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Photon self energy

  1. Mar 30, 2013 #1
    I am a beginner to QFT and I try to plot the Feynman diagram for the photon self-energy. Following Mandl-Shaw book (page 109 Eq. 7.22)
    [itex]
    \int d^4x_1 d^4x_2 (-1)\mathrm{Tr}(iS_F(x_2-x_1)\gamma A^-(x_1) iS_F(x_1-x_2) \gamma A^+(x_2))
    [/itex]
    but when I try to convert it to momentum space I get

    [itex]
    \int \mathrm{Tr}(d^4x_1 d^4x_2 \gamma\epsilon(k')e^{-ik'x_1}\frac{1}{(2\pi)^4}\int d^4p S_F(p)e^{-ip(x_2-x_1)}\gamma\epsilon(k)e^{-ikx_2}\frac{1}{(2\pi)^4}\int d^4p' S_F(p')e^{-ip'(x_1-x_2)})
    [/itex]
    this expression is technical but I basically assumed that the incoming and outgoing photons have momenta k and k' (which results in k=k' of course) and the electron and positron have pomenta p and p'. From this I got F(p)S_F(p-k). According to references this result is wrong and the correct result is S_F(p+k)S_F(p). Am I missing something fundamental or is it just algebra?

    Thank you.
     
    Last edited: Mar 30, 2013
  2. jcsd
  3. Mar 30, 2013 #2

    mfb

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    If I understand that correctly, you integrate over p anyway. Could both versions lead to the same integrals? It is just a shift p->p+k.
     
  4. Mar 30, 2013 #3
    I think you are right. I somehow did not pay attention to this... :-)
    Thank you.
     
    Last edited: Mar 30, 2013
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