Operator error for the finite sine-prefix rotation table

The real $Y$-rotation matrix is $R_y(\theta )=\left(\begin{array}{cc}\cos (\theta \mathrm{/}2)& -\sin (\theta \mathrm{/}2)\\ \sin (\theta \mathrm{/}2)& \cos (\theta \mathrm{/}2)\end{array}\right)$. For any unit vector $x$, [F5] and [F4] give $\mathrm{\parallel }(R_y(\beta )-R_y(\alpha ))x{\mathrm{\parallel }}^2=\mathrm{\parallel }(R_y(\beta -\alpha )-I)x{\mathrm{\parallel }}^2=4{\sin }^2((\beta -\alpha )\mathrm{/}4)\mathrm{.}$ Taking the supremum over unit $x$ and using [F3] and [F4] yields $\mathrm{\parallel }{R}_y(\beta )-R_y(\alpha )\mathrm{\parallel }\le \mathrm{\mid }\beta -\alpha \mathrm{\mid }$. A controlled rotation is block diagonal with identity and this rotation on its control subspaces, so its operator error obeys the same bound.

For each leaf $\tau$, put $x_{\tau }=\pi (\tau +1\mathrm{/}2)\mathrm{/}T\in (0,\pi )$. If $x_{\tau }\le \pi \mathrm{/}2$, [F3] gives $\sin x_{\tau }\ge 2x_{\tau }\mathrm{/}\pi >0$; if $x_{\tau }>\pi \mathrm{/}2$, [F8] reduces positivity to $\pi -x_{\tau }\in (0,\pi \mathrm{/}2)$ and the same bound. Thus each leaf weight is positive, and [F7] makes every split ratio lie in $(0,1)$. The rotation sends $\mathrm{\mid }0⟩$ to $\sqrt{1-r}\mathrm{\mid }0⟩+\sqrt{r}\mathrm{\mid }1⟩$, so its branch amplitudes are exactly the square roots of the right- and left-child mass fractions. Induction from the root shows that leaf $\tau$ has amplitude $\sqrt{w_{\tau }\mathrm{/}{W}_T(0,T)}=\sqrt{2\mathrm{/}T}\sin x_{\tau }$, by [F6] and [F7]. With left encoded by $1$, interval leaf $\tau$ has computational label $T-1-\tau$. Its amplitude equals that of label $\tau$ because $x_{T-1-\tau }=\pi -x_{\tau }$ and [F8] gives $\sin x_{T-1-\tau }=\sin x_{\tau }$. Hence the ideal circuit prepares $\mathrm{\mid }{\mathrm{\Psi }}_T⟩$.

For products ${\tilde{U}}_N\cdots {\tilde{U}}_1$ and $U_N\cdots U_1$, insert and subtract one factor at a time. This writes their difference as a sum of $N$ terms, each containing one ${\tilde{U}}_j-U_j$ between unitary products. By [F1] the outer products preserve norm, and [F2] bounds the sum by $N\delta$. For ${ϵ}_{\mathrm{\Psi }}>0$, $N=T-1$ and $\delta ={ϵ}_{\mathrm{\Psi }}\mathrm{/}(100T)$ this is less than ${ϵ}_{\mathrm{\Psi }}\mathrm{/}100$. Applying the same estimate to the inverse rotation list, whose ideal angles are negated and whose synthesis uses the same per-angle tolerance, gives the same bound for inverse preparation.