Thread: Nambu Goldstone Boson View Single Post
 P: 335 I'd like to ask some questions about the following example of broken symmetry and non-invariant vaccum. The basic argument goes as follows: $$\Cal L= \partial^\mu \phi \partial_\mu \phi \: - \: \mu^2 \phi^* \phi \: - \: \lambda (\phi^*\phi)^2$$ $$\frac{\partial V} {\partial \phi}=0$$ $$<\phi>^2=\frac{\mu^2}{2\lambda}$$ My first question regards the following term in the lagrangian: $$\lambda (\phi^*\phi)^2$$ does this term indicate that the field is self-interacting? What does that mean for a field to self interact? My next question regards the result. Now, the expectation value of the field is shown not to be equal to zero when V is minimized. I have read that "this implies that the vacuum is not invariant under the u(1) symmetry $$\phi \rightarrow e^{i\theta}\phi$$ therefore there must be a zero mass particle in the theory." I really don't understand the conclusion. Why is the vacuum not invariant as I don't see the term |0> anywhere and why does it imply that there is a zero mass particle??? I can follow the math but not what the math means. Any help appreciated.