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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?

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closed as not a real question by user1504, Manishearth Apr 10 '13 at 11:42

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

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 :) – Unlimited Dreamer Apr 9 '13 at 17:42
@nervxxx , I just want to get Equation (1) from equation (2) – Unlimited Dreamer 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

1 Answer 1

up vote 3 down vote accepted

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.

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So, I can transform the potential to $U(\phi)= \frac{1}{8} \phi^2 (\phi -2)^2$ ? If yes then how? – Unlimited Dreamer Apr 9 '13 at 18:30
More details on… – Unlimited Dreamer Apr 9 '13 at 18:31

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