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The Density matrix of a quantum system in the microcanonical ensemble is given by the \begin{equation} \rho_S = \frac{1}{dim \mathcal{H}_S} \mathbb{1}, \end{equation} which is a mixed state.This system can be thought of as a part of a bigger system $S-\epsilon$ (i.e. the purification of $\rho_S$) with Hilbert space $\mathcal{H}_{S-\epsilon}$ and (pure) state $|\Psi_{S-\epsilon}\rangle$ s.t. $tr_{\epsilon} |\Psi_{S-\epsilon}\rangle\langle\Psi_{S-\epsilon}| = \rho_S$. Since $\rho_S$ is of the form above, we can conclude that the system, $S$, is maximally entangled with the environment, $\epsilon$ (i.e. every quantum system in the microcanonical ensemble is maximally entangled with its environment).

This result is a bit strange to me and I don't seem to be fully understanding it. I was wondering if anyone ever came across the same question or has seen it being mentioned in a paper or has any opinion about it.

Note: We can regard the Hilbert space, $\mathcal{H}_S$, as the physical corner of the Hilbert space of the system spanned by the allowed energy eigenstates (i.e. only eigenstates with energy equal to $\bar{E}$).

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Your definition of the density matrix for a state in the microcanonical ensemble is incorrect. The microcanonical $\rho$ isn't actually proportional to the identity; it's proportional to $\sum_i' |E_i\rangle\langle E_i|$, where the $|E_i\rangle$ are energy eigenstates of the Hamiltonian and the sum $\sum_i'$ means that we are only summing over eigenstates whose energies either lie exactly at a fixed energy $\bar{E}$, or within a very narrow range of energies $[\bar{E} - \epsilon, \bar{E} + \epsilon]$ (possibly weighted), depending on the application. See here. Typically only an infintesimally tiny fraction of the energy eigenstates lie within this range, so the microcanonical ensemble is nowhere near maximally entanged. Indeed, in a finite system (which necessarily has discrete energy levels) with no degeneracy, the microcanonical ensemble is just an energy eigenstate and so is actually a pure state.

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  • $\begingroup$ Thanks for your answer. Although I agree with the last part of your answer (i.e. for finite non-degenerate systems, $\rho$ is pure, hence It's not maximally entangled), I think we can always confine ourselves to the physical corner of the Hilbert space spanned by $|\bar{E}_i\rangle$ where $\rho$ has non-zero diagonal elements. $\endgroup$ – B. T. Nov 24 '17 at 19:35
  • $\begingroup$ @BehradTaghavi The subspace of the Hilbert space spanned by the energy eigenvalues in a narrow energy window is not "the physical corner of the Hilbert space". The energy eigenstates are usually extremely nonlocal and unphysical, in the sense that it's experimentally extremely difficult to prepare a realistic system in an energy eigenstate (other than the ground state). By only considering the subspace containing those eigenstates, you typically lose the entire tensor product structure of the full Hilbert space, which in turn destroys the notion of locality. This is very unphysical. $\endgroup$ – tparker Nov 24 '17 at 20:00
  • $\begingroup$ @BehradTaghavi Moreover, it's not actually very important that the microcanonical density matrix is an equal-weighted mixture of the eigenstates in the narrow energy range. The important part is just that the energy range be narrow - physically important quantities usually end up being independent of the specific choice of weight function within that range. As the Wiki article says, different choices of weight function $f$ are possible, as long as they fall off to zero at large argument. $\endgroup$ – tparker Nov 24 '17 at 20:05
  • $\begingroup$ So the choice of the maximally mixed state over the subspace of eigenstates within the energy window is purely a matter of convenience, because it's easy to define. People often choose non-maximally-mixed states instead for calculational convenience. $\endgroup$ – tparker Nov 24 '17 at 20:06

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