- #1

- 970

- 3

## Main Question or Discussion Point

Can someone explain how one derives the 4-point Fermi interaction from the full Standard Model?

I understand you set up a cutoff [tex]\Lambda[/tex], and the cost of this is that the coefficients of all terms in your Lagrangian become functions of this cutoff, and you also have an infinite number of new, non-renormalizeable terms. So your new propagator should be of the form 1/(Z([tex]\Lambda[/tex])k^2-M^2), where Z([tex]\Lambda[/tex]) is the coefficient of your kinetic term, and since integrations are now only to [tex]\Lambda[/tex], this propagator can be replaced by (-1/M^2) since Z is at most log([tex]\Lambda[/tex])=log(M)?

But after doing all this, your Lagrangian still involves the heavy boson terms. The book says to solve their equations of motion, and substitute this back into the Lagrangian. What is the justification for this?

I understand you set up a cutoff [tex]\Lambda[/tex], and the cost of this is that the coefficients of all terms in your Lagrangian become functions of this cutoff, and you also have an infinite number of new, non-renormalizeable terms. So your new propagator should be of the form 1/(Z([tex]\Lambda[/tex])k^2-M^2), where Z([tex]\Lambda[/tex]) is the coefficient of your kinetic term, and since integrations are now only to [tex]\Lambda[/tex], this propagator can be replaced by (-1/M^2) since Z is at most log([tex]\Lambda[/tex])=log(M)?

But after doing all this, your Lagrangian still involves the heavy boson terms. The book says to solve their equations of motion, and substitute this back into the Lagrangian. What is the justification for this?