# 角动量

$\mathbf{L} = \mathbf{r} \times \mathbf{p}$

$\mathbf{L} = \mathbf{r}\times\mathbf{p} = \mathbf{r}\times(m\mathbf{v}) = mr^2\boldsymbol{\omega} = I\boldsymbol{\omega}\,\!$

## 角動量量子化

### 量子化角動量和不确定性原理

$[L_i, L_j] = i\hbar\epsilon_{ijk} L_k$

$\left[L_i, L^2\right] = 0$

$L^2 = -\frac{\hbar^2}{\sin\theta}\frac{\partial}{\partial \theta}\left( \sin\theta \frac{\partial}{\partial \theta}\right) - \frac{\hbar^2}{\sin^2\theta}\frac{\partial^2}{\partial \phi^2}$

$L^2 | l, m \rang = {\hbar}^2 l(l+1) | l, m \rang$
$L_z | l, m \rang = \hbar m | l, m \rang$

$\lang \theta , \phi | l, m \rang = Y_{l,m}(\theta,\phi)$

$l$是某非負整數。$-l\le m \le l$是絕對值不大於$l$的整數。

### 能量均分與角動量量子化

$E=\frac{L_z^2}{2 I}$

$L_z$是分子旋轉的角動量，$I$轉動慣量和原子的距离平方成正比。從運用統計力學的配分函數

$Z=\int_{-\infty}^{\infty} d L_z e^{-\beta \frac{L_z^2}{2I}} = \sqrt{\frac{2\pi I}{\beta}}$

$\beta=\frac{1}{k_{B}T}$是温度$T$的倒數）可以得到古典旋轉運動對平均能量的貢献：

$\frac{}{N} = - \frac{\partial \log Z}{\partial \beta} = \frac{1}{2\beta} = \frac{k_B T}{2}$

$Z=\sum_{n=-\infty}^{\infty} e^{-\beta \frac{n^2 \hbar^2}{2 I}}$

$Z \simeq 1 + e^{-\beta \frac{n^2 \hbar^2}{2 I}} + \cdots$

$\frac{}{N} = -\frac{\partial \log Z}{\partial \beta} \simeq \frac{n^2 \hbar^2}{2 I}$

$T^* \approx \frac{\hbar^2}{2I}$