# Interaction potential analysis from $\phi^4$ model

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In this paper, the authors consider a real scalar field theory in $d$-dimensional flat Minkowski space-time, with the action given by $$S=\int d^d\! x \left[\frac12(\partial_\mu\phi)^2-U(\phi)\right],$$ where $U(x)$ is a general self-interaction potential. Then, the authors proceed by saying that for the standard $\phi^4$ theory, the interaction potential can be written as $$U(\phi)= \frac{1}{8} \phi^2 (\phi -2)^2.$$

Why is this so? What is the significance of the cubic term present?

In this question Willie Wong answered by setting $\psi = \phi - 1$, why is that? Or why is this a gauge transformation? Does anyone have better argument to understand the interection potential?

This post imported from StackExchange Physics at 2014-03-22 17:08 (UCT), posted by SE-user Unlimited Dreamer
It's a field redefinition. You can use this $\psi$ in your calculations and if necessary switch back to $\phi$ in the end, if you need to. In other words, knowing the correlation functions $\langle \phi(x_1) \dotsm \phi(x_n) \rangle$ is completely equivalent to knowing $\langle \psi(x_1) \dotsm \psi(x_n) \rangle$.

This post imported from StackExchange Physics at 2014-03-22 17:08 (UCT), posted by SE-user Vibert

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It's not a gauge transformation, it's a field redefinition. Srednicki gives an example of this in exercise 10.5. In this exercise, a free field theory is turned into what looks like an interacting field theory by a field redefinition, however in perturbation theory, the scattering amplitudes vanish, confirming that the physics hasn't changed.

I suspect you will find the same here (though I haven't done it!) - if you compute scattering amplitudes for the 3-way vertices represented by the cubic terms resulting from this field redefinition, they should cancel.

This post imported from StackExchange Physics at 2014-03-22 17:08 (UCT), posted by SE-user twistor59
answered Mar 1, 2013 by (2,500 points)
That's precisely what I meant with my comment above.

This post imported from StackExchange Physics at 2014-03-22 17:08 (UCT), posted by SE-user Vibert
@Vibert Oh yes, I think we posted simultaneously!

This post imported from StackExchange Physics at 2014-03-22 17:08 (UCT), posted by SE-user twistor59
Twistor, i got the problem 10.5. can you please redefine this field so i will have more clear understanding.

This post imported from StackExchange Physics at 2014-03-22 17:08 (UCT), posted by SE-user Unlimited Dreamer
@QFTdreamer I'm not sure exactly what you mean. All I'm saying is that your $\phi \rightarrow \phi-1$ is analogous to Srednicki's $\phi \rightarrow \phi+\lambda\phi^2$.

This post imported from StackExchange Physics at 2014-03-22 17:08 (UCT), posted by SE-user twistor59
How will I confirm that my redefination is correct?

This post imported from StackExchange Physics at 2014-03-22 17:08 (UCT), posted by SE-user Unlimited Dreamer
@QFTdreamer Like in the Srednicki problem you could start by computing the tree level amplitudes and show that the three-point vertex contributions cancel.

This post imported from StackExchange Physics at 2014-03-22 17:08 (UCT), posted by SE-user twistor59
Or is there any way to validate the redefination?

This post imported from StackExchange Physics at 2014-03-22 17:08 (UCT), posted by SE-user Unlimited Dreamer
Can you show me that please? then I will have clear understanding.

This post imported from StackExchange Physics at 2014-03-22 17:08 (UCT), posted by SE-user Unlimited Dreamer
@QFTdreamer I would say that is a separate question. I'm not sure that asking for drill type questions like that is appropriate for phyiscs.se, but you could try, to see what response you get.

This post imported from StackExchange Physics at 2014-03-22 17:08 (UCT), posted by SE-user twistor59
Okay I'm trying to find out that.

This post imported from StackExchange Physics at 2014-03-22 17:08 (UCT), posted by SE-user Unlimited Dreamer
physics.stackexchange.com/questions/55532/…. can you check this question pleaee?

This post imported from StackExchange Physics at 2014-03-22 17:08 (UCT), posted by SE-user Unlimited Dreamer
A field redefinition cannot be 'correct' or 'wrong'. At this point, it's just bookkeeping. But you can interpret the resulting Lagrangian. If you get a positive mass term and a positive $\phi^4$ coupling, then you have a physical boson with quartic self-interactions. If you get a negative mass term, there will be SSB (as usual) and you should think of the field as "vev + excitations". However, formally the two theories are equal.

This post imported from StackExchange Physics at 2014-03-22 17:08 (UCT), posted by SE-user Vibert

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