# Is proton a Dirac fermion? If yes, does it also have a Lande-g factor $g=2$?

Proton is a spin-1/2 particle but composite i.e., it's a bound state of three quarks. Protons have partner called anti-proton which is also composite. Is it not a Dirac fermion? If not, why? In other words, why should a Dirac fermion always be elementary such as electron, positron or neutrino?

By Dirac fermion, I understand a quantum of the Dirac field and which also has a antiparticle partner. If it's a Dirac particle, does it mean protons also have Lange-g factor $g=2$ like an electron (apart from the anomalous contribution)?

Since the proton is made from quarks, its interaction with the electromagnetic field is more complicated than that of a pointlike fermion with charge $+1$. Consequently its $g$-factor is not two. Indeed, the proton $g$-factor has been measured to be about $5.6$.
• If we treat proton as a composite Dirac particle of 3 quarks (and no gluons) each of which interacts with the EM field, will the theory of QED predict correct value of $g$? I think there is a contribution from QCD as well. Am I correct? @gj255 – SRS Apr 27 '18 at 14:05