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I Spherical Harmonics

  1. Nov 1, 2016 #1
    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. jcsd
  3. Nov 2, 2016 #2

    DrClaude

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    Staff: Mentor

    Thats is not normalized in ##\phi##.

    What is the orthogonality relation for the Legendre polynomial?
     
  4. Dec 6, 2016 #3
    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.
    $$\int^{2\pi}_{0}e^{2im\phi}d\phi=[\frac{e^{2im\phi}}{2im}]^{2\pi}_0=\frac{1}{2im}-\frac{1}{2im}=0$$
     
  5. Dec 7, 2016 #4

    DrClaude

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    The spherical harmonics are complex, so the scalar product is
    $$
    \left( Y_l^m, Y_{l'}^{m'} \right) = \int_0^{2\pi} \int_0^\pi \left[ Y_l^{m} (\theta,\phi)\right]^* Y_{l'}^{m'} (\theta,\phi) \sin\theta d\theta d\phi
    $$
     
  6. Dec 7, 2016 #5
    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?
     
  7. Dec 7, 2016 #6

    DrClaude

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    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##.
     
  8. Dec 7, 2016 #7
    Thanks a lot!
     
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