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QM question, angular momentum operator and eigen functions

  1. Apr 10, 2010 #1
    For the operator L(z) = -i[STRIKE]h[/STRIKE][d/d(phi)]

    phi = azimuthal angle

    1) write the general form of the eigenfunctions and the eigenvalues.
    2) a particle has azimuthal wave function PHI = A*cos(phi)
    what are the possible results of a measurement of the observable L(z) and what is the probability of each.

    this is a past paper qu im doing for revision
    i think 1) is = A*R(r)*sin(theta)exp[i*phi] and the eigenvalue is [STRIKE]h[/STRIKE]
     
  2. jcsd
  3. Apr 10, 2010 #2

    gabbagabbahey

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    That's certainly an eigenfunction, but it isn't the most general form. What is the eigenvalue equation (expanded in the position basis) for the operator [itex]L_z[/itex]? What do you get if you assume that the eigenfunctions are separable?
     
  4. Apr 10, 2010 #3
    The question only ask for the eigenfunction of the operator L(z)
    So you should not write out one hydrogen wavefunction which may count as WRONG answer!

    Solve for [tex]L_{z}\Phi(\phi)=m\Phi(\phi)[/tex] and you will get the answer. Think about it
     
  5. Apr 10, 2010 #4
    i still dont really understand what i need to do...
    when you say [tex]
    L_{z}\Phi(\phi)=m\Phi(\phi)
    [/tex] is that you saying m is the eigenvalue? i thought that it was hbar.

    im just confused cause we didnt try to find the eigenfunctions in lectures.
     
  6. Apr 10, 2010 #5
    how about this?

    L(z)[tex]\Phi[/tex] = m[tex]\hbar[/tex]*exp[im[tex]\phi[/tex]]
    -i[tex]\hbar[/tex]*d/d[tex]\phi[/tex]*[tex]\Phi[/tex] = m[tex]\hbar[/tex]*exp[im[tex]\phi[/tex]]

    so if [tex]\phi[/tex] = Aexp[im[tex]\phi[/tex]]
    normalised A = 1/[tex]\sqrt{}2\pi[/tex]
     
    Last edited: Apr 10, 2010
  7. Apr 10, 2010 #6

    gabbagabbahey

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    That doesn't prove that [itex]Ae^{im\phi}[/itex] is the only eigenfunction. Assume that the state [itex]\psi(r,\theta,\phi)[/itex] is an eigenfunction of the operator [itex]L_z[/itex]. Furthermore, assume that [itex]\psi(r,\theta,\phi)[/itex] is separable (i.e. [itex]\psi(r,\theta,\phi)=f(r)g(\theta)h(\phi)[/itex]). Now apply the operator [itex]L_z[/itex] to that eigenfunction and solve the differential equation you get.
     
  8. Apr 10, 2010 #7
    okay i solved the partial diff. eqn. and got th same value for PHI,
     
  9. Apr 11, 2010 #8
    does this eigen value answer the question?
     
  10. Apr 11, 2010 #9

    gabbagabbahey

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    I'm not sure exactly what your final answer is, you haven't posted it.
     
  11. Apr 11, 2010 #10
    i got the eigen function being

    [tex]\frac{1}{\sqrt{2\pi}}[/tex]eim[tex]\phi[/tex]
     
  12. Apr 11, 2010 #11

    gabbagabbahey

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    Then no, that is not the general form of the eigenfunction. Your eigenfunction can also have some radial or polar angle dependence, can it not?
     
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