Fermi–Dirac Statistics

Fermi–Dirac statistics gives the equilibrium occupation of ideal fermionic modes. For a mode $i$ with one-particle energy $\epsilon_i$,

$$\overline n_i = \frac{1}{ e^{\beta(\epsilon_i-\mu)}+1 }, \qquad \beta = \frac{1}{k_{\mathrm B}T}.$$

The plus sign in the denominator follows from the fermionic occupation rule

$$n_i \in \{0,1\}.$$

Every complete one-particle mode is either empty or occupied by one fermion. The mean $\overline n_i$ can lie anywhere between zero and one because it is an ensemble average, not a fractional eigenvalue.

The formula card collects the result for quick lookup. Here the two possible mode occupations lead to the distribution, its fluctuations, and its temperature dependence.

Helpful background. The grand-canonical ensemble supplies equilibrium weights; Pauli exclusion supplies the two allowed mode occupations.

Independent fermionic modes

Take positive temperature and a number-conserving ideal Hamiltonian

$$H=\sum_i\epsilon_i n_i,\qquad N=\sum_i n_i.$$

A mode includes all quantum labels, such as momentum, spin, and band. Each complete fermionic mode has $n_i=0,1$. Two spin states in one spatial orbital are two modes, with a combined capacity of two electrons.

The thermal law is exact in the grand-canonical state $\rho=\Xi^{-1}e^{-\beta(H-\mu N)}$. Exclusion alone does not determine the mean: nonequilibrium states obey the same binary restriction but need not have Fermi–Dirac means. Conversely, matching the thermal means alone does not prove that a many-body state is the thermal density operator. Fixed-number ensembles and interacting bare modes require the qualifications below.

Grand-Canonical Derivation

For the grand Hamiltonian

$$K = H-\mu N,$$

independent modes give

$$K = \sum_i (\epsilon_i-\mu)n_i.$$

The grand partition function factorizes:

$$\Xi = \prod_i \xi_i.$$

One fermionic mode has only two allowed occupations, so

$$\begin{aligned} \xi_i &= \sum_{n_i=0}^{1} e^{-\beta(\epsilon_i-\mu)n_i} \\ &= 1+q_i, \end{aligned}$$

where

$$q_i \equiv e^{-\beta(\epsilon_i-\mu)}.$$

The mean occupation is

$$\begin{aligned} \overline n_i &= q_i \frac{\partial}{\partial q_i} \ln\xi_i \\ &= \frac{q_i}{1+q_i} \\ &= \frac{1}{ e^{\beta(\epsilon_i-\mu)}+1 }. \end{aligned}$$

Unlike a bosonic geometric sum, the one-mode fermionic sum is finite. For a finite set of fermionic modes, no convergence bound analogous to $\mu<\epsilon_0$ is needed for finite real $\mu$.

For infinitely many modes the trace is finite only when $\sum_i\ln(1+q_i)<\infty$. A confined gas with a sufficiently growing spectrum can satisfy this; an infinite-volume gas is described by thermodynamic limits of densities, not by assuming a finite global trace.

The ideal-mode grand potential is

$$\Omega_{\mathrm G} = -k_{\mathrm B}T \sum_i \ln \left[ 1+e^{-\beta(\epsilon_i-\mu)} \right].$$

The Full One-Mode Distribution

One ideal fermionic mode has a Bernoulli distribution:

$$P_i(0) = \frac{1}{1+q_i}, \qquad P_i(1) = \frac{q_i}{1+q_i}.$$

Therefore

$$P_i(1) = \overline n_i, \qquad P_i(0) = 1-\overline n_i.$$

The probability ratio is

$$\frac{P_i(1)}{P_i(0)} = q_i = e^{-\beta(\epsilon_i-\mu)}.$$

At $\epsilon_i=\mu$, the empty and occupied states have equal grand energy, so

$$P_i(0) = P_i(1) = \frac{1}{2}$$

for every finite positive temperature.

Dimensionless Profile

Define

$$x \equiv \beta(\epsilon-\mu).$$

The Fermi function is

$$f_{\mathrm F}(x) = \frac{1}{e^x+1}.$$

It is monotone, bounded, and centered at

$$f_{\mathrm F}(0) = \frac{1}{2}.$$

At low temperature it approaches a sharp step in energy. Finite temperature rounds that step over an energy interval of order $k_{\mathrm B}T$.

Cooling sharpens the Fermi–Dirac occupation toward a step at $\epsilon=\mu$. The one-mode variance $f_{\mathrm F}(1-f_{\mathrm F})$ is concentrated in the same thermal window and reaches its maximum $1/4$ at the chemical potential.

Pauli Blocking

For one complete fermionic mode,

$$0 \leq \overline n_i \leq 1.$$

The mean availability of an initially unoccupied complete mode is

$$1-\overline n_i.$$

In independent-quasiparticle kinetic equations, transition rates factor into incoming occupations and outgoing availability factors such as $1-f$. This product form additionally neglects suitable higher correlations; exclusion itself is exact. A fully occupied final mode cannot accept another identical fermion.

This is not a new repulsive force. The restriction comes from the antisymmetric state space and fermionic operator algebra. It changes the allowed many-particle configurations and therefore has major energetic and thermodynamic consequences.

Particle–Hole Symmetry of the Function

The dimensionless Fermi function obeys

$$f_{\mathrm F}(-x) = 1-f_{\mathrm F}(x).$$

Equivalently,

$$f_{\mathrm F}(x) + f_{\mathrm F}(-x) = 1.$$

An energy $\mu+\Delta$ above the chemical potential has the same particle occupation as the hole probability at $\mu-\Delta$ below it:

$$f_{\mathrm F} \left( \frac{\Delta}{k_{\mathrm B}T} \right) = 1- f_{\mathrm F} \left( -\frac{\Delta}{k_{\mathrm B}T} \right).$$

This identity belongs to the ideal Fermi function. A physical system need not possess an exact particle–hole symmetry in its density of states, dispersion, or interactions.

Number Fluctuations

Because $n_i^2=n_i$,

$$\langle n_i^2\rangle = \overline n_i.$$

In any state, not only in equilibrium, the variance is therefore

$$\operatorname{Var}(n_i) = \overline n_i \left( 1-\overline n_i \right).$$

This is smaller than the Poisson benchmark $\operatorname{Var}(n)=\overline n$ whenever $0<\overline n<1$. Fermionic exclusion suppresses occupation fluctuations.

The variance is largest at half occupation:

$$\operatorname{Var}(n_i)_{\max} = \frac{1}{4}, \qquad \epsilon_i = \mu.$$

Modes far below $\mu$ are almost certainly occupied, and modes far above $\mu$ are almost certainly empty. Both have small variance.

Fluctuation–Response Identity

Holding the one-particle energies and temperature fixed, differentiating with respect to chemical potential gives

$$\left( \frac{\partial\overline n_i}{\partial\mu} \right)_T = \beta \overline n_i \left( 1-\overline n_i \right).$$

Hence

$$\left( \frac{\partial\overline n_i}{\partial\mu} \right)_T = \beta \operatorname{Var}(n_i).$$

For independent modes,

$$\operatorname{Var}(N) = \sum_i \overline n_i \left( 1-\overline n_i \right),$$

and

$$\left( \frac{\partial\overline N}{\partial\mu} \right)_T = \beta \operatorname{Var}(N).$$

Fixed-$N$ canonical states do not have these independent grand-canonical mode fluctuations; the exact number constraint introduces correlations among occupations.

The Thermal Window

The energy derivative is

$$- \frac{\partial f_{\mathrm F}}{\partial\epsilon} = \frac{\beta}{ 4\cosh^2 \left[ \beta(\epsilon-\mu)/2 \right] }.$$

This nonnegative kernel is centered at $\epsilon=\mu$ and has unit area on the full mathematical energy line:

$$\int_{-\infty}^{\infty} d\epsilon \left( -\frac{\partial f_{\mathrm F}}{ \partial\epsilon} \right) = 1.$$

A physical band may occupy only part of this line. For a lower band edge $\epsilon_0$, the area over $[\epsilon_0,\infty)$ is $f_{\mathrm F}[\beta(\epsilon_0-\mu)]$, not always one. Its maximum is

$$\left. -\frac{\partial f_{\mathrm F}}{ \partial\epsilon} \right|_{\epsilon=\mu} = \frac{1}{4k_{\mathrm B}T}.$$

The full width at half maximum is

$$\Delta\epsilon_{\mathrm{FWHM}} = 4k_{\mathrm B}T \operatorname{arcosh}\sqrt{2} \simeq 3.53k_{\mathrm B}T.$$

Thus only states within a few $k_{\mathrm B}T$ of the chemical potential change occupation appreciably at low temperature. This fact underlies low-temperature Fermi-gas thermodynamics and the Sommerfeld expansion.

Zero-Temperature Limit

For fixed $\mu$ and $\epsilon\neq\mu$,

$$\lim_{T\to0^+} f_{\mathrm F} \left[ \beta(\epsilon-\mu) \right] = \begin{cases} 1, & \epsilon<\mu, \\ 0, & \epsilon>\mu. \end{cases}$$

Equivalently,

$$f_{\mathrm F}(\epsilon) \longrightarrow \Theta(\mu-\epsilon),$$

away from the discontinuity.

For a continuum integral, assigning a value to the step at one energy usually has no effect. For a discrete level exactly at $\mu$, the fixed-$\mu$ thermal limit specifically retains half occupation and its fluctuations; it is not an arbitrary convention. Along a path $\mu(T)=\epsilon+a k_{\mathrm B}T$, that level instead approaches $(1+e^{-a})^{-1}$. A fixed-number ground state imposes its own filling of any degenerate shell. The finite-temperature Fermi function has

$$f_{\mathrm F}(\mu) = \frac{1}{2}.$$

At zero temperature, ideal fermions fill the lowest available complete one-particle modes. The occupied region in momentum space is called the Fermi sea, and its boundary in a translationally invariant system is the Fermi surface.

Chemical Potential and Fermi Energy

For a uniform continuum ideal Fermi gas at fixed nonzero density, the zero-temperature limit of the chemical potential is the Fermi energy:

$$\mu(T=0) = \epsilon_{\mathrm F}.$$

At finite temperature, $\mu$ generally shifts with $T$ when particle number is held fixed. It need not equal the zero-temperature Fermi energy.

The dimension-by-dimension density conventions and formulas for $k_{\mathrm F}$ and $E_{\mathrm F}$ are collected in Fermi Momentum and Fermi Energy.

The distinction is even more important outside a simple ideal gas:

• in a finite system, addition and removal energies can bracket a range of chemical potentials;
• in an insulator, $\mu$ can lie in a spectral gap with no one-particle state at that energy;
• in an interacting system, $\mu$ is a many-body thermodynamic derivative;
• in nonequilibrium transport, different reservoirs can impose different chemical potentials.

The Fermi–Dirac formula uses $\epsilon_i-\mu$. It does not make $\mu$ the energy of a particular occupied particle.

Dilute Maxwell–Boltzmann Limit

When

$$x = \beta(\epsilon-\mu) \gg 1,$$

set $y=e^{-x}\ll1$. Then

$$\begin{aligned} f_{\mathrm F}(x) &= \frac{y}{1+y} \\ &= y-y^2+y^3-\cdots. \end{aligned}$$

Therefore

$$f_{\mathrm F}(x) \simeq e^{-x}.$$

The leading correction is negative:

$$f_{\mathrm F}(x) - e^{-x} = -e^{-2x} + O(e^{-3x}).$$

Fermionic exclusion reduces occupation relative to the dilute classical result. Bose and Fermi occupations both approach Maxwell–Boltzmann statistics when all relevant mode occupations are small. Classical Limit of Quantum Statistics develops the phase-space-density criterion and the leading equation-of-state correction.

Degenerate Levels

If an energy level $\epsilon$ contains $g_\epsilon$ independent fermionic modes, then

$$\overline N_\epsilon = g_\epsilon f_{\mathrm F} \left[ \beta(\epsilon-\mu) \right].$$

The maximum total occupation is $g_\epsilon$, not one.

If the modes fluctuate independently in the grand-canonical ensemble,

$$\operatorname{Var}(N_\epsilon) = g_\epsilon \overline n_\epsilon \left( 1-\overline n_\epsilon \right).$$

In a continuum approximation,

$$\overline N = \int d\epsilon\, g(\epsilon) \frac{1}{ e^{\beta(\epsilon-\mu)}+1 },$$

where $g(\epsilon)$ is the one-particle density of states including the chosen internal degeneracies.

Entropy Per Mode

The entropy of one fermionic thermal mode is the binary entropy

$$\frac{s_{\mathrm F}}{k_{\mathrm B}} = -\overline n \ln\overline n - \left( 1-\overline n \right) \ln \left( 1-\overline n \right).$$

It vanishes as $\overline n\to0$ or $\overline n\to1$ because the occupation becomes certain. It is maximal at half occupation:

$$s_{\mathrm F,max} = k_{\mathrm B}\ln2.$$

At low temperature, entropy is concentrated in modes near the chemical potential, where neither empty nor occupied is overwhelmingly certain.

Example: One Equilibrium Quantum-Dot Level

Consider one spinless level of energy $\epsilon_d$ weakly coupled to a single equilibrium fermionic reservoir. If coupling-induced level broadening is negligible on the thermal energy scale and the empty and occupied dot states equilibrate grand canonically, then

$$P(1) = f_{\mathrm F} \left[ \beta(\epsilon_d-\mu) \right],$$

and

$$P(0) = 1-P(1).$$

With appreciable broadening, even a noninteracting dot occupation is a Fermi-weighted integral over its spectral line rather than simply the Fermi function at the line center. In the sharp-level approximation, resonance $\epsilon_d=\mu$ gives half occupation. With several reservoirs at different chemical potentials, there is generally no single equilibrium Fermi function for the dot; rates, coupling asymmetry, and transport dynamics matter.

Example: Cold Fermionic Atoms

Ultracold fermionic atoms realize conserved fermion species with tunable density and interactions. For a noninteracting trapped or homogeneous gas, the mean occupation of each one-particle trap or momentum mode is Fermi–Dirac.

Multiple hyperfine states act as distinct internal species. Two atoms in different internal states may share the same spatial mode because their complete one-particle labels differ.

At low temperature, Pauli blocking suppresses available final states for scattering. The Degenerate Fermi Gas page connects that mode-level factor to the thin active shell, pressure, heat capacity, and cold-atom applications. Interactions, trapping, dimensionality, and pairing can move the system beyond the ideal independent-mode formula.

Finite Systems and Fixed Particle Number

In a finite grand-canonical system, $\overline N$ need not be an integer even though every number measurement yields an integer. The chemical potential is adjusted to obtain the desired mean.

For a strictly fixed-$N$ canonical system, the occupations satisfy

$$\sum_i n_i = N$$

in every microstate. One mode becoming occupied forces the allowed occupations of other modes to adjust. The factorized Bernoulli distribution is therefore not exact at finite fixed $N$.

Canonical and grand-canonical predictions often agree for local observables in a suitable thermodynamic limit, but finite-size shell structure, charging energies, exact parity, and small particle number can make their differences observable.

Interactions and Quasiparticles

In an interacting fermion system, bare mode occupations generally do not factorize. The microscopic Hamiltonian is not simply

$$H = \sum_i \epsilon_i n_i.$$

If a controlled effective theory has independent fermionic quasiparticles,

$$H_{\mathrm{eff}} = E_0 + \sum_\alpha E_\alpha \gamma_\alpha^\dagger\gamma_\alpha,$$

their equilibrium occupations may take the form

$$\overline n_\alpha = \frac{1}{ e^{\beta(E_\alpha-\mu_\alpha)}+1 }.$$

A nonzero $\mu_\alpha$ requires a corresponding conserved quasiparticle charge or an explicitly justified constrained ensemble. More commonly the quadratic operator being diagonalized is already $K=H-\mu N$, with excitation energies $E_\alpha$; then the occupation is $1/(e^{\beta E_\alpha}+1)$ and no second chemical-potential subtraction is made. This is the usual superconducting mean-field convention.

Strong correlations can invalidate a simple quasiparticle picture. Then spectral functions and many-body correlation functions, rather than a bare Fermi function alone, determine measurable occupations and response.

Exercises

Prove the particle–hole identity

Show that

$$f_{\mathrm F}(-x) = 1-f_{\mathrm F}(x).$$

Solution

Starting from the left side,

$$\begin{aligned} f_{\mathrm F}(-x) &= \frac{1}{e^{-x}+1} \\ &= \frac{e^x}{1+e^x}. \end{aligned}$$

Meanwhile,

$$\begin{aligned} 1-f_{\mathrm F}(x) &= 1- \frac{1}{e^x+1} \\ &= \frac{e^x}{1+e^x}. \end{aligned}$$

The expressions are equal. This functional identity does not require the physical density of states to be particle–hole symmetric.

Two spin modes in one orbital

A spatial orbital has two independent spin modes with the same energy. Let each have occupation probability $f$. Find the probabilities for total orbital occupation $N=0,1,2$, its mean, and its variance.

Solution

The two spin occupations are independent Bernoulli variables. Therefore

$$\begin{aligned} P(N=0) &= (1-f)^2, \\ P(N=1) &= 2f(1-f), \\ P(N=2) &= f^2. \end{aligned}$$

The mean is

$$\overline N = 2f,$$

and independence gives

$$\operatorname{Var}(N) = 2f(1-f).$$

The orbital can contain two electrons because the spin-up and spin-down states are distinct complete one-particle modes.

Find the thermal-window width

The derivative kernel is

$$- \frac{\partial f_{\mathrm F}}{\partial\epsilon} = \frac{1}{4k_{\mathrm B}T} \operatorname{sech}^2 \left( \frac{\epsilon-\mu}{2k_{\mathrm B}T} \right).$$

Find its full width at half maximum.

Solution

Half maximum requires

$$\operatorname{sech}^2 y = \frac{1}{2},$$

so

$$\cosh y = \sqrt2, \qquad |y| = \operatorname{arcosh}\sqrt2.$$

Since

$$y = \frac{\epsilon-\mu}{2k_{\mathrm B}T},$$

the two half-maximum points are separated by

$$\begin{aligned} \Delta\epsilon_{\mathrm{FWHM}} &= 4k_{\mathrm B}T \operatorname{arcosh}\sqrt2 \\ &\simeq 3.53k_{\mathrm B}T. \end{aligned}$$

A discrete level at zero temperature

A grand-canonical mode has energy $\epsilon$ and $\mu(T)=\epsilon+a k_{\mathrm B}T$. Find its occupation, variance, and entropy as $T\to0^+$. Does the limiting state have to be pure?

Solution

The exponent is $\beta(\epsilon-\mu)=-a$ at every temperature, so $f=(1+e^{-a})^{-1}$, variance $f(1-f)$, and entropy $-k_{\mathrm B}[f\ln f+(1-f)\ln(1-f)]$ remain constant. For finite $a$ the mode remains mixed. At $a=0$ the two grand-energy states are degenerate and the entropy is $k_{\mathrm B}\ln2$. A discrete degenerate mode therefore cannot be discarded as a measure-zero point in an energy integral.

References

• N. W. Ashcroft and N. D. Mermin, Solid State Physics, Holt, Rinehart and Winston (1976).
• A. L. Fetter and J. D. Walecka, Quantum Theory of Many-Particle Systems, Dover (2003).
• S. Giorgini, L. P. Pitaevskii, and S. Stringari, “Theory of ultracold atomic Fermi gases”, Reviews of Modern Physics 80, 1215–1274 (2008).
• K. Huang, Statistical Mechanics, 2nd ed., Wiley (1987).
• M. Kardar, Statistical Physics of Particles, Cambridge University Press (2007).
• L. D. Landau and E. M. Lifshitz, Statistical Physics, Part 1, 3rd ed., Butterworth–Heinemann (1980).
• G. D. Mahan, Many-Particle Physics, 3rd ed., Springer (2000).
• R. K. Pathria and P. D. Beale, Statistical Mechanics, 3rd ed., Elsevier (2011).