5
$\begingroup$

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)?

$\endgroup$
5
$\begingroup$

The proton can indeed be modeled as a Dirac fermion, as in the Yukawa theory that describes protons interacting with pions. However, this does not capture all of the behaviour of the proton, since Dirac fermions are pointlike, whereas protons are not.

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 $5.6$.

$\endgroup$
  • $\begingroup$ 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 $\endgroup$ – SRS Apr 27 '18 at 14:05
  • 2
    $\begingroup$ To the best of my knowledge, a complete calculation would require taking into account QCD, yes. $\endgroup$ – gj255 Apr 27 '18 at 14:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.