# 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. Jun 15, 2017 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, 2017 at 22:46
• It's not clear what you mean by "measure the state". We can measure various things like position, momentum, energy, etc. Jun 15, 2017 at 22:47
• Okay, so to be pragmatic, this formula "would work" only to predict the results of measurements of an observable which has the same eigenstates as the unperturbed Hamiltonian ?
– Pao
Jun 15, 2017 at 23:06
• You can think about it that way, yes. Jun 15, 2017 at 23:09