Robust one-step OAA including ancilla leakage

By [F10], $J^{†}J=I$ and $\mathrm{\Pi }$ is an orthogonal projection. By the circuit order in [L1], the factors in $C$ are unitary by [F2]–[F4], so $CJ$ is an isometry. Put $D:=J^{†}CJ$. By [F1] and the minus sign in $C$, $D=3M\mathrm{/}2-M{M}^{†}M\mathrm{/}2$. Write $E=M-V$. For every unit vector $x$, $\mathrm{\parallel }Mx\mathrm{\parallel }\le \mathrm{\parallel }Vx\mathrm{\parallel }+\mathrm{\parallel }Ex\mathrm{\parallel }\le 1+\eta$, so $\mathrm{\parallel }M\mathrm{\parallel }\le 1+\eta$ by [F5]. For each endomorphism $T$ of the nonzero space $S$, Cauchy–Schwarz gives $\mathrm{\parallel }Tx\mathrm{\parallel }={\max }_{\mathrm{\parallel }y\mathrm{\parallel }=1}\mathrm{\mid }⟨Tx,y⟩\mathrm{\mid }$ (with $Tx=0$ immediate), hence $\mathrm{\parallel }T\mathrm{\parallel }={\max }_{\mathrm{\parallel }x\mathrm{\parallel }=\mathrm{\parallel }y\mathrm{\parallel }=1}\mathrm{\mid }⟨Tx,y⟩\mathrm{\mid }$. The adjoint identity interchanges the unit vectors, so $\mathrm{\parallel }{T}^{†}\mathrm{\parallel }=\mathrm{\parallel }T\mathrm{\parallel }$; in particular $\mathrm{\parallel }{M}^{†}\mathrm{\parallel }\le 1+\eta$ and $\mathrm{\parallel }{E}^{†}\mathrm{\parallel }=\mathrm{\parallel }E\mathrm{\parallel }\le \eta$.

Since $V{V}^{†}V=V$, $M{M}^{†}M-V=(M-V)M^{†}M+V(M^{†}-V^{†})M+V{V}^{†}(M-V)$. For every unit vector, apply the vector triangle inequality [F6] and [F7] to these three terms; using $\mathrm{\parallel }{E}^{†}\mathrm{\parallel }=\mathrm{\parallel }E\mathrm{\parallel }\le \eta$ and $\mathrm{\parallel }V\mathrm{\parallel }=\mathrm{\parallel }{V}^{†}\mathrm{\parallel }=1$ from [F4] gives $\mathrm{\parallel }M{M}^{†}M-V\mathrm{\parallel }\le \eta ((1+\eta {)}^2+(1+\eta )+1)$. Maximizing over unit vectors by [F5], $\mathrm{\parallel }D-V\mathrm{\parallel }\le \frac{3}{2}\eta +\frac{1}{2}\eta ((1+\eta {)}^2+(1+\eta )+1)\le 4\eta$ for $\eta \le 1\mathrm{/}100$.

Let $L=(I-\mathrm{\Pi })CJ$. Since $\mathrm{\Pi }=J{J}^{†}$ and $J^{†}J=I$ by [F10], $CJ=JD+L$ and $J^{†}L=J^{†}(I-J{J}^{†})CJ=J^{†}CJ-(J^{†}J)J^{†}CJ=0$. Hence both cross terms vanish in $(CJ{)}^{†}CJ=(JD+L{)}^{†}(JD+L)$, and $D^{†}D+L^{†}L=(CJ{)}^{†}CJ=I$ because $C$ and $J$ are isometries. Thus $L^{†}L=I-D^{†}D$ and $\mathrm{\parallel }L{\mathrm{\parallel }}^2=\mathrm{\parallel }I-D^{†}D\mathrm{\parallel }$ by [F7].

Set $\delta =\mathrm{\parallel }D-V\mathrm{\parallel }\le 4\eta$ by step 2.1. For every unit vector $x$, $\mathrm{\parallel }Dx\mathrm{\parallel }\le \mathrm{\parallel }Vx\mathrm{\parallel }+\mathrm{\parallel }(D-V)x\mathrm{\parallel }\le 1+\delta$, so $\mathrm{\parallel }D\mathrm{\parallel }\le 1+\delta$. Since $V^{†}V=I$, $I-D^{†}D=(V-D{)}^{†}D+V^{†}(V-D)$; the vector triangle inequality [F6] and [F7] therefore give $\mathrm{\parallel }(I-D^{†}D)x\mathrm{\parallel }\le \delta \mathrm{\parallel }D\mathrm{\parallel }+\delta \le \delta (2+\delta )$ for unit $x$. Maximizing by [F5], $\mathrm{\parallel }I-D^{†}D\mathrm{\parallel }\le \delta (2+\delta )\le 4\eta (2+4\eta )\le 9\eta$. Thus $\mathrm{\parallel }L\mathrm{\parallel }\le 3\sqrt{\eta }$.

From $CJ-JV=J(D-V)+L$, for every unit $x$ the vector triangle inequality [F6] and the isometry of $J$ in [F10] give $\mathrm{\parallel }(CJ-JV)x\mathrm{\parallel }\le \mathrm{\parallel }D-V\mathrm{\parallel }+\mathrm{\parallel }L\mathrm{\parallel }\le 4\eta +3\sqrt{\eta }$. Maximizing by [F5] and using $\eta \le 1\mathrm{/}100$ yields $\mathrm{\parallel }CJ-JV\mathrm{\parallel }\le 10\sqrt{\eta }$.

For any finite reference space $R$, choose an orthonormal basis $(e_j)$ by [F9] and write $\xi ={\sum }_j{x}_j\otimes e_j$. By [F8], $\mathrm{\parallel }(CJ-JV)\otimes I_R\xi {\mathrm{\parallel }}^2={\sum }_j\mathrm{\parallel }(CJ-JV)x_j{\mathrm{\parallel }}^2\le (10\sqrt{\eta }{)}^2{\sum }_j\mathrm{\parallel }{x}_j{\mathrm{\parallel }}^2=(10\sqrt{\eta }{)}^2\mathrm{\parallel }\xi {\mathrm{\parallel }}^2$. Thus the same clean-input isometry bound holds with any finite reference, including entangled inputs.