I am studying Renormalization but I don't understand why theories which have a coupling constant with negative dimension of mass requires more and more counterterms going up with the perturbative order.

Let we use for simplicity the quartic Fermi Theory in four dimensions: $\mathcal{L}=G(\bar{\psi}\psi)(\bar{\psi}\psi)$, since $[\psi]=[M]^{3/2}$ and $[\mathcal{L}]=[M]^4$ we have $[G]=[M]^{-2}$, so far so good.

From what I understand is that the first loop correction has two vertecies, and so 1-Loop-Correction ~ $ G^{2} $ ~ $ [M]^{-4}$, this means that we could also add a counterterm with one vertex which has a coupling constant $\delta G$ with $\delta G=[M]^{-4}$.

This is my first problem, because the interaction $\delta G(\bar{\psi}\psi)(\bar{\psi}\psi)$ has not the right dimensions of $[M]^{4}$ but instead of $[M]^{2}$, so what I am doing wrong?

Secondly my professor said that we could also add another term: $\tilde{G}(\partial_{\mu}\bar{\psi}\partial^{\mu}\psi)(\partial_{\nu}\bar{\psi}\partial^{\nu}\psi)$ ($\tilde G=[M]^{0}$? I am not sure) because it has the same dimension of the one loop divergence; but again the thing that I don't understand is that $[(\partial_{\mu}\bar{\psi}\partial^{\mu}\psi)(\partial_{\nu}\bar{\psi}\partial^{\nu}\psi)]=[M]^{10}$.

In the ends this should demonstrate that we need all the possible counterterms (compatibles with the symmetries of our theory), but I don't understand why these countermens have different dimensions of mass from the Lagrangian.

I checked some books (Weinberg, Peskin, Maggiore) but I didn't find these explanations in detail, if someone can point a book and chapter where this (or any non renormalizable theory) is explained better I would be more than happy!

  • 1
    $\begingroup$ My rough understanding: A 1-loop diagram (with two $G$-vertices), will indeed give a diagram that depends like $G^2$, so the counterterm $\delta G = [M]^{-4}$, but this means the $\delta G$ is in front of a derivatively coupled operator, as in your counterterm will be $\delta G (\partial_\mu \bar{\psi} \partial^\mu \psi) (\partial_\nu \bar{\psi} \partial^\nu \psi)$. You seem to think that the counterterm has to be in front of the same operator as the original, but the point is that you generate divergences of different powers when you have non-renormalizable operators $\endgroup$ Jul 25, 2022 at 23:06
  • $\begingroup$ I see, but the problem is again that $[\delta G(\partial_{\mu}\bar{\psi}\partial^{\mu}\psi)(\partial_{\nu}\bar{\psi}\partial^{\nu}\psi)]=[M]^{6}$, so it should be something like $\delta G(\partial_{\mu}\bar{\psi}\partial^{\mu}\psi)(\bar{\psi}\psi)$ ? (This would have the correct dimensions of $[M]^{4}$) $\endgroup$ Jul 25, 2022 at 23:15

1 Answer 1

  1. The interaction term $G\psi^2\bar{\psi}^2$ has a coupling constant $G$ of mass-dimension $[G]=-2$, and the interaction term $\tilde{G}(\partial\psi)^2(\partial\bar{\psi})^2$ has a coupling constant $\tilde{G}$ of mass-dimension $[\tilde{G}]=-6$, i.e. both coupling constants are irrelevant/non-renormalizable, cf. e.g. this Phys.SE post.

  2. The 1-loop diagram $>\!\!O\!\!<$ with 4 external fermions contributes to the interaction term $G\psi^2\bar{\psi}^2$, and hence to the running of the coupling constant $G$. OP's mismatch in mass dimension seems caused by the omission of 2 internal fermion propagators and various integrations in the 1-loop diagram.


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