Spherical harmonics/Addendum: Difference between revisions

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{{subpages}}
==First few spherical harmonics==
The following functions are normalized to unity and have the Condon & Shortley phase. The functions are listed first in θ and φ and then in ''x'', ''y'', ''z'', and ''r''. These parameters are connected by
:<math>
x=r\sin\theta\cos\phi,\quad y=r\sin\theta\sin\phi,\quad z=r\cos\theta, \quad r=\sqrt{x^2+y^2+z^2}
</math>
----


<math>
\begin{align}
Y_{0}^{0}(\theta,\varphi)&=\sqrt{1\over 4\pi}\\
& \\
Y_{1}^{0}(\theta,\varphi)&=\sqrt{3\over 4\pi}\, \cos\theta = \sqrt{3\over 4\pi}\, \frac{z}{r} \\
Y_{1}^{\pm1 }(\theta,\varphi)&=\mp \sqrt{3\over 4\pi}\sqrt{\frac{1}{2}} \, \sin\theta \, e^{\pm i\varphi} = \mp \sqrt{3\over 4\pi}\sqrt{\frac{1}{2}} \,\frac{x\pm iy}{r} \\
& \\
Y_{2}^{0}(\theta,\varphi)&=\sqrt{5\over 4\pi}\, \frac{1}{2}(3\cos^{2}\theta-1)=\sqrt{5\over 4\pi}\, \frac{1}{2}\frac{3z^2-r^2}{r^2}\\
Y_{2}^{\pm1}(\theta,\varphi)&=\mp\sqrt{5\over 4\pi}\sqrt{\frac{3}{2}}\, \sin\theta\,\cos\theta\, e^{\pm i\varphi}=\mp\sqrt{5\over 4\pi}\sqrt{\frac{3}{2}}\, \frac{z(x\pm iy)}{r^2}\\
Y_{2}^{\pm2}(\theta,\varphi)&=\sqrt{5\over 4\pi}\,\sqrt{\frac{3}{8}} \sin^{2}\theta \, e^{\pm2i\varphi} = \sqrt{5\over 4\pi}\,\sqrt{\frac{3}{8}} \frac{(x\pm iy)^2}{r^2} \\
& \\
Y_{3}^{0}(\theta,\varphi)&=\sqrt{\frac{7}{4\pi}}\,\frac{1}{2}\, (5\cos^{3}\theta-3\cos\theta)= \sqrt{\frac{7}{4\pi}}\,\frac{1}{2}\, \frac{5z^3 -3zr^2}{r^3} \\
Y_{3}^{\pm1}(\theta,\varphi)&=\mp\sqrt{\frac{7}{4\pi}}\,\sqrt{\frac{3}{16}}\, \sin\theta(5\cos^{2}\theta-1)e^{\pm i\varphi}= \mp\sqrt{\frac{7}{4\pi}}\,\sqrt{\frac{3}{16}}\,\frac{(x\pm i y)(5z^2-r^2)}{r^3} \\
Y_{3}^{\pm2}(\theta,\varphi)&=\sqrt{\frac{7}{4\pi}}\,\sqrt{\frac{15}{8}}\, \sin^2\theta\cos\theta e^{\pm 2i\varphi}= \sqrt{\frac{7}{4\pi}}\,\sqrt{\frac{15}{8}}\,\frac{z(x\pm i y)^2}{r^3} \\
Y_{3}^{\pm3}(\theta,\varphi)&=\mp\sqrt{\frac{7}{4\pi}}\,\sqrt{\frac{5}{16}}\, \sin^3\theta e^{\pm 3i\varphi}= \mp\sqrt{\frac{7}{4\pi}}\,\sqrt{\frac{5}{16}}\,\frac{(x\pm i y)^3}{r^3} \\
&\\
Y_{4}^{0}(\theta,\varphi)&=\sqrt{\frac{9}{4\pi}}\,\frac{1}{8}\, (35\cos^{4}\theta-30\cos^2\theta+3)= \sqrt{\frac{9}{4\pi}}\,\frac{1}{8}\, \frac{35z^4 -30z^2r^2+3r^4}{r^4} \\
Y_{4}^{\pm1}(\theta,\varphi)&=\mp\sqrt{\frac{9}{4\pi}}\,\sqrt{\frac{5}{16}}\, \sin\theta(7\cos^3\theta-3\cos\theta) e^{\pm i\varphi}= \mp\sqrt{\frac{9}{4\pi}}\,\sqrt{\frac{5}{16}}\,\frac{(x\pm i y)(7z^3-3zr^2)}{r^4} \\
Y_{4}^{\pm2}(\theta,\varphi)&=\sqrt{\frac{9}{4\pi}}\,\sqrt{\frac{5}{32}}\, \sin^2\theta(7\cos^2\theta-1) e^{\pm 2i\varphi}= \sqrt{\frac{9}{4\pi}}\,\sqrt{\frac{5}{32}}\,\frac{(x\pm i y)^2(7z^2-r^2)}{r^4} \\
Y_{4}^{\pm3}(\theta,\varphi)&=\mp\sqrt{\frac{9}{4\pi}}\,\sqrt{\frac{35}{16}}\, \sin^3\theta\cos\theta e^{\pm 3i\varphi}= \mp\sqrt{\frac{9}{4\pi}}\,\sqrt{\frac{35}{16}}\,\frac{z(x\pm i y)^3}{r^4} \\
Y_{4}^{\pm4}(\theta,\varphi)&=\sqrt{\frac{9}{4\pi}}\,\sqrt{\frac{35}{128}}\, \sin^4 e^{\pm 4i\varphi}= \sqrt{\frac{9}{4\pi}}\,\sqrt{\frac{35}{128}}\,\frac{(x\pm i y)^4}{r^4} \\
\end{align}
</math>

Revision as of 05:07, 9 September 2009

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