35C (Statistical Physics)

In two dimensions, each allowed wavevector occupies area

(2 π)^{2} / A

k

-space. Including the two spin states, the number of states in an annulus

(k, k + d k)

d N = 2 \frac{A}{( 2 π ) ^{2}} 2 πk d k .

(1)

Since

E = \frac{ℏ ^{2} k ^{2}}{2 m}, k d k = \frac{m}{ℏ ^{2}} d E,

(2)

the two-dimensional free-electron density of states is

g (E) = \frac{d N}{d E} = \frac{A m}{π ℏ ^{2}}

(3)

for

E \geq 0

. The Fermi-Dirac distribution is

f (E) = \frac{1}{e ^{β (E - μ)} + 1}, β = \frac{1}{k _{B} T} .

(4)

b

 0  0

Solution

 0  0

At zero temperature all states up to

E_{F}

are occupied. Writing

g = \frac{A m}{π ℏ ^{2}},

(1)

we find

N = \int_{0}^{E_{F}} g d E = g E_{F}

(2)

and

E_{tot} = \int_{0}^{E_{F}} E g d E = \frac{1}{2} g E_{F}^{2} = \frac{1}{2} N E_{F} .

(3)

Because

g = C A

with

C = m / (π ℏ^{2})

, fixed particle number gives

E_{F} = N / (C A)

and

E_{tot} = \frac{N ^{2}}{2 C A} .

(4)

The two-dimensional degeneracy pressure of a two-dimensional Fermi gas is therefore

p = - (\frac{\partial E _{tot}}{\partial A})_{N} = \frac{N ^{2}}{2 C A ^{2}} = \frac{N E _{F}}{2 A} .

(5)

c

 0  0

Solution

 0  0

Keep the chemical potential at its zero-temperature value

μ = E_{F}

and put

E^{'} = E - E_{F}

. Away from

E^{'} = 0

f (E) - 1_{E < E_{F}} = \frac{1}{2} [sgn (\frac{β E ^{'}}{2}) - tanh (\frac{β E ^{'}}{2})] .

(1)

Consequently

Δ N = X \int_{- E_{F}}^{\infty} [sgn (\frac{β E ^{'}}{2}) - tanh (\frac{β E ^{'}}{2})] d E^{'},

(2)

where

X = \frac{g}{2} = \frac{A m}{2 π ℏ ^{2}} .

(3)

The bracket is an odd function localized to

∣ E^{'} ∣ ≲ k_{B} T

. When

β E_{F} ≫ 1

, extending the lower limit to

- \infty

makes the integral vanish exactly. The omitted tail is exponentially small; indeed direct integration gives

Δ N = \frac{g}{β} lo g (1 + e^{- β E_{F}}) .

(4)

This is the low-temperature particle-number cancellation for constant density of states.

For the energy,

Δ E_{tot} = g \int_{- E_{F}}^{\infty} (E_{F} + E^{'}) [f (E_{F} + E^{'}) - 1_{E^{'} < 0}] d E^{'} = E_{F} Δ N + X \int_{- E_{F}}^{\infty} E^{'} [sgn (\frac{β E ^{'}}{2}) - tanh (\frac{β E ^{'}}{2})] d E^{'} .

(5)

At low temperature the first term is exponentially small. In the second, extend the lower limit to

- \infty

and set

z = β E^{'} /2

Δ E_{tot} \sim \frac{4 X}{β ^{2}} \int_{- \infty}^{\infty} z [sgn (z) - tanh z] d z .

(6)

The remaining integral is a finite positive constant, proving the quadratic low-temperature energy correction for constant density of states:

Δ E_{tot} \propto β^{- 2} \propto T^{2} .

(7)

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Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 3 / 35C (Statistical Physics)

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