I am given the following lagrangian: $L=-\frac{1}{2}\phi\Box\phi\color{red}{ +} \frac{1}{2}m^2\phi^2-\frac{\lambda}{4!}\phi^4$ and the questions asks:
How many constants c can you find for which $\phi(x)=c$ is a solution to the equations of motion? Which solution has the lowest energy (ground state)?
My attempt: since lagrangian is second order we have the following for the equations of motion: $$\frac{\partial L}{\partial \phi}-\frac{\partial}{\partial x_\mu}\frac{\partial L}{\partial(\partial^\mu \phi)}+\frac{\partial^2}{\partial x_\mu \partial x_\nu}\frac{\partial^2 L}{\partial(\partial^\mu \phi)\partial(\partial^\nu \phi)}=0 $$ then the second term is zero since lagrangian is independent of the fist order derivative. so we will end up with:
$$\frac{\partial L}{\partial \phi}=-\frac{1}{2} \Box \phi+m^2\phi-\frac{\lambda}{3!}\phi^3$$ and:$$\frac{\partial^2}{\partial x_\mu \partial x_\nu}\frac{\partial^2 L}{\partial(\partial^\mu \phi)\partial(\partial^\nu \phi)}=-\frac{1}{2}\Box\phi$$ so altogether we have for the equations of motion: $$-\frac{1}{2}\Box\phi+m^2\phi-\frac{\lambda}{6}\phi^3-\frac{1}{2}\Box\phi=0$$ and if $\phi=c$ where "c" is a constant then $\Box\phi=0$ and then the equation reduces to $$m^2\phi-\frac{\lambda}{6}\phi^3=0$$ which for $\phi=c$ gives us 3 solutions:$$c=-m\sqrt{\frac{6}{\lambda}}\\c=0\\c=m\sqrt{\frac{6}{\lambda}}$$ My question is is my method and calculations right and how do I see which one has the lowest energy (ground state)? so I find the Hamiltonin for that?