在阅读该页内容之前,我们向量子计算的开创者费曼和Deutsch致敬,同时向三位量子信息学的奠基人Charles H. Bennett, David Deutsch, Peter Shor表示敬意.
问题：
Prove that two eigenvectors of a Hermitian operator with different eigenvalues are necessarily orthogonal.
证明Hermitian算子的不同特征值对应的特征向量是正交的.
解答
$ A|v_{1}\rangle=\lambda_{1}|v_{1}\rangle $，$ A|v_{2}\rangle=\lambda_{2}|v_{2}\rangle $.
则$ \left\langle v_{1}|A| v_{2}\right\rangle=\lambda_{2}\left\langle v_{1} \mid v_{2}\right\rangle $，$ \left\langle v_{2}|A| v_{1}\right\rangle=\lambda_{1}\left\langle v_{2} \mid v_{1}\right\rangle $.
由Hermitian性质得到$\left\langle v_{1}|A| v_{2}\right\rangle$=$\left\langle v_{2}|A| v_{1}\right\rangle$,即$\lambda_{2}\left\langle v_{1} \mid v_{2}\right\rangle $=$\lambda_{1}\left\langle v_{2} \mid v_{1}\right\rangle $.
由$\lambda_{1}$不等于$\lambda_{2}$，故$\left\langle v_{1} \mid v_{2}\right\rangle=0$.
参考
[1]www.qtumist.com
参与者
作者：HKL, W65
贡献者:Dingyan, Wjw,Wxw
