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

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

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：

\(\widehat L_zY_{\ell}^{m_{\ell}}(\theta, \varphi) = m_l\hbar Y_{\ell}^{m_{\ell}}(\theta, \varphi) = -i\hbar\frac{\partial Y_{\ell}^{m_{\ell}}(\theta, \varphi)}{\partial \phi}\)
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。