Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Does this commutator commute?

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

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

    and

    [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.

    Thanks.
     
  2. jcsd
  3. Jan 3, 2010 #2

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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.

    Yes.
     
  6. Jan 3, 2010 #5
    Well

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

    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:

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

    would also be zero in this case as well

    :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Does this commutator commute?
  1. Commutation integral (Replies: 2)

Loading...