---
title: The Ideal Fermi Gas at Zero Temperature
module: Degenerate Fermi Gas
moduleNumber: 9
lessonNumber: 1
order: 901
summary: >
  At absolute zero a gas of non-interacting fermions fills every single-particle
  state up to the Fermi energy and leaves the rest empty, a filled Fermi sphere in
  momentum space. This lesson computes the Fermi momentum, energy, and temperature
  from the density, the density of states, the total ground-state energy, and the
  degeneracy pressure that grows as $n^{5/3}$. Numerical Fermi energies for metals
  set the scale: they are electron-volts, so room temperature is deep in the
  degenerate regime.
topics: [Degenerate Fermi Gas]
sources:
  - book: Schroeder
    ref: "Ch. 7 — Quantum Statistics; §7.3 The Fermi Gas"
  - book: Reif
    ref: "Ch. 9 — Quantum Statistics of Ideal Gases; §9.16–9.17"
  - book: Pathria & Beale
    ref: "Ch. 8 — Ideal Fermi Systems; §8.1"
  - book: Kardar (Statistical Physics of Particles)
    ref: "Ch. 7 — Quantum Statistical Mechanics; §7.3"
draft: false
---

The Fermi-Dirac occupation of a single-particle state of energy $\epsilon$ is

$$
\bar n(\epsilon) = \frac{1}{e^{(\epsilon-\mu)/k_BT}+1},
$$

with $\mu$ the chemical potential. Every mode sits between empty and singly
occupied because the exclusion principle forbids two identical fermions in one
state. This lesson takes the limit $T\to 0$, where the exponential turns the
occupation into a step and the gas settles into its unique ground state. The
degenerate electron gas of a metal and the matter inside a white dwarf are both
governed by that ground state, because their Fermi temperatures are far above any
temperature at which they exist.

## The filled Fermi sea

As $T\to 0$ the argument $(\epsilon-\mu)/k_BT$ runs to $-\infty$ for
$\epsilon<\mu$ and to $+\infty$ for $\epsilon>\mu$. The occupation collapses to a
step function,

$$
\bar n(\epsilon) \xrightarrow{\,T\to 0\,}
\begin{cases}
1, & \epsilon < \epsilon_F,\\[2pt]
0, & \epsilon > \epsilon_F,
\end{cases}
$$

where the zero-temperature chemical potential is the **Fermi energy**,
$\epsilon_F \equiv \mu(T=0)$. All states below $\epsilon_F$ are filled, all above
are empty, and none is half-filled. The filled set is the **Fermi sea**.

$$
% caption: The zero-temperature occupation is a step of unit height that drops to zero at $\epsilon_F$; no state is partially filled.
\begin{tikzpicture}[>=Latex,font=\footnotesize,scale=1.0]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->,black] (0,0)--(5.8,0) node[right,black]{$E$};
\draw[->,black] (0,0)--(0,3.2) node[above,black]{occupation};
\draw[acc,very thick] (0,2.4)--(3.2,2.4)--(3.2,0)--(5.4,0);
\draw[black,dashed] (0,2.4) node[left,black]{$1$} -- (3.2,2.4);
\draw[black,dashed] (3.2,0) node[below,black]{$E_F$} -- (3.2,2.4);
\node[acc] at (1.5,1.4) {occupied};
\node[black] at (4.4,0.55) {empty};
\end{tikzpicture}
$$

For free particles in a cubic box of volume $V$, the single-particle energies are
$\epsilon = \hbar^2 k^2/2m$, and the allowed wavevectors form a lattice of density
$V/(2\pi)^3$ in $\vec k$-space. Including a spin degeneracy $g_s$ (with $g_s=2$ for
spin-$\tfrac12$ electrons or neutrons), the occupied states at $T=0$ fill a sphere
of radius $k_F$ in $\vec k$-space, the **Fermi sphere**. The corresponding momentum
$p_F=\hbar k_F$ is the **Fermi momentum**.

$$
% caption: In momentum space the ground state fills a sphere of radius $p_F$; states inside are occupied, states outside empty, and the surface $|\vec p|=p_F$ is the Fermi surface.
\begin{tikzpicture}[>=Latex,font=\footnotesize,scale=1.0]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->,black] (-2.9,0)--(2.9,0) node[right,black]{$p_x$};
\draw[->,black] (0,-2.9)--(0,2.9) node[above,black]{$p_y$};
\fill[acc!14] (0,0) circle (2.05);
\begin{scope}
\clip (0,0) circle (2.02);
\foreach \x in {-2,-1.5,-1,-0.5,0,0.5,1,1.5,2}
  \foreach \y in {-2,-1.5,-1,-0.5,0,0.5,1,1.5,2}
    \fill[acc] (\x,\y) circle (1.5pt);
\end{scope}
\draw[acc,very thick] (0,0) circle (2.05);
\draw[->,black!70,thick] (0,0)--(1.45,1.45);
\node[black!70] at (1.15,0.62) {$p_F$};
\node[acc] at (0,-2.45) {occupied};
\node[black] at (2.15,2.15) {empty};
\end{tikzpicture}
$$

## Fermi momentum, energy, and temperature

The total number of particles equals $g_s$ times the number of $\vec k$-lattice
points inside the Fermi sphere,

$$
N = g_s\,\frac{V}{(2\pi)^3}\,\frac{4}{3}\pi k_F^3 = \frac{g_s V}{6\pi^2}\,k_F^3 .
$$

Solving for $k_F$ in terms of the number density $n=N/V$ gives the **Fermi
wavevector** and momentum,

$$
k_F = \left(\frac{6\pi^2 n}{g_s}\right)^{1/3},\qquad
p_F = \hbar\left(\frac{6\pi^2 n}{g_s}\right)^{1/3}.
$$

For the standard electron gas ($g_s=2$) this reduces to $k_F=(3\pi^2 n)^{1/3}$.
The Fermi wavevector is set by the density alone: it is the inverse of the mean
interparticle spacing up to a factor of order unity. The **Fermi energy** is the
kinetic energy at the Fermi surface,

$$
\epsilon_F = \frac{p_F^2}{2m} = \frac{\hbar^2}{2m}\left(\frac{6\pi^2 n}{g_s}\right)^{2/3}
\;\xrightarrow{\,g_s=2\,}\;
\frac{\hbar^2}{2m}\left(3\pi^2 n\right)^{2/3},
$$

and the **Fermi temperature** $T_F=\epsilon_F/k_B$ marks the crossover between the
degenerate regime $T\ll T_F$, where the step is sharp, and the classical regime
$T\gg T_F$, where Maxwell-Boltzmann statistics apply. The key structural fact is
that $\epsilon_F\propto n^{2/3}$: compressing the gas raises the Fermi energy,
because the added particles must occupy higher momentum states.[^efscale]

[^efscale]: Schroeder, §7.3; Pathria & Beale, §8.1. Companion notes at
<https://physics.weber.edu/schroeder/thermal/>.

The crossover condition $T\sim T_F$ is the same one that separates classical from
quantum statistics in the earlier modules. Writing the thermal de Broglie
wavelength $\lambda=h/\sqrt{2\pi m k_BT}$, the ratio $T/T_F$ is, up to a numerical
factor, $(\lambda^3 n)^{-2/3}$: the gas is degenerate exactly when the thermal
wavelength exceeds the interparticle spacing, $n\lambda^3\gtrsim1$, so the
wavepackets overlap and the exclusion principle operates. The Fermi temperature is
therefore the temperature at which quantum concentration is reached. For electrons
in a metal the density is fixed by the lattice, so $T_F$ is a material constant of
order $10^4$–$10^5\,\mathrm{K}$, far above any temperature at which the solid
survives.

## The density of states

The number of single-particle states with energy below $\epsilon$ is $g_s$ times
the $\vec k$-space volume inside $k=\sqrt{2m\epsilon}/\hbar$,

$$
\mathcal N(\epsilon) = \frac{g_s V}{6\pi^2}\left(\frac{2m\epsilon}{\hbar^2}\right)^{3/2}.
$$

Its derivative is the **density of states** $g(\epsilon)=\d\mathcal N/\d\epsilon$,
the number of states per unit energy,

$$
g(\epsilon) = \frac{g_s V}{4\pi^2}\left(\frac{2m}{\hbar^2}\right)^{3/2}\epsilon^{1/2}
= \frac{3N}{2}\,\frac{\epsilon^{1/2}}{\epsilon_F^{3/2}}.
$$

The second form, valid once $N$ and $\epsilon_F$ are fixed, follows from
$N=\int_0^{\epsilon_F}g(\epsilon)\,\d\epsilon = \tfrac23\,\epsilon_F\,g(\epsilon_F)$
because $g\propto\epsilon^{1/2}$. Evaluated at the Fermi surface it gives the
much-used relation

$$
g(\epsilon_F) = \frac{3N}{2\epsilon_F},
$$

the density of states that controls the thermal and magnetic response of the next
lesson. The $\epsilon^{1/2}$ growth reflects the three-dimensional phase space: a
thin spherical shell in $\vec k$-space has area $\propto k^2$ and thickness
$\propto \d k \propto \epsilon^{-1/2}\d\epsilon$, so its volume scales as
$\epsilon^{1/2}\d\epsilon$.

$$
% caption: The density of states rises as $\epsilon^{1/2}$; at $T=0$ every level up to $\epsilon_F$ is filled (shaded) and every level above is empty. The filled count is the area under the curve to $\epsilon_F$.
\begin{tikzpicture}[>=Latex,font=\footnotesize,scale=1.0]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->,black] (0,0)--(6,0) node[right,black]{$E$};
\draw[->,black] (0,0)--(0,4) node[above,black]{$g(E)$};
\fill[acc!14] plot[domain=0:3.2,samples=60] (\x,{1.75*sqrt(\x)}) -- (3.2,0) -- cycle;
\draw[acc,very thick] plot[domain=0:5,samples=90] (\x,{1.75*sqrt(\x)});
\draw[black,dashed] (3.2,0) node[below,black]{$E_F$} -- (3.2,{1.75*sqrt(3.2)});
\node[acc] at (1.55,0.95) {occupied};
\node[black] at (4.35,2.75) {empty};
\end{tikzpicture}
$$

## Ground-state energy and degeneracy pressure

The total energy at $T=0$ is the first moment of the filled density of states,

$$
U = \int_0^{\epsilon_F}\epsilon\,g(\epsilon)\,\d\epsilon
= \frac{3N}{2\epsilon_F^{3/2}}\int_0^{\epsilon_F}\epsilon^{3/2}\,\d\epsilon
= \frac{3}{5}\,N\epsilon_F .
$$

The average energy per fermion is $\tfrac35\epsilon_F$, not zero: even at absolute
zero the exclusion principle forces most particles into states of high momentum.
This zero-point kinetic energy is the origin of the pressure that supports white
dwarfs and neutron stars.

$$
% caption: The ground-state energy is the area under $\epsilon\,g(\epsilon)\propto\epsilon^{3/2}$ up to $\epsilon_F$; the mean energy per particle works out to $\tfrac35\epsilon_F$.
\begin{tikzpicture}[>=Latex,font=\footnotesize,scale=1.0]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->,black] (0,0)--(6,0) node[right,black]{$E$};
\draw[->,black] (0,0)--(0,4) node[above,black]{$E\,g(E)$};
\fill[acc!14] plot[domain=0:3.2,samples=60] (\x,{0.62*(\x)^1.5}) -- (3.2,0) -- cycle;
\draw[acc,very thick] plot[domain=0:3.7,samples=90] (\x,{0.62*(\x)^1.5});
\draw[black,dashed] (3.2,0) node[below,black]{$E_F$} -- (3.2,{0.62*(3.2)^1.5});
\node[acc] at (2.0,0.85) {area $=U$};
\end{tikzpicture}
$$

Because the energy of a nonrelativistic ideal gas satisfies $PV=\tfrac23 U$
(a result independent of statistics, derived for the general quantum gas), the
zero-temperature **degeneracy pressure** is

$$
P = \frac{2U}{3V} = \frac{2}{5}\,n\,\epsilon_F
= \frac{(3\pi^2)^{2/3}}{5}\,\frac{\hbar^2}{m}\,n^{5/3}
\qquad (g_s=2).
$$

The pressure is finite at $T=0$ and depends only on the density, not the
temperature. It grows steeply with compression, $P\propto n^{5/3}$: on logarithmic
axes this is a straight line of slope $\tfrac53$.

$$
% caption: The zero-temperature degeneracy pressure follows $P\propto n^{5/3}$; on log axes it is a line of slope $\tfrac53$, marked by the run-and-rise triangle.
\begin{tikzpicture}[>=Latex,font=\footnotesize,scale=1.0]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->,black] (0,0)--(6,0) node[right,black]{$\log n$};
\draw[->,black] (0,0)--(0,4.6) node[above,black]{$\log P$};
\draw[acc,very thick] (0.7,0.4)--(2.9,4.07);
\draw[black] (1.8,2.235) -- (2.4,2.235) -- (2.4,3.236);
\node[black,below] at (2.1,2.2) {$3$};
\node[black,right] at (2.4,2.74) {$5$};
\node[acc] at (1.35,3.5) {slope $\tfrac{5}{3}$};
\end{tikzpicture}
$$

The **bulk modulus** of the degenerate gas follows from
$P\propto V^{-5/3}$ at fixed $N$,

$$
K = -V\left(\frac{\partial P}{\partial V}\right)_N = \frac{5}{3}P = \frac{2}{3}\,n\,\epsilon_F .
$$

This is the stiffness that resists compression of the conduction-electron gas and
sets the compressibility of a metal to the right order of magnitude.

The degeneracy pressure has no classical counterpart. A classical ideal gas at the
same density would exert $P_{\mathrm{cl}}=n k_B T$, which vanishes as $T\to0$; the
Fermi gas holds $P=\tfrac25 n\epsilon_F$ at absolute zero. Their ratio is
$P/P_{\mathrm{cl}}\sim\epsilon_F/k_BT=T_F/T$, so at room temperature the electron
gas is stiffer than a classical gas of the same density by four orders of
magnitude. This pressure is set entirely by the exclusion principle, not by
thermal motion, and it is what supports white dwarfs and neutron stars against
gravitational collapse in the later lessons.

## Numerical scales for metals

Treating the conduction electrons of a monovalent metal as a free Fermi gas gives
Fermi energies of a few electron-volts and Fermi temperatures of tens of thousands
of kelvin.[^ammetals] With the electron mass $m_e$ and the measured conduction-
electron densities, the free-electron model gives the following.

| Metal | $n$ ($10^{28}\,\mathrm{m^{-3}}$) | $\epsilon_F$ (eV) | $T_F$ ($10^4\,\mathrm{K}$) | $v_F$ ($10^6\,\mathrm{m\,s^{-1}}$) |
|-------|------|------|------|------|
| Na | $2.65$ | $3.24$ | $3.77$ | $1.07$ |
| Cu | $8.47$ | $7.00$ | $8.16$ | $1.57$ |
| Au | $5.90$ | $5.53$ | $6.42$ | $1.40$ |

Two consequences follow. First, room temperature ($T\approx 300\,\mathrm{K}$) sits
at $T/T_F\sim 4\times10^{-3}$, so the conduction electrons are extremely
degenerate and the zero-temperature picture is an excellent starting point.
Second, the **Fermi velocity** $v_F=p_F/m_e\sim 10^6\,\mathrm{m\,s^{-1}}$ is about
one percent of the speed of light: metallic electrons are fast but nonrelativistic,
which justifies the $\epsilon=p^2/2m$ dispersion used throughout. Relativistic
corrections enter only at the far higher densities of a white dwarf, treated in a
later lesson.

> **Worked example.** For copper, $n = 8.47\times 10^{28}\,\mathrm{m^{-3}}$. The
> Fermi wavevector is
>
> $$
> k_F = (3\pi^2 n)^{1/3} = \left(3\pi^2\cdot 8.47\times10^{28}\right)^{1/3}
> = 1.36\times 10^{10}\,\mathrm{m^{-1}},
> $$
>
> so the Fermi energy is
>
> $$
> \epsilon_F = \frac{\hbar^2 k_F^2}{2m_e}
> = \frac{(1.055\times10^{-34})^2(1.36\times10^{10})^2}{2\,(9.11\times10^{-31})}
> = 1.12\times10^{-18}\,\mathrm{J} = 7.0\,\mathrm{eV},
> $$
>
> matching the tabulated value, with $T_F=\epsilon_F/k_B = 8.2\times10^4\,\mathrm{K}$.
> The degeneracy pressure is
>
> $$
> P = \frac{2}{5}\,n\,\epsilon_F = \frac{2}{5}\,(8.47\times10^{28})(1.12\times10^{-18})
> = 3.8\times10^{10}\,\mathrm{Pa},
> $$
>
> about $4\times10^5$ atmospheres. This is the pressure the conduction electrons
> would exert if the ions did not confine them; it is balanced in the metal by the
> electrostatic attraction of the lattice.

## Summary

- At $T=0$ the Fermi-Dirac occupation is a step: states fill up to the Fermi
  energy $\epsilon_F$ and are empty above it, forming a filled Fermi sphere of
  radius $p_F=\hbar(6\pi^2 n/g_s)^{1/3}$ in momentum space.
- The Fermi energy scales with density as $\epsilon_F\propto n^{2/3}$; the density
  of states $g(\epsilon)\propto\epsilon^{1/2}$ satisfies $g(\epsilon_F)=3N/2\epsilon_F$.
- The ground-state energy is $U=\tfrac35 N\epsilon_F$, and with $PV=\tfrac23 U$ the
  degeneracy pressure is $P=\tfrac25 n\epsilon_F\propto n^{5/3}$, finite at absolute
  zero and independent of temperature.
- For metals $\epsilon_F$ is a few eV and $T_F\sim10^4$–$10^5\,\mathrm{K}$, so
  conduction electrons at room temperature are deep in the degenerate regime and
  are fast but nonrelativistic.

[^ammetals]: Ashcroft & Mermin, Ch. 2, tabulate free-electron parameters for the
elemental metals; the densities and derived $\epsilon_F$, $T_F$, $v_F$ here follow
that treatment. Constants from NIST, <https://physics.nist.gov/cuu/Constants/>.