Quantum occupations and limits of equipartition
| English | Português |
|---|---|
| mean occupation | mean occupation |
| equipartition | equipartition |
A decision before an answer
- A fermionic energy level can contain several particles when it contains several distinct states; the exclusion rule applies per complete state.
- Your goal: Compute mean occupation per complete quantum state for fermions and bosons.
Assign occupation to complete states
- 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.
At (ε−μ)/(kBT)=ln3, the mean occupation of one fermionic state is:
n_F=1/(3+1)=1/4; bosonic mean would be 1/(3−1)=1/2.
Count indistinguishable configurations
- 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.
Two identical fermions can occupy six distinct complete states. How many occupation patterns exist?
Select the two occupied states: choose(6,2)=15. Ordered assignments double-count identical fermions.
Check the dilute approximation
- 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.
At x=(ε−μ)/(kBT)=ln4, a single state has n_F=0.2 or n_B=1/3. A five-state degenerate level therefore has mean 1 fermion or 5/3 bosons; total fermionic level occupation can exceed one at other x because there are five distinct states. A quantum oscillator at ε/(kBT)=ln2 has mean excitation energy ε and C/kB=2(ln2)²=0.960906, close to but different from the classical value one.
At x=ln5, one bosonic state has mean occupation ____ (give an exact decimal).
n_B=1/(5−1)=1/4=0.25.
Test quantum versus classical modes
- 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=ℏω, 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.
Which answer fits this case?
Compute mean occupation per complete quantum state for fermions and bosons
A temperature-independent oscillator zero-point energy ε/2 contributes zero to its heat capacity.
Heat capacity differentiates energy with respect to temperature; a fixed additive constant has zero derivative.
Keep the distinctions
- mean occupation 平均占据数 — Ensemble average number of particles in one complete quantum state or a specified group of states.
- equipartition 能量均分 — Classical equilibrium rule assigning kBT/2 to each independent quadratic Hamiltonian term.
- Compute mean occupation per complete quantum state for fermions and bosons.
- Count allowed identical-particle occupations and identify the classical dilute limit.
- Compare classical quadratic-mode heat capacity with a quantum oscillator response.
Match each term with its precise meaning in this lesson.
Keep the distinctions stated in the teaching example.
Put this lesson’s reasoning or event sequence in order.
The order follows the stated process; check each stage before the next.