# Is there an inverse for Stinespring dilation?

Given a set of Kraus operators we can find a unitary that does the equivalent map on an extended space including the environment using Stinespring dilation. My question is how do we go about doing the opposite? I know that Kraus operation for the given unitary cannot be unique, as it depends on the ancilla state being used. But how do I go about proving the existence or non-existence of such an inverse of the Stinespring dilation?

• I might be misunderstanding, but are you asking how to find the (a) Kraus representation for a channel from its Stinespring representation?
– glS
Aug 12 at 16:10

To be more concrete, given a unitary $$U$$ on a composite system $$\mathcal H_A \otimes \mathcal H_B$$, the linear map $$\phi(X) := \mathrm{tr}_B ( U X \otimes |0 \rangle\langle 0| U^\dagger ),$$ is a completely positive trace-preserving map on $$\mathcal{H}_A$$ (i.e. a quantum channel). This is probably the "inverse" you're looking for (but TBH you have to start from here, otherwise there's no point in the discussion).
Now, Stinespring's dilation theorem says that for any CPTP map $$\phi$$, there is always an auxillary system $$\mathcal{H}_B$$ and a unitary $$U$$ on $$\mathcal H_A \otimes \mathcal H_B$$ such that the above reduction yields $$\phi$$.
You see, the mapping from $$U$$ to $$\phi$$ is well-defined, but it lacks an inverse since the unitary $$U$$ (and even the system $$\mathcal{H}_B$$) in Stinespring's theorem is not unique. Given a Kraus decomposition of $$\phi$$, one can explicitly construct a dilation as you mentioned. The non-uniqueness of the dilation is reflected in the one of the Kraus decomposition.