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Consider a scalar field doublet $(\phi_1, \phi_2)$ with a Mexican hat potential

$$V~=~\lambda (\phi_1^2+\phi_2^2-a^2)^2.$$

When $a=0$ this is a quartic potential and the symmetry is not spontaneously broken. However when the field acquire a VEV, the fields splits into a massive mode and a massless boson mode called the Goldstone boson.

I am wondering about the initial potential with $a=0$: does it have 2 massive modes?

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up vote 6 down vote accepted

AFAIR it has two massless modes, as there are no quadratic terms around the minimum.

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Sounds about right. – David Z May 17 '12 at 1:02
3  
You should say that this requires fine-tuning, as the mass-term will renormalize away from zero. You need to fine-tune the $m^2$ to some negative value to get to the critical point. – Ron Maimon May 17 '12 at 1:24
    
Indeed, two massless modes, but neither is a Goldstone boson, as each transforms linearly into each other by the, now, unbroken O(2) symmetry. – Cosmas Zachos Feb 16 at 15:45

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