I'm trying to understand a part of the Holevo bound, as found in Nielsen and Chuang, p. 533.

In an excercise, they write

Define $U_y$ to be the unitary operator acting on the system $M$ whose action on a basis is $U_y|y' \rangle = |y +y' \rangle$, where addition is done modulo $n+1$. Show that $\{\sqrt{E_y} \otimes U_y\}$ is a set of operation elements defining a trace-preserving quantum operation $\varepsilon$ whose action on states of the form $\sigma \otimes |0 \rangle \langle 0|$ satisfies $\varepsilon(\sigma \otimes |0 \rangle \langle 0|) = \sum_y \sqrt{E_y}\sigma\sqrt{E_y} \otimes |y \rangle\langle y|$.

In the above, $\{ E_y \}$ is a set of POVM elements.

I am able to prove the trace-preserving property, but I am confused about the terminology for the rest of the question. What exactly is meant by the "action on states"? Is this simply asking for an expression for $(\sqrt{E_Y} \otimes U_y) (\sigma \otimes |0 \rangle \langle 0|)$? I don't think this is the case, since I don't get the desired expression.

I understand how to get the associated probabilities of measurement outcomes for a set of POVM elements, but I'm unclear about how this is related. What is the action that I'm being asked about here?


$M_y=\sqrt{E_y}\otimes U_y$ defines a set of Kraus operators which describes a quantum channel, i.e., $$\varepsilon(\rho) = \sum_y M_y\rho M_y^\dagger\ .$$ You are asked to verify that it acts as specified on the input $\sigma\otimes |0\rangle\langle 0|$.

  • $\begingroup$ I think I can prove that, but I'm confused about where your expression for $\varepsilon(\rho)$ comes from. I understand that $M_y \rho M_y^+$ would be the unnormalized state after measuring the density operator $\rho$. Why are we summing over all possible $y$, and why do we ignore the associated outcome probabilities? p.s. I don't see anything about Kraus operators in the text. $\endgroup$ – theQman Jan 19 '17 at 17:33
  • $\begingroup$ I think that's exactly what they mean by "a set of operation elements defining a trace-preserving quantum operation ε" -- a CP map (=quantum operation) with Kraus operators given by those guys. $\endgroup$ – Norbert Schuch Jan 20 '17 at 1:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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