Sparse diagonal block-encoded linear system

The matrices $A,S$ are self-adjoint, commute, and satisfy $A^2+S^2=I$. Thus $U^{\ast }=U$ and $U^2=\left(\begin{array}{cc}{A}^2+S^2& AS-SA\\ SA-AS& S^2+A^2\end{array}\right)=I$. The all-zero ancilla compression of $U$ is its top-left block $A$, so [F1] gives an exact $(1,1,0)$ block encoding.

The diagonal entries of $A$ are its singular values. Both inputs are supported on singular values at least $\gamma =1\mathrm{/}2$, and $\mathrm{\parallel }A{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}=\alpha =1$, so $\delta =\gamma \mathrm{/}\alpha =1\mathrm{/}2$. For $x\in [1\mathrm{/}2,1]$, $-x\le 2x^2-1\le x$ because $2x^2+x-1=(2x-1)(x+1)\ge 0$ and $x-(2x^2-1)=(1-x)(2x+1)\ge 0$. Hence $\mathrm{\mid }p(x)-\delta \mathrm{/}(2x)\mathrm{\mid }=\mathrm{\mid }2x^2-1\mathrm{\mid }\mathrm{/}(4x)\le 1\mathrm{/}4$. Also $p$ is odd, has degree one, satisfies $p(0)=0$, and $\mathrm{\mid }p(x)\mathrm{\mid }\le 1\mathrm{/}2$ on $[-1,1]$.

For $\mathrm{\Phi }=(0,\pi \mathrm{/}3)$, [F2] gives upper-left entry $e^{i\pi \mathrm{/}3}x$; the conjugate list gives $e^{-i\pi \mathrm{/}3}x$. Their coherent real extraction is $\frac{1}{2}(e^{i\pi \mathrm{/}3}+e^{-i\pi \mathrm{/}3})x=\frac{1}{2}x=p(x)$ by [F3]. Thus the exact response multipliers on the two singular vectors are $p(1)=1\mathrm{/}2$ and $p(1\mathrm{/}2)=1\mathrm{/}4$.

For $\mathrm{\mid }b⟩=\mathrm{\mid }0⟩$, the QSVT success probability is $\mathrm{\mid }p(1){\mathrm{\mid }}^2=1\mathrm{/}4$ and the ideal inverse-block probability is $\mathrm{\mid }(\delta \mathrm{/}2)(1\mathrm{/}1){\mathrm{\mid }}^2=1\mathrm{/}16$. For $\mathrm{\mid }b⟩=\mathrm{\mid }1⟩$, these probabilities are $\mathrm{\mid }p(1\mathrm{/}2){\mathrm{\mid }}^2=1\mathrm{/}16$ and $\mathrm{\mid }(\delta \mathrm{/}2)(1\mathrm{/}(1\mathrm{/}2)){\mathrm{\mid }}^2=1\mathrm{/}4$. Each successful state is the same basis vector as the normalized QLSP solution because the singular space is one-dimensional and both multipliers are positive. For the optional Bernstein reciprocal check, set $x=(1+u)\mathrm{/}2$ for $0\le u\le 1$. Then the two polynomials are $\frac{1}{4}(2x^2+x-1)=\frac{1}{8}(3u+u^2)$ and $\frac{1}{4}(1+x-2x^2)=\frac{1}{8}(2-u-u^2)$, with degree-two Bernstein coefficient lists $(0,3\mathrm{/}16,1\mathrm{/}2)$ and $(1\mathrm{/}4,3\mathrm{/}16,0)$. All coefficients are nonnegative, so the finite reciprocal certificate is valid.