# How can a non-derivative interaction involve the derivative of a scalar field?

I was reading up on the paper "The Fate of the False Vacuum" by Sidney Coleman and the claim is made that a scalar field with standard Lagrangian density: $$L = \frac{1}{2} \delta_{\mu} \delta^{\mu} \phi - U(\phi)$$

has non- derivative interactions. How can this assumption hold with the 4-derivative implicitly taken in the "kinetic" part of the Lagrangian? Sorry for the basic question and thanks in advance.

• Are those deltas $\delta_\mu\delta^\mu$ supposed to be partials $\partial_\mu\partial^\mu$?
– hft
Commented Jul 17 at 23:24

First, just a small terminology note. People normally say "derivative" instead of "4-derivative" to refer to $$\partial_\mu$$. People might say "4-gradient" if they want to specify the number of dimensions. However, "4-derivative" might confuse people in this context, because when talking about Lagrangians one might want to talk about "4 derivative interactions", which would be terms like $$\partial_\mu \phi \partial^\mu \phi \partial_\nu \partial^\nu \phi$$ that involve four derivative symbols, $$\sim \partial^4$$ (as opposed to, say, a 6 derivative interaction, like $$(\partial_\mu \phi \partial^\mu \phi)^3$$). The fact that these are derivatives with respect to a 4-dimensional spacetime is implied from context so people usually don't specify the 4 associated with the spacetime dimension explicitly when referring to the derivative operator. Anyway, that is a minor note, but I'm just mentioning it since it initially confused me when I read your question.
Normally we split a Lagrangian into a "free" part and an "interacting" part $$L = L_{\rm free} + L_{\rm int}$$ where the free part of the Lagrangian is at most quadratic in the fields. For a single scalar field on Minkowski spacetime, typically $$L_{\rm free} = -\frac{1}{2} \partial_\mu \phi \partial^\mu \phi - \frac{1}{2} m^2 \phi^2$$ where the mass $$m$$ can be zero. In your example, normally one would split out the quadratic term in $$U(\phi)$$ and include it in the free part of the Lagrangian. The reason this is called the "free" part of the Lagrangian, is that it corresponds a harmonic oscillator, a problem we can solve exactly, and physically represents particles that do not scatter or interact with each other.
Then, the interaction terms consist of terms that involve higher orders in the fields. In principle, there can also be arbitrary numbers of derivatives. So, in general, $$L_{\rm int}$$ is some series that looks something like (schematically) $$L_{\rm int} \sim \sum_{n_d} \sum_{n_\phi \geq 3} \frac{g}{\Lambda^{n_d+n_\phi-4}} \partial^{n_d} \phi^{n_\phi}$$ where $$g$$ is some dimensionless coupling constant associated with each interaction term, and $$\Lambda$$ is a mass scale needed for dimensional consistency. Note this is just schematic, the derivatives would be contracted in a way that they didn't form a total derivative, which we usually can ignore in a Lagrangian. For example, $$\partial_\mu \phi \partial^\mu \phi \partial_\nu \phi \partial^\nu \phi$$ would be an interesting interaction, but $$\partial_\mu \partial^\mu \partial_\nu \partial^\nu \phi^4$$ would not, because it is a total derivative. Also, $$n_d$$ must be an even number, in order to be able to contract all the indices.
So, what Coleman is saying, is that he is considering an interaction Lagrangian $$L_{\rm int}$$ with $$n_d=0$$ for every term. Then $$L_{\rm int} = \sum_{n_\phi \geq 3} \frac{g}{\Lambda^{n_\phi - 4}} \phi^{n_\phi} = U(\phi)$$ is just some function of $$\phi$$, with no derivatives involved. The reason he is considering this case is that he is interested in vacuum (or false vacuum) states, or states with lowest energy (or at least a local minimum in energy), and any spatial or temporal variation in the field would tend to have higher energy, due to gradient energy coming from the kinetic term $$(\partial \phi)^2$$. Or, in simpler terms, it is an approximation that is expected to be good in the case he is interested in. Happily, this approximation also makes the mathematical problem tractable.
If $$U(\phi)$$ is not a quadratic function, then the higher-order terms (e.g. a $$\phi^4$$ term) can be interpreted as field self-interactions that don’t involve derivatives.