Theorem: Exact recovery criterion, locally restated

For an isometry $V:C\to R\otimes B$, the following are equivalent:

1. there is a CPTP map $\mathcal{D}:R\to C$ with $\mathcal{DN}={\mathrm{id}}_C$;

2. $\mathcal{C}={\mathcal{T}}_{\sigma }$ for some state $\sigma$;

3. $\mathcal{C}(E_{ij})={\delta }_{ij}\sigma$ for every matrix unit;

4. $$J_{\mathrm{n}}(\mathcal{C})=\frac{I_Q}{d}\otimes \sigma .$$

When these conditions hold, there are a separable Hilbert space $K$, a state ${\sigma }_K$, and an isometry $A:C\otimes K\to R$ such that

An exact decoder is

Proof. Linearity gives equivalence of (ii) and (iii). Expanding the Bell projector gives equivalence of (iii) and (iv).

Assume (ii), and write the nonzero spectral decomposition

The index set is finite or countable. The constancy condition puts every $V|i⟩$ in $R\otimes \mathrm{supp}\sigma$. Write

The complementary matrix-unit identities imply

Consequently

extends to an isometry. With ${\sigma }_K={\sum }_{\alpha }{p}_{\alpha }|\alpha ⟩⟨\alpha |$, tracing $B$ gives (eq:normalform).

Both terms of (eq:decoder) are completely positive, and their traces sum to $\mathrm{Tr}Y$. Substitution into (eq:normalform) gives exact recovery of every matrix unit.

Conversely, dilate an exact decoder by a Stinespring isometry $L:R\to C\otimes G$. The normal channel dilation is the standard mathematical structure theorem Stinespring,Watrous. Put $Z=(L\otimes I_B)V$. Since the decoded marginal for basis input $|i⟩$ is pure,

Exact recovery of $E_{ij}$ forces $⟨{\eta }_j,{\eta }_i⟩=1$. All the normalized vectors ${\eta }_i$ therefore coincide. Tracing the decoded register and $G$ gives

which is (iii).