# Why does the quantum expectation value depend only on the diagonals of the Hamiltonian in the long time limit?

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.

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.