# What physical situation will give a negative mass squared?

Consider the Lagrangian density of $\phi^4$ theory $$\mathcal{L}=-\frac{1}{2}\partial^\mu\phi\partial_\mu\phi-\frac{1}{2}m^2\phi^2-\frac{1}{24}\lambda\phi^4$$ the potential energy term is given by $$V(\phi)=\frac{1}{2}m^2\phi^2+\frac{1}{24}\lambda\phi^4$$ when $m^2<0$, the potential will turn to a "W"-like shape. This is a typical situation in spontaneous symmetry breaking.

My question is: under what physical circumstances will we have $m^2<0$? I heard that cooper coupling was a sort of example, is there any good reference on that?

• àny textbook on second order phase transition should do. You can start with en.wikipedia.org/wiki/Landau_theory – Adam May 24 '17 at 6:28
• This post (v2) seems like a list question. – Qmechanic May 24 '17 at 7:14

Before electroweak symmetry breaking, the scalar potential in the Standard Electroweak theory is supposed to be described by a potential (similar to $\phi^4$-theory) $$V(\Phi)=m^2(\Phi^\dagger\Phi)+\lambda(\Phi^\dagger\Phi)^2$$ with $m^2<0$ and $\Phi$ being the scalar Higgs doublet. When $m^2<0$, it is just a parameter of the Lagrangian; it does not represent the mass, and therefore, there is nothing wrong about it being negative. After the symmetry breaks spontaneously, the mass of the Higgs field will be given by $=2\lambda v^2=-2m^2$ which is positive. Here, $v$ stands for vacuum expectation value given by $v^2=\frac{-m^2}{\lambda}$. Please verify the numerical factors.
Addendum There is yet another way to understand this. Consider the Lagrangian $$\mathscr{L}=\frac{1}{2}(\partial_\mu\phi)(\partial^\mu\phi)-V(\phi).$$ Let us expand $V(\phi)$ in a Taylor series about a point $\phi=\phi_0$. This gives $$V(\phi)=V(\phi_0)+\frac{\partial V}{\partial\phi}\Big|_{\phi=\phi_0}(\phi-\phi_0)+\frac{1}{2!}\frac{\partial^2 V}{\partial\phi^2}\Big|_{\phi=\phi_0}(\phi-\phi_0)^2+...$$ For the potential $V(\phi)=\frac{1}{2}m^2\phi^2+\lambda\phi^4$, the second derivative of the potential is negative of $m^2<0$. This implies that the potential is expanded not about a minimum but about the maximum $\phi=0$.