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We define the thermal density operator as

$$\tau(\beta) = \frac{e^{-\beta H}}{\mathrm{Tr}(e^{-\beta H})}$$

where $H$ is the systems Hamiltonian. The thermal state is characterized by the fact that it maximizes the entropy for a given, fixed energy.

Now when we consider the second law of thermodynamics which states that

$$\Delta S \geq 0$$

which states that over time the entropy of systems either remains the same or increases.

Now I know that equilibration/thermalization is a process which is not yet well understood on the quantum scale. But if we would forget about this fact and just say that yes, quantum systems equilibrate aswell would this mean that they equilibrate towards the thermal state?

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There is a mistake in your description.

The thermal state is characterized by the fact that it maximizes the entropy for a given, fixed energy.

That statement is true for micro-canonical ensemble. The thermal density matrix $\rho = \frac{e^{-\beta H}}{{\rm Tr}\, e^{-\beta H}}$ is for canonical ensemble.

For most of the quantum systems, because of the interaction between the environment and the system (S+E), when $t$ is large enough the density matrix of the system $\rho_S = {\rm Tr}_E \rho \sim e^{-\beta H}$. The total density matrix $\rho$ is still evolving under a unitary transformation, but its partial trace is not.

In fact we do know some system do not have such property. It's called localization phase. In the article arXiv: 1404.0686, the authors claimed that:

We then focus on a class of systems which fail to quantum thermalize and whose eigen- states violate the ETH: These are the many-body Anderson localized systems; their long-time properties are not captured by the conventional ensembles of quantum statistical mechanics. These systems can locally remember forever information about their local initial conditions, and are thus of interest for possibilities of storing quantum information.

Hope these will be helpful.

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