Information
- 官方介绍: https://ocw.mit.edu/courses/physics/8-421-atomic-and-optical-physics-i-spring-2014/
- 视频列表地址: https://www.youtube.com/playlist?list=PLUl4u3cNGP62FPGcyFJkzhqq9c5cHCK32
Fourier limit?
Lorentz model for an atom
见另一独立的
Larmor Procession
经典磁矩可以假想电流得到
$$\begin{align} \vec{\mu} = I \vec{S} \end{align}$$其中 $I$ 为环形电流, $\vec{S}$ 为环形电流围成的面积. 若是带电 $-e$ 电子, 则
$$\begin{align} \mu = \frac{-e}{2\pi r / v} \cdot \pi r^2 = \frac{-e}{2m} \cdot m v r = -\frac{e}{2m} L \end{align}$$写成矢量形式, 并定义旋磁比 gyromagnetic ratio $\gamma = -\frac{e}{2m}$
$$\begin{align} \vec{\mu} = \gamma \vec{L} \end{align}$$磁矩在磁场中受到的力矩为
$$\begin{align} \vec{\tau} = \vec{\mu} \times \vec{B} \end{align}$$那么角动量的变化就是力矩
$$\begin{align} \dot{\vec{L}} = \vec{\mu}\times \vec{B} = -\gamma \vec{B} \times \vec{L} \end{align}$$可知, $\vec{L}$ 变化的方向与 $\vec{B}$ 和 $\vec{L}$ 都垂直, 且大小不变, 也就是说 在绕 $B$ 进动. 而 $-\gamma \vec{B}$ 是频率的量纲, 是进动的频率, Larmor 频率
$$\begin{align} \Omega_L = \frac{e}{2m}B \end{align}$$定义 Bohr magneton 为轨道角动量为 $L = - \hbar$ 时对应的磁矩
$$\begin{align} \mu_B = \gamma \cdot (-\hbar) =\frac{e\hbar}{2m_e} \approx 2\pi \times 1.4 \hbar \mathrm{MHz /G} \end{align}$$但是电子自旋的 gyromagnetic ratio 是经典值的 $g_s = -g_e = 2$ 倍. 即 如, 电子的 gyromagnetic ratio is
$$\begin{align} |\gamma_e| \approx 2 \times \frac{e}{2m_e} = 2\pi \times 2.8 \mathrm{MHz /G} \end{align}$$如果自旋 $z$ 分量的 量子数为 $m_s$ , 那么对应的磁矩为
$$\begin{align} \mu_z = g_s\cdot \frac{-e}{2 m_e}\cdot m_s\hbar = - g_s \mu_B m_s \end{align}$$总的来说, $L = \hbar, S = \frac{1}{2}\hbar$ 的电子, 对应的磁矩都是 $\mu_B$ , 但是磁矩的方 向与角动量的方向相反.
The angular precession frequency has an important physical meaning: It is the angular cyclotron frequency, the resonance frequency of an ionized plasma being under the influence of a static finite magnetic field, when we superimpose a high frequency electromagnetic field.
Rotating Coordinate Trans
$$\begin{align} \left.\frac{\mathrm{d}}{\mathrm{d}t}\right|_{\mathrm{rot}} = \left.\frac{\mathrm{d}}{\mathrm{d}t}\right|_{\mathrm{inertial}} - \vec{\Omega}\times \end{align}$$Exp: \begin{align} \dot{\mathrm{L}}_{\mathrm{rot}} = \dot{\mathrm{L}}_{\mathrm{inertial}} - \Omega\times \vec{L} = \gamma \vec{L} (\vec{B} + \frac{\vec{\Omega}}{\gamma}) \end{align}
If choose rotating frequency
$$\begin{align} \vec{\Omega} = \Omega_L = - \gamma \vec{B} \end{align}$$then
$$\begin{align} \vec{B}_{\mathrm{rot}} = 0 \end{align}$$so, you can transform away the effect of a magnetic field by going to rotating fram at the Larmor frequency.
Reference
- 杨福家, 原子物理学, 第四版
- Wikipedia: Gyromagnetic ratio
- Spin magnetic moment
- G-factor
- Foot, C. J. Atomic physics. (Oxford University Press, 2005).