# Symmetry breaking for 4 scalar fields

So I've been trying to get familiar with concept of symmetry breaking with minimal experience with field theory and I've got stuck on, I think, a simple problem. Let's consider four scalar fields $$\phi_i$$ and their Lagrangian: $$\mathcal{L}=\frac{1}{2}\partial_\mu\phi_i\partial^\mu\phi_i-\frac{1}{2}\mu^2(\phi_i\phi_i)-\frac{1}{4}\lambda(\phi_i\phi_i)^2.$$ The potential is quartic and, if $$\mu^2<0$$, the minima are at $$\phi_i=-\mu^2/\lambda\equiv v^2$$. To correctly(why?) interpret theory we have to expand around one of the minima by introducing new scalar field $$\phi_0=v+\sigma$$. Additionally the reference decided to rewrite, for $$i\neq0$$, $$\phi_i=\pi_i$$. Therefore they obtained new Lagrangian: $$\mathcal{L}=\frac{1}{2}\partial_\mu\sigma\partial^\mu\sigma-\frac{1}{2}(-2\mu^2)\sigma^2-\lambda v\sigma^3-\frac{\lambda}{4}\sigma^4$$ $$+\frac{1}{2}\partial_\mu\pi_i\partial^\mu\pi_i-\frac{\lambda}{4}(\pi_i\pi_i)^2-\lambda v\pi_i\pi_i\sigma-\frac{\lambda}{2}\pi_i\pi_i\sigma^2.$$ The first line is clear to me, but I struggle with the second one - I have no idea where do the terms mixing $$\pi_i$$ and $$\sigma$$ come from and why there is no $$\pi_i\pi_i$$ term, which is important for further discussion about fields' masses. I would be very grateful if somebody could lead me through the calculations. I've attached a picture with the whole example from the publication I'm reading trough: https://arxiv.org/abs/hep-ph/0503172 I hope it's clearer than what I've wrtitten.

• For the first "why" in your question this might be useful: physics.stackexchange.com/q/605486 Jan 6, 2021 at 22:53
• Your confusion is probably caused by the implicit summation over the index $i$. The last term $-1/4 \lambda(\Sigma_i \phi_i \phi_i)^2$ is mixing the fields. Jan 6, 2021 at 23:35
• Yeah, that's it - I suppose lagrangian without fields mixed in the first place wouldn't be really interesting, as there would be no interaction between them, right?
– Ream
Jan 7, 2021 at 10:37
• If it is interesting or not might be subjective :) , but yes no mixing terms results in no interactions. Jan 7, 2021 at 10:39

The easiest way to work out your algebra, in case you haven't, and are alarmed by the missing constant term, is to rewrite your potential, eliminating the pestiferous unphysical $$\mu^2$$ that has caused grief to generations of students. Recalling that summations over repeated indices is implied, the potential is $$\tfrac{\lambda}{4} (\phi_i \phi_i - v^2)^2 -\frac{\lambda}{4} v^2 \\ = \tfrac{\lambda}{4} (\pi_a \pi_a +(\sigma +v)^2- v^2)^2 -\frac{\lambda}{4} v^2 \\ =\tfrac{\lambda}{4} (\pi_a \pi_a +\sigma^2 +2v\sigma)^2 -\frac{\lambda}{4} v^2.$$ I have used indices a=1,2,3 for the 𝜋 s, to avoid confusion with the four i s.
You should be able to work out the kinetic terms, as $$\partial_\mu v=0$$.