Scalar decay to one-loop in Yukawa interaction

In summary, the author is trying to calculate the amplitude for a decay ##\phi \to e^+e^-## under a Yukawa interaction ##\mathcal{L}_I = -g\phi \bar{\psi}\psi## to one-loop order (with massless fermions for simplicity). The author has no problem calculating the integrals and using counterterms to cancel the infinities that arise, but is not sure if the conditions they use for renormalization are correct.
  • #1
Gaussian97
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One-loop correction for $\phi \to e^+e^-$ under a Yukawa interaction seems to vanish trivially.
I am trying to calculate the amplitude for a decay ##\phi \to e^+e^-## under a Yukawa interaction ##\mathcal{L}_I = -g\phi \bar{\psi}\psi## to one-loop order (with massless fermions for simplicity).

If I'm not wrong, there are 4 diagrams that contribute to 1 loop, three diagrams involving self-energy corrections (i.e. inserting a loop into the external lines) and an extra diagram with vertex correction (a ##\phi## field exchanged by ##e^+## and ##e^-##).

I have no problem calculating the integrals and using counterterms to cancel the infinities that arise, but I'm not sure if the conditions I use for renormalization are correct. Following the example of QED, to apply on-shell renormalization I used the following conditions;

The scalar propagator in the limit ##p^2 \to M^2## should be ##\frac{i}{p^2-M^2}##

The fermion propagator in the limit ##\not{\!p} \to 0## should be ##\frac{i}{\not{p}}##

The vertex function in the limit ##p^2 \to M^2## should be ##-ig##. (##p## is the momentum of the scalar particle.)

Now, because the self-energy diagrams are all in external legs, the first two corrections mean that those diagrams vanish.
But the third condition tells that the vertex correction must also vanish when the scalar particle is on-shell (as in my diagram). Therefore all the diagrams here vanish trivially due to renormalization conditions.

Is this analysis correct? Or did I make some mistake in the renormalization part?
 
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  • #2
Gaussian97 said:
Yukawa interaction,,,with massless fermions
By doing so, didn't you just set the coupling to zero?
 
  • #3
Vanadium 50 said:
By doing so, didn't you just set the coupling to zero?
Mmm... Not sure I follow you, maybe I'm saying something stupid. But how is the coupling constant ##g## in ##\mathcal{L}_I = -g\phi \bar{\psi}\psi## related to the mass of the fermions?
 
  • #4
I'm sorry. I saw "Yukawa" and my brain immediately jumped to "Higgs Yukawa".
 
  • #5
Oh, okay I understand now the confusion.
I'm doing this simply to practice (most textbooks deal with $\phi^4$ and QED), so I thought that Yukawa was a simple enough example to try to do it by myself.
There is no intention of this being applicable in the Standard Model or anything like that, just to have fun.
 

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