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I am currently trying to calculate scattering amplitudes of tree level QED processes and i am a bit confused due to the ordering of the factors yielded by the feynman spinor QED rules. Especially in processes having photons in the initial or final states.

Could someone write a explanation how and why to order the factors? All of the literature i looked at just showed the feynman rules but did not gave further explanation about the order of the factors ...

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  • $\begingroup$ Are you asking how do we go from Feynman diagram to the amplitude of a given process? $\endgroup$
    – Quiver
    Apr 6, 2020 at 8:22
  • $\begingroup$ That's what i am trying to do, yes. But my question is how to order the factors and the $\gamma$-matrices obtained by the feynman rules? $\endgroup$
    – AlmostClueless
    Apr 6, 2020 at 9:23

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Let't take a simple process for example $$e^-+e^+ \to \mu^-+\mu^+$$ The only tree level diagram for this process is the following enter image description here

To build up the matrix element of this process you just have to follow the fermion lines backwards, that's all. Beginning from the end of the process you have then $$(\mu^- \to \text{vertex}\to\mu^+)\text{ Photon propagator }(e^+\to\text{vertex}\to e^-)$$ which gives $$\mathcal{M}=\bar u(p_3,\sigma_3)(-ie\gamma^\nu)v(p_4\sigma_4)\frac{-ig_{\mu\nu}}{q^2+i\epsilon}\bar v(p_2,\sigma_2)(-ie\gamma^\mu)u(p_1,\sigma_1)$$ with $q=p_1+p_2$

This holds for every Feynman diagram in general, you have to follow backwards the fermion lines and use the appropriate spinors for particles and antiparticles. The convention of starting from the end does not matter, you could as well start from the beginning as long as you follow the fermion lines backwards.

Edit

Since the OP asked for the case of Compton scattering I add it. So the s-channel Compton scattering process is given by this diagram (and the one with the photons with exchanged positions) enter image description here

where the two fermion lines are just electrons. As we said before let's follow the fermion lines backwards $$(e^-\to\text{vertex}\to\text{photon})\text{Fermion propagator}(\text{photon}\to\text{vertex}\to e^-)$$ which gives

$$\bar u(p^\prime, \sigma^\prime)(-ie\gamma^\nu)\epsilon_\nu^{*}(k^\prime)\frac{\not p+\not k+m}{(p+k)^2-m^2+i\epsilon}\epsilon_\mu(k)(-ie\gamma^\mu)u(p, \sigma)$$

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  • $\begingroup$ And what for external photons ? Since the Polarisation vector has no dirac indices, i was confused how to order the terms in case of compton scattering. Do i still follow the fermion-lines ? $\endgroup$
    – AlmostClueless
    Apr 6, 2020 at 12:04
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    $\begingroup$ I'll add it to my answer then! $\endgroup$
    – Quiver
    Apr 6, 2020 at 12:45
  • $\begingroup$ @Jbag1212 I advise you to open a new question for that, I'll answer whenever I can. But in short gluons are the photons of QCD. You'll have the two polarization of the outgoing gluons, the propagator of the gluon coming from quark annihilation and then the two spinor fields for the quarks. Plus the t-channel diagram where the gluon propagator becomes a quark propagator. $\endgroup$
    – Quiver
    Mar 16, 2022 at 10:17

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