--- title: "The Sommerfeld Model: Ground State and Heat Capacity" module: Free-Electron Fermi Gas moduleNumber: 5 lessonNumber: 2 order: 502 summary: > Quantizing the free-electron gas in a box fills a Fermi sphere in k-space. The density of states grows as the square root of energy in three dimensions, and the Fermi energy, temperature, and wavevector follow for real metals. The Sommerfeld expansion shows only a thermal shell of width k_BT near E_F is excited, giving an electronic heat capacity linear in T that sits beneath the phonon T-cubed term. topics: [Free-Electron Fermi Gas] draft: false sources: - book: Ashcroft & Mermin ref: "Ch. 2 — The Sommerfeld Theory of Metals" - book: Kittel ref: "Ch. 6 — Free Electron Fermi Gas" - book: Hook & Hall ref: "Ch. 3 — Electrons in Metals" --- The [previous lesson](/condensed-matter/free-electron-fermi-gas/free-electron-gas-and-conduction) showed that treating the conduction electrons as a classical gas fails: the heat capacity is a hundred times too small, and the culprit is the exclusion principle. Arnold Sommerfeld's 1927 fix keeps Drude's picture of free electrons in a box but counts their states quantum-mechanically and fills them according to Fermi–Dirac statistics. Everything measurable about the electron gas — its energy, pressure, compressibility, and its tiny heat capacity — follows from one object: the **density of states** $g(E)$, the number of single-electron levels per unit energy. ## Quantizing the electron gas in a box Take $N$ electrons in a cube of side $L$ and volume $V = L^3$, ignoring both the electron–electron interaction and the periodic potential of the ions. Each electron obeys the free-particle Schrödinger equation $$ -\frac{\hbar^2}{2m}\nabla^2 \psi_{\vec k}(\vec r) = E\,\psi_{\vec k}(\vec r), $$ whose solutions are plane waves $\psi_{\vec k}(\vec r) = V^{-1/2}\,e^{i\vec k\cdot\vec r}$ with energy $$ E(\vec k) = \frac{\hbar^2 k^2}{2m}. $$ To count states cleanly we impose **Born–von Kármán periodic boundary conditions**, $\psi(\vec r + L\hat x) = \psi(\vec r)$ and likewise for $\hat y, \hat z$. This wraps the box onto a torus; it changes nothing about the bulk but makes the allowed wavevectors a discrete grid, $$ k_x = \frac{2\pi}{L}n_x, \qquad k_y = \frac{2\pi}{L}n_y, \qquad k_z = \frac{2\pi}{L}n_z, \qquad n_x,n_y,n_z \in \mathbb{Z}. $$ Each allowed $\vec k$ occupies a volume $(2\pi/L)^3 = 8\pi^3/V$ of $k$-space, so the number of $\vec k$-points per unit volume of $k$-space is $V/8\pi^3$. Because each spatial state holds two electrons (spin up and down), the number of electron states in a $k$-space region of volume $\Omega_k$ is $$ \d N = 2\cdot\frac{V}{8\pi^3}\,\Omega_k = \frac{V}{4\pi^3}\,\Omega_k. $$ ## The Fermi sphere In the ground state the electrons fill the lowest-energy levels, and since $E$ depends only on $|\vec k|$, the occupied region is a sphere in $k$-space — the **Fermi sphere** — of radius $k_F$, the **Fermi wavevector**. Its surface is the **Fermi surface**; for free electrons it is a perfect sphere. $$ % caption: In the ground state the allowed k-points fill a sphere of radius k_F. % Each grid cell has volume (2 pi / L) cubed and holds two electrons; the sphere % surface is the Fermi surface, and every occupied state lies inside it. \begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} % k-space grid \draw[black, very thin] (-2.6,-2.6) grid[step=0.4] (2.6,2.6); \draw[->, black] (-2.9,0) -- (2.9,0) node[below right] {$k_x$}; \draw[->, black] (0,-2.9) -- (0,2.9) node[left] {$k_y$}; % filled Fermi disk (2D slice) \fill[acc!14] (0,0) circle (2.0); \draw[acc, very thick] (0,0) circle (2.0); % k_F radius \draw[black, ->, thick] (0,0) -- (1.414,1.414); \node[black, anchor=south] at (0.7,1.12) {$k_F$}; % occupied dots inside \foreach \i in {-4,...,4}{ \foreach \j in {-4,...,4}{ \pgfmathsetmacro{\rr}{sqrt((\i*\i+\j*\j))*0.4} \ifdim\rr pt<2.0pt \fill[acc] (\i*0.4,\j*0.4) circle (1.1pt); \fi } } \node[acc, anchor=west] at (2.15,-2.3) {Fermi surface}; \end{tikzpicture} $$ Setting the total occupied volume $\tfrac43\pi k_F^3$ equal to the count of $N$ electrons through $\d N$ above, $$ N = \frac{V}{4\pi^3}\cdot\frac{4}{3}\pi k_F^3 = \frac{V k_F^3}{3\pi^2}, $$ so the Fermi wavevector is fixed entirely by the electron density $n = N/V$: $$ k_F = \left(3\pi^2 n\right)^{1/3}. $$ The **Fermi energy** is the energy of the topmost filled level and the **Fermi velocity** is the speed of an electron on the Fermi surface: $$ E_F = \frac{\hbar^2 k_F^2}{2m} = \frac{\hbar^2}{2m}\left(3\pi^2 n\right)^{2/3}, \qquad v_F = \frac{\hbar k_F}{m}. $$ > **Definition (Fermi wavevector, energy, and surface).** The **Fermi wavevector** > $k_F$ is the radius of the filled sphere in $k$-space, $k_F = (3\pi^2 n)^{1/3}$; > the **Fermi energy** $E_F = \hbar^2 k_F^2/2m$ is the energy of the highest > occupied level at $T = 0$; the **Fermi surface** is the constant-energy surface > $E(\vec k) = E_F$ separating filled from empty states. Every equilibrium > property of a metal is dominated by electrons within $k_BT$ of this surface. For a monovalent metal like sodium or copper, $n \sim 10^{29}\ \text{m}^{-3}$ gives $k_F \sim 1.4\ \text{Å}^{-1}$ (comparable to a reciprocal-lattice vector, a fact that returns in band theory), $E_F \sim 3$–$7\ \text{eV}$, and $v_F \sim 10^6\ \text{m/s}$, about $1\%$ of the speed of light. The corresponding **Fermi temperature** $T_F = E_F/k_B$ runs from $30{,}000$ to $80{,}000\ \text{K}$, so at any real temperature $T/T_F \lesssim 0.01$ and the gas is deeply **degenerate**. ## The density of states Most calculations need not the individual levels but their number per unit energy. Counting states inside a sphere of radius $k(E) = \sqrt{2mE}/\hbar$, $$ N(E) = \frac{V}{4\pi^3}\cdot\frac43\pi k^3 = \frac{V}{3\pi^2}\left(\frac{2mE}{\hbar^2}\right)^{3/2}, $$ and differentiating gives the **density of states** $g(E) = \d N/\d E$: $$ g(E) = \frac{V}{2\pi^2}\left(\frac{2m}{\hbar^2}\right)^{3/2}\sqrt{E} = \frac{3}{2}\,\frac{N}{E_F}\,\sqrt{\frac{E}{E_F}}. $$ The characteristic square-root growth is a three-dimensional signature: in $d$ dimensions $g(E) \propto E^{d/2-1}$, so a two-dimensional gas has a constant density of states and a one-dimensional gas diverges as $E^{-1/2}$ at each band bottom. A tidy relation follows by evaluating at the Fermi energy: $$ g(E_F) = \frac{3N}{2E_F} = \frac{3n}{2E_F}\,V. $$ $$ % caption: The three-dimensional density of states g(E) grows as the square root % of energy. At T = 0 all levels below E_F are full (shaded). At finite T the % Fermi-Dirac factor rounds the step over a width of a few k_B T, promoting % electrons from just below E_F to just above it. \begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[->, black] (0,0) -- (7.4,0) node[below] {$E$}; \draw[->, black] (0,0) -- (0,3.7) node[left] {$g(E)$}; % sqrt curve \draw[acc, very thick, domain=0:6.4, samples=100, variable=\x] plot ({\x},{1.15*sqrt(\x)}); % filled to E_F at x=4 \fill[acc!14] (0,0) -- plot[domain=0:4.0, samples=80, variable=\x] ({\x},{1.15*sqrt(\x)}) -- (4.0,0) -- cycle; \draw[black, dashed] (4.0,0) -- (4.0,2.3) node[above, black] {$E_F$}; % finite-T occupied region: shade the traded sliver above E_F lightly \fill[acc!8] (4.0,0) -- plot[domain=4.0:4.9, samples=40, variable=\x] ({\x},{1.15*sqrt(\x)}) -- (4.9,0) -- cycle; % annotate thermal width \draw[black, <->] (3.4,-0.55) -- (4.6,-0.55) node[midway, below, font=\scriptsize] {few $k_BT$}; \end{tikzpicture} $$ ## The ground-state energy and pressure Integrating $E\,g(E)$ over the filled sphere gives the total energy at $T = 0$: $$ U_0 = \int_0^{E_F} E\,g(E)\,\d E = \frac{3}{2}\frac{N}{E_F^{3/2}}\int_0^{E_F}E^{3/2}\,\d E = \frac{3}{5}N E_F, $$ so the average energy per electron is $\tfrac35 E_F$, not the classical $\tfrac32 k_BT$. Even at absolute zero the gas carries an enormous zero-point energy because the exclusion principle forces electrons into high-$k$ states. This energy depends on volume through $E_F \propto n^{2/3} \propto V^{-2/3}$, so it exerts a **degeneracy pressure** $$ P = -\frac{\partial U_0}{\partial V} = \frac{2}{3}\frac{U_0}{V} = \frac{2}{5}n E_F, $$ of order $10^{10}\ \text{Pa}$ ($10^5\ \text{atm}$) in a typical metal. It is this Fermi pressure, resisted by the electrostatic binding to the ions, that sets the compressibility (bulk modulus) of a metal; the free-electron estimate lands within a factor of a few of the measured values for the alkali metals. ## Finite temperature: the Sommerfeld expansion At $T > 0$ the occupation of a level of energy $E$ is the Fermi–Dirac function $$ f(E) = \frac{1}{e^{(E-\mu)/k_BT} + 1}, $$ with $\mu$ the chemical potential, fixed by holding $N$ constant. The function drops from $1$ to $0$ over a window of a few $k_BT$ centred on $\mu$; at $T = 0$ it is a step and $\mu = E_F$. Any thermodynamic average has the form $\int H(E)\,g(E)f(E)\,\d E$, and because $-\d f/\d E$ is a sharp spike at $\mu$ of width $k_BT$, such integrals are dominated by the immediate neighbourhood of the Fermi surface. Expanding a smooth function $H(E)$ about $\mu$ and integrating term by term gives the **Sommerfeld expansion**, $$ \int_{-\infty}^{\infty} H(E)\,f(E)\,\d E = \int_{-\infty}^{\mu} H(E)\,\d E + \frac{\pi^2}{6}(k_BT)^2 H'(\mu) + \frac{7\pi^4}{360}(k_BT)^4 H'''(\mu) + \cdots. $$ The expansion parameter is $(k_BT/E_F)^2 \sim 10^{-4}$, so the leading correction is already superb. Applying it to the number constraint fixes how $\mu$ drifts below $E_F$ as the gas warms, $$ \mu(T) = E_F\left[1 - \frac{\pi^2}{12}\left(\frac{k_BT}{E_F}\right)^2 + \cdots\right], $$ a shift of parts in $10^4$ at room temperature, so $\mu \approx E_F$ is an excellent approximation. $$ % caption: The Fermi-Dirac occupation f(E) at T = 0 (sharp step at E_F, solid) % and at finite T (rounded, dashed). Its derivative minus df/dE is a spike of % width a few k_B T centred on E_F: only electrons in that shell change their % occupation and can absorb heat. \begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[->, black] (0,0) -- (7.0,0) node[below] {$E$}; \draw[->, black] (0,0) -- (0,2.9) node[left] {$f(E)$}; % T=0 step at x=3.5 \draw[black, very thick] (0.2,2.3) -- (3.5,2.3) -- (3.5,0); \node[black, anchor=south, font=\scriptsize] at (1.6,2.32) {$T=0$}; % finite-T rounded (logistic), width scale 0.5; steepness kept low so exp stays % under the pgfmath overflow ceiling at the domain edge \draw[acc, very thick, domain=0.2:6.4, samples=120, variable=\x] plot ({\x},{2.3/(1+exp((\x-3.5)*2.6))}); \node[acc, anchor=west, font=\scriptsize] at (4.4,1.35) {f\/inite $T$}; \draw[black, dashed] (3.5,0) -- (3.5,-0.05) node[below] {$E_F$}; \draw[black, <->] (2.9,-0.6) -- (4.1,-0.6) node[midway, below, font=\scriptsize] {$\sim k_BT$}; \end{tikzpicture} $$ ## The electronic heat capacity The heat capacity is $C_V = (\partial U/\partial T)_V$. A quick physical estimate gets the answer up to a number: only the fraction $\sim k_BT/E_F$ of electrons lying within $k_BT$ of the Fermi surface can be thermally excited, and each gains energy $\sim k_BT$, so $$ U(T) - U_0 \sim N\frac{k_BT}{E_F}\,k_BT, \qquad C_V \sim N k_B\,\frac{k_BT}{E_F} = N k_B\,\frac{T}{T_F}. $$ The tiny factor $T/T_F$ supplies the suppression that repairs Drude's failure. The Sommerfeld expansion supplies the exact coefficient. Applying it to $U = \int E\,g(E)f(E)\,\d E$ with the number constraint gives $$ C_V = \frac{\pi^2}{3}\,g(E_F)\,k_B^2\,T \equiv \gamma T, \qquad \gamma = \frac{\pi^2}{3}k_B^2\,g(E_F) = \frac{\pi^2}{2}\frac{N k_B}{T_F}. $$ The linear coefficient $\gamma$ (the **Sommerfeld coefficient**) is directly proportional to the density of states at the Fermi surface — a fact that makes low-temperature heat-capacity measurement one of the cleanest probes of $g(E_F)$ in any metal. > **Theorem (Electronic heat capacity).** For a degenerate free-electron gas the > electronic heat capacity is linear in temperature, $C_V = \gamma T$ with > $\gamma = \tfrac{\pi^2}{3}k_B^2\,g(E_F)$. The coefficient measures the density > of states at the Fermi energy; the smallness of $C_V$ relative to the classical > $\tfrac32 N k_B$ is the ratio $\tfrac{\pi^2}{3}(T/T_F)$. > **Proof.** Write $U = \int E\,g\,f\,\d E$ and use the Sommerfeld expansion with > $H(E) = E\,g(E)$. The zero-temperature term is $U_0$; the $(k_BT)^2$ term is > $\tfrac{\pi^2}{6}(k_BT)^2\,[g(E_F) + E_F g'(E_F)]$. The number constraint > $N = \int g\,f\,\d E$, expanded the same way, forces $\mu$ to shift so that the > $E_F g'(E_F)$ piece is exactly cancelled, leaving > $U(T) = U_0 + \tfrac{\pi^2}{6}(k_BT)^2 g(E_F)$. Differentiating in $T$ gives > $C_V = \tfrac{\pi^2}{3}k_B^2 g(E_F)\,T$. $\qquad\blacksquare$ ## Separating electrons from phonons At low temperature the total heat capacity of a metal is the sum of the electron term and the [Debye phonon term](/condensed-matter/lattice-dynamics/debye-einstein-heat-capacity), which varies as $T^3$: $$ C_V = \gamma T + \beta T^3. $$ Dividing by $T$ linearizes the data: $$ \frac{C_V}{T} = \gamma + \beta T^2. $$ A plot of $C_V/T$ against $T^2$ is therefore a straight line whose **intercept** is the electronic $\gamma$ and whose **slope** is the phonon $\beta$ (from which the Debye temperature follows). This is the standard experimental separation of the two contributions, and it works because the electron and phonon terms carry different powers of $T$. $$ % caption: Plotting C over T against T squared gives a straight line. The % intercept is the electronic Sommerfeld coefficient gamma; the slope is the % phonon coefficient beta, from which the Debye temperature follows. Two metals % with different densities of states at E_F show different intercepts. \begin{tikzpicture}[scale=1.0, >=stealth, font=\footnotesize] \definecolor{acc}{HTML}{4A6FA5} \draw[->, black] (0,0) -- (6.8,0) node[below] {$T^2$}; \draw[->, black] (0,0) -- (0,3.9) node[left] {$\frac{C_V}{T}$}; % line A: higher intercept, moderate slope \draw[acc, very thick] (0,1.1) -- (6.2,3.5); \fill[acc] (0,1.1) circle (1.8pt); \node[acc, anchor=east] at (-0.05,1.1) {intercept}; \foreach \x in {1,2,3,4,5}{ \fill[acc] (\x,{1.1+0.387*\x}) circle (1.4pt); } \node[acc, anchor=south east] at (5.6,3.15) {metal A}; % line B: lower intercept, steeper slope \draw[black, very thick, dashed] (0,0.5) -- (6.2,3.1); \fill[black] (0,0.5) circle (1.8pt); \node[black, anchor=east] at (-0.05,0.5) {intercept}; \node[black, anchor=north west] at (4.3,2.35) {metal B}; \end{tikzpicture} $$ Two loose ends point ahead. First, measured values of $\gamma$ agree with the free-electron prediction only to within factors of order unity: in transition metals the discrepancy is large, and it is customary to absorb it into a **thermal effective mass** $m^\ast$ through $\gamma \propto m^\ast$, a first hint that the periodic lattice reshapes $g(E_F)$ — the subject of [band theory](/condensed-matter/band-theory/blochs-theorem-and-energy-bands). Second, the same thermal shell that carries the heat also carries the electric and thermal current; the [next lesson](/condensed-matter/free-electron-fermi-gas/transport-and-the-hall-effect) puts the Fermi surface in motion under applied fields and derives the Wiedemann–Franz law and the Hall effect from it.