Fermi–Dirac Statistics in Solids

Introduction

The density of states counts allowed electron states, but actual electrons occupy only some of them. Electrons are fermions, and the Pauli exclusion principle prevents two identical electrons from using the same complete state. At absolute zero, the free-electron model fills states in energy order up to a sharp boundary. At finite temperature, there is no single last occupied level: a narrow range around a chemical potential becomes partly occupied. Fermi–Dirac statistics quantifies that probability and underlies metal heat capacity, semiconductor carrier densities and device behaviour.

Core explanation

At thermal equilibrium the probability that an available one-electron state of energy E is occupied is f(E) = 1/[exp((E−μ)/(k BT)) + 1] , where μ is the electronic chemical potential, k B is Boltzmann's constant and T is absolute temperature. The formula assumes a suitable independent-quasiparticle state picture in equilibrium. At E = μ, the exponent is zero and f = 1/2. Well below μ by several k BT, f is close to one; well above, it is close to zero. This half-occupation statement refers to an individual allowed state, not to half the atoms in a specimen being ionised.

At T approaching zero, f(E) approaches a step: one below μ and zero above. For an ordinary free-electron metal at zero temperature, μ equals the Fermi energy E F, the highest occupied energy in the ground state. At finite temperature, μ can shift with T to keep the total electron count correct. People often call μ the Fermi level, but a careful calculation distinguishes a material's zero-temperature E F from μ(T). The electron count is N = ∫g(E)f(E)dE across allowed bands. Both DOS and f must be known to infer a population.

The occupation edge broadens over an energy scale of several k BT. At 300 K, k BT ≈ 0.0259 eV. In a typical metal with E F of several electronvolts, this is a small fraction of the filled energy range. Deeply occupied states remain nearly completely full, and high empty states remain nearly empty. Thermal response largely involves a thin shell around the Fermi surface. This is why applying classical equipartition to every conduction electron overestimates electronic heat capacity. The MIT OpenCourseWare materials lecture sequence treats DOS and the Fermi–Dirac distribution consecutively, while OpenStax's free-electron chapter shows their product as the occupied-energy distribution.

Semiconductors require special care. Their chemical potential may lie inside a band gap where the ideal DOS is zero, yet f(μ) = 1/2 remains mathematically true for a hypothetical state of energy μ. There may be no allowed state at exactly that energy to be half occupied. Electrons thermally promoted into conduction states leave holes in valence states. Doping shifts μ relative to band edges and changes carrier populations, but a temperature-dependent chemical potential inside a gap is not itself a conducting band. The formulas for electron and hole concentrations integrate appropriate band DOS with f or 1−f.

The familiar Boltzmann approximation f(E) ≈ exp[−(E−μ)/(k BT)] applies when E is many k BT above μ, so the exponential in the denominator dominates 1. It does not apply near or below μ. Likewise 1−f(E) has an exponential approximation well below μ. These approximations simplify nondegenerate semiconductor calculations, but degenerate doping requires the full Fermi–Dirac expression. Energy units must match: use k BT in joules if E and μ are in joules, or in electronvolts if those are electronvolts.

Step-by-step reasoning

1. Identify the energy E, chemical potential μ and absolute temperature T. 2. Express E−μ and k BT in the same units. 3. Compute x = (E−μ)/(k BT) and f = 1/(e^x+1). 4. Combine f with DOS when calculating a number of electrons. 5. Check whether a state at E actually exists and whether a Boltzmann approximation is justified.

Visual explanation

Plot occupation probability against energy for three temperatures. The zero-temperature line is a sharp step at E F. Warmer curves are smooth S shapes crossing one-half at their respective μ. Place a second plot of DOS nearby: it may be zero inside a semiconductor gap even where the mathematical occupation curve crosses one-half. The product g(E)f(E) shades only allowed occupied states.

Real-world analogy

Imagine an auditorium in which seats are ranked by price and people fill cheaper seats first. At an ideal zero-temperature limit there is a sharp boundary between filled and empty rows. A little thermal rearrangement swaps mainly people near that boundary, while far cheaper rows remain full and far dearer rows empty. The auditorium's layout is DOS; the probability of occupying each seat is f.

Real-world example

In a silicon diode, donor and acceptor doping alter the chemical potential and carrier concentration on opposite sides of a junction. It would be wrong to read carrier numbers from the Fermi function alone: the conduction- and valence-band DOS specify where electrons and holes can exist. At equilibrium, the chemical potential is spatially constant through the connected device even as band edges bend with position.

Why?

Why does f(μ) equal one-half at every nonzero temperature? At E = μ, the exponent is zero and exp(0) = 1, so the denominator is 2. This algebraic midpoint does not mean that precisely half of all possible states in the solid are occupied.

Common misconception

"Heating a metal excites every electron by k BT." The Fermi sea is constrained by occupied neighbouring states. Only electrons near unoccupied states at the Fermi boundary readily change occupation, so their thermal response is a small fraction of the total population at ordinary temperatures.

Worked example

Question: At 300 K, estimate the occupation of an allowed state 0.050 eV above μ, taking k BT = 0.0259 eV. Would the Boltzmann approximation be exact here?

Reasoning: x = 0.050/0.0259 ≈ 1.93. Therefore f = 1/[exp(1.93)+1] ≈ 1/(6.89+1) ≈ 0.127. The Boltzmann estimate exp(−1.93) ≈ 0.145 is moderately close but not exact because the discarded 1 in the denominator is still noticeable. Several additional k BT above μ would make the approximation better.

Answer: The full Fermi–Dirac occupation is about 0.13, and the Boltzmann estimate is only approximate.

Quick check

1. What is f(E) when E = μ at nonzero temperature? Answer: One-half, because the exponential equals one. This does not imply there is an allowed state at μ if μ lies in a band gap.

Exam focus

Use the full formula with a dimensionless exponent. State when μ equals E F and when it may shift. Distinguish occupancy from DOS, especially for energies in a gap. Explain temperature smearing with k BT, not with an arbitrary fixed percentage of total band width.

Advanced insight

In equilibrium μ is the thermodynamic cost of adding an electron to the whole system, subject to other conserved quantities. The one-electron energy reference can be shifted by a constant if μ is shifted with it, leaving occupations unchanged. Device band diagrams therefore emphasise differences between band edges and μ, not their absolute numerical values without a stated reference.

Summary

Fermi–Dirac statistics gives the equilibrium occupation f(E) = 1/[exp((E−μ)/k BT)+1]. It becomes a sharp step at zero temperature and is rounded over energies of order k BT at finite temperature. Electron counts require the DOS as well as f, and a chemical potential inside a semiconductor gap need not coincide with an actual allowed state.

Practice questions

1. What is f(E) at E = μ? Answer: One-half for any finite temperature under the Fermi–Dirac formula. 2. What happens to the occupation curve as T approaches zero? Answer: It approaches a step from occupied states below μ to empty states above it. 3. Why is f(μ) = 1/2 not enough to determine carrier concentration? Answer: Carrier concentration also depends on the number of allowed states, represented by the DOS. 4. When is f(E) ≈ exp[−(E−μ)/k BT] justified? Answer: When E lies many k BT above μ so exp[(E−μ)/k BT] is much greater than one.