证明酉算子的本征值是单位复数
设∣α⟩为酉算子的本征向量,a为本征值∣⟨α∣α⟩∣=U†U is an identity operator∣⟨α∣U†U∣α⟩∣=∣⟨α∣a2∣α⟩∣=∣a2∣⋅∣⟨α∣α⟩∣因此∣a2∣=∣a∣2=1即酉算子的本征值是单位复数 \text{设} |\alpha\rangle\text{为酉算子的本征向量,a为本征值}\\ \begin{vmatrix} \langle\alpha|\alpha\rangle \end{vmatrix}\overset{U^\dag U \text{ is an identity operator}}{=}\begin{vmatrix} \langle\alpha|U^\dag U|\alpha\rangle \end{vmatrix}\\ =|\langle\alpha|a^2|\alpha\rangle|\\ =|a^2|\cdot|\langle\alpha|\alpha\rangle|\\ \\ \text{因此} |a^2|=|a|^2=1\\ \text{即酉算子的本征值是单位复数} 设∣α⟩为酉算子的本征向量,a为本征值∣∣⟨α∣α⟩∣∣=U†U is an identity operator∣∣⟨α∣U†U∣α⟩∣∣=∣⟨α∣a2∣α⟩∣=∣a2∣⋅∣⟨α∣α⟩∣因此∣a2∣=∣a∣2=1即酉算子的本征值是单位复数