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：
Verify that $(·, ·)$ just defined is an inner product on $C^n$.
证明 $(·, ·)$是定义在空间$C^n$上的內积操作
解答
令空间$C^n$上的两个向量分别为$$\left|v\right\rangle=(v_{1},v_{2},…..,v_{n}),$$$$\left|w\right\rangle=(w_{1},w_{2},…..,w_{n}),$$
有$$(\left|v\right\rangle,\left|w\right\rangle)=v_{1}^{*} w_{1}+v_{2}^{*} w_{2}+\cdots+v_{n}^{*} w_{n}.$$
下面证明这个式子满足內积的三条定义：
1.由数学归纳的知识，这里只证明$i=2$的简单情况,
$$\left(\left|v_{1}\right\rangle, \lambda_{1}\left|w_{1}\right\rangle+\lambda_{2}\left|w_{2}\right\rangle\right)=\sum v_{i}^{*}\left(\lambda_{1} w_{1i}+\lambda_{2} w_{2 i}\right).$$
$$\lambda_{1}\left(|v\rangle,\left|w_{1}\right\rangle\right)+\lambda_{2}\left(|v\rangle,\left|w_{2}\right\rangle\right)=\sum v_{i}^{} \lambda_{1} w_{1 i}+\sum v_{i}^{} \lambda_{2} w_{2 i}.$$
故$$\left(\left|v_{1}\right\rangle, \lambda_{1}\left|w_{1}\right\rangle+\lambda_{2}\left|w_{2}\right\rangle\right)=\lambda_{1}\left(|v\rangle,\left|w_{1}\right\rangle\right)+\lambda_{2}\left(|v\rangle,\left|w_{2}\right\rangle\right).$$
2.$$(\left|v\right\rangle,\left|w\right\rangle)=v_{1}^{*} w_{1}+v_{2}^{*} w_{2}+\cdots+v_{n}^{*} w_{n}.$$
$$(|w\rangle,|v\rangle)=w_{1}^{*} v_{1}+w_{2}^{*} v_{2}+\cdots+w_{n}^{*} v_{n};$$
故$$(\left|v\right\rangle,\left|w\right\rangle)=(|w\rangle,|v\rangle)^{*}.$$
3.$$(\left|v\right\rangle,\left|v\right\rangle)=v_{1}^{*} v_{1}+v_{2}^{*} v_{2}+\cdots+v_{n}^{*} v_{n}=\left|v_{1}\right|^{2}+\left|v_{2}\right|^{2}+\cdots+\left|v_{n}\right|^{2}.$$
故$(\left|v\right\rangle,\left|v\right\rangle)\geq0$,当取等号时，$v_{1}=v_{2}=……=v_{n}=0$,即$\left|v\right\rangle=(0,0,…..,0)=O.$
参考
[1]www.qtumist.com
参与者
作者: HKL, W65
贡献者: Dingyan, Wjw,Wxw
