单比特门三件套:X、Z、H

⎣1 0⎦

演算 X|0⟩ = (0×1+1×0, 1×1+0×0)ᵀ = (0,1)ᵀ = |1⟩;同理 X|1⟩ = |0⟩;几何:绕布洛赫球 X 轴旋转 180°

⎣0 −1⎦

Z|0⟩ = |0⟩、Z|1⟩ = −|1⟩;概率不变(|−1|² = 1),|1⟩ 分量获得相位 −1;几何:绕 Z 轴旋转 180°

⎣1 −1⎦

H|0⟩ = (|0⟩+|1⟩)/√2 ≡ |+⟩、H|1⟩ = (|0⟩−|1⟩)/√2 ≡ |−⟩;几何:绕 X–Z 角平分轴旋转 180°

H² = (1/2)·⎡1+1 1−1⎤ = ⎡1 0⎤ = I;又 H 实对称 ⇒ H† = H ⇒ H†H = H² = I ✓

⎣1−1 1+1⎦ ⎣0 1⎦

类型definition-derivation(三门并列 + 逐元素演算)