# How to compute the tensor product of the depolarizing channel with the identity?

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!

• generalised to what?
– glS
May 19 '20 at 11:36
• how to get the general output of the tensor product of several depolarizing channels May 19 '20 at 11:42

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