I 1-loop Fermion mass correction in toy EFT

Siupa
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Where does the ##m## in ##(3.2)## come from? It doesn’t seem to enter anywhere in Feynman rules for the given diagram
 

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Source?

Anyway, it comes from the internal fermion propagator - you can not just look at the interaction Feynman rules.
 
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malawi_glenn said:
Source?

Anyway, it comes from the internal fermion propagator - you can not just look at the interaction Feynman rules.
Source is A. Pich, Effective Field Theory, beginning of chapter 3.

I understand that the momentum integral of the internal propagator comes from that loop, but why specifically is there a factor of ##m## in front of it? Or better, why isn‘t it inside the integral in the numerator summed with ##\not\! k##? Is it because the integral with ##k## in the numerator vanishes due to the integrand being odd? Isn’t this argument only valid for convergent integrals though?
 
in the numerator you have ##\not\! k - m ## in the fermion propagator.

Yes, it will vanish because of odd integrand see https://arxiv.org/abs/2006.16285 eq. 18 page 14
(note the typo, it should read ##(\gamma^5)^2 = 1##)

Look at it this way. How would you get something that has the dimension of mass, in that diagram, unless you had a factor ##m## in front? Can you come up with anything?
 
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malawi_glenn said:
in the numerator you have ##\not\! k - m ## in the fermion propagator.

Yes, it will vanish because of odd integrand see https://arxiv.org/abs/2006.16285 eq. 18 page 14
(note the typo, it should read ##(\gamma^5)^2 = 1##)

Look at it this way. How would you get something that has the dimension of mass, in that diagram, unless you had a factor ##m## in front? Can you come up with anything?
I understand, thank you!
 
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Seems to be a textbook making the "confusing issue" of divergencies even more confusing. Of course your integral must be regularized first, before you can make any sense of it. The most convenient, but a bit unintuitive, regularization is "dimensional regularization" since it keeps Poincare invariance, and then indeed it's correct to conclude that the part of the integrand ##\gamma_{\mu} k^{\mu}## can be set to 0.
 
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