- 命题框：max_z C(z) → 对角成本哈密顿量 H_C = Σ_z C(z)|z⟩⟨z|，满足 H_C|z⟩ = C(z)|z⟩
- 推导步骤（MaxCut 逐行）：
1. 切断指示符（自旋约定 b_i = (1−s_i)/2，s_i = ±1）：𝟙[b_i ≠ b_j] = b_i + b_j − 2·b_i·b_j = (1−s_i·s_j)/2
2. 代数验证：(1−s_i)/2 + (1−s_j)/2 − 2·(1−s_i)(1−s_j)/4 = (2−s_i−s_j)/2 − (1−s_i−s_j+s_i·s_j)/2 = (1−s_i·s_j)/2 ✓
3. 算符提升：Z_i|b⟩ = s_i|b⟩（Z|0⟩ = +|0⟩、Z|1⟩ = −|1⟩）→ Z_iZ_j 的本征值 = s_i·s_j
4. 逐边求和：H_C = Σ_{(i,j)∈E} (1−Z_iZ_j)/2 = (|E|/2)·I − ½·Σ_{(i,j)∈E} Z_iZ_j
- QUBO 卡（公式 + 代入核对）：b_i → (I−Z_i)/2、b_i·b_j → (I−Z_i)(I−Z_j)/4，归类得 H_C = c₀I + Σ h_iZ_i + Σ J_ij·Z_iZ_j（J_ij = Q_ij/4 等）；MaxCut 代入（Q_ii = deg(i)、Q_ij = −2）核对：h_i = −deg(i)/2 + deg(i)/2 = 0 ✓、J_ij = −2/4 = −½ ✓、c₀ = |E| − |E|/2 = |E|/2 ✓——无场项 ⇔ 切断数在同构翻转下不变
- 背景行（一句）：MaxCut、Max-SAT、TSP、图着色多为 NP-Hard；经典近似有理论上限（MaxCut 唯一博弈猜想下 0.878 对应 Goemans-Williamson）；QAOA 用小整数 p 层、多项式资源换随 p 提升的近似解