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I've often read that entanglement entropy in quantum field theory is ill-defined because local algebras are generally of type III, which implies that a trace doesn't exist. For a normal state $\omega_{\rho}$ in the folium of an arbitrary state $\omega$, we can consider the GNS triple $(H, \pi, \Omega)$ for $\omega$ and a density matrix $\rho \in B(H)$. The state is given by

$$ \omega_{\rho}(A) = \langle \Omega,\rho \pi(A)\Omega \rangle $$

for all $A \in \mathcal{A}$. Suppose we have a type III von Neumann algebra. The entropy would then be

$$ S(\omega_{\rho}) = -\operatorname{Tr}(\rho \log(\rho)). $$

As far as I understand, the algebras are associated with observables, not states (correct me if I'm wrong). So why is the type of algebra important? It seems to me that this should not affect the states. While $A$ is not trace class, $\rho \log(\rho)$ might be, because in principle $\rho$ doesn't need to be in $\mathcal{A}$.

Can someone clarify this point for me? With references if you have.

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  • $\begingroup$ I'm not terribly familiar with this, so take the following with a grain of salt. However, Srednicki's paper on entropy of quantum fields shows that the entanglement entropy requires regularization even for the case of free fields, where a region of linear dimension $R$ has an area-law $\propto (R/a)^2$ where $a$ is a lattice spacing that functions as an ultraviolet cutoff. $\endgroup$
    – user196574
    Commented Aug 2 at 18:29
  • $\begingroup$ My heuristic understanding is that we can view quantum field theories as infinite numbers of harmonic oscillators, and we need to have a finite number of oscillators to have a finite entropy, so regulators necessarily come into play. $\endgroup$
    – user196574
    Commented Aug 2 at 18:29
  • $\begingroup$ Yes, I know the physics behind ultraviolet divergences when using a cut-off, and it is exactly what you mentioned. However, I am looking for an answer within AQFT. Basically, I am interested in knowing whether the density matrix of a given state should belong to the type III algebra associated with the observables. Nowhere have I found anyone affirming this, but everyone mentions that the trace is not well-defined.... I understand that it is not well-defined for operators representing observables, but not for the density matrix of a state. $\endgroup$
    – Gabriel Palau
    Commented Aug 2 at 20:36
  • $\begingroup$ Thanks for the perspective! I'm afraid I won't be able to comment anything of use because of lack of knowledge of AQFT or even von Neumann algebras. Is at least $\rho$ guaranteed to be trace class? As someone who only works with finite-dimensional Hilbert space, I feel I would require that of a density matrix. $\endgroup$
    – user196574
    Commented Aug 3 at 18:35

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