Why is $V=(1/2) m^2 \phi^2$ for a free relativistic scalar field of mass $m$? Bit of a basic question here but how come for a free relativistic scalar field of mass $m$ such as Klein Gordon theory, we take the potential to be $$V=\frac{1}{2} m^2 \phi^2$$
Is the mass term squared just out of convention or is it purposeful? I also don't really understand why we're assuming a $\phi^2$ dependence.
Any help would be much appreciated
 A: 
Why $m^2$ in front of $\phi^2$ and why is $m$ the mass?

Fist of all, from dimensional analysis the prefactor to the $\phi^2$ term in the Lagrangian must have mass-dimension$^1$ $2$ in $3+1$ dimensions since the Lagrangian has mass-dimension $4$ and $\phi$ has mass-dimension $1$. This just tells us that we can write the term as $m^2\phi^2$ where $m$ is some mass-scale, but it does not give the relation ot the particle mass.
To get this connection recall the quantum-mechanical operator relations $E = i\frac{\partial}{\partial t}$ and $\vec{p} = -i\nabla$, and the free particle relativistic energy-momentum relation
$$E^2 = \vec{p}^2 + m^2$$
If the free scalar field is to be in agreement with relativity this operator relation must be satisfied when acting on $\phi$. This gives us
$$-\frac{\partial^2}{\partial t^2}\phi = -\nabla^2\phi + m^2\phi = 0 \implies \square \phi + m^2\phi = 0$$
which is exactly the equation of motion for a canonical scalar field with potential $V = \frac{1}{2}m^2\phi^2$.
Another reason for why $m$ is the mass can be found by looking at the relativistic Dirac equation
$$(i\gamma^\mu\partial_\mu - m)\psi = 0$$
By multiplying the equation above by $(i\gamma^\mu\partial_\mu + m)$ it turns into
$$\square \psi + m^2\psi = 0$$
thus the Dirac spinor $\psi$ satisfy the same equation as for a free scalar field with potential $V = \frac{1}{2}m^2\psi^2$.

Why $\phi^2$ and not something more complicated?

We can have more complicated potentials, but it would not lead to a free scalar field. For a free scalar field the equation of motion should be linear as otherwise the full field is no longer the superposition of individual exitations and the field will consequently self-interact. This restricts the potential to be on the form $V = V_0 + \mu^3\phi + \frac{1}{2}m^2\phi^2$. The $V_0$ term does not change the equation of motion and is only important cosmologically and there it's indistingvishable from a cosmological constant. The $\mu^3$ term can be removed by performing a field redefinition $\phi \to \phi  - \frac{\mu^3}{m^2}$. Thus we can without loss of generality take $\mu = 0$ unless $m^2 = 0$. However if $m^2=0$ then the field will be massless. The most general scalar field potential of a free and massive scalar particle is therefore $\frac{1}{2}m^2\phi^2$.

$^1$: I'm using Planck-units $\hbar=c=1$ in this answer to make it easy for myself
