# Time evolution of a projected mixed state

Suppose a quantum system (non-interacting) at finite temperature ($\beta^{-1}$). I want to know how to compute the transition probability between two degrees of freedom ($u$ and $v$) at two different times.

The system starts ($t=0$) in a mixed state, described by $$\hat \rho = \sum_l e^{-\lambda_l \beta} |\psi_l\rangle\langle \psi_l|/Z$$ I projected the mixed state into $|u\rangle$, applying the projector $$\hat P_u = |u\rangle\langle u|.$$ Therefore at $t=0$, I have $$\hat P_u \hat \rho \hat P_u.$$ Because I want the transition probability in the future, I used the evolution operator $$\hat U(t_f) = \sum_n e^{-i\lambda_n \beta} |\psi_n\rangle\langle \psi_n|$$ to evolve the mixed state $$\hat U(t_f)^\dagger\hat P_u \hat \rho \hat P_u \hat U(t_f).$$ Then I projected the last operator into $|v\rangle$,

$$\hat P_v \hat U(t_f)^\dagger\hat P_u \hat \rho \hat P_u \hat U(t_f)\hat P_v$$

Computing the trace of the above operator,

$$\mathrm{Tr}[\hat P_v \hat U(t_f)^\dagger\hat P_u \hat \rho \hat P_u \hat U(t_f)\hat P_v] ,$$

I get

$$(\sum\limits_l \frac{e^{-\beta \lambda_l}\langle \psi_l|u\rangle\langle u|\psi_l\rangle}{Z}) (\sum\limits_m e^{-i\lambda_m t}\langle \psi_m|v\rangle\langle u|\psi_m\rangle) (\sum\limits_n e^{i\lambda_n t}\langle \psi_n|u\rangle\langle v|\psi_n\rangle)$$

Did I make any mistakes?

I expected some mixing between the time and temperature in the last equation, but if everything is right there is no mixing in that case.

• Everything looks right, Jitendras suggestion is basically a time reversal and won’t dramatically change anything. What do you mean by a mixing of time and temperature? – Shane P Kelly Aug 4 '18 at 1:59

Your way of evolving is wrong. It should be $\hat U(t_f)\hat P_u \hat \rho \hat P_u \hat U(t_f)^\dagger$.