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Consider two quantum systems A and B, B goes through a depolarizing noise channel, while A is not changed, i.e., they go through the channel $\mathbb{I}_A \otimes \mathcal{E_{\text{depol}}} $. If the input is a density operator $\rho_{AB}$, is it correct that the output is $p \rho_A \otimes \mathbb{I}/2 + (1-p) \rho_{AB}$? How can this be generalized? Thanks in advance!

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    $\begingroup$ generalised to what? $\endgroup$ – glS May 19 at 11:36
  • $\begingroup$ how to get the general output of the tensor product of several depolarizing channels $\endgroup$ – Dina Abdelhadi May 19 at 11:42
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Let $\Phi_{dp}$ denote the fully depolarising channel: $\Phi_{dp}(\rho)=\operatorname{Tr}(\rho) I/d$ with $d$ the dimension of the space.

The depolarising channel $\mathcal E_{depol}$ in the OP can be written as $\mathcal E_{depol}=(1-p) \operatorname{Id} + p \Phi_{dp}$ with $\operatorname{Id}$ the identity channel. It follows that $$\operatorname{Id}\otimes \mathcal E_{depol} = (1-p) \operatorname{Id}\otimes \operatorname{Id} + p \operatorname{Id}\otimes \Phi_{dp}.$$ The only thing that might appear nontrivial is how the second term acts on states. There are several ways to compute this. For example:

$$(\operatorname{Id}\otimes\Phi_{dp} )\rho = \sum_{ijk\ell} \rho_{ijk\ell} (\operatorname{Id}\otimes\Phi_{dp} ) (|ij\rangle\!\langle k\ell|) = \sum_{ijk\ell} \rho_{ijk\ell} |i\rangle\!\langle k|\otimes \underbrace{\Phi_{dp} (|j\rangle\!\langle\ell|)}_{=\delta_{j\ell}/d} \\ = \sum_{ijk} \rho_{ijkj} |i\rangle\!\langle k|\otimes I / d \equiv \operatorname{Tr}_B(\rho) \otimes I/d, \equiv \rho_A\otimes I/d. $$

Similar calculations can be performed in different representations of the map (natural, Choi, etc) with analogous results.

Working out tensor products of more depolarising channels is analogous.

TL;DR: Yes, that's correct.

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