TS.5 · Quantum occupations and limits of equipartition

For noninteracting particles in thermal and particle exchange equilibrium, use chemical potential μ and x=(ε−μ)/(kBT). Mean occupation of one complete state is n_F=1/(e^x+1) for fermions and n_B=1/(e^x−1) for bosons. Fermionic occupation of that state is zero or one; its mean lies between them. Bosonic occupation can exceed one. A level of degeneracy g has total mean g times the single-state occupation when its states share the same energy. Count spin as part of a complete state. The Bose denominator requires ε>μ for the ordinary finite expression, with the ground-state limit treated separately. For equilibrium photons μ=0 because photon number is not conserved; do not set μ=0 for every material particle gas.

Count occupation patterns rather than labelling identical particles. Two identical fermions distributed among four distinct complete states have choose(4,2)=6 allowed patterns. Two identical bosons among those states have choose(4+2−1,2)=10 patterns because both may share one state. Two labelled distinguishable particles would have 4²=16 assignments. These are different counting models, not three interchangeable answers to the same specification. If a question supplies spin degeneracy, first decide whether its stated number counts complete states or just orbital levels. A Pauli prohibition on two identical complete states does not prohibit opposite-spin fermions in one spatial orbital.

When x is large and positive, occupation is small and both denominators are dominated by e^x: n_F≈n_B≈e^(−x), the Maxwell–Boltzmann dilute limit. At x=ln4 the means are 1/5 for fermions, 1/3 for bosons and 1/4 in the classical approximation; the difference is still significant. A large total particle number alone does not justify classical statistics: density, temperature and accessible states control occupation. For fermions at low T, states below μ become nearly occupied and those above nearly empty. At ε=μ a fermionic state has mean one half; inserting that value into the ordinary Bose formula would instead produce a divergent denominator and requires different limiting treatment.

Classical equipartition assigns kBT/2 of mean energy to each independent quadratic term in an equilibrated Hamiltonian. A monatomic ideal gas has three translational terms, giving U=3NkBT/2 and C_V=3NkB/2. One classical one-dimensional harmonic oscillator has kinetic and potential terms, giving mean kBT and C=kB. For a quantum oscillator with fixed spacing ε=$\hbar$ω, the thermal energy above its temperature-independent zero point is ε/(e^x−1), now x=ε/(kBT), and C/kB=x²e^x/(e^x−1)². At high T it approaches the classical value; at low T excitation freezes out and C→0. A zero-point energy ε/2 shifts U but not C. Molecular rotational/vibrational contributions likewise need their energy scales checked before assigning classical quadratic terms.

Exclude two fermions from one complete state, not from an entire degenerate energy level. State which x is used; chemical-potential occupations and fixed oscillator excitation formulas are different models.

mean occupation: Ensemble average number of particles in one complete quantum state or a specified group of states.

equipartition: Classical equilibrium rule assigning kBT/2 to each independent quadratic Hamiltonian term.