 Solution

Let $A$ be a register with orthonormal basis $\{|x⟩\}$ for Alice's classical symbol, $Q$ the transmitted system, and $B$ a register initially in a fixed blank state. The initial classical-quantum state isA POVM fixes probabilities but not a unique conditional state of $Q$. Choose the Lüders rule instrument $M_y=\sqrt{E_y}$, which satisfies ${\sum }_y{M}_y^{†}{M}_y=I$. Record outcome $y$ in orthogonal states of $B^{\prime }$ and retain the corresponding quantum output $Q^{\prime }$. The final nonselective state isThe positive operators in this sum are unnormalized: their traces already include the outcome probabilities. If $\mathrm{Tr}(E_y{\rho }_x)>0$, the normalized conditional state of $Q^{\prime }$ is $M_y{\rho }_x{M}_y^{†}/\mathrm{Tr}(E_y{\rho }_x)$.

More generally, an instrument may use operators $M_{y\mu }$ with ${\sum }_{\mu }{M}_{y\mu }^{†}{M}_{y\mu }=E_y$, replacing $M_y{\rho }_x{M}_y^{†}$ by ${\sum }_{\mu }{M}_{y\mu }{\rho }_x{M}_{y\mu }^{†}$. 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.