Graduate Validity of Scalar Field Lagrangian with Linear and Quadratic Terms

[Gaussian97]

TL;DR

Is a term proportional to $\phi$ valid in a scalar Lagrangian?

Hi, if I want to construct the most general Lagrangian of a single scalar field up to two fields and two derivatives, I usually see that is
$$\mathscr{L} = \phi \square \phi + c_2 \phi^2$$ i.e. the Klein-Gordon Lagrangian.
My question is, would be valid the Lagrangian
$$\mathscr{L} = \phi \square \phi + c_1 \phi + c_2 \phi^2$$
?

 

[nrqed]

With a linear term, the energy is not bounded from below. So this is usually not considered. This is a problem for applying perturbation theory, this indicates that one is expanding around a field configuration that cannot be used as a vacuum. So one then shifts $\phi \rightarrow \phi - \frac{c_1}{2c_2}$ which gets rid of the linear term.

 

[Haborix]

With a linear term, the energy is not bounded from below.

A linear term shifts the location of the minimum of the potential and shifts the overall potential by a finite constant. (speaking loosely about potential densities as potentials)

 

[vanhees71]

With the linear term and the quadratic term the energy is still bounded from below provided $c_2$ has the "right sign". So it's a fine theory (as $\phi^3$ theory is not though it's treated at length in some textbooks to have a simple model to explain perturbative renormalization theory, e.g., in Collin's Renormalization; adding a $\phi^4$ term makes it again a theory with the Hamiltonian bounded from below).

For the free field here you can just introduce a new field shifted field as explained in #2, and you are back at the usual free theory.

 

[Gaussian97]

Ok! Thank you!