Theorem: Exact recovery from a hybrid output

Let

where the ${\mathcal{F}}_y$ are CP and ${\sum }_y{\mathcal{F}}_y$ is TP. Then $\mathcal{H}$ exactly recovers the entire quantum code if and only if:

1. there are input-independent probabilities $p_y$ with

   $$\mathrm{Tr}{\mathcal{F}}_y(X)=p_y\mathrm{Tr}X;$$

2. for every $p_y>0$, the normalized channel ${\mathcal{N}}_y={\mathcal{F}}_y/p_y$ has an exact quantum-return decoder on $K$.

Thus an input-informative retained classical record is incompatible with exact whole-code recovery, even when a quantum factor is also returned.

Proof. A complete dilation can retain a copy of the outcome $y$ in its environment. If $\mathcal{H}$ is exactly recoverable, its complete complement is constant by Theorem thm:exact. Measuring the environment's outcome label proves (i).

Restrict an exact decoder to the $y$-th diagonal input block to obtain a CPTP map ${\mathcal{D}}_y:K\to C$. Then

The identity has rank-one Choi operator. Positivity implies that every normalized summand has that same Choi operator, hence ${\mathcal{D}}_y{\mathcal{N}}_y={\mathrm{id}}_C$. This proves (ii).

Conversely, read the classical label and apply its branch decoder. The resulting channel is ${\sum }_y{p}_y{\mathrm{id}}_C={\mathrm{id}}_C$. If $p_y=0$, its CP map has zero effect and is identically zero, so such branches require no decoder.