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Consider a time dependent Hamiltonian $$\hat{H}(t)=\hat{H}_0+\hat{V}\tag{1}\label{1}(t)$$ where $\hat{H}_0$ is the unperturbed Hamiltonian, for which the eigenvalue problem has been solved $$\hat{H_0}\lvert\phi_k\rangle=E_k\lvert\phi_k\rangle\tag{2}\label{2}$$ Suppose that at $t=0$ the system is in the $n$-th eigenstate of the unperturbed Hamiltonian, i.e. $\lvert\phi_k\rangle$. Then, as this is not an eigenstate of the total Hamiltonian \eqref{1} it will have non-trivial time evolution and at a time $t=t'$ the state will have (in general non-zero) components $c_i(t')$ along the other eigenstates \eqref{2}, which can be calculated at first order using the usual formulas. Up to this point everything is fine. What sounds weird to me is considering the modulus squared $\lvert c_i(t')\lvert^2$ as a transition probability to the $i$-th eigenstate \eqref{2}.

According to the postulates of quantum mechanics, after measuring an observable represented by $\hat{A}$, the state will collapse to an eigenstate of that operator and the outcome of the measurement will be the corresponding eigenvalue. In our case, the observable we're dealing with is energy, which is represented by the time dependent Hamiltonian \eqref{1}, so the outcome of the measurement at $t=t'$ will be an instantaneous eigenvalue of $\hat{H}(t')$ and the state will collapse to an instantenous eigenstate.

On the other hand, saying that $\lvert c_i(t')\lvert^2$ is the transition probability to the $i$-th eigenstate of \eqref{1} at time $t=t'$ makes it sound as if measuring energy at $t=t'$ could make the state collapse, with probability $\lvert c_i(t')\lvert^2$, to $\lvert\phi_k\rangle$ which is not an eigenstate of the total Hamiltonian \eqref{1}. This would only make sense if the Hamiltonian were just $\hat{H}_0$. Once again I want to clarify that the problem is in the probabilistic interpretation (collapse) we're giving to this coefficient because it apparently contradicts the postulates of QM. Are we measuring only the $\hat{H}_0$ part somehow? What is going on here?

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$|c_i(t)|^2$ is interpreted as transition probability in perturbative treatments, such as Fermi golden rule. That is, we are still looking at the states of the unperturbed Hamiltonian, and what interests us is how the population of these states changes with time (due to the presence of the perturbation.)

When perturbation is strong, i.e., cannot be considered perturbatively, as, e.g., in the case of Rabi oscillations, one usually does not talk about $|c_i(t)|^2$ as a transition probability, but rather something like probability to find system in state $i$ at time $t$. Note however, that there is nothing wrong with measuring the states of Hamiltonian $H_0$, even though the eigenstates of the system are not the eigenstates of this Hamiltonian - it simply means that our measurement operator commutes with $H_0$, but not with the full Hamiltonian. In other words, we are projecting the results on the eigenstates of $H_0$ - this does not deviate from the prescriptions of basic QM.

Finally, as a corollary, one could mention situations where the perturbation lasts only finite time (or vanishes for $t\rightarrow\pm\infty$). Then $H_0$ is the exact Hamiltonian before and after the perturbation is on, so speaking of $|c_i(t)|^2$ as a transition probability is meaningful even in non-perturbative case - see, e.g., Landau-Zener transition.

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  • $\begingroup$ "Note however, that there is nothing wrong with measuring the states of Hamiltonian $H_0$, even though the eigenstates of the system are not the eigenstates of this Hamiltonian - it simply means that our measurement operator commutes with $H_0$, but not with the full Hamiltonian." So practically we don't do energy measurements which would create the problem I described above but we measure another quantity that commutes with the original hamiltonian (e.g. angular momentum for a central potential)? $\endgroup$ Nov 9, 2022 at 13:38
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    $\begingroup$ @Feynman_00 that's correct. But note again that in perturbative treatment this doesn't make any difference to the required order of precision. $\endgroup$
    – Roger V.
    Nov 9, 2022 at 13:43
  • $\begingroup$ If I understand your last remark, you're saying that although there is a conceptual difference, it practically makes no difference numerically for weak perturbations, right? One more thing, could you clarify what you mean here "one usually does not talk about $|c_i(t)|^2$ as a transition probability, but rather something like probability to find system in state $i$ at time $t$"? this sounds like the same thing. $\endgroup$ Nov 9, 2022 at 14:05
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    $\begingroup$ @Feynman_00 Transition usually means that we have an initial and final state - that is, we have performed a measurement and found the system in state $i$. On the other hand, $|c_i(t)|^2$ is defined for all times, regardless of when we will measure it. In fact, it is useful to look at it from the point of view of density matrix: its diagonal elements are true probabilities... it is just that we cannot neglect the non-diagonal elements. $\endgroup$
    – Roger V.
    Nov 9, 2022 at 14:25

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