Energy fluctuations in the canonical ensemble

Remarks

Proof idea: $Z$ is a Laplace transform that converges on an open interval, so it can be differentiated term by term. Then $Z'/Z = -\langle H \rangle$ and $Z''/Z = \langle H^2 \rangle$, which give $(\ln Z)'' = \langle H^2 \rangle - \langle H \rangle^2$. Since a variance is nonnegative, the heat capacity is nonnegative: $C_V \ge 0$. Example: for $N$ independent two-level units with energies $0$ and $\varepsilon$, $\ln Z = N \ln (1 + e^{-\beta \varepsilon})$. So both $\langle H \rangle$ and $\operatorname{Var}(H)$ are proportional to $N$, and the relative fluctuation $\sqrt{\operatorname{Var}(H)} / \langle H \rangle$ decays like $N^{-1/2}$.

Going deeper

For a student

This is the simplest fluctuation–response relation: a response coefficient, the heat capacity, equals a variance in equilibrium. The smallness of relative fluctuations for large $N$ is the heuristic reason for the equivalence of ensembles. Similar computations in the grand canonical ensemble relate the variance of the particle number to the compressibility. At a critical point the heat capacity per particle can diverge, and then energy fluctuations are anomalously large.