球諧函數

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

函數的推導

本微分方程的推導

球坐標下的拉普拉斯方程式:

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

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 ${\displaystyle f(r,\ \theta ,\ \varphi )=R(r)Y(\theta ,\ \varphi )=R(r)\Theta (\theta )\Phi (\varphi )}$ 。其中${\displaystyle Y(\theta ,\ \varphi )}$代表角度部分的解，也就是球諧函數。

代入拉普拉斯方程，得到：

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

分離變量後得：

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

本徵方程的求解

這裡，${\displaystyle \Phi }$是一個以${\displaystyle 2\pi }$為周期的函數，即滿足周期性邊界條件${\displaystyle \Phi (\varphi )=\Phi (\varphi +2\pi )}$，因此${\displaystyle m}$必須為整數。而且可以解出：

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

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

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

故球諧函數可以表達為：

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

其中N 是歸一化因子。

經過歸一化後，球諧函數表達為：

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

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

其中${\displaystyle P_{\ell }^{m}(x)\,\!}$ 用方程式定義為：

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

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

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

前幾階球諧函數

主條目：球諧函數表

${\displaystyle l}$	${\displaystyle m}$	${\displaystyle \Phi (\varphi )}$	${\displaystyle \Theta (\theta )}$		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	${\displaystyle {\frac {1}{\sqrt {2\pi }}}}$	${\displaystyle {\frac {1}{\sqrt {2}}}}$		${\displaystyle {\frac {1}{2{\sqrt {\pi }}}}}$	${\displaystyle {\frac {1}{2{\sqrt {\pi }}}}}$	${\displaystyle {\mbox{s}}\,}$
1	0	${\displaystyle {\frac {1}{\sqrt {2\pi }}}}$	${\displaystyle {\sqrt {\frac {3}{2}}}\cos \theta }$		${\displaystyle {\frac {1}{2}}{\sqrt {\frac {3}{\pi }}}\cos \theta }$	${\displaystyle {\frac {1}{2}}{\sqrt {\frac {3}{\pi }}}{\frac {z}{r}}}$	${\displaystyle {\mbox{p}}_{z}\,}$
1	+1	${\displaystyle {\frac {1}{\sqrt {2\pi }}}\exp(i\varphi )}$	${\displaystyle {\frac {\sqrt {3}}{2}}\sin \theta }$	${\displaystyle {\Bigg \{}}$	${\displaystyle {\frac {1}{2}}{\sqrt {\frac {3}{\pi }}}\sin \theta \cos \varphi }$	${\displaystyle {\frac {1}{2}}{\sqrt {\frac {3}{\pi }}}{\frac {x}{r}}}$	${\displaystyle {\mbox{p}}_{x}\,}$
1	-1	${\displaystyle {\frac {1}{\sqrt {2\pi }}}\exp(-i\varphi )}$	${\displaystyle {\frac {\sqrt {3}}{2}}\sin \theta }$		${\displaystyle {\frac {1}{2}}{\sqrt {\frac {3}{\pi }}}\sin \theta \sin \varphi }$	${\displaystyle {\frac {1}{2}}{\sqrt {\frac {3}{\pi }}}{\frac {y}{r}}}$	${\displaystyle {\mbox{p}}_{y}\,}$
2	0	${\displaystyle {\frac {1}{\sqrt {2\pi }}}}$	${\displaystyle {\frac {1}{2}}{\sqrt {\frac {5}{2}}}(3\cos ^{2}\theta -1)}$		${\displaystyle {\frac {1}{4}}{\sqrt {\frac {5}{\pi }}}(3\cos ^{2}\theta -1)}$	${\displaystyle {\frac {1}{4}}{\sqrt {\frac {5}{\pi }}}{\frac {2z^{2}-x^{2}-y^{2}}{r^{2}}}}$	${\displaystyle {\mbox{d}}_{3z^{2}-r^{2}}}$
2	+1	${\displaystyle {\frac {1}{\sqrt {2\pi }}}\exp(i\varphi )}$	${\displaystyle {\frac {\sqrt {15}}{2}}\sin \theta \cos \theta }$	${\displaystyle {\Bigg \{}}$	${\displaystyle {\frac {1}{2}}{\sqrt {\frac {15}{\pi }}}\sin \theta \cos \theta \cos \varphi }$	${\displaystyle {\frac {1}{2}}{\sqrt {\frac {15}{\pi }}}{\frac {zx}{r^{2}}}}$	${\displaystyle {\mbox{d}}_{zx}\,}$
2	-1	${\displaystyle {\frac {1}{\sqrt {2\pi }}}\exp(-i\varphi )}$	${\displaystyle {\frac {\sqrt {15}}{2}}\sin \theta \cos \theta }$		${\displaystyle {\frac {1}{2}}{\sqrt {\frac {15}{\pi }}}\sin \theta \cos \theta \sin \varphi }$	${\displaystyle {\frac {1}{2}}{\sqrt {\frac {15}{\pi }}}{\frac {yz}{r^{2}}}}$	${\displaystyle {\mbox{d}}_{yz}\,}$
2	+2	${\displaystyle {\frac {1}{\sqrt {2\pi }}}\exp(2i\varphi )}$	${\displaystyle {\frac {\sqrt {15}}{4}}\sin ^{2}\theta }$	${\displaystyle {\Bigg \{}}$	${\displaystyle {\frac {1}{4}}{\sqrt {\frac {15}{\pi }}}\sin ^{2}\theta \cos 2\varphi }$	${\displaystyle {\frac {1}{4}}{\sqrt {\frac {15}{\pi }}}{\frac {x^{2}-y^{2}}{r^{2}}}}$	${\displaystyle {\mbox{d}}_{x^{2}-y^{2}}}$
2	-2	${\displaystyle {\frac {1}{\sqrt {2\pi }}}\exp(-2i\varphi )}$	${\displaystyle {\frac {\sqrt {15}}{4}}\sin ^{2}\theta }$		${\displaystyle {\frac {1}{4}}{\sqrt {\frac {15}{\pi }}}\sin ^{2}\theta \sin 2\varphi }$	${\displaystyle {\frac {1}{2}}{\sqrt {\frac {15}{\pi }}}{\frac {xy}{r^{2}}}}$	${\displaystyle {\mbox{d}}_{xy}\,}$
3	0	${\displaystyle {\frac {1}{\sqrt {2\pi }}}}$	${\displaystyle {\frac {1}{2}}{\sqrt {\frac {7}{2}}}(5\cos ^{3}\theta -3\cos \theta )}$		${\displaystyle {\frac {1}{4}}{\sqrt {\frac {7}{\pi }}}(5\cos ^{3}\theta -3\cos \theta )}$	${\displaystyle {\frac {1}{4}}{\sqrt {\frac {7}{\pi }}}{\frac {z(2z^{2}-3x^{2}-3y^{2})}{r^{3}}}}$	${\displaystyle {\mbox{f}}_{z(5z^{2}-3r^{2})}}$
3	+1	${\displaystyle {\frac {1}{\sqrt {2\pi }}}\exp(i\varphi )}$	${\displaystyle {\frac {1}{4}}{\sqrt {\frac {21}{2}}}(5\cos ^{2}\theta -1)\sin \theta }$	${\displaystyle {\Bigg \{}}$	${\displaystyle {\frac {1}{4}}{\sqrt {\frac {21}{2\pi }}}(5\cos ^{2}\theta -1)\sin \theta \cos \varphi }$	${\displaystyle {\frac {1}{4}}{\sqrt {\frac {21}{2\pi }}}{\frac {x(5z^{2}-r^{2})}{r^{3}}}}$	${\displaystyle {\mbox{f}}_{x(5z^{2}-r^{2})}}$
3	-1	${\displaystyle {\frac {1}{\sqrt {2\pi }}}\exp(-i\varphi )}$	${\displaystyle {\frac {1}{4}}{\sqrt {\frac {21}{2}}}(5\cos ^{2}\theta -1)\sin \theta }$		${\displaystyle {\frac {1}{4}}{\sqrt {\frac {21}{2\pi }}}(5\cos ^{2}\theta -1)\sin \theta \sin \varphi }$	${\displaystyle {\frac {1}{4}}{\sqrt {\frac {21}{2\pi }}}{\frac {y(5z^{2}-r^{2})}{r^{3}}}}$	${\displaystyle {\mbox{f}}_{y(5z^{2}-r^{2})}}$
3	+2	${\displaystyle {\frac {1}{\sqrt {2\pi }}}\exp(2i\varphi )}$	${\displaystyle {\frac {\sqrt {105}}{4}}\cos \theta \sin ^{2}\theta }$	${\displaystyle {\Bigg \{}}$	${\displaystyle {\frac {1}{4}}{\sqrt {\frac {105}{\pi }}}\cos \theta \sin ^{2}\theta \cos 2\varphi }$	${\displaystyle {\frac {1}{4}}{\sqrt {\frac {105}{\pi }}}{\frac {z(x^{2}-y^{2})}{r^{3}}}}$	${\displaystyle {\mbox{f}}_{z(x^{2}-y^{2})}}$
3	-2	${\displaystyle {\frac {1}{\sqrt {2\pi }}}\exp(-2i\varphi )}$	${\displaystyle {\frac {\sqrt {105}}{4}}\cos \theta \sin ^{2}\theta }$		${\displaystyle {\frac {1}{4}}{\sqrt {\frac {105}{\pi }}}\cos \theta \sin ^{2}\theta \sin 2\varphi }$	${\displaystyle {\frac {1}{2}}{\sqrt {\frac {105}{\pi }}}{\frac {xyz}{r^{3}}}}$	${\displaystyle {\mbox{f}}_{xyz}\,}$
3	+3	${\displaystyle {\frac {1}{\sqrt {2\pi }}}\exp(3i\varphi )}$	${\displaystyle {\frac {1}{4}}{\sqrt {\frac {35}{2}}}\sin ^{3}\theta }$	${\displaystyle {\Bigg \{}}$	${\displaystyle {\frac {1}{4}}{\sqrt {\frac {35}{2\pi }}}\sin ^{3}\theta \cos 3\varphi }$	${\displaystyle {\frac {1}{4}}{\sqrt {\frac {35}{2\pi }}}{\frac {x(x^{2}-3y^{2})}{r^{3}}}}$	${\displaystyle {\mbox{f}}_{x(x^{2}-3y^{2})}}$
3	-3	${\displaystyle {\frac {1}{\sqrt {2\pi }}}\exp(-3i\varphi )}$	${\displaystyle {\frac {1}{4}}{\sqrt {\frac {35}{2}}}\sin ^{3}\theta }$		${\displaystyle {\frac {1}{4}}{\sqrt {\frac {35}{2\pi }}}\sin ^{3}\theta \sin 3\varphi }$	${\displaystyle {\frac {1}{4}}{\sqrt {\frac {35}{2\pi }}}{\frac {y(3x^{2}-y^{2})}{r^{3}}}}$	${\displaystyle {\mbox{f}}_{y(3x^{2}-y^{2})}}$

${\displaystyle l=0}$

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

${\displaystyle l=1}$

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

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

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

${\displaystyle l=2}$

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

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

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

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

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

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