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Given a set of projectors $\{P_i\}$ acting on a space $\mathcal H_S$, let $\Phi$ be the LCPT map defined by $$\Phi(\rho)=\sum_i P_i\rho P_i.$$ The goal is to show that $S(\Phi(\rho))\ge S(\rho)$. The Klein inequality $$S(\rho\Vert\Phi(\rho))=\text{tr}[\rho\log\rho]-\text{tr}[\rho\log\Phi(\rho)]\ge 0 $$ already implies $-\text{tr}[\rho\log\Phi(\rho)]\ge S(\rho)$, so all that is left is to show that $$S(\Phi(\rho))\ge-\text{tr}[\rho\log\Phi(\rho)].$$

So, using the spectral decomposition of $\rho$ I obtained $$\rho\log\Phi(\rho)=\rho\sum_{ij}\log(p_j)P_i|j\rangle\langle j|P_i=\sum_i\rho P_i\sum_j\log(p_j)|j\rangle\langle j|P_i $$ and by applying the cyclic property of the trace $$-\text{tr}[\rho\log\Phi(\rho)]=-\text{tr}\left[\sum_iP_i\rho P_i\sum_j\log(p_j)|j\rangle\langle j|\right]=-\text{tr}[\Phi(\rho)\log\rho].$$ However I realized something disquieting: the Klein inequality should also hold for $S(\Phi(\rho)\Vert \rho)$, implying $$S(\Phi(\rho))\le -\text{tr}[\Phi(\rho)\log\rho]=-\text{tr}[\rho\log\Phi(\rho)] $$ so either something has gone wrong with my calculations and the proof is done in a completely different way, or $S(\Phi(\rho))=-\text{tr}[\rho\log\Phi(\rho)]$ and I'm not seeing why. Either way, I'd appreciate some clarifying help!

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    $\begingroup$ What does LCPT stand for? $\endgroup$
    – Rammus
    Jul 26 at 7:08
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Firstly you have made a mistake in your calculations, note that $$ \sum_i P_i \log (\rho) P_i \neq \log(\sum_i P_i \rho P_i) $$ in general. However, your working appears to assume that this equality is true.

Nevertheless, an equality that is true and is particularly useful here is $$ \log(\sum_i P_i \rho P_i) = \sum_i \log (P_i \rho P_i). $$ Moreover each term on the RHS is an operator whose support is contained within the subspace that $P_i$ projects onto. The easiest way to see this is to work in the basis in which the $P_i$ are simulataneously diagonalized. Then the operator $\sum_i P_i \rho P_i$ is block diagonal in this basis. Furthermore, we also have $$ \begin{aligned} \log(\sum_i P_i \rho P_i) &= \sum_{i} P_i \log (P_i \rho P_i) P_i \\ &= \sum_{i,j} P_j \log(P_i \rho P_i) P_j \end{aligned} $$ where on the second line we used the fact that $\sum_{i,j} P_i P_j = \delta_{ij}$ for a set of orthogonal projectors.

Thus by the cyclic property of the trace we find that $$ -\mathrm{tr}[\rho \log \Phi(\rho)] = - \mathrm{tr}[\Phi(\rho) \log \Phi(\rho)] $$ and so by Klein's inequality the result follows.

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  • $\begingroup$ Could you elaborate on why $$\log\Big(\sum_iP_i\rho P_i\Big)=\sum_i\log(P_i\rho P_i) $$ holds? $\endgroup$ Jul 26 at 13:33
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    $\begingroup$ @There'sStrangeStuffOutHere Work in the basis in which the $P_i$ are simultaneously diagonalized then the sum is a block diagonal operator. The log of a block diagonal operator is the log of its blocks. $\endgroup$
    – Rammus
    Jul 26 at 13:36
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    $\begingroup$ @There'sStrangeStuffOutHere Or alternatively you can probably prove this via the Taylor series using properties of the projectors. $\endgroup$
    – Rammus
    Jul 26 at 13:38
  • $\begingroup$ Right. Then $\log(\sum_i P_i \rho P_i) = \sum_{i} P_i \log (P_i \rho P_i) P_i$ should hold because those projectors act as the identity as $\log(P_i\rho P_i)$ already lives in the $i$-th subspace, and then you can extend the sum to any $P_j$ considering that only the one with $j=i$ is non vanishing. I think I get it, thank you :-) $\endgroup$ Jul 26 at 15:26

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