Question bank — Bose-Einstein statistics — bosons

True. $\bar{n}$ is the average occupancy of a level over the ensemble, so a value like $0.58$ or $9.5$ is fine even though any single measurement gives an integer count.

False. If $\mu >{\epsilon }_{\min }$ then $x=e^{(\mu -\epsilon )/k_{B}T}>1$ for the ground level and the geometric sum $\sum x^n$ diverges — the distribution has no meaning there. See Chemical potential.

True. Both have integer spin and symmetric total wavefunctions, so both allow unlimited occupancy; only their $\mu$ differs (photons $\mu =0$, He-4 $\mu <{\epsilon }_{\min }$).

True. When $\epsilon -\mu \gg k_{B}T$ the $-1$ is negligible against the huge $e^{(\epsilon -\mu )/k_{B}T}$, so $\bar{n}\approx e^{-(\epsilon -\mu )/k_{B}T}$, the classical MB tail.

False. The $-1$ is exact and is the whole boson signature; it is precisely near $\mu$ (denominator $\to 0$) that it dominates and produces the $\bar{n}\to \infty$ pile-up.

False. Low energy has nothing to do with a cap — bosons never have an occupancy limit. Low energy just makes the average occupancy large.

False. BEC sets in at a finite critical temperature $T_c=\frac{2\pi {\hslash }^2}{m}(\frac{N/V}{\zeta (3/2)}{)}^{2/3}/k_B$; below it a macroscopic fraction of particles occupies the ground state. See Bose-Einstein condensation.

Backwards. Pauli exclusion applies to fermions (antisymmetric wavefunction). Integer-spin bosons have no exclusion — that is why they can share.

That is the fermion sum (only $n=0,1$). Bosons sum $n=0\to \infty$: ${\mathcal{Z}}_i=1/(1-x)$, yielding $\bar{n}=x/(1-x)$ and the $-1$.

For fermions $\mu =\epsilon$ is harmless. For bosons $\mu ={\epsilon }_{\min }$ makes ${\mathcal{Z}}_0=1/(1-x)$ diverge — the mathematical breakdown is the onset of BEC, not a non-event.

The masslessness is not the reason. $\mu =0$ because photon number is not conserved (photons are created/destroyed freely), so no cost is charged for adding one. See Grand canonical ensemble.

The occupancy grows but each particle still carries ${\epsilon }_i$; total energy $\bar{n}{\epsilon }_i$ stays finite as long as $\mu <{\epsilon }_{\min }$ strictly. The divergence is avoided by that strict inequality.

The clean trick is $\bar{n}=x{\partial }_x\ln {\mathcal{Z}}_i$ (or $-\frac{1}{\beta }{\partial }_{\mu }\ln {\mathcal{Z}}_i$), which pulls the mean count out of the generating function; a raw ${\partial }_{\epsilon }$ without the right prefactors misses constants.

No — it is derived in the Grand canonical ensemble precisely because $N$ is not fixed; particles flow to/from a reservoir at chemical potential $\mu$.

It converges only for $x=e^{\beta (\mu -\epsilon )}<1$, i.e. ${\epsilon }_i-\mu >0$; physically this forces $\mu <{\epsilon }_i$ for every level, so $\mu$ must stay below the lowest one.

Symmetry means $\psi (1,2)=+\psi (2,1)$, so $\psi$ does not vanish when two particles share a state, unlike the antisymmetric fermion case where it forces $\psi =0$.

The $-1$ shrinks the denominator faster than the classical $e^{-\beta (\epsilon -\mu )}$ (recall $\beta =1/k_{B}T$), so ${\bar{n}}_{BE}\gg {\bar{n}}_{MB}$ at small $\epsilon -\mu$ — the seed of lasing and condensation.

With $\mu =0$, $\bar{n}=1/(e^{\beta \epsilon }-1)$; multiplying by photon energy and the density of states gives the blackbody spectral energy density.

As $\mu \to {\epsilon }_{\min }^{-}$ the ground occupancy grows without bound while the sum over excited states (weighted by $g(\epsilon )\propto \sqrt{\epsilon }$, which vanishes at $\epsilon =0$) saturates at a finite maximum, so the ground term must be extracted by hand to keep $N$ finite.

The density of states $g(\epsilon )\propto \sqrt{\epsilon }\to 0$ at $\epsilon =0$, so the integral gives zero weight to the ground state; the discrete sum still holds a macroscopic ${\bar{n}}_0$ there, so the integral undercounts and you must add the ground state back separately.

Both come from a geometric-style sum, but the fermion sum stops at $n=1$ (giving $+1$) while the boson sum runs to infinity (giving $-1$) — a single sign that flips exclusion into bunching.

Then $x=e^{\beta (\mu -\epsilon )}=1$ and the denominator is $e^0-1=0$, so $\bar{n}\to \infty$ — which is why $\mu =\epsilon$ is forbidden and marks the BEC threshold (see Figure 3).

$\beta =1/k_{B}T\to \infty$ so $\beta (\epsilon -\mu )\to \infty$, giving $\bar{n}\to 0$: excited levels empty out and particles collapse toward the ground state.

No. With $\mu <\epsilon$ we have $e^{\beta (\epsilon -\mu )}>1$, so the denominator is positive and $\bar{n}>0$ always.

$\bar{n}=1/(e^{\beta \epsilon }-1)\approx 1/(\beta \epsilon )=k_{B}T/\epsilon \to \infty$: low-frequency modes are heavily populated, the classical equipartition (Rayleigh-Jeans) limit.

No. Both corrections are dwarfed by the large exponential $e^{\beta (\epsilon -\mu )}$, so BE, FD and MB all collapse to $\bar{n}\approx e^{-\beta (\epsilon -\mu )}$.

Not intrinsically — each level's occupancy is unbounded; the total is fixed only when you impose a particle-number constraint that sets $\mu$.

Since $T_c\propto (N/V{)}^{2/3}$, doubling $N/V$ raises $T_c$ by a factor $2^{2/3}\approx 1.59$ — denser gases condense at higher temperature.