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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?

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  • $\begingroup$ 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. $\endgroup$
    – Rammus
    Sep 9, 2022 at 12:29
  • $\begingroup$ @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. $\endgroup$
    – Sherlock
    Sep 9, 2022 at 13:18
  • $\begingroup$ 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 $\endgroup$
    – Sherlock
    Sep 9, 2022 at 13:19
  • $\begingroup$ $\varepsilon (|0\rangle \langle 1|)=p\frac{I}{2}+(1-p)|0\rangle \langle 1|$. The answer is not the same. $\endgroup$
    – Sherlock
    Sep 9, 2022 at 13:20

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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. $$\mathcal E(|i\rangle\!\langle j|)=(1-p) |i\rangle\!\langle j|.$$

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  • $\begingroup$ 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? $\endgroup$
    – Sherlock
    Sep 9, 2022 at 13:35
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    $\begingroup$ @Sherlock depends on how you define the channel. If you define it in terms of Kraus operators, then sure, it comes from there. See e.g. quantumcomputing.stackexchange.com/a/12909/55. If you define the channel via its action, it's just usually defined with the trace from the start (or at least it should). This aside, you can check whether you need the term by computing the trace. Channels must be trace-preserving for any input, so checking that you can ensure the definition you have is sound $\endgroup$
    – glS
    Sep 9, 2022 at 14:04

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