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In Quantum Mechanics, a state vector $|\psi\rangle$ will evolve in time according to $$|\psi(t)\rangle=e^{-\frac{i}{\hbar}\hat H t}|\psi(0)\rangle$$ Imagine we have a system such that, for a short period of time $T$, the Hamiltonian increases by a constant and then returns to normal, such that $$\hat H=\hat H_0+ \begin{cases} 0 & \text{($t\lt0,\, t\gt T$)}\\ A & \text{($0\leq t\leq T$)}\\ \end{cases} $$ At $t=T$ we will have $$|\psi(T)\rangle=e^{-\frac{i}{\hbar}AT}e^{-\frac{i}{\hbar}\hat H_0T}|\psi(0)\rangle$$ Now, following the first equation, since after $t=T$ there is no $A$, it should just become $$|\psi(t)\rangle=e^{-\frac{i}{\hbar}\hat H_0 t}|\psi(0)\rangle$$ But this seems strange, it's as if that period of interaction with whatever caused the extra energy had no effect on the particle whatsoever. I think it makes more sense to apply the time evolution operator separately and obtain $$|\psi(t)\rangle=\hat U(t-T)|\psi(T)\rangle=e^{-\frac{i}{\hbar}\hat H_0(t-T)}e^{-\frac{i}{\hbar}AT}e^{-\frac{i}{\hbar}\hat H_0T}|\psi(0)\rangle=e^{-\frac{i}{\hbar}\hat H_0t}e^{-\frac{i}{\hbar}AT}|\psi(0)\rangle$$ Is my idea wrong or is the time evolutikon operator different in this case? If so, the what would be the case for a time-dependent Hamiltonian?

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marked as duplicate by Aaron Stevens, ZeroTheHero, Buzz, Qmechanic quantum-mechanics Dec 18 '18 at 7:14

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Your first equation does not hold for a time dependent Hamiltonian.

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    $\begingroup$ "Is my idea wrong ... in this case? If so, then what would be the case for a time-dependent Hamiltonian?" $\endgroup$ – Aaron Stevens Dec 18 '18 at 3:38
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    $\begingroup$ "Read the question carefully. What, specifically, is the question asking for? Make sure your answer provides that – or a viable alternative. The answer can be “don’t do that”, but it should also include “try this instead”. Any answer that gets the asker going in the right direction is helpful, but do try to mention any limitations, assumptions or simplifications in your answer. Brevity is acceptable, but fuller explanations are better. " $\endgroup$ – Aaron Stevens Dec 18 '18 at 4:36
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    $\begingroup$ All votes are subjective. Based on the entire section you reference (not just part of a sentence) I believe the answer could be better with barely any additional effort. Hence the -1. It's not personal, I just don't think it's a sufficient answer. My first comment was to let you know why do I didn't just leave you wondering why someone gave you a down vote. $\endgroup$ – Aaron Stevens Dec 18 '18 at 4:36
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    $\begingroup$ Unfortunately as written now this post is not an answer but a comment. Methinks it should be developed but with so little details it’s not super useful, although as a comment it is entirely correct, $\endgroup$ – ZeroTheHero Dec 18 '18 at 4:52
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    $\begingroup$ This is silly. You may be technically right in the narrow sense so take your win if you want, but it would still benefit from being expanded, although this would repeat an answer posted in a duplicate to the original question. $\endgroup$ – ZeroTheHero Dec 18 '18 at 4:58

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