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Eigenvalues of L operators

  1. Nov 14, 2007 #1
    Hi

    1. The problem statement, all variables and given/known data

    We're given the operators Lx, Ly and Lz in matrix form and asked to show that they have the correct eigenvalues for l=1. Obviously no problem determining the values and Lz comes out right, however we've never actually seen the e.v.s for Lx and Ly.

    2. Relevant equations

    I tried finding the eigenvalues in the wave formulation for Lx and Ly using the operators in polar spherical co-ords:

    [tex] L_{x} = i\hbar(sin\phi\frac{d}{d\theta} + cot \theta cos \phi \frac{d}{d\phi}) [/tex]
    [tex] L_{y} = i\hbar(-cos\phi\frac{d}{d\theta} + cot \theta sin \phi \frac{d}{d\phi}) [/tex]

    Dealing with l = 1, so my spherical harmonics are in one of two forms:

    [tex] Y_{1,0} = (\frac{3}{4\pi})^{1/2}cos\theta [/tex]
    [tex] Y_{1,\pm1} = \mp(\frac{3}{8\pi})^{1/2}sin\theta exp(\pm i \phi) [/tex]

    3. The attempt at a solution

    Well, applying the operators to the wfs gives me nothing like an eigenvalue. For instance:

    [tex] L_{x}Y_{1,0} = -i \hbar sin \phi tan \theta Y_{1,0} [/tex]

    Anyone see where I'm going wrong, or just happen to know offhand the eigenvalues for Lx and Ly?

    Cheers,

    El Hombre
     
    Last edited: Nov 15, 2007
  2. jcsd
  3. Nov 14, 2007 #2

    nrqed

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    It's not wrong because the Y_lm are simply NOT eigenstates of L_x and L_y! The eigenstates of L_x and L_y are linear combinations of the Y_lm. But if the operators were given to you as matrices in the first place, why do you use differential operators? Simply find the eigenvalues of the matrices you were given !
     
  4. Nov 15, 2007 #3
    Hi

    First off, those LaTex codes didn't quite come out the way I intended. Second, like I said I had no problem finding the eigenvalues of the matrix operators. The problem is in verifying them. I get the same eigenvalues for each matrix, i.e. mh, with the three possible values of m for l=1. The problem is I have nothing to check these against except common sense.
     
  5. Nov 15, 2007 #4

    nrqed

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    Oh, so you want to double check your calculation in position space!

    Is your L_z matrix diagonal? Then if you have the eigenstates of L_x, say, as a column vector, just reexpress that eigenstate in terms of the Y_lm and apply the differential operator expression of L_x on it as a check.

    I mean, if L_z is -1,0,1 on the diagonal then it means that the column vector with 1 at the top represents Y_(l=1, m=-1), the column vector with one in the middle represents Y(l=1, m=0) etc.
     
  6. Nov 15, 2007 #5
    Hi nrqed

    Yes, Lz is diagonal. In fact, I'll struggle with LaTex and tell you the matrices:

    [tex] L_{x} = \frac{\hbar}{\sqrt{2}} \left(\begin{array}{ccc}0&1&0\\1&0&1\\0&1&0\end{array}\right) [/tex]
    [tex] L_{y} = \frac{\hbar}{\sqrt{2}} \left(\begin{array}{ccc}0&i&0\\-i&0&i\\0&-i&0\end{array}\right) [/tex]
    [tex] L_{z} = \hbar \left(\begin{array}{ccc}1&0&0\\0&0&0\\0&0&-1\end{array}\right) [/tex]

    This gives my eigenvectors for Lx as:

    [tex] \frac{1}{2} \left(\begin{array}{c}1&\sqrt{2}&1\end{array}\right) \frac{1}{\sqrt{2}} \left(\begin{array}{c}1&0&-1\end{array}\right) \frac{1}{2} \left(\begin{array}{c}1&-\sqrt{2}&1\end{array}\right) [/tex]

    So if I get this right, the first eigenstate on the left corresponds to:

    [tex] \frac{1}{2} Y_{11} + \sqrt{2} Y_{10} + \frac{1}{2} Y_{1,-1} [/tex].

    Is that right?

    Cheers,

    El Hombre
     
  7. Nov 15, 2007 #6

    nrqed

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    Hi. Thanks for posting them, I did not feel like typing them all!

    You are riht except for one tiny mistake: the coefficient of the Y_10 is [itex] \frac{1}{\sqrt{2}} [/itex], not sqrt(2) .

    Now, if you apply L_x written as a differential operator on that expression (that will be long!) you should find that it's an eigenstate with the expected eigenvalue. But if you want to just make one double check, you should use the second eigenstate of L_x which at least has no Y_10 term!

    It's long to check this way but that's the way to do it if you really don't mind all the algebra. The fact that you got the correct eigenvalues for the matrices L_x and L_Y is already a convincing argument that you di dit right. If you want to be sure that you got the correct corresponding eigenstates, you coudl simply apply the matrices on your answer and check that it works. That would be much quicker than using the representations as differential operators.
     
  8. Nov 16, 2007 #7
    Hi

    Yeah, that's my latexing again. If you click on it you'll see I did put a \frac in there but to no avail. I don't know why they come out like that but I just don't spend enough time on here to get to grips with the grammar.

    Cheers. Yes, that does seem overcomplicated, to the point where I'm wondering if it can possibly have been what was intended. But even if it isn't I am at least wiser now. Just not about latex.

    Thanks again for all your help.

    El Hombre
     
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