Fermi Gas Formula Sheet

This sheet specializes the Quantum Gas Formula Sheet to ideal fermions. It collects the formulas most often needed for state counting, thermodynamics, response checks, and translation among one-, two-, and three-dimensional continuum conventions.

Required background. Fermi–Dirac statistics derives the mode distribution, and the ideal Fermi gas explains how state counting determines thermodynamics.

Quadratic fermions and their energy scales

Unless stated otherwise:

• the particles are noninteracting fermions of mass $m$;
• the continuum is homogeneous and $d$ dimensional;
• $\epsilon_{\mathbf k}=\hbar^2k^2/(2m)$;
• $g$ counts equally populated internal components;
• $n=N/V$ is total density across those components;
• the thermodynamic limit follows periodic finite-volume state counting;
• $\epsilon=0$ is the one-particle band bottom;
• $\beta=1/(k_{\mathrm B}T)$ and $z=e^{\beta\mu}$;
• all density-of-states functions below include the factor $g$.

For noninteracting lattice bands, traps, relativistic dispersions or spin imbalance, use the appropriate spectrum and component chemical potentials in the Fermi function. Interactions generally prevent the exact bare-mode occupation from having this simple form; an effective quasiparticle approximation must be stated separately.

Core Distribution

The Fermi–Dirac occupation is

$$f(\epsilon) \equiv \frac{1}{e^{\beta(\epsilon-\mu)}+1} = \frac{1}{z^{-1}e^{\beta\epsilon}+1}.$$

It obeys

$$0 \le f(\epsilon) \le 1.$$

At zero temperature,

$$f(\epsilon) \longrightarrow \theta(\epsilon_{\mathrm F}-\epsilon),$$

with $\mu(0)=\epsilon_{\mathrm F}$ for the ideal continuum gas at fixed density.

The particle–hole identity around $\mu$ is

$$f(\mu+\delta) = 1-f(\mu-\delta).$$

Thermal Window

The derivative that localizes low-temperature corrections is

$$\begin{aligned} -\frac{\partial f}{\partial\epsilon} &= \beta f(\epsilon) \left[ 1-f(\epsilon) \right] \\ &= \frac{\beta}{ 4\cosh^2\!\left[ \beta(\epsilon-\mu)/2 \right] }. \end{aligned}$$

Its width is of order $k_{\mathrm B}T$, its maximum is $\beta/4$, and

$$\int_{-\infty}^{\infty} d\epsilon\, \left( -\frac{\partial f}{\partial\epsilon} \right) = 1.$$

For a spectrum bounded below by zero, extending the integral to $-\infty$ is accurate in the degenerate regime $\mu\gg k_{\mathrm B}T$.

Exact Independent-Mode Formulas

For one-particle modes $\alpha$,

$$H = \sum_\alpha \epsilon_\alpha n_\alpha, \qquad n_\alpha\in\{0,1\}.$$

The grand partition function and grand potential are

$$\begin{aligned} \Xi &= \prod_\alpha \left[ 1+e^{-\beta(\epsilon_\alpha-\mu)} \right], \\ \Omega &= -k_{\mathrm B}T \sum_\alpha \ln\left[ 1+e^{-\beta(\epsilon_\alpha-\mu)} \right]. \end{aligned}$$

The mean and variance of one mode are

$$\begin{aligned} \langle n_\alpha\rangle &= f(\epsilon_\alpha), \\ \operatorname{Var}(n_\alpha) &= f(\epsilon_\alpha) \left[ 1-f(\epsilon_\alpha) \right]. \end{aligned}$$

The one-mode entropy is

$$\frac{S_\alpha}{k_{\mathrm B}} = -f_\alpha\ln f_\alpha - (1-f_\alpha) \ln(1-f_\alpha).$$

These independent-mode fluctuation formulas are grand canonical. Fixing the total particle number correlates the occupations.

Density-of-States Workflow

Let $D(\epsilon)$ be the total one-particle density of states, including internal degeneracy and physical volume. Then

$$N = \int_0^\infty d\epsilon\, D(\epsilon)f(\epsilon).$$

At fixed $N$, this equation determines $\mu(T)$. The internal energy is

$$U = \int_0^\infty d\epsilon\, \epsilon D(\epsilon)f(\epsilon),$$

and the grand potential is

$$\begin{aligned} \Omega &= -k_{\mathrm B}T \int_0^\infty d\epsilon\, D(\epsilon) \mathcal L(\epsilon), \\ \mathcal L(\epsilon) &\equiv \ln\left[ 1+e^{-\beta(\epsilon-\mu)} \right]. \end{aligned}$$

The entropy is

$$S = -k_{\mathrm B} \int_0^\infty d\epsilon\, D(\epsilon) \times \left[ f\ln f + (1-f)\ln(1-f) \right].$$

These formulas remain useful for a noninteracting lattice band or trap after inserting the correct $D(\epsilon)$. Identities such as $U=dPV/2$ do not survive an arbitrary dispersion or confining potential.

Free-Continuum Density of States

For $\epsilon_{\mathbf k}=\hbar^2k^2/(2m)$,

$$D_d(\epsilon) = \frac{gV}{\Gamma(d/2)} \left( \frac{m}{2\pi\hbar^2} \right)^{d/2} \times \epsilon^{d/2-1}.$$

In one, two, and three dimensions:

$$\begin{aligned} \frac{D_1(\epsilon)}{L} &= \frac{g}{\pi\hbar} \sqrt{ \frac{m}{2\epsilon} }, \\ \frac{D_2(\epsilon)}{A} &= \frac{gm}{2\pi\hbar^2}, \\ \frac{D_3(\epsilon)}{V} &= \frac{g}{4\pi^2} \left( \frac{2m}{\hbar^2} \right)^{3/2} \sqrt{\epsilon}. \end{aligned}$$

Here $L$, $A$, and $V$ denote one-, two-, and three-dimensional measures. The one-dimensional divergence at the band bottom is integrable in the particle-number integral away from singular limits.

Fermi-Sphere Counting

Let

$$\mathcal V_d \equiv \frac{\pi^{d/2}}{ \Gamma(d/2+1) }$$

be the volume of the unit $d$-ball. At $T=0$,

$$n = \frac{g\mathcal V_d}{(2\pi)^d} k_{\mathrm F}^d.$$

Equivalently,

Dimension	Density–wave-number relation	Fermi wave number
$d=1$	$n=gk_{\mathrm F}/\pi$	$k_{\mathrm F}=\pi n/g$
$d=2$	$n=gk_{\mathrm F}^2/(4\pi)$	$k_{\mathrm F}=\sqrt{4\pi n/g}$
$d=3$	$n=gk_{\mathrm F}^3/(6\pi^2)$	$k_{\mathrm F}=(6\pi^2n/g)^{1/3}$

The associated scales are

$$\begin{aligned} \epsilon_{\mathrm F} &= \frac{\hbar^2k_{\mathrm F}^2}{2m}, \\ T_{\mathrm F} &= \frac{\epsilon_{\mathrm F}}{k_{\mathrm B}}, \\ v_{\mathrm F} &= \frac{\hbar k_{\mathrm F}}{m}, \\ \lambda_{\mathrm F} &= \frac{2\pi}{k_{\mathrm F}}. \end{aligned}$$

For unpolarized spin-$1/2$ fermions, $g=2$. For a polarized gas, count each component with its own density and Fermi momentum rather than inserting one common factor $g=2$.

Zero-Temperature Thermodynamics

For the quadratic continuum in $d$ dimensions,

$$\begin{aligned} \mu(0) &= \epsilon_{\mathrm F}, \\ \frac{U_0}{N} &= \frac{d}{d+2} \epsilon_{\mathrm F}, \\ P_0 &= \frac{2}{d+2} n\epsilon_{\mathrm F}. \end{aligned}$$

The scaling identity is

$$P = \frac{2}{d} \frac{U}{V}$$

at any temperature for this homogeneous quadratic ideal gas.

At the Fermi energy,

$$\frac{D_d(\epsilon_{\mathrm F})}{V} = \frac{dn}{2\epsilon_{\mathrm F}}.$$

The zero-temperature isothermal compressibility is

$$\kappa_T(0) = \frac{1}{n^2} \left( \frac{\partial n}{\partial\mu} \right)_{T=0} = \frac{d}{2n\epsilon_{\mathrm F}}.$$

Dimension-specific values are:

$d$	$U_0/N$	$P_0$	$\kappa_T(0)$
$1$	$\epsilon_{\mathrm F}/3$	$2n\epsilon_{\mathrm F}/3$	$1/(2n\epsilon_{\mathrm F})$
$2$	$\epsilon_{\mathrm F}/2$	$n\epsilon_{\mathrm F}/2$	$1/(n\epsilon_{\mathrm F})$
$3$	$3\epsilon_{\mathrm F}/5$	$2n\epsilon_{\mathrm F}/5$	$3/(2n\epsilon_{\mathrm F})$

The nonzero pressure at $T=0$ is a consequence of state filling under Pauli exclusion. It is not a two-body repulsive force.

Finite-Temperature Continuum Formulas

For real $s>0$ and $z>0$, define

$$\begin{aligned} f_s(z) &\equiv -\operatorname{Li}_s(-z) \\ &= \frac{1}{\Gamma(s)} \int_0^\infty dt\, \frac{t^{s-1}}{z^{-1}e^t+1}. \end{aligned}$$

For

$$\lambda_T = \sqrt{ \frac{2\pi\hbar^2}{mk_{\mathrm B}T} },$$

the number density, pressure, and energy are

$$\begin{aligned} n &= \frac{g}{\lambda_T^d} f_{d/2}(z), \\ P &= \frac{gk_{\mathrm B}T}{\lambda_T^d} f_{d/2+1}(z), \\ U &= \frac d2PV. \end{aligned}$$

The entropy density is

$$\frac{S}{k_{\mathrm B}V} = \frac{g}{\lambda_T^d} \Bigg[ \left( \frac d2+1 \right) f_{d/2+1}(z) - \ln z\, f_{d/2}(z) \Bigg].$$

The number equation must be solved for $z$ or $\mu$ when density, rather than chemical potential, is fixed.

Degeneracy Diagnostics

Define

$$\vartheta \equiv \frac{T}{T_{\mathrm F}}$$

and the phase-space density per internal component

$$x \equiv \frac{n\lambda_T^d}{g}.$$

For the quadratic continuum,

$$x = \frac{1}{\Gamma(d/2+1)} \vartheta^{-d/2}.$$

Thus:

Regime	Equivalent diagnostics	Useful method
dilute classical	$x\ll1$, $z\ll1$, $T\gg T_{\mathrm F}$	fugacity or virial expansion
crossover	$x=O(1)$, $T=O(T_{\mathrm F})$	numerical Fermi integrals
degenerate	$x\gg1$, $z\gg1$, $T\ll T_{\mathrm F}$	Sommerfeld expansion

The condition “low temperature” is meaningful only relative to $T_{\mathrm F}$ and any other energy scales.

Sommerfeld Master Formula

For a sufficiently smooth function $\varphi(\epsilon)$ and $k_{\mathrm B}T\ll\mu$,

$$\begin{aligned} I_\varphi(T,\mu) &\equiv \int_0^\infty d\epsilon\, \varphi(\epsilon)f(\epsilon) \\ &= \int_0^\mu d\epsilon\, \varphi(\epsilon) + \frac{\pi^2}{6} (k_{\mathrm B}T)^2 \varphi'(\mu) + \frac{7\pi^4}{360} (k_{\mathrm B}T)^4 \varphi'''(\mu) + \cdots. \end{aligned}$$

The expansion assumes the lower spectral endpoint is far outside the thermal window and that $\varphi$ is smooth near $\mu$. Van Hove singularities, band edges, gaps, and rapidly varying matrix elements require separate treatment.

Fixed-Density Low-Temperature Results

For the homogeneous quadratic gas, let $\vartheta=T/T_{\mathrm F}\ll1$. At fixed density,

$$\frac{\mu(T)}{\epsilon_{\mathrm F}} = 1 - \frac{\pi^2}{12} (d-2)\vartheta^2 + O(\vartheta^4).$$

The energy and pressure are

$$\begin{aligned} \mathcal R_d(\vartheta) &\equiv 1+ \frac{\pi^2(d+2)}{12} \vartheta^2 + O(\vartheta^4), \\ \frac{U}{N} &= \frac{d}{d+2} \epsilon_{\mathrm F} \mathcal R_d(\vartheta), \\ P &= \frac{2}{d+2} n\epsilon_{\mathrm F} \mathcal R_d(\vartheta). \end{aligned}$$

The heat capacity and entropy are

$$\begin{aligned} \frac{C_V}{Nk_{\mathrm B}} &= \frac{\pi^2d}{6} \vartheta + O(\vartheta^3), \\ \frac{S}{Nk_{\mathrm B}} &= \frac{\pi^2d}{6} \vartheta + O(\vartheta^3). \end{aligned}$$

To the displayed leading orders, for $d=3$:

$$\begin{aligned} \mu(T) &\simeq \epsilon_{\mathrm F} \left( 1- \frac{\pi^2}{12}\vartheta^2 \right), \\ \frac{C_V}{Nk_{\mathrm B}} &\simeq \frac{\pi^2}{2}\vartheta. \end{aligned}$$

In $d=2$, the free-particle density of states is constant. The algebraic $T^2$ shift of $\mu$ vanishes; the leading fixed-density correction is exponentially small, as shown below.

Exact Two-Dimensional Chemical Potential

For a two-dimensional quadratic gas,

$$\frac{D_2}{A} = \frac{gm}{2\pi\hbar^2} \equiv \mathcal D_0.$$

The number equation is

$$n = \mathcal D_0 k_{\mathrm B}T \ln(1+z).$$

Since $\epsilon_{\mathrm F}=n/\mathcal D_0$,

$$\begin{aligned} \mu(T) &= k_{\mathrm B}T \ln\left( e^{\epsilon_{\mathrm F}/(k_{\mathrm B}T)}-1 \right) \\ &= \epsilon_{\mathrm F} + k_{\mathrm B}T \ln\left( 1-e^{-\epsilon_{\mathrm F}/(k_{\mathrm B}T)} \right). \end{aligned}$$

Therefore $\mu-\epsilon_{\mathrm F}$ is exponentially small when $T\ll T_{\mathrm F}$. A finite upper band edge or energy-dependent density of states changes this special result. For numerical evaluation at low temperature, the second expression avoids overflow; a stable implementation uses log1p and expm1 in the appropriate regimes.

Number Response and Fluctuations

The number derivative is

$$\left( \frac{\partial N}{\partial\mu} \right)_T = \beta \int_0^\infty d\epsilon\, D(\epsilon) f(\epsilon) \times \left[ 1-f(\epsilon) \right].$$

In the grand-canonical ensemble,

$$\operatorname{Var}(N) = k_{\mathrm B}T \left( \frac{\partial N}{\partial\mu} \right)_T.$$

At low temperature,

$$\begin{aligned} \left( \frac{\partial N}{\partial\mu} \right)_T &\simeq D(\epsilon_{\mathrm F}), \\ \operatorname{Var}(N) &\simeq k_{\mathrm B}T D(\epsilon_{\mathrm F}). \end{aligned}$$

For the quadratic continuum,

$$\frac{\operatorname{Var}(N)}{N} \simeq \frac d2 \frac{T}{T_{\mathrm F}}.$$

This is a variance-to-mean ratio, not the squared relative fluctuation. The relative root-mean-square fluctuation still vanishes with increasing $N$ at fixed thermodynamic conditions.

The isothermal compressibility is

$$\kappa_T = \frac{1}{n^2} \left( \frac{\partial n}{\partial\mu} \right)_T.$$

Dilute Classical Limit

For

$$x = \frac{n\lambda_T^d}{g} \ll 1,$$

the activity and chemical potential are

$$\begin{aligned} z &= x + \frac{x^2}{2^{d/2}} + O(x^3), \\ \frac{\mu}{k_{\mathrm B}T} &= \ln x + \frac{x}{2^{d/2}} + O(x^2). \end{aligned}$$

The pressure is

$$\frac{P}{nk_{\mathrm B}T} = 1 + \frac{x}{2^{d/2+1}} + O(x^2).$$

The positive correction is the leading ideal exchange effect for fermions. It does not imply a microscopic repulsive potential.

The leading classical entropy per particle is

$$\frac{S}{Nk_{\mathrm B}} = \frac d2+1 - \ln\left( \frac{n\lambda_T^d}{g} \right) + O(x).$$

Model-Boundary Translations

Lattice bands

Use the band dispersion $\epsilon_{a\mathbf k}$ and Brillouin-zone sum:

$$N = \sum_{a,\mathbf k} f(\epsilon_{a\mathbf k}).$$

The Fermi surface is defined by $\epsilon_{a\mathbf k}=\mu$ at low temperature. The free-space formulas for $k_{\mathrm F}$, pressure, and $D_d(\epsilon)$ generally do not apply.

Traps

Replace the homogeneous density of states by the trap density of states or use a controlled local-density approximation. The symbol $V$ should not be interpreted as a uniform box volume when no such volume exists.

Relativistic fermions

Replace the dispersion and state-counting measure consistently. In the ultrarelativistic limit, $P=U/(dV)$ rather than $P=2U/(dV)$.

Interacting fermions

The exact momentum occupation need not be a Fermi–Dirac function of the bare one-particle energy. Hartree–Fock theory, quasiparticle theory, and Fermi-liquid theory introduce effective spectra and renormalized response coefficients. State which spectrum enters $f$.

Exercises

1. Density of states at the Fermi level

Use $N\propto\epsilon_{\mathrm F}^{d/2}$ to show that

$$\frac{D_d(\epsilon_{\mathrm F})}{V} = \frac{dn}{2\epsilon_{\mathrm F}}$$

and derive the zero-temperature compressibility.

Solution

At zero temperature,

$$N = \int_0^{\epsilon_{\mathrm F}} d\epsilon\, D_d(\epsilon).$$

Because $D_d(\epsilon)\propto\epsilon^{d/2-1}$,

$$N = \frac{2}{d} D_d(\epsilon_{\mathrm F}) \epsilon_{\mathrm F}.$$

Divide by $V$ and rearrange to obtain

$$\frac{D_d(\epsilon_{\mathrm F})}{V} = \frac{dn}{2\epsilon_{\mathrm F}}.$$

Since $\mu(0)=\epsilon_{\mathrm F}$,

$$\begin{aligned} \kappa_T(0) &= \frac{1}{n^2} \left( \frac{\partial n}{\partial\mu} \right)_{T=0} \\ &= \frac{d}{2n\epsilon_{\mathrm F}}. \end{aligned}$$

2. Low-temperature chemical potential in dimension d

For $D_d(\epsilon)=C\epsilon^{d/2-1}$, use the Sommerfeld expansion at fixed $N$ to derive the leading shift of $\mu(T)$.

Solution

Let $a=d/2-1$. The number equation is

$$N = C \frac{\mu^{a+1}}{a+1} + \frac{\pi^2}{6} (k_{\mathrm B}T)^2 Ca\mu^{a-1} + O(T^4).$$

Write $\mu=\epsilon_{\mathrm F}+\delta\mu$ and compare with the zero-temperature value. To order $T^2$,

$$C\epsilon_{\mathrm F}^{a} \delta\mu + \frac{\pi^2}{6} (k_{\mathrm B}T)^2 Ca\epsilon_{\mathrm F}^{a-1} = 0.$$

Hence

$$\frac{\delta\mu}{\epsilon_{\mathrm F}} = -\frac{\pi^2}{6} a \left( \frac{T}{T_{\mathrm F}} \right)^2.$$

Using $a=(d-2)/2$ gives

$$\frac{\mu(T)}{\epsilon_{\mathrm F}} = 1 - \frac{\pi^2}{12} (d-2) \times \left( \frac{T}{T_{\mathrm F}} \right)^2 + O\!\left[ \left( \frac{T}{T_{\mathrm F}} \right)^4 \right].$$

3. Exact two-dimensional chemical potential

Starting from the constant two-dimensional density of states, derive the exact fixed-density expression for $\mu(T)$.

Solution

With $D_2/A=\mathcal D_0$,

$$\begin{aligned} n &= \mathcal D_0 \int_0^\infty d\epsilon\, \frac{1}{e^{\beta(\epsilon-\mu)}+1} \\ &= \mathcal D_0 k_{\mathrm B}T \ln(1+e^{\beta\mu}). \end{aligned}$$

Because $n=\mathcal D_0\epsilon_{\mathrm F}$,

$$e^{\beta\mu} = e^{\beta\epsilon_{\mathrm F}}-1.$$

Taking the logarithm gives

$$\mu(T) = k_{\mathrm B}T \ln\left( e^{\epsilon_{\mathrm F}/(k_{\mathrm B}T)}-1 \right).$$

Factoring out the large exponential yields the equivalent form

$$\mu(T) = \epsilon_{\mathrm F} + k_{\mathrm B}T \ln\left( 1-e^{-\epsilon_{\mathrm F}/(k_{\mathrm B}T)} \right).$$

4. Grand-canonical number variance

Show that a low-temperature quadratic gas satisfies

$$\frac{\operatorname{Var}(N)}{N} \simeq \frac d2 \frac{T}{T_{\mathrm F}}.$$

Solution

The fluctuation relation gives

$$\operatorname{Var}(N) = k_{\mathrm B}T \left( \frac{\partial N}{\partial\mu} \right)_T.$$

At low temperature, the derivative is localized at the Fermi energy:

$$\left( \frac{\partial N}{\partial\mu} \right)_T \simeq D_d(\epsilon_{\mathrm F}).$$

Using

$$D_d(\epsilon_{\mathrm F}) = \frac{dN}{2\epsilon_{\mathrm F}},$$

one obtains

$$\frac{\operatorname{Var}(N)}{N} \simeq \frac d2 \frac{k_{\mathrm B}T}{\epsilon_{\mathrm F}} = \frac d2 \frac{T}{T_{\mathrm F}}.$$

This result is grand canonical; it is identically zero for an exactly fixed total $N$.

References

• N. W. Ashcroft and N. D. Mermin, Solid State Physics, Harcourt (1976) — electron-gas state counting, Fermi surfaces, and low-temperature response.
• P. Coleman, Introduction to Many-Body Physics, Cambridge University Press (2015) — Fermi seas, quasiparticles, and many-body conventions.
• K. Huang, Statistical Mechanics, 2nd ed., Wiley (1987) — density-of-states integrals, virial expansion, and degenerate limits.
• L. D. Landau and E. M. Lifshitz, Statistical Physics, Part 1, 3rd ed., Butterworth–Heinemann (1980) — degenerate Fermi thermodynamics and low-temperature expansions.
• G. D. Mahan, Many-Particle Physics, 3rd ed., Springer (2000) — Fermi functions, response, and interacting extensions.
• R. K. Pathria and P. D. Beale, Statistical Mechanics, 3rd ed., Elsevier (2011) — ideal Fermi gases and thermodynamic ensembles.