Theorem: Exact noisy full-transfer comparison

For the full-transfer dephasing experiment,

The specified decoder attains this optimum. For its Bell-input decoded output,

The Bell-input accessible marginal is independent of $\lambda$:

Writing

with its continuous endpoint values, the mutual informations are

At complete dephasing, $\lambda =0$, occupation delivery is perfect, decoded concurrence is zero, and the optimal recovery error is $1/2$.

Proof. On the logical qubit, the specified decoded channel is

The channels ${\mathrm{Ad}}_Z$ and $\mathrm{id}$ have half-diamond distance one: input $|+⟩$ gives orthogonal outputs. Hence

For the lower bound valid for every decoder, use logical inputs $|+⟩$ and $|-⟩$. Their noisy accessible outputs have trace distance $\lambda$, whereas their ideal targets have distance one. A decoder of error $\epsilon$ must therefore satisfy

Thus $\epsilon \ge (1-\lambda )/2$, proving optimality.

The decoded Bell state is

Its overlap with $|\Phi ⟩$ is $(1+\lambda )/2$. Its spin-flip square-root eigenvalues are $(1+\lambda )/2,(1-\lambda )/2,0,0$, so its concurrence is $\lambda$.

Tracing the reference in (eq:fullNoisy) gives (eq:identicalMarginal). The $F$ marginal is

Thus $S(EF)=h_2(a_{\lambda })$, while $S(Q)=S(R)=1$. The joint state $QREF$ is pure, so the same pure-state entropy argument as in Theorem thm:information gives (eq:noiseMIr) and (eq:noiseMIe).