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I am pretty new to QED and Quantum field theory. I am currently learning Feynman Diagrams and I was interested in electron-neutron scattering, $e^-N -> e^-N $ specifically in vertex factors. In Griffiths, Introduction to Elementary Particles (2nd edition) for QED coupling constant he gives a general form for any 1/2 spin : ge = -q$\sqrt{4 \pi/\hbar c}$ for . Where he states q to be the charge of the particle.

Would that still apply to neutron as it has 0 charge? because it would just equate the ge to 0 resulting in vertex factor : ige $\gamma$$\mu$ to 0.

Thank you very much in advance!

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If you would consider a neutron as a single particle then indeed it does not interact via electromagnetism with the electron. However, we know that the neutron is made up of quarks and gluons and so a more correct way is to describe the interaction via a distribution functions of these quarks and gluons forming the neutron. This is called a parton distribution function.

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