问题：

If $|w\rangle$ and $|v\rangle$ are any two vectors, show that$(|w\rangle\langle v|)^{\dagger}=|v\rangle\langle w|$.

解答

$$\left|v\right\rangle=(v_{1},……,v_{n}).$$

$$|w\rangle\langle v|=\left[\begin{array}{cccc} v_{1}^* w_{1} & v_{1}^* w_{2} & \cdots & v_{1}^* w_{n} \\ v_{2}^* w_{1} & v_{2}^* w_{2} & & v_{2}^* w_{n} \\ \vdots & \vdots & & \vdots \\ v_{n}^* w_{1} & v_{n}^* w_{2} & \cdots & v_{n}^* w_{n} \end{array}\right].$$

$$(|w\rangle\langle v|)^{\dagger}=\left[\begin{array}{cccc} v_{1} w_{1}^* & v_{2} w_{1}^* & \cdots & v_{n} w_{1}^* \\ v_{1} w_{2}^* & v_{2} w_{2}^* & & v_{n} w_{2}^* \\ \vdots & \vdots & & \vdots \\ v_{1} w_{n}^* & v_{2} w_{n}^* & \cdots & v_{n} w_{n}^* \end{array}\right].$$

$$|v\rangle\langle w|=\left[\begin{array}{cccc}w_{1}^{*} v_{1} & w_{1}^{*} v_{2} & \cdots & w_{1}^{*} v_{n} \\ w_{2}^{*} v_{1} & w_{2}^{*} v_{2} & & w_{2}^{*} v_{n} \\ \vdots & \vdots & & \vdots \\ w_{n}^{*} v_{1} & w_{n}^{*} v_{2} & \cdots & w_{n}^{*} v_{n}\end{array}\right].$$

参考

[1]www.qtumist.com