# Interpretation of an "interaction" term

In QFT a polynomial (of degree >2) in the fields is said to be an interaction term, Ex.: $\lambda\phi^4$.

Question

Is it possible to give an interpretation to terms like $\frac{1}{\phi^n}$? (for $n\in\mathbb{N}$)

Cheers

• Make a lattice, and do a Monte-Carlo, and then all the issues of renormalization go away at finite lattice spacing, and you can understand all the field potential terms immediately. Nov 8, 2012 at 18:35
• @RonMaimon: But then you have to explain why your results are independent of the lattice chosen, and the renormalization issues reappear in full complexity. Nov 8, 2012 at 18:50
• @ArnoldNeumaier: Yes, of course, but they are obvious then. Nov 8, 2012 at 19:16

In principle, yes, but only if the expectation value of $\phi$ is nonzero, so $\phi$ would immediately be shifted. Moreover, the result would be badly nonrenormalizable, so nobody is using such terms.
An important case of a nonpolynomial interaction that received considerable attention in 1+1D is the interaction $\sin\phi(x)$ of the sine-Gordon model http://en.wikipedia.org/wiki/Sine-Gordon#Quantum_version
• Could you in principle have something like $1/(1-\phi)$ which then would just result in a simple power series expansion? Nov 8, 2012 at 18:18
Another problem: The interaction $V(\phi,x) = 1/\phi^n(x)$ isn't stable if $n$ is odd. This energy isn't bounded below. You can try to fix this by setting $V(\phi,x) = 1/|\phi|^n(x)$, but there's still a kind of stability problem. The minimum of this potential is at $\phi(x) = \infty$, so if you start in the naive vacuum, you'll generate a huge expectation value.