I am working through an exercise in QED from Halzen and Martin's textbook Quarks and Leptons.

For the QED scattering process $e^-(k)\mu^-(p)\to e^-(k')\mu^-(p')$, the absolute square of the Feynman amplitude averaged over the electron and muon spins can be expressed in short as $$\overline{|\mathcal{M}|}^2=\frac{e^4}{q^2}L_e^{\mu\nu}L^{\rm muon}_{\mu\nu}$$ where $q=k-k'=p'-p,$ and $$L_e^{\mu\nu}=\sum_{\text{e spins}}[\bar{u}(k')\gamma^\mu u(k)][\bar{u}(k')\gamma^\nu u(k)]^*\\ L^{\rm muon}_{\mu\nu}=\sum_{\mu ~\text{spins}}[\bar{u}(p')\gamma_\mu u(p)][\bar{u}(p')\gamma_\nu u(p)]^*$$ The book then requires us to justify (Exercise $6.8$) that if the electron scatters off a spin-0 particle, then all one has to do is to replace $L^{\rm muon}$ by $(p+p')_\mu(p+p')_\nu$, in order to find the corresponding $\overline{|\mathcal{M}|}^2$.

Which QFT vertex will cause an electron to scatter off a spin-0 charged particle electromagnetically?

We cannot write a Lorentz-invariant interaction vertex with one fermion field (the electron) and two boson fields (the spin-0 particle from which the electron scatters and the spin-1 photon mediator). If someone can point out, I'll highly appreciate that.

  • $\begingroup$ Alpha particles have spin zero. If your math makes you conclude that electrons cannot scatter electromagnetically off of helium nuclei, you need to re-examine the math. $\endgroup$ Sep 27, 2020 at 7:18
  • $\begingroup$ Sorry about the poor language. I do not doubt that such a scattering will take place. I wonder what will be QFT interaction vertex for such a process. $\endgroup$ Sep 27, 2020 at 7:22

1 Answer 1


When an electron scatters off of a muon, there is no muon-electron-photon vertex. Instead there are two vertices- one with two electrons and a photon and one with two muons and a photon.

Similarly if you have an electron scattering off a boson, you wouldn't have a single electron-boson-photon vertex, you would have the usual "two electrons and a photon" vertex and an additional "two bosons and a photon" vertex in the Feynman diagram.

  • $\begingroup$ Got the point. Thank you very much :-) So if I understood you then I'll have one normal QED vertex and one scalar QED vertex. $\endgroup$ Sep 27, 2020 at 7:29
  • 1
    $\begingroup$ Yes. It's unknown if any fundamental charged scalars actually exist in nature, but mathematically at least there is no problem. $\endgroup$
    – Chris
    Sep 27, 2020 at 7:46

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