What is the physical meaning of [H,ρ ]=0 for a stationary state?

let suppose you have of H:

[itex]\hat{H}|E_i\rangle=E_{H,i}|E_i\rangle[/itex]

If you can diagonalize H and rho together
there is no reason to think that the set of [itex]|E_i\rangle[/itex] that can be diagonalized at the same time with H are ALL the [itex]|E_i\rangle[/itex] that are diagonal for [itex]\hat{H}[/itex]. in general the eigenstates in common is a subset of the eigenstates of H.
But let consider only the sub set of eigenstates of H that are also eigenstates of rho, you have

[itex]\hat{\rho}|E_i\rangle=E_{p,i}|E_i\rangle[/itex]

and so you can write rho as

[itex]\hat{\rho}=\Sigma_i \lambda_i |E_i\rangle\langle E_i|[/itex]

that is the representation of your rho in the eigenstates of hamiltonian.

 

The density matrix being diagonal corresponds to a completely incoherent mixture. So the eigenvectors of the density matrix are the constituents of this mixture.

 

The complete equation of motion for the Statistical Operator is the von Neumann equation,

[tex]\left (\frac{\partial \hat{\rho}}{\partial t} \right )_{\text{exp}} + \frac{1}{\mathrm{i} \hbar} [\hat{\rho},\hat{H}]=0.[/tex]

Since a stationary state is defined as one for which the Statistical Operator is not explicitly dependent on time, for this case you arrive at the simplified equation

[tex][\hat{\rho},\hat{H}]=0.[/tex]

This means that for a stationary (or equilibrium) state, the Statistical Operator of the system must be a function of the conserved quantities and not explicitly dependent on time.