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
$$
\begin{aligned}
D(\rho\|Y(\rho))
&=\operatorname{Tr}(\rho\log\rho)-\sum_ip_i\log p_i\\
&=\boxed{S(Y(\rho))-S(\rho)}.
\end{aligned}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 323 / 2 / c / Solution

Choose a basis $|a\rangle$ of $H$ and define Kraus operators from $H$ to $\mathbb C^m$ by
$$
K_{ka}=|k\rangle\langle a|\sqrt{\Lambda_k}.
$$
Then
$$
\sum_{k,a}K_{ka}\rho K_{ka}^\dagger
=\sum_k|k\rangle\langle k|\operatorname{Tr}(\Lambda_k\rho),
$$
and
$$
\sum_{k,a}K_{ka}^\dagger K_{ka}
=\sum_k\Lambda_k=I.
$$
The displayed map is therefore a completely positive trace-preserving <measure-and-prepare channel>.

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 323 / 4 / c / Solution

A measure-and-prepare channel measures a POVM $\{M_x\}$ and outputs a prescribed state $\sigma_x$: $T(\rho)=\sum_x\operatorname{Tr}(M_x\rho)\sigma_x$. Such channels are exactly the <entanglement-breaking channels>.

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