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I was recently learning about re-normalization in quantum field theory (in particular I was looking at the re-normalization of phi to the fourth theory). The superficial degrees of divergence of a Feynman diagram for a theory involving one scalar field with a self interaction term that goes like gϕ^n is given by the following formula; D = 4 - [g]V - E where [g] is the mass dimension of the coupling constant g (which in our case is equal to 4-n, where n is the exponent of the interaction term), V is the number of vertices in the diagram, and E is the number of external lines in the Feynman diagram. This works out to be D = 4 -V*(4-n) - E.

Now, with ϕ^3 theory the superficial degree of divergence D = 4 - V - E. Any diagrams where D is greater than or equal to 0 have UV divergences that need to absorbed into some bare parameter in the Lagrangian. There are only a few diagrams that I can find in ϕ^3 theory with D greater than or equal to zero (to one loop order). Naively I'd expect that ϕ^3 theory should be 'more' re-normalizable than ϕ^4 theory, since for ϕ^4 theory D = 4 - E, whereas for ϕ^3 theory D = 4 - V - E (it get's lower for each additional vertex). But this is probably not the case since I can't seem to find anyone anywhere talking about the re-normalization of ϕ^3 theory in 3+1 dimensions (in 5+1 dimensions I believe I've seen someone tackle it). The only divergent diagrams that I can find to one loop order are some vacuum diagrams with no external legs and the one loop correction to the ϕ propagator (D = 4 - 2 - 2 = 0, so a log divergence). So am I just missing something here? Can ϕ^3 theory not be re-normalized in 3+1 dimensions?

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$\phi^3$ theory in 4 dimensions is what is called a super-renormalizable theory. This means that as yourself pointed out that the superficial degrees of divergence decrease at higher orders. The reason that people rarely speak of it is that $\phi^4$ theory is arguably more interesting.

A feature of super-renormalizable theories is that they are quite often very sensative to UV physics. A nice way to see this would be to add a heavy fermion with mass $M$ and coupling $g' \phi \bar{\Psi}\Psi $ to the spectrum. You would find a vertex correction to the $\phi^3$ coupling, g, of the form $$\delta g \sim M g'^3 \left( \mathcal{O}(1)+logs \right) $$ Which looks very strange if we try to take the $M\rightarrow \infty $ limit.

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