Complete renormalization in $\phi^4$-theory?

In the one-loop renormalization of $\phi^4$-theory, only 1PI vertex functions $\Gamma^{(2)}$ and $\Gamma^{(4)}$ are regularized and renormalized. But they do not exhaust all the irreducible connected diagrams at one loop. One can have a diagram, for example, with one-loop, 3 vertices and 6 external lines, or with one-loop, 4 vertices and 8 external lines and so on. What about these diagrams? They respectively correspond to $\Gamma^{(6)}$ and $\Gamma^{(8)}$. What about these 1PI diagrams with one-loop? Shouldn't they require renormalization as well? In fact these diagrams contribute to the effective potential.

EDIT : arxiv.org/abs/hep-ph/9901312 This might be an useful reference. Please look at the one-loop diagrams in the calculation of the effective potential in $ϕ^4$-theory.

• Alright, then this is a round-about way of asking how we know that $\phi^4$-theory is renormalizable (i.e., once when you take care of $\phi^2$ and $\phi^4$ operators, everything else is alright). – innisfree Sep 23 '15 at 6:58

The naive power counting approach for a $d$-dimensional theory with coupling constant $\lambda$ tells us that the amplitude of diagrams with $E$ external lines and $V$ vertices behaves with the cutoff $\Lambda$ as $\propto\Lambda^D$ with $$D = d - [\lambda]V - \frac{d-2}{2}E$$ where $[\dot{}]$ is the mass dimension. Since $\phi^4$ in four dimension has a dimensionless coupling, $$D = 4-E$$ and since only diagrams with $D \geq 0$ need renormalization, the only diagrams needing it in 4D $\phi^4$ are those with $E \leq 4$. All diagrams with an odd number of external lines vanish due to the $\phi\mapsto-\phi$ symmetry, so what's left to renormalize is $E=0,2,4$, which are the vacuum energy, the propagator, and the 4-vertex, respectively.
The diagrams you ask about exist, but have $D < 0$, and do not need to be renormalized, since they are not diverging when we take the cutoff to infinity.
• @ACM Can you please make a contrast with $\phi^6$ theory? In particular, how can I understand that infinitely many diagrams have to be renormalized here? – SRS Mar 1 '17 at 8:58
• @SRS All you need to do is plug in the different value of $[\lambda]$ and then think about which diagrams have $D\geq 0$ with this new formula. – ACuriousMind Mar 1 '17 at 12:16
• @ACuriousMind I think the formula $D = d - [\lambda]V - [\phi]E$ holds true only for $\lambda \phi^N$ theories. How does one get the corresponding formula for more complicated theories? Or, is this a general formula for all theories? – Nanashi No Gombe Jul 23 '17 at 13:02
• @NanashiNoGombe Of course it's not a general formula for all theories (how could it be, what would "$\lambda$" be in a theory with more than one coupling constant?), but the generalization is pretty straightforward (if awkward to write down) - you need to introduce different counts $V_i$ for different types of vertices with coupling constants $\lambda_i$. – ACuriousMind Jul 24 '17 at 8:01