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球谐函数 - 维基百科,自由的百科全书

球谐函数

维基百科,自由的百科全书
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球谐函数拉普拉斯方程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。

函数的推导[编辑]

本微分方程的推导[编辑]

球坐标下的拉普拉斯方程:

2 f = 1 r 2 r ( r 2 f r ) + 1 r 2 sin θ θ ( sin θ f θ ) + 1 r 2 sin 2 θ 2 f φ 2 = 0 {\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\,\!} \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 Ylm,l = 0 到 4 (由上至下),m=0 到 4(由左至右)。负数阶球谐函数 Yl,-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 f ( r , θ , φ ) = R ( r ) Y ( θ , φ ) = R ( r ) Θ ( θ ) Φ ( φ ) {\displaystyle f(r,\ \theta ,\ \varphi )=R(r)Y(\theta ,\ \varphi )=R(r)\Theta (\theta )\Phi (\varphi )} f(r,\ \theta ,\ \varphi )=R(r)Y(\theta ,\ \varphi )=R(r)\Theta (\theta )\Phi (\varphi ) 。其中 Y ( θ , φ ) {\displaystyle Y(\theta ,\ \varphi )} 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:

Θ Φ r 2 d d r ( r 2 d R d r ) + R Φ r 2 sin θ d d θ ( sin θ d Θ d θ ) + R Θ r 2 sin 2 θ d 2 Φ d φ 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\,\!} {\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:

{ 1 R d d r ( r 2 d R d r ) = λ 1 Φ d 2 Φ d φ 2 = m 2 λ + 1 Θ sin θ d d θ ( sin θ d Θ d θ ) m 2 sin 2 θ = 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}}}
{\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 { r 2 R + 2 r R λ R = 0 Φ + m 2 Φ = 0 sin θ d d θ ( sin θ Θ ) + ( λ sin 2 θ m 2 ) Θ = 0 {\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}}} {\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[编辑]

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

Φ = e i m ϕ {\displaystyle \Phi =e^{im\phi }} \Phi =e^{{im\phi }} m Z {\displaystyle m\in \mathbb {Z} } m\in {\mathbb  {Z}}

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

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

故球谐函数可以表达为:

Y m ( θ , φ ) = N Φ ( φ ) Θ ( θ ) = N e i m φ P m ( cos θ ) {\displaystyle Y_{\ell }^{m}(\theta ,\varphi )=N\Phi (\varphi )\Theta (\theta )=N\,e^{im\varphi }\,P_{\ell }^{m}(\cos {\theta })\,\!} Y_{\ell }^{m}(\theta ,\varphi )=N\Phi (\varphi )\Theta (\theta )=N\,e^{{im\varphi }}\,P_{\ell }^{m}(\cos {\theta })\,\! l N , m = 0 , ± 1 , ± 2 , ± l {\displaystyle l\in \mathbb {N} ,m=0,\pm 1,\pm 2,\ldots \pm l} 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:

Y m ( θ , φ ) = ( 1 ) m ( 2 + 1 ) 4 π ( | m | ) ! ( + | m | ) ! P m ( cos θ ) e i m φ {\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 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 Y m {\displaystyle Y_{\ell }^{m}\,\!} Y_{\ell }^{m}\,\! 称为 {\displaystyle \ell \,\!} \ell\,\! m {\displaystyle m\,\!} m\,\! 的球谐函数。以上推导过程中, i {\displaystyle i\,\!} i\,\!虚数单位 P m {\displaystyle P_{\ell }^{m}\,\!} P_{\ell }^{m}\,\!伴随勒让德多项式

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

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

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

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

前几阶球谐函数[编辑]

主条目:球谐函数表
l {\displaystyle l} l m {\displaystyle m} m Φ ( φ ) {\displaystyle \Phi (\varphi )} \Phi (\varphi ) Θ ( θ ) {\displaystyle \Theta (\theta )} \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 1 2 π {\displaystyle {\frac {1}{\sqrt {2\pi }}}} {\frac  {1}{{\sqrt  {2\pi }}}} 1 2 {\displaystyle {\frac {1}{\sqrt {2}}}} \frac{1}{\sqrt{2}} 1 2 π {\displaystyle {\frac {1}{2{\sqrt {\pi }}}}} {\frac  {1}{2{\sqrt  {\pi }}}} 1 2 π {\displaystyle {\frac {1}{2{\sqrt {\pi }}}}} {\frac  {1}{2{\sqrt  {\pi }}}} s {\displaystyle {\mbox{s}}\,} {\mbox{s}}\,
1 0 1 2 π {\displaystyle {\frac {1}{\sqrt {2\pi }}}} {\frac  {1}{{\sqrt  {2\pi }}}} 3 2 cos θ {\displaystyle {\sqrt {\frac {3}{2}}}\cos \theta } {\sqrt  {{\frac  {3}{2}}}}\cos \theta 1 2 3 π cos θ {\displaystyle {\frac {1}{2}}{\sqrt {\frac {3}{\pi }}}\cos \theta } {\frac  {1}{2}}{\sqrt  {{\frac  {3}{\pi }}}}\cos \theta 1 2 3 π z r {\displaystyle {\frac {1}{2}}{\sqrt {\frac {3}{\pi }}}{\frac {z}{r}}} {\frac  {1}{2}}{\sqrt  {{\frac  {3}{\pi }}}}{\frac  {z}{r}} p z {\displaystyle {\mbox{p}}_{z}\,} {\mbox{p}}_{{z}}\,
1 +1 1 2 π exp ( i φ ) {\displaystyle {\frac {1}{\sqrt {2\pi }}}\exp(i\varphi )} {\frac  {1}{{\sqrt  {2\pi }}}}\exp(i\varphi ) 3 2 sin θ {\displaystyle {\frac {\sqrt {3}}{2}}\sin \theta } {\frac  {{\sqrt  {3}}}{2}}\sin \theta { {\displaystyle {\Bigg \{}} {\Bigg \{} 1 2 3 π sin θ cos φ {\displaystyle {\frac {1}{2}}{\sqrt {\frac {3}{\pi }}}\sin \theta \cos \varphi } {\frac  {1}{2}}{\sqrt  {{\frac  {3}{\pi }}}}\sin \theta \cos \varphi 1 2 3 π x r {\displaystyle {\frac {1}{2}}{\sqrt {\frac {3}{\pi }}}{\frac {x}{r}}} {\frac  {1}{2}}{\sqrt  {{\frac  {3}{\pi }}}}{\frac  {x}{r}} p x {\displaystyle {\mbox{p}}_{x}\,} {\mbox{p}}_{{x}}\,
1 -1 1 2 π exp ( i φ ) {\displaystyle {\frac {1}{\sqrt {2\pi }}}\exp(-i\varphi )} {\frac  {1}{{\sqrt  {2\pi }}}}\exp(-i\varphi ) 3 2 sin θ {\displaystyle {\frac {\sqrt {3}}{2}}\sin \theta } {\frac  {{\sqrt  {3}}}{2}}\sin \theta 1 2 3 π sin θ sin φ {\displaystyle {\frac {1}{2}}{\sqrt {\frac {3}{\pi }}}\sin \theta \sin \varphi } {\frac  {1}{2}}{\sqrt  {{\frac  {3}{\pi }}}}\sin \theta \sin \varphi 1 2 3 π y r {\displaystyle {\frac {1}{2}}{\sqrt {\frac {3}{\pi }}}{\frac {y}{r}}} {\frac  {1}{2}}{\sqrt  {{\frac  {3}{\pi }}}}{\frac  {y}{r}} p y {\displaystyle {\mbox{p}}_{y}\,} {\mbox{p}}_{{y}}\,
2 0 1 2 π {\displaystyle {\frac {1}{\sqrt {2\pi }}}} {\frac  {1}{{\sqrt  {2\pi }}}} 1 2 5 2 ( 3 cos 2 θ 1 ) {\displaystyle {\frac {1}{2}}{\sqrt {\frac {5}{2}}}(3\cos ^{2}\theta -1)} {\frac  {1}{2}}{\sqrt  {{\frac  {5}{2}}}}(3\cos ^{2}\theta -1) 1 4 5 π ( 3 cos 2 θ 1 ) {\displaystyle {\frac {1}{4}}{\sqrt {\frac {5}{\pi }}}(3\cos ^{2}\theta -1)} {\frac  {1}{4}}{\sqrt  {{\frac  {5}{\pi }}}}(3\cos ^{2}\theta -1) 1 4 5 π 2 z 2 x 2 y 2 r 2 {\displaystyle {\frac {1}{4}}{\sqrt {\frac {5}{\pi }}}{\frac {2z^{2}-x^{2}-y^{2}}{r^{2}}}} {\frac  {1}{4}}{\sqrt  {{\frac  {5}{\pi }}}}{\frac  {2z^{2}-x^{2}-y^{2}}{r^{2}}} d 3 z 2 r 2 {\displaystyle {\mbox{d}}_{3z^{2}-r^{2}}} {\mbox{d}}_{{3z^{2}-r^{2}}}
2 +1 1 2 π exp ( i φ ) {\displaystyle {\frac {1}{\sqrt {2\pi }}}\exp(i\varphi )} {\frac  {1}{{\sqrt  {2\pi }}}}\exp(i\varphi ) 15 2 sin θ cos θ {\displaystyle {\frac {\sqrt {15}}{2}}\sin \theta \cos \theta } {\frac  {{\sqrt  {15}}}{2}}\sin \theta \cos \theta { {\displaystyle {\Bigg \{}} {\Bigg \{} 1 2 15 π sin θ cos θ cos φ {\displaystyle {\frac {1}{2}}{\sqrt {\frac {15}{\pi }}}\sin \theta \cos \theta \cos \varphi } {\frac  {1}{2}}{\sqrt  {{\frac  {15}{\pi }}}}\sin \theta \cos \theta \cos \varphi 1 2 15 π z x r 2 {\displaystyle {\frac {1}{2}}{\sqrt {\frac {15}{\pi }}}{\frac {zx}{r^{2}}}} {\frac  {1}{2}}{\sqrt  {{\frac  {15}{\pi }}}}{\frac  {zx}{r^{2}}} d z x {\displaystyle {\mbox{d}}_{zx}\,} {\mbox{d}}_{{zx}}\,
2 -1 1 2 π exp ( i φ ) {\displaystyle {\frac {1}{\sqrt {2\pi }}}\exp(-i\varphi )} {\frac  {1}{{\sqrt  {2\pi }}}}\exp(-i\varphi ) 15 2 sin θ cos θ {\displaystyle {\frac {\sqrt {15}}{2}}\sin \theta \cos \theta } {\frac  {{\sqrt  {15}}}{2}}\sin \theta \cos \theta 1 2 15 π sin θ cos θ sin φ {\displaystyle {\frac {1}{2}}{\sqrt {\frac {15}{\pi }}}\sin \theta \cos \theta \sin \varphi } {\frac  {1}{2}}{\sqrt  {{\frac  {15}{\pi }}}}\sin \theta \cos \theta \sin \varphi 1 2 15 π y z r 2 {\displaystyle {\frac {1}{2}}{\sqrt {\frac {15}{\pi }}}{\frac {yz}{r^{2}}}} {\frac  {1}{2}}{\sqrt  {{\frac  {15}{\pi }}}}{\frac  {yz}{r^{2}}} d y z {\displaystyle {\mbox{d}}_{yz}\,} {\mbox{d}}_{{yz}}\,
2 +2 1 2 π exp ( 2 i φ ) {\displaystyle {\frac {1}{\sqrt {2\pi }}}\exp(2i\varphi )} {\frac  {1}{{\sqrt  {2\pi }}}}\exp(2i\varphi ) 15 4 sin 2 θ {\displaystyle {\frac {\sqrt {15}}{4}}\sin ^{2}\theta } {\frac  {{\sqrt  {15}}}{4}}\sin ^{2}\theta { {\displaystyle {\Bigg \{}} {\Bigg \{} 1 4 15 π sin 2 θ cos 2 φ {\displaystyle {\frac {1}{4}}{\sqrt {\frac {15}{\pi }}}\sin ^{2}\theta \cos 2\varphi } {\frac  {1}{4}}{\sqrt  {{\frac  {15}{\pi }}}}\sin ^{2}\theta \cos 2\varphi 1 4 15 π x 2 y 2 r 2 {\displaystyle {\frac {1}{4}}{\sqrt {\frac {15}{\pi }}}{\frac {x^{2}-y^{2}}{r^{2}}}} {\frac  {1}{4}}{\sqrt  {{\frac  {15}{\pi }}}}{\frac  {x^{2}-y^{2}}{r^{2}}} d x 2 y 2 {\displaystyle {\mbox{d}}_{x^{2}-y^{2}}} {\mbox{d}}_{{x^{2}-y^{2}}}
2 -2 1 2 π exp ( 2 i φ ) {\displaystyle {\frac {1}{\sqrt {2\pi }}}\exp(-2i\varphi )} {\frac  {1}{{\sqrt  {2\pi }}}}\exp(-2i\varphi ) 15 4 sin 2 θ {\displaystyle {\frac {\sqrt {15}}{4}}\sin ^{2}\theta } {\frac  {{\sqrt  {15}}}{4}}\sin ^{2}\theta 1 4 15 π sin 2 θ sin 2 φ {\displaystyle {\frac {1}{4}}{\sqrt {\frac {15}{\pi }}}\sin ^{2}\theta \sin 2\varphi } {\frac  {1}{4}}{\sqrt  {{\frac  {15}{\pi }}}}\sin ^{2}\theta \sin 2\varphi 1 2 15 π x y r 2 {\displaystyle {\frac {1}{2}}{\sqrt {\frac {15}{\pi }}}{\frac {xy}{r^{2}}}} {\frac  {1}{2}}{\sqrt  {{\frac  {15}{\pi }}}}{\frac  {xy}{r^{2}}} d x y {\displaystyle {\mbox{d}}_{xy}\,} {\mbox{d}}_{{xy}}\,
3 0 1 2 π {\displaystyle {\frac {1}{\sqrt {2\pi }}}} {\frac  {1}{{\sqrt  {2\pi }}}} 1 2 7 2 ( 5 cos 3 θ 3 cos θ ) {\displaystyle {\frac {1}{2}}{\sqrt {\frac {7}{2}}}(5\cos ^{3}\theta -3\cos \theta )} {\frac  {1}{2}}{\sqrt  {{\frac  {7}{2}}}}(5\cos ^{3}\theta -3\cos \theta ) 1 4 7 π ( 5 cos 3 θ 3 cos θ ) {\displaystyle {\frac {1}{4}}{\sqrt {\frac {7}{\pi }}}(5\cos ^{3}\theta -3\cos \theta )} {\frac  {1}{4}}{\sqrt  {{\frac  {7}{\pi }}}}(5\cos ^{3}\theta -3\cos \theta ) 1 4 7 π z ( 2 z 2 3 x 2 3 y 2 ) r 3 {\displaystyle {\frac {1}{4}}{\sqrt {\frac {7}{\pi }}}{\frac {z(2z^{2}-3x^{2}-3y^{2})}{r^{3}}}} {\frac  {1}{4}}{\sqrt  {{\frac  {7}{\pi }}}}{\frac  {z(2z^{2}-3x^{2}-3y^{2})}{r^{3}}} f z ( 5 z 2 3 r 2 ) {\displaystyle {\mbox{f}}_{z(5z^{2}-3r^{2})}} {\mbox{f}}_{{z(5z^{2}-3r^{2})}}
3 +1 1 2 π exp ( i φ ) {\displaystyle {\frac {1}{\sqrt {2\pi }}}\exp(i\varphi )} {\frac  {1}{{\sqrt  {2\pi }}}}\exp(i\varphi ) 1 4 21 2 ( 5 cos 2 θ 1 ) sin θ {\displaystyle {\frac {1}{4}}{\sqrt {\frac {21}{2}}}(5\cos ^{2}\theta -1)\sin \theta } {\frac  {1}{4}}{\sqrt  {{\frac  {21}{2}}}}(5\cos ^{2}\theta -1)\sin \theta { {\displaystyle {\Bigg \{}} {\Bigg \{} 1 4 21 2 π ( 5 cos 2 θ 1 ) sin θ cos φ {\displaystyle {\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 }}}}(5\cos ^{2}\theta -1)\sin \theta \cos \varphi 1 4 21 2 π x ( 5 z 2 r 2 ) r 3 {\displaystyle {\frac {1}{4}}{\sqrt {\frac {21}{2\pi }}}{\frac {x(5z^{2}-r^{2})}{r^{3}}}} {\frac  {1}{4}}{\sqrt  {{\frac  {21}{2\pi }}}}{\frac  {x(5z^{2}-r^{2})}{r^{3}}} f x ( 5 z 2 r 2 ) {\displaystyle {\mbox{f}}_{x(5z^{2}-r^{2})}} {\mbox{f}}_{{x(5z^{2}-r^{2})}}
3 -1 1 2 π exp ( i φ ) {\displaystyle {\frac {1}{\sqrt {2\pi }}}\exp(-i\varphi )} {\frac  {1}{{\sqrt  {2\pi }}}}\exp(-i\varphi ) 1 4 21 2 ( 5 cos 2 θ 1 ) sin θ {\displaystyle {\frac {1}{4}}{\sqrt {\frac {21}{2}}}(5\cos ^{2}\theta -1)\sin \theta } {\frac  {1}{4}}{\sqrt  {{\frac  {21}{2}}}}(5\cos ^{2}\theta -1)\sin \theta 1 4 21 2 π ( 5 cos 2 θ 1 ) sin θ sin φ {\displaystyle {\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 }}}}(5\cos ^{2}\theta -1)\sin \theta \sin \varphi 1 4 21 2 π y ( 5 z 2 r 2 ) r 3 {\displaystyle {\frac {1}{4}}{\sqrt {\frac {21}{2\pi }}}{\frac {y(5z^{2}-r^{2})}{r^{3}}}} {\frac  {1}{4}}{\sqrt  {{\frac  {21}{2\pi }}}}{\frac  {y(5z^{2}-r^{2})}{r^{3}}} f y ( 5 z 2 r 2 ) {\displaystyle {\mbox{f}}_{y(5z^{2}-r^{2})}} {\mbox{f}}_{{y(5z^{2}-r^{2})}}
3 +2 1 2 π exp ( 2 i φ ) {\displaystyle {\frac {1}{\sqrt {2\pi }}}\exp(2i\varphi )} {\frac  {1}{{\sqrt  {2\pi }}}}\exp(2i\varphi ) 105 4 cos θ sin 2 θ {\displaystyle {\frac {\sqrt {105}}{4}}\cos \theta \sin ^{2}\theta } {\frac  {{\sqrt  {105}}}{4}}\cos \theta \sin ^{2}\theta { {\displaystyle {\Bigg \{}} {\Bigg \{} 1 4 105 π cos θ sin 2 θ cos 2 φ {\displaystyle {\frac {1}{4}}{\sqrt {\frac {105}{\pi }}}\cos \theta \sin ^{2}\theta \cos 2\varphi } {\frac  {1}{4}}{\sqrt  {{\frac  {105}{\pi }}}}\cos \theta \sin ^{2}\theta \cos 2\varphi 1 4 105 π z ( x 2 y 2 ) r 3 {\displaystyle {\frac {1}{4}}{\sqrt {\frac {105}{\pi }}}{\frac {z(x^{2}-y^{2})}{r^{3}}}} {\frac  {1}{4}}{\sqrt  {{\frac  {105}{\pi }}}}{\frac  {z(x^{2}-y^{2})}{r^{3}}} f z ( x 2 y 2 ) {\displaystyle {\mbox{f}}_{z(x^{2}-y^{2})}} {\mbox{f}}_{{z(x^{2}-y^{2})}}
3 -2 1 2 π exp ( 2 i φ ) {\displaystyle {\frac {1}{\sqrt {2\pi }}}\exp(-2i\varphi )} {\frac  {1}{{\sqrt  {2\pi }}}}\exp(-2i\varphi ) 105 4 cos θ sin 2 θ {\displaystyle {\frac {\sqrt {105}}{4}}\cos \theta \sin ^{2}\theta } {\frac  {{\sqrt  {105}}}{4}}\cos \theta \sin ^{2}\theta 1 4 105 π cos θ sin 2 θ sin 2 φ {\displaystyle {\frac {1}{4}}{\sqrt {\frac {105}{\pi }}}\cos \theta \sin ^{2}\theta \sin 2\varphi } {\frac  {1}{4}}{\sqrt  {{\frac  {105}{\pi }}}}\cos \theta \sin ^{2}\theta \sin 2\varphi 1 2 105 π x y z r 3 {\displaystyle {\frac {1}{2}}{\sqrt {\frac {105}{\pi }}}{\frac {xyz}{r^{3}}}} {\frac  {1}{2}}{\sqrt  {{\frac  {105}{\pi }}}}{\frac  {xyz}{r^{3}}} f x y z {\displaystyle {\mbox{f}}_{xyz}\,} {\mbox{f}}_{{xyz}}\,
3 +3 1 2 π exp ( 3 i φ ) {\displaystyle {\frac {1}{\sqrt {2\pi }}}\exp(3i\varphi )} {\frac  {1}{{\sqrt  {2\pi }}}}\exp(3i\varphi ) 1 4 35 2 sin 3 θ {\displaystyle {\frac {1}{4}}{\sqrt {\frac {35}{2}}}\sin ^{3}\theta } {\frac  {1}{4}}{\sqrt  {{\frac  {35}{2}}}}\sin ^{3}\theta { {\displaystyle {\Bigg \{}} {\Bigg \{} 1 4 35 2 π sin 3 θ cos 3 φ {\displaystyle {\frac {1}{4}}{\sqrt {\frac {35}{2\pi }}}\sin ^{3}\theta \cos 3\varphi } {\frac  {1}{4}}{\sqrt  {{\frac  {35}{2\pi }}}}\sin ^{3}\theta \cos 3\varphi 1 4 35 2 π x ( x 2 3 y 2 ) r 3 {\displaystyle {\frac {1}{4}}{\sqrt {\frac {35}{2\pi }}}{\frac {x(x^{2}-3y^{2})}{r^{3}}}} {\frac  {1}{4}}{\sqrt  {{\frac  {35}{2\pi }}}}{\frac  {x(x^{2}-3y^{2})}{r^{3}}} f x ( x 2 3 y 2 ) {\displaystyle {\mbox{f}}_{x(x^{2}-3y^{2})}} {\mbox{f}}_{{x(x^{2}-3y^{2})}}
3 -3 1 2 π exp ( 3 i φ ) {\displaystyle {\frac {1}{\sqrt {2\pi }}}\exp(-3i\varphi )} {\frac  {1}{{\sqrt  {2\pi }}}}\exp(-3i\varphi ) 1 4 35 2 sin 3 θ {\displaystyle {\frac {1}{4}}{\sqrt {\frac {35}{2}}}\sin ^{3}\theta } {\frac  {1}{4}}{\sqrt  {{\frac  {35}{2}}}}\sin ^{3}\theta 1 4 35 2 π sin 3 θ sin 3 φ {\displaystyle {\frac {1}{4}}{\sqrt {\frac {35}{2\pi }}}\sin ^{3}\theta \sin 3\varphi } {\frac  {1}{4}}{\sqrt  {{\frac  {35}{2\pi }}}}\sin ^{3}\theta \sin 3\varphi 1 4 35 2 π y ( 3 x 2 y 2 ) r 3 {\displaystyle {\frac {1}{4}}{\sqrt {\frac {35}{2\pi }}}{\frac {y(3x^{2}-y^{2})}{r^{3}}}} {\frac  {1}{4}}{\sqrt  {{\frac  {35}{2\pi }}}}{\frac  {y(3x^{2}-y^{2})}{r^{3}}} f y ( 3 x 2 y 2 ) {\displaystyle {\mbox{f}}_{y(3x^{2}-y^{2})}} {\mbox{f}}_{{y(3x^{2}-y^{2})}}

l = 0 {\displaystyle l=0} l=0

Y 0 0 ( θ , φ ) = 1 2 1 π {\displaystyle Y_{0}^{0}(\theta ,\varphi )={1 \over 2}{\sqrt {1 \over \pi }}} Y_{{0}}^{{0}}(\theta ,\varphi )={1 \over 2}{\sqrt  {1 \over \pi }}

l = 1 {\displaystyle l=1} l=1

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

l = 2 {\displaystyle l=2} l=2

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

参见[编辑]

Measure
Measure
Summary | 5 Annotations
关于 t {\displaystyle t} 的伴随勒让德方程
2021/01/25 02:49
可以解出
2021/01/25 02:50
即可以解出
2021/01/25 02:50
球谐函数表达为
2021/01/25 02:50
代表角度部分的解,也就是球谐函数
2021/01/25 02:52