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Possible outcomes of angular momentum state

  1. May 27, 2012 #1
    1. The problem statement, all variables and given/known data
    A particle is in the state [itex]\psi = R(r)(\sqrt{\frac{1}{3}}Y_{11} + i\sqrt{\frac{2}{3}}Y_{10})[/itex]]. If a measurement of the x component of angular momentum is made, what are the possible outcomes and what are the probabilites of each?


    2. Relevant equations
    [tex]L_{\pm}Y_{lm}=\sqrt{l(l+1)-m(m \pm 1)}Y_{l(m\pm 1)}[/tex]
    [tex]L_x = \frac{1}{2}(L_+ + L_-)[/tex]
    [tex]\psi = \sum \alpha_{lm} Y_{lm}[/tex]


    3. The attempt at a solution
    I understand how to get the expectation value of [itex]L_x[/itex] for the entire wavefunction through the inner product [itex]\langle \psi |L_x| \psi \rangle[/itex] and how to get the Fourier coefficients for the state probabilities, but I don't see how to get the "possible outcomes". Expectation values of individual eigenstates [itex]\langle Y_{lm} |L_x| L_{lm} \rangle[/itex] are always equal to 0, so I don't see how you can measure any outcome but 0 for definite eigenstates. Shouldn't the only outcome be the expectation value of the entire wavefunction?
     
    Last edited: May 27, 2012
  2. jcsd
  3. May 27, 2012 #2

    vela

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    You need to expand the state in terms of the eigenfunctions of Lx.
     
  4. May 27, 2012 #3
    Could you explain this a bit more? I was under the impression that there were no [itex]L_x Y_{lm}[/itex] eigenstates because the effect of the operator on the spherical harmonics is to raise and lower the "m" index, a la [itex]L_x Y_{lm} = \frac{1}{2}(L_+ +L_-)Y_{lm} = \frac{\hbar}{2}(\sqrt{l(l+1)-m(m+1)}Y_{l(m+1)}+\sqrt{l(l+1)-m(m-1)}Y_{l(m-1)})[/itex], changing the basis vectors. Taking the expectation value of the entire wavefunction only leads to 0 as well, so I'm starting to think that that's it.
     
  5. May 27, 2012 #4

    vela

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    That's right. The spherical harmonics are not eigenstates. You have to find linear combinations which are eigenstates of Lx.
     
  6. May 27, 2012 #5
    Is it valid to just take x as equivalent to the z direction (since the coordinates are arbitrary anyways) and define [itex]L_x Y_{lm} = m \hbar Y_{lm}[/itex] and just solve it that way?
     
  7. May 28, 2012 #6

    vela

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    No, it's not. The wave function as written implies a coordinate system, and the problem is asking you questions with respect to this coordinate system.

    Find the matrix representation of Lx, and then find the eigenvalues and eigenvectors of that matrix.
     
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