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QFT Wicks theorem contraction -- different fields terms of propagation

  1. Nov 30, 2016 #1
    1. The problem statement, all variables and given/known data

    I am trying to express ##T(\phi(x1)\Phi(x2)\phi(x3)\Phi(x4)\Phi(x5)\Phi(x6))## in terms of the Feynman propagators ##G_F^{\phi}(x-y)## and ##G_F^{\Phi}(x-y)##

    where ##G_F^{\phi}(x-y) =\int \frac{d^{4}k}{(2\pi)^{4}}e^{ik(x-y)} \frac{ih}{-k.k - m^2 -i\epsilon} ##
    and ##G_F^{\Phi}(x-y) =\int \frac{d^{4}k}{(2\pi)^{4}}e^{ik(x-y)} \frac{ih}{-k.k - M^2 -i\epsilon} ##

    2. Relevant equations

    ##<0|T(\phi(x1)\Phi(x2)\phi(x3)\phi(x4)\phi(x5)\phi(x6)) |0> = : non-fully contracted terms : + fully contracted terms = fully contracted terms ##

    where ##T## is the time ordering operator

    ##<0| ## being the vacuum state,

    since non-fully contracted terms, i.e. where every field is not involved in a contraction will vanish.

    3. The attempt at a solution

    My question is how/ can you contract over two different fields?

    I am fine with contractions over two of the same field, but can't find any notes/examples on what to do in the case of two different fields - here ##\phi(x)## and ##\Phi(x)##

    In particular, if you can't, the question ask to express the final answer in terms of ##G_F^{\phi}(x-y)## and ##G_F^{\Phi}(x-y)##, and I can't see how you could write the contraction over two different fields in terms of this.

    I know ##G^{\phi} (x1-x3) = contraction of the fields \phi(x1) , \phi(x3)##

    and similarly that ##G^{\Phi} (x4-x5) = contraction of the fields \Phi(x4) , \Phi(x5)##

    but no idea how you would write e.g contraction over the fields ## \phi(x1), \Phi(x5)## in terms of ##G_F^{\phi}(x-y)## and ##G_F^{\Phi}(x-y)##?

    Can you contract over two different fields?

    Help really appreciated,
    (Or a point to some notes on this)

    Thanks
     
  2. jcsd
  3. Nov 30, 2016 #2

    Orodruin

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    No.
     
  4. Nov 30, 2016 #3
    many thanks.

    a possible explanation or a point towards one?
     
  5. Nov 30, 2016 #4

    Orodruin

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    I can't type much (on my phone). Go back to how contractions were constructed, i.e., what does a contraction entail?
     
  6. Nov 30, 2016 #5
    oh right, no worries at all, thanks for your help,
    so i looked at a derivation of time ordering of a field by splitting a field ##\phi## into its annihilation and creation operator components, and then get normal ordered + commutator terms, the commutator terms being the contraction...oh so commutators of different fields is zero- is this why?
     
  7. Nov 30, 2016 #6

    Orodruin

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    Right.
     
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