1. 展开特征方程：λ² − 2λ·cos θ + (cos²θ + sin²θ) = λ² − 2λ·cos θ + 1 = 0
2. 求根：λ = cos θ ± i·sin θ = e^{±iθ}（欧拉公式收尾）
3. λ₁ = e^{iθ} 代回 (λI−G)v = 0：i·sin θ·v₁ = sin θ·v₂ → v₂ = i·v₁
4. 归一化：|u₁⟩ = (|g⟩ + i|b⟩)/√2；同理 |u₂⟩ = (|g⟩ − i|b⟩)/√2
5. 验证：G|u₁⟩ = (cos θ+i sin θ, −sin θ+i cos θ)ᵀ/√2 = e^{iθ}|u₁⟩ ✓；⟨u₁|u₂⟩ = ½(1 + (−i)(−i)) = ½(1−1) = 0 ✓