# Minimum of a potential with $SO(2)\times SO(2)$ symmetry

Given

$\Phi=\left(\begin{array}{c} \phi_1\\ \phi_2\\ \phi_3\\ \phi_4 \end{array}\right)$

where $\Phi$ is real, I have the following Lagrangian density:

$\mathcal{L}=\frac{1}{2}(\partial_\mu \Phi)^T\partial^\mu \Phi -\frac{1}{2} \Phi^T M^2 \Phi-\frac{\lambda}{2m^2}(\Phi^TM^2 \Phi)^2$

where

$M^2 =m^2\left(\begin{array}{cccc} 1 & 0 & 0& 0\\ 0 & 1 & 0& 0\\ 0 & 0 & -1& 0\\ 0 & 0 & 0& -1 \end{array}\right)$

I think that the fields have a symmetry $SO(2)\times SO(2)$, since I can rotate indipendently the first two fields or the second two fields. However then I expect a symmetry breaking only for the second two fields because they have the minus sign on the mass term. Is it this correct? Is there an easy way to find the minima of the potential?

• I think the symmetry should be O(2,2), that is, the one that leaves $M^2$ invariant ($SO(2)xSO(2)$ is a subgroup of O(2,2)), right ? – Adam Jun 9 '17 at 11:20
• I think you're right but I am not so an expert in group theory so any hint or clarification is welcomed – Alessandro Mininno Jun 9 '17 at 11:28
• I don't think I can help much more, unfortunately. What I meant is that if you find a minimum, say for $\Phi=(0,0,0,\phi)$, then you can find another minimum using an O(2,2) transformation, which is defined as the group of matrix $U$ such that $U M^2 U^T=M^2$. However, this group is not compact, and I don't know much about field theories in that case... – Adam Jun 9 '17 at 12:31
• If I were you, I would start with a simpler example, using a smaller group, such as O(1,1), that is, with two fields and $M^2=diag(1,-1)$. – Adam Jun 9 '17 at 12:32
• @Adam 's last point might help you. In this 2-field theory you see "symmetry breaking" in the potential, with one massive and one massless field, even though O(1,1) is not a symmetry of the kinetic term... It is a "custodial" symmetry of the potential only, so the tree level goldston will develop a mass in pert theory! In your original example, O(2,2) is only custodial, so only one of the 3 massless states you will find will be a true goldston (the one coming from the lower block) and the two upper massless fields will develop a mass radiatively (common mass, as their SO(2) is intact.) – Cosmas Zachos Jun 9 '17 at 15:24

You need to minimise

$$\left(\phi_1^2+\phi_2^2-\phi_3^2-\phi_4^2\right)+\lambda\left(\phi_1^2+\phi_2^2-\phi_3^2-\phi_4^2\right)^2$$

By differentiation with respect to any of the $\phi_i^2$:

$$1 + 2\lambda \left(\phi_1^2+\phi_2^2-\phi_3^2-\phi_4^2\right) = 0$$

So there is a set of degenerate extrema, whose equation is

$$\phi_1^2+\phi_2^2-\phi_3^2-\phi_4^2 = -\frac{1}{2\lambda}$$

The matrix of second-order derivatives is diagonal, and its diagonal is

$$2\lambda\begin{pmatrix}1&1&-1&-1\end{pmatrix}$$

from which you can read where the extrema are actually minima.