1. 解除第 3 页陷阱:|ψ⟩ = (1/√2, i/√2)ᵀ → ⟨ψ|ψ⟩ = |1/√2|² + |i/√2|² = 1/2 + 1/2 = 1 ✓(AMBER 结果)
2. 正交性验算(教材 §2.4):⟨+|−⟩ = (1/2)(⟨0|+⟨1|)(|0⟩−|1⟩) = (1/2)(⟨0|0⟩ − ⟨0|1⟩ + ⟨1|0⟩ − ⟨1|1⟩) = (1/2)(1 − 0 + 0 − 1) = 0
3. 归一性验算:‖|+⟩‖² = (1/√2)² + (1/√2)² = 1/2 + 1/2 = 1——{|+⟩, |−⟩} 与 {|0⟩, |1⟩} 一样是标准正交基