Theorem: Complete restoration and exact local recovery

For a complete interface, choose any ${\tau }_C\in \mathcal{D}(C)$. The map

is CPTP and satisfies ${\mathcal{L}}_V\mathcal{V}={\mathrm{id}}_C$. Isometric embedding preserves the trace norm of every trace-class operator, including after tensoring with $I_Q$.

Moreover, the following are equivalent:

1. a CPTP decoder satisfies $\mathcal{DN}={\mathrm{id}}_C$;

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

3. $\mathcal{E}(|i⟩⟨j|)={\delta }_{ij}{\sigma }_B$ on all matrix units;

4. $$J_{\mathrm{n}}(\mathcal{E})=I_Q/d\otimes {\sigma }_B,J_{\mathrm{n}}(\Phi ):=({\mathrm{id}}_Q\otimes \Phi )(|{\Phi }_d⟩⟨{\Phi }_d|),$$

   where $|{\Phi }_d⟩=d^{-1/2}{\sum }_i|i{⟩}_Q|i{⟩}_C$.

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

An exact decoder on all of $R$ is

If ${\sigma }_B$ has finite rank $r$, then $\dim R\ge dr$.

Proof. Compression and positive-effect measurement followed by preparation are CP. Their traces in (eq:restore) add to $\mathrm{Tr}Y$, and direct substitution gives the left inverse. An isometry preserves all nonzero singular values, proving trace-norm preservation.

Matrix units span $\mathcal{L}(C)$, and their Bell expansion proves the equivalence of (ii)–(iv). Assume (ii), and diagonalize the nonzero residual support:

The index set is finite or countable. Constancy implies that all code vectors are supported in $R\otimes \mathrm{supp}{\sigma }_B$, so write

The complementary matrix-unit identities give

Define $A(|i⟩|\alpha ⟩)=|u_{i\alpha }⟩$ and ${\sigma }_K={\sum }_{\alpha }{p}_{\alpha }|\alpha ⟩⟨\alpha |$. These vectors are orthonormal, proving the normal form and the dimension bound. The two terms in (eq:Schmidt-decoder) are CP and have traces summing to $\mathrm{Tr}Y$; substitution returns $X$.

Conversely, dilate an exact decoder by a Stinespring isometry $L:R\to C\otimes G$ Stinespring. The composite $Z=(L\otimes I_B)V$ has pure decoded output $|i⟩$ on input $|i⟩$, hence

Recovery of each off-diagonal matrix unit requires $⟨{\xi }_j,{\xi }_i⟩=1$. All ${\xi }_i$ are one vector $\xi$. Tracing the decoded output and $G$ gives

Every equality holds on all operators and therefore on all reference extensions.