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.

  • 1
    $\begingroup$ 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? $\endgroup$ Aug 4, 2018 at 1:59

1 Answer 1


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


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