# 角動量算符

## 數學定義

$\mathbf{L}\ \stackrel{def}{=}\ \mathbf{r}\times\mathbf{p}\,\!$

$\hat{\mathbf{L}}\ \stackrel{def}{=}\ \hat{\mathbf{r}}\times\hat{\mathbf{p}}\,\!$

$\hat{\mathbf{p}}= - i\hbar\nabla\,\!$

$\hat{\mathbf{L}}= - i\hbar(\hat{\mathbf{r}}\times\nabla) \,\!$

## 角動量是厄米算符

$\hat{L}_x=\hat{y}\hat{p}_z - \hat{z}\hat{p}_y\,\!$

$\hat{L}_x^{\dagger}=(\hat{y}\hat{p}_z - \hat{z}\hat{p}_y)^{\dagger}=\hat{p}_z^\dagger\hat{y}^{\dagger} - \hat{p}_y^+\hat{z}^{\dagger}\,\!$

$\hat{L}_x^{\dagger}=\hat{p}_z\hat{y} - \hat{p}_y\hat{z}\,\!$

$\hat{L}_x^{\dagger}=\hat{y}\hat{p}_z - \hat{z}\hat{p}_y=\hat{L}_x\,\!$

$\hat{L}^2=\hat{L}_x^2+\hat{L}_y^2+\hat{L}_z^2\,\!$

$(\hat{L}^2)^{\dagger}=(\hat{L}_x^2+\hat{L}_y^2+\hat{L}_z^2)^{\dagger}=(\hat{L}_x^2)^{\dagger}+(\hat{L}_y^2)^{\dagger}+(\hat{L}_z^2)^{\dagger}\,\!$

$(\hat{L}^2)^{\dagger}=\hat{L}_x^2+\hat{L}_y^2+\hat{L}_z^2=\hat{L}^2\,\!$

## 對易關係

### 角動量算符與自己的對易關係

\begin{align} \left.\right.[\hat{L}_x,\ \hat{L}_y] & = [\hat{y}\hat{p}_z - \hat{z}\hat{p}_y,\ \hat{z}\hat{p}_x - \hat{x}\hat{p}_z] \\ & =[\hat{y}\hat{p}_z,\ \hat{z}\hat{p}_x] - [\hat{z}\hat{p}_y,\ \hat{z}\hat{p}_x] - [\hat{y}\hat{p}_z,\ \hat{x}\hat{p}_z]+[\hat{z}\hat{p}_y,\ \hat{x}\hat{p}_z] \\ & =i\hbar (\hat{x}\hat{p}_y - \hat{y}\hat{p}_x) \\ & =i\hbar\hat{L}_z \\ \end{align}\,\!

$\Delta L_x\ \Delta L_y \ge \left|\frac{\langle[\hat{L}_x,\ \hat{L}_y]\rangle}{2i}\right|=\frac{\hbar |\langle \hat{L}_z\rangle|}{2}\,\!$

$L_x\,\!$ 的不確定性與 $L_y\,\!$ 的不確定性的乘積 $\Delta L_x\ \Delta L_y \,\!$ ，必定大於或等於 $\frac{\hbar |\langle L_z\rangle|}{2}\,\!$

$L_x\,\!$$L_z\,\!$ 之間，$L_y\,\!$$L_z\,\!$ 之間，也有類似的特性。

### 角動量平方算符與角動量算符之間的對易關係

\begin{align}\left.\right.[\hat{L}^2,\ \hat{L}_z] & = [\hat{L}_x^2+\hat{L}_y^2+\hat{L}_z^2,\ \hat{L}_z] \\ & = \hat{L}_x\hat{L}_x\hat{L}_z - \hat{L}_z\hat{L}_x\hat{L}_x+\hat{L}_y\hat{L}_y\hat{L}_z - \hat{L}_z\hat{L}_y\hat{L}_y \\ & = \hat{L}_x(\hat{L}_z\hat{L}_x - i\hbar\hat{L}_y) - (\hat{L}_x\hat{L}_z+i\hbar\hat{L}_y)\hat{L}_x+\hat{L}_y(\hat{L}_z\hat{L}_y+i\hbar\hat{L}_x) - (\hat{L}_y\hat{L}_z - i\hbar\hat{L}_x)\hat{L}_y \\ & = 0 \\ \end{align}\,\!

$\hat{L}^2 \,\!$$\hat{L}_z\,\!$對易的$L^2 \,\!$$L_z\,\!$ 彼此是相容可觀察量，兩個算符有共同的本徵態。根據不確定性原理，我們可以同時地測量到 $L^2 \,\!$$L_z\,\!$ 的本徵值。

$[\hat{L}^2,\ \hat{L}_x] =0\,\!$
$[\hat{L}^2,\ \hat{L}_y] =0\,\!$

$\hat{L}^2\,\!$$\hat{L}_x\,\!$ 之間、$\hat{L}^2\,\!$$\hat{L}_y\,\!$ 之間，都分別擁有類似的物理特性。

### 哈密頓算符與角動量算符之間的對易關係

$[\hat{H},\ \hat{L}_z]=\left[i\hbar\frac{\partial}{\partial t},\ \hat{x}\hat{p}_y - \hat{y}\hat{p}_x\right]=0\,\!$

$\hat{H}\,\!$$\hat{L}_z\,\!$對易的$H\,\!$$L_z\,\!$ 彼此是相容可觀察量，兩個算符擁有共同的本徵態。根據不確定性原理，我們可以同時地測量到 $H\,\!$$L_z\,\!$ 的同樣的本徵值。

$[\hat{H},\ \hat{L}_x] =0\,\!$
$[\hat{H},\ \hat{L}_y] =0\,\!$

$\hat{H}\,\!$$\hat{L}_x\,\!$ 之間，$\hat{H}\,\!$$\hat{L}_y\,\!$ 之間，都分別擁有類似的物理特性。

### 在經典力學裏的對易關係

$\{L_i,\ L_j\}=\epsilon_{ijk}L_k\,\!$

## 本徵值與本徵函數

\begin{align}\hat{\mathbf{L}} & = \frac{\hbar}{i}\hat{\mathbf{r}}\times\nabla \\ & = \frac{\hbar}{i} r\mathbf{e}_r \times \left(\mathbf{e}_r \frac{\partial}{\partial r}+\mathbf{e}_{\theta} \frac{1}{r}\frac{\partial}{\partial \theta}+ \mathbf{e}_{\phi} \frac{1}{r\sin\theta}\frac{\partial}{\partial\phi}\right) \\ & = \frac{\hbar}{i}\left( - \mathbf{e}_{\theta}\frac{1}{\sin\theta} \frac{\partial}{\partial \phi} +\mathbf{e}_{\phi}\frac{\partial}{\partial \theta}\right) \\ \end{align} \,\!

$\hat{\mathbf{L}}=\frac{\hbar}{i}\left[ \mathbf{e}_x \left( - \sin \phi\frac{\partial}{\partial \theta} - \cot\theta\cos\phi\frac{\partial}{\partial \phi}\right) +\mathbf{e}_y\left(\cos \phi\frac{\partial}{\partial \theta} - \cot\theta\sin\phi\frac{\partial}{\partial \phi}\right) +\mathbf{e}_z\frac{\partial}{\partial \phi}\right] \,\!$

$\hat{L}_x=\frac{\hbar}{i}\left( - \sin \phi\frac{\partial}{\partial \theta} - \cot\theta\cos\phi\frac{\partial}{\partial \phi}\right)\,\!$
$\hat{L}_y=\frac{\hbar}{i}\left(\cos \phi\frac{\partial}{\partial \theta} - \cot\theta\sin\phi\frac{\partial}{\partial \phi}\right)\,\!$
$\hat{L}_z=\frac{\hbar}{i}\frac{\partial}{\partial \phi}\,\!$

$\hat{L}^2=\hat{L}_x^2+\hat{L}_y^2+\hat{L}_z^2\,\!$

\begin{align}\hat{L}_x^2 & = - \hbar^2\left( - \sin \phi\frac{\partial}{\partial \theta} - \cot\theta\cos\phi\frac{\partial}{\partial \phi}\right)\left( - \sin \phi\frac{\partial}{\partial \theta} - \cot\theta\cos\phi\frac{\partial}{\partial \phi}\right) \\ & = - \hbar^2\left(\sin^2\phi\frac{\partial^2}{\partial \theta^2}+\cot\theta\cos^2\phi\frac{\partial}{\partial \theta}+\cot\theta\sin\phi\cos\phi\frac{\partial^2}{\partial \theta \partial \phi} - \csc^2\theta\sin\phi\cos\phi\frac{\partial}{\partial \phi}\right. \\ \end{align}\,\!
$\left.+\cot\theta\sin\phi\cos\phi\frac{\partial^2}{\partial \theta \partial \phi} - \cot^2\theta\sin\phi\cos\phi\frac{\partial}{\partial \phi}+\cot^2\theta\cos^2\phi\frac{\partial^2}{\partial \phi^2}\right)\,\!$
\begin{align}\hat{L}_y^2 & = - \hbar^2\left(\cos\phi\frac{\partial}{\partial \theta} - \cot\theta\sin\phi\frac{\partial}{\partial \phi}\right)\left(\cos \phi\frac{\partial}{\partial \theta} - \cot\theta\sin\phi\frac{\partial}{\partial \phi}\right) \\ & = - \hbar^2\left(\cos^2\phi\frac{\partial^2}{\partial \theta^2}+\cot\theta\sin^2\phi\frac{\partial}{\partial \theta} - \cot\theta\sin\phi\cos\phi\frac{\partial^2}{\partial \theta \partial \phi}+\csc^2\theta\sin\phi\cos\phi\frac{\partial}{\partial \phi}\right. \\ \end{align}\,\!
$\left. - \cot\theta\sin\phi\cos\phi\frac{\partial^2}{\partial \theta \partial \phi}+\cot^2\theta\sin\phi\cos\phi\frac{\partial}{\partial \phi}+\cot^2\theta\sin^2\phi\frac{\partial^2}{\partial \phi^2}\right)\,\!$
$\hat{L}_z^2= - \hbar^2\frac{\partial^2}{\partial \phi^2}\,\!$

$\hat{L}^2= - \hbar^2\left(\frac{\partial^2}{\partial \theta^2}+\cot\theta\frac{\partial}{\partial \theta}+(1+\cot^2\theta)\frac{\partial^2}{\partial \phi^2}\right) = - \hbar^2\left(\frac{1}{\sin\theta}\frac{\partial}{\partial \theta}\left(\sin\theta\frac{\partial}{\partial \theta}\right)+\frac{1}{\sin^2\theta}\frac{\partial^2}{\partial \phi^2}\right)\,\!$

$\hat{L}^2 Y_{\ell m}= - \ell(\ell+1)\hbar^2 Y_{\ell m}\,\!$

$\hat{L}_z Y_{\ell m}= m\hbar Y_{\ell m}\,\!$

$Y_{\ell m}(\theta,\ \phi) =(i)^{m+|m|} \sqrt{{(2\ell+1)\over 4\pi}{(\ell - m)!\over (\ell+m)!}} \, P_{\ell m} (\cos{\theta}) \, e^{im\phi}\,\!$

$P_{\ell m}(x) = (1 - x^2)^{|m|/2}\ \frac{d^{|m|}}{dx^{|m|}}P_\ell(x)\,$

$P_\ell(x)\,\!$$\ell$勒讓德多項式，可用羅德里格公式表示為：

$P_\ell(x) = {1 \over 2^\ell \ell!} {d^\ell \over dx^\ell }(x^2 - 1)^\ell$

$\int_{0}^{2\pi}\int_{0}^{\pi}\ Y_{\ell_1 m_1}Y_{\ell_2 m_2}\sin(\theta)d\theta d\phi=\delta_{\ell_1 \ell_2}\delta_{m_1m_2}\,\!$

$\psi(\theta,\,\phi)=\sum_{\ell ,m}\ A_{\ell m}Y_{\ell m}(\theta,\,\phi)\,\!$

## 參考文獻

1. ^ Introductory Quantum Mechanics, Richard L. Liboff, 2nd Edition, ISBN 0201547155

## 外部連結

• 圣地牙哥加州大学物理系量子力学視聽教學：角動量加法