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How would you go about showing that in the quantum canonical ensemble (that is, in the density matrix and operator formulation), the energy fluctuations, namely $\langle H^2\rangle - \langle H\rangle^2$ is the same as in the classical case, i.e. $k_b T^2 C_V$?

I've seen a lot of books asserting this, but can't find a formal derivation similar to the classical one, as given here, for example.

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up vote 5 down vote accepted

The quantum-mechanical proof is actually pretty much identical to the classical one given in the link. You simply replace the integral over the phase space with a trace over the states in the Hilbert space. The equilibrium density operator is $$ \rho = \frac{e^{-\beta H}}{Z},$$ where the partition function is $Z = \mathrm{Tr}(e^{-\beta H})$ and the inverse temperature is $\beta = 1/k_BT$. Explicitly, the expectation value of the energy is given by $$ \langle H \rangle = \frac{1}{Z}\mathrm{Tr}(H e^{-\beta H}) = -\frac{1}{Z} \frac{\partial}{\partial\beta} \mathrm{Tr}(e^{-\beta H}) = -\frac{1}{Z} \frac{\partial Z}{\partial\beta}. $$ Likewise, you should find $\langle H^2 \rangle = \frac{1}{Z}\frac{\partial^2 Z}{\partial\beta^2}$, and $$ C_V = \frac{\partial \langle H\rangle}{\partial T} = -\frac{1}{k_B T^2}\frac{\partial}{\partial\beta} \left[\frac{1}{Z} \mathrm{Tr}(H e^{-\beta H})\right] = \frac{1}{k_BT^2}\left(\langle H^2 \rangle - \langle H \rangle^2\right).$$ You will have to fill in a few steps for yourself to prove the last equality.

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