科训早期的乱七八糟玩意，鬼知道有啥用
一维谐振子
代数解法
\[\begin{equation} \left\{ \begin{aligned} &\mathrm{i} \hbar \frac{\partial \Psi}{\partial t} = \hat{H} \Psi \\ &\hat{H} = - \frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} + U(x) \\ &U(x) = \frac{1}{2} m {\omega}^2 {x}^2 \end{aligned} \right. \end{equation}\]通过分离变量法，可得时间因子$\mathrm{e}^{\mathrm{i}\frac{E}{\hbar}t}$，以及定态薛定谔方程$\frac { \hbar ^ { 2 } } { 2 m } \frac { d ^ { 2 } \Psi } { d x ^ { 2 } } + ( E - \frac { m \omega ^ { 2 } } { 2 } x ^ { 2 } ) \Psi = 0$ ，对定态薛定谔方程进行变量替换，可得： \(\begin{equation} \left\{ \begin{aligned} &\frac{d^2 \Psi}{d \xi^2} + (\lambda-\xi^2)\Psi = 0\\ &\xi = \sqrt{\frac{m \omega}{\hbar}} x = \alpha x , \lambda = \frac{2E}{\hbar \omega} \end{aligned} \right. \label{math:1} \end{equation}\)
常规解
由无穷远处性质，可以设解$\Psi(\xi) = \mathrm{e}^{-\frac{\xi^2}{2}} H(\xi)$，方程化为对$H(\xi)$的厄米方程： \begin{equation} \frac{d^2 H}{d \xi^2} - 2\xi \frac{dH}{d\xi} + (\lambda - 1 ) H = 0 \end{equation}
利用级数解法，可以知道当$\lambda = 2n+1 (n=0,1,2,…)$时方程在$(-\infty, +\infty)$上有有限解，此时方程称为n阶厄米方程，解为： \begin{equation} H_n(\xi) = (2\xi)^n - \frac{n(n-1)}{1!} (2\xi)^{n-2} + \frac{n(n-1)(n-2)(n-3)}{2!} (2\xi)^{n-4} - … \end{equation}
所以有$\Psi(\xi) = c_n \mathrm{e}^{-\frac{\xi^2}{2}} H_n(\xi), E_n = \left(n+\frac{1}{2}\right) \hbar \omega$。
解析解
使用Mathematica对方程\eqref{math:1}进行求解，得： \begin{equation} \Psi(\xi) = c_1 D_{\frac{\lambda -1}{2}}\left(\sqrt{2} \xi \right)+c_2 D_{\frac{-\lambda -1}{2}}\left(\mathrm{i} \sqrt{2} \xi \right) \end{equation} 其中，$D_{\nu}(z)$为抛物柱面函数（ParabolicCylinderD），由方程的对称性可知，将坐标$x$取反的解也是方程的解，上式给出的是线性无关的两个解，为使解具有对称性，取以下解形式： \begin{equation} \Psi(\xi) = c_1 D_{\frac{\lambda -1}{2}}\left(\sqrt{2} \xi \right)+c_2 D_{\frac{-\lambda -1}{2}}\left(\mathrm{i} \sqrt{2} \xi \right) + c_3 D_{\frac{\lambda -1}{2}}\left(-\sqrt{2} \xi \right)+c_4 D_{\frac{-\lambda -1}{2}}\left(-\mathrm{i} \sqrt{2} \xi \right) \end{equation} 此解在无穷远处与负无穷远处的一阶展开式为： \(\begin{equation} \begin{aligned} &\Psi(\xi \to +\infty) = \frac{1}{\sqrt{\pi }}2^{\frac{1}{4} (-\lambda -5)} \mathrm{e}^{-\frac{\xi^2}{2} - \frac{\mathrm{i} \pi \lambda }{4}} \xi^{\frac{1}{2} (-\lambda -1)} \\ &\Bigg\{(1-\mathrm{i}) 2^{\frac{\lambda}{2}} \xi^{\lambda} \left[(1+\mathrm{i}) \sqrt{\pi} \mathrm{e}^{\frac{\mathrm{i} \pi \lambda }{4}} \left(c_1-\mathrm{i} c_3 \mathrm{e}^{\frac{\mathrm{i} \pi \lambda }{2}}\right) + c_4 \left(1 + \mathrm{e}^{\mathrm{i} \pi \lambda }\right) \Gamma \left(\frac{1}{2}-\frac{\lambda }{2}\right)\right] \\ & + (1 + \mathrm{i}) \sqrt{2} \mathrm{e}^{\xi^2} \left[\sqrt{\pi} \left(c_4 \mathrm{e}^{\frac{\mathrm{i} \pi \lambda }{2}}- \mathrm{i} c_2 \right) + (1-\mathrm{i}) c_3 \mathrm{e}^{\frac{\mathrm{i} \pi \lambda }{4}} \cos \left(\frac{\pi \lambda }{2}\right) \Gamma \left(\frac{\lambda +1}{2}\right)\right] \Bigg\} \end{aligned} \end{equation}\) \(\begin{equation} \begin{aligned} &\Psi(\xi \to -\infty) = -\frac{\mathrm{i}}{\sqrt{\pi}} 2^{\frac{1}{4} (-\lambda -5)} \mathrm{e}^{-\frac{\xi^2}{2}-\mathrm{i} \pi \lambda } \left(\frac{1}{\xi}\right)^{\frac{1}{2}-\frac{\lambda }{2}} \\ &\Bigg\{(1-\mathrm{i}) 2^{\frac{\lambda}{2}} \mathrm{e}^{\frac{5 \mathrm{i} \pi \lambda }{4}} \left[(1+\mathrm{i}) \sqrt{\pi } \mathrm{e}^{\frac{\mathrm{i} \pi \lambda }{4}} \left(c_3 - \mathrm{i} c_1 \mathrm{e}^{\frac{\mathrm{i} \pi \lambda }{2}}\right) + c_2 \left(1+\mathrm{e}^{\mathrm{i} \pi \lambda }\right) \Gamma \left(\frac{1}{2}-\frac{\lambda }{2}\right)\right] \\ &+ \sqrt{2} \mathrm{e}^{\xi^2} \left(\frac{1}{\xi}\right)^{\lambda } \left[c_1 \left( 1 + \mathrm{e}^{\mathrm{i} \pi \lambda }\right) \Gamma \left(\frac{\lambda +1}{2}\right) + (1+\mathrm{i}) \sqrt{\pi } \mathrm{e}^{\frac{\mathrm{i} \pi \lambda }{4}} \left(c_2 \mathrm{e}^{\frac{\mathrm{i} \pi \lambda }{2}}- \mathrm{i} c_4 \right)\right]\Bigg\} \end{aligned} \end{equation}\)
算符解法
对哈密顿算符做分解： \(\begin{equation} \begin{aligned} \hat{H} &= \frac{\hat{p}^2}{2m} + \frac{1}{2} m \omega^2 \hat{x}^2 \\ &= \hbar\omega \left(\sqrt{\frac{m \omega}{2\hbar}}\hat{x} - \mathrm{i} \sqrt{\frac{1}{2m\hbar\omega}} \hat{p}\right) \left(\sqrt{\frac{m \omega}{2\hbar}}\hat{x} + \mathrm{i} \sqrt{\frac{1}{2m\hbar\omega}} \hat{p}\right) - \frac{\mathrm{i}\omega}{2} (\hat{x}\hat{p} - \hat{p}\hat{x}) \\ &= \hbar \omega \left(\hat{a}^\dagger \hat{a} + \frac{1}{2}\right) \end{aligned} \end{equation}\) 上式中定义了下降算符$\hat{a}$和它的厄米共轭上升算符$\hat{a}^\dagger$，令$\hat{N} = \hat{a}^\dagger \hat{a}$为粒子数算符，则有对易关系： \(\begin{equation} \left\{ \begin{aligned} & [\hat{a}, \hat{a}^\dagger] = 1 \\ & [\hat{H}, \hat{N}] = 0 \\ & [\hat{N}, \hat{a}] = - \hat{a} \\ & [\hat{N}, \hat{a}^\dagger] = \hat{a} \end{aligned} \right. \end{equation}\)
设粒子数算符的本征态为$\ket{n}$，则有$\hat{N} \ket{n} = n \ket{n}$，可得： \(\begin{equation} \left\{ \begin{aligned} & \hat{N} \hat{a} \ket{n} = (n-1) \hat{a} \ket{n} \\ & \hat{N} \hat{a}^\dagger \ket{n} = (n+1) \hat{a}^\dagger \ket{n} \end{aligned} \right. \end{equation}\) 上式表明，$\hat{a} \ket{n}$和$\hat{a}^\dagger \ket{n}$也是粒子数算符的本征态，且本征值分别为$n-1$和$n+1$。
由于$\bra{n} \hat{N} \ket{n} = \bra{n} \hat{a}^\dagger \hat{a} \ket{n} = n \braket{n}$， 所以$n = \frac{\abs{\hat{a} \ket{n}}^2}{\abs{\ket{n}}^2} \geqslant 0$，故存在最小的本征值极其对应的本征态$\ket{0}$，利用上升算符$\hat{a}^\dagger$，可以得到粒子数算符的归一化本征态$\ket{n} = \frac{1}{\sqrt{n!}} ({\hat{a}}^{\dagger})^{n} \ket{0},n=0,1,2,…$。
哈密顿算符和粒子数算符对易，所以它们拥有相同的本征态，故有： \begin{equation} \hat{H} \ket{n} = E_n \ket{n} = \hbar \omega \left( \hat{N} + \frac{1}{2} \right) \ket{n} = \hbar \omega \left( n + \frac{1}{2} \right) \ket{n} \end{equation} 即得到能级$E_n = \hbar \omega \left( n + \frac{1}{2} \right), n = 0,1,2,…$。
最后计算波函数，利用基态关系$\hat{a} \ket{0} = 0$，可得方程： \(\begin{equation} \begin{aligned} \left( \sqrt{\frac{m\omega}{2\hbar}} \hat{x} + \mathrm{i} \sqrt{\frac{1}{2m\hbar\omega}} \hat{p} \right) \Psi_0(x) &= 0 \\ \frac{1}{\Psi_0(x)} \frac{d\Psi_0(x)}{dx} + \frac{m\omega}{\hbar} x &= 0 \end{aligned} \end{equation}\) 这是一阶线性微分方程，积分可得解，利用归一化条件得到常数后，可得基态归一化波函数： \(\begin{equation} \Psi_0 (x) = \left( \frac{m\omega}{\hbar \pi}\right) ^\frac{1}{4} \mathrm{e} ^{-\frac{m\omega}{2\hbar}x^2} \end{equation}\) 再利用上升算符可得其它本征态： \(\begin{equation} \begin{aligned} \Psi_n(x) &= \left(\frac{1}{2^n n!}\right)^\frac{1}{2} \left( \sqrt{\frac{m\omega}{\hbar}}x - \sqrt{\frac{\hbar}{m\omega}} \frac{d}{dx} \right)^n \Psi_0(x) \\ &= \left(\frac{1}{2^n n!} \sqrt{\frac{m\omega}{\hbar \pi}} \right)^\frac{1}{2} \mathrm{e} ^{-\frac{m\omega}{2\hbar}x^2} \mathrm{e} ^{\frac{m\omega}{2\hbar}x^2} \left( \sqrt{\frac{m\omega}{\hbar}}x - \sqrt{\frac{\hbar}{m\omega}} \frac{d}{dx} \right)^n \mathrm{e} ^{-\frac{m\omega}{2\hbar}x^2} \\ &= N_n \mathrm{e} ^{-\frac{m\omega}{2\hbar}x^2} H_n(\sqrt{\frac{m\omega}{\hbar}}x) \end{aligned} \end{equation}\) 其中： \begin{equation} N_n = \left(\frac{1}{2^n n!} \sqrt{\frac{m\omega}{\hbar \pi}} \right)^\frac{1}{2} \end{equation} \begin{equation} H_n(\xi) = \mathrm{e}^{\frac{\xi^2}{2}} \left(\xi - \frac{d}{d\xi} \right)^n \mathrm{e}^{-\frac{\xi^2}{2}} \end{equation} $H_n(\xi)$为厄米多项式，对比此结果与代数解法结果，可以发现二者相同。
一维反谐振子
代数解法
\[\begin{equation} \left\{ \begin{aligned} &\mathrm{i} \hbar \frac{\partial \Psi}{\partial t} = \hat{H} \Psi \\ &\hat{H} = - \frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} + U(x) \\ &U(x) = - \frac{1}{2} m {\omega}^2 {x}^2 \end{aligned} \right. \end{equation}\]通过分离变量法，可得时间因子$\mathrm{e}^{\mathrm{i}\frac{E}{\hbar}t}$，以及定态薛定谔方程$\frac { \hbar ^ { 2 } } { 2 m } \frac { d ^ { 2 } \Psi } { d x ^ { 2 } } + ( E + \frac { m \omega ^ { 2 } } { 2 } x ^ { 2 } ) \Psi = 0$ ，对定态薛定谔方程进行变量替换，可得： \(\begin{equation} \left\{ \begin{aligned} &\frac{d^2 \Psi}{d \xi^2} + (\lambda+\xi^2)\Psi = 0\\ &\xi = \sqrt{\frac{m \omega}{\hbar}} x = \alpha x , \lambda = \frac{2E}{\hbar \omega} \end{aligned} \right. \label{math:2} \end{equation}\)
使用Mathematica对方程\eqref{math:2}进行求解，得： \begin{equation} \Psi(\xi) = c_1 D_{-\frac{1}{2} \mathrm{i} (-\mathrm{i}+\lambda )}[(1+\mathrm{i}) x]+c_2 D_{\frac{1}{2} \mathrm{i} (\mathrm{i}+\lambda )}[(-1+\mathrm{i}) x] \end{equation} 其中，$D_{\nu}(z)$为抛物柱面函数（ParabolicCylinderD），与正谐振子相同，此处取对称解： \begin{equation} \Psi(\xi) = c_1 D_{-\frac{1}{2} \mathrm{i} (-\mathrm{i}+\lambda )}[(1+\mathrm{i}) x]+c_2 D_{\frac{1}{2} \mathrm{i} (\mathrm{i}+\lambda )}[(-1+\mathrm{i}) x] + c_3 D_{-\frac{1}{2} \mathrm{i} (-\mathrm{i}+\lambda )}[-(1+\mathrm{i}) x] + c_4 D_{\frac{1}{2} \mathrm{i} (\mathrm{i}+\lambda )}[-(-1+\mathrm{i}) x] \end{equation} 此解在无穷远处与负无穷远处的一阶展开式为： \(\begin{equation} \begin{aligned} &\Psi(\xi \to +\infty) = \frac{\sqrt[8]{-1}}{\sqrt{\pi }} 2^{-\frac{3}{4}-\frac{\mathrm{i} \lambda }{4}} \mathrm{e}^{-\frac{3 \pi \lambda }{8}-\frac{\mathrm{i} \xi^2}{2}} \xi^{-\frac{1}{2}-\frac{\mathrm{i} \lambda }{2}} \\ &\Bigg\{(1+\mathrm{i}) \sqrt{\pi } \left(c_3 - \mathrm{i} c_1 \mathrm{e}^{\frac{\pi \lambda }{2}}\right) + 2^{\frac{1}{2}+\frac{\mathrm{i} \lambda }{2}} \mathrm{e}^{\mathrm{i} \xi^2} \xi^{\mathrm{i} \lambda } \bigg[\sqrt{\pi } \left(c_4 \mathrm{e}^{\frac{\pi \lambda }{2}}-\mathrm{i} c_2 \right) \\ &+(1-\mathrm{i}) c_3 \mathrm{e}^{\frac{\pi \lambda }{4}} \cosh \left(\frac{\pi \lambda }{2}\right) \Gamma \left(\frac{1}{2}-\frac{\mathrm{i} \lambda }{2}\right)\bigg] + c_2 \mathrm{e}^{-\frac{\pi \lambda }{4} } \left(\mathrm{e}^{\pi \lambda } + 1 \right) \Gamma \left(\frac{\mathrm{i} \lambda }{2}+\frac{1}{2}\right)\Bigg\} \end{aligned} \end{equation}\) \(\begin{equation} \begin{aligned} &\Psi(\xi \to -\infty) = \frac{-\sqrt[8]{-1}}{\sqrt{\pi }}(1+\mathrm{i}) 2^{-\frac{5}{4}-\frac{\mathrm{i} \lambda }{4}} \mathrm{e}^{-\frac{7 \pi \lambda }{8}-\frac{\mathrm{i} \xi^2}{2}} \left(\frac{1}{\xi}\right)^{\frac{1}{2}-\frac{\mathrm{i} \lambda }{2}} \\ &\Bigg\{(1+\mathrm{i}) 2^{\frac{\mathrm{i} \lambda }{2}} \mathrm{e}^{\mathrm{i} \xi^2} \left[(1-\mathrm{i}) c_1 \mathrm{e}^{\frac{\pi \lambda }{4}} \cosh \left(\frac{\pi \lambda }{2}\right) \Gamma \left(\frac{1}{2}-\frac{\mathrm{i} \lambda }{2}\right) +\sqrt{\pi } \left(c_2 \mathrm{e}^{\frac{\pi \lambda }{2}}-\mathrm{i} c_4 \right)\right] \\ &+\sqrt{2} \mathrm{e}^{\pi \lambda } \left(\frac{1}{\xi}\right)^{\mathrm{i} \lambda } \left[\sqrt{\pi } \left(c_3 \mathrm{e}^{\frac{\pi \lambda }{2}}+\mathrm{i} c_1 \right)+(1+\mathrm{i}) c_4 \mathrm{e}^{\frac{\pi \lambda }{4}} \cosh \left(\frac{\pi \lambda }{2}\right) \Gamma \left(\frac{\mathrm{i} \lambda }{2}+\frac{1}{2}\right)\right]\Bigg\} \end{aligned} \end{equation}\)
下面考虑散射问题，实际上通解只需考虑前两项，另外两项可以通过线性组合表示，故可令$c_3=c_4=0$，正无穷远处分解渐进解，得到两个方向相反的波： \(\begin{equation} \left\{ \begin{aligned} \Psi_{right}(\xi) &= -(-1)^{\frac{5}{8}} c_2 2^{\frac{1}{4} \mathrm{i} (\lambda + \mathrm{i})} \mathrm{e}^{-\frac{3 \pi \lambda }{8}} \mathrm{e}^{\frac{\mathrm{i} \xi^2}{2} + \frac{\mathrm{i} \lambda - 1}{2} \ln \xi }= F \mathrm{e}^{\frac{\mathrm{i} \xi^2}{2} + \frac{\mathrm{i} \lambda - 1}{2} \ln \xi} \\ \Psi_{left}(\xi) &= \frac{\sqrt[8]{-1}}{\sqrt{\pi }} 2^{-\frac{3}{4}-\frac{\mathrm{i} \lambda }{4}} \mathrm{e}^{-\frac{5 \pi \lambda }{8}} \left[(1-\mathrm{i}) \sqrt{\pi } c_1 \mathrm{e}^{\frac{3 \pi \lambda }{4}}+ c_2 \left(\mathrm{e}^{\pi \lambda }+1\right) \Gamma \left(\frac{\mathrm{i} \lambda }{2}+\frac{1}{2}\right)\right] \mathrm{e}^{-\frac{\mathrm{i} \xi^2}{2} - \frac{\mathrm{i} \lambda + 1}{2} \ln \xi} \\ &= G \mathrm{e}^{-\frac{\mathrm{i} \xi^2}{2} - \frac{\mathrm{i} \lambda + 1}{2} \ln \xi} \end{aligned} \right. \end{equation}\) 同样对负无穷远处，也可以分解为两个方向相反的波，这里需要注意$\xi$为负值，需要考虑复变对数函数： \(\begin{equation} \left\{ \begin{aligned} \Psi_{right}(\xi) &= -(-1)^{\frac{7}{8}} c_1 2^{-\frac{1}{4}-\frac{\mathrm{i} \lambda }{4}} \mathrm{e}^{\frac{\pi \lambda}{8}} \mathrm{e}^{\frac{\mathrm{i} - \lambda}{2} \pi}\mathrm{e}^{- \frac{ \mathrm{i} \xi^2}{2} + \frac{1 + \mathrm{i} \lambda}{2} \ln \left| \frac{1}{\xi} \right|} = A \mathrm{e}^{- \frac{ \mathrm{i} \xi^2}{2} + \frac{1 + \mathrm{i} \lambda}{2} \ln \left| \frac{1}{\xi} \right|}\\ \Psi_{left}(\xi) &= -\frac{(-1)^{\frac{5}{8}}}{\sqrt{\pi }} 2^{\frac{1}{4} \mathrm{i} (\lambda +\mathrm{i})} \mathrm{e}^{-\frac{5 \pi \lambda }{8}} \left[\sqrt{\pi } c_2 \mathrm{e}^{\frac{\pi \lambda }{4}}+(1-\mathrm{i}) c_1 \cosh \left(\frac{\pi \lambda }{2}\right) \Gamma \left(\frac{1}{2}-\frac{\mathrm{i} \lambda }{2}\right)\right] \mathrm{e}^{\frac{\mathrm{i} + \lambda}{2} \pi} \mathrm{e}^{\frac{\mathrm{i} \xi^2}{2} + \frac{1 - \mathrm{i} \lambda}{2} \ln \left| \frac{1}{\xi} \right|} \\ &= B \mathrm{e}^{\frac{\mathrm{i} \xi^2}{2} + \frac{1 - \mathrm{i} \lambda}{2} \ln \left| \frac{1}{\xi} \right|} \end{aligned} \right. \end{equation}\)
算符解法
对哈密顿算符做分解： \(\begin{equation} \begin{aligned} \hat{H} &= \frac{\hat{p}^2}{2m} - \frac{1}{2} m \omega^2 \hat{x}^2 \\ &= \hbar\omega \left(\sqrt{\frac{1}{2m\hbar\omega}} \hat{p} + \sqrt{\frac{m \omega}{2\hbar}}\hat{x} \right) \left(\sqrt{\frac{1}{2m\hbar\omega}} \hat{p} - \sqrt{\frac{m \omega}{2\hbar}}\hat{x}\right) + \frac{\omega}{2} (\hat{p}\hat{x} - \hat{x}\hat{p}) \\ &= \hbar \omega \left(\hat{u}^+ \hat{u}^- - \frac{\mathrm{i}}{2}\right) \\ &= \hbar \omega \left(\hat{u}^- \hat{u}^+ + \frac{\mathrm{i}}{2}\right) \\ &= \frac{\hbar \omega}{2} \left(\hat{u}^+ \hat{u}^- + \hat{u}^- \hat{u}^+ \right) \end{aligned} \end{equation}\) 上述定义了两个算符$\hat{u}^+$和$\hat{u}^-$，它们有如下对易关系： \(\begin{equation} \left\{ \begin{aligned} & [\hat{u}^+, \hat{u}^-] = \mathrm{i} \\ & [\hat{H}, \hat{u}^+] = -\mathrm{i} \hbar \omega \hat{u}^+ \\ & [\hat{H}, \hat{u}^-] = \mathrm{i} \hbar \omega \hat{u}^- \end{aligned} \right. \end{equation}\) 设哈密顿算符的本征态为$\ket{\mathscr{E}}$，则有$\hat{H} \ket{\mathscr{E}} = \mathcal{E} \ket{\mathscr{E}}$，可得： \(\begin{equation} \left\{ \begin{aligned} & \hat{H} \hat{u}^+ \ket{\mathscr{E}} = (\mathcal{E}-\mathrm{i} \hbar \omega) \hat{u}^+ \ket{\mathscr{E}} \\ & \hat{H} \hat{u}^- \ket{\mathscr{E}} = (\mathcal{E}+\mathrm{i} \hbar \omega) \hat{u}^- \ket{\mathscr{E}} \end{aligned} \right. \end{equation}\)
结构
自旋
考虑自旋中的泡利算符矩阵： \(\begin{equation} \left\{ \begin{aligned} & \sigma_x = \mqty[0 & 1 \\ 1 & 0] \\ & \sigma_y = \mqty[0 & -\ii \\ \ii & 0] \\ & \sigma_z = \mqty[1 & 0 \\ 0 & -1] \end{aligned} \right. \end{equation}\) 它们之间有对易关系： \(\begin{equation} \left\{ \begin{aligned} & [\sigma_x, \sigma_y] = 2 \sigma_z \\ & [\sigma_y, \sigma_z] = 2 \sigma_x \\ & [\sigma_z, \sigma_x] = 2 \sigma_y \end{aligned} \right. \end{equation}\) 显然，它们是无迹厄米矩阵的基，则它们生成$\mathfrak{su}(2)$李代数，也是$\mathrm{SU}(2)$李群的生成元，故全体群元的矩阵表示为： \(\begin{equation} U = \mathrm{e}^{\mathrm{i}\theta_x\sigma_x + \mathrm{i}\theta_y\sigma_y + \mathrm{i}\theta_z\sigma_z} = \begin{bmatrix} \cos \theta + \mathrm{i} \theta_z \frac{\sin \theta}{\theta} & (\mathrm{i} \theta_x + \theta_y) \frac{\sin \theta}{\theta} \\ (\mathrm{i} \theta_x - \theta_y) \frac{\sin \theta}{\theta} & \cos \theta - \mathrm{i} \theta_z \frac{\sin \theta}{\theta} \end{bmatrix} \end{equation}\) 其中，$\theta = \sqrt{\theta_x^2 + \theta_y^2 + \theta_z^2}$。
下面验证这确实构成一个群，首先注意到矩阵乘法自动满足结合律；其次当参量全为零时，$U$为二阶单位矩阵，正好满足幺元性质，即存在单位元；接着，将参量全取反，得到共轭转置矩阵$U^\dagger$，易验证$U^\dagger U = I$，故存在逆元；最后验证封闭性，定义两个群元： \begin{equation} U_k = \mathrm{e}^{\mathrm{i}\theta_{kx}\sigma_x + \mathrm{i}\theta_{ky}\sigma_y + \mathrm{i}\theta_{kz}\sigma_z} \quad (k = 1, 2) \end{equation} 则二者相乘为： \(\begin{equation} \left\{ \begin{aligned} & U = U_1 U_2 = \begin{bmatrix} a & b \\ -b^* & a^* \end{bmatrix} \\ & a = \frac{1}{\theta_1 \theta_2} \left[ (\theta_{1} \cos \theta_{1} + \mathrm{i} \theta_{1z} \sin \theta_{1}) (\theta_{2} \cos \theta_{2} + \mathrm{i} \theta_{2z} \sin \theta_{2} ) - (\theta_{1x} - \mathrm{i} \theta_{1y}) (\theta_{2x} + \mathrm{i} \theta_{2y}) \sin \theta_{1} \sin \theta_{2} \right] \\ & b = \frac{1}{\theta_1 \theta_2} \left[ \sin \theta_{1} (\theta_{1y} + \mathrm{i} \theta_{1x}) (\theta_{2} \cos \theta_{2} - \mathrm{i} \theta_{2z} \sin \theta_{2}) + \sin \theta_{2} (\theta_{2y} + \mathrm{i} \theta_{2x}) (\theta_{1} \cos \theta_{1} + \mathrm{i} \theta_{1z} \sin \theta_{1}) \right] \\ & \theta_k = \sqrt{\theta_{kx}^2 + \theta_{ky}^2 + \theta_{kz}^2} \quad (k = 1, 2) \\ \end{aligned} \right. \end{equation}\) 使用Mathematica易验证$\left| a \right|^2 + \left| b \right|^2 = 1$，则相乘元满足$\mathrm{SU}(2)$群群元定义，是群中元素。
利用BCH公式以及对易关系，或者直接进行矩阵运算，可以得到如下关系： \(\begin{equation} \left\{ \begin{aligned} & \mathrm{e}^{-\mathrm{i} \theta \sigma_x} \sigma_y \mathrm{e}^{\mathrm{i} \theta \sigma_x} = \sigma_y \cos (2 \theta) + \sigma_z \sin (2 \theta) \\ & \mathrm{e}^{-\mathrm{i} \theta \sigma_y} \sigma_z \mathrm{e}^{\mathrm{i} \theta \sigma_y} = \sigma_z \cos (2 \theta) + \sigma_x \sin (2 \theta) \\ & \mathrm{e}^{-\mathrm{i} \theta \sigma_z} \sigma_x \mathrm{e}^{\mathrm{i} \theta \sigma_z} = \sigma_x \cos (2 \theta) + \sigma_y \sin (2 \theta) \end{aligned} \right. \end{equation}\)
正负谐振子
考虑以下三个算符： \(\begin{equation} \left\{ \begin{aligned} & \hat{H}_0 = \frac{\hat{p}^2}{2m} + \frac{1}{2} m \omega^2 \hat{x}^2 \\ & \hat{H}_1 = \mathrm{i} \hbar \omega \left( x \frac{\partial}{\partial x} +\frac{1}{2}\right) = - \frac{\omega}{2} \left(\hat{p}\hat{x} + \hat{x}\hat{p}\right) \\ & \hat{H}_2 = \frac{\hat{p}^2}{2m} - \frac{1}{2} m \omega^2 \hat{x}^2 \end{aligned} \right. \end{equation}\) 它们有对易关系： \(\begin{equation} \left\{ \begin{aligned} & [\hat{H}_0, \hat{H}_1] = 2 \mathrm{i} \hbar \omega \hat{H}_2 \\ & [\hat{H}_1, \hat{H}_2] = - 2 \mathrm{i} \hbar \omega \hat{H}_0 \\ & [\hat{H}_2, \hat{H}_0] = 2 \mathrm{i} \hbar \omega \hat{H}_1 \end{aligned} \right. \end{equation}\) 上述对易关系满足$\mathfrak{su}(1,1)$李代数，故应可以生成$\mathrm{SU}(1,1)$李群。为方便运算，进行无量纲化处理，取$\hbar = \omega = m = 1$，并且取$\hat{K}_{i} = 2 \hat{H}_i \quad (i = 0,1,2)$，则算符与对易关系变为： \(\begin{equation} \left\{ \begin{aligned} & \hat{K}_0 = \hat{p}^2 + \hat{x}^2 \\ & \hat{K}_1 = - \left(\hat{p}\hat{x} + \hat{x}\hat{p}\right) \\ & \hat{K}_2 = \hat{p}^2 - \hat{x}^2 \\ & [\hat{K}_0, \hat{K}_1] = \mathrm{i} \hat{K}_2 \\ & [\hat{K}_1, \hat{K}_2] = - \mathrm{i} \hat{K}_0 \\ & [\hat{K}_2, \hat{K}_0] = \mathrm{i} \hat{K}_1 \end{aligned} \right. \end{equation}\)
做指数映射可得到全体群元的表示： \(\begin{equation} \hat{U} = \mathrm{e}^{\mathrm{i}\theta_0\hat{K}_0 + \mathrm{i}\theta_1\hat{K}_1 + \mathrm{i}\theta_2\hat{K}_2} \end{equation}\) 当参数为0时，易知存在幺元；当所有参数取反时，存在逆元；由算符运算的结合律可知群元乘法满足结合律；最后验证封闭性，考虑两个群元： \(\begin{equation} U_k = \mathrm{e}^{\mathrm{i}\theta_{k0}\hat{K}_0 + \mathrm{i}\theta_{k1}\hat{K}_1 + \mathrm{i}\theta_{k2}\hat{K}_2} \quad (k = 1, 2) \end{equation}\) 计算两者指数上算符线性组合的对易关系： \(\begin{equation} \left\{ \begin{aligned} A_k &= \mathrm{i}\theta_{k0}\hat{K}_0 + \mathrm{i}\theta_{k1}\hat{K}_1 + \mathrm{i}\theta_{k2}\hat{K}_2 \quad (k = 1, 2) \\ [A_1, A_2] &= \mathrm{i} (\theta_{11}\theta_{22}-\theta_{21}\theta_{12}) \hat{K}_0 + \mathrm{i} (\theta_{10}\theta_{22}-\theta_{12}\theta_{20}) \hat{K}_1 + \mathrm{i} (\theta_{11}\theta_{20}-\theta_{10}\theta_{21}) \hat{K}_2 \\ & \equiv \mathrm{i}\theta_{20}\hat{K}_0 + \mathrm{i}\theta_{21}\hat{K}_1 + \mathrm{i}\theta_{22}\hat{K}_2 \\ & \equiv A_2 \\ \end{aligned} \right. \end{equation}\) 可以发现两个算符对易子是一个参数不同但形式相同的算符，下面使用BCH公式进行展开： \(\begin{equation} \left\{ \begin{aligned} f(t) &= \mathrm{e}^{At} \mathrm{e}^{Bt} \\ \frac{\mathrm{d}f}{\mathrm{d}t} &= (A + \mathrm{e}^{At} B \mathrm{e}^{-At}) \mathrm{e}^{At}\mathrm{e}^{Bt} \\ &= \left(A + B + \frac{1}{1!} [A, B] t + \frac{1}{2!} [A, [A, B]] t^2 + \dots \right) \mathrm{e}^{At}\mathrm{e}^{Bt} \end{aligned} \right. \label{math:3} \end{equation}\) 两式相除后令$A = A_1, B = A_2$则： \(\begin{equation} \begin{aligned} \frac{\mathrm{d}f}{f} &= \left(A_1 + A_2 + \frac{1}{1!} [A_1, A_2] t + \frac{1}{2!} [A_1, [A_1, A_2]] t^2 + \dots\right) \mathrm{d}t \\ &\equiv \left(A_1 + A_2 + \frac{1}{1!} A_3 t + \frac{1}{2!} A_4 t^2 + \dots\right) \mathrm{d}t \end{aligned} \end{equation}\) 上式中${A_k}$算符集合具有相同的结构，这是由于之前\eqref{math:3}决定的，对式子积分并令$t=1$可以得到： \(\begin{equation} \begin{aligned} \ln f(1) &= A_1 + A_2 + \frac{1}{2!} A_3 + \frac{1}{3!} A_4 + \dots \\ & \equiv \mathrm{i}\theta_{0}\hat{K}_0 + \mathrm{i}\theta_{1}\hat{K}_1 + \mathrm{i}\theta_{2}\hat{K}_2 \\ & \equiv A \end{aligned} \end{equation}\) 故有： \begin{equation} U = U_1 U_2 = \mathrm{e}^{A_1} \mathrm{e}^{A_2} = \mathrm{e}^{A} \end{equation} 可以看到$U$也是群元，封闭性得证。
利用BCH公式以及对易关系，有： \(\begin{equation} \begin{aligned} \mathrm{e}^{-\mathrm{i}\theta \hat{K}_0} \hat{K}_1 \mathrm{e}^{\mathrm{i}\theta \hat{K}_0} &= \hat{K}_1 + \frac{1}{1!} (-\mathrm{i}\theta) [\hat{K}_0, \hat{K}_1] + \frac{1}{2!} (-\mathrm{i}\theta)^2 [\hat{K}_0, [\hat{K}_0, \hat{K}_1]] + \dots \\ &= \sum_{n=0}^{\infty} \frac{1}{n!} (-\mathrm{i}\theta)^n [(\hat{K}_0)^n, \hat{K}_1] \\ &= \sum_{k=0}^{\infty} \frac{1}{(2k+1)!} (-\mathrm{i}\theta)^{2k+1} \mathrm{i} \hat{K}_2 + \sum_{k=0}^{\infty} \frac{1}{(2k)!} (-\mathrm{i}\theta)^{2k} \hat{K}_1 \\ &= \cos \theta \hat{K}_1 + \sin \theta \hat{K}_2 \end{aligned} \end{equation}\) 同理可以得到如下关系： \(\begin{equation} \left\{ \begin{aligned} & \mathrm{e}^{-\mathrm{i}\theta \hat{K}_0} \hat{K}_1 \mathrm{e}^{\mathrm{i}\theta \hat{K}_0} = \hat{K}_1 \cos \theta + \hat{K}_2 \sin \theta \\ & \mathrm{e}^{-\mathrm{i}\theta \hat{K}_0} \hat{K}_2 \mathrm{e}^{\mathrm{i}\theta \hat{K}_0} = \hat{K}_2 \cos \theta - \hat{K}_1 \sin \theta \\ & \mathrm{e}^{-\mathrm{i}\theta \hat{K}_1} \hat{K}_2 \mathrm{e}^{\mathrm{i}\theta \hat{K}_1} = \hat{K}_2 \cosh \theta - \hat{K}_0 \sinh \theta \\ & \mathrm{e}^{-\mathrm{i}\theta \hat{K}_1} \hat{K}_0 \mathrm{e}^{\mathrm{i}\theta \hat{K}_1} = \hat{K}_0 \cosh \theta - \hat{K}_2 \sinh \theta \\ & \mathrm{e}^{-\mathrm{i}\theta \hat{K}_2} \hat{K}_0 \mathrm{e}^{\mathrm{i}\theta \hat{K}_2} = \hat{K}_0 \cosh \theta + \hat{K}_1 \sinh \theta \\ & \mathrm{e}^{-\mathrm{i}\theta \hat{K}_2} \hat{K}_1 \mathrm{e}^{\mathrm{i}\theta \hat{K}_2} = \hat{K}_1 \cosh \theta + \hat{K}_0 \sinh \theta \end{aligned} \right. \end{equation}\)