Question on Eigenstate thermalization hypothesis

Question

My understanding of a system that satisfies the eigenstate thermalization hypothesis (ETH) is the following. If we consider a local operator, $O_i$ we can relate its expectation value with respect to some eigenstate, $|\epsilon\rangle$ of the Hamiltonian, $H$ to a thermal expectation value via a thermal density matrix, $e^{-\beta H}$.

$\langle\epsilon| O_i | \epsilon \rangle = \frac{1}{Z} Tr(O_i e^{-\beta H})$.

The inverse temperature is presumably defined implicitly by the consistency condition

$\epsilon = \frac{1}{Z} Tr(H e^{-\beta H})$.

Let us now look at a system with a bounded spectrum, like the Hamiltonian of a spin chain with the lowest energy $\epsilon_{min}$ and highest energy, $\epsilon_{max}$. I will assume that there are no degeneracies for simplicity. I want to simply understand what temperature corresponds to different energies as defined by the above equation. That is $\beta(\epsilon)$. Zero temperature, $\beta \rightarrow \infty$ corresponds to $\epsilon = \epsilon_{min}$ as expected. Infinite temperature, $\beta = 0$, corresponds to the mean energy, $\bar{\epsilon} = \frac{\sum_i \epsilon_i}{\sum_i 1}$.

My question is how should I think about the other eigenstates above $\bar{\epsilon}$. This seems to need negative temperatures which is unphysical. ($\epsilon_{max}$ corresponds to $\beta \rightarrow - \infty$). Or maybe the way I am thinking about how to assign temperature itself if wrong?

Answer 1

@DominicElse answer in the comment was great. Just a bit of add up:

Consider $DoS(\epsilon)$ as the density of state for your system.

If you chose an eigenstate in part of the spectrum where $dDoS(\epsilon)/d\epsilon > 0$, the temperature associated with that eigenstate is positive. Otherwise, in the region with a negative slope for $DoS(\epsilon)$, $T$ is negative.

In everyday life, we are not dealing with systems with an upper bound on its energy spectrum. So we assume that $DoS$ increases exponentially and then, we always correctly expect a positive temperature. But it is different in discrete systems.