0
$\begingroup$

In the Schroedinger picture, we can calculate a time-evolution operator \begin{align} U_S(t_0, t_1) = T \bigg\{ e^{-i\int_{t_0}^{t_1} dt H_S(t) } \bigg\} \end{align}

This operator is pretty handy in describing the time evolution of states, and it can be used as well to change to the Heisenberg-picture, by: \begin{align} O_H(t) = U_S^{-1}(t_0, t) O_S(t) U_S(t_0, t) \\ |\Psi(t) \rangle_H = U^{-1}(t_0, t) |\Psi(t) \rangle_H \end{align}

However, as discussed for example in this question, $H_H(t)$ and $H_S(t)$ are in general not equal anymore. My question now is: Does that as well mean that we can't calculate the time-evolution operator in the same way as before? I tried to show that the equation given above, with a plugged in $H_H(t) = H_H(t) = U_S^{-1}(t_0, t) H_S(t) U_S(t_0, t) $ instead of $H_S(t)$, gives the same result as before. But I wasn't able to show that the outcoming operator is the same (which makes sense if the Operators $H_H$ and $H_S$ are not equal anymore.

Is it possible to calculate $U_S(t_0, t_1)$, when all I know is the Heisenberg Hamilton operator? And furthermore, is it useful?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.