Given von Neumann equation 
$$\frac{d}{dt} \rho(t) = -i [H, \rho(t)] = -i e^{-iHt}[H, \rho(0)]e^{iHt}.$$


If we know that $[H, \rho(0)] \neq 0$, how do we prove that the solution will fluctuate forever. (stabilization means the fluctuation approaching to zero, but the derivative can still be large, so here, we want to show that it will endlessly oscillate and never be damped.)


Mathematically, $H$ is an Hermitian matrix and $\rho(t)$ is a square matrix with trace one, positiveness and self-adjoint.


We need a mathematical proof that particularly by using the fact that $[H, \rho(0)] \neq 0$, i.e. $H\rho(0)\neq\rho(0)H$



Proof the solution of von Neumann equation will fluctuate forever if Hamiltonian and initial density matrix does not commute
In the eigenbasis of $H$ the density matrix has elements


$$\rho\_{nm}(t)=\rho\_{nm}(0)\exp[i(E\_m-E\_n)t]$$


where $E\_n$ are the eigenvalues of $H$; since you assume that $\rho(0)$ does not commute with $H$, there are nonzero matrix elements $\rho\_{nm}(0)$ with $n\neq m$. So these matrix elements will oscillate forever, without any damping.


