POVM used in proof of Holveo bound

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 $\sigma\otimes |0\rangle\langle 0|$.
• 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. – theQman Jan 19 '17 at 17:33