# Increasing entropy for projective LCPT mapping

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!

• What does LCPT stand for? Jul 26, 2021 at 7:08

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.
• Could you elaborate on why $$\log\Big(\sum_iP_i\rho P_i\Big)=\sum_i\log(P_i\rho P_i)$$ holds? Jul 26, 2021 at 13:33
• @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. Jul 26, 2021 at 13:36
• 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 :-) Jul 26, 2021 at 15:26