# $\phi ^4$ theory explaining [closed]

In $φ^4$ theory we often write the Lagrangian as $$\mathcal{L}=\frac{1}{2}\partial^\mu \phi \partial_\mu \phi -\frac{m^2}{2}\phi^2 -\frac{\lambda}{4!}\phi^4 \tag {1}$$

If I want to write from the Relativistic Lagrangian then it takes $$\mathcal{L}=\frac{1}{2}\partial^\mu \phi \partial_\mu \phi -V \tag{2}$$ but how will I convert this equation to equation like (1) ?

EDIT: I just want to get Equation (1) from equation (2)

EDIT by joshphysics: What motivates choosing to study the $\phi^4$ potential as opposed to other potentials?

• what does 'write down from the source' mean? – nervxxx Apr 9 '13 at 17:40
• Are you asking why the potential is chosen to be of that form in the sense of what do we gain by studying the $\phi$-fourth potential? – joshphysics Apr 9 '13 at 17:41
• Yup , mathematical details will be good for me :) – user12906 Apr 9 '13 at 17:42
• @nervxxx , I just want to get Equation (1) from equation (2) – user12906 Apr 9 '13 at 17:45
• if you just want to get (1) from (2) then er... set $V = \frac{m}{2} \phi^2 + \frac{\lambda}{4!} \phi^4$......? I don't really know what you're trying to ask. – nervxxx Apr 9 '13 at 17:49

As I understand it, you would like to know why we often choose to study the special potential $$V(\phi) = \frac{1}{2} m\phi^2 + \frac{1}{4!}\lambda \phi^4$$ Here are a couple of reasons
1. It yields a simple example of an interacting field theory. If you were to have chosen $V(\phi) = \frac{1}{2}m\phi^2$, then the corresponding Lagrangian is that of a free (non-interacting) massive scalar of mass $m$, and we want to go beyond free theory.
2. It's one of the simplest local functions of $\phi$ you can write down that gives you interactions since it's just a polynomial in $\phi$. Why not $\phi^3$? Well you can really just as well pick $\phi^3$ and learn a lot about how scalar field theory works (as in Srednicki), but people like using the $\phi^4$ potential because it is bounded below and doesn't have an unstable critical point at the origin like the $\phi^3$ potential does.
• So, I can transform the potential to $U(\phi)= \frac{1}{8} \phi^2 (\phi -2)^2$ ? If yes then how? – user12906 Apr 9 '13 at 18:30