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Angular Momentum Raising Operator

  1. Nov 27, 2007 #1
    1. The problem statement, all variables and given/known data
    In Problem 4.3 you showed that

    [tex]Y^{1}_{2}(\theta , \phi) = -\sqrt{15/8\pi} sin\theta cos\theta e^{i\phi}[/tex]

    Apply the raising operator to find [tex]Y^{2}_{2}(\theta , \phi)[/tex]. Use Equation 4.121 to get the normalization.

    2. Relevant equations

    [Eq. 4.121] [tex]A^{m}_{l} = \hbar \sqrt{l(l + 1) - m(m \pm 1)} = \hbar \sqrt{(l \mp m)(l \pm m +1)}[/tex].

    3. The attempt at a solution

    So, I think my problem, in part, is that I don't know what they mean when they say "Use Equation 4.121 to get the normalization." So, with that in mind I did try something; but it seems too simple:

    [tex]L_{+} Y^{1}_{2}(\theta , \phi) = \epsilon \sqrt{\frac{(2l + 1)(l - |(m+1)|)\fact}{4\pi(l + |(m+1)|)\fact}} e^{i(m+1)\phi} P^{(m+1)}_{l}(x)[/tex]

    So, everywhere I have just [tex]m[/tex], I add one to it, which really just gives me the formula for [tex]Y^{2}_{2}[/tex]. Then solving it is simple. So I must not be understanding the question properly. And I don't see what equation 4.121 has to do with anything. I'm used to noramlization being something like:

    [tex]1 = \int^{\infty}_{-\infty}|\psi|^{2} dx[/tex]
    Last edited: Nov 27, 2007
  2. jcsd
  3. Nov 27, 2007 #2


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    First step:

    [itex] L_+ Y^1_2 [/itex] is equal to what? It's equal to some constant times [itex] Y_2^2 [/itex], right? What is that constant?
  4. Nov 27, 2007 #3
    You mean the whole:

    [tex]\hat{Q}f(x) = qf(x)[/tex]

    If that's what you mean, then I guess it would just be:

    [tex]L_{+}Y^{1}_{2} = lY^{2}_{2}[/tex]

    ??? Is that right?
    Earlier on the chapter discusses

    [tex]L^{2}f_{t} = \lambda f_{t}[/tex]

    But it doesn't apply the [tex]L_{+}[/tex] operator.
    Last edited: Nov 27, 2007
  5. Nov 27, 2007 #4


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    Of course, Y^1_2 is NOT an eigenstate of L_+ so you can't use the eigenvalue equation.

    What I mean is : use equation 4.121 to figure out the constant! That's what 4.121 is for! To figure out the constant generated when applying the raising or lowering operators!
  6. Nov 27, 2007 #5
    Ah, ok. I understand (I hope).

    So, with Eq. 4.121 goes to:

    [tex]A^{2}_{2} = \hbar \sqrt{2(2+1) - 2(2\mp 1}[/tex] ---->

    [tex]A^{2}_{2} = \hbar \sqrt{6 - 2} = 2\hbar[/tex]

    The other leads to zero, so I guess I can throw that out.

    So, if I understood correctly the first step is done?

    [tex]L_{+} Y^{1}_{2} = 2\hbar Y^{2}_{2}[/tex] ?

    Or do I just say:

    [tex]\frac{L_{+} Y^{1}_{2}}{2\hbar} = Y^{2}_{2}[/tex]
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