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Angular momentum operator algebra

  1. Apr 12, 2015 #1

    Maylis

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    Gold Member

    1. The problem statement, all variables and given/known data
    upload_2015-4-12_18-18-23.png

    2. Relevant equations


    3. The attempt at a solution
    This whole thing about angular momentum has me totally confused and stumped, but I am trying this problem given in a youtube video lecture I watched.

    I know of this equation
    ##L^{2} = L_{\pm}L_{\mp} + L_{z}^{2} \mp \hbar L_{z}##
    ##L^{2}f = \lambda f##
    ##L_{z}f = \mu f##

    In the case ##L^{2}f = 2 \hbar^{2}f##, the eigenvalue ##\lambda = 2 \hbar^{2}##

    So I expand
    $$L^{2}f = (L_{-}L_{+} + L_{z}^{2} + \hbar L_{z})f$$
    But, I don't know what ##L_{z}## should be. Also, how do I know if I should pick ##L_{-}L_{+}## or ##L_{+}L_{-}##?
     

    Attached Files:

  2. jcsd
  3. Apr 12, 2015 #2

    stevendaryl

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    I'm a little puzzled by the examples, because normally [itex]L^2[/itex] has eigenvalues [itex]\hbar^2 l(l+1)[/itex], where [itex]l[/itex] is a nonnegative integer. For your second example, [itex]l[/itex] would have to be fractional. It's possible for the total angular momentum (which includes both spin angular momentum and orbital angular momentum) to be fractional, but usually [itex]L[/itex] refers to orbital angular momentum.

    But in any case, it's overkill to consider raising and lowering operators. The fact that's relevant is that if [itex]L^2[/itex] has the value [itex]\hbar^2 l(l+1)[/itex], then [itex]L_z[/itex] can take on any of the following values:
    • [itex]l[/itex]
    • [itex]l-1[/itex]
    • [itex]l-2[/itex]
    • ...
    • [itex]-l[/itex]
     
  4. Apr 18, 2015 #3

    Maylis

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    Gold Member

    Hi stevedaryl,

    sorry for my late response. I was able to go to my prof's office and get some help on this. Using your post alone, I was not equipped with the necessary understanding to answer this question. I needed to understand the sphere with cones inside to get some intuition for what ##l## and ##m## mean. After rereading your response, it makes sense what you were saying, I just needed to know about ##m##
     
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