= Solution
Choose $P_X$ uniform on $\mathcal A_M$ and zero outside it, so the marginal <density operator> of $Q$ is $\bar\rho_Q=k^{-1}\sum_m\rho(m)_Q$. Define the <binary test for quantum decoding success> by
$$
\boxed{F(0)=\sum_{m\in\mathcal A_M}|m\rangle\langle m|_{\widetilde X}\otimes E(m)_Q,\qquad
F(1)=I_{\widetilde XQ}-F(0).}
$$
Each $E(m)$ is a <positive contraction>. The <computational basis> blocks of $F(0)$ are these effects on $\mathcal A_M$ and zero elsewhere. Thus $0\leq F(0)\leq I$, making $F$ a <POVM> on the whole space, including labels outside $\mathcal A_M$.
On the correlated <classical-quantum state>, its acceptance probability is
$$
\operatorname{Tr}[F(0)\rho_{\widetilde XQ}]
=\frac1k\sum_m\operatorname{Tr}[E(m)\rho(m)]=1-\epsilon.
$$
On the product of the marginal <density operators>,
$$
\operatorname{Tr}[F(0)(\rho_{\widetilde X}\otimes\bar\rho_Q)]
=\frac1k\sum_m\operatorname{Tr}[E(m)\bar\rho_Q]
=\frac1k\operatorname{Tr}\bar\rho_Q=\frac1k,
$$
since $\sum_mE(m)=I_Q$. \b[The two requested probabilities are therefore $1-\epsilon$ and $1/k$.] In the notation for binary output <density operators>, the <measurement channel> sends the joint input to $\omega[\epsilon]$ and the product input to $\omega[1-1/k]$; their parameters describe outcome one, not outcome zero.
Back to article page
