Quantum Field Theory - Interacting Scalar Fields Is there an infinite number of interacting theories? Or is there a limit?
For example, I know about $\phi^6$ theory, which is non-renormalisable in 4D spacetime, but I've never really gone beyond $\phi^6$.
Is there say a $\phi^{10}$ or beyond?
 A: Let's do a small exercise of renormalization. The dimension of a scalar field $\phi$ is $[\phi] = \frac{d-2}{2}$, so the dimension of an operator $\mathcal{O}_p \equiv \phi^p$ is given by
$$
[\mathcal{O}_p] = p \,\frac{d-2}{2}\,.
$$
We want to see whether this operator is relevant ($[\mathcal{O}_p] \leq d$) or irrelevant ($[\mathcal{O}_p] > d$).${}^1$ Let's solve the inequality assuming $d>2$
$$
p > \frac{2d}{d-2}\;\Longleftrightarrow \mbox{$\mathcal{O}_p$ is irrelevant}\,.
$$
The case for $d = 2$ is special because $\phi$ is dimensionless, so you can have any powers of $\phi$. But let's stick to $d>2$. By plugging the values $d=3,4,5,\ldots$ you see that the only relevant operators are${}^2$


*

*For $d = 3$: $p=2,3,4,5,6$,

*For $d=4$: $p = 2,3,4$,

*For $d=5,6\hspace{1pt}$: $p=2,3$,

*For $d > 6$: Only $p=2$.


The operator $\mathcal{O}_2$ is the mass, so we obviously always have it.
Why do we care if the theory is renormalizable or not? Well, it depends on the question that you are asking. If you want to build effective field theories with a certain cutoff, then all $\mathcal{O}_p$ can be there in principle. But if you are trying to do one of these two things


*

*Building an UV complete theory

*Studying the low energy regime of a physical system


Then the irrelevant operators cannot appear, otherwise the theory is inconsistent (if you insist on 1.) or they simply vanish in the IR limit (if you are interested in 2.).

$\quad{}^1\;$If we are in the edge case $[\mathcal{O}_p] = d$ then $\mathcal{O}_p$ is actually said to be marginal. But for the purpose of this answer let's ignore this distinction.
$\quad{}^2\;$As noted in the comments, theories with an odd $p$ have a potential unbounded from below. They can be made sense of perturbatively, but not as a fully fledged QFT. We will also ignore this issue in this answer.
