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Commutation of (L^2)op and (Lz)op

  1. Nov 19, 2009 #1
    1. The problem statement, all variables and given/known data

    It has been shown that the operators (Lx)op and (Ly)op do not commute but satisfy the following equation:

    (Lx)op(Ly)op - (Ly)op(Lx)op = i(hbar)(Lz)op

    (a) Use this relation and the two similar equations obtained by cycling the coordinate labels to show that (L2)op(Lz)op = (Lz)op(L2)op, that is, these two operators commute. [Hint: You do not need to introduce the differential formulas for the operators. Use the fact that (AB)C = A(BC) where A, B, and C are operators]


    The question continues, but this is the part I am having trouble with.



    2. Relevant equations

    Relevant equations are included in the question

    3. The attempt at a solution

    My attempted solution is attached (Scan0004.jpg). I work it out, but I'm going wrong somewhere as I'm finding that they do not commute as they should.



    Thank you in advance for any help!

    Andrew
     

    Attached Files:

  2. jcsd
  3. Nov 19, 2009 #2

    Ben Niehoff

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    Science Advisor
    Gold Member

    This is a rather convoluted way to do this, so I'm not going to try to pick apart all of it. But I can tell you that you have a significant error in the very first three lines. You have attempted to set

    [tex]L_z = L_x L_y - L_y L_x[/tex]

    when the correct expression is

    [tex]L_z = \frac{1}{i \hbar} (L_x L_y - L_y L_x)[/tex]

    Now, if you like, you can set [itex]\hbar = 1[/itex], as this amounts to just choosing a special system of units. However, you cannot set [itex]i = 1[/itex]. The imaginary unit i is crucial to getting the cancellation you need.
     
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