此图片的alt属性为空;文件名为1625798919-6446d860dbbfe54.png

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$ (A^†)^ † = A$.

说明$ (A^†)^ † = A$.

解答

令$$A=\left(\begin{array}{cccc}
a_{11} & a_{12} & \cdots & a_{1n} \\
a_{21} & a_{22} & \cdots & a_{2n} \\
\vdots & \vdots & & \vdots \\
a_{n1} & a_{n2} & \cdots & a_{nn}
\end{array}\right),$$

$$A^†=\left(\begin{array}{cccc}
a_{11}^* & a_{21}^* & \cdots & a_{n1}^* \\
a_{21}^* & a_{22}^* & \cdots & a_{2n}^* \\
\vdots & \vdots & & \vdots \\
a_{1n}^* & a_{2n}^* & \cdots & a_{nn}^*
\end{array}\right),$$

$$\left(A^{\dagger}\right)^{\dagger}=\left(\begin{array}{cccc}
\left(a_{11}^{*}\right)^{*} & \left(a_{12}^{*}\right)^{*} & \cdots & \left(a_{1 n}^{*}\right)^{*} \\
\left(a_{21}^{*}\right)^{*} & \left(a_{22}^{*}\right)^{*} & \cdots & \left(a_{2 n}^{*}\right)^{*} \\
\vdots & \vdots & & \vdots \\
\left(a_{n 1}^{*}\right)^{*} & \left(a_{n 2}^{*}\right)^{*} & \cdots & \left(a_{n n}^{*}\right)^{*}
\end{array}\right).$$

故$ (A^†)^ † = A$.

参考

[1]www.qtumist.com

参与者

作者:HKL, W65

贡献者:Dingyan, Wjw,Wxw

发表评论

后才能评论