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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?

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  • $\begingroup$ 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? $\endgroup$ Commented Sep 3, 2019 at 4:39
  • $\begingroup$ @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? $\endgroup$ Commented Sep 3, 2019 at 18:48
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    $\begingroup$ 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. $\endgroup$ Commented Sep 3, 2019 at 23:24
  • $\begingroup$ @NorbertSchuch that's a great comment! I understand the link now. I'm happy to accept your comment as an answer if you wish. $\endgroup$ Commented Sep 4, 2019 at 18:56
  • $\begingroup$ Do you mean Stinespring dilation is used in quantum channels while Naimark's theorem is used in POVM? I'm a little bit confused since the two concepts are pure mathematics, how can they differ in two physical concepts? $\endgroup$
    – narip
    Commented Jun 1, 2023 at 13:41

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I would say it is the opposite: Channels are special cases of measurements (where you forget the outcome).

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.

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  • $\begingroup$ Revisiting this after a long time: Mark Wilde's book (Section 4.6.6 of arxiv.org/pdf/1106.1445.pdf) deals with the question but shows it in the other direction i.e. a way to write a POVM as a channel. He shows the Kraus operators explicitly so one can construct the Stinespring isometry as well from that. Is it then the case that one can see it both ways i.e. channels are special cases of measurements and measurements are a kind of channel? $\endgroup$ Commented Feb 18, 2021 at 11:31

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