# Why does the counterterm's propagator have inverse units of the propagator? $\phi^4$-theory

According to Peskin & Schroeder (page 325), the Feynman rule for the counterterm

 ------(x)-----


for

$$\frac12 \delta_Z(\partial_\mu\phi_r)^2-\frac12\delta_m \phi_r^2$$

being $\phi_r$ the renormalized field, is given by

$$i(p^2\delta_Z-\delta_m)$$

which resembles rather the (multiplicative) inverse of the propagator for the original Lagrangian (whith physical quantities). Why?

• because this term is taken as an interaction term rather than a free term. Nov 7, 2013 at 10:24

Thee diverging terms for the propagator come from the renormalization of the self-energy $\Sigma$, defined by $G^{-1}=G_0^{-1}-\Sigma$, where $G_0$ is the propagator defined by the Lagrangian (i.e. bare propagator + counterterms) :

$G^{-1}_0=(1+\delta Z)p^2+(m^2_0+\delta m^2)$.

One chooses the counterterms to cancel the divergences coming from $\Sigma$ order by order. If the theory is perturbatively renormalizable, only these two counterterms are sufficient at every order in perturbation theory.

Counter Term Feynman Rules

A counter term in a Lagrangian is interpreted as an interaction, and the Feynman rules may be derived by the Feynman path integral in the standard way. This accounts for the factor of $i$ and the functional derivatives will take care of the factor of $\frac12$.

Just like in $\frac{\lambda}{4!}\phi^4$ theory we have a $-i\lambda$ Feynman rule, with $\frac{\delta_m}{2}\phi^2$ we have a $-i\delta_m$ Feynman rule. In addition, derivatives of fields give rise to momenta: $(\partial \phi)^2 \mapsto p^2$. So the overall counter term is,

$$i(p^2\delta_Z-\delta_m).$$

The key is, if in doubt, always check things explicitly by treating the terms as interactions in the Feynman path integral.

Alternative Method

I highly recommend Collins' book on renormalisation for a method of renormalisation which dispenses with a Lagrangian and is ideal for systematically computing diagrams. This involves a combination of the BPHZ method and dimensional regularisation.

Essentially, the Zimmermann forest formula generates all the counter term graphs. Then instead of inserting Feynman rules, you insert a particular 'subtraction' operator to remove the divergences.

Of course, this can be related back to a Lagrangian by considering what the Feynman rules would read, and equating the two results to find the counter terms to whatever order you need.

• For anybody else reading this in the future, both the OP and @JamalS are using the $(+---)$ metric signature, which you can tell by the fact that in the original post the kinetic term is $+\frac{1}{2}(\partial\phi)^2$. The $(\partial_0\phi)^2$ term must be positive. Had they chosen the $(-+++)$ signature, the counterterm would be $-i\left( p^2\delta_Z +\delta_m \right)$. (sorry, that sign was causing me some issues :). Feb 26, 2019 at 22:03
• @ArturodonJuan It’s the signature used most in QFT (not counting string theory as QFT). Feb 28, 2019 at 13:34
• I've heard that before but honestly I don't think it's true. I'm pretty sure in modern formal QFT research (that actually works in Minkowski, or locally Minkowski, space) the convention is the mostly-plus metric. Probably the people that would use the mostly-minus metric are phenomenologists that aren't actually doing loop calculations, where the other metric is obviously better. Feb 28, 2019 at 18:42
• @ArturodonJuan "Actually works in Minkowski"? What? I have done plenty of loop calculations and either signature makes really no difference. Choices of gauge and regularisation are more impactful. Feb 28, 2019 at 23:02
• I'm talking about Euclideanizing momentum integrals of course, which has one less factor of $-1$ in the mostly plus metric. Also in quantum gravity or supergravity (not necessarily string theory), the mostly plus metric makes more sense - which I don't think you were debating. But you're definitely right about the choice of gauge and regularization scheme being much more impactful, which is why (I think) most people choose Feynman gauge and dimensional regularization when doing loop calculations. Mar 2, 2019 at 19:28

I also found it hard to find a rigorous explanation. By dimensional analysis and explicit checking you indeed find that this is the correct Feynman rule for the counterterm.

The closest thing I have to a derivation is that we can include the counterterm in the kinetic term and expand this full propagator for small $\delta Z$

$G = \frac{i}{(1+\delta Z)p^2+m^2} \approx \frac{1}{p^2+m^2}-\frac{\delta Z}{(p^2+m^2)^2} + ...$.

I think that not including the $(p^2+m^2)^2$ term in the Feynman rule is correct since this counterterm gets treated as an (amputated) vertex. A four point function similarly does not include 4 propagator factors in its feynman rule.

I do not know why I get a different sign than Peskin & Schroeder.

Also, the expansion is questionable because $\delta Z$ is in fact infinite as opposed to small.