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What is the Stinespring representation of the adjoint of a channel?

Quantum Computing Asked on January 7, 2021

For any completely positive trace nonincreasing map $N_{Arightarrow B}$, the adjoint map is the unique completely positive linear map $N^dagger_{Brightarrow A}$ that satisfies

$$langle N^dagger(sigma), rhorangle = langle sigma, N(rho)rangle$$

for all linear operators $sigma in mathcal{L}(mathcal{H}_B)$ and $rho in mathcal{L}(mathcal{H}_A)$.

Let $V_{Arightarrow BE}$ be any isometry such that $text{Tr}_E(Vrho V^dagger) = N(rho)$. This is the Stinespring representation of any completely positive map. Since $N^dagger$ is also a completely positive map, it also has a Stinespring representation.

Question: Given $V$, can one write down the Stinespring representation of $N^dagger$? Naively taking the transpose conjugate of $V$ to write down something like

$$text{Tr}_E(V^daggersigma V) = N^dagger(sigma)$$

doesn’t even make sense since $V^dagger_{BErightarrow A}$ whereas the isometry we are after should go from $B$ to $AE’$.

One Answer

An adjoint to partial trace is just tensoring by $I$, i.e. $text{Tr}_2^dagger(sigma) = sigma otimes I$.

So in this case we can write $$ N^dagger(sigma) = V^dagger (sigma otimes I_E)V. $$

It still can be written as $text{Tr}_{E'}(Usigma U^dagger)$ for an isometry $U:B rightarrow AE'$, but I don't think there is a simple expression for $U$ in terms of $V$.

Correct answer by Danylo Y on January 7, 2021

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