球谐函数

球谐函数是拉普拉斯方程的球坐标系形式解的角度部分。在古典场论、量子力学等领域广泛应用。

函数的推导

本微分方程的推导

球坐标下的拉普拉斯方程序:

\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\,\!

。

实值的球谐函数 Y_lm，l = 0 到 4 （由上至下），m=0 到 4（由左至右）。负数阶球谐函数 Y_l,-m 可由正数阶函数对 z-轴转 90/m 度得到。

利用分离变量法，设定 $f(r,\ \theta ,\ \varphi )=R(r)Y(\theta ,\ \varphi )=R(r)\Theta (\theta )\Phi (\varphi )$ 。其中 $Y(\theta ,\ \varphi )$ 代表角度部分的解，也就是球谐函数。

代入拉普拉斯方程，得到：

{\Theta \Phi  \over r^{2}}{d \over dr}\left(r^{2}{dR \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\,\!

分离变量后得：

{\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 \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^{2}R''+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

这里， $\Phi$ 是一个以 $2\pi$ 为周期的函数，即满足周期性边界条件 $\Phi (\varphi )=\Phi (\varphi +2\pi )$ ，因此 $m$ 必须为整数。而且可以解出：

\Phi =e^{im\phi }

，

m\in \mathbb {Z}

而对于 $\Theta$ 的方程，进行变量替换 $t=\cos \theta$ ， $dt=-\sin \theta d\theta$ ， $|t|\leqslant 1$ ，得到关于 $t$ 的伴随勒让德方程。方程的解应满足在 $[-1,1]$ 区间上取有限值，此时必须有 $\lambda =l(l+1)$ ，其中 $l$ 为自然数，且 $l\geqslant |m|$ 。对应方程的解为 $P_{\ell }^{m}(t)$ 。即可以解出：

\Theta =P_{\ell }^{m}(\cos \theta )

，

l\in \mathbb {N} ,l\geqslant |m|

故球谐函数可以表达为：

Y_{\ell }^{m}(\theta ,\varphi )=N\Phi (\varphi )\Theta (\theta )=N\,e^{im\varphi }\,P_{\ell }^{m}(\cos {\theta })\,\!

，

l\in \mathbb {N} ,m=0,\pm 1,\pm 2,\ldots \pm l

；

其中N 是归一化因子。

经过归一化后，球谐函数表达为：

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

；

这里的 $Y_{\ell }^{m}\,\!$ 称为 $\ell \,\!$ 和 $m\,\!$ 的球谐函数。以上推导过程中， $i\,\!$ 是虚数单位， $P_{\ell }^{m}\,\!$ 是伴随勒让德多项式。

其中 $P_{\ell }^{m}(x)\,\!$ 用方程序定义为：

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

；

而 $P_{\ell }(x)\,\!$ 是 $l$ 阶勒让德多项式，可用罗德里格公式表示为：

{\displaystyle P_{\ell }(x)={1 \over 2^{\ell }\ell

。

前几阶球谐函数


$l$	$m$	$\Phi (\varphi )$	$\Theta (\theta )$		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	${\frac {1}{\sqrt {2\pi }}}$	${\frac {1}{\sqrt {2}}}$		${\frac {1}{2{\sqrt {\pi }}}}$	${\frac {1}{2{\sqrt {\pi }}}}$	${\mbox{s}}\,$
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	${\frac {1}{\sqrt {2\pi }}}\exp(-i\varphi )$	${\frac {\sqrt {3}}{2}}\sin \theta$	${\Bigg \{}$	${\frac {1}{2}}{\sqrt {\frac {3}{\pi }}}\sin \theta \sin \varphi$	${\frac {1}{2}}{\sqrt {\frac {3}{\pi }}}{\frac {y}{r}}$	${\mbox{p}}_{y}\,$
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	${\frac {1}{\sqrt {2\pi }}}\exp(-i\varphi )$	${\frac {\sqrt {15}}{2}}\sin \theta \cos \theta$	${\Bigg \{}$	${\frac {1}{2}}{\sqrt {\frac {15}{\pi }}}\sin \theta \cos \theta \sin \varphi$	${\frac {1}{2}}{\sqrt {\frac {15}{\pi }}}{\frac {yz}{r^{2}}}$	${\mbox{d}}_{yz}\,$
2	+2	${\frac {1}{\sqrt {2\pi }}}\exp(2i\varphi )$	${\frac {\sqrt {15}}{4}}\sin ^{2}\theta$	${\Bigg \{}$	${\frac {1}{4}}{\sqrt {\frac {15}{\pi }}}\sin ^{2}\theta \cos 2\varphi$	${\frac {1}{4}}{\sqrt {\frac {15}{\pi }}}{\frac {x^{2}-y^{2}}{r^{2}}}$	${\mbox{d}}_{x^{2}-y^{2}}$
2	-2	${\frac {1}{\sqrt {2\pi }}}\exp(-2i\varphi )$	${\frac {\sqrt {15}}{4}}\sin ^{2}\theta$	${\Bigg \{}$	${\frac {1}{4}}{\sqrt {\frac {15}{\pi }}}\sin ^{2}\theta \sin 2\varphi$	${\frac {1}{2}}{\sqrt {\frac {15}{\pi }}}{\frac {xy}{r^{2}}}$	${\mbox{d}}_{xy}\,$
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	${\frac {1}{\sqrt {2\pi }}}\exp(-i\varphi )$	${\frac {1}{4}}{\sqrt {\frac {21}{2}}}(5\cos ^{2}\theta -1)\sin \theta$	${\Bigg \{}$	${\frac {1}{4}}{\sqrt {\frac {21}{2\pi }}}(5\cos ^{2}\theta -1)\sin \theta \sin \varphi$	${\frac {1}{4}}{\sqrt {\frac {21}{2\pi }}}{\frac {y(5z^{2}-r^{2})}{r^{3}}}$	${\mbox{f}}_{y(5z^{2}-r^{2})}$
3	+2	${\frac {1}{\sqrt {2\pi }}}\exp(2i\varphi )$	${\frac {\sqrt {105}}{4}}\cos \theta \sin ^{2}\theta$	${\Bigg \{}$	${\frac {1}{4}}{\sqrt {\frac {105}{\pi }}}\cos \theta \sin ^{2}\theta \cos 2\varphi$	${\frac {1}{4}}{\sqrt {\frac {105}{\pi }}}{\frac {z(x^{2}-y^{2})}{r^{3}}}$	${\mbox{f}}_{z(x^{2}-y^{2})}$
3	-2	${\frac {1}{\sqrt {2\pi }}}\exp(-2i\varphi )$	${\frac {\sqrt {105}}{4}}\cos \theta \sin ^{2}\theta$	${\Bigg \{}$	${\frac {1}{4}}{\sqrt {\frac {105}{\pi }}}\cos \theta \sin ^{2}\theta \sin 2\varphi$	${\frac {1}{2}}{\sqrt {\frac {105}{\pi }}}{\frac {xyz}{r^{3}}}$	${\mbox{f}}_{xyz}\,$
3	+3	${\frac {1}{\sqrt {2\pi }}}\exp(3i\varphi )$	${\frac {1}{4}}{\sqrt {\frac {35}{2}}}\sin ^{3}\theta$	${\Bigg \{}$	${\frac {1}{4}}{\sqrt {\frac {35}{2\pi }}}\sin ^{3}\theta \cos 3\varphi$	${\frac {1}{4}}{\sqrt {\frac {35}{2\pi }}}{\frac {x(x^{2}-3y^{2})}{r^{3}}}$	${\mbox{f}}_{x(x^{2}-3y^{2})}$
3	-3	${\frac {1}{\sqrt {2\pi }}}\exp(-3i\varphi )$	${\frac {1}{4}}{\sqrt {\frac {35}{2}}}\sin ^{3}\theta$	${\Bigg \{}$	${\frac {1}{4}}{\sqrt {\frac {35}{2\pi }}}\sin ^{3}\theta \sin 3\varphi$	${\frac {1}{4}}{\sqrt {\frac {35}{2\pi }}}{\frac {y(3x^{2}-y^{2})}{r^{3}}}$	${\mbox{f}}_{y(3x^{2}-y^{2})}$

$l=0$

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 }

参见

This article is issued from Wikipedia. The text is licensed under Creative Commons - Attribution - Sharealike. Additional terms may apply for the media files.