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Homework Help: Angular momentum eigenfunctions

  1. Apr 6, 2010 #1
    1. The problem statement, all variables and given/known data

    This is problem 18.1 from Merzbacher.

    "The hamiltonian of a rigid rotator in a magnetic field perpendicular to the x axis is of the form [itex]H=AL^2+BL_z+CL_y[/itex], if the term \that is quadratic in the field is neglected. Obtain the exact energy eigenvalues and eigenfunctions of the hamiltonian."

    2. Relevant equations

    [tex]L^2Y_l^m(\theta,\phi)=l(l+1)\hbar^2 Y_l^m(\theta,\phi)[/tex]


    [tex]L_Y Y_l^m(\theta,\phi)=\frac{1}{2\imath}(L_+-L_-)Y_l^m(\theta,\phi)[/tex]

    [tex]L_+ Y_l^m(\theta,\phi)=\sqrt{l(l+1)-m(m+1)}\hbar Y_l^{m+1}(\theta,\phi)[/tex]

    [tex]L_- Y_l^m(\theta,\phi)=\sqrt{l(l+1)-m(m-1)}\hbar Y_l^{m-1}(\theta,\phi)[/tex]

    3. The attempt at a solution

    For the first two parts of the hamiltonian the answer is easy:


    But what is the eigenvalue for the last part (of [itex]L_Y[/itex])? Is it just




    And for the eigenfunctions: for the first two parts of H it is just the spherical harmonics, but for the last part is it both





    Many days without sleep. THis is an easy question I know, please don't judge me I am sleep deprived :)

    Last edited: Apr 6, 2010
  2. jcsd
  3. Apr 6, 2010 #2


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    When you act on the state [itex]Y_{\ell}{}^{m}(\theta,\phi)[/itex] with the Hamiltonian, do you get a constant multiple of [itex]Y_{\ell}{}^{m}(\theta,\phi)[/itex] ? If not, it isn't an eigenfunction of the Hamiltonian.
  4. Apr 6, 2010 #3
    Right! This is what confused me. So I guess it cannot be an eigenfunction, and the only one is the regular old spherical harmonic.

    And the eigenvalue is just



    Of course this could be simplified..
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