I Spherical Harmonics

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1. Nov 1, 2016

TimeRip496

The normalized angular wave functions are called spherical harmonics: $$Y^m_l(\theta,\phi)=\epsilon\sqrt{\frac{(2l+1)}{4\pi}\frac{(l-|m|)!}{(l+|m|)!}}e^{im\phi}*P^m_l(cos\theta)$$
How do I obtain this from this(http://www.physics.udel.edu/~msafrono/424-2011/Lecture 17.pdf) (Page 8)?

The normalisation condition for Y is $$\int^{2\pi}_0\int^{\pi}_0|Y|^2\sin\theta d\theta d\phi=1$$ Note that 0<∅<2π and 0<θ<π.

$$Y=\Theta(\theta)*\Phi(\phi)$$.
$$\Theta(\theta)=A*P^m_l(cos\theta)$$ where A is the coefficient, x=cosθ, P is the associated Legendre Polynomial.
$$\Phi(\phi)=e^{im\phi}$$
$$Y^m_l(\theta,\phi)=A*e^{im\phi}P^m_l(cos\theta)$$
where $$A=\epsilon\sqrt{\frac{(2l+1)}{4\pi}\frac{(l-|m|)!}{(l+|m|)!}}$$.
I know how to obtain Φ but not A from Θ(θ), which is why I don't know how to obtain the spherical harmonics equation above.

2. Nov 2, 2016

Staff: Mentor

Thats is not normalized in $\phi$.

What is the orthogonality relation for the Legendre polynomial?

3. Dec 6, 2016

TimeRip496

They are orthogonal to each other.
$$\int^1_{-1}P^m_lP^m_ldx=\frac{2}{2p+1}\frac{(p+m)!}{(p-m)!}\delta_{pq}$$
$$\int^{2\pi}_{0}e^{2im\phi}\int^\pi_{0}A^2P^m_lP^m_lsin \theta d\theta d\phi=\int^{2\pi}_{0}e^{2im\phi}[A^2(\frac{4}{2p+1}\frac{(p+m)!}{(p-m)!})]d\phi$$
I am stuck at here cause when I integrate the e2im∅, I get a zero value which make the equation unsolvable.
$$\int^{2\pi}_{0}e^{2im\phi}d\phi=[\frac{e^{2im\phi}}{2im}]^{2\pi}_0=\frac{1}{2im}-\frac{1}{2im}=0$$

4. Dec 7, 2016

Staff: Mentor

The spherical harmonics are complex, so the scalar product is
$$\left( Y_l^m, Y_{l'}^{m'} \right) = \int_0^{2\pi} \int_0^\pi \left[ Y_l^{m} (\theta,\phi)\right]^* Y_{l'}^{m'} (\theta,\phi) \sin\theta d\theta d\phi$$

5. Dec 7, 2016

TimeRip496

Thanks but i still did not get the correct answer.
$$\int^{2\pi}_{0}e^{im\phi}*e^{-im\phi}d\phi=\int^{2\pi}_{0}e^{0}d\phi=2\pi$$
$$\int^{2\pi}_{0}e^{0}d\phi\int^\pi_{0}A^2P^m_lP^m_ld\theta\int^\pi_{0}sin \theta d\theta =(2\pi)[A^2(\frac{2}{2p+1}\frac{(p+m)!}{(p-m)!})](2)=[A^2(\frac{8}{2p+1}\frac{(p+m)!}{(p-m)!})]$$
Cause $$\int^{\pi}_0 sin\theta d\theta =2$$
The correct answer should be $$[A^2(\frac{4}{2p+1}\frac{(p+m)!}{(p-m)!})]$$ where it should be 4 instead of 8.
Where do I go wrong?

6. Dec 7, 2016

Staff: Mentor

There is no such integral in the problem. Start from orthogonality condition you had in post #3 and make the substitution $x = \cos \theta$.

7. Dec 7, 2016

TimeRip496

Thanks a lot!

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