Proposition: Flagged transfer and its complete destination

Let

where $C_R$ and $C_B$ are copies of $C$, and let $0\le s\le 1$. Define

Then $V_s$ is an isometry, and

Exactly,

The complete output recovers the code for every $s$.

Proof. The two flag sectors are orthogonal, so the squared norm is preserved and the displayed partial traces follow.

The decoder that returns the successful code block and prepares ${\tau }_C$ on the erasure block realizes

Its error is $s{a}_d$.

For any decoder, the failure block is independent of the input. On the Bell input, its decoded target overlap is $d^{-2}$. The successful block has overlap at most one. Therefore every decoder has Bell overlap at most $1-s+s/d^2$, giving error at least $s{a}_d$.

For leakage, dephase a candidate replacement state between code and flag and twirl its code block over the Weyl unitaries. The phase averaging leaves ${\mathcal{C}}_s$ unchanged; the Weyl averaging uses its covariance and input-unitary invariance of diamond distance. Convexity thus permits the replacement state

On the Bell input, the half trace-norm difference is

If $q\le s$, this equals $s-q/d^2\ge s{a}_d$. If $q\ge s$, replacing $\left|s-q/d^2\right|$ by its lower bound $s-q/d^2$ gives a value at least $q{a}_d\ge s{a}_d$. Thus every replacement has distance at least $s{a}_d$.

For $q=s$, the channel difference is the code-block embedding of $s({\mathrm{id}}_C-{\mathcal{T}}_{{\tau }_C})$. Lemma lem:unitary-mixture gives the matching upper bound. Complete recovery is Theorem thm:global.