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

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?

• Note that even if $\rho$ is entangled you can still expand it as $\sum_{ij} |i \rangle \langle j| \otimes \sigma_{ij}$. As channels are linear maps you could just apply it then to each term in the sum. Commented Sep 9, 2022 at 12:29
• @Rammus I agree if we use Kraus operators such as eq.(2). But for $\varepsilon$, I don't think $I\otimes \varepsilon \left( \sum_{ij}{|}i\rangle \langle j|\otimes \sigma _{ij} \right) =\sum_{ij}{|}i\rangle \langle j|\otimes \varepsilon \left( \sigma _{ij} \right)$ is right. Since domain of $\varepsilon$ is density matrices while $\sigma_{ij}$ generally is not a density matrix. And we can calculate a concert example to show this. Commented Sep 9, 2022 at 13:18
• We can use two formulas, $\varepsilon (\rho )\equiv p\frac{I}{2}+(1-p)\rho$ and $\mathcal{E} (\rho )=(1-3p/4)\rho +p/4(X\rho X+Y\rho Y+Z\rho Z)$. For $|0\rangle\langle 1|$, the second formula imply that $$\mathcal{E} (|0\rangle \langle 1|)=(1-3p/4)|0\rangle \langle 1|+p/4(X|0\rangle \langle 1|X+Y|0\rangle \langle 1|Y+Z|0\rangle \langle 1|Z) \\ =(1-3p/4)|0\rangle \langle 1|+p/4(|1\rangle \langle 0|-|1\rangle \langle 0|-|0\rangle \langle 1|) \\ =(1-3p/4)|0\rangle \langle 1|-p/4|0\rangle \langle 1|$$ while the first one imply that Commented Sep 9, 2022 at 13:19
• $\varepsilon (|0\rangle \langle 1|)=p\frac{I}{2}+(1-p)|0\rangle \langle 1|$. The answer is not the same. Commented Sep 9, 2022 at 13:20

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. $$\mathcal E(|i\rangle\!\langle j|)=(1-p) |i\rangle\!\langle j|.$$
• Did the $Tr(\cdot)$ come from Kraus operators? If I came across other channels, how can I do the same thing like spot the $Tr$ in this case? Commented Sep 9, 2022 at 13:35