Degenerate Fermi Gas and the Sommerfeld Expansion

Derivation

Degenerate Fermi Gas and the Sommerfeld Expansion

D-243 Home PU-302 Threads matter · chance · energy Depends on Bose-Einstein and Fermi-Dirac Occupation Numbers

Statement

For a gas of \(N\) non-interacting spin-\(\tfrac12\) fermions of mass \(m\) confined to volume \(V\) at low temperature, we derive the Fermi energy \(\varepsilon_F=\frac{\hbar^2}{2m}\!\left(3\pi^2 n\right)^{2/3}\), the ground-state degeneracy pressure \(P_0=\tfrac{2}{5}\,n\,\varepsilon_F\), and — through the Sommerfeld low-temperature expansion of the Fermi–Dirac integrals — the linear-in-\(T\) electronic heat capacity \(C_V=\frac{\pi^2}{3}k_B^2\,T\,g(\varepsilon_F)=\frac{\pi^2}{2}Nk_B\,\dfrac{T}{T_F}\), valid for \(k_BT\ll\varepsilon_F\).

Why it matters

The conduction electrons in a metal are the archetypal degenerate Fermi gas: their Fermi temperature \(T_F\sim 10^4\text{–}10^5\,\mathrm{K}\) dwarfs room temperature, so they are frozen into their quantum ground state and only a thin shell of width \(k_BT\) around the Fermi surface can be thermally excited. This single fact resolves the classical catastrophe of the Drude model, where equipartition predicted an electronic heat capacity \(\tfrac{3}{2}Nk_B\) that experiment never showed.

The Sommerfeld expansion is the mathematical machine that turns "only a shell near \(\varepsilon_F\) participates" into quantitative predictions. It yields the electronic \(C_V\propto T\) that dominates the specific heat of metals below a few kelvin, fixes the degeneracy pressure that supports white dwarfs against gravity, and provides the template for every low-temperature Fermi-liquid property.

Assumptions

Non-interacting particles.If dropped, Coulomb correlations renormalise \(m\to m^*\) and shift the coefficients; the free-electron result becomes the leading term of Landau Fermi-liquid theory rather than the exact answer.

Parabolic dispersion \(\varepsilon(\mathbf{k})=\hbar^2k^2/2m\).If dropped (real band structure), \(g(\varepsilon)\) is no longer \(\propto\sqrt{\varepsilon}\) and both \(\varepsilon_F\) and the effective mass in \(C_V\) must be read from the actual bands.

Macroscopic volume, quasi-continuous spectrum.If dropped (quantum dots, nanoparticles), the level spacing \(\Delta\varepsilon\) becomes comparable to \(k_BT\), sums cannot be replaced by integrals, and the smooth density of states fails.

Degenerate limit \(k_BT\ll\varepsilon_F\).If dropped, the Sommerfeld series in \((k_BT/\varepsilon_F)^2\) diverges in usefulness; one must use the full Fermi–Dirac integrals, and near \(T\gtrsim T_F\) the gas crosses over to the classical Maxwell–Boltzmann regime.

Smoothness of the density of states across the shell \(|\varepsilon-\mu|\lesssim k_BT\).If dropped (van Hove singularity or band edge pinned at \(\mu\)), the Taylor expansion underlying Sommerfeld is invalid because \(g(\varepsilon)\) is non-analytic there, and the \((k_BT)^2\) coefficient is not simply \(g'(\mu)\).

Derivation

\[ N=\sum_{\mathbf{k},\,s}\Theta(k_F-|\mathbf{k}|)=2\cdot\frac{V}{(2\pi)^3}\cdot\frac{4}{3}\pi k_F^3=\frac{V k_F^3}{3\pi^2} \]

Count occupied single-particle states in the \(T=0\) ground state: fill every plane-wave orbital up to the Fermi wavevector \(k_F\). The factor \(2\) is spin, \(V/(2\pi)^3\) is the \(\mathbf{k}\)-space density of states from periodic boundary conditions. A

\[ k_F=\left(3\pi^2 n\right)^{1/3},\qquad n\equiv\frac{N}{V} \]

Invert step 1 for \(k_F\) in terms of the number density \(n\). Symbols only; no numbers yet. A

\[ \varepsilon_F=\frac{\hbar^2 k_F^2}{2m}=\frac{\hbar^2}{2m}\left(3\pi^2 n\right)^{2/3} \]

Substitute \(k_F\) into the free-particle dispersion. The Fermi energy is the highest occupied single-particle energy at \(T=0\). A

\[ g(\varepsilon)\,d\varepsilon=2\cdot\frac{V}{(2\pi)^3}\,4\pi k^2\,dk\ \Longrightarrow\ g(\varepsilon)=\frac{V}{2\pi^2}\left(\frac{2m}{\hbar^2}\right)^{3/2}\!\sqrt{\varepsilon} \]

Convert the \(\mathbf{k}\)-space shell \(4\pi k^2\,dk\) to an energy interval using \(k=\sqrt{2m\varepsilon}/\hbar\) and \(dk/d\varepsilon\). This defines the density of states \(g(\varepsilon)\), the number of orbitals per unit energy. B

\[ g(\varepsilon_F)=\frac{V}{2\pi^2}\left(\frac{2m}{\hbar^2}\right)^{3/2}\!\sqrt{\varepsilon_F}=\frac{3N}{2\varepsilon_F} \]

Evaluate \(g\) at the Fermi level and eliminate the prefactor using \(N=\int_0^{\varepsilon_F}g\,d\varepsilon=\frac{2}{3}g(\varepsilon_F)\varepsilon_F\). This compact form \(g(\varepsilon_F)=3N/2\varepsilon_F\) recurs throughout. B

\[ U_0=\int_0^{\varepsilon_F}\varepsilon\,g(\varepsilon)\,d\varepsilon=g(\varepsilon_F)\varepsilon_F^{-1/2}\!\int_0^{\varepsilon_F}\!\varepsilon^{3/2}d\varepsilon=\frac{3}{5}N\varepsilon_F \]

Integrate energy against the density of states over the filled Fermi sea. The mean energy per particle is \(\tfrac35\varepsilon_F\), not \(\tfrac12\varepsilon_F\), because \(g\propto\sqrt\varepsilon\) weights the top of the band. B

\[ U_0=\frac{3}{5}N\,\frac{\hbar^2}{2m}\left(3\pi^2\right)^{2/3}N^{2/3}V^{-2/3}\ \Longrightarrow\ U_0\propto V^{-2/3} \]

Make the volume dependence explicit by writing \(n=N/V\) inside \(\varepsilon_F\). At fixed \(N\), the ground-state energy rises as the gas is compressed — the origin of degeneracy pressure. A

\[ P_0=-\left(\frac{\partial U_0}{\partial V}\right)_{N}=-\left(-\frac{2}{3}\frac{U_0}{V}\right)=\frac{2}{3}\frac{U_0}{V}=\frac{2}{5}\,n\,\varepsilon_F \]

The ground state has zero entropy, so at \(T=0\) the mechanical pressure is \(-\partial U_0/\partial V\). Differentiating \(U_0\propto V^{-2/3}\) gives the exponent \(-\tfrac23\); insert \(U_0=\tfrac35 N\varepsilon_F\). B

\[ I\equiv\int_{-\infty}^{\infty}H(\varepsilon)\,f(\varepsilon)\,d\varepsilon=-\int_{-\infty}^{\infty}K(\varepsilon)\,\frac{\partial f}{\partial\varepsilon}\,d\varepsilon,\qquad K(\varepsilon)=\int_{-\infty}^{\varepsilon}\!H(\varepsilon')\,d\varepsilon' \]

Now turn on temperature. For any smooth \(H\) that vanishes as \(\varepsilon\to-\infty\), integrate the Fermi-weighted integral by parts. The boundary term dies because \(f(\varepsilon\to\infty)=0\) and \(K(\varepsilon\to-\infty)=0\); this trades \(f\) for the sharply peaked \(-\partial f/\partial\varepsilon\). C

\[ -\frac{\partial f}{\partial\varepsilon}=\frac{1}{k_BT}\frac{e^{x}}{(e^{x}+1)^2},\quad x\equiv\frac{\varepsilon-\mu}{k_BT}\;;\qquad K(\varepsilon)=\sum_{r\ge0}\frac{K^{(r)}(\mu)}{r!}(\varepsilon-\mu)^r \]

The weight \(-\partial f/\partial\varepsilon\) is even in \(x\), normalised, and concentrated within a few \(k_BT\) of \(\mu\). Taylor-expand \(K(\varepsilon)\) about \(\varepsilon=\mu\): only this locality makes the expansion legitimate. C

\[ \int_{-\infty}^{\infty}\!\frac{e^{x}}{(e^{x}+1)^2}\,dx=1,\quad \int_{-\infty}^{\infty}\!\frac{x\,e^{x}}{(e^{x}+1)^2}\,dx=0,\quad \int_{-\infty}^{\infty}\!\frac{x^2\,e^{x}}{(e^{x}+1)^2}\,dx=\frac{\pi^2}{3} \]

Evaluate the moments of the weight. The zeroth is unity (normalisation), the first vanishes by symmetry, the second is the standard integral \(2\sum_{k\ge1}(-1)^{k+1}/k^2\cdot\ldots=\pi^2/3\) (related to \(\zeta(2)\)). Odd moments vanish, so corrections are in even powers of \(k_BT\). C

\[ \int_{0}^{\infty}H(\varepsilon)\,f(\varepsilon)\,d\varepsilon=\int_{0}^{\mu}H(\varepsilon)\,d\varepsilon+\frac{\pi^2}{6}\left(k_BT\right)^2 H'(\mu)+\frac{7\pi^4}{360}\left(k_BT\right)^4 H'''(\mu)+\cdots \]

Assemble steps 9–11: the \(r=0\) moment gives \(\int_0^\mu H\,d\varepsilon\) (lower limit pushed to \(0\) since \(H\) is negligible below), the \(r=2\) moment gives the \(\frac{\pi^2}{6}H'(\mu)\) term using \(K''=H'\). This is the Sommerfeld expansion. C

\[ N=\int_0^\infty g\,f\,d\varepsilon\approx\int_0^\mu g\,d\varepsilon+\frac{\pi^2}{6}(k_BT)^2 g'(\mu) \]

Apply Sommerfeld with \(H=g\) to enforce particle-number conservation. \(N\) is fixed, so this equation determines how \(\mu\) drifts with \(T\). B

\[ 0\approx g(\varepsilon_F)\,(\mu-\varepsilon_F)+\frac{\pi^2}{6}(k_BT)^2 g'(\varepsilon_F)\ \Longrightarrow\ \mu=\varepsilon_F\left[1-\frac{\pi^2}{12}\left(\frac{k_BT}{\varepsilon_F}\right)^2\right] \]

Subtract the \(T=0\) identity \(N=\int_0^{\varepsilon_F}g\,d\varepsilon\), approximate \(\int_{\varepsilon_F}^{\mu}g\,d\varepsilon\approx g(\varepsilon_F)(\mu-\varepsilon_F)\), and use \(g'/g=1/2\varepsilon\) for \(g\propto\sqrt\varepsilon\). The chemical potential falls quadratically with \(T\). B

\[ U=\int_0^\infty \varepsilon\,g\,f\,d\varepsilon\approx\int_0^\mu \varepsilon g\,d\varepsilon+\frac{\pi^2}{6}(k_BT)^2\Big[g(\varepsilon_F)+\varepsilon_F g'(\varepsilon_F)\Big] \]

Apply Sommerfeld again with \(H=\varepsilon g\), so \(H'=g+\varepsilon g'\) evaluated at the Fermi level. This is the internal energy at temperature \(T\). B

\[ \int_0^\mu \varepsilon g\,d\varepsilon=U_0+\varepsilon_F g(\varepsilon_F)(\mu-\varepsilon_F)=U_0-\frac{\pi^2}{6}(k_BT)^2\,\varepsilon_F g'(\varepsilon_F) \]

Split the first integral about \(\varepsilon_F\) and insert the \(\mu\)-shift from step 14. Crucially, the term \(\varepsilon_F g'(\varepsilon_F)\) generated here is equal and opposite to the one inside the bracket of step 15. C

\[ U=U_0+\frac{\pi^2}{6}(k_BT)^2\,g(\varepsilon_F) \]

Add steps 15 and 16: the \(\varepsilon_F g'\) contributions cancel exactly, leaving a clean thermal energy proportional to \(g(\varepsilon_F)\). The cancellation is the physical statement that only the shell near \(\varepsilon_F\) stores thermal energy. B

\[ C_V=\left(\frac{\partial U}{\partial T}\right)_{N,V}=\frac{\pi^2}{3}k_B^2\,T\,g(\varepsilon_F)=\frac{\pi^2}{2}Nk_B\,\frac{T}{T_F},\qquad T_F\equiv\frac{\varepsilon_F}{k_B} \]

Differentiate step 17 with respect to \(T\), then substitute \(g(\varepsilon_F)=3N/2\varepsilon_F\) from step 5 and define the Fermi temperature. The heat capacity is linear in \(T\) and suppressed by \(T/T_F\ll1\) relative to the classical value. A

Result

\[ \varepsilon_F=\frac{\hbar^2}{2m}\!\left(3\pi^2 n\right)^{2/3},\qquad P_0=\frac{2}{5}\,n\,\varepsilon_F,\qquad C_V=\frac{\pi^2}{3}k_B^2\,T\,g(\varepsilon_F)=\frac{\pi^2}{2}Nk_B\,\frac{T}{T_F} \]

Reading. The Fermi energy is set purely by density and mass — denser or lighter gases push electrons to higher momenta. Even at \(T=0\) the gas exerts a large pressure \(P_0=\tfrac25 n\varepsilon_F\) because the Pauli principle forbids the particles from all sitting at rest. As temperature rises, only the fraction \(\sim T/T_F\) of electrons within \(k_BT\) of the Fermi surface can absorb energy, so each contributes \(\sim k_B\) but only \(\sim(T/T_F)N\) of them participate, giving \(C_V\sim Nk_B(T/T_F)\) — linear, not constant, and tiny at room temperature.

Units check. \(\varepsilon_F\): \(\frac{(\mathrm{J\,s})^2}{\mathrm{kg}}(\mathrm{m^{-3}})^{2/3}=\frac{\mathrm{J^2 s^2}}{\mathrm{kg\,m^2}}=\mathrm{J}\) since \(\mathrm{J}=\mathrm{kg\,m^2\,s^{-2}}\). \(P_0\): \(\mathrm{m^{-3}\cdot J}=\mathrm{J\,m^{-3}}=\mathrm{Pa}\). \(C_V\): \(k_B^2 T\,g=\mathrm{(J\,K^{-1})^2\cdot K\cdot J^{-1}}=\mathrm{J\,K^{-1}}\), a heat capacity, and \(Nk_B(T/T_F)\) is likewise \(\mathrm{J\,K^{-1}}\) with the ratio dimensionless.

Limiting cases

\(T\to0\): \(C_V\to0\) linearly and \(\mu\to\varepsilon_F\); the gas occupies its unique ground state with energy \(U_0=\tfrac35 N\varepsilon_F\) and pressure \(P_0=\tfrac25 n\varepsilon_F\).
\(k_BT\ll\varepsilon_F\) (degenerate metal): leading Sommerfeld term dominates, \(C_V=\gamma T\) with \(\gamma=\frac{\pi^2}{2}Nk_B/T_F\); the total low-\(T\) specific heat is \(C=\gamma T+\beta T^3\) (electrons + phonons).
\(k_BT\gtrsim\varepsilon_F\) (non-degenerate): the Sommerfeld series fails; \(f\to e^{-(\varepsilon-\mu)/k_BT}\) and the gas recovers the classical ideal gas with \(C_V\to\tfrac32 Nk_B\) and \(PV=Nk_BT\).
Ultra-relativistic (\(\varepsilon=\hbar c k\), e.g. white-dwarf cores): \(g\propto\varepsilon^2\), \(\varepsilon_F=\hbar c(3\pi^2 n)^{1/3}\), and \(P_0=\tfrac14 n\varepsilon_F\propto n^{4/3}\) — the softer exponent that permits the Chandrasekhar mass.
Two dimensions: \(g(\varepsilon)=\text{const}\), so \(g'(\varepsilon_F)=0\); the chemical potential is temperature-independent to all orders in the Sommerfeld sense and \(\mu(T)\) closes exactly.

Breaks when

High temperature, \(k_BT\gtrsim\varepsilon_F\): the expansion parameter \((k_BT/\varepsilon_F)^2\) is no longer small, successive Sommerfeld terms grow, and the low-\(T\) forms for \(\mu\), \(U\), and \(C_V\) are quantitatively wrong. One must integrate the full Fermi–Dirac functions.
Density of states non-smooth near \(\mu\): at a band edge, van Hove singularity, or gap, \(g(\varepsilon)\) is non-analytic within the thermal shell, so the Taylor expansion of \(K(\varepsilon)\) (step 10) is invalid and the \((k_BT)^2 g'(\mu)\) coefficient is meaningless.
Strong interactions / low density: when the Coulomb energy \(\sim e^2 n^{1/3}\) is comparable to \(\varepsilon_F\) (large \(r_s\)), correlations dominate, the free-electron \(g(\varepsilon)\) is wrong, and Wigner crystallisation or Mott physics can intervene.
Nanoscale confinement: when the discrete level spacing \(\Delta\varepsilon\sim\varepsilon_F/N\) approaches \(k_BT\), the continuum density of states and the integral-for-sum replacement both fail.

Failure modes

Using the classical equipartition value \(C_V=\tfrac32 Nk_B\) for the electrons — the Drude error that Sommerfeld corrects by the factor \(\sim T/T_F\).
Forgetting the spin factor of \(2\) in the state count, which shifts \(k_F\) by \(2^{1/3}\) and \(\varepsilon_F\) by \(2^{2/3}\).
Confusing \(\varepsilon_F\) (a \(T=0\) constant) with \(\mu(T)\) (temperature-dependent); writing \(\mu=\varepsilon_F\) in the heat-capacity derivation misses the cancellation in step 16–17.
Dropping the \(g'(\varepsilon_F)\) terms too early and concluding \(C_V\) has the wrong coefficient; the cancellation only works if both the \(\mu\)-shift and the \(\varepsilon_F g'\) term are kept.
Taking mean energy per particle as \(\tfrac12\varepsilon_F\) (as if \(g\) were constant) instead of \(\tfrac35\varepsilon_F\); this corrupts \(U_0\) and hence \(P_0\).
Setting the lower limit of the Sommerfeld integral to \(-\infty\) for \(g\propto\sqrt\varepsilon\) without noting that the error is exponentially small only because \(\mu\gg k_BT\).

Discussion

The deep message of this derivation is that quantum statistics, not interactions, is what makes a metal a metal. Because electrons are fermions, filling \(N\) orbitals forces occupation up to \(\varepsilon_F\), and the resulting Fermi surface — the sphere \(|\mathbf{k}|=k_F\) in the free case — is the object around which all low-energy physics organises. Every transport and thermodynamic property at low \(T\) is a statement about how states within \(k_BT\) of that surface respond. The linear heat capacity is simply the count of thermally accessible states, \(g(\varepsilon_F)k_BT\), times a typical excitation energy \(k_BT\), divided by \(T\).

The degeneracy pressure is the same Pauli principle read mechanically. It is not a thermal pressure — it survives at \(T=0\) — and it is enormous: for a metal \(P_0\) is tens of gigapascals, comparable to the bulk modulus, which is why metals are hard to compress. Scaled up, this is what holds a white dwarf against gravitational collapse; the crossover from the \(n^{5/3}\) non-relativistic pressure to the \(n^{4/3}\) relativistic pressure is precisely what caps the stable mass at the Chandrasekhar limit.

The exact cancellation of the \(\varepsilon_F g'(\varepsilon_F)\) terms between the number-conservation shift of \(\mu\) and the energy integral (steps 16–17) is not an accident but a manifestation of a deeper structure: the entropy of the Fermi gas, \(S=\frac{\pi^2}{3}k_B^2 T\,g(\varepsilon_F)\), equals \(C_V\) to this order, so \(U-\mu N\) and \(F\) organise into a universal form \(\propto g(\varepsilon_F)(k_BT)^2\). In Landau Fermi-liquid theory this generalises: interactions renormalise \(g(\varepsilon_F)\to g^*(\varepsilon_F)=m^* k_F/\pi^2\hbar^2\), and the measured Sommerfeld coefficient \(\gamma\) becomes a direct probe of the quasiparticle effective mass \(m^*\), which in heavy-fermion compounds can exceed the bare mass by factors of hundreds.

Common misconceptions. The Fermi energy is often mistaken for a thermal energy scale — it is not; \(\varepsilon_F\) exists at absolute zero. Likewise, "degenerate" here means the states are maximally occupied per the Pauli principle, not that energy levels coincide. And the smallness of the electronic heat capacity does not mean electrons are inert: they carry essentially all the current and dominate the thermal conductivity; it is only their capacity to store thermal energy that is quenched, because almost all of them are Pauli-blocked.

Worked examples

\[ \text{Fermi energy and Fermi temperature of copper},\quad n=8.49\times10^{28}\,\mathrm{m^{-3}} \]

Copper donates one \(4s\) electron per atom; use the free-electron model with the bare electron mass \(m=9.109\times10^{-31}\,\mathrm{kg}\).

\[ k_F=(3\pi^2 n)^{1/3}=\big(3\pi^2\cdot8.49\times10^{28}\big)^{1/3}=\big(2.513\times10^{30}\big)^{1/3}=1.36\times10^{10}\,\mathrm{m^{-1}} \]

\[ \varepsilon_F=\frac{\hbar^2 k_F^2}{2m}=\frac{(1.055\times10^{-34})^2(1.36\times10^{10})^2}{2(9.109\times10^{-31})}=1.13\times10^{-18}\,\mathrm{J} \]

\[ \varepsilon_F=\frac{1.13\times10^{-18}}{1.602\times10^{-19}}\,\mathrm{eV}=7.04\,\mathrm{eV},\qquad T_F=\frac{\varepsilon_F}{k_B}=\frac{1.13\times10^{-18}}{1.381\times10^{-23}}=8.18\times10^{4}\,\mathrm{K} \]

\[ \varepsilon_F\approx7.0\ \mathrm{eV},\qquad T_F\approx8.2\times10^{4}\ \mathrm{K} \]

Reading. Room temperature (\(300\,\mathrm{K}\)) is only \(0.4\%\) of \(T_F\): the copper electron gas is deeply degenerate, justifying the leading Sommerfeld term. The value \(7\,\mathrm{eV}\) matches the measured Fermi energy of copper to within the free-electron model's accuracy.

\[ \text{Electronic heat capacity of one mole of copper at }T=300\,\mathrm{K},\ \text{vs the lattice} \]

Use the molar Sommerfeld coefficient \(\gamma=\frac{\pi^2}{2}\dfrac{R}{T_F}\) with \(R=8.314\,\mathrm{J\,mol^{-1}K^{-1}}\) and \(T_F=8.18\times10^{4}\,\mathrm{K}\) from Example 1.

\[ \gamma=\frac{\pi^2}{2}\frac{R}{T_F}=\frac{9.870}{2}\cdot\frac{8.314}{8.18\times10^{4}}=5.02\times10^{-4}\,\mathrm{J\,mol^{-1}K^{-2}} \]

\[ C_V^{\text{el}}=\gamma T=(5.02\times10^{-4})(300)=0.151\,\mathrm{J\,mol^{-1}K^{-1}} \]

\[ \frac{C_V^{\text{el}}}{C_V^{\text{lat}}}\approx\frac{0.151}{3R}=\frac{0.151}{24.9}=6.1\times10^{-3} \]

\[ \gamma\approx0.50\ \mathrm{mJ\,mol^{-1}K^{-2}},\qquad \frac{C_V^{\text{el}}}{C_V^{\text{lat}}}\approx0.6\%\ \text{at }300\,\mathrm{K} \]

Reading. The free-electron prediction \(\gamma\approx0.50\,\mathrm{mJ\,mol^{-1}K^{-2}}\) sits close to the measured \(0.695\,\mathrm{mJ\,mol^{-1}K^{-2}}\); the ratio \(\gamma_{\exp}/\gamma_{\text{free}}\approx1.4\) is exactly the effective-mass enhancement \(m^*/m\). At room temperature the electronic term is a sub-percent correction to the Dulong–Petit lattice value, which is why classical theory could ignore it — but below \(\sim5\,\mathrm{K}\) the \(\gamma T\) term overtakes the \(\beta T^3\) phonon term and dominates.

Problems

Sodium has one conduction electron per atom and number density \(n=2.65\times10^{28}\,\mathrm{m^{-3}}\). Compute \(k_F\) and \(\varepsilon_F\) in eV.
Solution
\(k_F=(3\pi^2 n)^{1/3}=(3\pi^2\cdot2.65\times10^{28})^{1/3}=(7.845\times10^{29})^{1/3}=9.22\times10^{9}\,\mathrm{m^{-1}}\). \(\varepsilon_F=\dfrac{\hbar^2 k_F^2}{2m}=\dfrac{(1.055\times10^{-34})^2(9.22\times10^{9})^2}{2(9.109\times10^{-31})}=5.19\times10^{-19}\,\mathrm{J}=3.24\,\mathrm{eV}\), matching the accepted value for sodium.
Prove that the total ground-state energy of a 3D free-electron gas is \(U_0=\tfrac35 N\varepsilon_F\), starting from \(g(\varepsilon)=A\sqrt\varepsilon\).
Solution
\(N=\int_0^{\varepsilon_F}A\sqrt\varepsilon\,d\varepsilon=\tfrac23 A\varepsilon_F^{3/2}\), so \(A=\tfrac32 N\varepsilon_F^{-3/2}\). \(U_0=\int_0^{\varepsilon_F}\varepsilon\,A\sqrt\varepsilon\,d\varepsilon=A\int_0^{\varepsilon_F}\varepsilon^{3/2}d\varepsilon=\tfrac25 A\varepsilon_F^{5/2}\). Substitute \(A\): \(U_0=\tfrac25\cdot\tfrac32 N\varepsilon_F^{-3/2}\cdot\varepsilon_F^{5/2}=\tfrac35 N\varepsilon_F\). The mean energy per electron is therefore \(\tfrac35\varepsilon_F\).
Estimate the zero-temperature degeneracy pressure of the copper electron gas (\(n=8.49\times10^{28}\,\mathrm{m^{-3}}\), \(\varepsilon_F=1.13\times10^{-18}\,\mathrm{J}\)) and compare it to atmospheric pressure.
Solution
\(P_0=\tfrac25 n\varepsilon_F=0.4\,(8.49\times10^{28})(1.13\times10^{-18})=3.83\times10^{10}\,\mathrm{Pa}\approx38\,\mathrm{GPa}\). This is \(\approx3.8\times10^{5}\) atmospheres — comparable to copper's bulk modulus (\(\sim140\,\mathrm{GPa}\)), which is why the electron gas contributes substantially to a metal's resistance to compression. The pressure is purely quantum-mechanical: it exists at \(T=0\).
Using \(\mu(T)=\varepsilon_F\big[1-\tfrac{\pi^2}{12}(k_BT/\varepsilon_F)^2\big]\), find the fractional shift \((\varepsilon_F-\mu)/\varepsilon_F\) for copper (\(T_F=8.18\times10^{4}\,\mathrm{K}\)) at \(T=300\,\mathrm{K}\).
Solution
\(\dfrac{\varepsilon_F-\mu}{\varepsilon_F}=\dfrac{\pi^2}{12}\left(\dfrac{T}{T_F}\right)^2=\dfrac{9.870}{12}\left(\dfrac{300}{8.18\times10^{4}}\right)^2=0.8225\,(3.67\times10^{-3})^2=1.1\times10^{-5}\). The chemical potential is essentially pinned at \(\varepsilon_F\) — a shift of about one part in \(10^{5}\) — confirming that treating \(\mu\approx\varepsilon_F\) in most estimates is excellent, while keeping the shift is essential only for the exact \(C_V\) cancellation.
At what temperature does the electronic heat capacity of a free-electron metal equal its classical lattice heat capacity \(3Nk_B\)? Express the answer as a multiple of \(T_F\) and evaluate for copper.
Solution
Set \(\tfrac{\pi^2}{2}Nk_B(T/T_F)=3Nk_B\Rightarrow T=\dfrac{6}{\pi^2}T_F=0.608\,T_F\). For copper \(T=0.608\,(8.18\times10^{4}\,\mathrm{K})=4.97\times10^{4}\,\mathrm{K}\) — far above the melting point, confirming that the electronic term never rivals the lattice term at any temperature where the solid exists. (In practice the electronic term only dominates the lattice term at the opposite extreme, below a few kelvin, where the phonon \(\beta T^3\) term is what it overtakes, not the classical \(3Nk_B\).)