In lectures on the standard model I recently saw, in writing down the SM Lagrangian the professor was careful to refer to fields like $e^c$ as a conjugate electron rather than an antielectron. He also mentioned that all the fields he wrote down were left-handed Weyl spinors.

First Question: What is the relationship between conjugate electrons and antielectrons?

Second Question: Are fields in the SM Lagrangian left-handed spinors by convention, or is this physically meaningful?

Third Question: Is there any connection between charge conjugation and handedness of a spinor?


1 Answer 1

  1. Given a Dirac spinor $\psi$, its "charge conjugate" spinor is given by $\psi^c = C\psi^\ast$, where $C$ is a charge conjugation matrix defined by a certain convention (e.g. $C^\dagger \gamma^\mu C = -(\gamma^\mu)^\ast$). Meanwhile, the anti-particles are associated with the Dirac adjoint (or Driac conjugate) spinor $\bar{\psi} = \psi^\dagger \gamma^0$.

  2. Electrons are not left-handed, they are massive Dirac spinors with both left- and right-handed components. The SM neutrinos are massless and usually thought to be only left-handed, but there are plenty of ideas for sterile right-handed neutrinos in formulations with massive neutrinos beyond the SM.

  3. No. You can charge conjugate a Dirac spinor as well as just a Weyl spinor of either chirality. The charge conjugation matrix for Weyl spinors is just given by projecting $C$ down to the subspace of the given chirality.

  • $\begingroup$ For 3., doesn't applying $C$ flip chirality? $P_L (\psi_L)^C = 0$? $\endgroup$
    – innisfree
    Apr 19, 2017 at 11:41
  • $\begingroup$ @innisfree No, $C$ commutes with the chiral projectors, since (in a particular basis) $C = \mathrm{i}\gamma^0\gamma^2$ and $P_{R/L} = \frac{1}{2}(1\pm\gamma^5)$ and $\gamma^5$ anticommutes with all $\gamma^\mu$, so $C$ and $P_{R/L}$ commute, so $P_L (\psi_L)^c = P_L C P_L \psi^\ast = P_L^2 C \psi^\ast = P_L C\psi^\ast = (\psi_L)^c$. $\endgroup$
    – ACuriousMind
    Apr 19, 2017 at 11:54
  • $\begingroup$ Oh I thought eg $C=i\gamma_2$. $\endgroup$
    – innisfree
    Apr 19, 2017 at 13:10
  • $\begingroup$ fermion "fields in the SM Lagrangian" are not massive. $\endgroup$
    – arivero
    Aug 2, 2017 at 1:48
  • $\begingroup$ @ACuriousMind If, say, a physical electron is represented by the spinor $\psi$, then is the positron represented by its adjoint, or its charge conjugate? I think it’s the former, since the c.c. would represent a ($E>0$) particle with the opposite charge, but I’m not sure. $\endgroup$
    – Bcpicao
    Jan 28 at 1:48

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