角动量算子

在本题中我们使用另一种方式定义角动量算子。
🐈‍⬛ 我们保留使用$\widehat{J}^2$和$\widehat{J}z$作为$ECOC$的思路。
🐈‍⬛ 他们的特征向量被写作：$|j, m\rangle$
🐈‍⬛ $\widehat{J}^2|j, m\rangle=\hbar^2 j(j+1)|j, m\rangle$
🐈‍⬛ $\widehat{J}_z|j, m\rangle=\hbar m|j, m\rangle$
🐈‍⬛ 并同样定义阶梯算子：$\widehat{J}{ \pm}=\widehat{J}_x \pm i \widehat{J}_y$

算子

阶梯算子的对易证明

$[\widehat J^2, \widehat J{\pm}] = [\widehat J^2, \widehat J{x}\pm i\widehat Jy] = [\widehat J^2, \widehat J{x}]\pm[\widehat J^2, \widehat J_{y}] = 0$
$\begin{aligned}{\left[\hat{J}z, \hat{J}{ \pm}\right]=\left[\hat{J}z, \hat{J}_x \pm i \hat{J}_y\right] } =&\left[\hat{J}_z, \hat{J}_x\right] \pm i\left[\hat{J}_z, \hat{J}_y\right] =i \hbar \hat{J} y+\hbar \hat{J}_x \& = \pm \hbar\left(\hat{J}_x \pm i \hat{J}_y\right)= \pm \hbar \hat{J}{ \pm}\end{aligned}$

证明$\widehat{J}{ \pm}|j, m\rangle$是$\hat{J}^2, \hat{J}{ \pm}$的特征向量

$\widehat{J}^2 \widehat{J}{ \pm}|j, m\rangle=\widehat{J}{ \pm} \widehat{J}^2|j, m\rangle=\hbar^2 j(j+1) \widehat{J}_{ \pm}|j, m\rangle$
$\widehat{J}z \widehat{J}{ \pm}|j, m\rangle=\left( \pm \hbar \widehat{J}{ \pm}+\widehat{J}{ \pm} \widehat{J}z\right)|j, m\rangle= \pm \hbar \widehat{J}{ \pm}|j, m\rangle+\hbar m \widehat{J}{ \pm}|j, m\rangle=\hbar(m \pm 1) \widehat{J}{ \pm}|j, m\rangle$

证明$-j \leqslant m \leqslant j$

计算$| \hat{J}_{ \pm}|j, m\rangle |$
- $| \widehat{J}{ \pm}|j, m\rangle |^2=\left\langle j, m\left|\widehat{J}{ \pm}^{\dagger} \widehat{J}{ \pm}\right| j, m\right\rangle = \left\langle j, m\left|\widehat{J}{\mp} \widehat{J}_{ \pm}\right| j, m\right\rangle$
- $\widehat{J}{\mp} \widehat{J}{ \pm}=\left(\widehat{J}_x \mp i \widehat{J}_y\right)\left(\widehat{J}_x \pm i \widehat{J}_y\right)=\widehat{J}_x^2+\widehat{J}_y^2 \pm i \widehat{J}_x \widehat{J}_y \mp i \widehat{J}_y \widehat{J}_x=\widehat{J}^2-\widehat{J}_z^2 \pm i\left[\widehat{J}_x, \widehat{J}_y\right]=\widehat{J}^2-\widehat{J}_z^2 \mp \hbar \widehat{J}_z$
- $| \widehat{J_{ \pm}}|j, m\rangle |^2=\left\langle j, m\left|\widehat{J}^2-\widehat{J}_z^2 \mp \hbar \widehat{J_z}\right| j, m\right\rangle=\hbar^2 j(j+1)-\hbar^2 m^2 \mp \hbar^2 m=\hbar^2[j(j+1)-m(m \pm 1)]$
- $| \widehat{J}_{ \pm}|j, m\rangle |=\sqrt{j(j+1)-m(m \pm 1)} \hbar$
解不等式
- $j(j+1)-m(m \pm 1) \geqslant 0 \Leftrightarrow j(j+1) \geqslant m(m \pm 1) \Leftrightarrow\left{\begin{array}{l}j(j+1) \geqslant m(m+1) \j(j+1) \geqslant m(m-1)\end{array}\right.$
- $-j \leqslant m \leqslant j$

证明$\widehat{J}_{ \pm}|j, m\rangle=\sqrt{j(j+1)-m(m \pm 1)} \hbar|j, m \pm 1\rangle$

$|j, m \pm 1\rangle$是标准的
$\begin{aligned}& \hat{J}^2 \hat{J}{ \pm}|j, m\rangle=\hbar^2 j(j+1) \hat{J}{ \pm}|j, m\rangle \& \hat{J}z \hat{J}{ \pm}|j, m\rangle=\hbar(m \pm \text { 1) } \hat{J}{ \pm}|j, m\rangle\end{aligned}\Rightarrow \widehat{J}{ \pm}|j, m\rangle \propto|j, m \pm 1\rangle$
$\widehat{J}_{ \pm}|j, m\rangle=\sqrt{j(j+1)-m(m \pm 1)} \hbar|j, m \pm 1\rangle$

证明$2j$和$2m$都是整数

$\widehat{J}^{2j}_{+}|j,-j\rangle=A|j,j\rangle$ 因此$2j$是整数
$m$的边界是$-j$和$j$，同理，$2m$是整数，且$m=-j,-j+1, \ldots, j-1, j$

球谐

接下来考虑球谐函数
🐈‍⬛ 球坐标系中：$\widehat{L}z=-i \hbar \frac{\partial}{\partial \varphi}$
🐈‍⬛ $Y{\ell}^{m{\ell}}(\theta, \varphi) \equiv F{\ell, m{\ell}}(\theta) e^{i m{\ell} \varphi}$是一个本征函数

证明$Y{\ell}^{m{\ell}}(\theta, \varphi) \equiv F{\ell, m{\ell}}(\theta) e^{i m_{\ell} \varphi}$

由于$Y{\ell}^{m{\ell}}(\theta, \varphi) \equiv F{\ell, m{\ell}}(\theta) e^{i m_{\ell} \varphi}$是一个本征函数，有：

$\widehat LzY{\ell}^{m{\ell}}(\theta, \varphi) = m_l\hbar Y{\ell}^{m{\ell}}(\theta, \varphi) = -i\hbar\frac{\partial Y{\ell}^{m_{\ell}}(\theta, \varphi)}{\partial \phi}$
解微分方程，可得球谐函数形式：$Y{\ell}^{m{\ell}}(\theta, \varphi)=F{\ell, m{\ell}}(\theta) e^{i m_{\ell \varphi}}$

证明对于轨道矩，$l$和$m_l$是整数

$Y{\ell}^{m{\ell}}(\theta, \varphi+2 \pi)=Y{\ell}^{m{\ell}}(\theta, \varphi) \Leftrightarrow F{\ell, m{\ell}}(\theta) e^{i m{\ell}(\varphi+2 \pi)}=F{\ell, m{\ell}}(\theta) e^{i m{\ell} \varphi} \Leftrightarrow e^{i 2 m{\ell} \pi}=1 \Leftrightarrow m{\ell} \in \mathbb{Z}$
根据$m_l$的可选取值，可得$l$也是整数。