1-loop Fermion mass correction in toy EFT

[Siupa]

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

 

[malawi_glenn]

Source?

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

 

[Siupa]

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?

 

[malawi_glenn]

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?

 

[Siupa]

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!

 

[vanhees71]

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.