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2

Here's an alternative proof: first note that any quantum map, $\Phi(\rho) \mapsto \sigma$ that can be written in the Kraus form, that is, as $\Phi(\cdot) = \sum_{j} K_j (\cdot) K_j^\dagger$, with, $K_j^\dagger K_j \geq 0, \sum_j K_j^\dagger K_j = \mathbb{I}$ is a CP map (see for example, Nielsen and Chuang, or Page 26 of https://arxiv.org/abs/1902.00967). ...

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Ok, so starting with Trace Preserving, since it's easier: $$Tr(I/2) = 1$$ $$Tr(\rho) = 1$$ $$Tr((1-\lambda)I/2 + \lambda\rho) = (1-\lambda)Tr(I/2) + \lambda Tr(\rho) = 1$$ Now for a map to be completely positive, it must take positive elements to positive elements. So since depolarizing noise is essentially just adding in a bit of the identity operator (who'...

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Adding my comment as an answer: I think this is a typo. The equation contains a scalar, namely $S(\cdot)$ and an operator, the output of the channel $\mathcal{E}(\cdot)$ in linear combination (their dimensions do not match). Therefore, this has to be a typo (there are no hidden Identity operators in the scalar terms as is easy to check).

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Here's a guess: they might be related to entanglement-breaking channels (also known as measure-and-prepare channels, quantum-classical channels, etc.). Any channel of the form, $$\Phi(\rho) = \sum\limits_{k} \operatorname{Tr}\left( M_{k} \rho \right) \sigma_{k} , \text{ where } M_{k}\geq0,\sum\limits_{k}^{} M_{k} = \mathbb{I},$$ are POVM elements and \$\{ \...

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