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robousy
robousy is offline
#1
Oct1-05, 03:31 PM
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:

[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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