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 1


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.

  • $\begingroup$ 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? $\endgroup$ Commented Nov 17, 2020 at 21:52
  • $\begingroup$ Sorry, just saw your edit. Thanks for clarifying! $\endgroup$ Commented Nov 17, 2020 at 21:53
  • $\begingroup$ 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. $\endgroup$
    – Condo
    Commented Oct 5, 2021 at 16:38
  • $\begingroup$ 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. $\endgroup$
    – Condo
    Commented Oct 5, 2021 at 16:51
  • $\begingroup$ @Condo, right, thanks for correction! $\endgroup$
    – Danylo Y
    Commented Oct 5, 2021 at 21:06

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