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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?

EDIT: Comparing with the scalar field theory the potential term involved with $\lambda$. What is the value we have inserted here?

Moreover I want to transform the potential to a new form $$U(\phi)= \frac{1}{8} \lambda(\phi ^2 -v^2)^2.$$ (I have got this from Mark Srednicki page no 576. )

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Standard $\phi^{4}$ theory doesn't have a cubic term. Are you trying to set up symmetry breaking? – Jerry Schirmer Jan 30 '13 at 20:14
your $U(\phi)$ term has a manifest cubic term in it. If you are doing the trick where you expand around a minimum or something, as in the Higgs mechanism, then you can do a transformation to get a term like you have by making manipulations to the action that don't change the EOM. – Jerry Schirmer Jan 30 '13 at 20:36
1 check the paper in the small amplitude expansion. I think you people sometimes close a good question without good reasoning. – Unlimited Dreamer Jan 31 '13 at 4:58
@Forhad_jnu, to be clear, QMechanic is not closing it permanently, it is closed so you'll edit it and address the issues that they have mentioned, and if it looks good, they'll reopen. It is just the way things are done here. – lurscher Jan 31 '13 at 6:04
I think everybody avoiding my simple question, I just want to know about interaction potential. Please help me if someone can elaborate the term., – Unlimited Dreamer Jan 31 '13 at 8:00
up vote 5 down vote accepted

Edit: Oh, I see that this is basically what twistor59 wrote in a comment.

Set $\psi = \phi - 1$. The Lagrangian written in terms of $\psi$ is

$$ S = \int |\nabla \psi|^2 + V(\psi) $$

where $V(\psi) = U(\phi) = U(\psi + 1)$, expanding the reparametrization we have

$$ V(\psi) = \frac18 (\psi + 1)^2 (\psi - 1)^2 = \frac18 (\psi^4 - 2 \psi^2 + 1) $$

The constant 1 is not dynamical so the potential is equivalent to

$$ \tilde{V}(\psi) = \frac18 (\psi^4 - 2 \psi^2) $$

which is equivalent to the standard $\psi^4$ potential without cubic term up to scaling constants.

That is to say: the $\phi^3$ term isn't really there: it can be gauged away.

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$U(\phi) = V(\psi)$ when $\psi = \phi -1$. Just plug it in. – Willie Wong Jan 31 '13 at 10:20
It is just a standard gauge transformation / change of variables. I really don't see what there is to elaborate (honest!). You graph the potential $U$, you graph the standard $\phi^4$ potential, you notice that the two potentials are exactly the same except for translations. You note that translations do not change the kinetic energy part, and voila! – Willie Wong Jan 31 '13 at 11:43
Minor notational comment: Scalar field theory does not have gauge symmetry, and in particular, $\phi\to \phi-1$ is not a gauge transformation. – Qmechanic Jan 31 '13 at 13:23
@Qmechanic: yes yes, I welcome suggestions on other colloquial names for "change of coordinate system on the target manifold". – Willie Wong Jan 31 '13 at 14:28
@WillieWong "Field redefinition" seems to be the standard term. – Michael Brown Feb 23 '13 at 15:38

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