Qubitization and Near-Optimal Hamiltonian Simulation — Examples

Since $s^2=1-x^2$, direct multiplication gives $U_x^{\ast }=U_x$ and $U_x^2=I$. Also $J^{\ast }{U}_{x}J=[x]=K\mathrm{/}\alpha$, so $U_x$ is unitary and has the claimed block encoding. Its clean and complementary components are $U_x\mathrm{\mid }0⟩=x\mathrm{\mid }0⟩+s\mathrm{\mid }1⟩$.

If $\mathrm{\mid }x\mathrm{\mid }<1$, then $s>0$, so the normalized complement defined in [F1] is $\mathrm{\mid }1⟩$; both input and output paired bases are $(\mathrm{\mid }0⟩,\mathrm{\mid }1⟩)$, and the matrix is exactly $U_x$. At $x=0$, $s=1$, hence $U_0=X$, confirming that zero is a regular two-dimensional signal plane.

If $x=1$ or $x=-1$, then $s=0$ and $U_x\mathrm{\mid }0⟩=x\mathrm{\mid }0⟩=U_x^{\ast }\mathrm{\mid }0⟩$. No complement vector is normalized in these cases; this verifies the endpoint convention and the encoded eigenvalue $\lambda \mathrm{/}\alpha =x$.

Substituting $n=2$ gives $w_1=w_2=\sqrt{2}\mathrm{/}2$. For $b\in \{0,1\}$, edge 1 joins $\mathrm{\mid }0,b⟩$ to $\mathrm{\mid }1,b\oplus x_1⟩$, and edge 2 joins $\mathrm{\mid }1,b⟩$ to $\mathrm{\mid }2,b\oplus x_2⟩$. Starting with $b=0$, these two edges produce the four paths in the table; each endpoint bit is $x_1\oplus x_2$. Each internal path vertex has degree two and each endpoint degree one.

Therefore $d_H=2$ and $h_{\max }=1\mathrm{/}\sqrt{2}$, so [F2] gives ${\alpha }_{\mathrm{s}\mathrm{p}}=d_H{h}_{\max }=\sqrt{2}$. This exact-weight example illustrates the path underlying [F3], whose lower bound instead uses rounded weights: at $n=2$ its grid exponent is $b=5$, so its two weights are $23\mathrm{/}32$ and its sparse normalization is $23\mathrm{/}16$. Thus [F3] does not identify this exact matrix with its rounded family or supply a lower bound for arbitrary block-encoding access.

In the parity-free basis, the path matrix is $H_0=\frac{1}{\sqrt{2}}\left(\begin{array}{ccc}0& 1& 0\\ 1& 0& 1\\ 0& 1& 0\end{array}\right),H_0^2=\frac{1}{2}\left(\begin{array}{ccc}1& 0& 1\\ 0& 2& 0\\ 1& 0& 1\end{array}\right)\mathrm{.}$ Its eigenvalues are $1,0,-1$, so spectral functional calculus gives $e^{-i\pi H_0}=I-2H_0^2$, which sends $\mathrm{\mid }0⟩$ to $-\mathrm{\mid }2⟩$. The prefix-parity basis permutation of [F1] restores the endpoint bit $x_1\oplus x_2$; equivalently, [F1] gives $(-i{)}^2=-1$ times the same endpoint.

The recurrence gives $T_2(x)=2x{T}_1(x)-T_0(x)=2x^2-1$. The $k=0$ term of the Chebyshev expansion is $J_0$; the $k=1$ coefficient is $2(-i)J_1=-2i{J}_1$; and the $k=2$ coefficient is $2(-i{)}^2{J}_2=-2J_2$. Therefore the degree-two truncation is exactly the displayed $P_2$.

The series in [F1] is real for real $\tau$, so $J_0,J_1,J_2$ are real. Since $T_2$ is even and $T_1$ is odd, $\mathrm{Re}{P}_2=J_0-2J_2{T}_2$ is even and $\mathrm{Im}{P}_2=-2J_1{T}_1$ is odd. Applying [F2] with $d=2$ gives the uniform tail $2{\sum }_{k>2}\mathrm{\mid }{J}_k(\tau )\mathrm{\mid }$; its factorial denominator condition is $\mathrm{\mid }\tau \mathrm{\mid }\mathrm{/}[2(2+2)]=\mathrm{\mid }\tau \mathrm{\mid }\mathrm{/}8<1$.

At $\tau =0$, the series gives $J_0(0)=1$ and $J_1(0)=J_2(0)=0$, so $P_2(x)=1=e^0$. At $x=\pm 1$, $T_1(x)=x$ and $T_2(x)=1$, and the same tail bound from [F2] applies without an endpoint exception.