πn
16
πn
x +
sin
x −
cos
x
2
(πn)2
2
(πn)3
2

cos

2
0

16 ((πn)2 − 2)(−1)n + 2
→ bn = −
(πn)3

Thus, the sine series is given by
∞
2

x = −16
n=1

((πn)2 − 2)(−1)n + 2
nπ
sin
x
3
(πn)
2

for 0 < x < 2
...

Now for the cosine series
...

an =
=

xo +L

2
L

dx f (x) cos
xo
2

1
2

dx x2 cos
−2
2

dx x2 cos

=
0

→ an =

ao =

2πn
x
L

πn
x
2

πn
x
2

16
(−1)2
(πn)2
2

1
2

dx x2
−2
2

dx x2

=
0

→ ao =

8
3

Thus,
4
x = + 16
3

∞

2

n=1

(−1)n
nπ
cos
x
(πn)2
2

This is the cosine series for x2 in the domain 0 < x < 2
...
3 Fourier Transforms
The Fourier transform of a function is the extension of this kind of Fourier series to an
innite domain
...
What are the coecients cn ?
cn =
=

Let kn =

2πn
L
...
This means that we can dene the Fourier transform (often abbreviated as FT)
of a function as
1
f (k) = √
2π

∞

dx e−ikx f (x)

−∞

(2
...
8)

The uses and consequences of these transforms will become more relevant when used in
quantum mechanics, as well as for Fourier Optics (see A2 notes for more details)
...

Consider a gaussian of the form
g(x) = √

x2
1
e− 2σ2
2πσ

Then, it's Fourier transform is given by
1
g(k) = √
2π
1
=
2πσ

∞

dx e−ikx √

−∞
∞

dx e

−

x2
1
e− 2σ2
2πσ

x2
+ikx
2σ 2

−∞

To solve this integral, we need to complete the square on the exponent
...
Thus, if a function is constrained in x-space, then it is spread out in
k -space
...


2
...
1 Properties of the Fourier Transform
The FT of a given function has some quite useful properties, as follows:
1
• Rescaling: f (ax) = a f (k/a)

• Shifting: f (a + x) = eika f (k)
• Exponential Multiplication: eiqx f (x) = f (k − q)
• Derivative:

∂f
∂x

= ik f (k)

• Polynomial Multiplication: xf (x) = i ∂ f
∂k

23

Toby Adkins

Mathematical Methods

The last two of these are particularly useful in the manipulation of dierential equations,
as they allow us to nd the functional form of the solution
...

∂2φ
− K 2φ
∂x2

F

= F (f (x))

(ik)2 φ(k) − K 2 φ(k) = f (x)
φ(k) = −

f (x)
+ K2

k2

To nd the solution, we need to apply the inverse Fourier transform to this expression for
φ(k)
...


2
...
2 The Convolution Theorem
The convolution f (x) of two functions h(x) and g(x) is given by
∞

dx h(x − x )g(x )

f (x) = (h g)(x) =
−∞

(2
...
But what is the Fourier transform of a convolution?
1
f (k) = √
2π
1
=√
2π
1
=√
2π

∞
−∞
∞
−∞
∞

∞

dx e−ikx

dx h(x − x )g(x )
−∞

dx dx e−ikx e−ik(x −x ) h(x − x )g(x )
dx dx e−ik(x−x ) e−ikx h(x − x )g(x )

−∞

Now let y = x − x such that dx dx = dy dx
...
10)

For example, considering a diraction grating in Fourier optics, the nal pattern will be
the product of the Fourier transforms of the aperture functions of both the nite slit and
innite double slits, in correspondence with the principle of superposition
...
4 The Dirac Delta Function
The Dirac Delta function is dened by the following properties:
δ(x − a) = 0

(2
...
12)

dx f (x)δ(x − a) = f (a)

(2
...

Evidently, no function exists that actually satises these properties
...
14)

Graphically, we can think of this as 'squashing' down a gaussian of nite width into this
innite 'spike', as illustrated below
...
3: The limiting case of a Gaussian with dispersion ε (left) to the Dirac Delta
function (right)
Let us now demonstrate that (2
...

∞

lim

ε→0 −∞

dx f (x) δε (x − a)
∞

= lim

ε→0 −∞

dx f (x) √

(x−a)2
1
e− 2ε2
2πε

∞
(x−a)2
1
dx f (x) e− 2ε2
ε→0
2πε −∞
∞
(x−a)2
1
1
= lim √
dx e− 2ε2
f (a) + (x − a)f (a) + (x − a)2 f (a) +
...

= lim √
ε→0
2πε
−∞
−∞
1
= lim f (a) + f (a) ε2 + O(ε4 )
ε→0
2
= f (a)

= lim √

25

Toby Adkins

Mathematical Methods

2
...
1 Properties of the Dirac Delta
The Dirac delta function has the following properties:
• Reversibility: δ(x)δ(x − a) = δ(a)δ(x − a)
• Rescaling: δ(ax) =

1
|a| δ(x)

• Symmetry: δ(−x) = δ(x)
• Dierence of Squares: δ(x2 − a2 ) =

1
2|a|

[δ(x + a) + δ(x − a)]

These can easily be proven by multiplying both sides by an arbitrary, well-dened function
f (x), and integrating both sides over some interval
...
4
...
This means that it behaves like the identity
operator in convolution
...
10):
f (k) =

√

2π f (k) δ(k)
1
→ δ(k) = √
2π

To nd the Fourier representation of the Dirac delta, we calculate the inverse Fourier
transform of this expression
...
15)

2
...
3 A basis for Fourier Space
The basis elements in Section (2
...
Supposee that the basis elements for Fourier space are of the form
1
ek (x) = √ eikx
2π

Let us check that this indeed form an orthonormal basis
...
Consequently, a Fourier transform
is simply the projection of this function onto that basis
...
We can subsequently write that f (x) = x|f and f (k) = k|f where x and
k are used as shorthand for the basis elements in both x-space and k -space
...
6), we can write the identity as
∞

∞

dk |k k|

dx |x x| =

I=

−∞

−∞

Let us now consider the inner product between two functions f (x) and g(x), and use this
denition of the identity
...
This is known as Parseval's formula
...


Sturm-Liouville Operators

This section will dene, and cover the basic concepts of, Sturm-Liouville Operators, including:
• The Eigenvalue Problem
• Denition and Properties
• Methods of Solution
• Known Dierential Equations
• Some Quantum Mechanics

The concept of the Sturm-Liouville operator plays a very large role in much of Physics as
it allows us to solve many second order dierential equations, the most common type of
dierential equation that we encounter
...
Note that the "•" symbol is used to show where the function
is placed when the operator is applied
...
1 The Eigenvalue Problem
Let us restrict our consideration to the functions in L2 [a, b]
...
This is because
the solutions to third order dierential equations can include disparate locations in space
eecting one another
...
1)

D |ψn = λn |ψn

as this allows us to solve for any state of the system by writing it as a linear superposition
of these eigenstates |ψn
...


3
...
Using (1
...
For two arbitrary
functions f (x) and g(x):
b

f |D|g =

dx f ∗ pg + rg + qg

a
b

=

dx f ∗ pg + f ∗ rq +f ∗ qg

a

(1)

(2)

Let us simplify each of these terms using integration by parts
...

b

(1) :

dx f ∗ pg = f ∗ pg

a

b

b
a

−

dx (f ∗ p) g

a

= f ∗ pg − (f ∗ p) g
b

(2) :
a

dx f ∗ pg = [f ∗ rg]b −
a

b

b
a

b

+

dx (f ∗ p) g

a

dx (f ∗ r) g

a

Putting this together, we nd
f |D|g
= f ∗ pg − (f ∗ p) g + f ∗ rg

b
a

b

+
a

29

dx (f ∗ g) g − (f ∗ r) g + f ∗ qg

Toby Adkins

Mathematical Methods

We note that
(f ∗ p) = p f ∗ + 2p (f )∗ + p(f )∗
(f ∗ r) = (f )∗ r + r f ∗

By inspection, it becomes clear that letting r = p will lead to some cancellation
...
This means that the condition simplies to
a
b

f ∗ pg

(3
...
As we have not specied
the boundaries a and b, we have freedom to choose these boundaries (within reason) such
that the operator is hermitian
...
For a general weight function,
an operator is said to a Sturm-Liouville operator if it is of the form:
DSL • =

1
d
w(x) dx

p(x)

d
dx

+ q(x) •

(3
...
It turns out that any second order operator can actually be
written in this form by imposing appropriate conditions on the weight function
...
4)

The conditions on w(x) and q(x) follow trivially from this:
p(x)
po (x)
q(x) = w(x) p2 (x)

w(x) =

30

(3
...
6)

Toby Adkins

Mathematical Methods

3
...
1 Eigenvalue Properties
SL operators, being hermitian, share the properties of Hermitian operators outlined in
Section (1
...
3)
...

Readers are welcome to skip this section if they are comfortable with these types of proofs
...

D un = λn un
D um = λm um

Multiplying the rst of these equations by u∗ , and taking the dierence with the complex
m
conjugate of the second equation multiplied by un :
b

b

dx w (u∗ D un − un D u∗ ) =
m
m

dx w (u∗ λn un − un λ∗ u∗ )
m
m m

a

a

b

0 = (λn − λ∗ )
m

dx w u∗ un
m

a

as D is hermitian by the denition of an SL operator
...
Letting n = m, the integrand on the right-hand side, and hence the integral, is
positive denitive, and so λn = λ∗
...
Letting n = m, we observe that the eigenvalues are distinct, and hence that the
integral must be zero
...
The eigenfunctions un (x) form a complete set, allowing us to express any function in
L2 as a linear combination of these eigenfunctions
w
The last of these results has not been proven here, but this is owing to the fact that it is
actually quite dicult to show; let us be content with just 'knowing' the result for now
...
3 Methods of Solution
There are multiple methods for solving the SL eigenvalue problem, which are detailed as
follows
...


3
...
1 Frobenius' Method
This is the most general method, involving subsituting
∞

un (x) =

(3
...
This will give, with careful treatment of summation notation,
a recurrence relation for the coecients ai
...
This gives the spectrum of eigenvalues λn and their corresponding
eigenfunctions un (x), though does not give the proper normalisation
...
3
...
The conditions are that
• q(x) = 0 in the operator
• p(x)/w(x) is at most a quadratic polynomial with real roots
• The product of the eigenfunctions and p(x)/w(x) are zero on the boundaries of the

space

If, and only if, these conditions are satised, then the eigenfunctions are given by
un (x) ∝

1 1
dn
kn w(x) dxn

pn (x)
wn−1 (x)

(3
...
Consequently, the eigenvalues are given by
λn = n

1
(n − 1)
2

p(x)
w(x)

+ k1 u1 (x)

(3
...
3
...
However, we
do need to normalise the eigenfunctions when they are obtained
...
3
...
10)

Given some explicit form of G(x, t), the eigenfunctions un (x) are dened to be the coecient of tn in the Taylor series expansion of G(t)
...

32

Toby Adkins

Mathematical Methods

3
...

Generally, we will just be required to recognise the dening dierential equation and then
quote the results, though we will do some solving of them here
...
4
...
11)

(1 − x2 )u (x) − 2xu (x) = λ u(x)

This is evidently an SL equation with p(x) = 1 − x2 , q(x) = 0 and w(x) = 1
...
Let us use Frobenius' method to nd a
solution
...
7) into the equation:
∞

∞

n(n − 1) an xn−2 −

∞

n(n − 1) an xn − 2
n=2

n=2

∞

n a n xn = λ
n=1

an xn
n=0

n=m, m=n+2
∞

∞
n

(n + 2)(n + 1) an+2 x −
n=0

∞
n

n(n − 1) an x − 2
n=0

∞
n

an xn

n an x = λ
n=0

n=0

∞

[(n + 2)(n + 1) an+2 − n(n + 1) an − λan ] xn = 0
n=0

Thus, we obtain a recurrence relation for the coecients an of
an+2 = an

n(n + 1) + λ
(n + 2)(n + 1)

However, we have a slight problem; as n → ∞, an+2 → an , meaning that the series does
not terminate
...
Suppose
that for some value n = l, an+2 = 0
...
We have thus found the eigenvalues required for the eigenfunctions
to exist (or else their dening equation would never terminate)
...
Then
• set ao = 1, a1 = 0 if l is even
• set ao = 0, a1 = 1 if l is odd

and apply the recurrence relation until n = l is reached
...


33

Toby Adkins

Mathematical Methods

Graphically, these are represented as:

Figure 3
...
12)

l=0

This is well worth remembering, as long as you can Taylor expand properly!

The Associated Legendre Equation
Legendre's dierential equation is simply the special case of the Associated Legendre Equation
m2
(1 − x2 )u (x) − 2xu (x) −
u(x) = λ u(x)
(3
...
3
...
As generally systems obey this condition, it is much rarer
to see this equation
...
11), but it's eigenfunctions are given by
Plm (x) = (1 − x2 )m/2

dm
Pl (x)
dxm

(3
...
These form the basis for the Spherical Harmonics, which are the natural
basis for two-dimensional functions that are constrained to the surface of a sphere
...
4
...
15)

Toby Adkins

Mathematical Methods

This can be written in Sturm-Liouville form as
ex

2

e−x

2

ex

(u (x) − 2xu (x)) = λu(x)
2

d
2 d
e−x
u(x) = λu(x)
dx
dx

This was weight function w(x) = e−x
...
5)), and so the space is bounded as [−∞, ∞]
...
Instead, let us consider the
generating function
∞
2tx−t2

G(x, t) = e

=

Hn (x)
n=0

tn
n!

(3
...

2!

t2
(4x2 − 2) +
...
16), we nd that the rst few Hermite polynomials
(shown in Figure (3
...


We can actually show that the Hermite Polynomials satisfy a recurrence relation
...
16) with respect to t:
d
2
e2tx−t =
dt
∞

2(x − t)
∞

−2

Hn
n=0
∞

−2n

tn+1
n!

Hn−1
n=0

tn
n!

Hn
n=0
∞

+ 2x

Hn
n=0
∞

+ 2x

Hn
n=0

tn
n!
tn
n!
tn
n!

∞

Hn
n=0
∞

=

Hn n
n=0
∞

=

Hn
n=1
∞

=

tn
n!

d
dt

tn−1
n!

tn−1
(n − 1)!

Hn+1
n=0

tn
n!

Equating powers of tn , we arrive at the recurrence relation that
Hn+1 (x) = 2xHn (x) − 2nHn−1 (x)

(3
...
18)

Toby Adkins

Mathematical Methods

Figure 3
...
Substitute (3
...
17) and dierentiate:
Hn+1 = 2xHn − Hn
Hn+1 = 2Hn + 2xHn − Hn
2(n + 1)Hn = 2Hn + 2xHn − Hn

Collecting terms, we obtain the desired result
...
19)

There will be more to follow on Hermite polynomials in Section (3
...


3
...
3 Bessel's Dierential Equation
Bessel's dierential equation is
x2 u (x) + xu (x) + (x2 − ν 2 )u(x) = 0

(3
...
This equation is
most common for cylindrical symmetries
...
The solutions to
x
this equation are Bessel functions
...
21)

n=−∞

Graphically, they are represented as

Figure 3
...
The
roots of the functions are particularly important for solving boundary value problems

37

Toby Adkins

Mathematical Methods

3
...

No knowledge of the QHO is required in order to nd these solutions, though of course it
informs the implications of the results obtained
...


(3
...
23)

+ (2 − x2 ) |ψ = 0

In the limit of x → ∞, it is easy to show by substitution that
x|ψ = Axk e−x

2 /2

≡ u(x)e−x

2 /2

is a valid solution to the the TISE
...
23):
d2
2
2
u(x)e−x /2 + (2 − x2 )u(x)e−x /2 = 0
2
dy
u e−x

2 /2

2

− 2xu e−x − ue−x

2 /2

2 /2

+ x2 ue−x

+ (2 − x2 )ue−x

2 /2

=0

u (x) − 2xu (x) + (2 − 1)u(x) = 0

We recognise the last of these as Hermite's dierential equation with λ = 1 − 2
...
7)
...

2

(3
...
The
time evolution of the system then follows trivially from this
...


Partial Dierential Equations

This section will cover the crucial concepts involved in solving partial dierential equations
(PDE's), including:
• Characteristics of the solution
• The Method of Separation of Variables
• General Solutions to common PDE's
• Green's Functions
• Lagrange Multipliers

Note that this Chapter has been included after Chapters (2) and (??) due to the fact that
it uses some of the results and methods from these previous Chapters
...


39

Toby Adkins

Mathematical Methods

4
...
1)

ˆ
for some linear operator D and arbitrary function f (x)
...
We have two further cases:
• f (x) = 0 - In this case, the equation is said to be inhomogeneous
...
When it is both linear and
homogeneous, we can super-impose solutions to the equation i
...


As we learnt during the CP3 dierential equations course, the solutions to these equations
will involve undetermined coecients
...
We have the following three types of boundary conditions:
• Dirichlet - When φ is specied on the boundary
• Neumann - When

φ · n is specied on the boundary

• Cauchy - When both of the above are specied on the boundary

The type of boundary conditions that we will use will depend entirely on our physical
system, and consequently the information available to us
...
1
...
For u = u(x, y), consider
a(t, x, u)

∂u
∂u
+ b(t, x, u)
= f (t, x, u)
∂x
∂y

Our method for interpreting this equation is similar to that for Lagrange Multipliers
...
We can remain on the
surface by making a small change ds = (dx, dy, dz) in the direction of c, and in this case
we obtain:
dx
dy
du
=
=
a
b
g

(4
...
Integrating it will
yield the characteristic curves of the solution, which can give us a solution for u(x, t) given
appropriate initial conditions
...
2 The Method of Separation of Variables
Generally referred to as separation of variables, this is a very powerful technique for solving
linear partial dierential equations
...

Suppose that your solution to the PDE is φ(x), where x = (x1 , x2 ,
...
Then, our trial
solution to the equation will be
n

φ(x) = f1 (x1 ) · f2 (x2 ) · f3 (x3 )
...
3)

i=1

Generally, substituting this into the equation will yield an equality involving n separated
equations that are functions of a single variable xi , meaning that they must all be equal
to a separation constant
...

When solving these types of problems, one generally follows the following methodology:
1
...
Substitute an appropriate trial solution of the form of (4
...
Solve the resultant single-variable equations with a separation constant
4
...
Note that if the boundary condition is dependant
on only one variable xi , this condition can be imposed on the corresponding equation
in isolation
...


4
...
2) by
solving some common types of partial dierential equations
...
3
...
It takes the form:
1 ∂2
−
c2 ∂t2

2

F =0

(4
...
We will drop the labelling of the variables for the
sake of notation from this point
...
Thus, we can write that
1 d2 fi
= λi
fi dqi

for i = 2, 3, 4 and qi = x, y, z , which has solution
fi = Ai cos λi x + Bi sin λi x

Now, if we treat the LHS, we nd that
1 d2 f1
= −c2 (λ2 + λ2 + λ2 )
2
3
4
f1 dt2
→ f1 = A1 cos ωt + B1 cos ωt

for ω 2 = c2 (λ2 + λ2 + λ2 )
...
Following are a
couple of examples
...
Find the normal modes of waves on the surface of a cube dened by the planes x = 0,
y = 0, z = 0, x = L, y = L and z = L given that there are no oscillations on the
edges
...

As the coordinates (x, y, z) are orthogonal, we can impose this boundary condition
on each of the functions fi separately
...

2
...

2L

42

Toby Adkins

Mathematical Methods

This is the case for the general solution above in two dimensions
...
At t = 0, we have that
∞

F =

an sin
n=1
∞

˙
F =

πnx
L

ωn bn sin
n=1

πnx
L

We can then nd the Fourier coecients by the normal method
...
Finding an :
3πx
πx
2 ∞
πnx
dx 2A sin
cos
sin
L 0
2L
2L
L
∞
1
πx
πnx
2πx
4A
dx
sin
sin
+ sin
=
L 0
2
L
L
L
∞
2A
πx
2πx
=
dx sin2
δn1 + sin2
δn2
L 0
L
L
= A (δn1 + δn2 )

an =

Here we have used the fact that sin x, sin 2x,
...

Thus, the nal displacement is given by
F (x, t) = A sin

πct
L

πx
cos
L

+ sin

2πx
L

cos

2πct
L

As expected, the subsequent solution is simply the superposition of the time evolution
of the initial excited modes, as we saw in CP4 last year
...
3
...

∂T
=α
∂t

2

T

(4
...
There will also
be detailed solutions to harder problems involving heat sources in said notes
...

At t < 0 the temperature throughout the bar is 0K , and at t = 0 the end at x = 0 is placed
in thermal contact with a heat bath at temperature To
...
9To
...
2 by letting T (x, t) =
f (t)g(x)
...
Note that we are cheating a little here; the substitution is of this form as we
know where we are going, but we could have let it be equal to any old constant
...

• The bar is homogeneous in at long times: t → ∞, T → To
ao = To
• The end in contact with the heat bath is always at a constant temperature: T (0, t) =
To , t > 0
...

∂T
=
∂x

∞

bk cos kx eαk

2t

k=1

cos kL = 0
1
π
2
(2n + 1)π
→k=
2L
kL =

n+

• The bar is at T = 0 initially: T (x, 0) = 0
...

Tav =
=

L

1
L

dx T (x, t)
0
∞

L

1
L

dx To −
0

n=0
∞

= To −
n=0
∞

= To −

4To
sin
(2n + 1)π

4To
−α
e
(2n + 1)π

(2n+1)π
2L

8To
−α
2 e
[(2n + 1)π]
n=0

(2n + 1)πx
2L

2

t

(2n+1)π
2L

1
L

·

e

L

−α

(2n+1)π
2L

2

t

(2n + 1)πx
2L

dx sin
0

2

t

Then, we can take the rst order term to approximate our solution
...
9To = To −
1
=
10

8To
−α
2 e
[(2n + 1)π]
n=0

∞

8To
−α
2 e
[(2n + 1)π]
n=0

(2n+1)π
2L

(2n+1)π
2L

2

t

2

t

8 −α π22 t
e 4L
π2
4L2
80
t=
log
2
απ
π2
≈

We can then put given values into this expression to nd a numerical result
...


4
...
3 Laplace's Equation
Laplace's equation is a very common partial dierential equation in Physics, particularly
in Electrostatics
...
6)

V =0

In fact, many scalar elds that are identically equal to zero on some closed boundary can
be shown to obey some form of Laplace's equation
...
3
...


In Spherical Polar Coordinates
Laplace's equation in this coordinate system takes the form
1 ∂
r2 ∂r

r2

∂V
∂r

+

1
∂
r2 sin θ ∂θ

sin θ

∂V
∂θ

+

1
∂2V
=0
r2 sin2 θ ∂φ2

Let V (r, θ, φ) = f (r)g(θ)h(φ) and divide through by f gh
...
This means that we can write that
1 d2 h
= −m2
h dφ2
→ h(φ) = Ceimφ

The reason that we have set the equation equal to a separation constant of −m2 is because
we require the function h(φ) to be periodic in [0; 2π], and so this is the only acceptable
form of the constant
...
Let this separation
constant be l(l + 1); the reason why we choose this form will soon become clear
...
7)
(4
...
For l = 0,
d
dr
r2

r2

df
dr

= l(l + 1)f

df
d2 f
+ 2r − l(l + 1)f = 0
2
dr
dr

We can assume that the solution will be of the form
∞

cn rn

f (r) =
n=0

Substituting this in, and being careful with our summation notation,
∞

r2

∞

∞

cn n(n − 1)rn−2 + 2r
n=2

cn nrn−1 − l(l + 1)
n=1

cn rn = 0
n=0

∞

cn rn (n(n − 1) + 2n − l(l + 1)) = 0
n=0

Each of these terms are orthogonal to one another as 1, r, r2 , r3 ,
...
8), let us make the substitution x = cos θ
...

If we have rotational azimuthal symmetry (i
...
9)

This equation is very useful due to the fact that it is formed from the product of two functions that have independent variables; this means that we can apply boundary conditions
to the individual parts of the equation
...

ˆ
Calculate the electric potential V given the fact that it satises Laplace's equation on the
boundary
...

If we then take the potential to be zero on the surface of the sphere, then we have the
following boundary conditions:
• V =0
al Rl +

bl
=0
Rl+1
→ bl = −al R2l+1

• V → −Eo z = −Eo r cos θ
∞

al rl Pl (cos θ) = −Eo r cos θ
l=0

a1 = −Eo

with all other al = 0
...

Thus, the nal solution becomes
V (r, θ) = −Eo r −

R3
r2

cos θ

= −Eo r cos θ + Eo
external eld

R3
cos θ
r2

induced eld

As expected, the charge distribution in the sphere will move such that it cancels the
external electric eld, ensuring that it remains in equipotential
...
Then, let V = f (r)g(φ)
...
We know that g has to be period in [0; 2π], i
...

• k=0
1 d2 g
=0
g dφ2
dg
=a
dφ
→ g(φ) = aφ + b

But for this to be periodic, we require that a = 0
...

r d
f dr

df
=0
dr
df
r
= c1
dr
df
c1
=
dr
r
→ f (r) = c1 log r + c2
r

• k == 0
d2 g
= −k 2 g
dφ2
→ g(φ) = c3 cos kφ + c4 sin kφ

Solving the radial equation using the same substitution as in the previous section:
r d
df
r
= k2
F dr
dr
d2 f
df
r2 2 + r − k2 f = 0
dr
dr
∞

n2 − k 2 an r n = 0
n=1

→ n = ±k

The nal solution is thus given by
∞

bn
an r − n
r
n

V (r, φ) = A log r + B +
n=1

48

∞

cn rn +

sin nφ +
n=1

dn
rn

cos nφ

(4
...
4 Green's Functions
In all of the above cases, we have been solving the homogeneous case of (4
...
We can
solve this by solving the similar equation of
ˆ
D G(x, t) = δ(x − t)

Note that both sides of this equation are transnationally invariant, such that G = G(x − t)
...
11)

G(x, t)f (t) dt

Essentially, we can think of f (x) as the 'source', and that the result that we obtain is the
convolution of G(x, t) and the source
...
This means that we require that D is of Sturm-Liouville form
...

∞

f (x) =

cn φn (x)
n=0

Now, we will suppose further that G(x, t) is of the form:
∞

φ∗ (t)φn (x)
n
λn

G(x, t) =
n=0

(4
...
Recall that
dt δ(x − t)f (t)

‘f (x) =

Multiplying both sides by f (t) and integrating
∞

ˆ
dt D G(x, t)f (t) =

φ∗ (t)φn (x) f (t)
n

dt
n=0
∞

=

∞

φn (x)

dt

φ∗ (t)
n

n=0
∞

=

φn (x)cm δmn
n=0
∞

=

cm φm (t)
m=0

cn φn (x)
n=0

= f (x)

49

Toby Adkins

Mathematical Methods

Thus, the functional form G(x, t) does indeed solve the equation, and thus must be the
only solution by the uniqueness theorem
...
4
...

• G(x, t) is hermitian
ˆ
ˆ
f, Dg = Df, g
ˆ
= g, Df

∗

→ G(x, t) = G∗ (t, x)
ˆ
• G(x, t) is a solution to the homogeneous equation D G(x, t) = 0 except at x = t
• G(x, t) is continuous at x = t
• The rst derivative of G(x, t) is discontinuous at x = t
ˆ
D G(x, t) = δ(x − t)

t+ε

dx
t−ε

p(t)

∂
∂
p(x)
+ q(x)
∂x
∂x
∂
∂
p(x)
+ q(x)
∂x
∂x
∂G
∂x

t+ε

G(x, t) = δ(x − t)
t+ε
t−ε
t+ε

t+ε

+

dx δ(x − t)

dx q(x)G(x, t) =
t−ε

t−ε

dx δ(x − t)

G(x, t) =

t−ε

∂G
∂x

→

=0
t+ε

=1

=
t−ε

1
p(t)

None of these are particularly useful or enlightening, but they have been included here for
the sake of completeness
...
4
...
13)

Toby Adkins

Mathematical Methods

Substituting these into (4
...
Then:
∞

1
(2π)n
1
=
(2π)n
1
=
(2π)n

dn k eik(x−y) ·

G(x − y) =

∞
∞

1
k2
∞

dn k eik(x−y)
∞
∞

dα e−α

2 k2

0
∞

dn k eik(x−y)−αk

dα

2

−∞

0

We can now complete the square, and change variables
...

51

Toby Adkins

Mathematical Methods

4
...
The reason is simple; there is no particular reason apart from
convenience
...

Consider the tangent t of g(x) at xo in a space of dimension spanned by {x}
...
Then, we can
write the above two conditions as a single constraint
(f + λg) = 0

(4
...
This
will give us n + 1 equations for the same number of unknowns without having to derive an
explicit form for g(x)
...

g(x, y) = 1 − x2 − y 2
h(x, y, λ) = f (x, y) + λg(x, y)
h = (5 − 2λx, 12 − 2λy, 1 − x2 − y 2 )
!
Equation 4
...
This means that we obtain three equations with three
unknowns that we now solve:

5
→λ= x
2
y
12 − 5
=0
x
5
x= y
12
1 − x2 − y 2 = 0
12
→y=
13
5
→x=
13

We have thus found the maximum point (5/13, 12/13)
...
5
...

Consider the action of a system S(q), dened by
tf

dt L q(t), q(t), t
˙

S(q) =
ti

We want to constrain this functional to some condition that Cα (q, t) = 0
...
15)

1
˙
Suppose now that T = 2 mq 2 and that V = V (q)
...


m¨ = − V + λα Cα
q

The constraint appears to act as a potential, where the force that results is perpendicular
to it
...


Probability and Statistics

This chapter aims to cover the basic probability concepts required to tackle probabilistic
Physics, including:
• Basic Probability Concepts
• Some Probability Distributions
• Basic Error Propagation

This is by no means a very comprehensive treatment of probability and statistics, more the
bare-minimum to get by
...
Note that throughout much of this chapter we will be using
x to refer to some general variable; it does not explicitly refer to displacement in this case
...
1 Basic Probability Concepts
Everyone reading these notes should be quite familiar with the concept of the 'probability'
of an outcome
...
For example, assuming that
one ips a perfectly circular coin of uniform density in exactly the same fashion many
times, then one would expect to obtain 'heads' or 'tails' roughly half of the time
...
In this context,
the word event takes on the same meaning as when it is used in Special Relativity
...
1
...
Let A1 , A2 ,
...
Then,
P (Ak |B) =

P (Ak ) P (B|Ak )
n
i=1 P (Ai ) P (B|Ai )

(5
...
As you might have guessed from the
wording of this, one has to be very careful about the probability that they actually want
to nd
...

Assuming that there is an equal chance of obtaining a boy or a girl at birth, consider the
following two scenarios:
1
...
What is the probability that
they have two boys?
2
...
What is the probability that they
have two boys?

At rst glance, it may seem that these two are asking the same question
...
Let us dene three events:
• A = both children are boys
• B = the older/rst child is a boy
• C = one of the children is a boy

The rst of these questions is asking for P (A|B)
...
Using conventional probability analysis on the
number of combinations, it follows that
P (A|B) =

P (A)P (B|A)
=
P (B)

1
4
1
2

=

1
2

The second question is asking for P (A|C)
...
Then
P (A|C) =

P (A)P (C|A)
=
P (C)

55

1
4
3
4

=

1
3

Toby Adkins

Mathematical Methods

This is a clear demonstration of the care that must be taken in probability; one needs to
be very clear about the actual event that we want to nd the probability of
...
1
...
A random variable is one whose value is subject to variations due to chance,
and can take on a set of possible dierent values, each with an associated probability
...
This is usually an integer e
...

Calculations dealing with these generally involve discrete sums

• Continuous Variables - If it can take on two particular real values such that it can also

take on all real values between them (even values that are arbitrarily close together),
the variable is said to be continuous in that interval
...
g the possible heights of
people between 1
...
Calculations dealing with these generally
involve integrals

Evidently, the rst of these screams non-physical in many scenarios as even 'Popular Science' has a caught onto the idea that nothing can be precisely determined
...

It is worth introducing at this stage the concept of a probability distribution, though we will
go into more detail on specic distributions in Section (5
...
A probability distribution
assigns a probability to each measurable subset of the possible outcomes of a random
experiment
...
There may be an innite set of of these discrete values, but the associated
probabilities decline to zero quickly enough such that they converge to a nite value

• Continuous Probability Distributions - Corresponding to continuous variables, this

is characterised by a probability density function
...
So talking about the probability
of a particular value is meaningless
...


These are probability distribution functions (PDF)
...
These exist
for both discrete and continuous PDF's
...
Generally, we are only interested in the
PDF of a random variable, as this is usually the most useful
...
As such, if the reader requires further examples of the manipulation of
these functions, it is recommended that they consult the A1 or A3 notes
...
1
...
They are as follows:
• Expectation Value - The average value of some variable weighted by the PDF
...
2)

i=1
continuous

discrete

Note that the expectation value may not actually be a value that the function can
take; it merely tells you how much you will obtain this value after many trials
...
5, despite the fact that all faces of
the die are of integer values
• Mean - This is simply the term used to refer to the expectation value of the random

variable itself
...
If there is not middle value, it is usually dened as the
mean of the middle two values
...
3)
(5
...

• Variance - This is a measure of how much the data points in a data sample or

distribution are spread out
...
Mathematically:
Var[x] = (x − µ)2
= x2 + µ2 − 2µ x

But µ = x
...
5)

A small variance indicates that the data points tend to be very close to the mean
and hence to each other, while a high variance indicates that the data points are very
spread out around the mean and from each other
...
6)
Evidently, this measures a very similar thing to the variance
...

σ 2 = Var[x]

57

Toby Adkins

Mathematical Methods

5
...
4 Estimators
All of the quantities dened in the previous section assumed that we knew the PDF of
the system; in many cases, we do not have this information, and so must use other formulae
...
In general, we want the expectation value of an estimator to
be equal to the true value; this is a measure of an unbiased estimator
...
The most common estimators are those for the mean and
the variance
...

For the mean, let us suppose that the estimator is of the form
n

1
µ=
n

xi
i=1

Once again, we require that µ = µ for the estimator to be unbiased
...
7)

xi
i=1

This is the same expression that readers would have been using for years to calculate the
mean of a sample
...

1
n

σ2 =

(xi − µ)2
i

2


1
=
n
=

=

1
n
1
n

i

xi − 1
n
x2 −
i

i

x2 −
i
i

2
xi
n

xj 
j

xj +
j

2 2 2
x −
n i
n

1
n

xj xk
jk

xi xj +
i=j

58

1
n2

x2 +
j
j

xj xk
j

k=j

Toby Adkins

Mathematical Methods

Here we have split up the sums so then we can more easily introduce some of the quantities
that we know, such as the mean
...

σ2 =
i

=
=
=

1
n


1
2
1−
n
n
1−
i

2
n
1
1−
n
1−

2
n


x2
i

2
−
n

xi xj
i=j

σ 2 + µ2 − 2 1 −

σ 2 + µ2 − 2 1 −

1
n

1
+ 2
n
1
n

µ2 +

x2
j
j

µ2 +

1
+ 2
n

xj xk 
j

k=j

1 2
1
σ + µ2 + 1 −
n
n

1 2
1
σ + µ2 + 1 −
n
n

µ2

µ2

σ2

In this case, we have found that the estimator is actually based
...
8)

i=1

This is the case as there are only n−1 independent data-points that you can use to measure
the variance, not n
...


59

Toby Adkins

Mathematical Methods

5
...
As expected, the majority are continuous probability distributions
...
2
...
The PDF is given by:
f (k) =

n k
p (1 − p)n−k
k

Figure 5
...
9)

Toby Adkins

Mathematical Methods

The simple re-labelling of summation indices l = k − 1 and m = n − 1 has been used
...
10)

µ = np

Now for the variance
...

n

x2 =

n k
p (1 − p)n−k
k

k2
k=0
n

n − 1 k−1
p (1 − p)(n−1)−(k−1)
k−1

= np
k=1
m

= np

(l + 1)
l=0

m l
p (1 − p)m−l
l
n

= np (n − 1)p
l=1

m − 1 l−1
p (1 − p)(m−1)−(j−1) +
l−1
m−1

= np (n − 1)p(p + (1 − p))

m

m l
p (1 − p)m−l
l

l=0

m

+ (p + (1 − p))

2 2

= n p + np(1 − p)

Again, we have used the same trick with re-labelling indices
...
11)

σ 2 = np(1 − p)

5
...
2 Poisson
Consider the limiting case of the Binomial distribution in which n depends on some continuous variable (such as time), and can be arbitrary small
...
9) and simplify using this large parameter n
...

(1) :

(2) :
(3) :

lim

n!
n(n − 1)(n − 2)
...

n→∞
n
n
n
=1
n
µ
= e−µ
1−
n
µ k
1−
=1
n

n→∞ nk (n

lim

n→∞

lim

n→∞

Thus, the nal expression for the Binomial distribution in this limit is (nothing that µ has
been re-labelled to λ)
f (k) =

λk e−λ
k!

(5
...
It expresses the probability of a given number
of events occurring in a xed interval if these events occur with a known average rate and
independently of the interval since the last event
...


Figure 5
...
Suppose that λ corresponds to
the average decay rate of a radioactive sample
...


62

Toby Adkins

Mathematical Methods

Let us calculate the mean of the Poisson distribution
...
It also makes sense from a statistical point of view; if the entire
distribution depends on some mean value, one would expect values to cluster around said
value
...
This is again what one would expect; the more one increases
the mean value, the greater the uncertainty expected, and so the greater the spread around
the mean value
...
13)

µ = σ2 = λ

5
...
3 Exponential
The Exponential distribution describes the interval (such as time) elapsed between independent events described by a process that obeys the Poisson distribution, and as such is
a continuous PDF
...
12), λ represents the average number of events in
some interval x
...

∞

dx e
0

−λx

1
= − e−λx
λ
1
=
λ

63

∞
0

Toby Adkins

Mathematical Methods

Hence,
f (x) = λe−λx

(5
...
3: The Exponential Distribution (showing the PDF and CDF)
It is worth thinking about whether this actually makes sense
...
Conversely, the probability that an event does occur will tend to one
...

∞

µ=
0

dx x λe−λx

∞

=λ
0
∞

=

dx x e−λx

dx e−λx

0

1
=
λ

Thus,
µ=

1
λ

(5
...

∞

x2 =
0

dx x2 λe−λx

∞

=2
0

=

2
λ2

64

dx x e−λx

Toby Adkins

Mathematical Methods

Thus, we obtain the variance as
1
λ2

σ2 =

(5
...


5
...
4 Normal
Let us consider the case of the Poisson distribution for large values of k
...
Thus for k 1, k can be approximated
as being roughly on the order of λ
...
12)
...

h(δ) =

µ(1 + δ) +

1
2

log(1 + δ)

µ(1 + δ) + 1/2
1+δ
2µ
µ(1 + δ) + 1/2
h (δ) =
−
1+δ
(1 + δ)2
1
h(δ) ≈ h(0) + h (0)δ + h (0)δ 2 +
...

2
2
2
h (δ) = µ log(1 + δ) +

However, as we are considering the case where µ
µ+

1, it follows that

1
1
≈µ− ≈µ
2
2

Substituting back in for f (δ),
eµδ −(µδ+ 1 µδ2 )
2
f (k) = √
e
2πµ
1
1
2
=√
e− 2 µδ
2πµ
(µδ)2
1
−
=√
e 2µ
2πµ

65

Toby Adkins

Mathematical Methods

Using the fact that µδ = k − µ and that µ = σ 2 for the Poisson distribution, we arrive at
the expression
f (x) = √

(x−µ)2
1
e− 2σ2
2πσ

(5
...
In the case
where µ = 0 and σ = 1, this is known as the unit normal distribution
...
4: The Normal Distribution with µ = 0 and varying σ
Note that the normal distribution is by denition symmetric around it's mean, and increases in width for higher dispersion
...
2), it is clear that for higher
values of k/x, the Poisson distribution would tend towards the Normal distribution
...

The reader may have guessed what the mean and the variance of this distribution are
...

∞

dx x √

x =
−∞
∞

(x−µ)2
1
e− 2σ2
2πσ

dx [(x − µ) + µ] √

=
−∞

σ
=√
2π

∞

dx
∞

(x−µ)2
1
e− 2σ2
2πσ

(x − µ) − (x−µ)2
e 2σ2 + µ
σ2

∞

dx √
−∞

(x−µ)2
1
e− 2σ2
2πσ

We now see that the right integral is the complete integral over a normalised PDF
...
The left-hand integral will evaluate to zero, and so we conrm the fact
that the mean is indeed µ
...

2σ 2
(x − µ)2 = √
π
2σ 2
= √
π
σ2

=√
=σ

pi

∞

2

du u2 e−u

−∞

1
2
− ue−u
2

∞

+
−∞

1
2

∞

2

du e−u

−∞

∞
−∞

2

We have hence conrmed out expectation that the variance is σ 2
...
This
can be used as a measure of the statistical signicance of a result; if we can say that a
result is of 4σ signicance, this means that it is located at or above 4σ from the centre
...

Probability to be within
±1σ : 0
...
9545
±3σ : 0
...
9999

Probability to be
> 1σ : 0
...
0228
> 3σ : 1
...
2
...
This is the reason why the Normal distribution is considered (certainly
in Physics) to be the most important distribution as it underpins so many others
...

Suppose we pick n random numbers xi from a arbitrary distribution, and dene the quantity
x=
¯

1
n

N

xi
i=1

We want to nd the probability distribution of x
...
Dene fn (k) = f (k) , where f (k) is the Fourier
x
¯
transform of f (x)
...

2!
2π
n
(ik)2 2
1
=√
1 − ik x +
x +
...

2!
2π

67

Toby Adkins

Mathematical Methods

We now want to Taylor expand the logarithm
...
− (ik x +
...

2!
2
2π
2
(ik)
1
x2 − x 2 + O (ik)3
= √ exp n −ik x +
2!
2π
k2 σ2
1
+ O (ik)3
= √ exp n −ikµ −
2
2π

We now want to take the inverse Fourier transform
...

1
2π
1
≈
2π

∞

dk exp n −ikµ −

fn (¯) =
x

=
=

1
2π

∞
∞

dk exp −
∞
∞

n
2

kσ −

i
(¯ − µ)
x
σ

2

∞

1
n
exp − 2 (¯ − µ)
x
2π
2σ

dk exp −
−∞

1
n
=
exp − 2 (¯ − µ)2
x
2π
2σ
→ fn (¯) = √
x

x
eikn¯

n 2 2
k σ − 2ik (¯ − µ)
x
2

dk exp −
∞

k2 σ2
+ O (ik)3
2

−
n
2

n
(¯ − µ)2
x
2σ 2
kσ −

i
(¯ − µ)
x
σ

2

√
n
√
2πσ

1
(¯ − µ)2
x
exp −
2
2σn
2πσn

Thus, we have shown that for suciently large n (or else all the terms would disappear in
the Fourier transform) that all distributions will tend towards a Normal distribution
...
18)

5
...
6 Chi-Squared
The PDF of the Chi-squared distribution is quite nasty (involving gamma functions) and
so we will not treat it explicitly
...
This is used a variety of Chi-squared
tests for accuracy of t, which will now be examined
...
Then the coecients that we require for
this function to give the best t to the curve are given by minimising
n

χ2 =
i=1

(yi − f (xi |ai ))2
σi

(5
...

68

Toby Adkins

Mathematical Methods

Figure 5
...
We can compute the value of χ2 , and then compare it to a take
of known χ2 values of the correct degrees of freedom; this will give the probability of
acceptance/rejection of the model (depending on how the table is set-up)
...
In this case, there are 10 degrees of freedom
...
4 rejection probability, where as a value of ≈ 29 would yield a rejection
probability of 0
...
Evidently, the rst curve would be a better t
...
3 Basic Error Propogation
Readers will already most likely be familiar with the majority of the material in this section
from the Practical Course
...
, xn )

Assuming that g(x) is unknown, we want to come up with some estimate
...
In this case, we shall assume
that the estimator given by (5
...
Then:
n

∂g
∂xi

g(x) ≈ g(µ) ±
i=1

(5
...
, µn )
...

What about the variance?
n

g(x) ≈ g(µ) +
i=1

∂g
∂xi

xi − µ i
x=µ

= g(µ)

as the expectation value of a particular variable xi from it's mean is zero by denition
...

The entity Vij (known as the covariance matrix) has the property that
for i = j
for i = j

0
2
σi

Vij =

assuming that the variables are uncorrelated
...
21)

x=µ

These are the rst order, and thus dominant uncertainties
...

70



---