与标准用法有两处替换：受控算子取 $U := G$，下方输入不用难以制备的本征态，改用易制备的 $|\psi\rangle$。由本征态定义反解 $|g\rangle = (|u_1\rangle+|u_2\rangle)/\sqrt{2}$、$|b\rangle = (-i/\sqrt{2})(|u_1\rangle-|u_2\rangle)$，代入展开得 $|\psi\rangle = C_1|u_1\rangle + C_2|u_2\rangle$，其中 $C_1 = (\sqrt{p}-i\sqrt{1-p})/\sqrt{2}$、$C_2 = (\sqrt{p}+i\sqrt{1-p})/\sqrt{2}$，模方 $|C_1|^2 = |C_2|^2 = 1/2$，总和为 1。相位估计是线性演化，对每个本征分量独立作用：$|0\rangle|\psi\rangle \to C_1|\tilde{\phi}_1\rangle|u_1\rangle + C_2|\tilde{\phi}_2\rangle|u_2\rangle$；本征值 $e^{\pm i\theta}$ 对应目标整数 $\phi_1 = N\theta/2\pi$、$\phi_2 = N - N\theta/2\pi$——$\theta$ 小时一峰近 0、一峰近 $N$，读数等概率落在两峰附近。反解链：$\hat{\phi} = 2\pi\tilde{\phi}/N$，若 $\hat{\phi} > \pi$ 取 $\theta = 2\pi - \hat{\phi}$，再由 $p = (1-\cos\theta)/2 = \sin^2(\theta/2)$ 还原。演算 $\theta = \pi/4$、$N = 8$：$p = \sin^2(\pi/8) = (2-\sqrt{2})/4 \approx 0.146$，$\phi_1 = 1$、$\phi_2 = 7$ 各以 $1/2$ 概率读出；读到 7 时 $\hat{\phi} = 7\pi/4 > \pi$，$\theta = 2\pi - 7\pi/4 = \pi/4$，还原同一个 $p$，往返一致。量子计数至此完成：两个峰任取其一都能还原 $p = \sin^2(\theta/2)$。