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Expansion in spherical harmonics

  1. Nov 24, 2007 #1
    1. The problem statement, all variables and given/known data
    I want to expand 1+sin(phi)sin(theta) in the spherical harmonics. I am not sure if this will be an infinite series or not? If it were infinite that would seem rather difficult because the spherical harmonics get really complicated when l > 3. Also, all of the sine terms in the spherical harmonics are imaginary, so how is this possible?


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 25, 2007 #2

    malawi_glenn

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    what have you tried so far?

    a hint is tho rewrite [tex] e^{i \phi} [/tex] as [tex] \cos \phi + i \sin \phi [/tex]

    And then you can get rid of the [tex] i [/tex] by doing a linear combination of two spherical harmonics with same [tex] l [/tex] but different [tex] m [/tex].

    The series will be finite, it will contain 3 different [tex] Y_{l,m} [/tex]

    Now go ahead and try
     
  4. Nov 25, 2007 #3
    OK. We can get take care of the constant term with Y0,0. For the other term I tried Y2,2 and Y2,1 but that does not get rid of the i. Are you saying I can find two spherical harmonics with exactly the same imaginary term up to a constant? Don't each of the spherical harmonics with the same l have different imaginary parts?
    Do I need to use negative m values?
     
  5. Nov 25, 2007 #4

    malawi_glenn

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    why try l = 2 first?.. why not l = 1?..
     
  6. Nov 25, 2007 #5
    Because Y1,0 does not have an imaginary part, it could not possibly cancel with the imaginary part of Y1,1.
     
  7. Nov 26, 2007 #6

    malawi_glenn

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    Take a look again at the l = 1 spherical harmonics. For each l, you have 2l + 1 spherical harmonics.

    Spherical harmonics with l = 2 have sin(theta)^2 ; so can IMPOSSIBLE do anything with those. Spherical harmonics with l = 1 have sin(theta); so I think it is quite obvious..
     
    Last edited: Nov 26, 2007
  8. Nov 26, 2007 #7
    How stupid of me! I see now. Thanks.
     
  9. Nov 28, 2007 #8

    malawi_glenn

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    The answer will be:

    [tex] 1 + \sin \theta \sin \phi = \sqrt{4\pi} + \sqrt{\frac{2\pi}{3}}i \cdot (Y_{1,+1} + Y_{1,-1}) [/tex]

    If i did it correct, you got the same?

    [edited]
     
    Last edited: Nov 28, 2007
  10. Nov 28, 2007 #9
    I got
    [tex] 1 + \sin \theta \sin \phi = \sqrt{4\pi} + \sqrt{\frac{2\pi}{3}}i \cdot (Y_{1,+1} + Y_{1,-1}) [/tex]
     
  11. Nov 28, 2007 #10

    malawi_glenn

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    yes that is correct, because I had i in the denominator on my papers, but wrote wrong here in TeX. Good job!
     
  12. Dec 5, 2007 #11
    I could see that one, but there's something silly I'm missing :(

    How would I expand
    sin(theta)?

    To me it seems only Y^0_1 has the right order, but its a cos(theta) that I cannot convert?!
     
  13. Dec 5, 2007 #12

    malawi_glenn

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    silly? now are you asking just for curiosity or is it homework-related?
     
  14. Dec 5, 2007 #13
    Well, its homework. But given by myself.
    I was wondering about this splitting of cubic symmetry along the 3 fold diagonal axis. It splits into 2+2+1 energylevels, but the basis system is is along the diagonal and not the z axis anymore so I was trying to project
    (3cos^2(theta+45°)-1)*exp(-i*(psi+45°))
    back into the usual system.
    But now I'm stuck with it. Maybe it's been a hard day today. Just can't see what I'm doing wrong :(
     
  15. Dec 5, 2007 #14

    malawi_glenn

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    I dont know either. Try posting the whole problem in its entire lenght in a new thread. Showing some of the relations you know of and your attempt to solve it. Then someone might be able to help you =)

    Btw, I just think you can write [tex] \sin ^2\theta = 1 - \cos ^2\theta [/tex] and take it from there.
     
  16. Dec 5, 2007 #15
    OK. Maybe its something more fundamental.
    I can switch between sin^2(a) and cos^2(a), but not to sin(a)cos(a)
    But I think I know all trigonometric relations. Yesterday I derived for fun:
    sin(a+b+c+d)/cos(a)/cos/(b)/cos(c)/cos(d)=tan(a)+tan(b)+tan(c)+tan(d)+tan(a)tan(b)tan(c)tan(d)(1/tan(a)+1/tan(b)+1/tan(c)+1/tan(d))
    Its pretty useless but nice.
     
  17. Dec 5, 2007 #16

    malawi_glenn

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    I was rather thinking of

    [tex] \sin ^2\theta = 1 - \cos ^2\theta [/tex]

    [tex] \sin \theta = \sqrt{4\pi Y_{0,0}^2-\frac{4\pi }{3}Y_{1,0}^2} [/tex]
     
  18. Dec 5, 2007 #17
    Yes, OK. But thats not linear in the spherical harmonics.
    Mr Wiki Pedia says
    http://en.wikipedia.org/wiki/Spherical_harmonics
    one can expand linearly. There is also an equation for the coefficients. I'm not in the state of doing integrals now, but I suspect I get zero for all of them.
     
  19. Dec 5, 2007 #18

    malawi_glenn

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    no ofcourse its not a linear expansion.

    You should in principle be able to expand ANY angular function as a combination of spherical harms-
     
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