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Does this commutator commute?

  1. Jan 3, 2010 #1
    If we define:

    [tex]A_{j}=\omega \hat{x}_{j}+i \hat{p}_{j}[/tex]


    [tex]A^{+}_{j}=\omega \hat{x}_{j}-i \hat{p}_{j}[/tex]

    Would it be true to say:

    [tex][A_k , (A^{+}_{i}+A_i)(A^{+}_{j}-A_j)]=0[/tex]

    My reasoning is that, because

    [tex][\hat{x}_{j}, \hat{p}_{i}]=0[/tex]

    the the ordering of the contents of commutation bracket shouldn't matter (as [tex]\hat{x}_{j} \hat{p}_{i}=\hat{p}_{i}\hat{x}_{j}[/tex]), so we simply get that:

    [tex][A_k , (A^{+}_{i}+A_i)(A^{+}_{j}-A_j)]=A_{k}(A^{+}_{i}+A_i)(A^{+}_{j}-A_j)-(A^{+}_{i}+A_i)(A^{+}_{j}-A_j)A_{k}= A_{k}(A^{+}_{i}+A_i)(A^{+}_{j}-A_j)-A_{k}(A^{+}_{i}+A_i)(A^{+}_{j}-A_j)=0[/tex]

    This seems obvious to me, but it would make a 10 mark exam question too easy! Would be grateful if someone could confirm whether this is right or not.

  2. jcsd
  3. Jan 3, 2010 #2


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    What about the case where i=j?
  4. Jan 3, 2010 #3
    The question, specifically, is to find:

    [tex][A_k , \hat{L}_{ij}]=[A_k, (A^{+}_{i}+A_i)(A^{+}_{j}-A_j)][/tex]

    So I am not sure we need to consider the case where i=j.

    Ofcourse it would get *really* messy if any of the subscripts are the same. But if i,j and k are not the same, the commutator would be zero, right?
  5. Jan 3, 2010 #4


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    Sounds like you do. You said the question specifically asks you to find that commutator -- not to find that commutator in the special case where i,j,k are all distinct!

    It might not be as bad as you think.

  6. Jan 3, 2010 #5

    [tex]\hat L_{ij}:=\hat{x}_{i}\hat{p}_{j} - \hat{x}_{j}\hat{p}_{i}

    So if i=j

    [tex]\hat L_{ii}:=\hat{x}_{i}\hat{p}_{i} - \hat{x}_{i}\hat{p}_{i}=0[/tex]

    So the commutator:

    [A^{+}_{j},\hat L_{ij}]

    would also be zero in this case as well

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