Heat capacity and fluctuation-dissipation theorem, meaning of energy fluctuations? I have read that from the fluctuation-dissipation theorem that the heat capacity is proportional to energy fluctuations (or populations fluctuations). In this context what is the meaning of 'energy fluctuations' (since a well defined state has constant energy) and why are they zero at $T=0$?
 A: For a system at thermal equilibrium with a thermostat at temperature $T$ (i.e. in the so-called canonical ensemble), the average energy of the system is
$$\langle E\rangle={1\over{\cal Z}}\sum Ee^{-\beta E}
=-{\partial\ln{\cal Z}\over\partial\beta}$$
where $\beta=1/k_BT$ and the sum extends over all possible micro-states of the system (for a quantum system, replace the sum by a trace over the Hilbert space of the system and $E$ by the Hamiltonian $H$). It is a simple exercise to show that $\langle E^2\rangle$ satisfies
$$\langle E^2\rangle-\langle E\rangle^2=k_BT^2C$$
where $C={\partial\over\partial T}\langle E\rangle$ is the specific heat. The later is therefore proportional to (thermal) energy fluctuations.
A $T=0$ ($\beta\rightarrow +\infty$), the Boltzmann weight $e^{-\beta E}$ imposes to the system in the micro-states with the lower energy $E_0$. Therefore, $\langle E\rangle=E_0$ and $\langle E^2\rangle=E_0^2$, i.e. energy fluctations vanish.
A: For heat transport in solid, the instantaneous heat flux (resulted from thermal fluctuations) is the sum of atom-resolved energy-density fluctuations times the instantaneous position of the atom,
$\textbf{J}(t)=\sum_{I=1}^{N} \dot{e}_I \textbf{R}_{I}$
In the phonon quasi-particle picture, this means the phonon occupations $n_{s}(\textbf{q})$ (or $E_s(\textbf{q},t)$) of each modes $(s,\textbf{q})$ are fluctuating, and
$\textbf{J}(t)=\sum_{s\textbf{q}} E_s(\textbf{q},t)\textbf{v}_{s}(\textbf{q})$
You know $n_{s}(\textbf{q})$ and $E_s(\textbf{q},t)$ are related with each other by Bose-Einstein distributions
