0
$\begingroup$

I am given that a system is defined by two independent states |+> defined as: \begin{bmatrix}1 \\ 0\end{bmatrix} and |-> defined as: \begin{bmatrix}0 \\1\end{bmatrix} With respect to these two states, the time-independent Hamiltonian is represented in matrix form as: \begin{pmatrix}E & U\\ U & E\end{pmatrix} Now the question asks me to calculate the transition probability of the system initially in state 1 to transition to state 2 in some time $t$ using time-dependent perturbation theory.

My question: Our hamiltonian is clearly time-independent. Then why do I need to use time-dependent perturbation theory. Isn't it true that the perturbation is time-independent too?

$\endgroup$
  • $\begingroup$ That's kind of strange, but technically time dependent perturbation theory should work fine. Nothing goes wrong in the equations if you "drive at frequency $\omega = 0$". $\endgroup$ – knzhou Mar 23 '18 at 0:38
  • $\begingroup$ It's probably just meant as a simple practice exercise and consistency check. Of course realistically in this situation you wouldn't use perturbation theory at all, you'd just do it exactly. $\endgroup$ – knzhou Mar 23 '18 at 0:38
  • $\begingroup$ The amplitudes of the two states will change because they are not eigenstates of the Hamiltonian. So you have to treat it as a dynamical system. $\endgroup$ – PeaBrane Mar 23 '18 at 5:33

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.