Three Quantum Statistics

Same trap, same N particles, same temperature — but three different rules for how identical particles can share energy levels. Each statistic solves its own particle-number constraint for its own μ, then predicts a different population for every level.

Temperature (T, in ħω/k_B)

Particle Number (N)

10.00

T_c (Bose)

T_F (Fermi)

μ Boltzmann

μ Bose

μ Fermi

N₀ Boltzmann

N₀ Bose

N₀ Fermi

The Three Distributions

$$\langle N_n\rangle_{\text{MB}} = g_n\, e^{-(E_n-\mu)/k_BT}$$ $$\langle N_n\rangle_{\text{BE}} = \frac{g_n}{e^{(E_n-\mu)/k_BT}-1}$$ $$\langle N_n\rangle_{\text{FD}} = \frac{g_n}{e^{(E_n-\mu)/k_BT}+1}$$

MB: Maxwell-Boltzmann — the classical limit; no restriction on how many particles share a state
BE: Bose-Einstein — bosons; unlimited sharing, μ must stay below E₀=0
FD: Fermi-Dirac — fermions; Pauli exclusion caps ⟨N_n⟩ at g_n per level, μ is unrestricted in sign

Why Do They Differ?

All three formulas share the same numerator (g_n) and the same Boltzmann factor e^{−(E_n−μ)/k_BT} in spirit — they only differ in what's added to the denominator's exponential: nothing (MB), −1 (BE), or +1 (FD).

That single sign is the entire content of quantum statistics. +1 (Fermi-Dirac) comes from the Pauli exclusion principle: at most one fermion per quantum state, which caps ⟨N_n⟩ at g_n no matter how cold the gas gets. −1 (Bose-Einstein) does the opposite: bosons are gregarious, and as μ → E₀⁻ the ground-state term diverges, driving condensation.

When e^{(E_n−μ)/k_BT} ≫ 1 (dilute gas, or high T) — both ±1 corrections become negligible next to that huge exponential, and all three formulas collapse to the same Maxwell-Boltzmann curve. That's the classical limit: particles are so unlikely to compete for the same state that it stops mattering whether they're bosons, fermions, or classical.

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