# 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

## 1 Answer

This is a very interesting and important question. The point is that in practice, one doesn't measure the full state of the fully interacting system. For example, one measures the energy of the photon emitted from an excited state of the hydrogen atom due to the interaction of the electromagnetic field with the orbiting electron.

And the magic that occurs is that when the various bits and pieces of the interacting system are well-separated enough, e.g. when the emitted photon is far away from the now de-excited hydrogen atom, one can show (see the Gell-Man--Low theorem) that the eigenstates of the fully interacting Hamiltonian are the same as the eigenstates of the unperturbed Hamiltonian. (Or, more correctly, are the same up to an overall normalization and a slight shifting of the particle's mass that one can compute.)