In this paper by Don Page, https://arxiv.org/pdf/gr-qc/9305007.pdf, He conjectures average entropy of a substem of dimension m with Hilbert space dimension mn, $m \leq n$. to be : $ S_{mn} = \sum_{n+1}^{mn} \dfrac{1}{k} - \dfrac{m-1}{2n} $. Suppose I have a spin chain with 8 spins. Then the dimension of the Hilbert space would be $ 2^8$. If I divide this into equal subsystems, then the split up would be $ m= 2^4, n=2^4$, Hence I would have to calculate $S_{16,16} = \sum_{17}^{256} \dfrac{1}{k} - \dfrac{15}{32}$. Unfortunately, this computation of the summation is giving different results than what was expected from numerical calculations. I think I am making some mistake with the dimension number or some silly mistake with the series. Can someone please point out what mistake I am making?

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    $\begingroup$ Equation (7) in that paper says $\langle \text{trace}(\rho_A^2)\rangle=(m+n)/(mn+1)$, where $\rho_A$ is the reduced density matrix for one subsystem. Have you tried computing $\langle \text{trace}(\rho_A^2)\rangle$ numerically to see if it agrees with equation (7)? $\endgroup$ – Chiral Anomaly Nov 12 '18 at 11:43

The formula you cite from Page is correct. Perhaps you are making a mistake when averaging over states? Have you checked that you averaged over enough states for the numerical average to have converged?

Let me illustrate/confirm Page's prediction for an $8$-site spin chain, where we bipartition it into a region A and B, each containing $4$ sites. There is no universal result for the entanglement entropy $S_A(|\psi\rangle)$ of a single random state $|\psi\rangle$, but Page predicts that if we average $S_A$ over random states (he notes he uses the Haar measure in his paper), then $\langle S_A \rangle$ should obey the formula you quoted. So I went ahead and did this for the $8$-site chain, obtaining:

enter image description here

More precisely, I average over only 2000 states (whose entropies are shown in the blue-ish histogram), yet there is already a nice approximate agreement between Page's prediction and the center-of-mass of this histogram: $$ S_\textrm{Page} \approx 2.275 \qquad \textrm{and} \qquad \langle S_A \rangle \approx 2.276 $$


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