# Stinespring dilation of a channel vs Naimark's theorem

I'm trying to understand the connection between the Stinespring dilation of a quantum channel and Naimark's theorem that shows that POVMs can be written as projective measurements in a larger Hilbert space. On Wikipedia, the proof of Naimark's theorem is given as a special case of the Stinespring dilation - however, I cannot follow this proof.

1. Given any channel $$N$$ from $$\mathcal{H}_A$$ to $$\mathcal{H}_B$$ given in terms of Kraus operators $$\{A_i\}$$ with $$\sum_i A_i^\dagger A_i = I_A$$, we can instead consider an isometry $$V_{A\rightarrow BE} = \sum_i A_i\otimes \vert i\rangle_E$$. Now the channel can be described as

$$N(\rho) = \sum_i A_i\rho_A A_i^\dagger = \text{Tr}_E(V\rho V^\dagger)$$

This is the Stinesping dilation of the channel.

2. Consider a POVM that consists of elements $$\{F_i\}$$. Since we know that $$\sum_i F_i = I$$ and $$F_i > 0$$, let us rewrite it as $$F_i = A^\dagger_iA_i$$. Now, we can consider the isometry $$V_{A\rightarrow BE} = \sum_i A_i\otimes \vert i\rangle_E$$. It is then true that the probability of the $$i^{th}$$ outcome is

$$p_i = \text{Tr}(A_i\rho_A A^\dagger_i) = \text{Tr}(V^\dagger(I\otimes\vert i\rangle\langle i\vert)V \rho_A)$$

Thus, we have a projective measurement on $$V\rho V^\dagger$$.

The mathematical description of the two things above seem very similar except that POVMs involve the additional step of picking a specific outcome $$i$$. I'm not sure how to express this as a channel. Moreover, if I could write the POVM as a channel, then how exactly does Naimark's theorem become a special case of the Stinespring dilation?

TL;DR

1) How do I write the POVM above as a quantum channel?

2) If I can do 1), what is the connection between the Stinespring dilation of this channel and Naimark's theorem?

• What exactly is your question? Would "just sum over $i$" be an answer, if in addition you consider the post-measurement state of the POVM? – Norbert Schuch Sep 3 at 4:39
• @NorbertSchuch my questions are - Can one see Naimark's theorem as a special case of the Stinespring dilation of a channel? If yes, can one write the POVM as a channel and show that its Stinespring representation is equivalent to a projective measurement in the larger space? – user1936752 Sep 3 at 18:48
• I would say it is the opposite: Channels are special cases of measurements (where you forget the outcome). And yes, Stinespring uses a partial trace, which can be seen as measuring the ancilla (Naimark!) and then forgetting the result. – Norbert Schuch Sep 3 at 23:24
• @NorbertSchuch that's a great comment! I understand the link now. I'm happy to accept your comment as an answer if you wish. – user1936752 Sep 4 at 18:56

In particular, in the Stinespring you have a partial trace $$\mathrm{Tr}_E$$, which you can think of as measuring the ancilla system $$E$$ -- this gives exactly the construction from Naimark's theorem -- and then forgetting the result, that is, summing over all (unnormalized) post-measurement states.