Two-term LCU normalization and access count

The coordinate formulas give $\mathrm{\parallel }{A}_0{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}=\mathrm{\parallel }{A}_1{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}=1$, so the chosen ${\alpha }_j$ satisfy the normalization promises. For a positive diagonal contraction $C$, put $S:=\sqrt{I-C^2}$ and $D(C):=\left(\begin{array}{cc}C& S\\ S& -C\end{array}\right)$. Since $C$ and $S$ are commuting self-adjoint diagonal matrices and $C^2+S^2=I$, direct block multiplication gives $D(C{)}^{\ast }=D(C)$ and $D(C{)}^2=I$, hence $D(C{)}^{\ast }D(C)=I$. The adjoint identity [F5] shows $\mathrm{\parallel }D(C)x{\mathrm{\parallel }}^2=⟨x,D(C{)}^{\ast }D(C)x⟩=\mathrm{\parallel }x{\mathrm{\parallel }}^2$, and $D(C{)}^2=I$ makes it invertible, so [F6] makes it unitary. Its top-left block is $C$. Thus $D(A_0\mathrm{/}{\alpha }_0)$ and $D(A_1\mathrm{/}{\alpha }_1)$ are exact unitary block encodings with one component ancilla each.

Here $\mathrm{\Lambda }=\mathrm{\mid }1\mathrm{\mid }\cdot 1+\mathrm{\mid }i\mathrm{\mid }\cdot 2=3$, so $\mathrm{\mid }\chi ⟩=\sqrt{1\mathrm{/}3}\mathrm{\mid }0⟩+\sqrt{2\mathrm{/}3}\mathrm{\mid }1⟩$. By [F1] and [F2], the projected block is $\frac{1}{3}{A}_0+\frac{2}{3}i(A_1\mathrm{/}2)=(A_0+i{A}_1)\mathrm{/}3=\mathrm{diag}(1\mathrm{/}3,i\mathrm{/}3)$; the normalization is $3$ and the error is zero.

With PREPARE, SELECT, and PREPARE${}^{†}$ separately supplied, the construction uses one call to each. If SELECT is assembled from component calls, its declaration must include one controlled call for each component branch and the relative phase operation; if those circuits are compiled from matrices, their data access and gate costs are added separately.