# Comparing interaction potential in standard $ϕ^4$theory

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

$$V(\phi)=\frac{1}{8}\phi^2(\phi-2)^2=\frac{1}{2}\phi^2-\frac{1}{2}\phi^3+\frac{1}{8}\phi^4$$

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.

• Can you show the mathematics for transforming the equation (4) into (2)? – user12906 Apr 14 '13 at 18:26
• 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. – levitopher Apr 15 '13 at 4:50