When testing a theory for its renormalizability, in practice one always calculates the mass dimension of the coupling constants $g_i$. If $[g_i]<0$ for any $i$ the theory is not renormalizable. I am wondering where this criterion/trick comes from? Is there an easy way to see that a coupling constant with negative mass dimension will yield a non-renormalizable theory?
Quantum Field Theory – Renormalizability by Power Counting Explained
dimensional analysisinteractionslagrangian-formalismquantum-field-theoryrenormalization
Related Solutions
At least at the operational level, whether an operator is relevant or irrelevant (in the IR) tells you about it's canonical scaling dimension.
I think the BPHZ picture of renormalization might help here. Given a physical theory, we'd like to estimate which Feynman graphs of that theory diverge. You can estimate a "superficial degree of divergence" for each Feynman diagram by considering the canonical scaling dimension of each vertex or edge in the diagram. Say you have a diagram with a vertex corresponding to an IR-irrelevant (UV-relevant) interaction term. Then, in the UV (for large momenta), it will cause the amplitude to diverge. Now, as you go to higher loop order, or insert more of those vertices at same loop order, the divergence becomes worse. And as you add more and more such vertices, the corresponding diagrams become more and more important in the UV, due to the couplings being UV-relevant! The essence is that there are infinitely many ("independent") Feynman diagrams which diverge, so you can't cook up enough counterterms to cancel all these divergences. So these terms in the lagrangian give rise to non-renormalizable theories.
See, for eg. Peskin & Schroeder section 10.1, or the link above for how to calculate the superficial degree of divergence of any diagram. Another technicality is that you don't actually consider Feynman graphs, you instead consider subgraphs. What that means is that the external (uncontracted) legs can be off-shell, as if they were sitting inside a bigger Feynman diagram.
Given your question, maybe you've already come across the stuff I've said and are asking for something else, in which case your question is not clear to me.
The key thing is that you need to be working with canonically normalized fields in order to use the power counting arguments.
Let's expand GR around flat space \begin{equation} g_{\mu\nu} = \eta_{\mu\nu} + \tilde{h}_{\mu\nu} \end{equation} The reason for the tilde will become clear in a second. So long as $\tilde{h}$ is "small" (or more precisely so long as the curvature $R\sim (\partial^2 \tilde{h})$ is "small"), we can view GR as an effective field theory of a massless spin two particle living on flat Minkowski space.
Then the Einstein Hilbert action takes the schematic form \begin{equation} S_{EH}=\frac{M_{pl}^2}{2}\int d^4x \sqrt{-g} R = \frac{M_{pl}^2}{2} \int d^4x \ (\partial \tilde{h})^2 + (\partial \tilde{h})^2\tilde{h}+\cdots \end{equation} where $M_{pl}\sim 1/\sqrt{G}$ in units with $\hbar=c=1$. $M_{pl}$ has units of mass. In this form you might thing that the interaction $(\partial \tilde{h})^2 \tilde{h}$ comes with a scale $M^2_{pl}$ with a positive power. However this is too fast--all the QFT arguments you have seen have assumed that the kinetic term had a coefficient of -1/2, not $M_{pl}^2$. Relatedly, given that $M_{pl}$ has units of mass and the action has units of $(mass)^4$, the field $\tilde{h}$ is dimensionless, so it is clearly not normalized the same way as the standard field used in QFT textbooks.
Now classically, the action is only defined up to an overall constant, so we are free to think of $M_{pl}^2$ as being an arbitrary constant. However, in QFT, the action appears in the path integral $Z=\int D\tilde{h}e^{iS[\tilde{h}]/\hbar}$ (note the notational distinction between $\tilde{h}$ and $\hbar$). Thus the overall constant of the action is not a free parameter in QFT, it is fixed and has physical meaning. Alternatively, you have to remember that the Einstein Hilbert action will ultimately be coupled to matter; when we do that, the scale $M_{pl}$ sitting in front of $S_{EH}$ will not multiply the matter action, and so $M_{pl}$ sets the relative scale between the gravitational action and the matter action.
The punchline is that we can't simply ignore the overall scale $M_{pl}^2$, it has physical meaning (ie, we can't absorb $M_{pl}$ into an overall coefficient multiplying the action). On the other hand, we want to put the action into a "standard" form where the overall scale isn't there, so we can apply the normal intuition about power counting. The solution is to work with a "canononically normalized field" $h$, related to $\tilde{h}$ by
\begin{equation} \tilde{h}_{\mu\nu} = \frac{h_{\mu\nu}}{M_{pl}} \end{equation}
Then the Einstein Hilbert action takes the form
\begin{equation} S_{EH} = \int d^4 x \ (\partial h)^2 + \frac{1}{M_{pl}} (\partial h)^2 h + \cdots \end{equation}
In this form it is clear that the interactions of the form $(\partial h)^2 h$ have a "coupling constant" $1/M_{pl}$ with dimensions 1/mass, which is non-renormalizable by power counting in the usual way.
Best Answer
Yes, suppose $[g] = \delta$. By dimensional analysis only we can write that a loop diagram contributes $$ \sim g^{n} \int \frac{d^4 k}{k^{4-n\delta}} $$ If $\delta=0$, this diverges logarithmically, but can be re-normalized. If $\delta$ is less than zero, it diverges by simple power counting.
This is VERY informal. Technically, you should study the superficial degree of divergence of a diagram. But that's called superficial for a reason. So for the whole story I think you need Weinberg's theorem, which is a rule for telling exactly if a diagram diverges.