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QM: More Spherical Harmonics

  1. Feb 14, 2009 #1
    1. The problem statement, all variables and given/known data
    A system's wavefunction is proportional to sin^2p. What are the possible results of measurements of Lz and L^2? Give the probabilities of each possible outcome.

    I'm using p for theta and q for phi.
    2. Relevant equations



    3. The attempt at a solution
    So I believe that the value of L^2 is 6 (as l is 2), and the possible values of Lz are -2 and +2. But I can't find the probabilities. I tried:

    |psi> = a|2,2> + b|2,-2>

    where the kets are |l,m> and the spherical harmonice are Y(l,m)

    <2,2|psi> = int dpdq <2,2|p,q><p,q|psi>
    <2,2|psi> = int dpdq Y(2,2)* [aY(2,2) + bY(2,-2)]

    And then inserted the spherical harmonic expressions in, but I just get a=a, so I guess this isn't the way.

    I also thought about using the ladder operators, applying L+|psi> = 2a|2, -1> but I don't think this is helpful.

    Any help wouldbe much appreciated!!

    EDIT: Another thing I tried was saying Lz|l,m>=m|l,m>. So I thought if I had a matrix representation of Lz, I could maybe have

    Lz (a*, b*) = 2(a*, -b*)

    Where ( , ) is a column matrix. But I don't know the matrix representation of Lz or what basis it should be in. Please help!
     
    Last edited: Feb 15, 2009
  2. jcsd
  3. Feb 16, 2009 #2

    malawi_glenn

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    the wavefunction has NO phi dependence?
     
  4. Feb 16, 2009 #3
    Yeah that was my first reaction. But I don't think it is possible to have no phi dependence and be proportional to sin^2, so I assumed that it meant it was proportional to sin^2 as well as some function of phi. I may be wrong though?
     
  5. Feb 17, 2009 #4

    malawi_glenn

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    The spherical harmonics are basis functions for angular functions, so one should be able to find a linear combination of them to build up sin^2 theta.

    Now if I give you this hint, you should be able to do it.

    [tex]-3\sin ^2 \theta +2 = 3\cos ^2 \theta - 1[/tex]

    Also recall that one of the S. harm's does not depend on any angle at all.
     
  6. Feb 17, 2009 #5
    Ahhhh I forgot about that one.

    I see now, I just need Y(2,0) and then need to subtract an amount of Y(0,0) to get rid of the constant, and normalise.

    Thank you!!!
     
  7. Feb 17, 2009 #6

    malawi_glenn

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    correct observation :-)

    Good luck
     
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