Finite braid compilation and adiabatic error budget

Put $S=U_T(1)A_0$ and $T_c=A_1{W}_c$, both isometries. With $Q=I-P(1)$, decompose $S-T_c=QS+(P(1)S-A_{1}V)+A_1(V-W_c)\mathrm{.}$ Since $A_{1}V$ lies in $\mathrm{ran}P(1)$, this identity is exact. The three terms have operator norms at most $A_m(1)$ by [F1], ${ϵ}_{\mathrm{t}\mathrm{r}}$ by the transport premise, and ${ϵ}_c$ because $A_1$ is an isometry. The triangle inequality gives $\mathrm{\parallel }S-T_c{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le A_m(1)+{ϵ}_{\mathrm{t}\mathrm{r}}+{ϵ}_c$.

For a pure logical input $\mathrm{\mid }\psi ⟩$, the output vectors $S\psi$ and $T_c\psi$ are unit vectors and differ by at most the step 1.1 operator bound. [F2] therefore bounds their output trace distance by that same number. For mixed $\rho$, decompose it as ${\sum }_j{p}_j\mathrm{\mid }{\psi }_j⟩⟨{\psi }_j\mathrm{\mid }$ by [F5]; trace-norm triangle and homogeneity [F4] give the same bound after mixing. Applying [F3] bounds exact-measurement total variation by $A_m(1)+{ϵ}_{\mathrm{t}\mathrm{r}}+{ϵ}_c$.

The implemented measurement differs from the exact output measurement by at most ${ϵ}_m$ by hypothesis. The total-variation triangle inequality adds this amount to step 2.1 and gives the claimed budget. The phase of each compiled braid is included in $W_c$ or corrected according to the phase-reference record; no phase is omitted.