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Tricky commutator

  1. Oct 29, 2009 #1
    1. The problem statement, all variables and given/known data

    Part of a much larger problem dealing with the heisenberg picture. I am not remembering how to start evaluating the following commutator:

    [tex]\left [ a_k(t),\left(\sum_{k,\ell}a_k^\dagger <k|h|\ell>a_\ell\right)\right][/tex]

    2. Relevant equations

    See (a)

    3. The attempt at a solution

    Just need some help getting started on this one, after that I'm good. What I do know is that when you do the commutator you cannot just lump the first term ([itex]a_k(t)[/itex]) into the sum.. any hints on how to go about breaking this down? Halp!

    Thanks

    IHateMayonnaise
     
  2. jcsd
  3. Oct 29, 2009 #2

    gabbagabbahey

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    To start with, the commutator is distributive (i.e. [Itex][A,B+C]=[A,B]+[A,C][/itex] ), so you take the sum out front. Also, like any inner product, [itex]\langle k|H|l\rangle[/itex] will be a scalar, and so can be pulled outside the commutator....
     
  4. Oct 29, 2009 #3
    Thanks for the reply!

    So you said that because the commutator is distributive I can take the sum out front. Do you mean I can do this:

    [tex]
    \left [ a_k(t),\left(\sum_{k,\ell}a_k^\dagger a_\ell\right)\right]=\sum_{k,\ell}<k|h|\ell> \left [ a_k(t),a_k^\dagger a_\ell\right]
    [/tex]
     
  5. Oct 29, 2009 #4

    gabbagabbahey

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    Yes, exactly...now keep going....simplify the commutator [itex]\left [ a_k(t),a_k^\dagger a_\ell\right][/itex]
     
  6. Oct 29, 2009 #5
    I think I got it from here, thank you so much for your help!
     
  7. Oct 29, 2009 #6

    gabbagabbahey

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    Wait, what does the index [itex]k[/itex] represent in the [itex]a_k(t)[/itex]?....If it is not being summed over you should use a different letter for the dummy index in your sum.

    [tex]\left [ a_k(t),\left(\sum_{n,\ell} a_n^\dagger a_\ell\right)\right]=\sum_{n,\ell} \left [ a_k(t),a_n^\dagger a_\ell\right][/tex]
     
    Last edited: Oct 29, 2009
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