I'm trying to undertand the following vertex:

Initial state of up and anti-down quarks with finalk state made of $W^+$ boson. Does it go with left or right projector? I think that from Lagrangian it should go with left projector but the vertex with $e^+$ and $W^-$ in initial state and $\bar{\nu}_e$ in the final one goes with right projector and this is not read from Lagrangian.

Accordingly to my professor, the case R-positron with $W^-$ to give R-antineutrino has a vertex that goes with $\gamma^\mu P_R$, so what I want to know is how extract this vertex if the Lagrangian does not contain that term, just $P_R\gamma^\mu P_L = \gamma^\mu P_L$


1 Answer 1


It would be helpful if you just wrote the Lagrangian terms you are looking at. Destroying a Left-handed u and R antidown to create a W+ corresponds to the vertices $$ W_\mu^- \overline {d} P_R\gamma^\mu P_L u + W_\mu^+ \overline {u} P_R\gamma^\mu P_L d , $$ while destroying a R-positron and a W- to yield a R-antineutrino to $$ W_\mu^-\overline {e} P_R\gamma^\mu P_L \nu + W_\mu^+ \overline{\nu} P_R\gamma^\mu P_L e ~~~, $$ where you focus on the first term in each line.

Note half the species, the R leptons/quarks and L anti leptons/antiquarks are simply missing from these couplings. (In an, impossible, notional world with no masses, these components would be missing everywhere and all spinors would be Weyl, and projectors would be superfluous.)

Further note L is in no way privileged over R: it is simply a convention of us mooring chirality on leptons/quarks instead of antileptons/antiquarks. (Sometimes this convention is subverted in QFT texts or GUT arraying.)

  • $\begingroup$ Destroying a positron requires of $\bar{e}$ while destroying $W^-$ needs of $W^-$ field. Plus creating $\bar{\nu}$ suppose the necessity of $\nu$ field, so the corresponding Lagrangian term that leads to this proicess is the one that goes as $\gamma^\mu P_L$, not $\gamma^\mu P_R$ as books say. What am I misunderstanding? $\endgroup$
    – Vicky
    Mar 16, 2019 at 1:20
  • $\begingroup$ I'm not sure what your undisclosed books says where. I suspect the best thing to do is to go review the Dirac equation and single out the creation and destruction and helicity pieces of the spinors. The field that destroys a positron creates an electron, as well. $\endgroup$ Mar 16, 2019 at 2:07
  • $\begingroup$ Yes, but I don't see how the vertex in $\gamma^\mu P_R$ appears since Lagrangian has not right projector. I've always been told that to get vertices you need to rib out fields of interaction Lagrangian and the remaining is the vertex. Nevertheless, there is no $P_R$ in Lagrangian, but there it is in vertex. I'm stuck $\endgroup$
    – Vicky
    Mar 16, 2019 at 2:14
  • $\begingroup$ You are telling me this: $\bar{\psi_L}\gamma^\mu = (P_L\psi)^\dagger\gamma^0\gamma^\mu = \bar{\psi}P_R\gamma^\mu = \bar{\psi}\gamma^\mu P_L$. So we don't get $P_R$ what does appear in vertex, this is why I don't understand the vertex with $P_R$, I don't see where this right projector comes $\endgroup$
    – Vicky
    Mar 16, 2019 at 2:20
  • $\begingroup$ I've made an edit in order to be more clear about what I want to know $\endgroup$
    – Vicky
    Mar 16, 2019 at 4:19

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