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Why eigenvalues of L_x^2 and L_z^2 identical?

  1. Oct 6, 2013 #1
    1. The problem statement, all variables and given/known data

    Calculate the eigenvalues of the [itex]L_x^2[/itex] matrix.
    Calculate the eigenvalues of the [itex]L_z^2[/itex] matrix.
    Compare these and comment on the result.

    2. Relevant equations

    [itex]L_x=\frac{1}{2}(L_+ + L_- )[/itex]

    3. The attempt at a solution

    I have derived eigenvalues for each: [itex]0[/itex] and [itex]\hbar^2[/itex] for both [itex]L_x^2[/itex] and [itex]L_z^2[/itex]. But why are they identical? I'm finding it difficult in qualitatively explaining why the eigenvalues are the same for both.
     
  2. jcsd
  3. Oct 6, 2013 #2
    Because the God of Physics does not care which direction you call the [itex]x[/itex]-direction and which direction you call the [itex]z[/itex]-direction.
     
  4. Oct 6, 2013 #3
    Sorry I forgot to mention this is for [itex]l=1[/itex].

    Okay, but I used [itex]L_z[/itex] eigenvalues of [itex]m\hbar[/itex], where [itex]m=-1,0,1[/itex] in this case, and used [itex]L_x=\frac{1}{2}(L_+ + L_- )[/itex]. I have called the z component the one in which is certain, so how can the x component squared in this case have the same eigenvalues as the z component squared?
     
  5. Oct 6, 2013 #4

    vela

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    The operators will have the same eigenvalues for the reason Oxvillian said, but that's not saying anything about the state a particle is in. If a particle is in an eigenstate of ##\hat{L}_z##, it's not in an eigenstate of ##\hat{L}_x##.
     
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