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Disconnected diagram Feynman rules momentum space

  1. Aug 6, 2008 #1

    I wonder if the Feynman rules in momentum space can also be applied to disconnected diagrams. So aussume I have a disconnected Feynman diagram 1 , i.e. the product of two connected Feynman diagrams 2 and 3.
    I can translate my diagrams with the position space Feynman rules to explicit mathematical expressions:

    D=C_2 * C_3

    I can compute the Fourier transforms F(C_2) and F(C_3) by using the Feynman rules in momentum space.

    Now if the FR in moentum space would be applicable for disconnected diagrams, this would mean

    This means the inverse Fourier transform yields
    D= F^-1 [ F(C_2)*F(C_3) ]

    => C_2 * C_3 = F^-1 [ F(C_2)*F(C_3) ]

    How can that be?? This rule is not generally true for Fourier transformations...

    Best regards, Martin
  2. jcsd
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