Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 3 / 35C / c

Solution

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})] .

Consequently

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

where

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

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}}) .

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^{'} .

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 .

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

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