How Do You Approach Angular Momentum Operator Algebra in Quantum Mechanics?

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Homework Statement


upload_2015-4-12_18-18-23.png


Homework Equations

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_{-}##?
 

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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]
 
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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##