 Solution

For noninteracting particles, a many-particle configuration has $E={\sum }_j{n}_j{ϵ}_j$ and $N={\sum }_j{n}_j$. Summing independently over allowed occupations therefore factorizes the grand partition function:At fixed $T,V$, differentiating the original state sum gives ${\partial }_{\mu }\ln \Theta =\beta ⟨N⟩$ and ${\partial }_{\mu }^2\ln \Theta ={\beta }^2\mathrm{Var}N$, soFor one level, write $q=e^{-\beta (ϵ-\mu )}$. Fermions allow $n=0,1$, giving ${\Theta }_F=1+q$ and $\bar{n}=q/(1+q)$. Bosons allow all nonnegative integers, giving ${\Theta }_B=1/(1-q)$ and $\bar{n}=q/(1-q)$, with $q<1$. ThusThe grand-canonical single-level occupation variance follows from the same derivative identity applied to one level:In the high-temperature nondegenerate regime, $\bar{n}\ll 1$ and both approach classical Poisson-like fluctuations, $\mathrm{Var}n\simeq \bar{n}$. At low temperature, nearly filled or empty fermion levels have strongly suppressed fluctuations; only the Fermi-surface window changes occupation. Highly occupied low-energy boson levels can instead have $\mathrm{Var}n\sim {\bar{n}}^2$, showing bosonic enhancement. This last statement depends on occupation and reservoir chemical potential, not temperature alone; grand-canonical condensate fluctuations must not be confused with a fixed-total-number ensemble.