Theorem: Exact recovery and a constant complete complement

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

1. one CPTP decoder on $R$ exactly recovers the entire code;

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

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

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, defined on all trace-class operators on $R$, is

Proof. Linearity proves (ii) equivalent to (iii). Expansion of the Bell projector proves (iii) equivalent to (iv).

Assume (ii), and write the nonzero spectral decomposition

The index set is finite or countable. Constancy of the residual implies that every code vector is supported in $R\otimes \mathrm{supp}\sigma$. Write

The matrix-unit condition gives

Hence

extends to an isometry, and ${\sigma }_K={\sum }_{\alpha }{p}_{\alpha }|\alpha ⟩⟨\alpha |$ gives (eq:normalform). Both terms of (eq:decoder) are CP; their traces sum to $\mathrm{Tr}Y$. The decoder recovers every $E_{ij}$.

Conversely, let $L:R\to C\otimes G$ be a Stinespring isometry for an exact decoder. The normal Stinespring structure is the standard external dilation theorem KSW,Watrous. Set $Z=(L\otimes I_B)V$. For basis input $|i⟩$, the decoded marginal is the pure state $|i⟩⟨i|$, so

Exact recovery of $E_{ij}$ requires $⟨{\eta }_j,{\eta }_i⟩=1$. All ${\eta }_i$ are therefore one common normalized vector. Tracing the decoded register and $G$ yields $\mathcal{C}(E_{ij})={\delta }_{ij}\sigma$.