Fermi-Dirac Distribution

The Fermi-Dirac distribution answers one sharply posed question: for a gas of non-interacting fermions in thermal and diffusive contact with a reservoir at temperature $T$ and chemical potential $\mu$ , what is the average number of particles occupying a single-particle level of energy $\varepsilon$ ? The answer, $\bar n(\varepsilon)=1/(e^{\beta(\varepsilon-\mu)}+1)$ with $\beta=1/k_BT$ , is bounded strictly between $0$ and $1$ — it is at once a mean occupation and the probability that the level is filled. That ceiling is the statistical face of the Pauli principle, and it reshapes the thermodynamics of electrons in metals, white dwarfs, neutron stars, and semiconductors. We take the microscopic input from the quantum-mechanics track. In aq-qm-pauli-exclusion-principle the exclusion rule "falls out of antisymmetry in two lines": the antisymmetric two-particle state "vanishes when the labels coincide," so "the fermion occupancy is forced to $n\in\{0,1\}$ ." The census of which particles obey it is owned by aq-qm-bosons-fermions, whose rule is "integer spin means boson, half-integer spin means fermion," with fermions exclusive because "put two in the same state and the antisymmetric combination cancels itself to zero." This article does not re-derive antisymmetry; it takes the cap $n\in\{0,1\}$ as given and builds the thermodynamic consequence — the grand-canonical mean occupation $\bar n(\varepsilon)$ as a function of $\mu$ and $T$ . The mechanism is a two-line grand-canonical calculation. Because exclusion truncates the single-level occupancy sum at $n=1$ , the per-mode grand partition function has only two terms, $\Xi_\varepsilon=1+e^{-\beta(\varepsilon-\mu)}$ , and differentiating $\ln\Xi_\varepsilon$ gives the Fermi function directly, with the $0\le\bar n\le1$ bound built in. Physically the distribution is a step at $T=0$ — the Fermi sea, filled to $\varepsilon_F$ — that softens to a shelf of thermal width $\sim k_BT$ around $\mu$ at finite temperature. Only that thin active shell responds to heat, which is why metals have a linear-in- $T$ electronic heat capacity rather than the classical $\tfrac32 Nk_B$ .

Key takeaways

The Fermi-Dirac distribution gives the mean occupation of a single-particle level $\varepsilon$ for non-interacting fermions in equilibrium: $\bar n(\varepsilon)=1/(e^{\beta(\varepsilon-\mu)}+1)$ .
Because Pauli exclusion forbids double occupancy, $\bar n$ is strictly bounded, $0\le\bar n\le1$ — it is simultaneously a mean occupation and a filling probability.
The microscopic cap $n\in\{0,1\}$ is owned by aq-qm-pauli-exclusion-principle (occupancy "forced to $n\in\{0,1\}$ "); this article derives the thermodynamic $\bar n(\varepsilon)$ .
At $T=0$ it is a step $\theta(\mu-\varepsilon)$ : all states below $\varepsilon_F=\mu(0)$ filled, all above empty — the Fermi sea.
Particle-hole symmetry $\bar n(\mu+\delta)+\bar n(\mu-\delta)=1$ pins $\mu$ as the exact half-filling energy; the variance $\bar n(1-\bar n)$ (antibunching) is peaked at $\varepsilon=\mu$ .
Key idea: exclusion truncates the single-level sum at $n=1$ , so $\Xi_\varepsilon=1+e^{-\beta(\varepsilon-\mu)}$ and $\bar n=z\,\partial_z\ln\Xi_\varepsilon=1/(e^{\beta(\varepsilon-\mu)}+1)$ — the $+1$ in the denominator is the entire content of Fermi statistics, and it guarantees $\bar n\le1$ .

The core insight

Fermi statistics lives entirely in the $+1$ . Exclusion caps the single-level occupancy at $n=1$ , so the per-mode grand partition function has just two terms, $\Xi_\varepsilon=1+e^{-\beta(\varepsilon-\mu)}$ . Differentiating $\ln\Xi_\varepsilon$ gives $\bar n(\varepsilon)=1/(e^{\beta(\varepsilon-\mu)}+1)$ , a number that never exceeds one — the statistical shadow of the microscopic rule $n\in\{0,1\}$ that quantum mechanics owns.

Where this lives today

The Fermi-Dirac distribution is the backbone of condensed-matter and astrophysics. In metals it fixes the Fermi sea whose surface controls conduction, magnetism, and the linear electronic heat capacity $C_e=\tfrac{\pi^2}{3}g(\varepsilon_F)k_B^2 T$ (via the Sommerfeld expansion) and the temperature-independent Pauli paramagnetic susceptibility $\chi\propto g(\varepsilon_F)$ . In semiconductors the position of $\mu$ relative to the band edges sets carrier concentrations, doping, and junction behavior; where $\varepsilon_c-\mu\gg k_BT$ the Fermi function reduces to a Boltzmann tail $n\propto e^{-(\varepsilon_c-\mu)/k_BT}$ . In astrophysics electron and neutron degeneracy pressure — read off the $T\approx0$ step — supports white dwarfs and neutron stars, with the Chandrasekhar mass $\approx1.4\,M_\odot$ marking where it fails. In the laboratory, trapped $^6$ Li and $^{40}$ K gases realize tunable ideal Fermi gases whose momentum-space step is imaged directly in time-of-flight, and "Pauli blocking" — suppressed scattering into filled states — is seen in quantum-gas microscopes.

The physical picture

Picture each single-particle level as a switch with exactly two settings: empty or occupied. The reservoir at $(\mu,T)$ sets an energy price $\varepsilon-\mu$ to flip it on. If filling lowers the grand potential ( $\varepsilon<\mu$ ) the level is almost always on ( $\bar n\to1$ ); if it costs energy ( $\varepsilon>\mu$ ) it is almost always off ( $\bar n\to0$ ). Right at the balance point $\varepsilon=\mu$ the switch is on exactly half the time, $\bar n=\tfrac12$ , for every temperature. The crossover region has thermal width $\sim k_BT$ . Because the cap is one, fermions cannot condense: even at $T=0$ they stack into successively higher levels, filling a Fermi sea up to $\varepsilon_F$ . That filled sea stores an enormous zero-point kinetic energy and, when squeezed, pushes back — the degeneracy pressure that supports white dwarfs, developed thermodynamically in aq-sm-degeneracy-pressure-fermions. The first figure is the hero: the Fermi step at $T=0$ softening into a thermal shelf as $T$ rises.

Figure

Fermi-Dirac occupation: a T=0 step at the chemical potential softening into a thermal shelf of width kT

The punchline: the whole distribution is pinned to $\bar n=\tfrac12$ at $\varepsilon=\mu$ and never exceeds $1$ — the exclusion ceiling makes it a filling probability, not an unbounded count.

Only the thin shell within $\sim k_BT$ of $\mu$ can change its occupation; everything deep below is locked at $\bar n=1$ and everything far above is locked at $\bar n=0$ . That is why so few electrons are thermally "active." The second figure isolates that active window — the derivative $-\partial\bar n/\partial\varepsilon$ , a bell centered at $\mu$ that becomes a delta function as $T\to0$ .

Figure

The thermally active window: minus the derivative of the Fermi function is a bell of width kT at mu

Electrons in this bell are the only ones that carry heat capacity, paramagnetism, and conductivity — the deep sea is inert. This window is what makes $C_e\propto T$ .

The core equations

The central result gives the mean occupation of a level as a function of its energy relative to the chemical potential. It follows from a two-term grand-canonical sum; the $+1$ is the entire imprint of Fermi statistics.

\bar n(\varepsilon)=\frac{1}{e^{\beta(\varepsilon-\mu)}+1} \qquad (1)

where: $\bar n(\varepsilon)$ — mean occupation of a single-particle level of energy $\varepsilon$ , bounded $0\le\bar n\le1$ ; $\beta=1/k_BT$ ; $\mu$ — chemical potential fixed by particle-number conservation; $\varepsilon$ — single-particle energy. Physical reading: the level is almost full when $\varepsilon\ll\mu$ , almost empty when $\varepsilon\gg\mu$ , and exactly half-full at $\varepsilon=\mu$ . Exclusion enters through the single-level grand partition function. Because the reservoir may add at most one fermion to a given level, the occupancy sum has only the two terms $n=0$ and $n=1$ .

\Xi_\varepsilon=\sum_{n=0}^{1}\big(z\,e^{-\beta\varepsilon}\big)^{n}=1+z\,e^{-\beta\varepsilon}=1+e^{-\beta(\varepsilon-\mu)} \qquad (2)

where: $\Xi_\varepsilon$ — grand partition function of one mode; $z=e^{\beta\mu}$ — fugacity; the sum is truncated at $n=1$ by Pauli exclusion. Physical reading: contrast the bosonic case, where the same sum runs to $n=\infty$ and becomes a geometric series — the single stopped term is the whole difference between Fermi and Bose statistics. The mean occupation is the logarithmic fugacity-derivative of $\Xi_\varepsilon$ , the standard grand-canonical relation for a single independent mode.

\bar n=z\left(\frac{\partial\ln\Xi_\varepsilon}{\partial z}\right)=k_BT\left(\frac{\partial\ln\Xi_\varepsilon}{\partial\mu}\right)_T \qquad (3)

where: $\ln\Xi_\varepsilon$ — grand potential of the mode (up to $-k_BT$ ). Physical reading: differentiating the two-term $\ln\Xi_\varepsilon$ immediately returns Eq. (1). The fluctuation of the occupation follows from a second derivative, and it is bounded and binary — vanishing when a level is certainly full or certainly empty.

\langle(\Delta n)^2\rangle=k_BT\,\frac{\partial\bar n}{\partial\mu}=\bar n(1-\bar n) \qquad (4)

where: $\langle(\Delta n)^2\rangle$ — variance of the occupation. Physical reading: maximal ( $=\tfrac14$ ) at $\bar n=\tfrac12$ , zero at $\bar n=0,1$ — fermions "antibunch," the opposite of boson bunching. Two structural identities express the symmetry of the distribution about $\mu$ . The first is particle-hole symmetry; the second is the thermal-broadening kernel that becomes a delta function at $T=0$ .

\bar n(\mu+\delta)+\bar n(\mu-\delta)=1,\qquad -\frac{\partial\bar n}{\partial\varepsilon}=\frac{\beta}{4\cosh^2\!\big[\tfrac{\beta(\varepsilon-\mu)}{2}\big]} \qquad (5)

where: $\delta$ — energy offset from $\mu$ . Physical reading: a filled state $\delta$ above $\mu$ mirrors an empty state $\delta$ below it (electrons and holes), and the response window $-\partial\bar n/\partial\varepsilon$ is a bell of width $\sim k_BT$ centered on $\mu$ . Summing the occupation over all levels, converted to an integral with the density of states, fixes $\mu(T,n)$ and yields the thermodynamics.

N=\sum_i\bar n(\varepsilon_i)=\int_0^\infty\!\mathrm{d}\varepsilon\,g(\varepsilon)\,\frac{1}{e^{\beta(\varepsilon-\mu)}+1} \qquad (6)

where: $N$ — total particle number; $g(\varepsilon)$ — single-particle density of states (owned by aq-sm-density-of-states-energy, including spin degeneracy). Physical reading: this implicit equation is what determines $\mu(T)$ ; its low- $T$ Sommerfeld expansion, owned by aq-sm-sommerfeld-expansion, powers the theory of metals.

Derivations

Derivation AThe Fermi-Dirac distribution from the grand canonical ensemble

The goal is to derive $\bar n(\varepsilon)=1/(e^{\beta(\varepsilon-\mu)}+1)$ from first principles in the grand canonical ensemble, taking as given only the microscopic occupation cap $n\in\{0,1\}$ that quantum mechanics owns (aq-qm-pauli-exclusion-principle: the antisymmetric state "vanishes when the labels coincide," occupancy "forced to $n\in\{0,1\}$ "). The engine is that in the occupation-number representation the ideal-gas microstate is just the list $\{n_i\}$ , so the grand sum factorizes over independent modes and each factor is a two-term sum. This card owns the thermodynamic distribution; it does not re-derive antisymmetry.

Step 1. Write the grand partition function in the occupation-number basis. A microstate is $\{n_i\}$ with energy $E=\sum_i n_i\varepsilon_i$ and number $N=\sum_i n_i$ , so $\Xi=\sum_{\{n_i\}}e^{-\beta\sum_i n_i(\varepsilon_i-\mu)}=\prod_i\sum_{n_i}e^{-\beta n_i(\varepsilon_i-\mu)}$ . Why: the exponential of a sum factorizes, and the modes are independent in the ideal gas, so the many-body sum becomes a product of single-mode sums.

Step 2. Apply exclusion to each factor. For fermions $n_i$ runs only over $0$ and $1$ , so each single-mode sum has exactly two terms: $\Xi=\prod_i\big(1+e^{-\beta(\varepsilon_i-\mu)}\big)$ . Why: the microscopic cap $n\in\{0,1\}$ stops the sum after the $n=1$ term — the point where the bosonic geometric series would continue to infinity.

Step 3. Extract the mean occupation of one mode $j$ . Weighting each occupancy by $n_j$ , $\bar n_j=\dfrac{\sum_{n_j}n_j e^{-\beta n_j(\varepsilon_j-\mu)}}{1+e^{-\beta(\varepsilon_j-\mu)}}=\dfrac{e^{-\beta(\varepsilon_j-\mu)}}{1+e^{-\beta(\varepsilon_j-\mu)}}$ , since only $n_j=1$ contributes to the numerator. Why: the mean is the probability-weighted occupancy, and $n_j=0$ contributes nothing to the average.

Step 4. Simplify to the Fermi form. Multiplying numerator and denominator by $e^{\beta(\varepsilon_j-\mu)}$ gives $\bar n_j=1/(e^{\beta(\varepsilon_j-\mu)}+1)$ . Dropping the subscript is Eq. (1). Why: the algebra converts the ratio into the canonical Fermi function, whose denominator $\ge1$ guarantees $\bar n\le1$ . Verified: the two-term $\Xi_\varepsilon=1+ze^{-\beta\varepsilon}$ , the derivative $z\,\partial_z\ln\Xi_\varepsilon=1/(e^{\beta(\varepsilon-\mu)}+1)$ , and the sign contrast with the Bose $-1$ are checked symbolically in verify_fermi-dirac-distribution.py.

Go deeper · Derivation B · 1 minThe occupation is bounded in the unit interval and particle-hole symmetric

Derivation BThe occupation is bounded in the unit interval and particle-hole symmetric

The goal is to prove the two structural facts that make the Fermi function the statistical face of exclusion: that $0\le\bar n\le1$ for every real value of the argument $\varepsilon-\mu$ (so $\bar n$ doubles as a filling probability), and that the distribution is particle-hole symmetric about $\mu$ , $\bar n(\mu+\delta)+\bar n(\mu-\delta)=1$ . These are the owned structural results of this article — the thermodynamic echo of the microscopic $n\in\{0,1\}$ — and neither requires knowing $\mu(T)$ or the spectrum.

Step 1. Bound the function. Write $y=\beta(\varepsilon-\mu)$ , any real number. Then $\bar n=1/(e^{y}+1)$ ; since $e^{y}>0$ the denominator exceeds $1$ , so $0<\bar n<1$ , with $\bar n\to1$ as $y\to-\infty$ and $\bar n\to0$ as $y\to+\infty$ . Why: a positive exponential added to one keeps the reciprocal strictly inside the unit interval — the exclusion ceiling is automatic, not imposed by hand.

Step 2. Evaluate at the chemical potential. At $\varepsilon=\mu$ , $y=0$ and $\bar n=1/(1+1)=\tfrac12$ , independent of $T$ . Why: $\mu$ is the exact half-filling energy, the symmetry point of the distribution for every temperature.

Step 3. Prove particle-hole symmetry. Let $f(\delta)=1/(e^{\beta\delta}+1)$ . Then $f(\delta)+f(-\delta)=\dfrac{1}{e^{\beta\delta}+1}+\dfrac{1}{e^{-\beta\delta}+1}$ ; multiplying the second term top and bottom by $e^{\beta\delta}$ gives $\dfrac{1}{e^{\beta\delta}+1}+\dfrac{e^{\beta\delta}}{e^{\beta\delta}+1}=1$ . Why: a filled state $\delta$ above $\mu$ is mirrored by an empty state $\delta$ below it — the electron-hole symmetry that underlies band-structure intuition.

Step 4. Read off the variance and the antibunching. From Eq. (4), $\langle(\Delta n)^2\rangle=\bar n(1-\bar n)$ , a downward parabola in $\bar n$ that is maximal at $\bar n=\tfrac12$ (i.e. $\varepsilon=\mu$ ) and zero at $\bar n=0,1$ . Why: a level that is certainly empty or certainly full cannot fluctuate; only the active shell near $\mu$ carries occupation noise, and its Fano-like variance $\bar n(1-\bar n)$ is the opposite sign of the boson enhancement $\bar n(1+\bar n)$ . Verified: $1/(e^{y}+1)\in(0,1)$ scanned over $|y|\le200$ , the exact $\bar n(\mu)=\tfrac12$ , the identity $f(\delta)+f(-\delta)=1$ , and the variance $\bar n(1-\bar n)=k_BT\,\partial_\mu\bar n$ are all checked in verify_fermi-dirac-distribution.py.

Limiting cases

The single formula $\bar n=1/(e^{\beta(\varepsilon-\mu)}+1)$ has four diagnostic limits.

Regime	Result	Meaning
Degenerate, $T\to0$	$\bar n(\varepsilon)\to\theta(\mu-\varepsilon)=\theta(\varepsilon_F-\varepsilon)$	Sharp step: states below $\varepsilon_F$ filled, above empty — the Fermi sea
At the chemical potential	$\bar n(\mu)=\tfrac12$ for all $T>0$	The level $\varepsilon=\mu$ is always half-filled; symmetry point
Classical, $\varepsilon-\mu\gg k_BT$	$\bar n\approx e^{-\beta(\varepsilon-\mu)}\ll1$	The $+1$ is negligible; Maxwell-Boltzmann tail, fugacity $z\ll1$
Bose comparison	$\bar n^{\mathrm{BE}}=1/(e^{\beta(\varepsilon-\mu)}-1)$	Same expression with $-1$ ; can diverge as $\varepsilon\to\mu^+$ , unlike the bounded Fermi case

Common trapThe chemical potential is not the Fermi energy at finite temperature

It is tempting to use $\mu$ and $\varepsilon_F$ interchangeably, but they coincide only at $T=0$ , where $\varepsilon_F\equiv\mu(T{=}0)$ . At finite temperature $\mu(T)$ shifts — downward for a 3D free gas, $\mu\approx\varepsilon_F[1-\tfrac{\pi^2}{12}(T/T_F)^2]$ — and in the nondegenerate limit $\mu$ can become negative, $\mu\approx k_BT\ln(n\lambda_T^3)<0$ , while $\varepsilon_F$ is a fixed property of the density. A second trap: $\bar n$ is a mean occupation (a pure number), not an energy, and $\mu$ is the full chemical potential including any potential-energy offset, not the kinetic Fermi energy alone.

Worked examples

Example 1 — Fill a small set of levels and read off the sea.

Step 1. Setup. Take five spinless levels at energies $\varepsilon=0,1,2,3,4$ (in units of $k_BT$ ) and put $\mu=2$ . Compute $\bar n=1/(e^{\varepsilon-2}+1)$ for each.

Step 2. Evaluate. $\bar n(0)=1/(e^{-2}+1)=\mathbf{0.881}$ ; $\bar n(1)=1/(e^{-1}+1)=\mathbf{0.731}$ ; $\bar n(2)=1/(e^{0}+1)=\mathbf{0.500}$ ; $\bar n(3)=1/(e^{1}+1)=\mathbf{0.269}$ ; $\bar n(4)=1/(e^{2}+1)=\mathbf{0.119}$ .

Step 3. Check the symmetry. The pair straddling $\mu$ satisfies $\bar n(1)+\bar n(3)=0.731+0.269=\mathbf{1.000}$ and $\bar n(0)+\bar n(4)=0.881+0.119=\mathbf{1.000}$ — particle-hole symmetry confirmed numerically. The half-filled level sits exactly at $\mu$ .

Step 4. Interpret. Even three units of energy is only $3k_BT$ ; a level at $\mu-3k_BT$ is $88\%$ full and one at $\mu+3k_BT$ is $12\%$ full, so the whole "action" is compressed into a window a few $k_BT$ wide around $\mu$ . Deep levels ( $\varepsilon\ll\mu$ ) would be locked at $\bar n\approx1$ .

Example 2 — Fermi energy and temperature of copper.

Step 1. Setup. Copper has one conduction electron per atom, giving $n=8.49\times10^{28}\,\mathrm{m^{-3}}$ . For a free spin-½ gas $\varepsilon_F=\dfrac{\hbar^2}{2m}(3\pi^2 n)^{2/3}$ .

Step 2. Fermi wavevector. $3\pi^2 n=3\pi^2(8.49\times10^{28})=2.514\times10^{30}\,\mathrm{m^{-3}}$ , so $k_F=(2.514\times10^{30})^{1/3}=\mathbf{1.36\times10^{10}\,m^{-1}}$ .

Step 3. Fermi energy. $\varepsilon_F=\dfrac{(1.0546\times10^{-34})^2(1.36\times10^{10})^2}{2(9.109\times10^{-31})}=1.13\times10^{-18}\,\mathrm{J}=\mathbf{7.0\ eV}$ , matching the measured value for copper.

Step 4. Fermi temperature. $T_F=\varepsilon_F/k_B=(1.13\times10^{-18})/(1.381\times10^{-23})=\mathbf{8.2\times10^{4}\,K}$ . Since room temperature is $\sim300\,\mathrm{K}\ll T_F$ , the electron gas is deeply degenerate: $T/T_F\approx0.004$ , so only $\sim0.4\%$ of the electrons — those within $k_BT$ of $\mu$ — are thermally active. This is precisely why the electronic heat capacity is a small linear-in- $T$ term rather than the classical $\tfrac32 Nk_B$ .

Origins · 1 minfermi-dirac statistics and how this was discovered

fermi-dirac statistics

formulated 1926 · Enrico Fermi (Rendiconti Lincei / Zeitschrift für Physik, early 1926) and, independently, Paul Dirac (Proceedings of the Royal Society, August 1926) derived the statistics of particles obeying Pauli exclusion. Fermi built it to fix the ideal-gas entropy; Dirac connected it to antisymmetric wavefunctions. The two papers were genuinely independent — the distribution carries both names, and its microscopic cap n∈{0,1} is owned by the quantum-mechanics track.

History

1925

— Pauli states the exclusion principle to explain atomic spectra (before spin)

1926

— Fermi derives the statistics of an ideal gas of exclusion-obeying particles (Rendiconti Lincei)

1926

— Dirac independently derives the same statistics from antisymmetric wavefunctions (Proc. Roy. Soc., August)

1927

— Sommerfeld applies Fermi-Dirac statistics to electrons in metals, explaining the small electronic heat capacity

1928

— Bloch extends the free-electron gas to periodic lattices, founding band theory

1931

— Chandrasekhar computes the white-dwarf mass limit from electron degeneracy pressure

Going further

Texts: Pathria & Beale, Statistical Mechanics (3rd ed.), Ch. 8; Kardar, Statistical Physics of Particles, Ch. 7; Reif, Fundamentals of Statistical and Thermal Physics, Ch. 9; Sethna, Statistical Mechanics, Ch. 7; Ashcroft & Mermin, Solid State Physics, Ch. 2.
Microscopic owner: the occupation cap $n\in\{0,1\}$ and the antisymmetry argument are owned by aq-qm-pauli-exclusion-principle; the boson/fermion census by aq-qm-bosons-fermions. This article develops the thermodynamic distribution $\bar n(\varepsilon)$ .
Builds on: Grand Canonical Ensemble for Ideal Quantum Gases · Pauli Exclusion Principle · Occupation Number Representation · Bosons, Fermions, and the Spin-Statistics Connection.
Leads to: Bose-Einstein Distribution · Maxwell-Boltzmann Classical Limit · Density of States in k-Space · Quantum Concentration and the Classicality Criterion; and downstream the Sommerfeld expansion, electronic heat capacity, and degeneracy pressure (SM5).
Provenance: AllQFT `sm` export (Pathria & Beale, Ch. 8), aq-sm-fermi-dirac-distribution, ref STAT.QSTAT.FDDIS.1206. All distribution, bound, symmetry, variance, and Fermi-energy claims python-verified (18 checks, all PASS). The copper values $\varepsilon_F\approx7.0$ eV, $T_F\approx8.2\times10^4$ K reproduce the measured Ashcroft-Mermin numbers.

Why is the Fermi function bounded by one while the Bose function is not?

Pauli exclusion truncates the single-level occupancy sum at $n=1$ , so $\Xi_\varepsilon=1+e^{-\beta(\varepsilon-\mu)}$ has only two terms and the mean occupation $\bar n=e^{-\beta(\varepsilon-\mu)}/(1+e^{-\beta(\varepsilon-\mu)})$ can never exceed one — the denominator always beats the numerator. The bosonic sum instead runs to $n=\infty$ and gives a denominator $e^{\beta(\varepsilon-\mu)}-1$ that can approach zero as $\varepsilon\to\mu^+$ , allowing macroscopic occupation (Bose-Einstein condensation). The single stopped term is the whole difference between the two statistics.

Is the chemical potential the same as the Fermi energy?

Only at $T=0$ , where by definition $\varepsilon_F=\mu(T{=}0)$ . At finite temperature $\mu(T)$ moves — for a 3D free gas it drops as $\mu\approx\varepsilon_F[1-\tfrac{\pi^2}{12}(T/T_F)^2]$ and eventually goes negative in the nondegenerate regime — whereas $\varepsilon_F$ is fixed by the density through $\varepsilon_F=(\hbar^2/2m)(3\pi^2 n)^{2/3}$ . Using them interchangeably at room temperature is usually harmless for a metal (because $T\ll T_F$ ) but wrong in principle and badly wrong for a nondegenerate gas.

Where does this article stop and aq-qm-pauli-exclusion-principle begin?

The quantum-mechanics article owns the microscopic rule: antisymmetry makes the two-particle state "vanish when the labels coincide," so the occupancy is "forced to $n\in\{0,1\}$ ." This statistical-mechanics article takes that cap as input and derives its thermodynamic consequence — the grand-canonical mean occupation $\bar n(\varepsilon)=1/(e^{\beta(\varepsilon-\mu)}+1)$ as a function of $\mu$ and $T$ . We do not re-derive antisymmetry; we translate its one-line output into a distribution.

Why is the electronic heat capacity linear in T rather than the classical (3/2)Nk?

Only electrons within $\sim k_BT$ of $\mu$ can absorb thermal energy: an electron deep in the sea has no empty state nearby to move into, so it is "Pauli-blocked." The number of active electrons is $\sim g(\varepsilon_F)k_BT$ , each gaining $\sim k_BT$ , so the excitation energy grows as $T^2$ and the heat capacity as $C\propto T$ . The sharp response window $-\partial\bar n/\partial\varepsilon$ (a bell of width $k_BT$ at $\mu$ ) is exactly what selects that active shell; the Sommerfeld expansion turns this into $C_e=\tfrac{\pi^2}{3}g(\varepsilon_F)k_B^2 T$ .

What is the meaning of the occupation being both a mean and a probability?

For a fermion level, $n$ can only be $0$ or $1$ , so its average $\bar n=0\cdot P(0)+1\cdot P(1)=P(1)$ is literally the probability that the level is occupied. This is special to fermions: the two-valued occupancy makes mean and probability coincide. For bosons, where $n$ ranges over all non-negative integers, the mean occupation can exceed one and is no longer a probability — you would need the full distribution to recover occupancy probabilities.

Does the Fermi-Dirac distribution assume the particles do not interact?

Yes — it is derived for an ideal gas, where the modes are independent and the grand sum factorizes. Remarkably, Landau's Fermi-liquid theory shows the form survives interactions: near the Fermi surface the low-energy excitations are long-lived "quasiparticles" that obey a Fermi-Dirac distribution in their renormalized energies, which is why the free-electron picture works so well for real metals. Strong correlations (Mott insulators, heavy fermions) can break this, but the ideal-gas distribution remains the reference point.

How does this connect to the density of states?

The distribution gives the occupation of a single level; to get thermodynamics you sum it over all levels, which becomes $N=\int g(\varepsilon)\bar n(\varepsilon)\,d\varepsilon$ once the sum is converted to an integral with the density of states $g(\varepsilon)$ (owned by aq-sm-density-of-states-energy). That single equation fixes $\mu(T,n)$ implicitly, and its low-temperature Sommerfeld expansion produces the heat capacity, susceptibility, and equation of state of the degenerate Fermi gas. The Fermi function supplies the "how full," the density of states supplies the "how many levels."