# On the Eigenstate Thermalization Hypothesis: what does "energy eigenstates are thermal" mean?

I'm trying to understand what exactly is the ETH, and miserably failing. Here's what I'm reading everywhere: isolated systems are supposed to thermalize, hence "forget about their initial condition", but the infinite time average of an observable $$O$$ is

$$\langle O\rangle_\infty =\int_0^\infty dt \langle\psi| e^{iHt}Oe^{-iHt}|\psi\rangle=\sum_\alpha |c_\alpha|^2 \langle\alpha|O|\alpha\rangle$$

where $$|\psi\rangle=\sum_\alpha c_\alpha |\alpha\rangle$$ and the $$|\alpha\rangle$$ are the energy eigenstates. This is a problem, because it means that expectation values of observables depend on the initial conditions at all times. A solution to this is supposing that the energy eigenstates are thermal, the trouble is, in every paper on the subject people seem to avoid explaining exactly what they mean by thermal. Most people say something like

$$\langle \alpha|O|\alpha\rangle=\langle O\rangle_{\mathrm{mc}}$$

where $$\langle O\rangle_{\mathrm{mc}}$$ is the microcanonical average. I have two main problems with this

• as far as I know $$\langle O\rangle_{\mathrm{mc}}$$ is something like$$\langle O\rangle_{\mathrm{mc}}=\lim_{\Delta E\to 0}\frac{1}{N(E,\Delta E)}\sum_{E_\beta\in(E_{\alpha}-\Delta E, E_{\alpha}+\Delta E)} O(E_\beta)$$ where $$O(E_\beta)$$ is the observable in the energy $$E_\beta$$, which surely is just $$\langle \beta| O|\beta\rangle$$. It seems then to me that $$\langle \alpha|O|\alpha\rangle=\langle O\rangle_{\mathrm{mc}}$$ is a tautology, especially with discrete energy levels.

• Even if that made sense, the microcanonical average still depends on the energy, so I don't see how we get rid of the initial condition dependence in the time average, which now is just $$\sum_{\alpha}|c_\alpha|^2 \langle O\rangle_{\mathrm{mc}}(E_\alpha)$$

Other people say something I can understand a little better, they say that if $$O$$ is local on a region of the lattice $$A$$, then the average must look like a Gibbs ensemble thermal average.

$$\langle \alpha|O_A|\alpha\rangle=\mathrm{Tr}\left(O_A e^{-\beta H_A}\right)$$

but this has the same problem as before, the time average still depends on the initial conditions. Moreover, I don't understand what is the difference between considering microcanonical averages and Gibbs averages.

TL;DR: what does "the energy eigenstates are thermal" mean?

• The ETH can be stated as an ansatz for the matrix elements of local operators in the energy eigenbasis. For diagonal elements, the ansatz is $\langle E_{\alpha} | O | E_{\alpha} \rangle = f(E_{\alpha})$, where $f$ is a smooth function. Note that the smoothness and the dependence only on energy is not a given, and indeed it fails in certain cases (e.g. many-body localized systems). This smooth dependence of diagonal matrix elements only on energy is the resolution to the dependence on initial conditions. Jun 8 '21 at 16:05
• Specifically, suppose the initial state has "well-defined energy", in the sense that it has overlap only with energy eigenstates in some small energy window $E_{\alpha} \in [E - \Delta, E + \Delta]$. Then we can approximate $\sum_{\alpha} |c_{\alpha}|^{2} \langle E_{\alpha} | O | E_{\alpha} \rangle \approx \sum_{\alpha} |c_{\alpha}|^{2} f(E) = f(E)$, where the first step uses smoothness in a small energy window, and the second step follows from state normalization $\sum_{\alpha} |c_{\alpha}|^{2} = 1$. Hence we recover the prediction of equilibrium stat-mech. Jun 8 '21 at 16:05

A thermal state describes a system at non-zero temperature. For "temperature" to have a thermodynamic meaning, as usual, you assume your system in in thermal equilibrium with a large reservoir at temperature $$T$$. So thermal states are the states of a system in thermal equilibrium with a reservoir.

I find the best way to introduce people to the topic is the intro paragraph of this paper, Thermalization and its mechanism for generic isolated quantum systems:

If we pierce an inflated balloon inside a vacuum chamber, very soon we find that the released air has uniformly filled the enclosure and that the air molecules have attained the Maxwell velocity distribution, the width of which depends only on the total number and energy of the air molecules. Different balloon shapes, placements, or piercing points all lead to the same spatial and velocity distributions.Classical physics explains this ‘thermodynamical universality’ as follows: almost all particle trajectories quickly begin to look alike, even if their initial points are very different, because nonlinear equations drive them to explore the constant-energy manifold ergodically, covering it uniformly with respect to precisely the microcanonical measure. However, if the system possesses further conserved quantities that are functionally independent of the hamiltonian and each other,then time evolution is confined to a highly restricted hypersurface of the energy manifold. Hence, microcanonical predictions fail and the system does not thermalize.

So basically systems that thermalise are also ergodic (they explore the available phase space so that their spatial and temporal averages look the same and are independent of the initial state), but if there are conserved quantities then these systems would be integrable systems that cannot thermalise and that remain limited to a subspace of the available phase space.

The ETH essentially says that you do not need to know exactly what state $$|\Psi\rangle$$ the system ends up being, because to compute the average of an operator $$\langle \hat A \rangle$$ you can just the statistical mechanical ensemble: $$\langle \Psi |\hat A|\Psi \rangle = \langle \hat A \rangle_{\text{ensemble}}.$$

Now from this you can ask all your quantitative questions about initial conditions etc. Which cannot really be answered because the ETH is a hypothesis with no definite proof (yet).

However, if you read the paper I link above, they do provide an explanation on how the system "forgets" about the initial conditions. They claim that an initial state in quantum mechanics is actually a superposition of thermal eigenstates, and that it's only the coherences between the latter that make it look like "non-thermal". With time, dephasing causes these coherences to wash away giving you a thermal state:

They provide numerical examples (both for thermal and integrable systems) along with a qualitative discussion of how the coherences tend to 0 because the off-diagonal entries of $$\hat A$$ end up being much smaller than the diagonal ones: $$A_{\alpha \beta} \ll A_{\alpha\alpha}$$.

I conclude by saying that the ETH rose to prominence with the advent of disorder-induced localised phases of matter, like Anderson localisation and Many-Body localisation (MBL) where thermodynamics does not work and hence all the usual phase-transition tell-signs fail. The ETH here has been proven to be a good discriminant between a thermal (ETH works) and localised (ETH fails) state.

• Thanks for the answer. I guess my problem is more about how these thermal averages $\langle A \rangle_{\mathrm{ensemble}}$ are defined Nov 14 '20 at 17:33