Theorem: Exact complete-complement criterion

For the complete interface (eq:interface), the following are equivalent:

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

2. $\mathcal{E}={\mathcal{T}}_{{\sigma }_B}$ for one ${\sigma }_B\in \mathcal{D}(B)$.

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

4. $J_{\mathrm{n}}(\mathcal{E})=I_Q/d\otimes {\sigma }_B$.

5. For every $O\in \mathcal{L}(B)$,

   $$V^{†}(I_R\otimes O)V=\mathrm{Tr}(O{\sigma }_B)I_C.$$

6. For a complete Kraus family $\{K_a\}$ of $\mathcal{N}$, every ${K_a}^{†}{K}_b$ is a scalar multiple of $I_C$.

If $r=\mathrm{rank}{\sigma }_B$, there are a space $K\simeq {\mathbb{C}}^r$, a state ${\sigma }_K$, and an isometry $A:C\otimes K\to R$ such that

An exact decoder on all of $\mathcal{L}(R)$ is

In particular,

Every recovery identity in the theorem retains arbitrary references.

Proof. Matrix units span $\mathcal{L}(C)$, so (ii) and (iii) are equivalent. The displayed Choi expansion proves equivalence with (iv). Trace duality gives

which proves equivalence between (ii) and (v).

In a basis $\{|a⟩\}$ of $B$, choose $K_a=(I_R\otimes ⟨a|)V$. Then

Consequently (ii) implies (vi). Conversely, if these products are scalar, evaluation at $I_C/d$ shows that their scalar entries form exactly the density operator $\mathcal{E}(I_C/d)$. The same equality for every $X$ proves (ii).

Assume (ii). Diagonalize the nonzero part of the residual state:

Each $V|i⟩$ is supported on $R\otimes \mathrm{supp}{\sigma }_B$. Indeed, its squared component in the complementary residual support equals the expectation of that support projection in ${\sigma }_B$, which is zero. We may therefore write

Comparing the coefficient of $|b_{\alpha }⟩⟨b_{\beta }|$ in $\mathcal{E}(E_{ij})={\delta }_{ij}{\sigma }_B$ gives

These are $dr$ orthonormal vectors in $R$, proving (eq:rank-dimension). Define

The orthogonality makes $A$ an isometry, and tracing $B$ proves (eq:normal-form).

Both terms in (eq:exact-decoder) are completely positive. Their traces sum to $\mathrm{Tr}Y$. Substitution of (eq:normal-form) gives

Thus (ii) implies (i), with a decoder defined off the encoded range as well as on it.

Conversely, suppose (i). Take Kraus operators $D_{\gamma }$ of the decoder and form the isometry

This is the finite Kraus form of a Stinespring dilation Stinespring. Set $Z=(L\otimes I_B)V$. On basis input $|i⟩$, its decoded marginal is pure, so

for a normalized vector ${\xi }_i$. Exact recovery of $E_{ij}$ requires

Hence all the ${\xi }_i$ are one vector $\xi$. Because $L$ acts only on $R$, tracing its outputs cannot change the $B$-marginal. It follows that

This is (iii).

The identities hold on all matrix units and therefore on every reference extension by linearity.