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I'm trying to understand the eigenstate thermalization hypothesis more and I keep coming across a limit I don't understand.

If the initial state of a system is in it's energy eigenstate basis as $|\Psi(0)\rangle = \sum\limits_{\alpha} c_{\alpha} | E_\alpha \rangle$, then by unitary time evolution $ | \Psi(t) \rangle = \sum\limits_{\alpha} c_{\alpha} e^{-E_\alpha i t/\hbar} | E_\alpha \rangle$. Therefore the expectation of any observable at any time is

$$\langle \hat{A} \rangle_t = \sum\limits_{\alpha,\beta} c_{\alpha}c_{\beta}^* \ A_{\alpha,\beta} \ e^{-(E_\alpha-E_\beta) i t/\hbar} \, .$$

This makes sense. However, as a few papers I've read and Wikipedia have stated that when you take the long time limit this converges to

$$\langle \hat{A}\rangle_{t \to \inf} = \sum\limits_\alpha |c_\alpha|^2 A_{\alpha,\alpha} \, .$$

Why is this the case? Should $\ e^{-(E_\alpha-E_\beta) i t/\hbar}$ oscillate forever, making the expectation value never converge? I suspect I am just missing some math, but this seems to imply that in the long time limit, the expectation only depends on the diagonal elements of the observable.

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You just need to keep scrolling down in the Wikipedia article, I will just add the mention of the Wick rotation.

First of all this is a hypothesis, so you can understand it as a "well motivated guess", the reasoning coming from (as stated in the Wiki): $$\bar{A} = \lim_{T\rightarrow \infty} \frac{1}{T} \int_0^T {\rm d}t\, \langle \psi(t)|\hat{A}|\psi(t)\rangle$$

To make sense of the limit a way to understand it is by performing a Wick rotation, that is $it\rightarrow \tau$, which is not very emphasized in the article but is usually encountered when making contact with thermal physics. After you plug in the expression you have for $\langle\hat{A}\rangle_t=\langle\psi(t)|\hat{A}|\psi(t)\rangle$, you get a split into a diagonal part and extra terms that fall of as $1/\tau$.

So the claim is that $\bar{A}$ becomes diagonal and retains some memory of the initial state, however the "raw" expectation value, $\langle\hat{A}\rangle_t$, does not.

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