球谐函数

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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。在量子力学等领域广泛应用。

函数的推导[编辑]

本征方程的推导[编辑]

球坐标下的拉普拉斯方程是:

 \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 Ylm,l = 0 到 4 (由上至下),m=0 到 4(由左至右)。負數階球諧函數 Yl,-m 可由正數階函數對 z-軸轉 90/m 度得到。

利用分离变量法,设定 f(r,\ \theta,\ \varphi)=R(r)Y(\theta,\ \varphi)=R(r)\Theta(\theta)\Phi(\varphi) 。其中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:

{\Theta \Phi \over r^2}{d \over dr}\left(r^2 {d R \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:

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

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

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

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

故球谐函数可以表达为:

 Y_\ell^m (\theta, \varphi ) = N \Phi(\varphi)\Theta(\theta) = N \, e^{i m \varphi } \, P_\ell^m (\cos{\theta} )\,\!l\in\mathbb{N},m=0,\pm1,\pm2,\ldots\pm l

其中N 是归一化因子。

經過歸一化後,球谐函数表達為:

 Y_\ell^m(\theta,\ \varphi) =(i)^{m+|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_\ell^m\,\! 称为 \ell\,\!m\,\! 的球谐函数。以上推导过程中,i\,\!虛數單位P_\ell^m\,\!伴随勒让德多项式

其中P_\ell^m(x)\,\! 用方程式定義為:

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

P_\ell(x)\,\!l勒讓德多項式,可用羅德里格公式表示為:

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]

l m \Phi(\varphi) \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 \frac{1}{\sqrt{2\pi}} \frac{1}{\sqrt{2}} \frac{1}{2\sqrt{\pi}} \frac{1}{2\sqrt{\pi}} \mbox{s} \,
1 0 \frac{1}{\sqrt{2\pi}} \sqrt{\frac{3}{2}}\cos\theta \frac{1}{2}\sqrt{\frac{3}{\pi}}\cos\theta \frac{1}{2}\sqrt{\frac{3}{\pi}}\frac{z}{r} \mbox{p}_{z} \,
1 +1 \frac{1}{\sqrt{2\pi}}\exp(i\varphi) \frac{\sqrt{3}}{2}\sin\theta \Bigg\{ \frac{1}{2}\sqrt{\frac{3}{\pi}}\sin\theta\cos\varphi \frac{1}{2}\sqrt{\frac{3}{\pi}}\frac{x}{r} \mbox{p}_{x} \,
1 -1 \frac{1}{\sqrt{2\pi}}\exp(-i\varphi) \frac{\sqrt{3}}{2}\sin\theta \frac{1}{2}\sqrt{\frac{3}{\pi}}\sin\theta\sin\varphi \frac{1}{2}\sqrt{\frac{3}{\pi}}\frac{y}{r} \mbox{p}_{y} \,
2 0 \frac{1}{\sqrt{2\pi}} \frac{1}{2}\sqrt{\frac{5}{2}}(3\cos^2\theta-1) \frac{1}{4}\sqrt{\frac{5}{\pi}}(3\cos^2\theta-1) \frac{1}{4}\sqrt{\frac{5}{\pi}}\frac{2z^2-x^2-y^2}{r^2} \mbox{d}_{3z^2-r^2}
2 +1 \frac{1}{\sqrt{2\pi}}\exp(i\varphi) \frac{\sqrt{15}}{2}\sin\theta\cos\theta \Bigg\{ \frac{1}{2}\sqrt{\frac{15}{\pi}}\sin\theta\cos\theta\cos\varphi \frac{1}{2}\sqrt{\frac{15}{\pi}}\frac{zx}{r^2} \mbox{d}_{zx} \,
2 -1 \frac{1}{\sqrt{2\pi}}\exp(-i\varphi) \frac{\sqrt{15}}{2}\sin\theta\cos\theta \frac{1}{2}\sqrt{\frac{15}{\pi}}\sin\theta\cos\theta\sin\varphi \frac{1}{2}\sqrt{\frac{15}{\pi}}\frac{yz}{r^2} \mbox{d}_{yz} \,
2 +2 \frac{1}{\sqrt{2\pi}}\exp(2i\varphi) \frac{\sqrt{15}}{4}\sin^2\theta \Bigg\{ \frac{1}{4}\sqrt{\frac{15}{\pi}}\sin^2\theta\cos2\varphi \frac{1}{4}\sqrt{\frac{15}{\pi}}\frac{x^2-y^2}{r^2} \mbox{d}_{x^2-y^2}
2 -2 \frac{1}{\sqrt{2\pi}}\exp(-2i\varphi) \frac{\sqrt{15}}{4}\sin^2\theta \frac{1}{4}\sqrt{\frac{15}{\pi}}\sin^2\theta\sin2\varphi \frac{1}{2}\sqrt{\frac{15}{\pi}}\frac{xy}{r^2} \mbox{d}_{xy} \,
3 0 \frac{1}{\sqrt{2\pi}} \frac{1}{2}\sqrt{\frac{7}{2}}(5\cos^3\theta-3\cos\theta) \frac{1}{4}\sqrt{\frac{7}{\pi}}(5\cos^3\theta-3\cos\theta) \frac{1}{4}\sqrt{\frac{7}{\pi}}\frac{z(2z^2-3x^2-3y^2)}{r^3} \mbox{f}_{z(5z^2-3r^2)}
3 +1 \frac{1}{\sqrt{2\pi}}\exp(i\varphi) \frac{1}{4}\sqrt{\frac{21}{2}}(5\cos^2\theta-1)\sin\theta \Bigg\{ \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}}\frac{x(5z^2-r^2)}{r^3} \mbox{f}_{x(5z^2-r^2)}
3 -1 \frac{1}{\sqrt{2\pi}}\exp(-i\varphi) \frac{1}{4}\sqrt{\frac{21}{2}}(5\cos^2\theta-1)\sin\theta \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}}\frac{y(5z^2-r^2)}{r^3} \mbox{f}_{y(5z^2-r^2)}
3 +2 \frac{1}{\sqrt{2\pi}}\exp(2i\varphi) \frac{\sqrt{105}}{4}\cos\theta\sin^2\theta \Bigg\{ \frac{1}{4}\sqrt{\frac{105}{\pi}}\cos\theta\sin^2\theta\cos2\varphi \frac{1}{4}\sqrt{\frac{105}{\pi}}\frac{z(x^2-y^2)}{r^3} \mbox{f}_{z(x^2-y^2)}
3 -2 \frac{1}{\sqrt{2\pi}}\exp(-2i\varphi) \frac{\sqrt{105}}{4}\cos\theta\sin^2\theta \frac{1}{4}\sqrt{\frac{105}{\pi}}\cos\theta\sin^2\theta\sin2\varphi \frac{1}{2}\sqrt{\frac{105}{\pi}}\frac{xyz}{r^3} \mbox{f}_{xyz} \,
3 +3 \frac{1}{\sqrt{2\pi}}\exp(3i\varphi) \frac{1}{4}\sqrt{\frac{35}{2}}\sin^3\theta \Bigg\{ \frac{1}{4}\sqrt{\frac{35}{2\pi}}\sin^3\theta\cos3\varphi \frac{1}{4}\sqrt{\frac{35}{2\pi}}\frac{x(x^2-3y^2)}{r^3} \mbox{f}_{x(x^2-3y^2)}
3 -3 \frac{1}{\sqrt{2\pi}}\exp(-3i\varphi) \frac{1}{4}\sqrt{\frac{35}{2}}\sin^3\theta \frac{1}{4}\sqrt{\frac{35}{2\pi}}\sin^3\theta\sin3\varphi \frac{1}{4}\sqrt{\frac{35}{2\pi}}\frac{y(3x^2-y^2)}{r^3} \mbox{f}_{y(3x^2-y^2)}

l=0

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

l=1

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

l=2

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

参见[编辑]