Theorem: Sharp coefficient-access verdicts

For each sector, uniformly over gauge states and references:

1. both coefficient legs admit zero-error subsystem recovery;

2. one coefficient leg has optimal half-diamond error $1/2$ for complete qubit return;

3. no coefficient leg has optimal half-diamond error $3/4$;

4. one coefficient leg still exactly recovers the named effect $P_{+}=(I+X)/2$.

Items two and three are bounds over all receiver-only CPTP decoders, not merely over the printed programmes.

Proof. Item one is Theorem thm:exact.

The channel retaining one leg is measure-and-prepare after discarding $K$, hence entanglement breaking on $C$. Fixing $K$ to a pure state and using a Bell input on $QC$, every receiver-only decoded output is separable across $Q$ and the returned qubit. Its Bell overlap is at most $1/2$: for a pure product vector this is Cauchy–Schwarz, and convexity gives the mixed separable case. The Bell projector therefore gives error at least $1/2$.

Equation (eq:one-leg) attains the bound. Indeed ${\Delta }_X=(\mathrm{id}+{\mathrm{Ad}}_X)/2$, so the triangle inequality gives error at most $1/2$, while the Bell output has orthogonal Bell components of weights $1/2,1/2$. Composition with ${\mathrm{Tr}}_K$ gives the same uniform upper bound.

Without coefficient legs, the accessible channel is independent of the qubit. For a fixed sector every decoded qubit is a replacement state $\tau$. Its Bell-input output has overlap $1/4$, so the error is at least $3/4$. Returning $I/2$ attains $3/4$, because qubit depolarization is the equal mixture of the four Pauli conjugations. Again composition with ${\mathrm{Tr}}_K$ gives a uniform bound.

Finally, ${\Delta }_X^{\ast }(P_{+})=P_{+}$. Before the one-leg inverse, the corresponding received effect is

Its pullback is $P_{+}\otimes I_K$.