Consider $\phi^4$-theory with counter-terms and renormalized mass, field and coupling constant, described by the Lagrangian $$\mathcal{L}= \underbrace{\frac{1}{2} \partial_{\mu} \phi_{r} \partial^{\mu} \phi_{r}-\frac{1}{2} m^{2} \phi_{r}^{2}}_{\mathcal{L}_{free}}\underbrace{-\frac{\lambda}{4 !} \phi_{r}^{4}+\left[\frac{1}{2} \delta_{Z}\left(\partial_{\mu} \phi_{r}\right)^{2}-\frac{1}{2} \delta_{m} \phi_{r}^{2}-\frac{\delta_{\lambda}}{4 !} \phi_{r}^{4}\right]}_{\mathcal{L}_{int}},$$ where $$\sqrt{Z}\phi_r = \phi,\quad \delta Z=Z-1, \quad \delta_{m}=m_{0}^{2} Z-m^{2}, \quad \delta_{\lambda}=\lambda_{0} Z^{2}-\lambda.$$ We get two new Feynman-rules due to the additional terms in the Lagrangian, given by enter image description here

If I understand correctly, the idea of these counterterms, and consequently the Feynman diagrams, is to make one-loop diagrams finite. The issue is that I don't understand how to apply the new Feynman rules to a problem. Let us consider the following amplitude of a 2-point function with a counterterm.

enter image description here

Using the Feynmanrules we find

$$\frac{1}{(p^{2}-m^{2})^{2}} i I\left(-p^{2}\right) + \frac{1}{(p^{2}-m^{2})^{2}} i(p^2\delta_Z-\delta_m),$$ where $$I\left(-p^{2}\right)=\frac{1}{2} \int \frac{-i \mathrm{~d} \ell^{4}}{(2 \pi)^{4}} \frac{1}{\ell^{2}-m^{2}} \frac{1}{(p-\ell)^{2}-m^{2}}.$$ How exactly does this counterterm help now to make the amplitude finite?

  • 1
    $\begingroup$ you don't have a $3$-point vertex in $\varphi^4$ theory. $\endgroup$ Jan 5, 2021 at 18:05
  • $\begingroup$ @Oбжорoв where do you see a 3-point vertex? $\endgroup$
    – Sito
    Jan 5, 2021 at 18:22
  • 1
    $\begingroup$ in your propagator $\endgroup$ Jan 5, 2021 at 18:41

1 Answer 1


Heuristically, the counterterms are also infinite, and are engineered so as to cancel out the infinities from the bare amplitude. How these values are chosen depends on your choice of regulator and subtraction scheme, but the same general principle holds - rather than the unphysical bare parameters that we used when we made our first stab at writing the Lagrangian, we want to "upgrade" them to describe the real world. I'll write the renormalised correlator with the counterterms at one-loop in a slightly more motivating form:

$$ \frac{1}{p^{2}-m^{2}}i \left(I\left(-p^{2}\right) +(p^2\delta_Z+\delta_m)\right)\frac{1}{p^{2}-m^{2}} $$

(I'm playing a bit fast and loose with the signs in front of the counterterms, hopefully that doesn't make too much of a difference). Now $I(-p^2)$ will have the general form $$ I(-p^2) = \text{Finite}+p^2D_1(\tilde{r})+D_2(\tilde{r}), $$ where $\tilde{r}$ is the regulator variable, and the two divergences $D_i$ are formally infinite in the limit of the regulator.

However, you can see that through a clever choice of $\delta_Z$, we can eliminate the $p^2$ infinity (renormalise it, as it were), and likewise for $\delta_m$ and $D_2$ - this is done by making the counterterms dependent on the regulator variable as well. So by modifying the parameters in the Lagrangian, we have made the (observable) one-loop correction to the 2-point function finite: $I(-p^2) = \text{Finite}$.

This stuff might be a little daunting the first time you see it, so I'll do a concrete example. The 1-loop amplitude for the massless $\phi^3$ theory is $$ \Sigma(-p^2)=-\frac{g^2}{32\pi^2}\ln\frac{-p^2}{\Lambda^2}, $$

[from Schwartz, Ch. 16] where, rather than dimensional regularisation (which usually employed here, slightly more powerful, but a bit of a hassle to work with) we have used the Pauli-Villars regulator, which introduces a fictitious scalar having mass $\Lambda$ that is designed to be taken to infinity. The coupling constant $g$ in $\phi^3$ theory is a bit annoying in that it is not dimensionless, so defining $\tilde{g} = \frac{g^2}{-p^2}$ implies $$ \Sigma(-p^2)=p^2\frac{\tilde{g}^2}{32\pi^2}\ln\frac{-p^2}{\Lambda^2}. $$

Inserting this into the full 1PI expansion yields $$ \langle \phi\phi\rangle = \frac{i}{p^2-m^2} + \frac{i}{p^2-m^2}\left(-i\Sigma-i\delta_m+ip^2\delta_Z\right)\frac{i}{p^2-m^2} + \ \dots $$

where the $\dots$ represent higher-order terms. This is just a geometric series, so: $$ \langle \phi\phi\rangle = \frac{i}{p^2-m^2-\Sigma-\delta_m+p^2\delta_Z} = \frac{i}{p^2-m^2-(\Sigma+\delta_m-p^2\delta_Z)} $$

Since there is a divergence in a multiplicative factor of $p^2$ ($D_1$ in our previous notation), taking $\delta_m$ as 0 and $\delta_Z$ to be, say, $ \frac{\tilde{g}^2}{32\pi^2}\ln\frac{f^2}{\Lambda^2}$, then

$$ \Sigma+\delta_m-p^2\delta_Z = p^2\frac{\tilde{g}^2}{32\pi^2}\ln\frac{-p^2}{\Lambda^2} + 0 - p^2\frac{\tilde{g}^2}{32\pi^2}\ln\frac{f^2}{\Lambda^2} \\ = p^2\frac{\tilde{g}^2}{32\pi^2}\ln\frac{-p^2}{f^2} $$

(due to the properties of the logarithm) resulting in a finite correlator. Naturally there is no renormalisation of the mass term in this case, but extending to massive $\phi^3$ will set up a simple system of equations to determine $\delta_Z$ and $\delta_m$ simultaneously.

  • $\begingroup$ Thanks for the answer, but the issue is that I understand what should theoretically happen (i.e. what you just nicely described), but I can't reproduce it explicitly in any concrete example.. $\endgroup$
    – Sito
    Jan 5, 2021 at 18:39
  • $\begingroup$ Sure, I've updated the answer $\endgroup$ Jan 6, 2021 at 13:54
  • 1
    $\begingroup$ @Sito Although note that there is actually a discrepancy in the theory that you've mentioned at the beginning of your question (i.e. $\phi^4$) and the one that your Feynman diagrams correspond to ($\phi^3$). I've provided a concrete example for the latter, since it's slightly easier, but hopefully you'll be able to extrapolate from that for other theories too. $\endgroup$ Jan 6, 2021 at 16:40
  • $\begingroup$ Thanks a lot for the explanations. I see, I messed up when writing down the Feynman diagrams. If I'm not misunderstanding, we have the situation shown in this picture. $\endgroup$
    – Sito
    Jan 6, 2021 at 20:13
  • $\begingroup$ @Sito that's correct. Glad my answer helped! $\endgroup$ Jan 7, 2021 at 3:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.