Lagrangian and finding equations of motion 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?
 A: Looks good so far.  To find the Hamiltonian you just use that if $L = T - U$ then $H = T + U$ (technically there are some extra assumptions there, but if your case it works out fine).  Since $T = 0$ if $\phi$ is constant, you just need to find out which of those values $c$ minimize(s) the potential energy $-1/2 m^2 \phi^2 + \lambda/4! \phi^4$.
A: You have a minor error from a missing minus sign here:$$\frac{1}{2}\Box\phi+m^2\phi-\frac{\lambda}{6}\phi^3-\frac{1}{2}\Box\phi.$$ It should be (after combining terms):
$$-\Box\phi+m^2\phi-\frac{\lambda}{6}\phi^3.$$
Now, for finding the Hamiltonian you might find it easier to integrate the term $\phi \Box \phi / 2$ by parts to get $-\partial_\mu \phi \partial^\mu \phi / 2 + \mathrm{surface\ term}$. That way you can use standard formulae for constructing the Hamiltonian using canonical momenta. That is, assuming you want to construct the Hamiltonian. This Lagrangian has a fairly simple structure with a kinetic energy term (time derivatives of $\phi$), and every other term is potential energy. So, since these states are constant in time and space, their energy will be just potential energy:$$E = \int \left[-\frac{m^2}{2} \phi^2 + \frac{\lambda}{4!} \phi^4\right] \operatorname{d}^3 x.$$
Edit: fix my own sign error.
A: Thanks to all you guys I have found that my mistake was at confusing the kinetic and interaction terms. so here is my answer to this question:
this problem is basically finding the values for $\phi$ that minimizes the effective potential and I have found them above named $c_1$,$c_2$ and $c_3$ considering those are correct, I have for my effective potential now:
$$V(\phi)=-\frac{1}{2}m^2 \phi^2+\frac{\lambda}{4!}\phi^4$$since $L=KE-V$ then my Hamiltonian will be $$H=-\frac{1}{2}\phi\Box\phi -\frac{1}{2}m^2\phi^2+\frac{\lambda}{4!}\phi^4$$ for c=0 its just gonna give me zero but for $c=\pm \sqrt{\frac{6m^2}{\lambda}}$ now substituting this into the hamiltonain:$$<H>=E=0-\frac{1}{2}m^2(\sqrt{\frac{6m^2}{\lambda}})^2+\frac{\lambda}{4!}(\sqrt{\frac{6m^2}{\lambda}})^4\\E=\frac{m^4}{\lambda}(-\frac{6}{2}+\frac{36}{4!})\\E=\frac{m^4}{\lambda}(-\frac{6}{2}+\frac{3}{2})\\E=-\frac{3}{2}\frac{m^4}{\lambda}$$ so there are two solutions that have the lowest energy which is $c=\phi=\pm\sqrt{\frac{6m^2}{\lambda}}$.
A: is just want to add to this discussion that book has no type error according to page 30 kinetic terms are billinear meaning that they have exactly two fields so kinetic terms in this case are:
T=−1/2(ϕ□ϕ)+1/2(m2ϕ2)

