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Many recent papers study entanglement in eigenstates of fermionic free hamiltonians (normally on a lattice) using the basic assumption that the reduced density matrices are thermal (e.g. Peschel 2003). As I understand it, the theorem that we need in order to proceed with the computation of entanglement is the following:

Consider a fermionic state, pure or mixed, which fulfills Wick's theorem (i.e.: all N-point functions can be expressed from the 2-point correlation function). Then, the reduced density matrix for a block can always be expressed as $\rho_B \propto \exp(-\beta H)$, where $H=\sum_k c_k d_k^\dagger d_k$. The $d_k$ and $d^\dagger_k$ are suitable fermionic operators which can be expressed as linear combinations of the original ones.

The reference given is to Gaudin 1960. But that article only contains the reverse theorem: if a fermionic density matrix is thermal, then Wick's theorem follows.

So, my main question: Does anybody know about a proof of this theorem?

Moreover: I guess this theorem is related to the thermalization theorem of QFT on curved spacetimes. Is this true?

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1 Answer 1

This will be a sketch but I think that the completion to a full proof is straightforward.

First, the conclusion of the theorem is that $\rho=C\exp(-\beta H)$ where $H=\sum\dots$

The latter statement may be rephrased by saying that $$ H = \sum_{K,L} c_{KL} d^\dagger_L d_L $$ i.e. the Hamiltonian is the most general bilinear function of the original fermionic operators (preserving the number of excitations, i.e. containing one operator with a dagger and one without it). Your form is nothing else than the diagonalization of mine.

However, the former statement is pretty trivial: one may always write $\rho$ as the exponential of some operator $L=-\beta H$: $L$ is just the logarithm of $\rho$ which is calculable (at least for most $\rho$).

Now, the assumption of the theorem says something about all $N$-point functions. In particular, it addresses 2-point functions. From those 2-point functions, one may extract the coefficient $c_{KL}$ of the operator $H$ related to $\log(\rho)$. The 2-point functions which have either 0 or 2 daggered operators instead of 1 must identically vanish, otherwise Wick's theorem is violated immediately, contradicting the assumptions of the theorem we want to prove.

The reverse theorem tells you that the thermal density matrix will preserve the Wick's theorem. The only additional thing you have to notice is that if the logarithm $L$ or $H$ failed to be bilinear in the operators, i.e. if there were additional terms in a Taylor expansion (which always terminates for a finite number of fermionic operators), that would modify the higher-point functions, and because Wick's theorem unambiguously determines all higher-point functions in terms of 2-point functions, Wick's theorem would be violated.

The thermalization theorem (Fulling-Davies theorem) I know says that the vacuum state of a quantum field theory may be rewritten in terms of the left and right Rindler spaces' Hilbert spaces, as a thermal-like entangled state (with Boltzmann's coefficients).

It is not clear why you think that this thermalization theorem has any relationship with a proof of Wick's theorem. Wick's structure of $N$-point functions is always linked to thermality, because of the theorem we started with, but this link doesn't seem to have anything to do with the decomposition into Rindler spaces. So I think that your guess is not right.

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Luboš, thanks for your answer, but it is not enough. Of course, the density matrix will be the logarithm of "something" (unless it has some zero eigenvalues). The fact that this "something" has this particular form is non-trivial for me, and requires more work. –  Javier Rodriguez Laguna Jan 31 '12 at 17:44
    
About the relation to the Fulling-Davies theorem, the link that I see is the following. You only get thermalization when a horizon appears (Rindler, Schwartzschild, etc.), which means that you're tracing out some degrees of freedom: those linked to the regions of spacetime that are now disconnected from you. Thermalization is due to this "tracing out", right? –  Javier Rodriguez Laguna Jan 31 '12 at 17:45

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