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In the research field of Many-body Localization (MBL), people are always talking about the eigenstate thermalization hypothesis (ETH). ETH asserts that for a isolated quantum system, all many-body eigenstates of the Hamiltonian are thermal, which means all sub-systems can involve to thermalzation in the end. ETH is not always true and violation to it means MBL for an interacting quantum many-body system. Well, my puzzle is as follows:

For a isolated quantum system $A$ and a space-specified sub-system $B\in{A}$. It is assumed the initial state of $A$ is one of the eigenstates $|\psi(t=0)\rangle_{A}$ of its Hamiltonian $H$. Of course it is a pure state. Note the the initial state $|\psi(t=0)\rangle_{B}$ of $B$ is not a pure state unless $|\psi(t=0)\rangle_{A}$ is the direct product state of $|\psi(t=0)\rangle_{B}$ and the state of $A/B$, which means there $B$ is disentangled with the rest part $A/B$. Since $B$ is chosen arbitrarily, mixed initial state of $B$ is the most general case and its state cannot be described by a single state but a density matrix $\rho_{B}(t=0)$.

Now let the system $A$ evolve along time. There are two ways to check $\rho_{B}$ at arbitrary time $t$.

1) I can partially trace $\rho_{A}$ by $\rho_{B}=\text{tr}_{A/B}\rho_{A}$. While $\rho_{A}=|\psi|\rangle_{A}\langle\psi|_{A}$ will not change because $|\psi\rangle_{A}$ is the eigenstate and it will not evolve under the time evolution operator thus $\rho_{B}$ will not change forever.

2) The mixed state $\rho_{B}(t=0)$ evolves along time and it may thermalize to Gibbs density matrix $\tilde{\rho}_{B}=\frac{1}{Z}e^{-\beta{H}}$ where $Z$ is its statistical partition function. This is indeed the statement of ETH.

What's wrong for the paradoxical results viewed from two different perspectives for the same thing?

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3 Answers 3

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The simplest way to resolve this paradox is to require

$$\rho_B(t=0)\simeq \tilde{\rho}_B=\frac{1}{Z}e^{-\beta H_B}.$$

That is to say, if you start from an eigenstate, you do not need any time evolution to reach thermal equilibrium, because the eigenstate itself is already thermalized, and this is the statement of eigenstate thermalization hypothesis (ETH). This is very different from a classical chaotic system, where you need some time to explore the phase space and to reach equilibrium. For a quantum chaotic system, the system can serve as its own heat bath: although each eigenstate is a pure state, the reduced density matrix of a subsystem looks thermal.

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  • $\begingroup$ I see...Quantum Chaotic system does not need time to thermalize. But I wonder how general is this requirement you provided here? Does it imply that if ETH is valid, then this requirement is fulfilled? $\endgroup$
    – Wayne Zheng
    Commented Dec 7, 2015 at 2:20
  • $\begingroup$ Yes, this requirement is general for any ETH state, as long as the region B is small compared to the whole system. You may even think $\rho_B=Z^{-1}e^{-\beta H_B}$ as an equivalent statement of ETH. $\endgroup$
    – Everett You
    Commented Dec 7, 2015 at 20:33
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At the risk of being too obvious, let me first point out: any state that changes with time is not an eigenstate of the Hamiltonian. So if you are describing a system that is at equilibrium for all time, then you may indeed assume that system A is in an eigenstate of the many-body Hamiltonian, but for any other situation (including anything in the real world!) it must be in some superposition of these states.

So in most interesting cases, your first assumption is incorrect. In particular, if you start with an equilibrium state, then take subsystem B locally out of equilibrium to watch it relax back to a new equilibrium, whatever perturbation you apply is necessarily putting the total system into a superposition of several many-body eigenstates. Then it relaxes to a new equilibrium. However, since the system is isolated, energy was not lost and the system is still in a superposition of eigenstates. It must therefore still be evolving in time, but the difference is that this evolution is no longer changing any local values for $\rho_B$ (or any other subsystem) and so the system appears locally time-independent again.

So the point is this: in a system satisfying the ETH, all eigenstates of the Hamiltonian are thermal, but the converse is not true: there are many states that are not eigenstates of the Hamiltonian but are, nonetheless, still thermal states. I believe that when you consider this the "paradox" is resolved.

For more information, I recommend the introduction to this paper.

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    $\begingroup$ I am reading your recommendation paper. It is lucid. $\endgroup$
    – Wayne Zheng
    Commented Dec 7, 2015 at 2:46
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The initial state does not need to be one of eigenstates of the hamiltonian, it could be superposition. Therefore time evolution will change it. I don't think your first assumption is correct.

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  • $\begingroup$ This is the correct answer as the OPs question is the result of a misunderstanding of ETH. Strong ETH states that any pure state of the system evolves so that all subsystems look thermal. A trivial corollary of this is that any static solution to the equations of motion (read eigenstate) is already thermal. $\endgroup$
    – ComptonScattering
    Commented Dec 30, 2016 at 15:29

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