Finite capped amplification succeeds with a supplied lower bound

For $0<\sin (2\varphi )$, the addition identity in [F6] gives $\sin ((4j+4)\varphi )-\sin (4j\varphi )=2\cos ((4j+2)\varphi )\sin (2\varphi )$. Summing over $j=0,\dots ,K-1$ telescopes, so ${\sum }_j\cos ((4j+2)\varphi )=\sin (4K\varphi )\mathrm{/}(2\sin (2\varphi ))$. The double-angle identities in [F6] therefore give ${\sum }_{j=0}^{K-1}{\sin }^2((2j+1)\varphi )=K\mathrm{/}2-\sin (4K\varphi )\mathrm{/}(4\sin (2\varphi ))$, and the corresponding cosine-square sum has a plus sign. Here the denominators are positive except at the separately treated endpoint $p=1$. Put $\theta =\arcsin \sqrt{p}$. For $p\le 1\mathrm{/}2$, $\sin (2\theta )=2\sqrt{p(1-p)}\ge \sqrt{2p}\ge 1\mathrm{/}(2\sqrt{2}\kappa )$. The finite sine-square identity from [F6] gives average success $\frac{1}{K}{\sum }_{j=0}^{K-1}{\sin }^2((2j+1)\theta )=\frac{1}{2}-\frac{\sin (4K\theta )}{4K\sin (2\theta )}\ge \frac{1}{2}-\frac{1}{4K\sin (2\theta )}\ge \frac{1}{2}-\frac{1}{4\sqrt{2}}>\frac{1}{4}\mathrm{.}$ For $1\mathrm{/}2<p<1$, put ${\theta }^{\mathrm{\prime }}=\pi \mathrm{/}2-\theta \in [0,\pi \mathrm{/}4)$. The cosine-square version of [F6] gives average success $\frac{1}{2}+\frac{\sin (4K{\theta }^{\mathrm{\prime }})}{4K\sin (2{\theta }^{\mathrm{\prime }})}$. If its sine numerator is nonnegative this is at least $1\mathrm{/}2$. If it is negative, then $4K{\theta }^{\mathrm{\prime }}>\pi$, and [F5] gives $\sin (2{\theta }^{\mathrm{\prime }})\ge 4{\theta }^{\mathrm{\prime }}\mathrm{/}\pi$, so the magnitude of the fraction is at most $\pi \mathrm{/}(16K{\theta }^{\mathrm{\prime }})<1\mathrm{/}4$. The average is again greater than $1\mathrm{/}4$. At $p=1$, every iteration succeeds. Thus uniform accepted $j$ has average success at least $1\mathrm{/}4$ for every $p\ge p_0$.

Since $K\le 2^b<2K$, the random-bit draw is accepted with probability $K\mathrm{/}{2}^b>1\mathrm{/}2$ and, conditioned on acceptance, is uniform on $\{0,\dots ,K-1\}$. Each fresh round therefore succeeds with probability at least $1\mathrm{/}8$. Independence of fresh preparation, bits and verification gives total failure at most $(7\mathrm{/}8{)}^R\le e^{-R\mathrm{/}8}$, since [F7] gives $\ln (8\mathrm{/}7)={\int }_1^{8\mathrm{/}7}dt\mathrm{/}t\ge (1\mathrm{/}7)(7\mathrm{/}8)=1\mathrm{/}8$ and $R\ge 16\ln (1\mathrm{/}\rho )$. Therefore the failure is at most $e^{-2\ln (1\mathrm{/}\rho )}={\rho }^2\le \rho$.

At most $K-1$ iterations occur in one round, each using one $V$ and one $V^{†}$ besides the initial $V$; hence each call count is at most $RK$, and their sum is at most $2RK$. The good component remains collinear by [F1, F3]; its scalar rotation coefficient may be negative, but every positive-probability well branch therefore has the same pure-state ray as the normalized initial well component.