Theorem: Exact full-transfer noise comparison

At full transfer,

Both the specified and balanced decoders attain the optimum. For their decoded Bell output,

The accessible Bell-input marginal is

independently of $\lambda$.

With

and continuous endpoint values,

Proof. The decoded channel is ${\Delta }_{\lambda }=a_{\lambda }{\mathrm{id}}_M+b_{\lambda }{\mathrm{Ad}}_Z$. The Pauli formula gives exact error $b_{\lambda }$. Equation (eq:noisyBellOpt) gives every decoder Bell overlap at most $a_{\lambda }$, so the Bell witness gives the matching error lower bound.

Alternatively, the noisy accessible outputs of logical $|+⟩$ and $|-⟩$ have distance $\lambda$. Contraction gives $1\le 2\epsilon +\lambda$, yielding the same lower bound for every decoder.

The decoded Bell state is

Its Bell overlap is $a_{\lambda }$. Its spin-flip square-root eigenvalues are $a_{\lambda },b_{\lambda },0,0$, giving concurrence $\lambda$.

Tracing the reference gives (eq:fullSameMarginal). The environment has vacuum $E$, and

Hence $S(EF)=h_2(a_{\lambda })$, while $S(Q)=S(R)=1$. Purity of $QREF$ gives the two information identities.