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。在经典场论、量子力学等领域广泛应用。

历史

皮埃尔-西蒙·拉普拉斯，1749–1827

球谐函数最初是与牛顿万有引力定律的引力势相关的，并被应用于三维空间的研究。1782年，皮埃尔-西蒙·德·拉普拉斯在其著作《天体力学》中确定了引力势${\displaystyle \mathbb {R} ^{3}\to \mathbb {R} }$在点x处，与位于点x_i的一组点质量m_i相关联的点 x 处，由下式给出

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

上述求和式中的每一项都是一个点质量的引力势。在此之前，阿德里安-玛丽·勒让德研究了以r = |x|和r₁ = |x₁|的幂展开引力势。他发现，如果r ≤ r₁则

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

其中γ是矢量x和x₁之间的夹角。函数${\displaystyle P_{i}:[-1,1]\to \mathbb {R} }$是勒让德多项式，它们可以作为球谐函数的特例推导而来。随后，拉普拉斯在其1782年的回忆录中，研究了这些系数，他使用球坐标系来表示x₁和x之间的夹角γ 。（参见Legendre polynomials § Applications了解更多详情。）

1867年，威廉·汤姆孙（开尔文勋爵）和彼得·格思里·泰特在其著作《自然哲学论》中引入了固体球谐函数，并首次将这类函数命名为“球谐函数”。固体球谐函数是齐次多项式解${\displaystyle \mathbb {R} ^{3}\to \mathbb {R} }$拉普拉斯方程$${\displaystyle {\frac {\partial ^{2}u}{\partial x^{2}}}+{\frac {\partial ^{2}u}{\partial y^{2}}}+{\frac {\partial ^{2}u}{\partial z^{2}}}=0.}$$通过在球坐标系下考察拉普拉斯方程，汤姆孙和泰特恢复了拉普拉斯球谐函数。（参见调和多项式表示。）威廉·惠威尔使用“拉普拉斯系数”一词来描述由此引入的特定解系，而其他人则将此名称保留给拉普拉斯和勒让德正确引入的区域球谐函数。

19 世纪傅里叶级数的发展使得解决矩形域中的各种物理问题成为可能，例如热方程和波动方程的解。这可以通过将函数展开为三角函数级数来实现。傅里叶级数中的三角函数表示弦振动的基本模式，而球谐函数则以非常相同的方式表示球振动的基本模式。傅里叶级数理论的许多方面可以通过在球谐函数中展开而不是三角函数中展开来推广。此外，类似于三角函数可以等效地写成复指数，球谐函数也具有与复值函数等价的形式。这对于具有球对称性的问题来说是一个福音，例如拉普拉斯和勒让德最初研究的天体力学问题。

球谐函数在物理学中的广泛应用，为其在20世纪量子力学诞生中的重要性奠定了基础。(复值)球谐函数${\displaystyle S^{2}\to \mathbb {C} }$是轨道角动量算符平方的特征函数$${\displaystyle -i\hbar \mathbf {r} \times \nabla ,}$$因此它们代表了原子轨道的不同量子化配置。

函数的推导

本微分方程的推导

球坐标下的拉普拉斯方程:

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

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

代入拉普拉斯方程，得到：

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

分离变量后得：

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

本征方程的求解

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

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

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

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

故球谐函数可以表达为：

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

其中N 是归一化因子。

经过归一化后，球谐函数表达为：

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

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

其中${\displaystyle P_{\ell }^{m}(x)\,\!}$ 用方程定义为：

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

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

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

前几阶球谐函数

主条目：球谐函数表

${\displaystyle l}$	${\displaystyle m}$	${\displaystyle \Phi (\varphi )}$	${\displaystyle \Theta (\theta )}$		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	${\displaystyle {\frac {1}{\sqrt {2\pi }}}}$	${\displaystyle {\frac {1}{\sqrt {2}}}}$		${\displaystyle {\frac {1}{2{\sqrt {\pi }}}}}$	${\displaystyle {\frac {1}{2{\sqrt {\pi }}}}}$	${\displaystyle {\mbox{s}}\,}$
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	${\displaystyle {\frac {1}{\sqrt {2\pi }}}\exp(-3i\varphi )}$	${\displaystyle {\frac {1}{4}}{\sqrt {\frac {35}{2}}}\sin ^{3}\theta }$		${\displaystyle {\frac {1}{4}}{\sqrt {\frac {35}{2\pi }}}\sin ^{3}\theta \sin 3\varphi }$	${\displaystyle {\frac {1}{4}}{\sqrt {\frac {35}{2\pi }}}{\frac {y(3x^{2}-y^{2})}{r^{3}}}}$	${\displaystyle {\mbox{f}}_{y(3x^{2}-y^{2})}}$

${\displaystyle l=0}$

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

${\displaystyle l=1}$

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

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

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

${\displaystyle l=2}$

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

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

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

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

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

参见

• 勒让德多项式
• 伴随勒让德多项式
• 施图姆-刘维尔理论
• 柱谐函数
• 矢量球谐函数