How identical fermions fill quantum states when no two may share one. The occupation of a state at energy E is n = 1/(e^((E−μ)/kT) + 1) — capped at one particle per state by the Pauli exclusion principle. Rendered, not quoted: the curve below is computed live, its four defining properties asserted at boot.
source E. Fermi, Zur Quantelung des idealen einatomigen Gases, Zeitschrift für Physik 36 (1926) 902; P. A. M. Dirac, On the Theory of Quantum Mechanics, Proc. R. Soc. London A112 (1926) 661. overview no stable open scan — cited by journal / year
A single quantum state of energy E, in contact with a reservoir at temperature T and chemical potential μ (the Fermi level). Because fermions obey Pauli exclusion, that state holds either 0 or 1 particle — its average occupation is a number in [0,1].
n(E) = 1 / ( e^{(E−μ)/kT} + 1 )The lone +1 in the denominator is the whole story. Bose–Einstein statistics put a −1 there (unbounded occupation); Maxwell–Boltzmann drops it entirely (classical). The +1 is what makes matter take up space. Natural units: k = 1, energies measured in the same units as T.
No two fermions alike — Fermi–Dirac statistics, 1926, derived independently by Fermi (Feb) and Dirac (Aug). The 1/(e+1) occupation is the exclusion-principle sibling of the-boltzmann-distribution: at high energy (E−μ » kT) the +1 becomes negligible and n → e−(E−μ)/kT, the Boltzmann tail. The step at the Fermi level explains why metals conduct, why white dwarfs resist collapse, and why atoms have shells.
Re-runs the full self-check against the live engine on demand, including the planted tamper in window 6. Green = every closed-form property holds; flips red if the occupation ever leaves [0,1].
witness idleA sweep of energies E across the Fermi level, plus the reservoir temperature T and Fermi level μ. Drag the sliders — the engine recomputes n(E) for every energy.
Dashed white = the T=0 step (all states below μ full, all above empty). The pink curve is n(E) at the chosen T; every temperature crosses exactly 1/2 at E=μ.
booting…Proven, live: 0 ≤ n(E) ≤ 1 for all E,T (Pauli); a step at T=0; n(μ)=1/2 exactly for every T>0; and the Boltzmann tail e−(E−μ)/kT recovered when E−μ » kT. Four asserts, checked to 1e−9 at boot.
That is precisely Bose–Einstein statistics, and it is wrong for fermions. Remove the +1 (or flip it to −1) and the occupation diverges as E→μ — a single electron state holding a million electrons. The Pauli exclusion principle is the physical wall: measured electron heat capacity, degeneracy pressure, and the periodic table all demand n≤1. Window 6 performs exactly this attack; window 7 catches it.
"At T=0 the distribution is smooth."
→ Correction: at T=0 it is a hard step — n=1 for E<μ, n=0 for E>μ. Smoothing only appears at T>0, over a width ~kT.
"The Fermi level is where n=1."
→ Correction: E=μ is where n=1/2, for every T>0. That half-filling defines the chemical potential.
"Fermi–Dirac and Boltzmann never agree."
→ Correction (AMBER, limit): for E−μ » kT they do — the +1 is negligible and FD collapses to the classical tail. They differ only near and below μ.
The disclosed planted void. Pressing tamper flips the denominator sign from +1 (Fermi–Dirac) to −1 (Bose–Einstein). The occupation then diverges and exceeds 1 near E=μ — Pauli is violated. The witness in window 7 detects it live.