Fermi Gas at Finite Temperature

At finite temperature, the Fermi-Dirac distribution “smears out” the step function at $\varepsilon_F$. The Sommerfeld expansion provides an asymptotic series for integrals of the form

$I = \int_0^\infty \frac{f(\varepsilon)}{e^{\beta(\varepsilon - \mu)} + 1}\, d\varepsilon$

When $k_BT \ll \varepsilon_F$ (the degenerate limit).

Theorem 4.1 (Sommerfeld Expansion). To leading order in $T/T_F$:

$I = \int_0^\mu f(\varepsilon)\, d\varepsilon + \frac{\pi^2}{6}(k_BT)^2 f"(\mu) + \mathcal{O}(T^4)$

Proof (sketch). Write $f(\varepsilon) = f(\mu) + f'(\mu)(\varepsilon - \mu) + \cdots$ and use the exact results:

$\int_0^\infty \frac{d\varepsilon}{e^{\beta(\varepsilon - \mu)} + 1} = \mu + \mathcal{O}(T^4)$

$\int_0^\infty \frac{(\varepsilon - \mu)\, d\varepsilon}{e^{\beta(\varepsilon - \mu)} + 1} = \frac{\pi^2}{6}(k_BT)^2$

$\int_0^\infty \frac{(\varepsilon - \mu)^2\, d\varepsilon}{e^{\beta(\varepsilon - \mu)} + 1} = \mathcal{O}(T^4)$

Combining these with the Taylor expansion of $f(\varepsilon)$ gives the result. The key integral identities follow from the substitution $x = \beta(\varepsilon - \mu)$ and the fact that the integrand is an odd function of $x$ to leading order. $\blacksquare$

Applying the Sommerfeld expansion to the number equation $N = \int_0^\infty g(\varepsilon) f_{\mathrm{FD}(\varepsilon)\, d\varepsilon}$ with $g(\varepsilon) = C\sqrt{\varepsilon}$:

$N = \frac{2}{3}C\mu^{3/2} + \frac{\pi^2}{6}(k_BT)^2 \cdot \frac{C}{2\sqrt{\mu}} + \mathcal{O}(T^4)$

At $T = 0$: $N = \frac{2}{3}C\varepsilon_F^{3/2}$. Expanding $\mu = \varepsilon_F + \delta\mu$ and keeping terms to $\mathcal{O}(T^2)$:

$\mu(T) \approx \varepsilon_F\left[1 - \frac{\pi^2}{12}\left(\frac{k_BT}{\varepsilon_F}\right)^2\right]$

The chemical potential decreases slightly with temperature.

Applying the Sommerfeld expansion to the energy:

$U = \int_0^\infty \varepsilon\, g(\varepsilon)\, f_{\mathrm{FD}(\varepsilon)\, d\varepsilon = \frac{2}{5}C\mu^{5/2} + \frac{\pi^2}{6}(k_BT)^2 \cdot \frac{3}{2}C\mu^{1/2} + \cdots}$

Substituting $\mu \approx \varepsilon_F$:

$U \approx \frac{3}{5}N\varepsilon_F\left[1 + \frac{5\pi^2}{12}\left(\frac{k_BT}{\varepsilon_F}\right)^2\right]$

$C_V = \frac{\partial U}{\partial T} = Nk_B \cdot \frac{\pi^2}{2}\frac{k_BT}{\varepsilon_F} = Nk_B \cdot \frac{\pi^2}{2}\frac{T}{T_F}$

Physical insight. At room temperature ($T \approx 300$ K), $T/T_F \approx 0.006$ for copper, so $C_V \approx 0.03 Nk_B$Which is negligible compared to the lattice contribution $\approx 3Nk_B$. This explains why the Dulong-Petit law works for metals despite the presence of conduction electrons.

Problem. Calculate the electronic contribution to $C_V$ for copper at $T = 300$ K. Compare with the lattice contribution. Given: $\varepsilon_F = 7.0$ eV, Debye temperature $\Theta_D = 343$ K.

• Forgetting the $T^2$ correction sign. The chemical potential decreases with temperature, not increases. Fix: The Sommerfeld expansion gives $\mu < \varepsilon_F$ because thermal excitations populate states above $\varepsilon_F$ while leaving holes below, shifting the average.
• Applying Sommerfeld expansion outside the degenerate regime. When $T \sim T_F$, the expansion parameter $(T/T_F)^2 \sim 1$ and the series diverges. Fix: The expansion only converges for $k_BT \ll \varepsilon_F$; use full numerical integration otherwise.
• Confusing $T_F$ with $T_c$ (critical temperature). Fermi temperature is a property of the ground state, unrelated to phase transitions. Fix: $T_F = \varepsilon_F/k_B$ is the degeneracy temperature scale, not a transition temperature.
• Electronic vs. lattice heat capacity crossover. At room temperature the electronic contribution is negligible, but below $\sim 10$ K it dominates. Fix: Compare $C_V^{\rm el} \propto T$ with $C_V^{\rm lat} \propto T^3$ at low $T$.

• Specific heat of metals: The linear $T$ term in $C_V$ at low temperatures is a hallmark of Fermi liquid behaviour and is used to extract the density of states at $\varepsilon_F$.
• Thermoelectric effect: The Sommerfeld expansion explains the Mott formula for thermopower, relating $S \propto T \cdot d(\ln \sigma)/d\varepsilon$ at $\varepsilon_F$.
• White dwarf cooling: Degenerate electron gas thermodynamics determines the heat capacity and cooling rate of white dwarfs, with $T_F \sim 10^8$ K.
• Heavy fermion systems: Materials with strongly renormalised effective masses show an enhanced Sommerfeld coefficient $\gamma = C_V/T$, signalling strong correlations.

Problem. For a 3D free electron gas at $T = 100$ K with $\varepsilon_F = 5$ eV, compute the fractional change in the chemical potential $\mu(T)$ relative to $\varepsilon_F$.

Solution. Using $\mu(T) \approx \varepsilon_F[1 - (\pi^2/12)(T/T_F)^2]$:

$T_F = \varepsilon_F/k_B = 5\;\mathrm{eV} / (8.617 \times 10^{-5}\;\mathrm{eV/K}) \approx 5.8 \times 10^4\;\mathrm{K}$

$\frac{T}{T_F} = \frac{100}{5.8 \times 10^4} \approx 1.72 \times 10^{-3}$

$\frac{\mu - \varepsilon_F}{\varepsilon_F} \approx -\frac{\pi^2}{12}(1.72 \times 10^{-3})^2 \approx -2.4 \times 10^{-6}$

The chemical potential decreases by only about $2.4 \times 10^{-4}\%$, confirming that $\mu \approx \varepsilon_F$ is an excellent approximation at ordinary temperatures.

$\blacksquare$