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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?

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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 $\langle \psi(x_1) \dotsm \psi(x_n) \rangle$. – Vibert Mar 1 at 8:46

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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.

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That's precisely what I meant with my comment above. – Vibert Mar 1 at 8:47
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@Vibert Oh yes, I think we posted simultaneously! – twistor59 Mar 1 at 8:54
Twistor, i got the problem 10.5. can you please redefine this field so i will have more clear understanding. – Unlimited Dreamer Mar 1 at 9:04
@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 $\phi \rightarrow \phi+\lambda\phi^2$. – twistor59 Mar 1 at 9:26
How will I confirm that my redefination is correct? – Unlimited Dreamer Mar 1 at 9:27
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