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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?

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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 $E_m-E_n$.

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  • $\begingroup$ Thank you for the solution. But I am not very sure where it was used the condition $[H, \rho(0)]\neq 0$ to prove that the solution $\rho(t)$ will be stabilized, where it is not related to the zero derivative, since there can be small fluctuation even with large derivative. $\endgroup$ – Xingdong Aug 21 '14 at 21:04
  • $\begingroup$ Hi @XingdongZuo, I'm not sure I understand what you mean by "stabilised". Do you mean reaching thermal equilibrium? $\endgroup$ – Mark Mitchison Aug 22 '14 at 8:17
  • $\begingroup$ Yes, and in the solution function graph the fluctuation of the curve to be smaller and smaller as time to infinity. The point is not against the proof with the eigenstates, but the place where we can use the condition that $H$ and $\rho(0)$ does not commutes, since the the form above can always be shown even if $[H, \rho(0)]=0$ $\endgroup$ – Xingdong Aug 22 '14 at 8:33
  • $\begingroup$ @XingdongZuo But the state of a system does not reach thermal equilibrium as $t\to 0$ under unitary evolution. If $[\rho,H]\neq 0$ you will just see coherent oscillations forever. There is no damping of fluctuations without some external influence. $\endgroup$ – Mark Mitchison Aug 22 '14 at 8:35
  • $\begingroup$ I agree with it that for the closed quantum system it will endlessly oscillate. Well, I'd like to know if it is possible to explicitly to show $[H, \rho(0)]\neq 0$ in your proof of the form of expansion of density operator above, i.e. $$ \rho(t) = \sum_{m,n} \rho_{mn}(0) e^{-iHt}\lvert m\rangle\langle n\rvert e^{-iHt} = \sum_{m,n} \rho_{mn}(0) e^{-i(E_m-E_n)t}\lvert m\rangle\langle n\rvert = \sum_{m,n} \rho_{mn}(t) \lvert m\rangle\langle n\rvert,$$ $\endgroup$ – Xingdong Aug 22 '14 at 8:42

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