湮灭算符：$\hat{a}=(\frac{m\omega}{2\hbar})^{\frac{1}{2}}(\hat{x}+\frac{i}{m\omega}\hat{p})$
产生算符：$\hat{a}^\dagger=(\frac{m\omega}{2\hbar})^{\frac{1}{2}}(\hat{x}-\frac{i}{m\omega}\hat{p})$
证明对易关系：$[\hat{a},\hat{a}^\dagger]=1$
$$\begin{aligned}[\hat{a},\hat{a}^\dagger]&=[(\frac{m\omega}{2\hbar})^{\frac{1}{2}}(\hat{x}+\frac{i}{m\omega}\hat{p}),(\frac{m\omega}{2\hbar})^{\frac{1}{2}}(\hat{x}-\frac{i}{m\omega}\hat{p})] \\ &=\frac{m\hbar}{2\hbar}[\hat{x}+\frac{i}{m\omega}\hat{p},\hat{x}-\frac{i}{m\omega}\hat{p}] \\ &=\frac{m\hbar}{2\hbar}\{[\hat{x},\hat{x}]-[\hat{x},\frac{i}{m\omega}\hat{p}]+[\frac{i}{m\omega}\hat{p},\hat{x}]-[\frac{i}{m\omega}\hat{p},\frac{i}{m\omega}\hat{p}]\} \\ &=\frac{m\hbar}{2\hbar}\{-\frac{i}{m\omega}[\hat{x},\hat{p}]+\frac{i}{m\omega}[\hat{p},\hat{x}]\} \\ &=\frac{m\hbar}{2\hbar}\cdot \frac{2\hbar}{m\omega} \\ &=1\end{aligned}$$
对于一维线性谐振子：
$$\xi =\sqrt{\frac{m\omega}{\hbar}}x=\alpha x$$
从而得出：
$$\hat{a}=\frac{1}{\sqrt{2}}(\xi+\frac{\partial}{\partial \xi})$$
$$\hat{a}^\dagger=\frac{1}{\sqrt{2}}(\xi-\frac{\partial}{\partial \xi})$$
将$\hat{a}$作用于谐振子哈密顿算符的第$n$个本征态$\psi_{n}$，可以得到：
$$\hat{a}\psi_{n}=\frac{1}{\sqrt{2}}(\xi+\frac{\partial}{\partial \xi})\psi_n$$
再根据：
$$\xi \psi_n=\sqrt{\frac{n}{2}}\psi_{n-1}+\sqrt{\frac{n+1}{2}}\psi_{n+1}$$
$$\frac{d}{d\xi}\psi_{n}=\sqrt{\frac{n}{2}}\psi_{n-1}-\sqrt{\frac{n+1}{2}}\psi_{n+1}$$
最终得到：
$$\begin{aligned}\hat{a}\psi_n &=\frac{1}{\sqrt{2}}\xi \psi_n +\frac{1}{\sqrt{2}}\frac{\partial}{\partial \xi}\psi_n \\ &=\frac{1}{\sqrt{2}}(\sqrt{\frac{n}{2}}\psi_{n-1}+\sqrt{\frac{n+1}{2}}\psi_{n+1}+\sqrt{\frac{n}{2}}\psi_{n-1}-\sqrt{\frac{n+1}{2}}\psi_{n+1}) \\ &=\frac{1}{\sqrt{2}}\cdot2\cdot\sqrt{\frac{n}{2}}\psi_{n-1} \\ &=\sqrt{n}\psi_{n-1}\end{aligned}$$
同理：
$$\hat{a}^\dagger \psi_n=\sqrt{n+1}\psi_{n+1}$$