# 球谐函数

## 函数的推导

### 本征方程的推导

$\nabla^2 f = {1 \over r^2}{\partial \over \partial r}\left(r^2 {\partial f \over \partial r}\right) + {1 \over r^2\sin\theta}{\partial \over \partial \theta}\left(\sin\theta {\partial f \over \partial \theta}\right) + {1 \over r^2\sin^2\theta}{\partial^2 f \over \partial \varphi^2} = 0\,\!$

${\Theta \Phi \over r^2}{d \over dr}\left(r^2 {d R \over dr}\right) + {R\Phi \over r^2\sin\theta}{d \over d \theta}\left(\sin\theta {d \Theta \over d \theta}\right) + {R\Theta \over r^2\sin^2\theta}{d^2 \Phi \over d \varphi^2} = 0\,\!$

$\begin{cases} \dfrac{1}{R}\dfrac{d}{dr}\left(r^2\dfrac{dR}{dr}\right) = \lambda \\ \dfrac{1}{\Phi} \dfrac{d^2 \Phi}{d\varphi^2} = -m^2 \\ \lambda + \dfrac{1}{\Theta\sin ^2\theta} \dfrac{d}{d\theta} \left ( \sin\theta \dfrac{d\Theta}{d\theta} \right ) -\dfrac{m^2}{\sin ^2\theta} = 0 \end{cases}$ ，整理得$\begin{cases} r^2R''+2rR'-\lambda R = 0\\ \Phi''+m^2\Phi = 0 \\ \sin\theta \dfrac{d}{d\theta} (\sin\theta \Theta') +(\lambda\sin ^2\theta-m^2)\Theta = 0 \end{cases}$

### 本征方程的求解

$\Phi = e^{im\phi}$$m\in\mathbb{Z}$

$\Theta = P_\ell^m(\cos\theta)$$l\in\mathbb{N},l\geqslant |m|$

$Y_\ell^m (\theta, \varphi ) = N \Phi(\varphi)\Theta(\theta) = N \, e^{i m \varphi } \, P_\ell^m (\cos{\theta} )\,\!$$l\in\mathbb{N},m=0,\pm1,\pm2,\ldots\pm l$

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

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

$P_\ell(x)\,\!$$l$勒讓德多項式，可用羅德里格公式表示為：

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

## 前几阶球谐函数

$l$ $m$ $\Phi(\varphi)$ $\Theta(\theta)$ 極坐標中的表達式 直角坐標中的表達式 量子力學中的記号
0 0 $\frac{1}{\sqrt{2\pi}}$ $\frac{1}{\sqrt{2}}$ $\frac{1}{2\sqrt{\pi}}$ $\frac{1}{2\sqrt{\pi}}$ $\mbox{s} \,$
1 0 $\frac{1}{\sqrt{2\pi}}$ $\sqrt{\frac{3}{2}}\cos\theta$ $\frac{1}{2}\sqrt{\frac{3}{\pi}}\cos\theta$ $\frac{1}{2}\sqrt{\frac{3}{\pi}}\frac{z}{r}$ $\mbox{p}_{z} \,$
1 +1 $\frac{1}{\sqrt{2\pi}}\exp(i\varphi)$ $\frac{\sqrt{3}}{2}\sin\theta$ $\Bigg\{$ $\frac{1}{2}\sqrt{\frac{3}{\pi}}\sin\theta\cos\varphi$ $\frac{1}{2}\sqrt{\frac{3}{\pi}}\frac{x}{r}$ $\mbox{p}_{x} \,$
1 -1 $\frac{1}{\sqrt{2\pi}}\exp(-i\varphi)$ $\frac{\sqrt{3}}{2}\sin\theta$ $\frac{1}{2}\sqrt{\frac{3}{\pi}}\sin\theta\sin\varphi$ $\frac{1}{2}\sqrt{\frac{3}{\pi}}\frac{y}{r}$ $\mbox{p}_{y} \,$
2 0 $\frac{1}{\sqrt{2\pi}}$ $\frac{1}{2}\sqrt{\frac{5}{2}}(3\cos^2\theta-1)$ $\frac{1}{4}\sqrt{\frac{5}{\pi}}(3\cos^2\theta-1)$ $\frac{1}{4}\sqrt{\frac{5}{\pi}}\frac{2z^2-x^2-y^2}{r^2}$ $\mbox{d}_{3z^2-r^2}$
2 +1 $\frac{1}{\sqrt{2\pi}}\exp(i\varphi)$ $\frac{\sqrt{15}}{2}\sin\theta\cos\theta$ $\Bigg\{$ $\frac{1}{2}\sqrt{\frac{15}{\pi}}\sin\theta\cos\theta\cos\varphi$ $\frac{1}{2}\sqrt{\frac{15}{\pi}}\frac{zx}{r^2}$ $\mbox{d}_{zx} \,$
2 -1 $\frac{1}{\sqrt{2\pi}}\exp(-i\varphi)$ $\frac{\sqrt{15}}{2}\sin\theta\cos\theta$ $\frac{1}{2}\sqrt{\frac{15}{\pi}}\sin\theta\cos\theta\sin\varphi$ $\frac{1}{2}\sqrt{\frac{15}{\pi}}\frac{yz}{r^2}$ $\mbox{d}_{yz} \,$
2 +2 $\frac{1}{\sqrt{2\pi}}\exp(2i\varphi)$ $\frac{\sqrt{15}}{4}\sin^2\theta$ $\Bigg\{$ $\frac{1}{4}\sqrt{\frac{15}{\pi}}\sin^2\theta\cos2\varphi$ $\frac{1}{4}\sqrt{\frac{15}{\pi}}\frac{x^2-y^2}{r^2}$ $\mbox{d}_{x^2-y^2}$
2 -2 $\frac{1}{\sqrt{2\pi}}\exp(-2i\varphi)$ $\frac{\sqrt{15}}{4}\sin^2\theta$ $\frac{1}{4}\sqrt{\frac{15}{\pi}}\sin^2\theta\sin2\varphi$ $\frac{1}{2}\sqrt{\frac{15}{\pi}}\frac{xy}{r^2}$ $\mbox{d}_{xy} \,$
3 0 $\frac{1}{\sqrt{2\pi}}$ $\frac{1}{2}\sqrt{\frac{7}{2}}(5\cos^3\theta-3\cos\theta)$ $\frac{1}{4}\sqrt{\frac{7}{\pi}}(5\cos^3\theta-3\cos\theta)$ $\frac{1}{4}\sqrt{\frac{7}{\pi}}\frac{z(2z^2-3x^2-3y^2)}{r^3}$ $\mbox{f}_{z(5z^2-3r^2)}$
3 +1 $\frac{1}{\sqrt{2\pi}}\exp(i\varphi)$ $\frac{1}{4}\sqrt{\frac{21}{2}}(5\cos^2\theta-1)\sin\theta$ $\Bigg\{$ $\frac{1}{4}\sqrt{\frac{21}{2\pi}}(5\cos^2\theta-1)\sin\theta\cos\varphi$ $\frac{1}{4}\sqrt{\frac{21}{2\pi}}\frac{x(5z^2-r^2)}{r^3}$ $\mbox{f}_{x(5z^2-r^2)}$
3 -1 $\frac{1}{\sqrt{2\pi}}\exp(-i\varphi)$ $\frac{1}{4}\sqrt{\frac{21}{2}}(5\cos^2\theta-1)\sin\theta$ $\frac{1}{4}\sqrt{\frac{21}{2\pi}}(5\cos^2\theta-1)\sin\theta\sin\varphi$ $\frac{1}{4}\sqrt{\frac{21}{2\pi}}\frac{y(5z^2-r^2)}{r^3}$ $\mbox{f}_{y(5z^2-r^2)}$
3 +2 $\frac{1}{\sqrt{2\pi}}\exp(2i\varphi)$ $\frac{\sqrt{105}}{4}\cos\theta\sin^2\theta$ $\Bigg\{$ $\frac{1}{4}\sqrt{\frac{105}{\pi}}\cos\theta\sin^2\theta\cos2\varphi$ $\frac{1}{4}\sqrt{\frac{105}{\pi}}\frac{z(x^2-y^2)}{r^3}$ $\mbox{f}_{z(x^2-y^2)}$
3 -2 $\frac{1}{\sqrt{2\pi}}\exp(-2i\varphi)$ $\frac{\sqrt{105}}{4}\cos\theta\sin^2\theta$ $\frac{1}{4}\sqrt{\frac{105}{\pi}}\cos\theta\sin^2\theta\sin2\varphi$ $\frac{1}{2}\sqrt{\frac{105}{\pi}}\frac{xyz}{r^3}$ $\mbox{f}_{xyz} \,$
3 +3 $\frac{1}{\sqrt{2\pi}}\exp(3i\varphi)$ $\frac{1}{4}\sqrt{\frac{35}{2}}\sin^3\theta$ $\Bigg\{$ $\frac{1}{4}\sqrt{\frac{35}{2\pi}}\sin^3\theta\cos3\varphi$ $\frac{1}{4}\sqrt{\frac{35}{2\pi}}\frac{x(x^2-3y^2)}{r^3}$ $\mbox{f}_{x(x^2-3y^2)}$
3 -3 $\frac{1}{\sqrt{2\pi}}\exp(-3i\varphi)$ $\frac{1}{4}\sqrt{\frac{35}{2}}\sin^3\theta$ $\frac{1}{4}\sqrt{\frac{35}{2\pi}}\sin^3\theta\sin3\varphi$ $\frac{1}{4}\sqrt{\frac{35}{2\pi}}\frac{y(3x^2-y^2)}{r^3}$ $\mbox{f}_{y(3x^2-y^2)}$

$l=0$

$Y_{0}^{0}(\theta,\varphi)={1\over 2}\sqrt{1\over \pi}$

$l=1$

$Y_{1}^{-1}(\theta,\varphi)={1\over 2}\sqrt{3\over 2\pi} \, \sin\theta \, e^{-i\varphi}$
$Y_{1}^{0}(\theta,\varphi)={1\over 2}\sqrt{3\over \pi}\, \cos\theta$
$Y_{1}^{1}(\theta,\varphi)={-1\over 2}\sqrt{3\over 2\pi}\, \sin\theta\, e^{i\varphi}$

$l=2$

$Y_{2}^{-2}(\theta,\varphi)={1\over 4}\sqrt{15\over 2\pi} \, \sin^{2}\theta \, e^{-2i\varphi}$
$Y_{2}^{-1}(\theta,\varphi)={1\over 2}\sqrt{15\over 2\pi}\, \sin\theta\, \cos\theta\, e^{-i\varphi}$
$Y_{2}^{0}(\theta,\varphi)={1\over 4}\sqrt{5\over \pi}\, (3\cos^{2}\theta-1)$
$Y_{2}^{1}(\theta,\varphi)={-1\over 2}\sqrt{15\over 2\pi}\, \sin\theta\,\cos\theta\, e^{i\varphi}$
$Y_{2}^{2}(\theta,\varphi)={1\over 4}\sqrt{15\over 2\pi}\, \sin^{2}\theta \, e^{2i\varphi}$