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I have some doubts regarding the allegedly different procedures used in $\lambda\phi^4$ and QED. First of all, I am more familiar with bare perturbation theory (no counterterms), so I would be grateful if you sticked to it as well. Furthermore, I won't write most calculations as the discussed topics are pretty standard and the calculations are not the focus of my question. Finally, I will only consider the one-loop level in both theories.

QED

In the case of QED there are three kinds of corrections we have to take into account:

  • Fermion self energy (Fermion propagator & mass renormalization)
  • Vacuum polarization (Photon propagator)
  • Vertex correction

From each of these, a $Z$ factor arises; since it is the coupling constant, the electric charge is the defined as the low momentum tranfer $2\to2$ scattering amplitude. To evaluate this at the one loop level, we have to use the corrected propagators instead of the "free" propagators we would use at the tree level. The overall result is the appearance of the field-strenght renormalization factors: each fermion leg contributes a factor $\sqrt{Z_2}$, the photon propagator contributes with a $\sqrt{Z_3}$ factor and finally the vertex correction with a $Z_1^{-1}$. Using the Ward identity, $Z_1$ and $Z_2$ cancel and we are left with $$\sqrt{Z_3}e_0=:e\tag{renormalized charge}$$ $e_0$ being the bare charge. So this was a procedure of multiplicative renormalization.

Quartic interaction

The first part is similar, just simpler. We only have to consider the tadpole (self-loop) diagrams as corrections and proceed with a resummation. We now have the complete propagator, which once again has a pole at the physical mass by definition thereof and we get as usual the field strength renormalization factor $Z$. Alright, now for the coupling constant we want to do the same thing, namely consider a low momentum transfer (or low energy) $2\to 2$ process, which at one loop level is given by the so-called fish diagram

fish diagram

What happens now? Cf. e.g. ref 1, I see that people just evaluate and regularize (with a cutoff below) this diagram, which diverges logarithmically and impose the renormalization condition so that $\lambda_0(\Lambda)$ compensates the logarithmic divergence. Now, this looks different from QED, where the $Z$ factors are absorbed to multiplicatively renormalize the charge. Having 4 legs at each vertex, I would expect to have $Z^2=\sqrt{Z}^4$ multiplying $\lambda_0$. I understand that in this case it's different from QED as we have a brand new loop (fish) that comes in with a new divergent part, say $f(\Lambda)=\mathcal{O}(\log\Lambda)$ but I don't see why we ignore the corrections we've made on the propagators. So why the renormalization reads like $$\lambda_R=\lambda_0+f(\Lambda)\tag{1}$$ instead of $$\lambda_R=Z^2\lambda_0+f(\Lambda)~?\tag{2}$$ As we do in QED (where there is only the multiplicative part, not the additive one).

Also, if at the one-loop level one uses bare propators, there is the bare mass - which is divergent at this loop level - at the denominator. Is it possible that they're putting all of this under the rug and "ignoring" the propagator correction and thus the bare mass divergence caused by renormalization? Another possible idea would be that my source instead cancels the field-strength factors of the external legs with those in the LSZ, but the problem for the internal lines is still there.

References

  1. A Modern Introduction to Quantum Field Theory, Michele Maggiore. 2005. Chapter 5.
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  • $\begingroup$ To avoid awkwardness like you describe (especially in the last paragraph) is precisely why we introduce renormalized perturbation theory. It's easy to show it is equivalent to the bare version, but the bookkeeping is much easier. $\endgroup$
    – Buzz
    Aug 31, 2023 at 1:09
  • $\begingroup$ Is eqs. (1) & (2) from a reference? Eq. (1) is presumably just the leading correction in a perturbative series. $\endgroup$
    – Qmechanic
    Aug 31, 2023 at 5:59
  • $\begingroup$ @Qmechanic eq. $(1)$ is a compact notation I've used for $(5.139)$ (maybe not so good notation as $f$ depends on $\lambda_0$, but that depends on the cutoff scale too). Eq. $(2)$ is what I would consider reasonable. More specifically, given that the framework is not so different from QED, I don't see why we wouldn't get multiplicative renormalization here as well. $\endgroup$ Aug 31, 2023 at 6:31
  • $\begingroup$ When you say leading order do you mean the first order in the expansion? Thinking about it now, at one-loop level in the $\phi^4$ theory, we have $Z=1$, which could explain what is going on here but to say so I need to know whether $(2)$ is conceptually wrong at higher loop levels. $\endgroup$ Aug 31, 2023 at 6:41

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In $\lambda\phi^4$ theory the first non-zero correction to the field strenght-renormalization $Z$ is of order $\lambda^2$: $$Z=1+\mathcal{O}(\lambda^2)\implies Z=1\quad\text{ at the 1-loop level}\tag{A}\label{A}$$ and so the corrected momentum-space propagator is (up to finite terms at $p^2=m^2$) $$\tilde{D}(p)=\frac{iZ}{p^2-m^2}\overset{\eqref{A}}{=}\frac{i}{p^2-m^2}\tag{B}\label{B}$$ So the point was precisely there, when I wrote:

Also, if at the one-loop level one uses bare propators, there is the bare mass - which is divergent at this loop level - at the denominator. Is it possible that they're putting all of this under the rug and "ignoring" the propagator correction and thus the bare mass divergence caused by renormalization?

They aren't putting anything under the rug and as I noted the renormalized mass is used, but it's not ad hoc. They are indeed using the correct propagator \eqref{B}, but thanks to \eqref{A}, that is no different from the bare propagator with the renormalized mass replacing the bare mass and. So that's what happened evaluating the only 1-loop correction to the 4-point diagram is the fish diagram above. Anyways, the logic was right and eq. $(10.17)$ in P&S (in the renormalized perturbation theory section, but that doesn't matter) confirms it $$\delta\lambda:=Z^2\lambda_0-\lambda_R\tag{10.17}$$ and $f(\Lambda)$ in my question would be $-\delta\lambda$ of course.

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