Criterion for stationary density matrix A density matrix $\rho$ is time independent iff it commutes with the Hamiltonian $H$. I am wondering if there is a criterion to test whether $[\rho, H] =0$ using some trace condition. Specifically, I want to know if $[\rho, H] =0$ as matrices is equivalent to the condition $\mathrm{tr} [\rho,H]^2 =0$. How can I prove or disprove it?
 A: Choose a basis in which $H$ is diagonal and assume for its eigenvalues $H_i\neq 0$. Then $C:= [\rho,H]$ has matrix elements
$$ C_{ij} = (H_i - H_j)\rho_{ij},$$
and so
\begin{align}\mathrm{tr}([\rho,H]^2) & = \sum_i (C^2)_{ii} = \sum_{i,j}C_{ij}C_{ji} = \sum_{ij} (H_i - H_j)\rho_{ij}(H_j - H_i)\rho_{ji} \\ & = - \sum_{i,j} (H_i - H_j)^2\lvert \rho_{ij}\rvert^2,\end{align}
which clearly is only zero when $\rho_{ij} = 0$ for all $i,j$ where $H_i\neq H_j$. So when $H$ is non-degenerate, this implies $\rho$ is diagonal in the diagonal basis of $H$, and therefore $[\rho,H] = 0$, too.
A more abstract way to see this is to observe that $C$ is skew-Hermitian, so $\mathrm{i}C$ is Hermitian and thus diagonalizable with real eigenvalues. So $-C^2$ has real, non-negative eigenvalues, and hence its trace can only be zero when all the eigenvalues of $C$ were zero to begin with, i.e. it was the zero matrix.
A: Assume that the (complex) Hilbert space $H$ is finite-dimensional. For two hermitian operators $A$, $B$ on $H$ it holds that
$$[A,B]^\dagger = - [A,B]$$ and thus
$$\forall \psi \in H:\, \left(\psi,[A,B]^2 \,\psi\right) = \left([A,B]^\dagger \,\psi,[A,B]\psi\right) = - \left([A,B]\psi,[A,B]\psi\right) = - ||[A,B]\psi||^2 \leq 0 \quad . $$
This means that $[A,B]^2$ is negative semi-definite, i.e. hermitian with only non-positive eigenvalues. In particular, since the trace of a hermitian operator is the sum of its eigenvalues, we have
$$\mathrm{Tr}[A,B]^2 \leq 0 $$
with equality if and only if $[A,B]^2=0$, which in turn is equivalent to $[A,B]=0$:
$$ \mathrm{Tr}[A,B]^2 = 0 \Longleftrightarrow[A,B]^2= 0 \Longleftrightarrow  \forall \psi \in  H: \,\left(\psi, [A,B]^2\,\psi\right)  = - ||[A,B]\psi||^2 =0 \Longleftrightarrow [A,B]= 0  \quad .$$
The proof of the above chain of equivalences is more or less trivial.
