= Solution
Let $A$ be a register with <orthonormal basis> $\{|x\rangle\}$ for Alice's classical symbol, $Q$ the transmitted system, and $B$ a register initially in a fixed blank state. The initial <classical-quantum state> is
$$
\boxed{\rho_{AQB}=\sum_xp(x)|x\rangle\langle x|_A\otimes\rho_x\otimes|0\rangle\langle0|_B.}
$$
A <POVM> fixes probabilities but not a unique conditional state of $Q$. Choose the <Lüders rule> instrument $M_y=\sqrt{E_y}$, which satisfies $\sum_yM_y^\dagger M_y=I$. Record outcome $y$ in orthogonal states of $B'$ and retain the corresponding quantum output $Q'$. The final nonselective state is
$$
\boxed{\rho_{A'Q'B'}=\sum_{x,y}p(x)|x\rangle\langle x|_{A'}\otimes M_y\rho_xM_y^\dagger\otimes|y\rangle\langle y|_{B'}.}
$$
The positive operators in this sum are unnormalized: their traces already include the outcome probabilities. If $\operatorname{Tr}(E_y\rho_x)>0$, the normalized conditional state of $Q'$ is $M_y\rho_xM_y^\dagger/\operatorname{Tr}(E_y\rho_x)$.
More generally, an instrument may use operators $M_{y\mu}$ with $\sum_\mu M_{y\mu}^\dagger M_{y\mu}=E_y$, replacing $M_y\rho_xM_y^\dagger$ by $\sum_\mu M_{y\mu}\rho_xM_{y\mu}^\dagger$. All subsequent classical information bounds are unchanged. This explicitly accounts for the fact that a <POVM does not determine the post-measurement state> of a retained quantum system.
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