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!

  • 1
    $\begingroup$ generalised to what? $\endgroup$
    – glS
    May 19, 2020 at 11:36
  • 1
    $\begingroup$ how to get the general output of the tensor product of several depolarizing channels $\endgroup$ May 19, 2020 at 11:42

1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.