Trouble Finding Renormalization Conditions in Yukawa Theory

Click For Summary
The discussion centers on calculating the beta functions in massless pseudoscalar Yukawa theory, specifically addressing the one-loop correction to the electron propagator. The Lagrangian includes a Yukawa coupling, and the participant seeks to establish the correct renormalization conditions at an unphysical momentum. There is confusion regarding the relationship between the momentum terms, particularly the need to set ##\not{\!p} = iM## to satisfy the renormalization conditions. Clarifications are made regarding the correct interpretation of momentum squared, leading to the conclusion that ##p^2 = \not{\!p}^2##. The conversation emphasizes the importance of accurately applying the anticommutator relations between gamma matrices in this context.
gobbles
Messages
16
Reaction score
1
I am trying to calculate the ##\beta## functions of the massless pseudoscalar Yukawa theory, following Peskin & Schroeder, chapter 12.2. The Lagrangian is

##{L}=\frac{1}{2}(\partial_\mu \phi)^2-\frac{\lambda}{4!}\phi^4+\bar{\psi}(i\gamma^\mu \partial_\mu)\psi-ig\bar{\psi}\gamma^5\psi\phi.##

When calculating the one-loop correction to the electron (##\psi##) propagator, there is one diagram, the expression for which is of the form

##g^2\not{\!p}\left[\mbox{logarithmic divergence} + \mbox{finite terms that depend on } \log(-p^2)\right].##

In order to calculate the ##\beta(g)## function, we now need to find the counterterm ##\delta_\psi## at the renormalization conditions given at an unphysical momentum ##p^2=−M^2##, where ##M## defines the scale we're working at. The renormalization conditions, if I understand right, are chosen to make the ##\log(−p^2)## term finite, but there is also the ##\not{\!p}## term which should be set. If I set

##\not{\!p}=M,##

I would get

##p^2=\not{\!p}^2=M^2,##

instead of ##p^2=−M^2##, as required. The remaining thing to do is to set ##\not{\!p}=iM##, but I'm having trouble justifying that.
What are the correct renormalization conditions in this case?
 
Last edited:
Physics news on Phys.org
You got it wrong, ##p^2 = p^\mu p_\mu##, what you wrote is square of p slash, which I am not sure how to LaTexed it.
 
I'm posting only for helping with the LaTeX.
MathematicalPhysicist said:
You got it wrong, ##p^2 = p^\mu p_\mu##, what you wrote is square of p slash, which I am not sure how to LaTexed it.
\not{\!p}
 
  • Like
Likes MathematicalPhysicist
##\not{\!p}## works, thanks.
 
Thanks for the reply!
##\not{\!p}^2=p_\mu \gamma^\mu p_\nu \gamma^\nu=p_\mu p_\nu(2g^{\mu\nu}-\gamma^\nu \gamma^\mu)=2p^2-\not{\!p}^2##
and therefore
##p^2=\not{\!p}^2##
 
  • Like
Likes MathematicalPhysicist
Thank you for correcting me, forgot the anticommutator relation between the gammas.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K