High order self-interacting potential terms in QFT

Question

In this answer on another Physics StackExchange thread, the self-interaction potential terms of a classical field theory ($ \phi^n $ terms with $ n>2 $) are said to correspond to higher-order allowed vertices in Feynman diagrams.

I think this is interesting, because a priori, is there any reason to believe that the potential term of a field theory should be a polynomial at all, let alone end after just a few terms? Of course, the existence of particles at all suggests that we should restrict ourselves to $ \phi^2 $ terms, but any potential will look like that in the vicinity of a minimum anyway, so I'm not sure that that is an argument.

If we allow for (analytic over $\mathbb{R}$) potentials, we could Taylor expand and get an infinite family of possible self-interacting vertices for our Feynman diagrams. Seems like a problem, but if the coefficients of the expansion drop off suitable quickly, I feel like that should restrict the contributions from diagrams including those vertices.

So my specific questions would be:

• Why do we assume that field potentials stop after at most a couple of terms? Are there technical theoretical reasons, or is it just because we don't need to based on experimental results?
• Is there any way for us to experimentally probe the field potentials in a more detailed manner? Do we need to, and why/why not?

For context, I'm a third year undergraduate, so please excuse me if these questions are obvious/show a misunderstanding of the theory :)

Answer 1

Weinberg's view (in The Quantum Theory of Fields) is particularly illuminating for your questions. As an undergrad, it wouldn't be the best idea to spend time deciphering Weinberg's words but basically it goes like this. You start with creation and annihilation operators. Then, there is a theorem that says all operators can be expressed as a sum of product of creation and annihilation operators.

To construct the interaction Hamiltonian, a basic requirement is that it has to be Hermitian. Then if the interaction Hamiltonian can be expressed as a space integral of a Hamiltonian density, another requirement is that the Hamiltonian density at spacelike separations have to commute (for the S-matrix to be Lorentz invariant, also known as the causality condition). There is a third requirement from the cluster decomposition principle but let's not worry about that here.

Now we need a sum of product of creation and annihilation operators that simultaneously satisfy these conditions. The easiest way to do it is to construct a field with simple commutation relations from the creation and annihilation operators and then express everything else in terms of the field. That is the motivation for the construction of quantum fields.

Thus, the Hamiltonian as a sum of product of creation and annihilation operators can now be expressed as a polynomial of the field. Based on the condition that the Hamiltonian has to be bounded below (a physical theory must have a ground state) and the symmetry of the theory, one can decide which terms are allowed. A condition on renormalizability puts further restrictions on which terms one can use.

All things considered, if a theory is renormalizable, there are only a finite number of terms in the interaction.

In a sense, all the experiments in particles and high energy physics are attempts at "probing" the structure of a theory. But the "field potentials" as you mentioned are operators, we cannot probe its structure directly and can only measure a very limited number of observables.