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Angular momentum operators and eigenfunctions

  1. May 3, 2014 #1
    1. The problem statement, all variables and given/known data

    yVt5RSm.png

    2. Relevant equations



    3. The attempt at a solution

    I have tried inserting the first wavefunction into Lz which gets me 0 for the eigenvalue for the first wavefunction. Is this correct?

    For the second wavefunction, I inserted it in to Lz and this gets me -i*hbar*xAe^-r/a which is not equal to an eigenvalue*Ay*e-r/a, although I am not sure if this is the correct justification.

    Any help is much appreciated, thank you.
     
    Last edited: May 3, 2014
  2. jcsd
  3. May 3, 2014 #2
    Seems right to me.
     
  4. May 3, 2014 #3
    I'm also stuck on how to convert the wavefunctions into spherical polar coordinates, having tried to use the formula for spherical polar coordinates, I notice that the wavefunctions do not depend on the variables phi and theta, so I am unsure on how to approach the first problem.
     
  5. May 3, 2014 #4

    BvU

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    Hello fan, and welcome to PF.

    You get 0 for ##\Psi_0## because it is symmetric under rotations around the z axis: there is no angular momentum. So I think you are right. It is an eigenfunction.

    For ##\Psi_1## you basically want to show that ##{\bf L}_z \left( \Psi_1 \right ) = L_z\, \Psi_1 ## with ##{\bf L}_z## the operator and ##L_z## an eigenvalue (i.e. a number, possibly complex), has no solutions.

    You may assume |A| is not equal to zero (because of the normalization) so the eigenvalue zero is already excluded. So if you multiply the right hand side from the left by ##\Psi_1^*## and integrate, that yields ##L_z##, which is nonzero.

    Do the same thing with the lefthand side and show that it does give zero. That way you prove that the assumption that ##\Psi_1## is an eigenfunction leads to a false equation.
     
  6. May 3, 2014 #5

    BvU

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    With ##z = r\cos\theta## you sure have a theta dependence!
     
  7. May 3, 2014 #6

    BvU

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    But the exercise wants you to write out the z-component of the operator cross product ##{\bf r} \times {\bf p}##, not the wave functions!
    [edit] sorry, it wants you to do both. In the first part the Lz, later on the ##\Psi##
     
  8. May 3, 2014 #7
    That has really cleared things up, thank you!
     
    Last edited: May 3, 2014
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