The Ideal Fermi Gas at Zero Temperature

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

\overset{n}{ˉ} (ϵ) = \frac{1}{e ^{(ϵ - μ) / k_{B} T} + 1},

with $μ$ 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 $(ϵ - μ) / k_{B} T$ runs to $- \infty$ for $ϵ < μ$ and to $+ \infty$ for $ϵ > μ$ . The occupation collapses to a step function,

\overset{n}{ˉ} (ϵ) T \to 0 {1, 0, ϵ < ϵ_{F}, ϵ > ϵ_{F},

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

The zero-temperature occupation is a step of unit height that drops to zero at

ϵ_{F}

; no state is partially filled.

For free particles in a cubic box of volume $V$ , the single-particle energies are $ϵ = ℏ^{2} k^{2} /2 m$ , and the allowed wavevectors form a lattice of density $V / (2 π)^{3}$ in $k$ -space. Including a spin degeneracy $g_{s}$ (with $g_{s} = 2$ for spin- $\frac{1}{2}$ electrons or neutrons), the occupied states at $T = 0$ fill a sphere of radius $k_{F}$ in $k$ -space, the Fermi sphere. The corresponding momentum $p_{F} = ℏ k_{F}$ is the Fermi momentum.

In momentum space the ground state fills a sphere of radius

p_{F}

; states inside are occupied, states outside empty, and the surface

∣ p ∣ = p_{F}

is the Fermi surface.

Fermi momentum, energy, and temperature

The total number of particles equals $g_{s}$ times the number of $k$ -lattice points inside the Fermi sphere,

N = g_{s} \frac{V}{( 2 π ) ^{3}} \frac{4}{3} π k_{F}^{3} = \frac{g _{s} V}{6 π ^{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} = (\frac{6 π ^{2} n}{g _{s}})^{1/3}, p_{F} = ℏ (\frac{6 π ^{2} n}{g _{s}})^{1/3} .

For the standard electron gas ( $g_{s} = 2$ ) this reduces to $k_{F} = (3 π^{2} n)^{1/3}$ . 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,

ϵ_{F} = \frac{p _{F}^{2}}{2 m} = \frac{ℏ ^{2}}{2 m} (\frac{6 π ^{2} n}{g _{s}})^{2/3} g_{s} = 2 \frac{ℏ ^{2}}{2 m} (3 π^{2} n)^{2/3},

and the Fermi temperature $T_{F} = ϵ_{F} / k_{B}$ marks the crossover between the degenerate regime $T ≪ T_{F}$ , where the step is sharp, and the classical regime $T ≫ T_{F}$ , where Maxwell-Boltzmann statistics apply. The key structural fact is that $ϵ_{F} \propto n^{2/3}$ : compressing the gas raises the Fermi energy, because the added particles must occupy higher momentum states.¹

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 $λ = h / 2 π m k_{B} T$ , the ratio $T / T_{F}$ is, up to a numerical factor, $(λ^{3} n)^{- 2/3}$ : the gas is degenerate exactly when the thermal wavelength exceeds the interparticle spacing, $n λ^{3} ≳ 1$ , 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 $1 0^{4}$ – $1 0^{5} K$ , far above any temperature at which the solid survives.

The density of states

The number of single-particle states with energy below $ϵ$ is $g_{s}$ times the $k$ -space volume inside $k = 2 m ϵ /ℏ$ ,

N (ϵ) = \frac{g _{s} V}{6 π ^{2}} (\frac{2 m ϵ}{ℏ ^{2}})^{3/2} .

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

g (ϵ) = \frac{g _{s} V}{4 π ^{2}} (\frac{2 m}{ℏ ^{2}})^{3/2} ϵ^{1/2} = \frac{3 N}{2} \frac{ϵ ^{1/2}}{ϵ _{F}^{3/2}} .

The second form, valid once $N$ and $ϵ_{F}$ are fixed, follows from $N = \int_{0}^{ϵ_{F}} g (ϵ) d ϵ = \frac{2}{3} ϵ_{F} g (ϵ_{F})$ because $g \propto ϵ^{1/2}$ . Evaluated at the Fermi surface it gives the much-used relation

g (ϵ_{F}) = \frac{3 N}{2 ϵ _{F}},

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

The density of states rises as

ϵ^{1/2}

; at

T = 0

every level up to

ϵ_{F}

is filled (shaded) and every level above is empty. The filled count is the area under the curve to

ϵ_{F}

Ground-state energy and degeneracy pressure

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

U = 0 \int ϵ_{F} ϵ g (ϵ) d ϵ = \frac{3 N}{2 ϵ _{F}^{3/2}} 0 \int ϵ_{F} ϵ^{3/2} d ϵ = \frac{3}{5} N ϵ_{F} .

The average energy per fermion is $\frac{3}{5} ϵ_{F}$ , 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.

The ground-state energy is the area under

ϵ g (ϵ) \propto ϵ^{3/2}

up to

ϵ_{F}

; the mean energy per particle works out to

\frac{3}{5} ϵ_{F}

Because the energy of a nonrelativistic ideal gas satisfies $P V = \frac{2}{3} U$ (a result independent of statistics, derived for the general quantum gas), the zero-temperature degeneracy pressure is

P = \frac{2 U}{3 V} = \frac{2}{5} n ϵ_{F} = \frac{( 3 π ^{2} ) ^{2/3}}{5} \frac{ℏ ^{2}}{m} n^{5/3} (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 $\frac{5}{3}$ .

The zero-temperature degeneracy pressure follows

P \propto n^{5/3}

; on log axes it is a line of slope

\frac{5}{3}

, marked by the run-and-rise triangle.

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

K = - V (\frac{\partial P}{\partial V})_{N} = \frac{5}{3} P = \frac{2}{3} n ϵ_{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 $P_{cl} = n k_{B} T$ , which vanishes as $T \to 0$ ; the Fermi gas holds $P = \frac{2}{5} n ϵ_{F}$ at absolute zero. Their ratio is $P / P_{cl} \sim ϵ_{F} / k_{B} T = T_{F} / T$ , 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.² With the electron mass $m_{e}$ and the measured conduction- electron densities, the free-electron model gives the following.

Metal	$n$ ( $1 0^{28} m^{- 3}$ )	$ϵ_{F}$ (eV)	$T_{F}$ ( $1 0^{4} K$ )	$v_{F}$ ( $1 0^{6} m s^{- 1}$ )
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 ( $T \approx 300 K$ ) sits at $T / T_{F} \sim 4 \times 1 0^{- 3}$ , so the conduction electrons are extremely degenerate and the zero-temperature picture is an excellent starting point. Second, the Fermi velocity $v_{F} = p_{F} / m_{e} \sim 1 0^{6} m s^{- 1}$ is about one percent of the speed of light: metallic electrons are fast but nonrelativistic, which justifies the $ϵ = p^{2} /2 m$ dispersion used throughout. Relativistic corrections enter only at the far higher densities of a white dwarf, treated in a later lesson.

Summary

At $T = 0$ the Fermi-Dirac occupation is a step: states fill up to the Fermi energy $ϵ_{F}$ and are empty above it, forming a filled Fermi sphere of radius $p_{F} = ℏ (6 π^{2} n / g_{s})^{1/3}$ in momentum space.
The Fermi energy scales with density as $ϵ_{F} \propto n^{2/3}$ ; the density of states $g (ϵ) \propto ϵ^{1/2}$ satisfies $g (ϵ_{F}) = 3 N /2 ϵ_{F}$ .
The ground-state energy is $U = \frac{3}{5} N ϵ_{F}$ , and with $P V = \frac{2}{3} U$ the degeneracy pressure is $P = \frac{2}{5} n ϵ_{F} \propto n^{5/3}$ , finite at absolute zero and independent of temperature.
For metals $ϵ_{F}$ is a few eV and $T_{F} \sim 1 0^{4}$ – $1 0^{5} K$ , so conduction electrons at room temperature are deep in the degenerate regime and are fast but nonrelativistic.

Schroeder, §7.3; Pathria & Beale, §8.1. Companion notes at https://physics.weber.edu/schroeder/thermal/. ↩
Ashcroft & Mermin, Ch. 2, tabulate free-electron parameters for the elemental metals; the densities and derived $ϵ_{F}$ , $T_{F}$ , $v_{F}$ here follow that treatment. Constants from NIST, https://physics.nist.gov/cuu/Constants/. ↩