Theorem: Exact recovery, constancy, and observable compression

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

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

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

3. for every code matrix unit $E_{ij}=|i⟩⟨j|$,

   $$\mathcal{C}(E_{ij})={\delta }_{ij}\sigma ;$$
4. $$J_{\mathrm{n}}(\mathcal{C})=\frac{I_Q}{d}\otimes \sigma ;$$

5. for every bounded residual observable $O$,

   $$V^{†}(I_R\otimes O)V\text{is a scalar multiple of }{I}_C.$$

Here

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

An exact complete decoder is

Proof. Linearity and expansion of the Bell projector give equivalence of (ii)–(iv). Duality gives

Thus (ii) implies (v). Conversely, in (v), evaluate at ${\tau }_C$ to identify the scalar as $\mathrm{Tr}(O\mathcal{C}({\tau }_C))$. Since bounded observables separate trace-class operators, $\mathcal{C}(X)=\mathrm{Tr}(X)\mathcal{C}({\tau }_C)$, proving (ii).

Assume (ii), and diagonalize

The support condition permits the expansion

Constancy on every matrix unit gives

Consequently $A(|i⟩|\alpha ⟩)=|r_{i\alpha }⟩$ is an isometry, and ${\sigma }_K={\sum }_{\alpha }{p}_{\alpha }|\alpha ⟩⟨\alpha |$ gives (eq:normalform). Formula (eq:decoder) is CPTP and recovers every matrix unit.

Conversely, dilate an exact decoder by an isometry $L:R\to C\otimes G$. Stinespring dilation is the standard channel structure theorem Watrous,KSW. The map $Z=(L\otimes I_B)V$ has

because the decoded marginal is pure. Exact recovery of $E_{ij}$ forces $⟨{\eta }_j,{\eta }_i⟩=1$. All ${\eta }_i$ coincide. Tracing the decoded register and $G$ therefore gives (iii).