How to calculate the action of a channel on part of a quantum state?

Question

As the title shows, but I think we can restrict ourselves into a more specific example. Let's consider depolarizing channel $\varepsilon$: $$\varepsilon(\rho)\equiv p\frac{I}{d}+(1-p)\rho\tag{1}$$ where $d$ is the dimension of quantum state $\rho$ of hilbert space $H_2$, and $p$ is the real number $\in[0,1]$.

I want to know how can I calculate $(I\otimes \varepsilon) (\rho)$ where $I\otimes \varepsilon$ acts on $\rho\in H_{12}$. I know I can calculate this by using Kraus operators, i.e. $$\sum_i{I\otimes E_i\rho I\otimes {E_i}^{\dagger}},\tag{2}$$ but the Kraus operators for depolarizing channel are different when the dimension of $H_2$ changes, refer to this question for details. Also if $\rho$ is a separable state, i.e. $\rho =\sum_i{p_i{\rho _i}^{\left( 1 \right)}\otimes {\rho _i}^{\left( 2 \right)}}$, we can calculate $(I\otimes \varepsilon) (\rho)$ as $\sum_i{p_i{\rho _i}^{\left( 1 \right)}\otimes \varepsilon \left( {\rho _i}^{\left( 2 \right)} \right)}$. But when $\rho$ is entangled I don't know how can I calculate it?

Answer 1

you can absolutely do the calculation in this comment. That is, you can just compute $$(I\otimes \mathcal E)\rho = \sum_{ij} (I\otimes \mathcal E)(|i\rangle\!\langle j|\otimes \sigma_{ij}) = \sum_{ij} |i\rangle\!\langle j|\otimes \left( p\operatorname{Tr}(\sigma_{ij}) \frac{I}{d}+(1-p) \sigma_{ij} \right).$$ Note that here $\sigma_{ij}$ are not necessarily states, so it's not always true that $\operatorname{Tr}(\sigma_{ij})=1$.

The domain of a channel can always be considered to be the full set of linear operators, even though physically you'd only consider its action on density matrices. The reason you got a wrong result is that the channel is better written as $\mathcal E(\rho)={\rm Tr}(\rho)I/d +(1-p)\rho$. You can neglect the trace term only when you restrict its action of unit-trace operators. So e.g. $$\mathcal E(|i\rangle\!\langle j|)=(1-p) |i\rangle\!\langle j|.$$