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。

歷史

皮埃爾-西蒙·拉普拉斯，1749–1827

球諧函數最初是與牛頓萬有引力定律的引力勢相關的，並被應用於三維空間的研究。1782年，皮埃爾-西蒙·德·拉普拉斯在其著作《天體力學》中確定了引力勢${\displaystyle \mathbb {R} ^{3}\to \mathbb {R} }$在點x處，與位於點x_i的一組點質量m_i相關聯的點 x 處，由下式給出

$${\displaystyle V(\mathbf {x} )=\sum _{i}{\frac {m_{i}}{|\mathbf {x} _{i}-\mathbf {x} |}}.}$$

上述求和式中的每一項都是一個點質量的引力勢。在此之前，阿德里安-瑪麗·勒讓德研究了以r = |x|和r₁ = |x₁|的冪展開引力勢。他發現，如果r ≤ r₁則

$${\displaystyle {\frac {1}{|\mathbf {x} _{1}-\mathbf {x} |}}=P_{0}(\cos \gamma ){\frac {1}{r_{1}}}+P_{1}(\cos \gamma ){\frac {r}{r_{1}^{2}}}+P_{2}(\cos \gamma ){\frac {r^{2}}{r_{1}^{3}}}+\cdots }$$

其中γ是向量x和x₁之間的夾角。函數${\displaystyle P_{i}:[-1,1]\to \mathbb {R} }$是勒讓德多項式，它們可以作為球諧函數的特例推導而來。隨後，拉普拉斯在其1782年的回憶錄中，研究了這些係數，他使用球坐標系來表示x₁和x之間的夾角γ 。（參見Legendre polynomials § Applications了解更多詳情。）

1867年，威廉·湯姆森（開爾文勳爵）和彼得·格思里·泰特在其著作《自然哲學論》中引入了固體球諧函數，並首次將這類函數命名為「球諧函數」。固體球諧函數是齊次多項式解${\displaystyle \mathbb {R} ^{3}\to \mathbb {R} }$拉普拉斯方程$${\displaystyle {\frac {\partial ^{2}u}{\partial x^{2}}}+{\frac {\partial ^{2}u}{\partial y^{2}}}+{\frac {\partial ^{2}u}{\partial z^{2}}}=0.}$$通過在球坐標系下考察拉普拉斯方程，湯姆森和泰特恢復了拉普拉斯球諧函數。（參見調和多項式表示。）威廉·惠威爾使用「拉普拉斯係數」一詞來描述由此引入的特定解系，而其他人則將此名稱保留給拉普拉斯和勒讓德正確引入的區域球諧函數。

19 世紀傅里葉級數的發展使得解決矩形域中的各種物理問題成為可能，例如熱方程和波動方程的解。這可以通過將函數展開為三角函數級數來實現。傅里葉級數中的三角函數表示弦振動的基本模式，而球諧函數則以非常相同的方式表示球振動的基本模式。傅里葉級數理論的許多方面可以通過在球諧函數中展開而不是三角函數中展開來推廣。此外，類似於三角函數可以等效地寫成復指數，球諧函數也具有與復值函數等價的形式。這對於具有球對稱性的問題來說是一個福音，例如拉普拉斯和勒讓德最初研究的天體力學問題。

球諧函數在物理學中的廣泛應用，為其在20世紀量子力學誕生中的重要性奠定了基礎。(復值)球諧函數${\displaystyle S^{2}\to \mathbb {C} }$是軌道角動量算符平方的特徵函數$${\displaystyle -i\hbar \mathbf {r} \times \nabla ,}$$因此它們代表了原子軌道的不同量子化配置。

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