球諧函數

球諧函數是拉普拉斯方程式的球坐標系形式解的角度部分。在經典場論、量子力學等領域廣泛應用。

函數的推導

本微分方程式的推導

球坐標下的拉普拉斯方程式:

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

實值的球諧函數 Y_lm，l = 0 到 4 （由上至下），m=0 到 4（由左至右）。負數階球諧函數 Y_l,-m 可由正數階函數對 z-軸轉 90/m 度得到。

利用分離變量法，設定 $f(r,\ \theta ,\ \varphi )=R(r)Y(\theta ,\ \varphi )=R(r)\Theta (\theta )\Phi (\varphi )$ 。其中$Y(\theta ,\ \varphi )$代表角度部分的解，也就是球諧函數。

代入拉普拉斯方程式，得到：

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

分離變量後得：

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

本徵方程式的求解

這裏，$\Phi$是一個以$2\pi$為週期的函數，即滿足週期性邊界條件$\Phi (\varphi )=\Phi (\varphi +2\pi )$，因此$m$必須為整數。而且可以解出：

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

而對於$\Theta$的方程式，進行變量替換 $t=\cos \theta$，$dt=-\sin \theta d\theta$，$|t|\leqslant 1$，得到關於$t$的伴隨勒讓德方程式。方程式的解應滿足在$[-1,1]$區間上取有限值，此時必須有$\lambda =l(l+1)$，其中$l$為自然數，且$l\geqslant |m|$。對應方程式的解為$P_{\ell }^{m}(t)$。即可以解出：

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

故球諧函數可以表達為：

$Y_{\ell }^{m}(\theta ,\varphi )=N\Phi (\varphi )\Theta (\theta )=N\,e^{im\varphi }\,P_{\ell }^{m}(\cos {\theta })\,\!$ ，$l\in \mathbb {N} ,m=0,\pm 1,\pm 2,\ldots \pm l$；

其中N 是歸一化因子。

經過歸一化後，球諧函數表達為：

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

這裏的 $Y_{\ell }^{m}\,\!$ 稱為 $\ell \,\!$ 和 $m\,\!$ 的球諧函數。以上推導過程中，$i\,\!$ 是虛數單位， $P_{\ell }^{m}\,\!$ 是伴隨勒讓德多項式 。

其中$P_{\ell }^{m}(x)\,\!$ 用方程式定義為：

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

而 $P_{\ell }(x)\,\!$ 是 $l$ 階勒讓德多項式，可用羅德里格公式表示為：

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

前幾階球諧函數

主條目：球諧函數表

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

$l=0$

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

$l=1$

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

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

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

$l=2$

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

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

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

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

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

參見

• 勒讓德多項式
• 伴隨勒讓德多項式
• 施圖姆-劉維爾理論
• 柱諧函數
• 向量球諧函數