- 命题框（Lemma 1–2）：$\mathrm{span}\{|x_0\rangle, |s\rangle\}$ 是 $U_O$ 与 $U_S$ 的公共不变子空间；但 $\langle x_0|s\rangle = 1/\sqrt{N} \ne 0$，生成元不正交，需 Gram–Schmidt 构造正交基
- 推导步骤（逐行）：
- $U_O|x_0\rangle = 2|x_0\rangle\langle x_0|x_0\rangle - |x_0\rangle = |x_0\rangle$；同理 $U_S|s\rangle = |s\rangle$（生成元映回自身）
- $U_O|s\rangle = 2|x_0\rangle\langle x_0|s\rangle - |s\rangle = (2/\sqrt{N})|x_0\rangle - |s\rangle \in$ 平面；同理 $U_S|x_0\rangle = (2/\sqrt{N})|s\rangle - |x_0\rangle \in$ 平面
- 构造：$|x_0^\perp\rangle = (\sqrt{N}\cdot|s\rangle - |x_0\rangle)/\sqrt{N-1}$；正交性演算 $\langle x_0|(\sqrt{N}|s\rangle - |x_0\rangle) = \sqrt{N}\cdot(1/\sqrt{N}) - 1 = 0$ ✓
- 范数演算：$\lVert\sqrt{N}|s\rangle - |x_0\rangle\rVert^2 = N - 1$ → $|x_0^\perp\rangle$ 归一 ✓
- 反解分解式：$|s\rangle = \sin\theta\cdot|x_0\rangle + \cos\theta\cdot|x_0^\perp\rangle$，其中 $\langle s|x_0\rangle = 1/\sqrt{N} = \sin\theta$、$\langle s|x_0^\perp\rangle = \sqrt{(N-1)/N} = \cos\theta$——$\theta = \arcsin(1/\sqrt{N})$ 的来历
- 正交补（$N-2$ 维）：$\langle x_0|v\rangle = \langle s|v\rangle = 0$ → $U_O|v\rangle = -|v\rangle$、$U_S|v\rangle = -|v\rangle$ → $G|v\rangle = (+1)|v\rangle$：其余维度从头到尾不动
- 结果框：初态 $|s\rangle$ 完全落在平面内且演化永不离开——$N$ 维问题约化为 $2\times 2$ 矩阵
类型definition-derivation