QFT Beginner Seeks Help: Photon Self Energy Feynman Diagram

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SUMMARY

The discussion centers on the photon self-energy Feynman diagram as outlined in the Mandl-Shaw textbook, specifically referencing Equation 7.22. The user attempts to convert the integral from position space to momentum space but encounters discrepancies in the results. The user initially derives F(p)S_F(p-k), while the correct result is S_F(p+k)S_F(p). The confusion arises from the integration over momenta and the algebraic manipulation involved in the transformation.

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PineApple2
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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)
<br /> \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))<br />
but when I try to convert it to momentum space I get

<br /> \int \mathrm{Tr}(d^4x_1 d^4x_2 \gamma\epsilon(k&#039;)e^{-ik&#039;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&#039; S_F(p&#039;)e^{-ip&#039;(x_1-x_2)})<br />
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:
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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.
 
I think you are right. I somehow did not pay attention to this... :-)
Thank you.
 
Last edited:

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