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QFT Klein Gordon Theory, momentum commutator computation

  1. Jul 19, 2017 #1
    1. The problem statement, all variables and given/known data

    momentum[].png
    2. Relevant equations


    3. The attempt at a solution

    I think I understand part b) . The idea is to move the operator that annihilates to the RHS via the commutator relation.

    However I can't seem to get part a.

    I have:

    ## [ P^u, P^v]= \int \int \frac{1}{(2\pi)^6} d^3k d^3 k' (a^{\dagger}(k)a(k)a^{\dagger}(k')a(k') k^u k'^v - a^{\dagger}(k')a(k')a^{\dagger}(k)a(k) k'^vk^u)##

    And I can't even think of a first move here, and I don't see anyway to include the commutator expression?

    Many thanks in advance
     
  2. jcsd
  3. Jul 19, 2017 #2

    Dr Transport

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    expand the commutator and insert it into the integral
     
  4. Jul 20, 2017 #3
    which commutator? i dont understand
     
  5. Jul 20, 2017 #4

    Dr Transport

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    annihilation commutation relation...
     
  6. Jul 20, 2017 #5
    As I said in my intiial post I can't see how you can factorise to get where to insert this commutator.
    I.e i can't spot a factorisation of the form ## f(a^{\dagger}(k'),a(k')) [a(k),a^{\dagger}(k)] ##

    Because ##a^{\dagger}(k'),a(k')## is on the rhs in the first term, but on the lhs in the second term...
     
  7. Jul 20, 2017 #6

    Dr Transport

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    expand the commutation relation and substitute in for [itex] a(k)a^\dagger(k')[/itex] in the first expression and similarly for the right hand equation, the delta function will integrate out and you can simplify from there.
     
  8. Jul 20, 2017 #7
    ahh ofc,thanks
     
  9. Jul 20, 2017 #8
    Nope actually I'm still struggling

    Doing this I have:

    ## \int \int \frac{1}{(2\pi)^6} d^3k d^3 k' k'^vk^u (a^{\dagger}(k)^{\dagger}(k')a(k)a(k') + a^{\dagger}(k)(2\pi)^3\delta^3(k-k')a(k') - a^{\dagger}(k')^{\dagger}(k)a(k')a(k) - a^{\dagger}(k')(2\pi)^3\delta^3(k-k')a(k))##

    ##= \int \int \frac{1}{(2\pi)^6} d^3k d^3 k' k'^vk^u (a^{\dagger}(k)^{\dagger}(k')a(k) a(k')) +\int \int \frac{1}{(2\pi)^6} d^3k d^3 k' k'^vk^u (a^{\dagger}(k)(2\pi)^3\delta^3(k-k')a(k')) - \int \int \frac{1}{(2\pi)^6} d^3k d^3 k' k'^vk^u (a^{\dagger}(k')^{\dagger}(k)a(k')a(k) ) - \int \int \frac{1}{(2\pi)^6} d^3k d^3 k' k'^vk^u (a^{\dagger}(k')(2\pi)^3\delta^3(k-k')a(k))##

    ##= \int \int \frac{1}{(2\pi)^6} d^3k d^3 k' k'^vk^u (a^{\dagger}(k)^{\dagger}(k')a(k) a(k')) + \int \int \frac{1}{(2\pi)^6} d^3k d^3 k' k'^vk^u (a^{\dagger}(k')^{\dagger}(k)a(k')a(k) )##

    so the two delta terms have cancelled but I dont know what to do with these two terms.

    Thanks
     
  10. Jul 22, 2017 #9
    I'm aware that the two annihilator operators and the two creation operators commute and so i suspect the answer is we can just swap these and then the two expressions are identical and cancel.

    However with the operator being swapped having another operator on both the left and right side , I am confused whether this is valid because , if we were only acting on a bra, then I agree its fine to swap the annihilator operators, and if only acting on a ket to swap the creation operators. However I believe we should make it general, so acting on a bra and a ket

    I can't get my head around, to me if you are swapping the operator in the middle of a string of three operators, you are not just assuming commutation between two of them, but all three... ? thanks
     
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