Imperial College London

Stat Phys Révision

Ari’s cute notes : Important formulae and concepts

Alonso Bizzi, Ari

13/04/2022
 0. Prelims
Useful tingz

To-do
- Revise examples :
o SHO, white dwarves, électrons in metal
o Fermi gases and fermi temp

1. Micro and Macrostates

Goal of statistical physics: obtain macroscopic properties from the microscopic view.
(predict the emergent/collective behaviour)
Single particle state Microstate Macrostate
Describes The possible quantum states The single-particle state of Ensemble effect
available to each individual each of the particles it
particle contains.
Variables Momentum, position, energy Same as left but iterated P,V,U,N
for a group of particles
Detail Form a complete, orthogonal set A different combination of Incomplete  know
single-particle states  nothing about individual
different microstates. particle properties
Notation Particle x is in the single particle Ψ = (j1, j2, …jn)
state jx Number of particles in
Number of particles in single state 1, in state 2… in state
particle state n: jn n.
Statistical weight Ω of a macrostate
- Number of microstates it contains
- Other names: multiplicity, thermodynamic probability (not a prob, a number)

Assumptions for single particle states

- Particles are independent
- An identical particle is described with the same single particle states
- A microstate is defined

Isolated system
Fixed U, N and V
Fundamental postulate Each accessible microstate is equally likely (in isolated system)
 Because we don’t know enough about the system to believe otherwise
1
 Prob of a microstate: p= (total prob (1)/number of parts to that prob)
Ω
Notation
 N identical distinguishable particles
 Particle x, where x can be from 1 to N
 Single-particle state (j) for each particle: jx
 Microstate (Ψ) defines the single-particle states of each particle: Ψ = (j 1, j2, …jn)
o One microstate =one possible configuration (of particles in single particle states)
  Occupancy (n): number of particles in the jth single particle state  nj
 Configuration: specifies the number of particles in each single-particle state (list of occupancies) [n0,
n1, …, nj,… ]
 Statistical weight of a configuration (n): number of different microstates that can be formed from the
configuration
 Probability of a configuration ∝ its statistical weight (as all microstates equally likely)

Equilibrium configuration
=most likely  most microstates (highest statistical weight)

Boltzmann entropy

k just a proportionality constant to be invesitgated

Why this form? Einstein argument

Imagine two macrosystems A and B with statistical weights (number of microstates) Ω A and ΩB

Total number of microstates: Ω AB =Ω A × Ω B

Total entropy must be the sum of the two: SAB=SA+SB

So S must satisfy:

 Logarithm satisfies this

2. Eqm and max S

e.g. Spin ½ particles
- two possible single particle states (j=1,2)
o Ψ1  spin up
o Ψ2  spin down
 Total number of possible microstates: Ω=2N
 Boltzmann entropy:

- How many partiles are spin up and how many spin down? Given by the configuration n=[n1, n2]=[m, N-
m]
- For each value of m (each possible configuration), the number of different microstates possible (Ω) is
given by the number of combinations:
o Number of ways of picking out m particles to be spin up from a bag of N particles = NPm (order
matters as the first category is spin up, second spin down – this is a bad example as there are
only two but order matters when more possible single-particle states)

 division by m! to get rid of spin up repeats and by (N-m)! 

spin down  all spin-down are indistinguishable from each other and likewise for spin up
(only thing that you can distinguish is whether a particle is spin up or down)
So
 ≈
Aside: Stirling approx.:

For most likely case

This gives S=

=
Same as entropy of the large system (not exactly the same but we have used the str approx.)
As N inf, Seqm confSwhole system

Considering energy and constraints

- Isolated system with each atom described by

- Constraints

(sum over all possible single-particle states)

- For a given config: [n1, n2, …, nj,…]

- Boltzmann entropy S

- Factor in constraints with Lagrange method of undetermined multipliers

 ∂
∂ nj
[ ln Ω−α n j−β ϵ j n j + ( constats wrt n j ) ]=0
∂ ∂
∂ nj
ln Ω=
∂nj
(α n j + β ϵ j nj )
∂
∂ nj[ '
j ]
lnN !−∑ ln n'j ! =α + β ϵ j

After using Stirling:

i.e. Most likely distribution among particles:

THE BOLTZMANN DISTRIBUTION

SUMMARY: FINDING THE MOST LIKELY CONFIG

1. Write the statistical weight
2. Plop into Boltzmann entropy formula
3. Differentiate and =0 for the peak (most likely)
4. Add constraints
5. Solve

3. Boltzmann distribution
Temperature
Thermodynamics T
dU =TdS−PdV + μdN

Stat Phys T

- Change the internal energy U whilst keeping V and N fixed. (i.e. weak thermal contact)

 S changes:
 ∂ ln Ω
δS=k B ∑ δnj
j ∂n j

¿ k B ∑ ( α + β n j ) δ n j=k B α ∑ δ n j+ β ∑ ϵ j δ n j =k B ( α × 0+ βδU )
j ( j j )
As N is fixed, the first term =0. Overall:
δS=k B βδU
So

Comparing to the thermodynamic relationship, we get:

Finding α

Partition function

sum over single-particle states (the sum over energy levels includes a gj factor to
account for the degeneracy of each level).
So Boltzmann dist becomes:

Divide by N to get p

0th law
Two systems in weak themal contact (heat can flow between them but assume available energy levels don’t
change)
Max entropy at dS=0 so B1=B2  same temp at eqm

4. Systems in heat baths

Ensemble
= probability distribution for microstates of a system
- = very large assembly of identical systems
Microcanonical
Uniform PDF

Canonical
- System A in thermal eqm with heat bath B
- Rearrange Boltzmann entropy

- Small change in heat bath U

- New Omega for the heat bath

For the combined system

Canonical ensemble probability distribution

General entropy (Gibbs)

- A generalisation of Boltzmann entropy
- Reduces to Boltzmann entropy when the probs of all systems are equal
- N identical, independent systems (distinguishable)
Entropy of assembly

Single system

5. Partition function and thermodynamics

Mean energy

Fluctuations
Canonical ensemble  P dist over states with diff energies energy will fluctuate
(ΔE)2 =
Bridge equation
F=U-TS = - kBlnZ
Derivation:

So can then calculate

And plug into U=F+TS

Factorisation of partition function

- Weakly interacting components can be treated as independent
- E.g.
o Weakly interacting particles

For N particles


Which means that the total Helmholtz free energy and internal energies are given by:

o Energy components:
Examples
- Quantum harmonic oscillators
- Einstein solid

6. Variable particle number

Grand canonical ensemble
= prob dist for a system in eqm w a particle and heat bath
- Imagine an assembly of systems between which particles and energy can flows
o NA  total number of systems
o nj number of systems in microstate Psij

- Constraints
o Total energy fixed

o Total #systems fixed

o Total #particles fixed

sum of number of systems in a microstate x number of particles in

that microstate = number of systems x total number of particles.

Eqm distribution: Method of Lagrange multipliers

Grand partition function

Temp and chemical potential

Thermodynamics way

Stats way

V and U fixed

Compare the two ways

Mu  energy required to add a particle to the system

Sub into the Grand canonical ensemble:

Grand potential
Returning to stats entropy
Plug in beta and lambda (derived by comparing stats and thermo)
 both sides equal to grand potential Φ

Ensembles summary
Microcanonical Canonical Grand canonical
Probability of a
microstate
Isolated system Thermal eqm with a heat Heat and particle bath
bath
Fundamental postulate: Each microstate involves
for an isolated system, the system exchanging
all microstates are energy Ej with the heat
equally likely bath.

7. Indistinguishable particles and density of states

- For all the above we have been assuming distinguishable particles (e.g. in a lattice)
o Discrete states
−ϵ j

Z=∑ e
kB T

j
- For indistinguishable particles, (gas, liquid…)
o Possible states form a continuum
−ϵ
kB T
Z=∫ e dg
Where g is the density of states
- For an integral over energy dg= g(
ϵ ¿ dϵ → g ( ϵ ) is the number of states withenergies ¿ ϵ ¿ ϵ+ dϵ .
o Density of states in energy
o Similar to degeneracy for discrete energy levels

3 1
2 πV
g ( ϵ )= 3
( 2m ) 2 ϵ 2
h
Density of states g (derivation)

 (l,m,n are any ints)

 ( )
3
2π 2π
orthogonal states uniformly spread out in k-space, every as the minimum jump is in all 3
L L
directions.
Density g:

3 2
In an elemental vol of k-space , whered k=4 π k dk
2
V 3 V 2 Vk
dg= 3
d k= 3
4 π k dk= 2 dk
(2 π ) (2 π ) 2π
2
Vk
So g(k) = 2
density of states in k (wavenumber)
2π
To transform to the density of states in energy:

Differentiating the dispersion relationship and plugging:

- This is just for translational motion

o For each translational state there will be several internal states (e.g. spin)
  represent by including degeneracy factor D

3 1
2 πV
g ( ϵ )=D 3
(2 m)2 ϵ2
h

8. Classical ideal gas

Dilute limit
- Assume pjN <<1 most of the time the gas is so dilute that there are no particles in state j
o Negligible likelihood of particles competing for the same state
Partition function
If we treat the particles as independent (i.e. dilute limit), the partition function of the system is the product of the
individual partition functions.
However, as the particles are indistinguishable  divide by N!

Energy

Internal

Helmholtz

Pressure

 we have recovered the ideal gas law  triumph!!

Entropy – The Sackur-Tetrode equation
S for monatomic ideal gas

Maxwell-Boltzmann speed dist

Number of particles with speed v to v+dv

#particles at v = density of states at v x mean occupancy of states at v

Mean occupancy (f) of a state = Npj

Maxwell-Boltzmann dist
Thermal speed:

9. Quantum: Fermions and bosons

Maths
- System with set of single particle states shared by the particles
- Indistinguishable particles  only need the occupancy of the single-particle states to specify microstate
Each single particle state (j)
- Energy per particle in that state εj (energy level)
- Total energy of j = njεj
- Can treat each j as a system in a heat and particle bath
o Weakly interacting  energy levels (εj ) not affected
o (usually when you add particles, the energy levels shift  e.g. He. We are basically taking the
0th order energy approximation)
- Each microstate defined by different values of n
 - To get the normalisation, you need to sum over all microstates (and thus over all values of n)

- Mean occupancy of a single particle state:

Fermions Bosons
Spin Half-integer Integer
Wavefunction (wrt particle antisymmetric symmetric
exchange)
 Max occupancy per 1 (Pauli exclusion principle) No limit
single particle state
Partition function

(standard summation valid if

)
Mean occupancy

These distributions therefore tell you about the mean number of particles in a state.

FFD  <n> approx. 1 if e-u<<kT  full occupancy

10. Photon gas

Cavity(black-body) radiation
Photons
- Spin 1 bosons
- #particles not conserved µ=0
- So:
Photon density of states
Wave equation (light)

Solutions E=E0ei(k.r – ckt)  ck=w

As before:

Number of photons

- D=2 for travelling waves  2 independent polarisations

2
V 2 1 V ω
n ( ω ) dω=2 × 2 3
ω × ℏω
dω= 2 3
× ℏω
dω
2π c kB T π c kB T
e −1 e −1
Planck and Wien
The Planck radiation law
The Planck radiation law
(solves UV catastrophe)
Spectral energy density
 Energy/V/w (energy per unit volume and frequency range)
= Energy per photon x number of photons per frequency range per volume
n ( ω ) dω ℏ ω
3
u ( ω ) dω=ℏω × = 2 3 ℏω
dω
V π c k BT
e −1
Convert to frequency
ω=2 πν → dω=2 π dν
3
h 1 (2 πν ) 8 πh ν 3
u ( ν ) dν= 2 πdν= dν
2 π π 2 c 3 ℏk2 πν
T c 3 hν
k T
e B
−1 e −1 B

Wien’s displacement law

Stefan-Boltzmann Law
- Total energy density (energy per unit volume)  u(f) over all fs
 - Radiance = power per unit solid angle (P/A)
=energy/volume*speed/solid angle
1
L= Uc
4π
- Radiant exitance = radiation flux (power) leaving the cavity  integrate L over area
o Yellow Jacobian
o cos θ →Black bodies obey lambert’s cosine law (Lambertian)
π
2π 2

j ¿ =∫ ∫ L cos θ sin θ dθ dϕ=2 π

0 0
( L2 )=Lπ = Uc4 ∝T 4

- Plug in U  Stephan-Boltzmann law:

¿ 4
j =σ T
(Derive Stephan-Boltzmann from Planck: https://gotohaggstrom.com/Deriving%20the%20Stefan-Boltzmann%20law%20from%20Plancks%20law.pdf)

11. Bose-Einstein condensates

- Non-relativistic, massive bosons ( μ ≠ 0 ¿
- Sum of the occupancies of each state = total number of particles N

- If closely spaced and taking D=1

- If N is fixed, changing T will cause mu to adjust to maintain constant n (occupancy)

- Nµ=0 max num particles in a system at a T
- Suggests that a system of N particles can’t have a lower T that that above 

Error! The approximation that summing over the density of states becomes an integral is not valid at
low T (near ε=0)
o only valid when kBT>>Δε
Solution: Add extra term for ground state occupancy when T<TB

At n0 take ε=0
Using a Taylor expansion for the denominator:

So mu is close to 0.  for T<TB use Nmu=0

Fraction of particles in excited states

Below TB the number of particles in the ground state becomes significant.

 12. Fermi gases
At low temperatures the occupancy of states becomes a step function

Fermi energy
µ at T=0 (the energy it takes to add a particle to the system at 0K)
- Degenerate = when the fermi gas is at T=0 (not the other def -> multiple single particle states per
energy level)
- Total number of single-particle states up to εF = total particles in the system
o (as each particle fills one state)

D=2 for non-relativistic free electrons (here the usual def of

degeneracy)

Rearrange:
Fermi temperature

- Just a definition nothing to do with the T=0 used to specify eF

Cases:
1. T>>TF  fFD v small  dilute  Boltzmann
2. T~TF  quantum statistics
3. T<<TF  approximate as degenerate (T=0)

Degenerate fermi gas properties

As fFD 1 as T0

Number of particles
∞ ϵF ϵF 3
2A 2
N=∫ g ( ϵ ) f ❑DE ( ϵ ) dϵ =∫ g ( ϵ ) dϵ =∫ A ϵ 1/ 2 dϵ= ϵ
0 0 0
3

Pressure

Examples
Electrons in metal
White dwarfs