数值实例：2×2 全流程手算

• 题面框：A = diag(1, 2)，b = (1, 1)ᵀ；经典解 x = A⁻¹b = (1, 1/2)ᵀ；‖x‖² = 1 + 1/4 = 5/4，归一化 (2/√5, 1/√5)ᵀ
• 演算（五行，符号式 → 代数值 → 结果）：

1. 制备：|b⟩ = (|0⟩ + |1⟩)/√2（一个 H 门）

2. QPE：m = 2、t = π/2；φ₀ = λ₀·t/2π = (1·π/2)/2π = 1/4 = 0.01₂，φ₁ = (2·π/2)/2π = 1/2 = 0.10₂；换算系数 2π/(t·2^m) = 2π/(π/2 × 4) = 1 ⇒ 时钟整数即特征值；输出 (|0⟩|01⟩ + |1⟩|10⟩)/√2

3. 受控旋转（C = λ_min = 1）：λ̃ = 1 ⇒ θ = 2·arcsin(1) = π，R_y(π)|0⟩ = |1⟩ 必成功；λ̃ = 2 ⇒ θ = 2·arcsin(1/2) = π/3，|1⟩ 振幅 = sin(π/6) = 1/2

4. 成功概率：P₁ = (1/2)·1² + (1/2)·(1/2)² = 1/2 + 1/8 = 5/8 ≥ 1/κ² = 1/4 ✓（κ = 2/1 = 2）

5. 输出态：|x⟩ = (1/√(5/8))·(1/√2)·(|0⟩ + (1/2)|1⟩) = (2/√5)|0⟩ + (1/√5)|1⟩；P(|0⟩) = 4/5、P(|1⟩) = 1/5，和 = 1 ✓

• 验证行：振幅平方比 4 : 1 = 1² : (1/2)²——与经典解分量比 x₀ : x₁ = 1 : 1/2 一致；|x⟩ ∝ (1, 1/2)ᵀ = A⁻¹b