I'd like to ask some questions about the following example of broken symmetry and non-invariant vaccum.
The basic argument goes as follows:
[tex] \Cal L= \partial^\mu \phi \partial_\mu \phi \: - \: \mu^2 \phi^* \phi \: - \: \lambda (\phi^*\phi)^2
[/tex]
[tex]
\frac{\partial V} {\partial \phi}=0
[/tex]
[tex]
<\phi>^2=\frac{\mu^2}{2\lambda}
[/tex]
My first question regards the following term in the lagrangian:
[tex]\lambda (\phi^*\phi)^2[/tex]
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 [tex] \phi \rightarrow e^{i\theta}\phi [/tex] 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.
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