$m^2$ term in quantum field theory Consider the following Hamiltonian :
$$ \mathcal{H} = \frac{1}{2} \left ( \partial_{0} \phi \right )^2 + \frac{1}{2} m^2 \phi^2 +\frac{1}{2} \left ( \nabla \phi \right )^2
$$
Recently I faced some difficulty with this Hamiltonian. If we try to use the discretized version of this lagrangian we consider a lattice whit N-Sites where on each site we put a particle then the Hamiltonian will contain the kinetic term + each site potential term and the coupling term between NN sites, the point is, in that case, the Hamiltonian would be:
$$ \mathcal{H} = \frac{1}{2m} \left ( \partial_{0} \phi \right )^2 + \frac{1}{2} m \omega^2 \phi^2 +\frac{1}{2} m\Omega^2 \left ( \nabla \phi \right )^2
$$
where $\Omega$ is the strength of coupling. My question is two fold


*

*how these two terms are related? I mean how can I start from the second one and find the first one?

*In the second Hamiltonian, how can I relate $\Omega$ to $ \omega$?

 A: Ok, this is the free Klein-Gordon Hamiltonian. Strictly, you should be writing this in terms of the momentum canonically conjugate to $\phi$. By convention, we use the symbol $\pi$ for that. So, your original Hamiltonian density is
\begin{array}
    \mathcal{H} = \frac{1}{2} \pi^2 + \frac{m^2}{2}\phi^2 + \frac{1}{2}(\nabla\phi)^2.
\end{array}
There's a good reason for being a stickler about notation in this way. You'll see, presently.
First, make the substitution $m\rightarrow \omega$ to get
\begin{array}
    \mathcal{H} = \frac{1}{2} \pi^2 + \frac{\omega^2}{2}\phi^2 + \frac{1}{2}(\nabla\phi)^2.
\end{array}
Next, make the change of field coordinate $\phi\rightarrow \sqrt{m}\phi$. The reason for being such a stickler about notation is that, if you work it out carefully (for instance, by doing the change of variables using the action/Lagrangian density and then transitioning back to the Hamiltonian picture) you'll find that rescaling $\phi$ in this way also requires you to make the substitution $\pi\rightarrow \pi/\sqrt{m}$. That gives you
\begin{array}
   \mathcal{H} = \frac{1}{2m} \pi^2 + \frac{m\omega^2}{2}\phi^2 + \frac{m}{2}(\nabla\phi)^2.
\end{array}
The addition of $\Omega$ is just something that is done by hand, afterward. It is totally unrelated to $\omega$. 
What you will find, though, is that the lattice supports modes that have frequencies that look like $$\omega_{\mathrm{mode}} = \sqrt{\omega^2 + \Omega^2 \mathbf{k}^2},$$
with $k$ some version of the wave number appropriate to your lattice.
