Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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).enter image description here

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 ?

share|improve this question
    
Related questions by OP: physics.stackexchange.com/q/52590/2451 and links therein. –  Qmechanic Apr 9 '13 at 18:03
add comment

1 Answer

up vote 2 down vote accepted

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.

share|improve this answer
1  
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.