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

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.

• 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. Sep 27, 2020 at 7:18
• 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. Sep 27, 2020 at 7:22