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Angular Momentum, L_x eigenvalues and eigenfunctions

  1. Jul 18, 2006 #1
    This is a very simple question, but I can't seem to get it right, there's probably something silly that I'm missing here. Here's the question:

    I have A system in the l=1 state, and I have L_z|\ket{lm} = \hbar m\ket{lm}and L^2 \ket{lm} = \hbar^2 l(l+1)\ket{lm}

    I need to find the eigenvalues and eigenvectors of L_x and L_y using the eigenvectors of L_z and L^2, assuming they are \ket{1,0}, \ket{1,-1} and \ket{1,1}.

    I use that L_x = \half (L_{+}+L_{-}) and get this:

    L_x(A\ket{1,0}+B\ket{1,1}+C{\ket{1,-1}) = \half \hbar \sqrt{2}( A\ket{1,-1}+A\ket{1,1} + B\ket{1,0} + C\ket{1,0})

    Ignoring the \half \hbar \sqrt{2} constant, I equate and get:
    A = B, A=C, and B+C=A. Which is obviously wrong... so what am I missing here? I feel like I'm missing a 1/2 or 2 somewhere....

    Thanks in advance, this question has been annoying me for ages...
    Last edited: Jul 18, 2006
  2. jcsd
  3. Jul 22, 2006 #2
    You're missing one very important thing: L^2 can have simultaneous eigenstates with one and only one component of angular momentum. Finding eigenvectors of Lx and Ly in this basis is a waste of time.

    To help see this, remember what you're doing when you're equating sides. You assume that each of the vectors are orthogonal. Be a bit more rigorous about it, apply <1,1| to both sides. You get L1*B = constants*A (assuming Lx |1,1> = L1|1,1>). You can repeat the same procedure for the others, and if worked out I believe you'd get equations similar to the ones you found above, for which no non-trivial solutions exist. The assumption that those vectors are eigenvectors is flawed, and thus so are the steps after it.

    Incidentally, where did you get that [itex] \sqrt{2} [/itex] on the right?
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