# 10.5: The quantum equilibrium ensembles


At equilibrium, the density operator does not evolve in time; thus, $${\partial \rho \over \partial t} = 0$$. Thus, from the equation of motion, if this holds, then $$[H,\rho]=0$$, and $$\rho (t)$$ is a constant of the motion. This means that it can be simultaneously diagonalized with the Hamiltonian and can be expressed as a pure function of the Hamiltonian

$\rho = f(H) \nonumber$

Therefore, the eigenstates of $${\rho}$$, the vectors, we called $$\vert w_k \rangle$$ are the eigenvectors $$\vert E_i \rangle$$ of the Hamiltonian, and we can write $$H$$ and $${\rho}$$ as

\begin{align*} H &= \sum_i E_i \vert E_i\rangle \langle E_i\vert \\[4pt] \rho &= \sum_i f(E_i)\vert E_i\rangle \langle E_i\vert \end{align*}

The choice of the function $$f$$ determines the ensemble.

This page titled 10.5: The quantum equilibrium ensembles is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Mark Tuckerman.