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I understand that an interaction like

$$g\overline\psi_L\psi_R\phi$$

is allowed in the SM (for the sake of this discussion lets ignore gauge charges and focus only on Lorentz invariance).

I want to build a Feynman diagram that has a mass vertex, so I can write an effective coupling, $$g'\overline\psi_L\psi_L\phi$$

However, dealing with the mass vertex has been a pain.

The way I would like to solve this is by use of appropriate projection operators. As I understand this would change the Lagrangian term to something like,

$$ h (P_L\psi)^\dagger \gamma^0 P_R\psi\phi $$

$$ P_L = \frac{1}{2}(1-\gamma^5) , P_R = \frac{1}{2}(1+\gamma^5)$$

The issue arises when I try to calculate this vertex on a Feynman diagram, since it seems like the vertex factor would be $$ ihP_LP_R $$ which is 0 since, $$ P_LP_R = 0 $$.

One way to solve this is by redefining h to have a gamma matrix in it, and then using a vertex factor like $$ P_L h'\gamma^\nu P_R $$ which seems to solve the problem, but also seems to change the scalar nature of $$ \phi $$?

How do I resolve these two things? What is the appropriate vertex factor?

Said another way;

How do you properly include left and right handed projection operators into a Yukawa-type vertex?

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  • $\begingroup$ your proposed vertex is not Lorentz invariant, it is not a valid interaction. $\endgroup$ Commented May 11 at 15:05
  • $\begingroup$ @AccidentalFourierTransform You mean the effective coupling? Yes, I dont care about its invariance right now. But if you are referring to something else, being more clear is more helpful! If something isn't Lorentz invariant, could you point out why, and what you might do to fix it? $\endgroup$
    – SamuelFGC
    Commented May 11 at 15:21
  • $\begingroup$ $\psi_L^\dagger\psi_L$ is not a Lorentz scalar. It is not a valid term in a Lagrangian. $\endgroup$ Commented May 11 at 15:24
  • $\begingroup$ Yes, that is not a term in my Lagrangian. This is an effective coupling I would like to probe. I can do this via a mass vertex. I have an incoming LH particle, that has a mass vertex which changes to RH, interacts with scalar, and leaves as LH. This is the process I need to compute. $\endgroup$
    – SamuelFGC
    Commented May 11 at 15:27
  • $\begingroup$ You are connecting a mass vertex to a Yukawa vertex with a fermion propagator. To leading order in m/p, you get the term in my answer. $\endgroup$ Commented May 13 at 16:29

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I suspect you are trapped in bad notation, which, sadly, is quite common, even though you appreciate the crucial chirality-flipping $\gamma^0$ involved in the overbar.

The more sensible/tasteful representation of your terms, utilized by the better authors, is $$g\overline{\psi_L}\psi_R\phi=g\overline{\psi}P_R\psi_R\phi= g\overline{\psi} \psi_R\phi,$$ just like a mass term, which it produces in the SM, $$ m ~\overline{\psi_L}\psi_R. $$ One hermitianizes terms in the action in the end.

I want to build a Feynman diagram that has a mass vertex, so I can write an effective coupling, $g'\overline{\psi_L}\psi_L\phi$

As the commenters pointed out, your term, if this is what you wrote, is automatically zero, and mercifully does not produce Lorentz-invariant mass terms; I'm not sure why you connect it to a mass vertex. A mass vertex is what is written above!

My first guess for your two-chirality-flip effective interaction, so, then no chirality-flip interaction, is $$igm\phi\overline{\psi_L}(/\!\!\partial ~/\Box) ~~\psi_L= igm \phi~~\overline{\psi}(/\!\!\partial ~/\Box) \psi_L,$$ which doesn't vanish. It only vanishes for infinite momenta, since the chirality-flipping mass is suppressed. This term flips LH to RH, and then, to LH again, in the same breath, so to speak.

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  • $\begingroup$ I have some gauge singlet fermion fields. LH and RH chiralities. They share some mass term in the Lagrangian (This term is Lorentz invariant) There is also a chargeless scalar field, $\phi$ Thus, one can DIRECTLY couple LH and RH chiral fields to this scalar (Lorentz invariant) Using these two interaction terms, we can INDIRECTLY couple LH fields to scalar, by use of a mass term; LH -> mass -> RH -> scalar -> LH I want to express this as an effective coupling strength. This is the computation I want to do. My best guess is -igP_R $\endgroup$
    – SamuelFGC
    Commented May 12 at 10:18
  • $\begingroup$ ..so you want your original LH to chirality flip by the mass in the propagator and then chirality flip through the Yukawa? The propagator will insert a $\gamma^\mu $ in the effective vertex... $\endgroup$ Commented May 12 at 11:42
  • $\begingroup$ yes! Could you elaborate more on this gamma factor insertion? This is the key part I dont understand $\endgroup$
    – SamuelFGC
    Commented May 12 at 11:44
  • $\begingroup$ I updated my answer. You connect the mass vertex which flips chirality to RH, propagate the RH freely through the p-slash to a RH, and then emit a scalar flipping chirality again, to a LH. To chirality flips amount to none such, so LH to LH... $\endgroup$ Commented May 12 at 21:25
  • $\begingroup$ There are some issues with this answer. For one, while a mass vertex is allowed, one must sum up contributions from all mass insertions. The resulting effective coupling has no momentum dependence. Instead, one can consider a new field, \psi = \psi_L + \psi_R and add a projection operator to select the desired chirality. At face value this seems correct, but Ive been told its not so easy. The proper gamma insertions depend on if particles are Majorana or Dirac, and if the scalar is real or complex. This is part I do not understand. $\endgroup$
    – SamuelFGC
    Commented May 15 at 11:11

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