Sommerfeld Expansion

Almost every property of a metal's electrons — how much heat they store, how their pressure resists compression, how their magnetism responds to a field — is an integral of some smooth function $H(\varepsilon)$ against the Fermi-Dirac occupation $f(\varepsilon)=1/(e^{\beta(\varepsilon-\mu)}+1)$ . At $T=0$ that occupation is a vertical cliff at the Fermi energy and the integral is trivial: fill everything below $\varepsilon_F$ , nothing above. The Sommerfeld expansion is the machine that computes the corrections as the cliff softens with warming, and it delivers them as a clean power series in the tiny number $(k_BT/\mu)^2$ . The whole trick is that warming only touches a thin shell. The function $-\partial f/\partial\varepsilon$ is a sharp, unit-area bump sitting at $\varepsilon=\mu$ with width $\sim k_BT$ ; only states within a few $k_BT$ of the chemical potential are thermally active. Integrating a Fermi integral by parts turns it into that bump convolved with the antiderivative of $H$ , and Taylor-expanding $H$ about $\mu$ leaves a sum of the bump's moments. Because the bump is even, odd moments die and the series contains only even powers of temperature — the reason metallic properties depend on $T^2$ , not $T$ . What makes the result quotable is that those even moments are universal numbers. The leading one is $\tfrac{\pi^2}{6}$ — the same $\zeta(2)=\pi^2/6$ that Euler found summing $1/n^2$ — and it fixes the linear-in- $T$ electronic heat capacity that first told physicists the electrons in a metal are a degenerate Fermi gas, not a classical one. This article owns the expansion, its coefficient $\tfrac{\pi^2}{6}$ , and the resulting downward shift of the chemical potential; the heat-capacity payoff it enables is developed in aq-sm-electronic-heat-capacity-metals.

Key takeaways

The Sommerfeld expansion is the low-temperature asymptotic expansion of Fermi integrals $\int_0^\infty H(\varepsilon)f(\varepsilon)\,\mathrm{d}\varepsilon$ in powers of $k_BT/\mu$ , valid in the degenerate regime $T\ll T_F$ .
Master formula: $\int_0^\infty H f\,\mathrm{d}\varepsilon=\int_0^\mu H\,\mathrm{d}\varepsilon+\tfrac{\pi^2}{6}(k_BT)^2 H'(\mu)+\tfrac{7\pi^4}{360}(k_BT)^4 H'''(\mu)+\cdots$
Only the shell of width $\sim k_BT$ around $\mu$ matters: the corrections are moments of the even bump $-\partial f/\partial\varepsilon$ , so only even powers of $T$ appear.
The coefficients are universal — $\tfrac{\pi^2}{6}$ and $\tfrac{7\pi^4}{360}$ — set by the Dirichlet eta values $\eta(2)=\pi^2/12$ , $\eta(4)=7\pi^4/720$ .
Holding $N$ fixed forces the chemical potential to drop: $\mu(T)=\varepsilon_F\big[1-\tfrac{\pi^2}{12}(k_BT/\varepsilon_F)^2\big]$ for a 3D free gas.
Key idea: a Fermi integral equals its $T=0$ value plus corrections set by the derivatives of the integrand at the Fermi level, weighted by the universal numbers $\tfrac{\pi^2}{6},\tfrac{7\pi^4}{360},\dots$ — only the immediate neighbourhood of the Fermi surface responds to heat.

The core insight

At low temperature a Fermi integral equals its $T=0$ value plus corrections set by the derivatives of the integrand at the Fermi level, weighted by the universal numbers $\tfrac{\pi^2}{6},\tfrac{7\pi^4}{360},\dots$ . Physically, only electrons within $\sim k_BT$ of the Fermi surface can absorb energy, so every thermal property of a degenerate Fermi gas is controlled by the density of states and its slope right at $\varepsilon_F$ .

Where this lives today

The Sommerfeld expansion is the daily workhorse of metal physics and degenerate-matter astrophysics. In the lab, the low-temperature specific heat of a metal is fit to $C=\gamma T+\beta T^3$ ; plotting $C/T$ against $T^2$ gives a straight line whose intercept is the electronic Sommerfeld coefficient $\gamma=\tfrac{\pi^2}{3}k_B^2g(\varepsilon_F)$ and whose slope is the Debye phonon term. That intercept is a direct measurement of the density of states at the Fermi level, and comparing it to the free-electron value defines the thermal effective mass $m^*$ — in heavy-fermion compounds like CeCu $_6$ , $\gamma$ is hundreds of times the free-electron value, the fingerprint of enormously enhanced quasiparticle masses. The same machinery gives the temperature corrections to the Pauli spin susceptibility, the thermopower through the Mott formula, and the finite-temperature corrections to degeneracy pressure that refine white-dwarf and neutron-star mass–radius relations. In every case the message is identical: because the gas is degenerate, its thermodynamics is dictated by the density of states and its derivatives right at the Fermi surface — precisely the quantities the Sommerfeld expansion isolates.

The physical picture

Picture the occupation $f(\varepsilon)$ as a melting ice cliff. At $T=0$ it is a vertical wall at $\varepsilon_F$ : every level below is full, every level above is empty. Warming rounds off the top over a band of thermal width $\sim k_BT$ — a few electrons just below the wall hop to a few empty states just above. Everything deep in the sea is inert, locked at $\bar n=1$ with no empty neighbour to move into, and everything far above is locked at $\bar n=0$ . All thermal response therefore lives in that thin active band, and it is governed by how the density of states behaves at the cliff edge. The mathematical engine is that $-\partial f/\partial\varepsilon$ is a normalized, nearly symmetric bump centred on $\mu$ . Convolving a smooth function with a symmetric bump returns the function's value at the centre plus curvature corrections weighted by the bump's even moments; the odd moments vanish by symmetry. That single fact — a symmetric weight — is why the answer contains only even powers of $T$ and why the first correction is quadratic. The first figure is the hero: the Fermi cliff softening, with the response bump that selects the active shell drawn beneath it.

Figure

The Fermi cliff softens over a band of width kT and the response bump selects the active shell

The punchline: warming does not shift the whole sea, it only ruffles the surface. The even bump $-\partial f/\partial\varepsilon$ picks out $\varepsilon=\mu$ and its curvature, so the correction is $\propto(k_BT)^2$ , not $k_BT$ .

Because the density of states of a 3D free gas rises with energy, $g\propto\sqrt\varepsilon$ , the softening is not symmetric in its effect: there are more empty states just above $\mu$ than filled states just below, so thermal smearing lifts slightly more electrons up than it drops down. To keep the particle number fixed, the chemical potential must sag. The second figure isolates that sag — the parabolic droop of $\mu(T)$ that the naïve assumption $\mu=\varepsilon_F$ completely misses.

Figure

The chemical potential sags parabolically with temperature away from the Fermi energy

For a real metal $T/T_F\sim10^{-2}$ , so the sag is minute — but keeping it is exactly what makes the heat capacity come out right.

The core equations

The object we expand is a Fermi integral: a smooth weight $H(\varepsilon)$ , usually built from the density of states, against the occupation. The occupation itself is the Fermi-Dirac function, whose only temperature-dependent feature at low $T$ is the softening of its step.

f(\varepsilon)=\frac{1}{e^{\beta(\varepsilon-\mu)}+1},\qquad \beta=\frac{1}{k_BT} \qquad (1)

where: $\varepsilon$ — single-particle energy; $\mu$ — chemical potential, itself temperature dependent; $\beta=1/k_BT$ . Physical reading: at $T=0$ this is the step $\theta(\mu-\varepsilon)$ (the Fermi sea, owned by aq-sm-degenerate-fermi-gas-zero-temperature); at finite $T$ it rounds off over a band of width $\sim k_BT$ centred on $\mu$ . The central result is the master expansion. Two or three terms give essentially exact answers throughout the metallic regime, because the expansion parameter $k_BT/\mu$ is of order $10^{-2}$ .

\int_0^\infty H(\varepsilon)\,f(\varepsilon)\,\mathrm{d}\varepsilon=\int_0^\mu H(\varepsilon)\,\mathrm{d}\varepsilon+\frac{\pi^2}{6}(k_BT)^2\,H'(\mu)+\frac{7\pi^4}{360}(k_BT)^4\,H'''(\mu)+\cdots \qquad (2)

where: $H'(\mu),H'''(\mu)$ — odd derivatives of the smooth weight, evaluated at the chemical potential. Physical reading: the leading term is just "fill up to $\mu$ "; every correction is a curvature of $H$ at the Fermi surface, weighted by a universal number times an even power of $T$ . Only odd derivatives appear because the response bump is even. The universal weights are the even moments of the response bump, and they are exactly twice the Dirichlet eta values — the alternating cousins of Euler's zeta numbers.

a_{2n}=\frac{1}{(2n)!}\int_{-\infty}^{\infty}\frac{x^{2n}\,e^{x}}{(e^{x}+1)^2}\,\mathrm{d}x=2\,\eta(2n)=2\big(1-2^{1-2n}\big)\zeta(2n),\qquad a_2=\frac{\pi^2}{6},\;\; a_4=\frac{7\pi^4}{360} \qquad (3)

where: $\zeta(2n)$ — Riemann zeta, $\zeta(2)=\pi^2/6$ , $\zeta(4)=\pi^4/90$ ; $\eta(2n)$ — Dirichlet eta, $\eta(2)=\pi^2/12$ , $\eta(4)=7\pi^4/720$ . Physical reading: the alternating sign of the Fermi factor $e^x/(e^x+1)^2$ is what turns $\zeta$ into $\eta$ ; the same $\pi^2/6$ that sums $\sum 1/n^2$ sets the leading metallic heat capacity. Applying the expansion to the number integral gives the equation that pins the chemical potential. Because $N$ is fixed, this is a constraint that must be solved before computing any energy-dependent quantity.

N=\int_0^\infty g(\varepsilon)\,f(\varepsilon)\,\mathrm{d}\varepsilon=\int_0^\mu g\,\mathrm{d}\varepsilon+\frac{\pi^2}{6}(k_BT)^2\,g'(\mu)+\cdots \qquad (4)

where: $N$ — total particle number (held fixed); $g(\varepsilon)$ — single-particle density of states (owned by aq-sm-density-of-states-free-particles). Physical reading: since $g'(\mu)>0$ for a 3D free gas, the correction pushes $N$ up unless $\mu$ drops — so $\mu(T)$ must decrease. Solving Eq. (4) for the 3D free gas, where $g(\varepsilon)=C\sqrt\varepsilon$ , gives the chemical-potential shift in closed form to leading order.

\mu(T)=\varepsilon_F\left[1-\frac{\pi^2}{12}\left(\frac{k_BT}{\varepsilon_F}\right)^2+\cdots\right]=\varepsilon_F\left[1-\frac{\pi^2}{12}\left(\frac{T}{T_F}\right)^2\right] \qquad (5)

where: $\varepsilon_F=\mu(T{=}0)$ — the Fermi energy (owned by aq-sm-fermi-energy-momentum-temperature); $T_F=\varepsilon_F/k_B$ — the Fermi temperature. Physical reading: the droop is quadratic and downward, with the $\tfrac{\pi^2}{12}$ set by the $\sqrt\varepsilon$ shape of the density of states; a flat $g$ would give no shift, a falling $g$ would raise $\mu$ . Feeding the self-consistent $\mu(T)$ back into the energy integral produces the payoff of the whole apparatus: an internal energy that grows as $T^2$ , whose derivative is the famous linear heat capacity.

U(T)=U_0+\frac{\pi^2}{6}(k_BT)^2\,g(\varepsilon_F)\quad\Longrightarrow\quad C_V=\left(\frac{\partial U}{\partial T}\right)_N=\frac{\pi^2}{3}k_B^2\,g(\varepsilon_F)\,T\equiv\gamma T \qquad (6)

where: $U_0=\tfrac{3}{5}N\varepsilon_F$ — the $T=0$ energy; $\gamma=\tfrac{\pi^2}{3}k_B^2 g(\varepsilon_F)$ — the Sommerfeld coefficient. Physical reading: the number of active electrons is $\sim g(\varepsilon_F)k_BT$ , each gaining $\sim k_BT$ , so $U-U_0\sim(k_BT)^2 g(\varepsilon_F)$ and $C_V\propto T$ ; the full heat-capacity story is owned by aq-sm-electronic-heat-capacity-metals.

Derivations

Derivation AThe Sommerfeld lemma from the moments of the Fermi derivative

The goal is to derive the master expansion, Eq. (2), and pin down its universal coefficients $a_{2n}=2\eta(2n)$ from first principles, taking as given only the Fermi-Dirac function of aq-sm-fermi-dirac-distribution and that $H(\varepsilon)$ is smooth and grows no faster than a power. The route is to integrate by parts so the integral becomes the sharp bump $-\partial f/\partial\varepsilon$ convolved with the antiderivative of $H$ , change variables to make the bump dimensionless and even, then Taylor-expand and read off the even moments. The one subtlety to watch is that extending the lower limit from $0$ to $-\infty$ is an approximation, correct only up to exponentially small terms $\sim e^{-\beta\mu}$ — which is exactly why the result is an asymptotic, not convergent, series.

Step 1. Integrate by parts. Define $K(\varepsilon)=\int_0^\varepsilon H\,\mathrm{d}\varepsilon'$ , so $K'=H$ , $K(0)=0$ . Then $\int_0^\infty Hf\,\mathrm{d}\varepsilon=[Kf]_0^\infty-\int_0^\infty K\,\tfrac{\partial f}{\partial\varepsilon}\,\mathrm{d}\varepsilon=\int_0^\infty K(\varepsilon)\big(-\tfrac{\partial f}{\partial\varepsilon}\big)\mathrm{d}\varepsilon$ . Why: the boundary terms vanish because $K(0)=0$ and $f(\infty)=0$ , and the weight $-\partial f/\partial\varepsilon$ is sharply peaked at $\mu$ while $K$ is smooth — the setup for a moment expansion.

Step 2. Make the weight dimensionless and even. Substitute $x=\beta(\varepsilon-\mu)$ : the weight becomes $-\partial f/\partial\varepsilon=\beta\,e^{x}/(e^{x}+1)^2$ , a unit-area bump that is manifestly even in $x$ (replacing $x\to-x$ leaves $e^{x}/(e^{x}+1)^2$ unchanged, since it equals $1/[4\cosh^2(x/2)]$ ). The integral is $\int_{-\beta\mu}^{\infty}K(\mu+k_BT\,x)\,\dfrac{e^{x}}{(e^{x}+1)^2}\,\mathrm{d}x$ . Why: the evenness is the whole reason only even powers of $T$ survive — it is worth isolating before expanding.

Step 3. Extend the limit and Taylor-expand. Because the weight is exponentially small for $x<0$ far from the origin, replacing the lower limit $-\beta\mu$ by $-\infty$ costs only $O(e^{-\beta\mu})$ . Expanding $K(\mu+k_BT\,x)=\sum_{m\ge0}\tfrac{(k_BT)^m x^m}{m!}K^{(m)}(\mu)$ and integrating term by term, the odd- $m$ terms vanish by evenness, leaving $\int_0^\infty Hf\,\mathrm{d}\varepsilon=\sum_{n\ge0}\tfrac{(k_BT)^{2n}}{(2n)!}K^{(2n)}(\mu)\,M_{2n}$ with $M_{2n}=\int_{-\infty}^\infty x^{2n}\,\tfrac{e^x}{(e^x+1)^2}\mathrm{d}x$ . Why: the $m=0$ term $K(\mu)=\int_0^\mu H$ is the "fill up to $\mu$ " piece; $m=2$ gives the first correction with $K''=H'$ .

Step 4. Evaluate the moments. The zeroth moment is $M_0=1$ (unit area). The second is $M_2=\pi^2/3$ , and in general $M_{2n}=(2n)!\cdot 2\eta(2n)=(2n)!\cdot 2(1-2^{1-2n})\zeta(2n)$ . Dividing by $(2n)!$ gives $a_2=\tfrac{1}{2!}\cdot\tfrac{\pi^2}{3}=\tfrac{\pi^2}{6}$ and $a_4=\tfrac{1}{4!}\cdot\tfrac{7\pi^4}{15}=\tfrac{7\pi^4}{360}$ , reproducing Eq. (2) with $H'(\mu)=K''(\mu)$ , $H'''(\mu)=K^{(4)}(\mu)$ . Why: the alternating Fermi factor is what converts the naive $\zeta$ into the eta function $\eta(2n)=(1-2^{1-2n})\zeta(2n)$ — the arithmetic fingerprint of Fermi statistics on thermodynamics. Verified: the moments $M_0=1$ , $M_2=\pi^2/3$ , $M_4=7\pi^4/15$ , the vanishing odd moment, the coefficients $a_2=\pi^2/6$ and $a_4=7\pi^4/360$ (both as moments and as $2\eta(2n)$ ), and the lead term against a fully numeric Fermi integral, are all checked in verify_sommerfeld-expansion.py.

Go deeper · Derivation B · 1 minThe chemical potential shift of the free Fermi gas

Derivation BThe chemical potential shift of the free Fermi gas

The goal is to derive Eq. (5), $\mu(T)=\varepsilon_F[1-\tfrac{\pi^2}{12}(k_BT/\varepsilon_F)^2]$ , by imposing particle-number conservation on the number expansion, Eq. (4), for a 3D free gas with $g(\varepsilon)=C\sqrt\varepsilon$ . The strategy is to write $N$ at temperature $T$ using the expansion, set it equal to the same $N$ evaluated at $T=0$ (which defines $\varepsilon_F$ ), and solve the resulting equation perturbatively in the small parameter $k_BT/\varepsilon_F$ . The point to watch is that $\mu$ and $\varepsilon_F$ are genuinely different quantities: $\varepsilon_F$ is fixed by the density once and for all, while $\mu(T)$ is whatever value makes the warm gas hold the same number of particles.

Step 1. Write both sides. With $g=C\sqrt\varepsilon$ , $\int_0^\mu g\,\mathrm{d}\varepsilon=\tfrac{2}{3}C\mu^{3/2}$ and $g'(\mu)=\tfrac{1}{2}C\mu^{-1/2}$ , so Eq. (4) gives $N=\tfrac{2}{3}C\mu^{3/2}+\tfrac{\pi^2}{6}(k_BT)^2\cdot\tfrac{1}{2}C\mu^{-1/2}=\tfrac{2}{3}C\mu^{3/2}\big[1+\tfrac{\pi^2}{8}(k_BT/\mu)^2\big]$ . Why: factoring out $\tfrac{2}{3}C\mu^{3/2}$ exposes the correction as a pure number times $(k_BT/\mu)^2$ .

Step 2. Fix $N$ at $T=0$ . The same integral at $T=0$ is $N=\tfrac{2}{3}C\varepsilon_F^{3/2}$ , which is the definition of $\varepsilon_F$ . Why: this is the constraint — the density does not change when we warm the gas, so the two expressions for $N$ must agree.

Step 3. Equate and solve perturbatively. Setting the two equal, $\mu^{3/2}[1+\tfrac{\pi^2}{8}t^2]=\varepsilon_F^{3/2}$ with $t=k_BT/\mu$ . Writing $\mu=\varepsilon_F(1+c\,t_F^2)$ with $t_F=k_BT/\varepsilon_F$ and keeping terms to $O(t_F^2)$ , the bracket forces $\tfrac{3}{2}c+\tfrac{\pi^2}{8}=0$ . Why: to lowest order $t\approx t_F$ , and $(1+c t_F^2)^{3/2}\approx1+\tfrac{3}{2}c\,t_F^2$ , so the $O(t_F^2)$ coefficient must cancel.

Step 4. Read off the coefficient. Solving $\tfrac{3}{2}c=-\tfrac{\pi^2}{8}$ gives $c=-\tfrac{\pi^2}{12}$ , so $\mu(T)=\varepsilon_F[1-\tfrac{\pi^2}{12}(k_BT/\varepsilon_F)^2]$ . The sign is negative because $g'(\mu)>0$ (states are denser above $\mu$ ), so $\mu$ must fall to keep $N$ fixed. Why: a rising density of states means thermal smearing promotes more electrons than it demotes, and only a lower $\mu$ restores the balance. Verified: the number expansion with $g=C\sqrt\varepsilon$ and the perturbative solution giving $c=-\pi^2/12$ are checked symbolically (series in $k_BT$ ) in verify_sommerfeld-expansion.py.

Limiting cases

The expansion is an asymptotic series in $t\equiv k_BT/\mu\approx T/T_F$ ; its structure and regime of validity are:

Case	Result	Meaning
Strongly degenerate, $T\ll T_F$	$k_BT/\mu\ll1$	Two or three terms are essentially exact — the metallic regime where the expansion is the tool of choice
Leading correction only	$\int_0^\infty Hf=\int_0^\mu H+\tfrac{\pi^2}{6}(k_BT)^2H'(\mu)$	Enough for the heat capacity and the $\mu(T)$ shift to lowest nontrivial order
Asymptotic breakdown, $T\sim T_F$	series diverges	Must use full Fermi-Dirac integrals (polylogarithms) or cross to the classical Maxwell-Boltzmann regime
Exponentially small remainder	$\sim e^{-\mu/k_BT}$	Extending the lower limit to $-\infty$ drops terms invisible to any finite order — the hallmark of an asymptotic series

Common trapThe chemical potential is not the Fermi energy at finite temperature

The single most common Sommerfeld error is treating $\mu$ as if it stayed pinned at $\varepsilon_F$ . It does not: holding $N$ fixed forces $\mu(T)=\varepsilon_F[1-\tfrac{\pi^2}{12}(T/T_F)^2]$ to sag, and in the nondegenerate limit $\mu$ eventually goes negative. Skipping this and evaluating the energy expansion at $\mu=\varepsilon_F$ leaves a spurious $\mu g'(\mu)$ term and gives the wrong heat-capacity coefficient. The correct procedure is always: expand $N(\mu,T)$ , invert for $\mu(T)$ , then substitute into the energy. A second, subtler trap is treating the series as convergent — it is asymptotic, valid strictly for $k_BT\ll\mu$ , and its high-order terms eventually blow up.

Worked examples

Example 1 — The coefficient $\pi^2/6$ from the second moment.

Step 1. Setup. The leading Sommerfeld coefficient is $a_2=\tfrac{1}{2!}M_2$ , where $M_2=\int_{-\infty}^\infty x^2\,e^x/(e^x+1)^2\,\mathrm{d}x$ is the second moment of the response bump. Evaluate $M_2$ and form $a_2$ .

Step 2. Evaluate the moment. Using $e^x/(e^x+1)^2=1/[4\cosh^2(x/2)]$ and the standard result $\int_{-\infty}^\infty x^2/[4\cosh^2(x/2)]\,\mathrm{d}x=\pi^2/3$ , we get $M_2=\pi^2/3$ . Equivalently $M_2=2\cdot2!\cdot\eta(2)=4\cdot\tfrac{\pi^2}{12}=\tfrac{\pi^2}{3}$ .

Step 3. Form the coefficient. $a_2=\tfrac{1}{2!}M_2=\tfrac12\cdot\tfrac{\pi^2}{3}=\boxed{\dfrac{\pi^2}{6}}$ . The next one follows identically from the fourth moment $M_4=7\pi^4/15$ : $a_4=\tfrac{1}{4!}\cdot\tfrac{7\pi^4}{15}=\mathbf{\dfrac{7\pi^4}{360}}$ .

Step 4. Interpret. The number $\pi^2/6$ is not a metallic accident — it is $\zeta(2)$ , the sum $\sum_{n\ge1}1/n^2$ that Euler found in 1735, entering here through the moment of the Fermi bump. Every metal's linear heat-capacity coefficient carries this factor.

Example 2 — The chemical-potential shift and heat capacity of copper.

Step 1. Setup. Copper has $\varepsilon_F=7.0\ \mathrm{eV}$ , so its Fermi temperature is $T_F=\varepsilon_F/k_B=(7.0\times1.602\times10^{-19})/(1.381\times10^{-23})$ .

Step 2. Fermi temperature. $T_F=(1.1215\times10^{-18})/(1.381\times10^{-23})=\mathbf{8.12\times10^{4}\ K}$ . At room temperature, $T/T_F=300/8.12\times10^{4}=\mathbf{3.7\times10^{-3}}$ — the electron gas is deeply degenerate.

Step 3. Chemical-potential shift. From Eq. (5), the fractional drop is $\tfrac{\pi^2}{12}(T/T_F)^2=\tfrac{9.87}{12}(3.7\times10^{-3})^2=\mathbf{1.1\times10^{-5}}$ . So at $300\ \mathrm{K}$ , $\mu$ differs from $\varepsilon_F$ by about eleven parts per million — utterly negligible for $\mu$ itself, yet its cancellation is what fixes the heat-capacity coefficient.

Step 4. Sommerfeld heat capacity. Per mole, $C_e/T=\gamma=\tfrac{\pi^2}{2}R/T_F=\tfrac{9.87}{2}(8.314)/(8.12\times10^{4})=\mathbf{5.05\times10^{-4}\ J\,mol^{-1}K^{-2}}=0.505\ \mathrm{mJ\,mol^{-1}K^{-2}}$ . The measured value for copper is $\approx0.695\ \mathrm{mJ\,mol^{-1}K^{-2}}$ , so the thermal effective mass is $m^*/m\approx0.695/0.505=\mathbf{1.4}$ — the deviation is a real, measurable signature of band structure and interactions, extracted purely from this expansion.

Origins · 1 minarnold sommerfeld and how this was discovered

arnold sommerfeld

1868–1951 · Arnold Sommerfeld built the modern quantum theory of metals: in 1927–28 he replaced Drude's classical electron gas with a Fermi-Dirac gas, and the low-temperature expansion of the resulting Fermi integrals — the machine derived here — carries his name. It immediately explained the long-standing puzzle of why metallic electrons contribute almost nothing to the specific heat: only the thin shell near the Fermi surface is active. Sommerfeld never won a Nobel Prize despite a record number of nominations, but he trained an extraordinary lineage of laureates, including Heisenberg, Pauli, Debye, and Bethe.

History

1735

— Euler evaluates the Basel sum,

\sum 1/n^2=\pi^2/6=\zeta(2)

, the number that reappears as the leading coefficient

1926

— Fermi and, independently, Dirac derive the statistics of an exclusion-obeying ideal gas

1927

— Pauli applies Fermi-Dirac statistics to explain the weak, temperature-independent paramagnetism of metals

1928

— Sommerfeld publishes the free-electron theory of metals, using the low-T expansion to resolve the electronic heat-capacity puzzle

1933

— Bloch and Wilson embed the free-electron gas in a periodic lattice, launching band theory

Going further

Texts: Pathria & Beale, Statistical Mechanics (3rd ed.), §8.1 and Appendix E (the Sommerfeld lemma); Kardar, Statistical Physics of Particles, Ch. 7; Reif, Fundamentals of Statistical and Thermal Physics, §9.16; Sethna, Statistical Mechanics, Ch. 7; Ashcroft & Mermin, Solid State Physics, Ch. 2; Landau & Lifshitz, Statistical Physics Part 1, §58.
Builds on: Quantum Statistics of Ideal Gases: Occupation Numbers · Degenerate Fermi Gas at Zero Temperature · Density of States for Free Particles · Fermi Energy, Momentum and Fermi Temperature. The Fermi function and its bell-shaped window $-\partial f/\partial\varepsilon$ come from aq-sm-fermi-dirac-distribution.
Leads to: Electronic Heat Capacity of Metals (which owns the full $C_e=\gamma T$ story this expansion enables) · Chemical Potential of a Fermi Gas vs Temperature · Degeneracy Pressure of Fermions · Fermi Surface and Fermi Sphere.
Provenance: AllQFT `sm` export (Pathria & Beale §8.1, Appendix E), aq-sm-sommerfeld-expansion, ref STAT.IQGAS.SOMM.1222. All moments, coefficients ( $a_2=\pi^2/6$ , $a_4=7\pi^4/360$ verified both as moment integrals and as $2\eta(2n)$ ), the $\mu(T)$ shift coefficient $-\pi^2/12$ , the heat-capacity algebra, and the copper numbers ( $T_F=8.12\times10^4$ K, $\mu$ shift $1.1\times10^{-5}$ , $\gamma=0.505$ mJ mol $^{-1}$ K $^{-2}$ , $m^*/m\approx1.4$ ) are python-verified (19 checks, all PASS).

Why does the expansion have only even powers of temperature?

Because the response weight $-\partial f/\partial\varepsilon=\beta\,e^x/(e^x+1)^2$ is an even function of $x=\beta(\varepsilon-\mu)$ once the integration is extended to $x\in(-\infty,\infty)$ — it is literally $\beta/[4\cosh^2(x/2)]$ , symmetric about $x=0$ . When you Taylor-expand the smooth part of the integrand about $\mu$ and integrate term by term, every odd power of $x$ integrates to zero against an even weight. Only the even moments survive, and they multiply $(k_BT)^{2n}$ . Physically, warming rounds off the Fermi step symmetrically to leading order, so there is no linear-in- $T$ effect.

Is the Sommerfeld series convergent?

No — it is an asymptotic series. The step that extends the lower limit from $0$ to $-\infty$ discards contributions of order $e^{-\beta\mu}=e^{-\mu/k_BT}$ , which are exponentially small in the degenerate regime but are completely invisible to any finite order of the power series. As a result, adding more terms improves the answer only up to a point; the high-order coefficients eventually grow and the series diverges. In practice, for a metal with $k_BT/\mu\sim10^{-2}$ , the first one or two terms already give more than enough accuracy, so the asymptotic nature never bites.

Why must I solve for μ(T) before computing the heat capacity?

Because both the energy and the number are integrals over the same occupation, and $N$ is fixed while the energy is not. If you naïvely set $\mu=\varepsilon_F$ in the energy expansion, you keep a spurious $\mu g'(\mu)$ term and get the wrong coefficient. The correct route is to expand $N(\mu,T)$ , impose $N(T)=N(0)$ to find $\mu(T)=\varepsilon_F[1-\tfrac{\pi^2}{12}(T/T_F)^2]$ , and only then substitute into the energy integral. When you do, the $\mu$ -shift terms cancel exactly, leaving the clean $U=U_0+\tfrac{\pi^2}{6}(k_BT)^2g(\varepsilon_F)$ and hence $C_V=\tfrac{\pi^2}{3}k_B^2g(\varepsilon_F)T$ .

Where does the Dirichlet eta function come from?

The Sommerfeld moment integrals $\int_{-\infty}^\infty x^{2n}e^x/(e^x+1)^2\,\mathrm{d}x$ evaluate to $(2n)!\cdot2\eta(2n)$ , where $\eta(s)=\sum_{k\ge1}(-1)^{k-1}/k^s=(1-2^{1-s})\zeta(s)$ is the alternating cousin of the Riemann zeta function. The alternating sign is the fingerprint of the Fermi factor: expanding $1/(e^x+1)$ produces a series with alternating signs, whereas the Bose factor $1/(e^x-1)$ gives all-positive terms and hence plain $\zeta$ . So $\eta(2)=\pi^2/12$ gives $a_2=\pi^2/6$ , and $\eta(4)=7\pi^4/720$ gives $a_4=7\pi^4/360$ .

Why does the chemical potential go down and not up?

It comes down to the shape of the density of states. For a 3D free gas $g(\varepsilon)=C\sqrt\varepsilon$ rises with energy, so just above the Fermi level there are more empty states than there are filled states just below it. When temperature smears the step, slightly more electrons are lifted up than are dropped down, which would increase the particle number. To keep $N$ fixed the chemical potential must fall, and the number expansion pins the drop to $\tfrac{\pi^2}{12}(T/T_F)^2$ . If $g$ were flat, $\mu$ would not move; if $g$ fell with energy (as in some bands), $\mu$ would rise.

How does this connect to the electronic heat capacity of metals?

The Sommerfeld expansion supplies the energy $U(T)=U_0+\tfrac{\pi^2}{6}(k_BT)^2g(\varepsilon_F)$ ; differentiating gives $C_V=\tfrac{\pi^2}{3}k_B^2g(\varepsilon_F)T=\gamma T$ , the linear-in- $T$ electronic term that aq-sm-electronic-heat-capacity-metals develops in full. The intuition is the "active shell" picture: only $\sim g(\varepsilon_F)k_BT$ electrons near the Fermi surface can absorb heat, each taking up $\sim k_BT$ , so the thermal energy scales as $T^2$ and the heat capacity as $T$ . This article owns the expansion and the coefficient; the heat-capacity article owns the measurement, the $C/T$ -vs- $T^2$ plot, and the effective-mass interpretation.

What replaces the expansion when the gas is not degenerate?

When $k_BT$ becomes comparable to $\mu$ , the Fermi step is no longer sharp and the moment expansion breaks down. The exact object is a Fermi–Dirac integral $F_j(\eta)=\tfrac{1}{\Gamma(j+1)}\int_0^\infty x^j/(e^{x-\eta}+1)\,\mathrm{d}x$ , a polylogarithm $-\mathrm{Li}_{j+1}(-e^\eta)$ , which interpolates smoothly all the way to the classical Maxwell-Boltzmann limit $\eta\to-\infty$ (where it reduces to $e^\eta$ ). In that dilute, high-temperature regime the gas obeys ordinary Boltzmann statistics and the whole degenerate-expansion apparatus is simply the wrong tool.