# Mass parameters in the spontaneous symmetry breaking

In Ashok: Lectures on Quantum Field Theory Sec. 7.5 p.279 he discusses the mass parameters for the potential

$$V = -\frac{m^2}{2}(\sigma^2+\xi^2) + \frac{\lambda}{16}(\sigma^2 + \xi^2)^2.$$

My (pretty simple) question is the claim:

We note that the second derivatives of the potential (with respect to field variables) give the mass (squared) parameters for the field operators (particles) in a theory...

I think this should come from the Taylor expansion of the potential, but I cannot see it clearly.

## 1 Answer

Ashok should say "We note that the second derivatives of the potential with respect to field variables computed at the minimum of the potential give the mass (squared) parameters for the field operators (particles) in a theory..."

• (I understand the minimum part, but) Why? – user2820579 Aug 8 '19 at 21:48
• You do undertstand that the Lagrangian $L= \frac 12 (\nabla \phi)^2+\frac 12 m^2 \phi^2$ has particles of mass $m$? – mike stone Aug 8 '19 at 22:37
• Don't worry, I see now Eq. (7.79), there is very clear that one of the fields do not carry mass. – user2820579 Aug 8 '19 at 23:03