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：
Show that any inner product (·, ·) is conjugate-linear in the first argument, $\left(\sum_{i} \lambda_{i}\left|w_{i}\right\rangle,|v\rangle\right)=\sum_{i} \lambda_{i}^{*}\left(\left|w_{i}\right\rangle,|v\rangle\right)$.
说明內积$(·, ·)$对第一个分量是共轭线性的
解答：
运用数学归纳的思想这里只证明$i=2$的情况
$$\left(\lambda_{1} | w_{1}\rangle+\lambda_{2}\left|w_{2}\right\rangle,|v\rangle\right)=$$$$\left(\lambda_{1} w_{11}+\lambda_{2} w_{21}\right)^{*} v_{1}+\left(\lambda_{1} w_{12}+\lambda_{2} w_{22}\right)^{*} v_{2}+\cdots+\left(\lambda_{1} w_{1n}+\lambda_{2} w_{2n}\right)^{*} v_{n}.$$
$$\lambda_{1}^{*}\left(\left|w_{1}\right\rangle,|v\rangle\right)+\lambda_{2}^{*}\left(\left|w_{2}\right\rangle,|v\rangle\right)=\sum_{i}\left[\lambda_{1}^{*}\left(\omega_{1i}^{*} v_{i}\right)+\lambda_{2}^{*}\left(w_{2i}^ * v_{i}\right)\right].$$
故$$\left(\lambda_{1} |w_{1}\rangle+\lambda_{2}\left|w_{2}\right\rangle,|v\rangle\right)=\lambda_{1}^{*}\left(\left|w_{1}\right\rangle,|v\rangle\right)+\lambda_{2}^{*}\left(\left|w_{2}\right\rangle,|v\rangle\right).$$
参考
[1]www.qtumist.com
参与者
作者：HKL, W65
贡献者:Dingyan, Wjw,Wxw
