Theorem: Exact whole-code recovery

An accessible output exactly recovers the whole code if and only if its complete complement is replacement:

In that case there are a state ${\sigma }_K$ and an isometry $T:C\otimes K\to R$ such that

One may take $\dim K=\mathrm{rank}{\tau }_B$. For a pure logical input,

Thus exact recovery does not force purity of the raw accessible output.

Proof. If the complement is constant, diagonalize ${\tau }_B={\sum }_{\alpha }{\lambda }_{\alpha }|\alpha ⟩⟨\alpha |$ on its nonzero support and write

The complementary equations on matrix units imply $⟨u_{j\beta }\mid u_{i\alpha }⟩={\delta }_{ij}{\delta }_{\alpha \beta }$. They define $T(|i⟩|\alpha ⟩)=|u_{i\alpha }⟩$. Compression by $T^{†}$, tracing $K$, and preparing a fixed state on the unused accessible subspace gives a deterministic decoder.

Conversely, dilate an exact decoder by $W:R\to CG$. The composite $(W\otimes I_B)V$ dilates the identity channel. Its Choi operator has rank one, so each Kraus operator is a scalar multiple of $I_C$. Hence the composite is $|\psi ⟩\mapsto |\psi {⟩}_C|\chi {⟩}_{GB}$ for fixed $\chi$. Tracing $CG$ proves complement constancy. The entropy statement follows from (eq:normalform).