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$\Phi(\phi)=A e^{-im \phi}+B e^{im \phi}$ and $\Theta(\theta)=P_l^m(\cos \theta)$ are combined to form

$$P_l^m(\cos \theta) e^{im \phi}$$ How?

Where did $A,B, e^{-im \phi}$ go?

Does it read implicitly that one chooses $A,B$ s.t. they and $e^{-im \phi}$ are eliminated?

http://mathworld.wolfram.com/SphericalHarmonic.html

mavavilj
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1 Answers1

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The idea is to build a solution $u(\theta,\phi) = \Theta(\theta)\Phi(\phi)$ for the equation

$$ \frac{\Phi(\phi)}{\sin \theta}\frac{{\rm d}}{{\rm d}\theta}\left( \sin\theta \frac{{\rm d}\Theta}{{\rm d}\theta}\right) + \frac{\Theta(\theta)}{\sin^2\theta} \frac{{\rm d}^2\Phi}{{\rm d}\phi^2} + l(l + 1)\Theta(\theta)\Phi(\phi) = 0 $$

and you just found that $\Phi(\phi) = e^{im\phi}$, $\Theta(\theta) = P^m_l(\cos\theta)$ is one. That is

$$ u(\theta, \phi) = e^{im\phi} P^m_l(\cos\theta) $$

is a solution. But so is $u^*(\theta,\phi)$. So in that sense they are combined

caverac
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  • What you mean found out that $\Phi(\phi)=e^{im \phi}$ is solution? The gen. sol is $Ae^{-im \phi}+Be^{im \phi}$. – mavavilj Dec 27 '18 at 15:11