取 $A = \mathrm{diag}(1, 2)$、$b = (1, 1)^{\top}$：经典解 $x = A^{-1}b = (1, 1/2)^{\top}$，范数平方 $1 + 1/4 = 5/4$，归一化 $(2/\sqrt{5}, 1/\sqrt{5})^{\top}$。第一步制备 $|b\rangle = (|0\rangle + |1\rangle)/\sqrt{2}$（一个 H 门）；第二步 QPE 取 $m = 2$、$t = \pi/2$，相位 $\phi_0 = (1\cdot\pi/2)/2\pi = 1/4 = 0.01_2$、$\phi_1 = (2\cdot\pi/2)/2\pi = 1/2 = 0.10_2$ 精确可表示，换算系数 $2\pi/(t\cdot 2^m) = 1$，时钟整数即特征值，输出 $(|0\rangle|01\rangle + |1\rangle|10\rangle)/\sqrt{2}$。第三步受控旋转取 $C = \lambda_{\min} = 1$：$\tilde{\lambda} = 1$ 时 $\theta = 2\cdot\arcsin(1) = \pi$，$R_y(\pi)|0\rangle = |1\rangle$ 必成功；$\tilde{\lambda} = 2$ 时 $\theta = 2\cdot\arcsin(1/2) = \pi/3$，$|1\rangle$ 振幅 $\sin(\pi/6) = 1/2$。第四步逆 QPE 还原时钟，第五步测辅助比特：$P_1 = (1/2)\cdot 1^2 + (1/2)\cdot(1/2)^2 = 1/2 + 1/8 = 5/8 \ge 1/\kappa^2 = 1/4$（$\kappa = 2$）。测得 $|1\rangle$ 后 B 坍缩为 $|x\rangle = (1/\sqrt{5/8})\cdot(1/\sqrt{2})\cdot(|0\rangle + (1/2)|1\rangle) = (2/\sqrt{5})|0\rangle + (1/\sqrt{5})|1\rangle$，$P(|0\rangle) = 4/5$、$P(|1\rangle) = 1/5$——振幅平方比 $4 : 1$ 与经典解分量比 $x_0 : x_1 = 1 : 1/2$ 一致，即 $|x\rangle \propto (1, 1/2)^{\top} = A^{-1}b$。