Theorem: Exact recovery if and only if the complement is constant

Let $V:{\mathcal{H}}_M\to {\mathcal{H}}_A\otimes {\mathcal{H}}_E$ be a finite-dimensional isometry, with accessible channel $\mathcal{N}$ and complement $\mathcal{C}$. The following are equivalent:

1. There is a CPTP map $\mathcal{D}:A\to M$ with $\mathcal{DN}={\mathrm{id}}_M$.

2. For some density operator ${\tau }_E$,

   $$\mathcal{C}(X)=\mathrm{Tr}(X){\tau }_E\text{for every }X\in \mathcal{B}({\mathcal{H}}_M).$$

3. For a Kraus family $\{A_a\}$ read from $V$,

   $${A_a}^{†}{A}_b=c_{ab}{I}_M$$

   for scalar coefficients $c_{ab}$.

Proof. Suppose (i). Compose $V$ with a Stinespring isometry $T:A\to M\otimes G$ for $\mathcal{D}$. The composite dilates the identity channel.

Every Kraus operator $B_j$ of the identity channel is proportional to $I_M$: its Choi matrix is rank one and is the sum of the positive rank-one matrices formed from the vectorized $B_j$, so each vectorized $B_j$ lies in the span of the vectorized identity. Therefore the composite isometry factors as

for a fixed normalized $\chi$. Tracing $MG$ leaves the original complement unchanged, hence proves (ii).

Conversely, suppose (ii). Diagonalize ${\tau }_E={\sum }_j{\lambda }_j|j⟩⟨j|$, omitting zero eigenvalues. The isometry has no component in the kernel of ${\tau }_E$. Write

The constant-complement identity on every matrix unit gives

Let $P_S$ project onto their span, and define an isometry $U:S\to M\otimes J$ by $U{\varphi }_{ij}=|i⟩|j⟩$. For an arbitrary fixed state ${\sigma }_M$, the map

is CPTP. Its action on $\mathcal{N}(X)$ is $X$, proving (i).

Finally,

This is a scalar multiple of $\mathrm{Tr}(X)$ for every $X$ exactly when every ${A_b}^{†}{A}_a$ is scalar. Thus (ii) and (iii) are equivalent.