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I'm considering a quantum system with $N$ bosonic degrees of freedom labelled by pairs $(x^i,p^i)$, such that the Hilbert space is $\mathcal{H}=L^2(\mathbb{R}^N)$. Given an arbitrary quantum state $|\psi\rangle$, I can compute its 2-point function \begin{align} C^{ab}=\langle\psi|\xi^a\xi^b|\psi\rangle\,, \end{align} where $\xi^a$ labels $(x^1,\cdots,x^N,p^1,\cdots, p^N)$. There always exists a (possibly mixed) Gaussian state $\rho$, such that the $2$-point function of $\rho$ is given by $C^{ab}$, this means $\mathrm{tr}(\rho\xi^a\xi^b)=C^{ab}$.

It is a well-known fact that the entanglement entropy $S_A(|\psi\rangle)$ associated to a subsystem $A$ is bounded from above by the entropy $S(\rho_A)$ of $\rho_A$, where $\rho_A$ is the Gaussian state restricted to the subsystem. This means, we always have \begin{align} S(\rho_A)\geq S_A(|\psi\rangle) \end{align}

What I'm looking for: Is there a known relation that allows me to bound $S_A(|\psi\rangle)$ from below? I'm looking for something like \begin{align} S_A(|\psi\rangle)\geq S(\rho_A)-f(|\psi\rangle)\,, \end{align} where $f$ can be computed from the higher $n$-point functions of $|\psi\rangle$ or something similar.

Why I'm interested in this: I know a class of systems for which $S(\rho_A)$ increases linearly in time and from numerical computations, I find evidence that also the true entanglement entropy of an arbitrary initial state $S_A(|\psi\rangle)$ grows linearly. So far, I could only prove it for Gaussians, but such an inequality would allow me to generalize the result...

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  • $\begingroup$ There is no reason why entropy should always grow linearly in time. (It's very good that you write why you are interested in this, but you might want to be more specific.) -- Also, your initial question is underspecified. Such as $f$ obviously trivially exists by setting the two sides equal. $\endgroup$ – Norbert Schuch Feb 23 '17 at 22:43
  • $\begingroup$ Thank you for your comment. Essentially, I'm looking for a lower bound on the entanglement entropy that can be easily computed from the $n$-point functions. Setting both sides equal will give such an $f$ in principal, but is very difficult to compute. $\endgroup$ – LFH Feb 24 '17 at 2:24
  • $\begingroup$ I still think that with more details from your actual proof, you might have a better chance of getting something which is useful for you. $\endgroup$ – Norbert Schuch Feb 26 '17 at 15:11

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