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As discussed here, the quantum state $S$ that corresponds to the solution $\psi(t,x)$ of a time-independent Schrödinger equation remains constant wrt. time. I assume that the the solution of a time-dependent Schrödinger equation may then produce a state $S_t$ that does depend on time.

What is the physical significance of $S$ versus $S_t$? More specifically, does (in some sense) $S_t \to S$ as $t \to \infty$? Heuristically, this would seem logical, since states like $S$ are occasionally described to be "standing waves".

A motivating example: Consider a particle in a box, whose time independent states are well known. Measuring the location of the particle would collapse the wave function which produces a state that is not an eigenstate of the Hamiltionian. The particle is then again delocalised, hence, the state clearly also evolves wrt. time. Question 1: Is this evolution $S_t$ a solution to a time-dependent Schrödinger equation? Question 2: Does $S_t$ approach an eigenstate of the Hamiltionian as $t \to \infty$?

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  • $\begingroup$ The answer to question 1 is yes; that's the idea of the time-dependent Schrödinger equation, it gives the time evolution of a system in arbitrary state. For Question 2, I don't have time to write a full answer, but, seeing you are a PhD student in mathematics, I guess you can work with these: 1. The initial state (in your example, after the wavefunction collapse) can be written as superposition of eigenstates of the Hamiltonian. 2. The time-dependent Schrödinger equation is linear, so the answer can be written as a linear combination of answers for each eigenstate. ... $\endgroup$
    – JiK
    Sep 13, 2016 at 10:11
  • $\begingroup$ ... 3. If you write the answer, you'll see that there's no limiting behaviour; the state remains a similar superposition of the eigenstates, only the phases, and not magnitudes, of the constants change. $\endgroup$
    – JiK
    Sep 13, 2016 at 10:11
  • $\begingroup$ @JiK: Thank you for you comments. However I'm a bit confused. Wouldn't "the state remains a similar superposition of the eigenstates, only the phases, and not magnitudes, of the constants change" imply that the state remains constant. If so, I don't see how the distribution of the position observable could change in time, since the distributions of observables are determined by the state. $\endgroup$
    – Matias Heikkilä
    Sep 13, 2016 at 10:24
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    $\begingroup$ It doesn't imply that. The quantum states described by wave functions $c_1 \phi_1 (x) + c_2 \phi_2 (x)$ and $e^{i\varphi} c_1 \phi_1 (x) + e^{i\varphi} c_2 \phi_2 (x)$ are same, but the state $e^{i\varphi_1} c_1 \phi_1 (x) + e^{i\varphi_2} c_2 \phi_2 (x)$ is different in general. $\endgroup$
    – JiK
    Sep 13, 2016 at 10:26

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