Example: A diary-independent bit can assist recovery

For a qubit, consider

The radiation channel is ${\Delta }_Z=(\mathrm{id}+{\mathrm{Ad}}_Z)/2$, with optimal recovery error $1/2$. Computational readout of $B$ gives a uniform bit $x$. Supplying $x$ to the observer and applying $Z^x$ to $R$ gives exact quantum recovery. The bit alone contains no diary information.

Proof. The unnormalized branch is

Correction by $Z^x$, followed by summation, returns $\rho$. Without correction, summation returns ${\Delta }_Z(\rho )$. Each outcome has probability $1/2$, independently of the input.

The two inputs $|+⟩$ and $|-⟩$ have identical dephased outputs. If a decoder has error $\epsilon$, the triangle inequality and contraction imply $1\le 2\epsilon$. The identity decoder attains $1/2$ by Lemma lem:unitary-mixture.