The Sommerfeld Expansion and Electrons in Metals

At $T=0$ the Fermi gas is frozen into its ground state and contributes nothing to the heat capacity. The classical equipartition estimate would give each electron $\frac{3}{2}{k}_B$, so a metal should carry a large electronic heat capacity on top of the lattice term. Experiment finds almost none at room temperature. The resolution is that only a thin shell of electrons near the Fermi surface can absorb thermal energy; the rest are locked below by fully occupied states above them. The Sommerfeld expansion makes this quantitative.

The thermally active shell

At finite temperature the occupation

departs from the step only within a few $k_{B}T$ of $\mu$. Its energy derivative $-\partial \bar{n}/\partial ϵ$ is a peak of width $\sim k_{B}T$ centered on $\mu$, normalized to unit area. Electrons more than a few $k_{B}T$ below the Fermi energy cannot be excited: the states just above them are already full, so the exclusion principle blocks any transition. Only the fraction $\sim k_{B}T/{ϵ}_F$ of electrons lying within the shell can move, and each that does gains energy $\sim k_{B}T$.

This picture gives the heat capacity up to a numerical factor without any integral. The thermal energy above the ground state is

so the heat capacity is $C_V=\partial \Delta U/\partial T\sim N{k}_{B}T/T_F$, linear in $T$ and smaller than the classical $\frac{3}{2}N{k}_B$ by the factor $T/T_F$. The Sommerfeld expansion supplies the exact coefficient.

The Sommerfeld expansion

Thermodynamic quantities are integrals of a smooth function $H(ϵ)$ against the Fermi function. For any $H$ that varies slowly on the scale $k_{B}T$ and vanishes fast enough at $ϵ=0$,¹

The leading term is the zero-temperature result with ${ϵ}_F$ replaced by $\mu$; the corrections are even powers of $k_{B}T$ weighted by odd derivatives of $H$ at the Fermi level. The expansion parameter is $(k_{B}T/{ϵ}_F{)}^2$, which is $\sim 10^{-5}$ for a metal at room temperature.

Fixing the density fixes $\mu (T)$ through $N={\int }_0^{\infty }g(ϵ)\bar{n}(ϵ)\mathbf{d}ϵ$. Applying the expansion with $H=g$ and using $g\propto {ϵ}^{1/2}$ so that $g^{\prime }/g=1/2ϵ$,

The chemical potential falls slightly below ${ϵ}_F$ as the gas warms, because the density of empty states just above ${ϵ}_F$ exceeds the density of filled states just below, and $\mu$ must drop to hold the particle number fixed.

The electronic heat capacity

Applying the expansion to the energy, $H=ϵg(ϵ)$, and eliminating $\mu (T)$ in favor of ${ϵ}_F$ gives

Differentiating at fixed volume,

The last form, written through $g({ϵ}_F)=3N/2{ϵ}_F$, is the general statement: the electronic heat capacity is set by the density of states at the Fermi level. Define the Sommerfeld coefficient

The linear law replaces the constant classical prediction. At room temperature $C_{\mathrm{el}}/\left(\frac{3}{2}N{k}_B\right)\sim T/T_F\sim 10^{-2}$: the electronic heat capacity is smaller than the classical value by two orders of magnitude, which is why it went unnoticed before quantum statistics.

Electron and phonon contributions in a metal

At low temperature a metal carries two heat-capacity terms: the linear electronic part $\gamma T$ and the Debye lattice part $A{T}^3$ from phonons. Their sum is

with ${\Theta }_D$ the Debye temperature. Dividing by $T$ linearizes the two contributions,

Plotting $C/T$ against $T^2$ gives a straight line whose intercept is $\gamma$ and whose slope is $A$. This extracts the density of states at the Fermi level from the intercept and the Debye temperature from the slope, and is the standard way to separate the electronic and lattice contributions in a low-temperature measurement.²

Pauli paramagnetism

A magnetic field $B$ splits the electron gas by spin. An electron's energy shifts by $\mp {\mu }_{B}B$ for spin aligned or anti-aligned with the field, where ${\mu }_B$ is the Bohr magneton. The two spin populations then fill to a common chemical potential, so the band lowered by the field holds more electrons than the raised band. The imbalance is set by the density of states at the Fermi level: each spin subband contributes $g({ϵ}_F)/2$, and its edge moves by ${\mu }_{B}B$, so the number transferred is $\frac{1}{2}g({ϵ}_F){\mu }_{B}B$ per band and the net magnetization is

The resulting Pauli susceptibility

is independent of temperature, in contrast to the Curie law $\chi \propto 1/T$ of localized moments. The reason is the same shell argument: only the electrons within $k_{B}T$ of the Fermi surface are free to flip, and their number is fixed by $g({ϵ}_F)$ rather than by the total $N$. Localized-moment paramagnetism uses every spin and therefore diverges as $T\to 0$; the degenerate gas uses only the Fermi-surface shell and stays finite.

A complementary orbital effect, Landau diamagnetism, arises from the quantization of cyclotron orbits and gives a negative contribution ${\chi }_L=-\frac{1}{3}{\chi }_P$ for free electrons, so the net susceptibility of an ideal gas is $\frac{2}{3}{\chi }_P$. Band-structure effects modify both terms in real metals, but the temperature-independent Pauli scale survives as the dominant feature of the conduction-electron magnetism.³

Summary

• Finite temperature blurs the Fermi step over a shell of width $k_{B}T$ around ${ϵ}_F$; only the fraction $\sim T/T_F$ of electrons in that shell respond thermally.
• The Sommerfeld expansion turns Fermi-function integrals into a series in $(k_{B}T/{ϵ}_F{)}^2$, giving $\mu (T)={ϵ}_F[1-\frac{{\pi }^2}{12}(k_{B}T/{ϵ}_F{)}^2]$ and $U(T)=\frac{3}{5}N{ϵ}_F[1+\frac{5{\pi }^2}{12}(k_{B}T/{ϵ}_F{)}^2]$.
• The electronic heat capacity is linear, $C_{\mathrm{el}}=\gamma T$ with $\gamma =\frac{{\pi }^2}{3}{k}_B^{2}g({ϵ}_F)$, resolving the missing-heat-capacity puzzle; the combined metal obeys $C=\gamma T+A{T}^3$, a line in $C/T$ versus $T^2$.
• Pauli paramagnetism gives a temperature-independent susceptibility ${\chi }_P={\mu }_0{\mu }_B^{2}g({ϵ}_F)$, with a Landau diamagnetic correction $-\frac{1}{3}{\chi }_P$ for the free gas.