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

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

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

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

Let $$V_{A\rightarrow BE}$$ be any isometry such that $$\text{Tr}_E(V\rho V^\dagger) = N(\rho)$$. 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

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

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

## 1 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 $$N^\dagger(\sigma) = V^\dagger (\sigma \otimes I_E)V.$$

If $$N^\dagger$$ is trace-preserving then it can be written as $$\text{Tr}_{E'}(U\sigma U^\dagger)$$ for some isometry $$U:B \rightarrow AE'$$. But in general $$N^\dagger$$ is not trace-preserving, but unital. So such form $$\text{Tr}_{E'}(U\sigma U^\dagger)$$ may not be possible.

• Thank you for the answer. Do correct me if I'm wrong but this isn't really a Stinespring representation of the adjoint channel since that should be written using an isometry from $B$ to $AE'$, right? Related to that is the fact that $V^\dagger$ is not an isometry (c.f. physics.stackexchange.com/q/550075/52363). So can one write a Stinespring representation for $N^\dagger$ in terms of $V$ somehow? Nov 17 '20 at 21:52
• Sorry, just saw your edit. Thanks for clarifying! Nov 17 '20 at 21:53
• It's worth mentioning that if the channel $N$ is CPTP (completely-positive trace-preserving) then the adjoint $N^\dagger$ is UCP (unital completely-positive) since $V^\dagger V$ is the identity while $VV^\dagger$ is a projection. Oct 5 '21 at 16:38
• In fact, I think it's incorrect (or at least not obvious) to state that $N^\dagger$ it can be written as $Tr_{E'}(U\sigma U^\dagger)$ for some isometries $U$ because that is the general form of a CPTP map and not a UCP map. Oct 5 '21 at 16:51
• @Condo, right, thanks for correction! Oct 5 '21 at 21:06