# 球谐函数

## 函数的推导[编辑]

### 本微分方程的推导[编辑]

${\nabla }^{2}f=\frac{1}{{r}^{2}}\frac{\partial }{\partial r}\left({r}^{2}\frac{\partial f}{\partial r}\right)+\frac{1}{{r}^{2}\mathrm{sin}\theta }\frac{\partial }{\partial \theta }\left(\mathrm{sin}\theta \frac{\partial f}{\partial \theta }\right)+\frac{1}{{r}^{2}{\mathrm{sin}}^{2}\theta }\frac{{\partial }^{2}f}{\partial {\phi }^{2}}=0$

$\frac{\Theta \Phi }{{r}^{2}}\frac{d}{dr}\left({r}^{2}\frac{dR}{dr}\right)+\frac{R\Phi }{{r}^{2}\mathrm{sin}\theta }\frac{d}{d\theta }\left(\mathrm{sin}\theta \frac{d\Theta }{d\theta }\right)+\frac{R\Theta }{{r}^{2}{\mathrm{sin}}^{2}\theta }\frac{{d}^{2}\Phi }{d{\phi }^{2}}=0$

$\left\{\begin{array}{c}\frac{1}{R}\frac{d}{dr}\left({r}^{2}\frac{dR}{dr}\right)=\lambda \\ \frac{1}{\Phi }\frac{{d}^{2}\Phi }{d{\phi }^{2}}=-{m}^{2}\\ \lambda +\frac{1}{\Theta \mathrm{sin}\theta }\frac{d}{d\theta }\left(\mathrm{sin}\theta \frac{d\Theta }{d\theta }\right)-\frac{{m}^{2}}{{\mathrm{sin}}^{2}\theta }=0\end{array}$
，整理得$\left\{\begin{array}{c}{r}^{2}{R}^{″}+2r{R}^{\prime }-\lambda R=0\\ {\Phi }^{″}+{m}^{2}\Phi =0\\ \mathrm{sin}\theta \frac{d}{d\theta }\left(\mathrm{sin}\theta {\Theta }^{\prime }\right)+\left(\lambda {\mathrm{sin}}^{2}\theta -{m}^{2}\right)\Theta =0\end{array}$

### 本征方程的求解[编辑]

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

$\Theta ={P}_{\ell }^{m}\left(\mathrm{cos}\theta \right)$$l\in N,l⩾|m|$

${Y}_{\ell }^{m}\left(\theta ,\phi \right)=N\Phi \left(\phi \right)\Theta \left(\theta \right)=N{e}^{im\phi }{P}_{\ell }^{m}\left(\mathrm{cos}\theta \right)$$l\in N,m=0,±1,±2,\dots ±l$

${Y}_{\ell }^{m}\left(\theta ,\phi \right)=\left(-1{\right)}^{m}\sqrt{\frac{\left(2\ell +1\right)}{4\pi }\frac{\left(\ell -|m|\right)!}{\left(\ell +|m|\right)!}}{P}_{\ell }^{m}\left(\mathrm{cos}\theta \right){e}^{im\phi }$

${P}_{\ell }^{m}\left(x\right)=\left(1-{x}^{2}{\right)}^{|m|/2}\frac{{d}^{|m|}}{d{x}^{|m|}}{P}_{\ell }\left(x\right)$

${P}_{\ell }\left(x\right)$$l$勒让德多项式，可用罗德里格公式表示为：

${P}_{\ell }\left(x\right)=\frac{1}{{2}^{\ell }\ell !}\frac{{d}^{\ell }}{d{x}^{\ell }}\left({x}^{2}-1{\right)}^{l}$

## 前几阶球谐函数[编辑]

$l$ $m$ $\Phi \left(\phi \right)$ $\Theta \left(\theta \right)$ 极坐标中的表达式 直角坐标中的表达式 量子力学中的记号
0 0 $\frac{1}{\sqrt{2\pi }}$ $\frac{1}{\sqrt{2}}$ $\frac{1}{2\sqrt{\pi }}$ $\frac{1}{2\sqrt{\pi }}$ $\text{s}$
1 0 $\frac{1}{\sqrt{2\pi }}$ $\sqrt{\frac{3}{2}}\mathrm{cos}\theta$ $\frac{1}{2}\sqrt{\frac{3}{\pi }}\mathrm{cos}\theta$ $\frac{1}{2}\sqrt{\frac{3}{\pi }}\frac{z}{r}$ ${\text{p}}_{z}$
1 +1 $\frac{1}{\sqrt{2\pi }}\mathrm{exp}\left(i\phi \right)$ $\frac{\sqrt{3}}{2}\mathrm{sin}\theta$ $\left\{$ $\frac{1}{2}\sqrt{\frac{3}{\pi }}\mathrm{sin}\theta \mathrm{cos}\phi$ $\frac{1}{2}\sqrt{\frac{3}{\pi }}\frac{x}{r}$ ${\text{p}}_{x}$
1 -1 $\frac{1}{\sqrt{2\pi }}\mathrm{exp}\left(-i\phi \right)$ $\frac{\sqrt{3}}{2}\mathrm{sin}\theta$ $\frac{1}{2}\sqrt{\frac{3}{\pi }}\mathrm{sin}\theta \mathrm{sin}\phi$ $\frac{1}{2}\sqrt{\frac{3}{\pi }}\frac{y}{r}$ ${\text{p}}_{y}$
2 0 $\frac{1}{\sqrt{2\pi }}$ $\frac{1}{2}\sqrt{\frac{5}{2}}\left(3{\mathrm{cos}}^{2}\theta -1\right)$ $\frac{1}{4}\sqrt{\frac{5}{\pi }}\left(3{\mathrm{cos}}^{2}\theta -1\right)$ $\frac{1}{4}\sqrt{\frac{5}{\pi }}\frac{2{z}^{2}-{x}^{2}-{y}^{2}}{{r}^{2}}$ ${\text{d}}_{3{z}^{2}-{r}^{2}}$
2 +1 $\frac{1}{\sqrt{2\pi }}\mathrm{exp}\left(i\phi \right)$ $\frac{\sqrt{15}}{2}\mathrm{sin}\theta \mathrm{cos}\theta$ $\left\{$ $\frac{1}{2}\sqrt{\frac{15}{\pi }}\mathrm{sin}\theta \mathrm{cos}\theta \mathrm{cos}\phi$ $\frac{1}{2}\sqrt{\frac{15}{\pi }}\frac{zx}{{r}^{2}}$ ${\text{d}}_{zx}$
2 -1 $\frac{1}{\sqrt{2\pi }}\mathrm{exp}\left(-i\phi \right)$ $\frac{\sqrt{15}}{2}\mathrm{sin}\theta \mathrm{cos}\theta$ $\frac{1}{2}\sqrt{\frac{15}{\pi }}\mathrm{sin}\theta \mathrm{cos}\theta \mathrm{sin}\phi$ $\frac{1}{2}\sqrt{\frac{15}{\pi }}\frac{yz}{{r}^{2}}$ ${\text{d}}_{yz}$
2 +2 $\frac{1}{\sqrt{2\pi }}\mathrm{exp}\left(2i\phi \right)$ $\frac{\sqrt{15}}{4}{\mathrm{sin}}^{2}\theta$ $\left\{$ $\frac{1}{4}\sqrt{\frac{15}{\pi }}{\mathrm{sin}}^{2}\theta \mathrm{cos}2\phi$ $\frac{1}{4}\sqrt{\frac{15}{\pi }}\frac{{x}^{2}-{y}^{2}}{{r}^{2}}$ ${\text{d}}_{{x}^{2}-{y}^{2}}$
2 -2 $\frac{1}{\sqrt{2\pi }}\mathrm{exp}\left(-2i\phi \right)$ $\frac{\sqrt{15}}{4}{\mathrm{sin}}^{2}\theta$ $\frac{1}{4}\sqrt{\frac{15}{\pi }}{\mathrm{sin}}^{2}\theta \mathrm{sin}2\phi$ $\frac{1}{2}\sqrt{\frac{15}{\pi }}\frac{xy}{{r}^{2}}$ ${\text{d}}_{xy}$
3 0 $\frac{1}{\sqrt{2\pi }}$ $\frac{1}{2}\sqrt{\frac{7}{2}}\left(5{\mathrm{cos}}^{3}\theta -3\mathrm{cos}\theta \right)$ $\frac{1}{4}\sqrt{\frac{7}{\pi }}\left(5{\mathrm{cos}}^{3}\theta -3\mathrm{cos}\theta \right)$ $\frac{1}{4}\sqrt{\frac{7}{\pi }}\frac{z\left(2{z}^{2}-3{x}^{2}-3{y}^{2}\right)}{{r}^{3}}$ ${\text{f}}_{z\left(5{z}^{2}-3{r}^{2}\right)}$
3 +1 $\frac{1}{\sqrt{2\pi }}\mathrm{exp}\left(i\phi \right)$ $\frac{1}{4}\sqrt{\frac{21}{2}}\left(5{\mathrm{cos}}^{2}\theta -1\right)\mathrm{sin}\theta$ $\left\{$ $\frac{1}{4}\sqrt{\frac{21}{2\pi }}\left(5{\mathrm{cos}}^{2}\theta -1\right)\mathrm{sin}\theta \mathrm{cos}\phi$ $\frac{1}{4}\sqrt{\frac{21}{2\pi }}\frac{x\left(5{z}^{2}-{r}^{2}\right)}{{r}^{3}}$ ${\text{f}}_{x\left(5{z}^{2}-{r}^{2}\right)}$
3 -1 $\frac{1}{\sqrt{2\pi }}\mathrm{exp}\left(-i\phi \right)$ $\frac{1}{4}\sqrt{\frac{21}{2}}\left(5{\mathrm{cos}}^{2}\theta -1\right)\mathrm{sin}\theta$ $\frac{1}{4}\sqrt{\frac{21}{2\pi }}\left(5{\mathrm{cos}}^{2}\theta -1\right)\mathrm{sin}\theta \mathrm{sin}\phi$ $\frac{1}{4}\sqrt{\frac{21}{2\pi }}\frac{y\left(5{z}^{2}-{r}^{2}\right)}{{r}^{3}}$ ${\text{f}}_{y\left(5{z}^{2}-{r}^{2}\right)}$
3 +2 $\frac{1}{\sqrt{2\pi }}\mathrm{exp}\left(2i\phi \right)$ $\frac{\sqrt{105}}{4}\mathrm{cos}\theta {\mathrm{sin}}^{2}\theta$ $\left\{$ $\frac{1}{4}\sqrt{\frac{105}{\pi }}\mathrm{cos}\theta {\mathrm{sin}}^{2}\theta \mathrm{cos}2\phi$ $\frac{1}{4}\sqrt{\frac{105}{\pi }}\frac{z\left({x}^{2}-{y}^{2}\right)}{{r}^{3}}$ ${\text{f}}_{z\left({x}^{2}-{y}^{2}\right)}$
3 -2 $\frac{1}{\sqrt{2\pi }}\mathrm{exp}\left(-2i\phi \right)$ $\frac{\sqrt{105}}{4}\mathrm{cos}\theta {\mathrm{sin}}^{2}\theta$ $\frac{1}{4}\sqrt{\frac{105}{\pi }}\mathrm{cos}\theta {\mathrm{sin}}^{2}\theta \mathrm{sin}2\phi$ $\frac{1}{2}\sqrt{\frac{105}{\pi }}\frac{xyz}{{r}^{3}}$ ${\text{f}}_{xyz}$
3 +3 $\frac{1}{\sqrt{2\pi }}\mathrm{exp}\left(3i\phi \right)$ $\frac{1}{4}\sqrt{\frac{35}{2}}{\mathrm{sin}}^{3}\theta$ $\left\{$ $\frac{1}{4}\sqrt{\frac{35}{2\pi }}{\mathrm{sin}}^{3}\theta \mathrm{cos}3\phi$ $\frac{1}{4}\sqrt{\frac{35}{2\pi }}\frac{x\left({x}^{2}-3{y}^{2}\right)}{{r}^{3}}$ ${\text{f}}_{x\left({x}^{2}-3{y}^{2}\right)}$
3 -3 $\frac{1}{\sqrt{2\pi }}\mathrm{exp}\left(-3i\phi \right)$ $\frac{1}{4}\sqrt{\frac{35}{2}}{\mathrm{sin}}^{3}\theta$ $\frac{1}{4}\sqrt{\frac{35}{2\pi }}{\mathrm{sin}}^{3}\theta \mathrm{sin}3\phi$ $\frac{1}{4}\sqrt{\frac{35}{2\pi }}\frac{y\left(3{x}^{2}-{y}^{2}\right)}{{r}^{3}}$ ${\text{f}}_{y\left(3{x}^{2}-{y}^{2}\right)}$

$l=0$

${Y}_{0}^{0}\left(\theta ,\phi \right)=\frac{1}{2}\sqrt{\frac{1}{\pi }}$

$l=1$

${Y}_{1}^{-1}\left(\theta ,\phi \right)=\frac{1}{2}\sqrt{\frac{3}{2\pi }}\mathrm{sin}\theta {e}^{-i\phi }$
${Y}_{1}^{0}\left(\theta ,\phi \right)=\frac{1}{2}\sqrt{\frac{3}{\pi }}\mathrm{cos}\theta$
${Y}_{1}^{1}\left(\theta ,\phi \right)=\frac{-1}{2}\sqrt{\frac{3}{2\pi }}\mathrm{sin}\theta {e}^{i\phi }$

$l=2$

${Y}_{2}^{-2}\left(\theta ,\phi \right)=\frac{1}{4}\sqrt{\frac{15}{2\pi }}{\mathrm{sin}}^{2}\theta {e}^{-2i\phi }$
${Y}_{2}^{-1}\left(\theta ,\phi \right)=\frac{1}{2}\sqrt{\frac{15}{2\pi }}\mathrm{sin}\theta \mathrm{cos}\theta {e}^{-i\phi }$
${Y}_{2}^{0}\left(\theta ,\phi \right)=\frac{1}{4}\sqrt{\frac{5}{\pi }}\left(3{\mathrm{cos}}^{2}\theta -1\right)$
${Y}_{2}^{1}\left(\theta ,\phi \right)=\frac{-1}{2}\sqrt{\frac{15}{2\pi }}\mathrm{sin}\theta \mathrm{cos}\theta {e}^{i\phi }$
${Y}_{2}^{2}\left(\theta ,\phi \right)=\frac{1}{4}\sqrt{\frac{15}{2\pi }}{\mathrm{sin}}^{2}\theta {e}^{2i\phi }$

## 参见[编辑]

Measure
Measure
Summary | 5 Annotations

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