Let $T: \mathcal{H}_A \rightarrow \mathcal{H}_B$ be a CPTP map with Stinespring extension $U: \mathcal{H}_{A} \rightarrow \mathcal{H}_{B} \otimes \mathcal{H}_E$. That is $U$ is an isometry such that for all states $\rho_A$, we have $Tr_{E} \left( U \rho_A U^\dagger\right)= T(\rho_A)$.

I am interested in Stinespring dilation of $T \otimes \mathbb{I}_C$ where $C$ is some additional register. My guess is $U \otimes \mathbb{I}_C$ should work but I am unable to prove.

That is, consider a state $\rho_{AC}$ and $\sigma_{BC}= T_A \otimes \mathbb{I}_C (\rho_{AC})$. Will the following equality hold:

$Tr_{E} \left( (U\otimes \mathbb{I}_C) \rho_{AC} (U\otimes \mathbb{I}_C)^\dagger\right)= \sigma_{BC}$.

Ultimately, I want to extend this to $U \otimes V$ but I suppose I can break it up to ($U \otimes \mathbb{I}) (\mathbb{I} \otimes V)$.


1 Answer 1


Yes. Note that in general ${\rm Tr}_1(\rho_{12} \otimes \rho_3) = {\rm Tr}_1(\rho_{12}) \otimes \rho_3$.

It's easy to verify your equality for $\rho_{AC} = \rho_A \otimes \rho_C$ :

$$ {\rm Tr}_{E} \left( (U\otimes \mathbb{I}_C) \rho_{AC} (U\otimes \mathbb{I}_C)^\dagger\right) = {\rm Tr}_{E} \left( U \rho_{A} U^\dagger \otimes \rho_C \right) = {\rm Tr}_{E} \left( U \rho_{A} U^\dagger \right) \otimes \rho_C = (T \otimes \mathbb{I}_C)(\rho_{AC}). $$

But since both parts are linear over $\rho_{AC}$ (it's common in QM) the equality is true for any $\rho_{AC}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.