Theorem: Exact whole-code recovery

For $V:C\to AE$, the following are equivalent:

1. There is a CPTP decoder $\mathcal{D}:A\to C$ with $\mathcal{DN}={\mathrm{id}}_C$.

2. $\mathcal{C}={\mathcal{T}}_{\tau }$ for a state ${\tau }_E$.

3. In a code basis, $\mathcal{C}(|i⟩⟨j|)={\delta }_{ij}{\tau }_E$.

If $r=\mathrm{rank}{\tau }_E$, there are a state ${\sigma }_K$, $\dim K=r$, and an isometry $T:C\otimes K\to A$ such that

An exact decoder on the entire accessible input space is

All equalities preserve arbitrary references.

Proof. Matrix units span $\mathcal{B}({\mathcal{H}}_C)$, proving (ii)–(iii).

Assume (ii), and diagonalize the nonzero part of the residual state:

Write

The complementary matrix-unit equations imply

Thus $T(|i⟩|\alpha ⟩)=|u_{i\alpha }⟩$ is an isometry. Taking ${\sigma }_K={\sum }_{\alpha }{\lambda }_{\alpha }|\alpha ⟩⟨\alpha |$ gives (eq:normalform). Formula (eq:exactdecoder) is CP, trace preserving, and recovers each matrix unit.

Conversely, dilate an exact decoder by an isometry $W:A\to CG$, and set

This is a dilation of the identity channel. The identity channel has rank-one Choi operator. Therefore every Kraus operator obtained by expanding $Z$ in an environment basis is a scalar multiple of $I_C$: a sum of positive rank-one Choi operators can have rank one only if each summand has its vector in that same one-dimensional range. Consequently

for a fixed normalized vector $\chi$. Tracing $CG$ gives a fixed state on $E$, proving (ii).

Equality of linear maps on all matrix units remains equality after tensoring with ${\mathrm{id}}_Q$.