Sommerfeld Free-Electron Heat Capacity
Statement
For a three-dimensional gas of non-interacting electrons in the degenerate limit \(k_{\mathrm B}T \ll \varepsilon_{\mathrm F}\), the Sommerfeld expansion of the Fermi-Dirac occupation applied to the internal energy yields an electronic heat capacity that is linear in temperature, \(C_{\mathrm{el}} = \gamma T\) with \(\gamma = \tfrac{\pi^2}{3}k_{\mathrm B}^2\, g(\varepsilon_{\mathrm F})\), equivalently \(C_{\mathrm{el}} = \tfrac{\pi^2}{2}Nk_{\mathrm B}\,(T/T_{\mathrm F})\).
Why it matters
Classical equipartition assigns every electron \(\tfrac{3}{2}k_{\mathrm B}\), predicting a temperature-independent electronic heat capacity of order \(\tfrac{3}{2}Nk_{\mathrm B}\). Measured electronic heat capacities of metals are smaller than this by a factor of order \(T/T_{\mathrm F}\sim 10^{-2}\) at room temperature, and they vanish linearly as \(T\to 0\). The Sommerfeld result resolves this discrepancy: only the thin shell of states within \(\sim k_{\mathrm B}T\) of the Fermi surface can be thermally excited, so a fraction \(\sim T/T_{\mathrm F}\) of the electrons each absorb \(\sim k_{\mathrm B}T\).
The linear term \(\gamma T\) is the experimental fingerprint of the Fermi surface. It dominates the lattice \(\propto T^3\) contribution below a few kelvin, so plotting \(C/T\) against \(T^2\) isolates \(\gamma\), from which the density of states at the Fermi level — and, through many-body enhancement, effective masses — is extracted.
Assumptions
Derivation
Result
Reading. The electronic heat capacity rises linearly from zero as the metal is warmed. Its slope \(\gamma\) (the Sommerfeld coefficient) is a direct measure of the density of single-particle states at the Fermi energy. Physically, only electrons within an energy shell \(\sim k_{\mathrm B}T\) of \(\varepsilon_{\mathrm F}\) — a fraction \(\sim T/T_{\mathrm F}\) of the total — are thermally active, and each carries \(\sim k_{\mathrm B}T\) of excess energy, so \(U-U_0\sim g(\varepsilon_{\mathrm F})(k_{\mathrm B}T)^2\) and \(C\sim k_{\mathrm B}^2 g(\varepsilon_{\mathrm F})T\).
Units check. \([g(\varepsilon_{\mathrm F})]=\mathrm{J^{-1}}\) (states per joule), \([k_{\mathrm B}^2]=\mathrm{J^2\,K^{-2}}\), \([T]=\mathrm K\), so \([\gamma T]=\mathrm{J^{-1}\cdot J^2 K^{-2}\cdot K}=\mathrm{J\,K^{-1}}\), a heat capacity. In the second form \([N k_{\mathrm B}]=\mathrm{J\,K^{-1}}\) and \(T/T_{\mathrm F}\) is dimensionless. Consistent.
Limiting cases
- \(T\to 0\): \(C_{\mathrm{el}}\to 0\) linearly, satisfying the third law (unlike the classical \(\tfrac32 Nk_{\mathrm B}\)).
- \(T\ll T_{\mathrm F}\) (metals at ordinary \(T\)): \(C_{\mathrm{el}}/(\tfrac32 Nk_{\mathrm B}) = \tfrac{\pi^2}{3}(T/T_{\mathrm F})\ll 1\) — heat capacity strongly suppressed below the classical value.
- \(k_{\mathrm B}T \gtrsim \varepsilon_{\mathrm F}\): the expansion fails; the gas becomes non-degenerate and \(C_{\mathrm{el}}\to \tfrac32 Nk_{\mathrm B}\) (classical equipartition).
- Low-\(T\) total: \(C = \gamma T + \beta T^3\); the electronic term dominates the phonon \(T^3\) term below \(T^\ast=\sqrt{\gamma/\beta}\), typically a few kelvin.
Breaks when
- Fermi level at a van Hove singularity or band edge. If \(g(\varepsilon)\) is non-analytic within \(k_{\mathrm B}T\) of \(\varepsilon_{\mathrm F}\), the Taylor step (2) is invalid and \(C_{\mathrm{el}}\) is no longer linear — one sees enhanced, temperature-dependent \(\gamma(T)\).
- Non-degenerate / high-temperature regime, \(k_{\mathrm B}T\gtrsim\varepsilon_{\mathrm F}\). The Sommerfeld series diverges; the chemical potential leaves the band, and the crossover to the classical ideal gas must be treated with the full Fermi-Dirac integral.
- Strong correlations / non-Fermi-liquid behaviour. Near a quantum critical point or in heavy-fermion and some cuprate systems, \(C_{\mathrm{el}}/T\) diverges logarithmically or as a power law, so no constant \(\gamma\) exists.
- Gapped or superconducting state. Opening an energy gap \(\Delta\) at \(\varepsilon_{\mathrm F}\) replaces the linear law with an exponential \(\sim e^{-\Delta/k_{\mathrm B}T}\) (s-wave) or a power law (nodal gaps).
Failure modes
- Forgetting the \(\mu(T)\) shift. Dropping Step 5 leaves a spurious \(\varepsilon_{\mathrm F}g'(\varepsilon_{\mathrm F})\) term; the exact cancellation in Step 9 is missed and the coefficient comes out wrong.
- Using equipartition. Assigning \(\tfrac32 k_{\mathrm B}\) per electron overestimates \(C_{\mathrm{el}}\) by \(\sim T_{\mathrm F}/T\sim 100\) at room temperature.
- Confusing \(g(\varepsilon_{\mathrm F})\) with \(N/\varepsilon_{\mathrm F}\). The free-electron result is \(g(\varepsilon_{\mathrm F})=\tfrac{3N}{2\varepsilon_{\mathrm F}}\); dropping the factor \(\tfrac32\) gives a \(33\%\) error in \(\gamma\).
- Per-spin vs total density of states. Failing to include both spin orientations halves \(g(\varepsilon_{\mathrm F})\) and hence \(\gamma\).
- Keeping the \(\tfrac{\pi^2}{6}\) but writing \(\tfrac{\pi^2}{6}\) instead of \(\tfrac{\pi^2}{3}\) in \(C\). The derivative of \(T^2\) supplies the factor 2; a common slip leaves \(\gamma\) too small by half.
- Treating \(\gamma\) as measuring the bare band mass. The measured \(\gamma\) is enhanced by electron-phonon and electron-electron interactions, so \(\gamma_{\mathrm{exp}}=(1+\lambda)\gamma_{\mathrm{band}}\).
Discussion
The physical heart of the result is the exact cancellation in Steps 8-9. Naively both the number and energy integrals produce \(O(T^2)\) terms proportional to \(g'(\varepsilon_{\mathrm F})\), yet the requirement that particle number be conserved forces \(\mu\) to drift by exactly the amount that annihilates the \(g'\) contribution to the energy. What survives depends only on \(g(\varepsilon_{\mathrm F})\): the heat capacity is a pure probe of how many states sit at the Fermi surface, blind to the slope of the band. This is why \(\gamma\) is such a clean experimental quantity.
Because \(C_{\mathrm{el}}=\gamma T\) while the Debye lattice contribution is \(\beta T^3\), the standard low-temperature analysis plots \(C/T = \gamma + \beta T^2\) against \(T^2\): the intercept gives \(\gamma\) and the slope gives \(\beta\) (hence the Debye temperature). Comparing the measured \(\gamma\) with the free-electron prediction defines the thermodynamic effective mass \(m^\ast/m = \gamma_{\mathrm{exp}}/\gamma_{\mathrm{free}}\), which lumps together band-structure curvature and many-body renormalisation.
The same Sommerfeld machinery gives the electronic entropy \(S_{\mathrm{el}}=\gamma T\) (identical coefficient, since \(C=T\,\partial S/\partial T\) with \(S\propto T\)) and, through the analogous expansion of the grand potential, the \(T^2\) correction to the chemical potential and the Pauli paramagnetic susceptibility \(\chi_{\mathrm P}=\mu_0\mu_{\mathrm B}^2 g(\varepsilon_{\mathrm F})\). The ratio \(\chi_{\mathrm P}/\gamma\) (the Wilson ratio) is a fixed number for free electrons and a sensitive diagnostic of correlations when it deviates.
At the next order the expansion yields a \(T^3\ln T\) electronic term and \(O(T^2)\) corrections to \(\gamma\) that carry information about \(g''(\varepsilon_{\mathrm F})\); in a genuine Fermi liquid these are reorganised into quasiparticle interaction (Landau) parameters, and the leading linear \(\gamma T\) survives with a renormalised coefficient. The Sommerfeld derivation is thus the \(T=0\) fixed-point statement of Fermi-liquid thermodynamics: interactions dress the quasiparticles but preserve the linear law until the Fermi-liquid description itself breaks down.
Common misconceptions. The linear-in-\(T\) heat capacity does not mean electrons individually gain energy linearly in \(T\); it reflects that the number of thermally excited electrons grows \(\propto T\) while each gains \(\propto T\), and the product \(\propto T^2\) in \(U\) differentiates to \(\propto T\). Also, \(\gamma\) is not "the heat capacity of one electron" — it is a collective property fixed by the Fermi-surface density of states, not by the total electron count alone.
Worked examples
Reading. The measured value for copper is \(\gamma_{\mathrm{exp}}=0.695\ \mathrm{mJ\,mol^{-1}K^{-2}}\), giving \(m^\ast/m\approx1.37\) — modest enhancement from band structure and electron-phonon coupling, confirming copper is a good nearly-free-electron metal. Units. \(\mathrm{J\,mol^{-1}K^{-2}}\), correct for a molar \(\gamma\).
Reading. Below about \(0.9\ \mathrm K\) the electronic term dominates the specific heat of potassium; above it phonons take over. This is why \(\gamma\) is measured in the sub-kelvin regime by plotting \(C/T\) vs \(T^2\). Units. \(\sqrt{(\mathrm{J\,mol^{-1}K^{-2}})/(\mathrm{J\,mol^{-1}K^{-4}})}=\sqrt{\mathrm K^2}=\mathrm K\). Correct.
Problems
- Show that the fraction of conduction electrons thermally excited at temperature \(T\) is of order \(T/T_{\mathrm F}\), and use it to argue \(C_{\mathrm{el}}\sim Nk_{\mathrm B}(T/T_{\mathrm F})\).
Solution
Electrons within \(\sim k_{\mathrm B}T\) of \(\varepsilon_{\mathrm F}\) can be excited. Their number is \(\Delta N\approx g(\varepsilon_{\mathrm F})\,k_{\mathrm B}T = \tfrac{3N}{2\varepsilon_{\mathrm F}}k_{\mathrm B}T = \tfrac32 N\,(T/T_{\mathrm F})\), so the excited fraction is \(\sim T/T_{\mathrm F}\). Each gains \(\sim k_{\mathrm B}T\), so \(U-U_0\sim \Delta N\,k_{\mathrm B}T\sim Nk_{\mathrm B}T^2/T_{\mathrm F}\), and \(C=\partial U/\partial T\sim Nk_{\mathrm B}(T/T_{\mathrm F})\), matching the exact \(\tfrac{\pi^2}{2}Nk_{\mathrm B}(T/T_{\mathrm F})\) up to the \(O(1)\) factor. - Silver has \(\varepsilon_{\mathrm F}=5.49\ \mathrm{eV}\). Compute the free-electron molar Sommerfeld coefficient and the ratio \(C_{\mathrm{el}}/(\tfrac32 R)\) at \(T=300\ \mathrm K\).
Solution
\(T_{\mathrm F}=5.49\ \mathrm{eV}/k_{\mathrm B}=6.37\times10^{4}\ \mathrm K\). \(\gamma=\tfrac{\pi^2}{2}R/T_{\mathrm F}=4.935\times8.314/6.37\times10^{4}=6.44\times10^{-4}\ \mathrm{J\,mol^{-1}K^{-2}}\approx0.64\ \mathrm{mJ\,mol^{-1}K^{-2}}\). At \(300\ \mathrm K\), \(C_{\mathrm{el}}=\gamma T=0.193\ \mathrm{J\,mol^{-1}K^{-1}}\). Ratio to classical: \(C_{\mathrm{el}}/(\tfrac32 R)=\tfrac{\pi^2}{3}(T/T_{\mathrm F})=3.29\times(300/6.37\times10^{4})=1.55\times10^{-2}\), i.e. the electronic heat capacity is about \(1.5\%\) of the equipartition value. - Starting from \(U-U_0=\tfrac{\pi^2}{6}(k_{\mathrm B}T)^2 g(\varepsilon_{\mathrm F})\), derive the electronic entropy \(S_{\mathrm{el}}(T)\) and verify \(S_{\mathrm{el}}=C_{\mathrm{el}}\).
Solution
\(C_{\mathrm{el}}=\partial U/\partial T=\tfrac{\pi^2}{3}k_{\mathrm B}^2 g(\varepsilon_{\mathrm F})T=\gamma T\). Then \(S_{\mathrm{el}}=\int_0^T \tfrac{C_{\mathrm{el}}}{T'}dT'=\int_0^T \gamma\,dT'=\gamma T\). Hence \(S_{\mathrm{el}}=\gamma T = C_{\mathrm{el}}\); both vanish linearly as \(T\to0\), consistent with the third law. The Helmholtz free energy contribution is \(F_{\mathrm{el}}=U-U_0-TS_{\mathrm{el}}=\tfrac12\gamma T^2-\gamma T^2=-\tfrac12\gamma T^2\). - A metal shows \(C/T = 1.35 + 12.0\,T^2\) in \(\mathrm{mJ\,mol^{-1}K^{-2}}\) (with \(T\) in K). Extract \(\gamma\), \(\beta\), the Debye temperature, and the crossover temperature \(T^\ast\).
Solution
Intercept: \(\gamma=1.35\ \mathrm{mJ\,mol^{-1}K^{-2}}\). Slope: \(\beta=12.0\ \mathrm{mJ\,mol^{-1}K^{-4}}=1.20\times10^{-2}\ \mathrm{J\,mol^{-1}K^{-4}}\). From \(\beta=\tfrac{12\pi^4}{5}R/\Theta_{\mathrm D}^3\): \(\Theta_{\mathrm D}^3=233.8\times8.314/1.20\times10^{-2}=1.62\times10^{5}\), so \(\Theta_{\mathrm D}=(1.62\times10^5)^{1/3}=54.6\ \mathrm K\). Crossover: \(T^\ast=\sqrt{\gamma/\beta}=\sqrt{1.35/12.0}=\sqrt{0.1125}=0.335\ \mathrm K\). - Estimate the ground-state (zero-point) total energy \(U_0=\tfrac35 N\varepsilon_{\mathrm F}\) and compare the \(T=300\ \mathrm K\) thermal excess \(U-U_0\) with it for copper (\(\varepsilon_{\mathrm F}=7.00\ \mathrm{eV}\)), per mole.
Solution
\(U_0=\tfrac35 N_{\mathrm A}\varepsilon_{\mathrm F}=0.6\times6.022\times10^{23}\times7.00\times1.602\times10^{-19}\ \mathrm J=4.05\times10^{5}\ \mathrm{J\,mol^{-1}}\). Thermal excess: \(U-U_0=\tfrac{\pi^2}{6}(k_{\mathrm B}T)^2 g(\varepsilon_{\mathrm F})=\tfrac12\gamma T^2\) with \(\gamma=0.51\ \mathrm{mJ\,mol^{-1}K^{-2}}\): \(U-U_0=0.5\times5.1\times10^{-4}\times(300)^2=23.0\ \mathrm{J\,mol^{-1}}\). Ratio \((U-U_0)/U_0=23.0/4.05\times10^5=5.7\times10^{-5}\): even at room temperature the thermal energy is a tiny fraction of the degenerate ground-state energy, confirming deep degeneracy.