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Time evolution of spherical harmonics

  1. Dec 8, 2007 #1
    1. The problem statement, all variables and given/known data
    At t=0, a given wavefunction is:

    [tex]\left\langle\theta,\phi|\psi(0)\right\rangle = \frac{\imath}{\sqrt{2}}(Y_{1,1}+Y_{1,-1})[/tex]

    Find [tex]\left\langle\theta,\phi|\psi(t)\right\rangle[/tex].

    2. Relevant equations
    [tex]\hat{U}(t)\left|\psi(0)\right\rangle = e^{-\imath\hat{H}t/\hbar}\left|\psi(t)\right\rangle[/tex]

    [tex]
    \hat{H}\left|\ E,l,m\right\rangle = E\left|\ E,l,m\right\rangle
    [/tex]
    [tex]
    \hat{L^{2}}\left|\ E,l,m\right\rangle = l(l+1)\hbar^{2}\left|\ E,l,m\right\rangle
    [/tex]
    [tex]
    \hat{L_{z}}\left|\ E,l,m\right\rangle = m\hbar\left|\ E,l,m\right\rangle
    [/tex]

    3. The attempt at a solution
    I know that you can use the above operator to make time evolution of an energy eigenstate, but I can't figure out what energy to use for the two spherical harmonics in the given state at t=0.
     
    Last edited: Dec 8, 2007
  2. jcsd
  3. Dec 8, 2007 #2

    Avodyne

    User Avatar
    Science Advisor

    It depends on what the hamiltonian is.
     
  4. Dec 8, 2007 #3
    Oh, okay, I should have realized this.

    This is for a rigid rotator with Hamiltonian

    [tex]\hat{H}=\hat{L^{2}}/2I[/tex]

    So this means that the energies are both:

    [tex]E =1(1+1)\hbar^{2}/2I = \hbar^{2}/I[/tex]

    And

    [tex]\left\langle\theta,\phi|\psi(t)\right\rangle=\frac{i}{\sqrt{2}}e^{-\imath\hbar t/I}(Y_{1,1}+Y_{1,-1})[/tex]
     
    Last edited: Dec 8, 2007
  5. Dec 8, 2007 #4

    Avodyne

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    Science Advisor

    Yep.
     
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