# 球谐函数

## 函数的推导

### 本微分方程的推导

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

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

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

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

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

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

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

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

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

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

## 前几阶球谐函数

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

${\displaystyle l=0}$

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

${\displaystyle l=1}$

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

${\displaystyle l=2}$

${\displaystyle Y_{2}^{-2}(\theta ,\varphi )={1 \over 4}{\sqrt {15 \over 2\pi }}\,\sin ^{2}\theta \,e^{-2i\varphi }}$
${\displaystyle Y_{2}^{-1}(\theta ,\varphi )={1 \over 2}{\sqrt {15 \over 2\pi }}\,\sin \theta \,\cos \theta \,e^{-i\varphi }}$
${\displaystyle Y_{2}^{0}(\theta ,\varphi )={1 \over 4}{\sqrt {5 \over \pi }}\,(3\cos ^{2}\theta -1)}$
${\displaystyle Y_{2}^{1}(\theta ,\varphi )={-1 \over 2}{\sqrt {15 \over 2\pi }}\,\sin \theta \,\cos \theta \,e^{i\varphi }}$
${\displaystyle Y_{2}^{2}(\theta ,\varphi )={1 \over 4}{\sqrt {15 \over 2\pi }}\,\sin ^{2}\theta \,e^{2i\varphi }}$