I am posting this question again because, Willie Wong asked me to do it. So it is a continuing post of the Interaction potential in standard ϕ4 theory.

I have been studying about solitions so I had to deal with scalar field theory. The problem I faced in Lagrangian of Scalar field is interacting potential.

According to Scalar field theory we can write: $$\mathcal{L}=\frac{1}{2}\partial^\mu \phi \partial_\mu \phi -\frac{m^2}{2}\phi^2 -\frac{\lambda}{4!}\phi^4 \tag {1}$$ The potential can be written separately $$ V(\phi)= \frac{m^2}{2}\phi^2 +\frac{\lambda}{24}\phi^4 \tag {2}$$ I found on Srednicki (Quantum Field Theory, page 188) that, the author wrote the potential as, $$ V(\phi)= \frac{1}{24} \lambda (\phi^2- v^2 )^2-\frac{\lambda}{24} v^4 \tag {3}$$ After that the author excluded the term $-\frac{\lambda}{24} v^4 $.

why is that?

In another paper an author wrote the potential as $$V(\phi)= \frac{1}{8} \phi^2 (\phi -2)^2 \tag{4}$$
I don't see coupling constant $\lambda$ in the equation (4).

What I'm trying to find is to get the potential in equation (4) from the equation (2)


Both (4) and (2) are $\phi$4 theories, but they are not the $same$ $\phi$4 theory. In equation (2), there is an explicit mass term and an interaction term, whereas in equation (4) the "other author" has set these the coupling and mass to specific values. Note


So actually this theory has a cubic interaction $and$ a quartic. These can be related by symmetry breaking. Starting with (2), define the shifted field $\rho(x)=\phi(x)-v$ (This is all in Srednicki $\S$30). Plug that into your potential and you find

$$V(\rho)=\frac{1}{6}\lambda v^2\rho^2+\frac{1}{6}\lambda v \rho^3 +\frac{1}{24}\lambda \rho^4$$

And with some redefinition of variables you can get (4). The different sign makes these really represent different theories.

As an answer to your first question, the term $\frac{\lambda}{24}v^4$ is a constant, and has no effect on the dynamics. If you minimize the Lagrangian, the derivatives will kill that.

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  • $\begingroup$ Can you show the mathematics for transforming the equation (4) into (2)? $\endgroup$ – user12906 Apr 14 '13 at 18:26
  • $\begingroup$ Well, the trivial transformation is choosing specific values for the coupling constants $\lambda=3$, $v=-1$, $\phi=\rho /\sqrt{2}$, but it is certainly not unique. That point is that they are different potentials, both $\phi$4 theories. $\endgroup$ – levitopher Apr 15 '13 at 4:50

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