Renormalization of scalar field theory

  • #1
69
1
I was reading about the renormalization of ##\phi^4## theory and it was mentioned that in order to renormalize the 2-point function ##\Gamma^{(2)}(p)## we add the counterterm :

[tex]\delta \mathcal{L}_1 = -\dfrac{gm^2}{32\pi \epsilon^2}\phi^2[/tex]

to the Lagrangian, which should give rise to a propagator term of the form:

[tex]-\dfrac{igm^2}{16\pi^2 \epsilon}[/tex]

that will cancel the divergent term in ##\Gamma^{(2)}(p)##. My problem is with the expression above, it's unclear to me how to reach this propagator correction just from the Lagrangian. How can this be achieved?
 

Answers and Replies

  • #3
ChrisVer
Gold Member
3,352
452
I haven't done the calculations, but wouldn't that term give rise to a propagator as the one in the mass term?
[itex] \frac{i}{p^2 - \mu^2 }[/itex]
But with [itex]\mu^2 = m^2 - \frac{gm^2}{32\pi^2 \epsilon}[/itex]
?
 
  • #4
vanhees71
Science Advisor
Insights Author
Gold Member
2019 Award
15,961
7,262
Yes, it's a constant term (because it's a one-point function). It's a self-energy diagram, i.e., it just contributes an additive divergent constant contributing to the mass (squared) term, i.e., you have to add a counter term to the Lagrangian. The minimal subtraction scheme just subtracts the divergent piece ##\propto 1/\epsilon##. The physical mass of the particle is given by the pole of the propagator, ##G=1/(p^2-m^2-\Sigma)##.
 
  • Like
Likes ChrisVer

Related Threads on Renormalization of scalar field theory

  • Last Post
Replies
6
Views
2K
Replies
0
Views
3K
Replies
5
Views
6K
Replies
5
Views
1K
Replies
5
Views
1K
Replies
5
Views
1K
Replies
2
Views
745
Replies
3
Views
730
Replies
8
Views
3K
Replies
1
Views
879
Top