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Angular momentum operators on matrix form

  1. Jun 12, 2013 #1
    1. The problem statement, all variables and given/known data
    Hi.

    I'm given a 3-dimensional subspace [itex]H[/itex] that is made up of the states [itex]|1,-1\rangle[/itex], [itex]|1,0\rangle[/itex] and [itex]|1,1\rangle[/itex] with the states defined as [itex]|l,m\rangle[/itex] and [itex]l=1[/itex] as you can see.

    The usual operator relations for [itex]L_{z}[/itex] and [itex]L^{2}[/itex] applies, and also:
    [tex]L_{+} = L_{x}+iL_{y}[/tex]
    [tex]L_{-} = L_{x}-iL_{y}[/tex]

    Then I'm told to express the operators [itex]L_{+}[/itex] and [itex]L_{-}[/itex] in [itex]H[/itex].

    The answer for [itex]L_{+}[/itex] is supposed to be:
    [tex]{{L}_{+}}=\sqrt{2}\hbar \left[ \begin{matrix}
    0 & 0 & 0 \\
    1 & 0 & 0 \\
    0 & 1 & 0 \\
    \end{matrix} \right]
    [/tex]
    And the transposed for [itex]L_{-}[/itex]

    But I'm really not sure how that is found.

    2. Relevant equations


    3. The attempt at a solution
    The [itex]\sqrt{2}\hbar[/itex] probably comes from the fact that:
    [tex]{{L}_{\pm }}\left| l,m \right\rangle =\sqrt{l\left( l+1 \right)-m\left( m\pm 1 \right)}\hbar \left| l,m \right\rangle[/tex]

    But I can't figure out why the entries in the matrix is like that.
    My first thought was that it should be diagonal, as it was made up of the 3 bases, but as you can see, it is not diagonal.


    So I was hoping someone could explain what I'm missing out ?


    Thanks in advance.
     
  2. jcsd
  3. Jun 13, 2013 #2
    I don't think that is right; it should change the value of m, and also m may not be greater than [itex]\ell[/itex].

    This acts on column vector that has the first row for m = -1, the second row for m =0, and the third row for m= 1.

    The first row of the matrix is all zero because no matter what, when [itex]L_{+}[/itex] acts on a state, there is never anything remaining in the m = -1 position. The second row takes whatever was in the m = -1 position and moves it to the m=0 position (multiplied by a constant). The third row of the the matrix takes whatever is in the m=0 and moves it to m =1 position, multiplied by a constant.

    This was a conceptual question and not a homework exercise, right?
     
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