# Renormalization of scalar field theory

• I
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 :

$$\delta \mathcal{L}_1 = -\dfrac{gm^2}{32\pi \epsilon^2}\phi^2$$

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

$$-\dfrac{igm^2}{16\pi^2 \epsilon}$$

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

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vanhees71
Science Advisor
Gold Member
2019 Award
You don't get it directly from the Lagrangian but by calculating the "one-loop tadpole diagram" for the self-energy of the boson in dimensional regularization. For a detailed treatment, see

https://itp.uni-frankfurt.de/~hees/publ/lect.pdf

Wledig
ChrisVer
Gold Member
I haven't done the calculations, but wouldn't that term give rise to a propagator as the one in the mass term?
$\frac{i}{p^2 - \mu^2 }$
But with $\mu^2 = m^2 - \frac{gm^2}{32\pi^2 \epsilon}$
?

vanhees71
Science Advisor
Gold Member
2019 Award
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)##.

ChrisVer