When is the density matrix real & symmetric? Book: Statistical Mechanics (3rd ed.) by R K Pathria
Page 118, Chapter 5, Sec 5.1, Eq. 13
The author says that the density matrix is real & symmetric if the system is in equilibrium. Can somebody give me a mathematical proof of this statement?
The density matrix is Hermitian by construction. How is it real & symmetric when the system is in equilibrium?
 A: A ready counterexample for 2x2 quantum-mechanical entities, if that is what is under consideration:
Consider Hermitian complex hamiltonians and density matrices
$$
H=\sigma_y , \qquad \rho =\tfrac{1}{2} (1\!\!1 + \sigma_y).
$$
They commute with each other, so the density matrix is stationary. 
They are Hermitian and diagonalizable to real diagonal matrices (by the same transformation), but not real, by inspection. To be sure, they can be transformed to real non diagonal matrices by an obvious rotation, but... why bother not diagonalizing them? 
A: Equilibrium means that
$ \frac{d \rho}{d t}  = 0$
for the density operator $\rho$ dependent on time $t$. By Ehrenfest's theorem this is equivalent to $[H,\rho] = 0$. Therefore, in equilibrium, $\rho$ is a function of $H$. This Hamiltonian $H$ must be Hermitian and therefore its eigenstates and eigenvectors must exist. Now, the density matrix can be expressed in terms of the matrix elements
$ \langle n|H|m \rangle ~\in \mathbb{R}$.
$\implies~ \langle n|\rho(H)|m \rangle \in \mathbb{R}~~~~~~~~~~$(i.e. it is not changed by complex conjugation).
Finally, $ \langle n|\rho(H)|m \rangle^* = \langle m|\rho(H)^\dagger|n \rangle = \langle m|\rho(H^\dagger)|n \rangle = \langle m|\rho(H)|n \rangle = \langle n|\rho(H)|m \rangle$.
