Theorem: Exact calibrated recovery and memory optima

For $a=a(K)$, both parities satisfy

For even $L$,

For odd $L$, ${\epsilon }_{\mathrm{op}}(2,3)=3(1-a)/4$.

Proof. Conjugation gives

Hence ${\mathcal{A}}_L(\rho )=(I+x{P}_X+y{P}_Y+z{P}_Z)/8$. Every proper even subset is constant. Applying $U{F}^L$ to the full triple and discarding factors $1,3$ recovers the message.

The occupied Bell overlap cannot exceed one and the vacuum-branch overlap is $1/4$. Lemma lem:bell therefore gives the unrestricted lower bound $3(1-a)/4$. Recover on the occupied branch and prepare $I/2$ on vacuum to attain it.

At capacity two, no quantum slot remains from the first packet at the second atomic arrival. An arbitrary first-packet instrument therefore leaves only a classical outcome. On the occupied rail, its effect is $E=(\mathrm{tr}E)(I+v_{x}X+v_{y}Y+v_{z}Z)/2$, $|v|\le 1$. The normalized conditional reference–future state is

where $A=X_2{Z}_3$, $B=Y_2{X}_3$, $C=Z_2{Y}_3$, $AB=-C$. It equals

The reference factors are positive and the joint projectors are rank one, so the state is separable across $R\mid 23$. This also covers arbitrary first-packet label mixing: the input branch blocks are already incoherent and orthogonal. Later operations preserve separability. The Bell overlap is at most $a/2+(1-a)/4$, giving $3/4-a/4$.

Measuring $Z_1,X_2,Z_3$ and preparing the logical $X$ eigenstate of their negative parity realizes

which attains the bound. The explicit decoders below attain unrestricted recovery at the remaining capacities.