Proposition: Full-code recovery and record-only obstruction

Let $\mathcal{N}:L(\mathcal{H})\to L(\mathcal{Q})$ be CPTP.

1. A CPTP left inverse on the entire input code exists if and only if a complete Stinespring complementary channel is a replacement channel

   $$\hat{\mathcal{N}}(X)=\mathrm{Tr}(X)\omega$$

   for a fixed state $\omega$. Equivalently, for Kraus operators $E_i$, every ${E_j}^{†}{E}_i$ is a scalar multiple of $I_{\mathcal{H}}$.

2. Let $\dim \mathcal{H}=D\ge 2$. If $\mathcal{E}$ is measure-and-prepare and $\mathcal{D}$ is any CPTP decoder to $\mathcal{H}$, then in half-diamond normalization,

   $$\frac{1}{2}{\Vert \mathcal{D}\mathcal{E}-{\mathrm{id}}_{\mathcal{H}}\Vert }_{\diamond }\ge 1-\frac{1}{D}.$$

   In particular, the bound applies when only a finite classical record is retained. Computational dephasing attains the bound.

The first statement concerns the full input code, not arbitrary subsystem or subalgebra recovery.

Proof. A standard complement has entries

It is a replacement channel exactly when ${E_j}^{†}{E}_i$ is scalar for every $i,j$. If $\mathcal{DN}=\mathrm{id}$, with decoder Kraus operators $K_a$, the rank-one Choi matrix of the identity implies $K_a{E}_i=c_{ai}I$. Hence

Conversely, diagonalizing the scalar Gram matrix gives

With $\Pi ={\sum }_i{W}_i{W}_i^{†}$, the channel

for any fixed state $\tau$ is a CPTP left inverse.

For the second assertion, a normalized maximally entangled input $|{\Phi }_D⟩$ gives a separable reference–decoded-output state under every measure-and-prepare channel followed by a decoder. Its overlap with ${\Phi }_D$ is at most $1/D$: this follows for product vectors by Cauchy–Schwarz and for separable states by convexity. Testing the Bell projector gives eq:classical-gap.

A channel to a finite classical register has form

a decoder prepares a state for each $x$, so the composite is measure-and-prepare. A finite adaptive process with only a final classical output has the same channel form.

For attainability, if $Z_D={{\sum }_{j=0}}^{D-1}{e}^{2\pi ij/D}|j⟩⟨j|$, computational dephasing is

The triangle inequality gives half-diamond distance at most $(D-1)/D$ from the identity, while the Bell test gives the matching lower bound.