# Potential in Relativistic Scalar Field Theory

My intention is to establish a Soliton equation. I have cropped a page from Mark Srednicki page no 576. I have understand the equation (92.1) but don't understand that how they guessed the potential in equation (92.2).

EDIT: Contrast the above potential the author used this potential in equation (2) $$U(\phi)= \frac{1}{8} \phi^2 (\phi -2)^2.$$ My query is , are we using different potential just for our convenient using? or I need to redefine the $\phi^4$ theory to get the potential ?

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Related questions by OP: physics.stackexchange.com/q/52590/2451 and links therein. –  Qmechanic Apr 9 '13 at 18:03

Any term in the action that does not transform under any symmetries is allowed. This means for a scalar field you can have any power $\phi^n$ in your Lagrangian. In equation (92.2) Scrednicki just choses a particular potential (that is, he chose (1) a polynomial in the scalar field along with (2)particular coefficents) that additionally gives you a vev for the scalar field of $\langle \phi \rangle = \pm v$. There are many potentials that will give your scalar field a vev, this is just the simplest.

EDIT: Like I said before 'There are many potentials that will give your scalar field a vev, this is just the simplest.' A bit more detail:

You can start with a general $\phi^4$ theory:

$V(\phi) = a_1 \phi + a_2 \phi^2 + a_3 \phi^3 +a_4 \phi^4$

and go from there. For example if you impose a discrete symmetry $\phi \rightarrow - \phi$ then $a_1 = a_3 = 0$. Moreover if you want $\phi$ to get a vev, one of the ways to do this is to choose $a_2 < 0$. But there are other potentials that will give the field a vev.

Also, judging from this and other questions to have posted, you seem to be obsessing over the 2 potentials

$V(\phi) = \frac{1}{8} \lambda (\phi^2 - v^2)^2$

and

$U(\phi) = \frac{1}{8}\phi^2 (\phi - v)^2$.

This is a bit like obsessing over whether vanilla or chocolate ice cream is better. It depends on your taste and what you are going to serve the ice cream with.

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The world would be a lot more interesting if you could turn vanilla into chocolate via a short sequence of elementary algebraic operations. –  user1504 Apr 11 '13 at 15:24
So you are saying that, we just giving the redefinition? Can we transform the later potential to the previous one? I have little confusions that, why discrete symmetry is the reason for vanishing $a_1 = a_3= 0$? –  Unlimited Dreamer Apr 11 '13 at 15:42
My intention is to get the original $\phi^4$ theory equation by adding these potential with Lagrangian. –  Unlimited Dreamer Apr 11 '13 at 15:48
@QFTdreamer - the transformation you are asking for was giving in a prior answer to another question of yours. See physics.stackexchange.com/questions/52590/… –  DJBunk Apr 11 '13 at 16:28