View Single Post
robousy
#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.
Phys.Org News Partner Physics news on Phys.org
'Squid skin' metamaterials project yields vivid color display
Team finds elusive quantum transformations near absolute zero
Scientists control surface tension to manipulate liquid metals (w/ Video)