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  Entanglement entropy for stabilizer states

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A stabilizer state for a stabilizer set S for a system of qubits can be written as $\rho=\frac{1}{2^{n}}\sum_{g\epsilon S}g$ . If we take a bipartition A-B of our system and partial trace over A, we get $\rho_{B}=\frac{1}{2^{n_{B}}}\sum_{g\epsilon S_{B}}g$ where $S_{B}$ is the subet of elements in S which are non-zero when traced over A . Entanglement Entropy of $\rho_{B}$ is proven as equal to $n_{B}-|{S_{B}}|$ where $|{S_{B}}|$ is the rank of $S_{B}$ or the size of its minimal generating set. Its trivial to see the first part of the result $n_{B}$ but the complete result $n_{B}-|{S_{B}}|$ requires to prove that ${Tr}[\frac{1}{2^{n_{B}}} {\sum}_{g\epsilon S_{B}}g (\log{\sum}_{g\epsilon S_{B}}g)]=|{S_{B}}|$ . Is there an easy way to prove this ? If it helps, I found this result here http://arxiv.org/abs/quant-ph/0406168

This post imported from StackExchange at 2015-10-14 16:58 (UTC), posted by SE-user user56199

asked Sep 5, 2015 in Theoretical Physics by anonymous [ revision history ]
edited Oct 14, 2015 by Dilaton

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