Meaning of coefficients in polynomial potential for scalar field

[PhysicsRock]

Homework Statement

For a potential ##V(\phi) = \sum_{k=0}^N c_k \phi^k##, we set ##c_k = 0## for all ##k \neq 2##. What is the physical meaning of ##c_2##?

Relevant Equations

Equation of motion ##\partial_\mu \partial^\mu \phi = 2c_2 \phi##.

For the solution of the equation of motion, we take a plane wave ##\phi(x) = e^{ik_\mu x^\mu}##. Plugged in, we obtain

$$
-(k_0)^2 + (\vec{k})^2 = 2c_2 \Rightarrow k_\mu k^\mu = 2c_2
$$

One can then find the group velocity (using ##(k_0)^2 = \omega^2##) to be

$$
\vec{v}_g = \frac{\vec{k}}{\sqrt{ \vec{k}^2 - 2c_2 }}
$$

which does not break causality only if ##c_2 \leq 0##. This leads to the assumption, at least from my perspective, that ##c_2## must be related to the mass / be the mass of the field, since if ##m = 0##, the field would propagate at the speed of light. However, using ##2c_2 = k_\mu k^\mu##, we can see that ##c_2## must have the same unit as ##k_\mu k^\mu##, i.e. m##^{-2}## in S.I. units.

Did I make a mistake along the way or am I misinterpreting the meaning of ##c_2 \leq 0##? Help is highly appreciated.

 

[PhysicsRock]

I think I've figured it out. We can set ##c = \hbar = 1##. Then ##p^\mu = k^\mu##, which implies ##k_\mu k^\mu = p_\mu p^\mu = p^2 = -m^2##. Thus, we get ##2c_2 = -m^2 \Leftrightarrow c_2 = -\frac{m^2}{2}##.