Eigenstate thermalization hypothesis (in progress)

 Why statistical mechanics works for classical physics and why it should not work for quantum physics

Source: Wikipedia

The assumption of statistical physics in a many body (10^23 particles) isolated classical system with given fixed energy is that every micro-state is equally probable. Why should this be the case? Because a system with 10^23 particles is highly likely to be chaotic. And a chaotic system will in general spend equal time in equal areas of its phase space. Hence, the expectation value of an observable will be the average of the expectation value corresponding to each of the micro-state.

$$\bar{A}_{\mathrm{classical}} = \frac{1}{N}\sum_{i=1}^N A_i$$

But in the quantum case, there appears to be a problem. Even for a many body quantum system, if we go in the energy eigenbasis, then time evolution of the state just rotates the phases in the energy eigenbasis. That is, if the system is prepared an eigenstate, it will stay there.

Hence, long time expectation value of a general observable depends upon the initial state of the system. This does NOT happen in the classical case. See the following calculation.

$$|\psi(0)\rangle=\sum _\alpha c_\alpha |E_\alpha \rangle$$

$$\langle \hat{A}  \rangle_t \equiv \langle \Psi(t) | \hat{A} |\Psi(t) \rangle = \sum_{\alpha, \beta} c_{\alpha}^{*} c_{\beta} A_{\alpha \beta} e^{-i \left ( E_{\beta} - E_{\alpha} \right )t / \hbar }$$

The time average is then,

$$\langle \hat{A}  \rangle_t \overset{t\to\infty}{\approx} \sum_{\alpha} \vert c_{\alpha}\vert^2 A_{\alpha \alpha}$$

This is surprising because on one hand, the expectation value of general observable depends on the initial state of the system and on the other hand, we intuitively know that a sub system of this many body state is going to thermalize in finite time for a large class of initial state, and hence so will thermalize the expectation values of the local operators on this subsystem.

How to resolve this paradox?