Fermi-Dirac Distribution Calculator

The fundamental equation of quantum statistics for fermions

The Fermi-Dirac distribution gives the probability that a quantum state at energy E is occupied by a fermion (electron, proton, neutron, or any half-integer spin particle) at thermal equilibrium:

f(E) = 1 / (exp((E − E_F) / k_B T) + 1)

Where E is the energy of the state in question, E_F is the Fermi energy (the chemical potential at T = 0), k_B is the Boltzmann constant, and T is the absolute temperature in Kelvin. f(E) ranges from 0 (state empty) to 1 (state fully occupied). This single equation underpins every theory of metals, semiconductors, and electron transport.

Four behaviors you should know

At T = 0 K the distribution is a perfect step: f(E) = 1 for E below E_F, f(E) = 0 for E above E_F. Every state below the Fermi level is full; every state above is empty. Pauli exclusion forces this: fermions stack from the bottom up until you run out of fermions.

At T greater than 0 K, the step blurs over a width of roughly ±2 k_B T around E_F. At room temperature k_B T ≈ 0.02585 eV, so the smearing is only about 50 meV wide. For a metal where E_F is several eV, this is a small fraction of the total Fermi energy.

At E = E_F exactly, f(E) = 1/2 at every temperature above absolute zero. The Fermi level is, by definition, the half-occupation energy. This is one of the cleanest invariants in solid-state physics.

For energies far above E_F (specifically E − E_F much greater than k_B T), f(E) reduces to the classical Maxwell-Boltzmann distribution: f(E) ≈ exp(−(E − E_F) / k_B T). Deep in the conduction band tail of a semiconductor, quantum statistics matters less and classical Boltzmann is a perfectly good approximation.

Worked example: copper at room temperature

Copper has E_F = 7.04 eV. At T = 300 K, k_B T = 0.02585 eV. For a state 0.1 eV above the Fermi level:

f = 1 / (exp(0.1 / 0.02585) + 1) = 1 / (exp(3.868) + 1) = 1 / 48.8 ≈ 0.0205

Only about 2 percent of states 0.1 eV above E_F are occupied. For a state 0.1 eV below the Fermi level, the same arithmetic gives f ≈ 0.9795 (the curve is antisymmetric about E_F). This is why electrical conduction in a metal is so dominated by the thin slice of electrons within k_B T of E_F: states far below are completely full and have no empty neighbors to scatter into, states far above are essentially empty. Only the smeared region near E_F has both occupied states and nearby empty states.

Where this matters in real materials

In semiconductors the Fermi level usually sits in the bandgap. The exponential tail of f(E) determines how many electrons leak into the conduction band, which is what controls intrinsic carrier concentration. Doping shifts E_F toward one of the band edges, drastically changing carrier density without changing temperature. The reason silicon at 300 K has about 10^10 intrinsic carriers per cubic centimeter while germanium has 10^13 is almost entirely the Fermi-Dirac tail evaluated at different bandgaps.

In white dwarf stars, electrons are so dense that E_F reaches MeV scale, and the temperature is effectively zero in dimensionless k_B T / E_F terms. The pressure that holds the star up against gravity is pure Fermi-Dirac degeneracy pressure, not thermal pressure. Chandrasekhar’s 1.4 solar mass limit is set by where this degeneracy pressure can no longer support the star.

In a scanning tunneling microscope, the tunneling current depends on the difference between the Fermi-Dirac distributions on either side of the tip-sample junction. STM imaging at finite temperature is essentially a map of the local f(E) near E_F.

Comparison with the other two distributions

For bosons (integer spin: photons, phonons, helium-4), use Bose-Einstein: f(E) = 1 / (exp((E − μ) / k_B T) − 1). Note the minus sign in the denominator. Bosons can stack arbitrarily many in one state, so the distribution diverges as E approaches μ. That divergence is what makes Bose-Einstein condensation possible at low temperatures.

For classical distinguishable particles (ideal gas, dilute solutions), use Maxwell-Boltzmann: f(E) = exp(−(E − μ) / k_B T). No quantum +1 or −1 in the denominator. All three distributions converge when E − E_F is much larger than k_B T. Quantum statistics matter only when occupation numbers approach 1 (fermions) or when thermal de Broglie wavelength is comparable to inter-particle spacing (bosons).

Choosing the right distribution

If your particles have half-integer spin (electrons, protons, neutrons, quarks), use Fermi-Dirac. If they have integer spin (photons, phonons, mesons, alpha particles), use Bose-Einstein. If quantum effects are negligible (the gas is dilute and warm), Maxwell-Boltzmann is a fine approximation regardless. The classical limit kicks in for all three when occupations are small.