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 $2c_2 = -m^2 \Leftrightarrow c_2 = -\frac{m^2}{2}$.