# 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?

• I have also found that since the first two terms look like kinetic terms and the last terms looks like interaction term my hamiltonian looks like: $$H=-\frac{1}{2}\phi\Box\phi +\frac{1}{2}m^2 \phi^2 +\frac{\lambda}{4!}\phi^4$$ and for energies I have :$$c=+/- m\sqrt{\frac{6}{\lambda}}--> E=\frac{9}{2}\frac{m^4}{\lambda}\\c=0--> E=0$$ is it right? so that c=0 has the lowest energy. – MSB Sep 18 '16 at 16:42
• This is Problem 3.5 in Schwartz's QFT textbook. – Qmechanic Sep 18 '16 at 17:38
• yes it is I have mentioned that I'm using Schwartz book in the comments – MSB Sep 18 '16 at 18:08

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 $-1/2 m^2 \phi^2 + \lambda/4! \phi^4$.

• You made an important sign error as well: in writing your equation of motion, you went from having the $\phi^2$ and $\phi^4$ terms having the opposite sign to having the same sign. So your final answer is incorrect. – tparker Sep 18 '16 at 17:04
• there was a minus sign typing error but for the rest of calculation i cant see where I did mistake. – MSB Sep 18 '16 at 18:07
• @tparker Not just any derivatives, time derivatives. Terms with only space derivatives are potential energy stored in the variations of the field with space. – Sean E. Lake Sep 18 '16 at 18:22
• Also, @MSB is right about that sign, I mixed it up in my post. The problem he's looking at is the prototype for looking at kink solitons, so it needs to have two minima in the potential. – Sean E. Lake Sep 18 '16 at 18:29
• @MSB Your $c$'s are fine, but in going from Lagrangian to Hamiltonian you need to change the sign of the $1/2 m^2 \phi^2$ term because it's a potential term. – tparker Sep 18 '16 at 18:55

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.

• Thank you Sean, Is that minus sign gonna affect my constant "c" values? – MSB Sep 18 '16 at 16:38
• Nope, because that term drops out, that's why it's minor. – Sean E. Lake Sep 18 '16 at 16:39

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 $c=\phi=\pm\sqrt{\frac{6m^2}{\lambda}}$.