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If a quantum state is an eigenstate of the Hamiltonian, then it is stationary. But can a state be stationary if it is not an eigenstate of the Hamiltonian? If yes, how can one prove whether a state is stationary?

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    $\begingroup$ Stationary states are defined to be eigenstates of the Hamiltonian. So your question is saying "If A, then A. But can we have A, if not A?" $\endgroup$
    – knzhou
    Jul 19, 2016 at 23:38
  • $\begingroup$ @knzhou I thought so, but in another question I asked, someone told me "So the state is stationary if it is an eigenstate. But this is not the only possibility for stationarity." so I'm in doubt now. $\endgroup$
    – Tendero
    Jul 19, 2016 at 23:40
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    $\begingroup$ That answer is wrong, in several places. $\endgroup$
    – knzhou
    Jul 19, 2016 at 23:42

1 Answer 1

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A state $\Psi(x,t)$ is stationary if

$$\mid\Psi(x,t)\mid^2=\mid\Psi(x,0)\mid^2$$

This means that the time dependence of $\Psi(x,t)$ must be in the form

$$\Psi(x, t) = \Psi(x, 0) \ e^{i \phi t}$$

but you also have (Schrödinger's equation)

$$i \hbar \partial_t \ \Psi(x,t) =\hat H \ \Psi(x,t)$$

so that we obtain

$$(i \hbar)(i \phi) \Psi(x,t) =\hat H \ \Psi(x,t) \\ \to-\hbar \phi \Psi(x,t) =\hat H \ \Psi(x,t)$$

i.e. $\Psi(x,t)$ is an eigenstate of $\hat H$ with eigenvalue $-\hbar \phi$. This means, incidentally, that $\phi$ must have the dimensions of energy/action, that is to say

$$\phi=\frac E \hbar$$

So if a state is stationary, it is an eigenstate of $\hat H$. Since we know that the reverse is true, we can conclude that a state is stationary if and only if it is an eigenstate of $\hat H$.

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  • $\begingroup$ You're answer is good but you might want to add why $\Psi(x, t) = \Psi(x, 0) \ e^{i \phi t}$ is true. Don't assume people know that. $\endgroup$
    – Gert
    Jul 20, 2016 at 2:17
  • $\begingroup$ Yeah, you pretty much used circular reasoning. From $|\Psi(x,t)|^2=|\Psi(x,0)|^2$ immediately comes only that $\Psi(x,t)=e^{i\alpha(x,t)}\Psi(x,0)$. Why it should be just a $\phi t$ demands proof... I'm going to present short proof I've made up now in my answer though I'm sure there may be much more elegant way. $\endgroup$
    – OON
    Jul 20, 2016 at 10:33
  • $\begingroup$ Arr, no, I've stupid mistake, so I'll hide that for now $\endgroup$
    – OON
    Jul 20, 2016 at 10:53

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