Theorem: Exact flagged calibration

For the schedule above,

For a pure referenced input ${\rho }_{QC}$, with marginal ${\rho }_C$,

All recovery claims are deterministic and reference preserving.

Proof. Let $P=J{J}^{†}$. The decoder

is CPTP and satisfies

Lemma lem:replacement gives the upper bound $q{a}_d$. On a Bell input, the occupied branch contributes at most $1-q$ to decoded Bell overlap. The vacant branch is input independent and contributes exactly $q/d^2$. The Bell test proves the matching lower bound. Endpoint $q=1$ is replacement.

For leakage, compare ${\mathcal{C}}_q$ to the replacement state

The difference is $q({\mathrm{id}}_C-{\mathcal{T}}_{I_C/d})$ on the occupied block, so the distance is $q{a}_d$, but this concerns the channel with occupied weight $q$, namely ${\mathcal{C}}_q$. Its distance to replacement is therefore $q{a}_d$. Accordingly, the asserted expression (eq:flagleak) would have the weights reversed. The correct complete leakage is

To prove the lower bound for (eq:correctflagleak), use the Bell input and the following effect: Bell projection on the occupied reference–memory block, together with the full vacant reference block. The actual output has probability one. A replacement output whose occupied probability is $s$ has probability

This alone gives a lower bound $s{a}_d$, not $q{a}_d$ uniformly in $s$, so we use a more precise block trace-norm calculation.

Pinching a candidate replacement state into the occupied and vacant blocks cannot increase distance. Write it as $s\sigma \oplus (1-s)$, and twirl its occupied state using logical Weyl unitaries. Covariance of the channel and convexity show that one can take $\sigma =I_C/d$. On the Bell input, the occupied block difference is

Its trace norm is

The vacant block difference has trace norm $|s-q|$. If $s\le q$, the sum is $2q-2s/d^2\ge 2q{a}_d$. If $q\le s\le q{d}^2$, it is $2s{a}_d\ge 2q{a}_d$. If $s\ge q{d}^2$, it is $2(s-q)\ge 2q{a}_d$. Thus every replacement has half-trace distance at least $q{a}_d$, and the upper bound proves (eq:correctflagleak).

The entropy formulas follow from orthogonal blocks. For a pure referenced input, the occupied $QR$ branch is pure, whereas its vacant branch has state ${\rho }_Q$, giving $S(QR)=h_2(q)+qS({\rho }_C)$. Substitution in mutual information gives (eq:referenceinformation).