Theorem: Exact complete-complement criterion

For a complete finite interface $V:C\to R\otimes K$, the following conditions are equivalent:

1. A deterministic CPTP decoder satisfies $\mathcal{DN}={\mathrm{id}}_C$.

2. There is a state ${\tau }_K$ such that ${\mathcal{N}}^c(X)=\mathrm{Tr}(X){\tau }_K$ for every $X\in \mathcal{B}({\mathcal{H}}_C)$.

3. In any orthonormal code basis $\{|i⟩\}$,

   $${\mathcal{N}}^c(|i⟩⟨j|)={\delta }_{ij}{\tau }_K.$$

The resulting recovery preserves every external reference.

Proof. Conditions 2 and 3 are equivalent because matrix units span $\mathcal{B}({\mathcal{H}}_C)$.

Assume condition 3, and diagonalize ${\tau }_K={{\sum }_{a=1}}^r{\lambda }_a|a⟩⟨a|$ with ${\lambda }_a>0$. Write

The complementary matrix-unit identities imply

Thus the $u_{ia}$ form an orthonormal family. On their span, define an isometry

The channel that applies this isometry, traces $S$, and assigns an arbitrary fixed code state on the orthogonal unused subspace is CPTP. It maps $\mathcal{N}(X)$ to $X$ for every matrix unit and hence every $X$.

Conversely, dilate an exact decoder by an isometry $W_D:R\to C\otimes S$. The composite

is a dilation of ${\mathrm{id}}_C$. Purity of the decoded output for $|i⟩$ implies $T|i⟩=|i⟩|{\chi }_i{⟩}_{SK}$. Applying exact recovery to superpositions of any two basis vectors implies $⟨{\chi }_j|{\chi }_i⟩=1$. Hence all ${\chi }_i$ are the same vector $\chi$, and

Tracing $C$ and $S$ proves condition 2.

Finally, equality of channels on all matrix units remains equality after tensoring with ${\mathrm{id}}_Q$. This establishes reference preservation.