Theorem: Exact complete-complement criterion

The accessible output exactly recovers the entire code if and only if

for one state ${\tau }_B$. If this holds and $r=\mathrm{rank}{\tau }_B$, there are a state ${\sigma }_K$, $\dim K=r$, and an isometry $T:C\otimes K\to R$ such that

An exact decoder on all of $R$ is

In particular, $\dim R\ge d\mathrm{rank}{\tau }_B$. Both complementary factors cannot independently recover the whole code when $d\ge 2$.

Proof. Suppose $\mathcal{C}={\mathcal{T}}_{{\tau }_B}$, and diagonalize its nonzero support:

Write

The complementary identities on all matrix units give

Consequently $T(|i⟩|\alpha ⟩)=|u_{i\alpha }⟩$ is isometric and yields (eq:normalform). The displayed decoder is CPTP and recovers every matrix unit.

Conversely, dilate an exact decoder by $W:R\to CG$. The composite $(W\otimes I_B)V$ dilates the identity channel. The Choi operator of the identity has rank one. Every Kraus operator of that composite is therefore proportional to $I_C$, so the composite has the form

with fixed $\chi$. Its $B$ channel is replacement. All identities extend to references by linearity. If both factors recovered, the complement of either recoverable factor would be both replacement and recoverable, which is impossible on two distinct inputs.