- 命题框:按书写顺序编号,$Z_0 = Z\otimes I$(Z 作用在最左比特);一般地 $Z_i = I\otimes\cdots\otimes Z\otimes\cdots\otimes I$(Z 在第 $i$ 位)
- 推导/演算步骤($|\Phi^+\rangle$ 例,逐行):
- $(Z\otimes I)|\Phi^+\rangle = (1/\sqrt{2})[(Z|0\rangle)\otimes|0\rangle + (Z|1\rangle)\otimes|1\rangle] = (1/\sqrt{2})(|00\rangle - |11\rangle)$
- $\langle Z_0\rangle = \langle\Phi^+|(Z\otimes I)|\Phi^+\rangle = (1/2)(\langle 00| + \langle 11|)(|00\rangle - |11\rangle) = (1/2)(1 - 1) = 0$(交叉项 $\langle 00|11\rangle$ 正交归零)
- 概率法复核(正是第 8 页部分测量):$P(0) = 1/2$(值 $+1$)、$P(1) = 1/2$(值 $-1$)→ $(+1)(1/2) + (-1)(1/2) = 0$
- 线性性:$\langle Z_0+Z_1\rangle = \langle\psi|Z_0|\psi\rangle + \langle\psi|Z_1|\psi\rangle$(矩阵乘法分配律);同法算 $(I\otimes Z)|\Phi^+\rangle = (1/\sqrt{2})(|00\rangle - |11\rangle)$ → $\langle Z_1\rangle = 0$
- 结果框:$\langle Z_0\rangle + \langle Z_1\rangle = 0 + 0 = 0$——期望值对算符线性,可逐项算再相加
类型definition-derivation