Theorem: Exact SWAP and record-only calibration

For this whole-code target,

If only the computational record on $C_A$ is returned, the channel is

and its optimal recovery error is

Thus, for $n$ qubits,

These are analytic identities, not reported numerical executions.

Proof. The decoder returning $C_A$ and preparing $I_{C_D}/m$ has decoded channel ${\mathrm{id}}_{C_A}\otimes {\mathcal{T}}_{I/m}$, with error $1-m^{-2}$. For the lower bound, use independent Bell states for $C_A,C_D$. After ${\mathrm{Tr}}_{C_D}$, the reference $Q_D$ is $I/m$ and independent of everything available to the decoder. The decoded full Bell overlap is at most $m^{-2}$: tracing $Q_D$ against its Bell projector gives $I_{C_D}/m^2$, and the remaining Bell-projector expectation is at most one. This proves the recovery lower bound.

The complementary channel is ${\mathrm{Tr}}_{C_A}$ with output $C_D$. Choosing replacement $I/m$ gives leakage at most $1-m^{-2}$. Fixing $C_A$ and testing a Bell state on $C_D$ gives the reverse bound by Lemma lem:benchmarks.

For the record channel, write $F_x=|x⟩⟨x|\otimes I_m$ and let ${\sigma }_x$ be the decoder output on record $x$. Its Bell overlap is

For attainability, prepare $|x⟩⟨x|\otimes I_m/m$. The decoded channel is ${\Delta }_a\otimes {\mathcal{T}}_{I/m}$, the uniform mixture of $a{m}^2$ orthogonal-Bell-output unitary channels, one of which is the identity. Its distance from the identity is $1-1/(a{m}^2)$.