Radical Potential
Schrodinger EQ
变形为
\begin{align} \nabla^2 \psi = \left[ \frac{2m}{\hbar^2} V(r) - k^2 \right]\psi \end{align}其中 \(k^2 = E/\frac{\hbar^2}{2m}\) . 写出 \(\nabla^2\) 在球 坐标系中的形式, 方程变为
\begin{align} \left[\frac{1}{r^2}\frac{\partial}{\partial r}r^2 \frac{\partial}{\partial r} + \frac{1}{r^2 \sin \theta}\frac{\partial}{\partial \theta}\sin \theta \frac{\partial}{\partial \theta} + \frac{1}{r^2\sin^2 \theta}\frac{\partial^2}{\partial\phi^2} \right] \psi = \left[ \frac{2m}{\hbar^2} V(r) - k^2 \right]\psi \end{align}Separation of variables
分离变量, 令
\begin{align} \psi(r,\theta,\phi) = R(r)\Theta(\theta)\Phi(\phi) \end{align}代回方程, 并且两边同时除以 \(R(r)\Theta(\theta)\Phi(\phi)\) 可得
\begin{align} \frac{1}{R}\frac{1}{r^2}\frac{\mathrm{d}}{\mathrm{d} r}r^2 \frac{\mathrm{d}}{\mathrm{d} r} R+ \frac{1}{\Theta}\frac{1}{r^2 \sin \theta}\frac{\mathrm{d}}{\mathrm{d} \theta}\sin \theta \frac{\mathrm{d}}{\mathrm{d} \theta}\Theta +\frac{1}{\Phi} \frac{1}{r^2\sin^2 \theta}\frac{\mathrm{d}^2}{\mathrm{d}\phi^2}\Phi = \frac{2m}{\hbar^2} V(r) - k^2 \end{align}所有含有 \(\phi\) 的项都在方程左边第三项, 与 \(r,\theta\) 没有耦合, 所以 它一定是一个常数. 因为其它的项中都与 \(\theta\) 无关, \(\theta\) 改变时其它 项保持不变. 记这个常数为 \(-m^2\) 即
\begin{align} \frac{1}{\Phi}\frac{\mathrm{d}^2}{\mathrm{d}\phi^2}\Phi = -m^2 \end{align}它的解为
\begin{align} \Phi (\phi) = e^{\mathrm{i}m\phi} \end{align}将它代回原方程, 原方程变为
\begin{align} \frac{1}{R}\frac{1}{r^2}\frac{\mathrm{d}}{\mathrm{d} r}r^2 \frac{\mathrm{d}}{\mathrm{d} r} R+ \frac{1}{\Theta}\frac{1}{r^2 \sin \theta}\frac{\mathrm{d}}{\mathrm{d} \theta}\sin \theta \frac{\mathrm{d}}{\mathrm{d} \theta}\Theta - \frac{m^2}{r^2\sin^2 \theta} = \frac{2m}{\hbar^2} V(r) - k^2 \end{align}\(\theta\) 和 \(r\) 也没有耦合, 同理可以把与 \(\theta\) 有关的项单拎出来, 是一个常数, 记为 \(-\lambda\) ,即
\begin{align} \frac{1}{\Theta}\frac{1}{ \sin \theta}\frac{\mathrm{d}}{\mathrm{d} \theta}\sin \theta \frac{\mathrm{d}}{\mathrm{d} \theta}\Theta - \frac{m^2}{\sin^2 \theta} = -\lambda \end{align}上述关于 \(\theta\) 的方程的解是连带 Legendre 多项式 \(P_l^m(\cos\theta)\), 本征值 \(\lambda\) 只能取值为 \(l(l+1)\) , \(l\) 是个 整数. 剩下的只有 \(r\) 的方程为
\begin{align} \frac{1}{R}\frac{1}{r^2}\frac{\mathrm{d}}{\mathrm{d} r}r^2 \frac{\mathrm{d}}{\mathrm{d} r} R = \frac{2m}{\hbar^2} V(r) - k^2 +\frac{l(l + 1)}{r^2} \end{align}即
\begin{align} \frac{1}{r^2}\frac{\mathrm{d}}{\mathrm{d} r}r^2 \frac{\mathrm{d}}{\mathrm{d} r} R -\left[ \frac{2m}{\hbar^2} V(r) - k^2 +\frac{l(l + 1)}{r^2}\right]R =0 \end{align}若 \(V(r) = 0\) 它的一般解的形式是球 Bessel 函数
\begin{align} R_l(kr) = a_{lm}j_l(kr) + b_{lm}y_l(kr) \end{align}Final Results
\(\Theta\) 和 \(\Phi\) 的解合在一起, 添加归一化条件为 Spherical harmonics \(Y_l^m(\theta,\phi)\) ,所以对于 \(V(r) = 0\) 的自由粒子, 总的解就是
\begin{align} \psi(r,\theta,\phi) = \sum_{l=0}^{\infty} \sum_{m=-l}^l \left[ a_{lm}j_l(kr) + b_{lm}y_l(kr) \right] Y_l^m(\theta,\phi) \end{align}Partial Wave Expansion
Free Particle
自由粒子只有动能项, Hamiltonian 与角动量 \(\vec{L}\) 和角动量 \(z\) 分量 \(L_z\) 都对易, 所以它们有共同的本征态, 记为 \(|E,l,m\rangle\) , 它在 动量和坐标表象下的矩阵元分别为
\begin{align} \langle \vec{k}|E,l,m\rangle =& \frac{\hbar}{\sqrt{mk}} \delta \left( E - \frac{\hbar^2k^2}{2m} \right) Y_l^m(\hat{k})\\ \langle \vec{r}|E,l,m\rangle =& \frac{\mathrm{i}^l}{\hbar} \sqrt{\frac{2mk}{\pi}}j_l(kr) Y_l^m(\hat{r}) \end{align}Rayleigh equation
它等价于
\begin{align} e^{\mathrm{i}\vec{k}\vec{r}} = 4\pi \sum_{l=0}^{\infty}\sum_{m=-l}^l \mathrm{i}^l Y_l^m(\hat{r})\left[ Y_l^m (\hat{k}) \right]^{*}j_l(kr) \end{align}Integral representation for \(j_l(kr)\)
\begin{align} j_l(kr) = \frac{1}{2\mathrm{i}^l}\int_{-1}^1 e^{\mathrm{i}kr\cos \theta}P_l(\cos\theta)\mathrm{d}(\cos\theta) \end{align}Summary
Summary of A Free Partical in Spherical Coordinate
自由粒子的 Schrodinger equation 是一个 Helmholtz 方程
\begin{align} \left(\nabla^2 \psi + k^2 \right)\psi = 0 \end{align}在球坐标系中可以分离变量
\begin{align} \psi(r,\theta,\phi) = R(r)\Theta(\theta)\Phi(\phi) \end{align}代回原式得
\begin{align} \frac{1}{R}\frac{1}{r^2}\frac{\mathrm{d}}{\mathrm{d} r}r^2 \frac{\mathrm{d}}{\mathrm{d} r} R+ \frac{1}{\Theta}\frac{1}{r^2 \sin \theta}\frac{\mathrm{d}}{\mathrm{d} \theta}\sin \theta \frac{\mathrm{d}}{\mathrm{d} \theta}\Theta +\frac{1}{\Phi} \frac{1}{r^2\sin^2 \theta}\frac{\mathrm{d}^2}{\mathrm{d}\phi^2}\Phi + k^2 = 0 \end{align}Schrodinger equation 化为三个方程, 分别是
关于 \(\phi\) 的方程
\begin{align} \frac{1}{\Phi}\frac{\mathrm{d}^2}{\mathrm{d}\phi^2}\Phi = -m^2 \end{align}解为
\begin{align} \Phi (\phi) = \sum_{m=-\infty}^{\infty} e^{\mathrm{i}m\phi} \end{align}关于 \(\theta\) 的方程 ( \(x = \cos \theta\) )( associated Legendre equation )
\begin{align} (1-x^2)\frac{\mathrm{d}^2}{\mathrm{d}x^2}\Theta - 2x \frac{\mathrm{d}}{\mathrm{d}x} \Theta+\left[ l(l + 1) - \frac{m^2}{1-x^2} \right] \Theta = 0 \end{align}解为 associated Legendre function
\begin{align} \Theta = \sum_{l=|m|}^{\infty}P_l^m(\cos\theta) \end{align}关于 \(r\) 的方程 ( spherical Bessel equation )
\begin{align} r^2 \frac{\mathrm{d}^2}{\mathrm{d}r^2}R +2r \frac{\mathrm{d}}{\mathrm{d}r}R +\left[ k^2r^2 - l(l + 1 ) \right]R = 0 \end{align}解为 spherical Bessel function
\begin{align} R = \sum_{l=|m|}^{\infty} a_{lm} j_l(kr) + b_{lm} y_l(kr) \end{align}总的解为 ( 能量为 \(E = \frac{\hbar^2k^2}{2m}\) 的一个本征态 ) (已舍去 在原点发散的 \(y_l(kr)\) ) ( 归一化为 \(\delta_{ll'} \delta_{mm'} \delta(E-E')\) )
\begin{align} \psi_k(r,\theta,\phi) = \sum_{l=0}^{\infty}\sum_{m = -l}^l \frac{\mathrm{i}^l}{\hbar}\sqrt{\frac{2mk}{\pi}} j_l(kr)Y_l^m(\theta,\phi) \end{align}其中 \(Y_l^m(\theta,\phi)\) 为 spherical harmonics
\begin{align} Y_l^m (\theta,\phi) \equiv \sqrt{\frac{2l+1}{4\pi}\frac{(l-m)!}{(l+m)!}} P_l^m(\cos\theta)e^{\mathrm{i}m\phi} \end{align}Summary of Some Useful Equations
Closure Relation for spherical Bessel functions
\begin{align} \int_0^{\infty} r^2 j_l(k_1r)j_l(k_2r) \mathrm{d}r = \frac{\pi}{2k_1^2}\delta(k_1-k_2) \end{align}Orthogonality of spherical harmonics
\begin{align} \int_0^{2\pi} \mathrm{d}\phi \int_0^{\pi}\sin \theta \mathrm{d}\theta \left[ Y_{l_1}^{m_1} (\theta,\phi) \right]^{*} Y_{l_2}^{m_2} (\theta,\phi) =\delta_{l_1l_2} \delta_{m_1m_2} \end{align}Orthogonality of Legendre Functions
\begin{align} \int_{-1}^{1} P_l(x)P_m(x) \cdot \mathrm{d}x = \frac{2 \delta_{lm}}{2l+1} \end{align}\(x \to \infty\) asymptotic behavior of spherical functions
\begin{align} j_n(x) =& \frac{1}{x}\sin (x - \frac{n\pi}{2}) \\ y_n(x) =& - \frac{1}{x}\cos (x - \frac{n\pi}{2}) \end{align}Reference
梁昆淼编, 刘法,缪国庆修订, 数学物理方法(第四版)
- 11.5 球贝塞尔方程
Arfken, Weber, Harris, Mathematical Methods for Physicists 7ed:
- Chap 7.5 Series Solutions-Frobenius' Methods
- Example 8.3.1 Legendre Equation
- Chap 14 Bessel Functions
- Chap 15.1 Legendre Polynomials
J. J Sakurai, Jim Napolitano, Modern Quantum Mechanics 2ed:
- Chap 6.4 Phase Shifts and Partial Waves
R. Mehrem, arXiv:0909.0494v4: https://arxiv.org/pdf/0909.0494.pdf