= Solution
With $P_i=|y_i\rangle\langle y_i|$, the channel has Kraus form $Y(\rho)=\sum_iP_i\rho P_i$ and $\sum_iP_i^\dagger P_i=I$, so it is completely positive and trace preserving. It measures in the $y$ basis and prepares the observed basis state, hence is a <measure-and-prepare channel>. If $p_i=\langle y_i|\rho|y_i\rangle$, then $Y(\rho)=\sum_ip_iP_i$ and
$$
\boxed{S(Y(\rho))=H(p)}.
$$
Moreover $\log Y(\rho)=\sum_i(\log p_i)P_i$ on its support, so
$$
\begin{aligned}
D(\rho\|Y(\rho))
&=\operatorname{Tr}(\rho\log\rho)-\sum_ip_i\log p_i\\
&=\boxed{S(Y(\rho))-S(\rho)}.
\end{aligned}
$$
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