Theorem: Sharp one-slot recovery error

Under Assumption ass:stream,

In particular exact recovery in this interface requires peak at least two, and Proposition prop:streamdecoder attains it.

Proof. The Bell-input state is, by (eq:N123),

Before carrier $2$ enters a one-slot apparatus, carrier $1$ must be consumed. Each complete classical history from that carrier has a POVM effect

For a positive-probability outcome, the conditional state is

where

The independent commuting operators $A,B$ have rank-one joint projectors

Consequently

Every first factor is a state because its Bloch vector has length at most one. The conditional state is separable across $Q\mid 23$. Subsequent decoder operations, including classical conditioning, preserve this separability.

A separable two-qubit state has Bell overlap at most $1/2$: the claim holds for pure product states by Cauchy–Schwarz and extends by convexity. The Bell-projector test therefore gives half-diamond error at least $1/2$.

For attainment, destructively measure $Z_1,X_2,Z_3$ sequentially, report the negative product of their eigenvalues, and prepare the corresponding logical $X$-eigenstate. This realizes

Its Bell overlap is $1/2$, and the displayed mixture gives half-diamond error at most $1/2$. The program uses one slot, including its final state preparation.