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Spherical Harmonics

  1. May 12, 2009 #1
    So, I was reading about the quantum http://en.wikipedia.org/wiki/Rigid_rotor" [Broken] and apparently its solutions are the so-called spherical harmonic functions, which are the solution to the angular portion of Laplace's equation. The way I see it, the rigid rotor Schroedinger equation is not the angular part of Laplace's equation (because of that E[tex]\Psi[/tex]part of the rigid rotor equation). Am I right? Am I missing something? Am I completely wrong? Maybe these functions are solutions to both equations?



    2. Relevant equations

    [tex]\Delta[/tex][tex]\Psi[/tex]=0 (Laplace's equation)

    (-h-bar2/2[tex]\mu[/tex])[tex]\Delta[/tex][tex]\Psi[/tex]=E[tex]\Psi[/tex]
    1. The problem statement, all variables and given/known data



    Errrr... I'm have a hard time writing these equations here, it's just worth mentioning that it's not 2 to the power of mu, but 2 times mu.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 12, 2009 #2
    If you actually solve the Laplace's equation you'll realize that the angular part is of the same form as the Shrödinger's equation. Start with the Laplace equation:

    [tex]\nabla^2 \Psi = 0[/tex]

    Assume the solution is of the form [tex]\Psi(r,\theta,\phi) = R(r) Y(\theta,\phi)[/tex] so that you can separate radius and angular dependence. You'll get:

    [tex]\frac{1}{R}\nabla_r^2R + \frac{1}{Y}\nabla^2_\theta_, _\phi Y = 0[/tex]

    Now the first term only depends on r while the second term depends only on the angles which means that both must be constant (denote C) with opposite signs. That means that you can solve them separately and the angular part becomes

    [tex]\nabla^2_\theta_, _\phi Y = CY[/tex]

    Which is of the same form as the Schrödinger equation.
     
  4. May 12, 2009 #3

    CompuChip

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    It's been a while for me, so please correct me if I am wrong here.
    If you want to solve an equation such as the second, it is common to divide the problem in finding a particular solution and the general solution to the homogeneous equation. In this case, the solutions to the homogeneous equation
    [tex]-\frac{\hbar^2}{2\mu} \Delta\Psi = 0[/tex]
    are the spherical harmonic functions (or a multiple thereof) because the homogeneous equation simply is Laplace's equation. After you have found a particular solution [itex]\Psi_0[/itex] satisfying [itex]-\frac{\hbar^2}{2\mu} \Delta\Psi_0 = E \Psi_0[/itex] you can construct the general solution as the sum of this particular solution and the general solution, which is in turn a linear combination of spherical harmonics.

    As for the equations, it would help if you all put it in just one pair of tags:
    [tex](-hbar^2 / 2\mu) \Delta\Psi = E \Psi[/tex]
    already looks much better (it'd be perfect if I'd written \hbar instead of hbar).
     
  5. May 12, 2009 #4
    I think the equation is homogenous anyway:


    [tex](-hbar^2 / 2\mu) \Delta\Psi - E \Psi= 0[/tex]


    Ok, so C is a constant with respect to the radius (and everything else actually), which means I had forgotten this step in the separation of variables. Thanks a lot.
     
    Last edited: May 12, 2009
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