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Consider two channels $N_1$ and $N_2$ which have Kraus operators $\{A_k\}$ and $\{B_k\}$ respectively, where the index $k$ runs over the number of Kraus operators. Define a third channel $N$ as a convex combination of $N_1$ and $N_2$. Specifically, for $0\leq p \leq 1$

$$N(\rho) = pN_1(\rho) + (1-p)N_2(\rho)$$

The Kraus operators corresponding to $N$ are $\{\sqrt{p}A_k\}\cup\{\sqrt{1-p}B_k\}$.

I would now like to look at the complementary channel of these channels. We have

$$N_1^c(\rho) = \sum_{k,j}\text{Tr}(A_k\rho A_j^\dagger)\otimes \vert k\rangle\langle j\vert$$

$$N_2^c(\rho) = \sum_{k,j}\text{Tr}(B_k\rho B_j^\dagger)\otimes \vert k\rangle\langle j\vert$$

The channel $N$ however picks up some cross terms when I work out the complemetary channel. \begin{align} N^c(\rho) &= \sum_{k,j}p\text{Tr}(A_k\rho A_j^\dagger)\otimes \vert k\rangle\langle j\vert + \sum_{k,j}\sqrt{p(1-p)}\text{Tr}(A_k\rho B_j^\dagger)\otimes \vert k\rangle\langle j\vert \\ &+ \sum_{k,j}\sqrt{p(1-p)}\text{Tr}(B_k\rho A_j^\dagger)\otimes \vert k\rangle\langle j\vert + \sum_{k,j}(1-p)\text{Tr}(B_k\rho B_j^\dagger)\otimes \vert k\rangle\langle j\vert \end{align}

What is a bit weird is if I now let $N_1 = N_2 = N$. Then, obviously $N_1^c = N_2^c = N^c$. However, the "cross terms" in the expression above seem like extra terms after setting all the $B_k = A_k$. It does not seem that they go to zero. What is the error here?

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There is more than one error.

First, for the complementary channel you need to attach a state (the $\otimes \vert k\rangle\langle j\vert$ in your formulas) for every possible Kraus operators. Your new channel has twice as many Kraus operators, therefore, you need an extra label for that degree of freedom.

What you then get for $N_1=N_2$ (with identical Kraus operators) however still has these cross-terms, just with the extra label.

Here comes the second error: The complementary channel is not uniquely defined. It is rather only defined up to basis transformations of the dual system. The cross-terms are exactly necessary to un-rotate the two-copy representation of $N^c$ with operators $A$ and $B$ to the original version. Without those cross-terms, the map would actually go onto that bigger space, and a rotation on the original $N^c$ would not be possible.

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  • $\begingroup$ Correct me if I am wrong but to rephrase what you have said, there exists an isometry on the dual space between the two expressions I have (once the range over which $k$ and $j$ run is corrected). $\endgroup$ – user1936752 Sep 8 '19 at 17:05
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    $\begingroup$ @user1936752 Precisely. (Basically, if $A_k=B_k$ then all four coefficients in your (cross-)terms will be equal for AA and AB and BB pairs; and the additional "tag" you add to the label $k$ should be exactly in a |+> state.) $\endgroup$ – Norbert Schuch Sep 9 '19 at 0:55

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