PhysicsStatistical MechanicsUniversity

AQAAPOntarioNSWCBSEGCE O-LevelMoECAPS

Fermi-Dirac Distribution

Calculate occupancy for fermions.

Understand the formulaSee the free derivationOpen the full walkthrough

f (E) = \frac{1}{e ^{(E - μ) / k_{B} T} + 1}

Open Full Walkthrough Try Calculator

This public page keeps the free explanation visible and leaves premium worked solving, advanced walkthroughs, and saved study tools inside the app.

Core idea

Overview

The Fermi-Dirac distribution is a probability function used in statistical mechanics to describe the occupancy of energy states by fermions in thermal equilibrium. It incorporates the Pauli Exclusion Principle, which prevents multiple fermions from occupying the same quantum state simultaneously.

When to use: Apply this equation to systems of identical particles with half-integer spin, such as electrons, protons, or neutrons. It is required when particle density is high or temperatures are low enough that quantum exclusion effects become dominant over classical Boltzmann behavior.

Why it matters: This distribution explains the electronic properties of metals and semiconductors, forming the basis for modern solid-state electronics. It is also crucial in astrophysics for describing the internal pressure of degenerate matter in white dwarf stars and neutron stars.

Symbols

Variables

E = Energy State, \mu = Fermi Level, T = Temperature, f(E) = Occupancy

E

Energy State

e V

μ

Fermi Level

e V

T

Temperature

K

f (E)

Occupancy

V a r iab l e

Walkthrough

Derivation

Derivation of the Fermi-Dirac Distribution

Gives the mean occupation of energy states for indistinguishable fermions in thermal equilibrium.

Particles are fermions (half-integer spin).
Pauli exclusion holds: each state has occupation number 0 or 1.
The system is described by the grand canonical ensemble.

Write the Single-State Grand Partition Sum:

Because $n_{s}$ can only be 0 or 1, the sum contains only two terms.

Z_{s} = n_{s} = 0 \sum 1 e^{- n_{s} (ϵ_{s} - μ) / (k T)} = 1 + e^{- (ϵ_{s} - μ) / (k T)}

Compute the Mean Occupation:

The average occupancy is the probability of being occupied divided by the total probability for 0 or 1 particle.

⟨ n_{s} ⟩ = \frac{e ^{- (ϵ_{s} - μ) / (k T)}}{1 + e ^{- (ϵ_{s} - μ) / (k T)}}

State the Fermi-Dirac Result:

This distribution ensures no state has average occupation exceeding 1.

\overset{n}{ˉ}_{s} = \frac{1}{e ^{(ϵ_{s} - μ) / (k T)} + 1}

Result

\overset{n}{ˉ}_{s} = \frac{1}{e ^{(ϵ_{s} - μ) / (k T)} + 1}

Source: Statistical Mechanics — Pathria

Free formulas

Rearrangements

Solve for $E$

Make E the subject

E = k_{B} T \ln\left(\frac{\left(1 - f(E)\right) e^{\frac{\mu}{k_{B} T}}}{f(E)} \right)}

Exact symbolic rearrangement generated deterministically for E.

Difficulty: 3/5

Solve for $μ$

Make u the subject

\mu = k_{B} T \ln\left(- \frac{f(E) e^{\frac{E}{k_{B} T}}}{f(E) - 1} \right)}

Exact symbolic rearrangement generated deterministically for u.

Difficulty: 3/5

Solve for $T$

Make T the subject

T = \frac{E - \mu}{k_{B} \ln\left(\frac{1 - f(E)}{f(E)} \right)}}

Exact symbolic rearrangement generated deterministically for T.

Difficulty: 3/5

Solve for $f (E)$

Make f the subject

f (E) = \frac{1}{e ^{\frac{E - μ}{k _{B} T}} + 1}

Exact symbolic rearrangement generated deterministically for f.

Difficulty: 3/5

The static page shows the finished rearrangements. The app keeps the full worked algebra walkthrough.

Visual intuition

Graph

The Fermi-Dirac distribution is a sigmoid-shaped curve that represents the probability of a quantum state being occupied by a fermion at a given energy level. The curve features a horizontal asymptote at a probability of 1 as energy decreases below the chemical potential and approaches 0 as energy increases above it, reflecting the Pauli exclusion principle. At the chemical potential, the distribution shows a turning point with a probability of exactly 0.5, illustrating how thermal energy influences the occupancy of states near the Fermi level.

Graph type: sigmoid

Why it behaves this way

Intuition

A smooth, S-shaped curve that transitions from a probability of 1 (for low energies) to 0 (for high energies) over a narrow energy range centered at the chemical potential, with the sharpness of the transition determined

f(E)

The average number of fermions occupying a single quantum state with energy E. Due to the Pauli Exclusion Principle, this is equivalent to the probability (between 0 and 1)

How 'full' an energy level is with fermions. A value of 1 means it's definitely occupied, 0 means definitely empty, and 0.5 means it's equally likely to be occupied or empty.

The specific energy value of the quantum state for which the occupancy probability is being calculated.

The 'height' on the energy ladder for which we want to know the chance of finding a fermion.

The chemical potential, which at finite temperature T is the energy level where the probability of occupation f(E) is exactly 1/2.

The 'dividing line' or 'fill line' in the energy spectrum. States below μ are mostly occupied, and states above μ are mostly empty. Its position depends on the total number of fermions in the system.

k_{B}

Boltzmann constant, a fundamental physical constant that relates temperature to energy.

Converts temperature into an energy scale, allowing T to be compared directly with energy differences like (E - μ). It quantifies the thermal 'blurring' of the Fermi surface.

Absolute temperature of the system in Kelvin.

A measure of the average thermal energy available to the particles. Higher T 'smears out' the distribution more, allowing more particles to occupy states above μ and leaving more states below μ empty.

Signs and relationships

(E-μ)/k_BT: This dimensionless ratio compares the energy difference (E-μ) to the thermal energy $k_{B}$ T. When E > μ, the exponent is positive, leading to a large exponential term, and f(E) approaches 0.
+1: This constant term in the denominator is a direct consequence of the Pauli Exclusion Principle, which states that no two identical fermions can occupy the same quantum state. It ensures that the probability f(E)

Free study cues

Insight

Canonical usage

The Fermi-Dirac distribution f(E) yields a dimensionless probability. All energy terms in the exponent (E, μ, and $k_{B}$ T) must be expressed in consistent energy units.

Common confusion

A common mistake is using inconsistent energy units for E, μ, and $k_{B}$ T (e.g., mixing Joules and electronvolts) or not ensuring $k_{B}$ 's units are compatible with the chosen energy unit.

Dimension note

The Fermi-Dirac distribution f(E) is a probability and is inherently dimensionless. The exponent (E-μ)/ $k_{B}$ T must also be dimensionless, requiring E, μ, and $k_{B}$ T to have consistent energy dimensions.

Unit systems

$f (E)$ dimensionless · Represents the probability of a quantum state with energy E being occupied by a fermion.

$E$ J or eV · Energy of the quantum state. Must be in consistent units with μ and k_BT.

$μ$ J or eV · Chemical potential, which is also an energy. Must be in consistent units with E and k_BT.

$k_{B}$ J K^-1 or eV K^-1 · Boltzmann constant. Its value is 1.380649 × 10^-23 J K^-1 (exact value, SI) or approximately 8.617333262 × 10^-5 eV K^-1. The unit must be consistent with the chosen units of E and μ.

$T$ K · Absolute temperature.

Ballpark figures

Quantity:

One free problem

Practice Problem

Calculate the occupancy probability (f) for an energy state at 0.10 eV in a system where the Fermi level (u) is 0.05 eV and the temperature is 300 K. Use Boltzmann constant kB = 8.617 × 10⁻⁵ eV/K.

Energy State0.1 eV

Fermi Level0.05 eV

Temperature300 K

Solve for: $f$

Hint: First calculate the exponent (E-u)/(kB × T) and then evaluate the exponential function.

The full worked solution stays in the interactive walkthrough.

Where it shows up

Real-World Context

Electrons in a metal.

Study smarter

Tips

The value of f(E) is strictly bounded between 0 and 1.
At the chemical potential (E = u), the probability of occupancy is always exactly 0.5.
In many solid-state problems, use $k_{B}$ = 8.617 × 10⁻⁵ eV/K for energies measured in electron-volts.
The distribution behaves like a step function at absolute zero (T = 0 K).

Avoid these traps

Common Mistakes

Confusing with Bose (denominator -1).
Using wrong units for k.

Common questions

Frequently Asked Questions

Gives the mean occupation of energy states for indistinguishable fermions in thermal equilibrium.

Apply this equation to systems of identical particles with half-integer spin, such as electrons, protons, or neutrons. It is required when particle density is high or temperatures are low enough that quantum exclusion effects become dominant over classical Boltzmann behavior.

This distribution explains the electronic properties of metals and semiconductors, forming the basis for modern solid-state electronics. It is also crucial in astrophysics for describing the internal pressure of degenerate matter in white dwarf stars and neutron stars.

Confusing with Bose (denominator -1). Using wrong units for k.

Electrons in a metal.

The value of f(E) is strictly bounded between 0 and 1. At the chemical potential (E = u), the probability of occupancy is always exactly 0.5. In many solid-state problems, use k_B = 8.617 × 10⁻⁵ eV/K for energies measured in electron-volts. The distribution behaves like a step function at absolute zero (T = 0 K).

References

Sources

Statistical Mechanics by Donald A. McQuarrie
Thermodynamics and an Introduction to Thermostatistics by Herbert B. Callen
Physical Chemistry by Peter Atkins and Julio de Paula
Wikipedia: Fermi-Dirac statistics
NIST CODATA
Atkins' Physical Chemistry
Callen, Thermodynamics and an Introduction to Thermostatistics
McQuarrie Statistical Mechanics

Fermi-Dirac Distribution

Overview

Variables

Derivation

Write the Single-State Grand Partition Sum:

Compute the Mean Occupation:

State the Fermi-Dirac Result:

Rearrangements

Graph

Intuition

Insight

Practice Problem

Real-World Context

Tips

Common Mistakes

Related Formulas

Bose⁻Einstein Distribution

Frequently Asked Questions

Sources