Stabilization of von Neumann equation Given the solution of the von Neumann equation 
$\rho(t) = e^{-i H t/\hbar} \rho(0) e^{i H t/\hbar}$
How can we justify if it will be stabilized as $t\rightarrow\infty$ in general?
For example, consider $H=\begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix} \\
\ \rho(0)=\begin{pmatrix}\frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2}\end{pmatrix}$ 
Is possible to fix for this case? Also, is it normal? Shouldn't there be some wave dynamics before reaching equilibrium? 
 A: A density matrix will never approach a steady state under purely unitary evolution. It is possible for the density matrix to be initialised in a steady state, if this state is already diagonal in the energy representation. That is, if the initial state satisfies $[H,\rho(0)]=0$ then the density matrix does not evolve, as in the OP's example. Otherwise, there will always be coherent oscillations. One can see this by examining the condition that the derivative of the density matrix be zero at some point in time:
$$ \dot{\rho}(t) = -i[H,\rho(t)] = -i[H,e^{-iHt}\rho(0)e^{iHt}] = -i e^{-iHt}[H,\rho(0)]e^{iHt} = 0.$$
This shows that if $\dot{\rho}(t)=0$ at any time, then the initial density matrix satisfies $[H,\rho(0)]$, which implies that $\dot{\rho}(t) = 0$ at every time.
Assuming that $\dot{\rho}(t)\neq 0$, one can expand the density matrix in eigenstates $\lvert n\rangle$ of the Hamiltonian satisfying $H|n\rangle = E_n|n\rangle$:
$$ \rho(t) = \sum_{m,n} \rho_{mn}(0) e^{-iHt}\lvert m\rangle\langle n\rvert e^{-iHt} = \sum_{m,n} \rho_{mn}(0) e^{-i(E_m-E_n)t}\lvert m\rangle\langle n\rvert = \sum_{m,n} \rho_{mn}(t) \lvert m\rangle\langle n\rvert,$$
where $\rho_{mn}(t)$ are the matrix elements at time $t$. Therefore, the populations $\rho_{nn}(t) = \rho_{nn}(0)$ stay the same while the coherences $\rho_{mn}(t)$ oscillate at the frequency $E_m-E_n$.
