# Meaning of probabilities in time-dependent perturbation theory

Consider a Hamiltonian $$H = H_0 + V(t)$$ where $V(t)$ is small enough to be considered as a perturbation.

One looks at the probability of transition between two states of the unperturbed hamiltonian, $$P_{i\rightarrow n} = \lvert c_n(t) \rvert ^2$$ with the state of the system: $\lvert \psi(t) \rangle = \sum_n c_n(t) \lvert n \rangle$, where $\lvert n \rangle$ are the eigenstates of the unperturbed Hamiltonian $H_0$.

What does this really mean?

Is that supposed to represent the probability to measure the system in this state? If so, how is that possible, given that the eigenstates are no more the same as the unperturbed ones?

• "Is that supposed to represent the probability to measure the system in this state?" Yes. "If so, how is that possible, given that the eigenstates are no more the same as the unperturbed ones?" I don't understand what you're asking. What's wrong with finding the system in an eigenstate of $H_0$? Nobody ever said a physical system has to always be found in eigenstates of the full Hamiltonian. – DanielSank Jun 15 '17 at 22:40
• But how do we measure the state then ? Are we supposed to measure the energy $E_n$ associated with the eigenstate $\vert n>$ ? This case is impossible as E_n is no more an eigenvalue of H.. – Pao Jun 15 '17 at 22:46
• It's not clear what you mean by "measure the state". We can measure various things like position, momentum, energy, etc. – DanielSank Jun 15 '17 at 22:47
• Yes, that was a possible interpretation of the "transition probability", but it does not make much sense.. – Pao Jun 15 '17 at 22:51
• I am still not sure what you mean to say. When you say "that was a possible interpretation..." what is "that"? Please use precise, unambiguous language. – DanielSank Jun 15 '17 at 22:52