Proposition: Unitary dephasing retains the code globally but not locally

The isometry

has both reduced channels equal to computational dephasing:

Neither output alone recovers an arbitrary qubit. Its optimal local recovery error is $1/2$, whereas the complete output is exactly recoverable.

Proof. Tracing either output annihilates $E_{01}$ and $E_{10}$. The two inputs $|+⟩$ and $|-⟩$ have identical dephased outputs but orthogonal ideal targets. If a decoder has error $\epsilon$, the triangle inequality therefore gives $1\le 2\epsilon$.

The identity decoder attains $1/2$, since

and $d_{\diamond }({\mathrm{Ad}}_Z,\mathrm{id})=1$, witnessed by $|+⟩$. Global recovery follows from Theorem thm:global.