Energy fluctuations in quantum canonical ensemble 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.
 A: 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. 
