Hi, I'm going to quote a lot of a book so that I can get some help, brace yourselves...(adsbygoogle = window.adsbygoogle || []).push({});

First, [itex]\phi_{a}[/itex] is my field with [itex]a=0,1[/itex] as internal components and my lagrangian is

[tex]

L=\frac{1}{2}\partial_{\mu}\phi_a \partial^\mu \phi_a +\frac{1}{2}\mu^2 \phi_a \phi_a +\frac{1}{4}\lambda (\phi_a \phi_a )^2

[/tex]

with [itex]U=\frac{1}{2}\mu^2 \phi_a \phi_a +\frac{1}{4}\lambda (\phi_a \phi_a )^2 [/itex]

now in Jan Smit's book, "Introduction to Quantum Fields on a Lattice" the page has the following:

My first question: Is that [itex]v[/itex] in the symmetry breaking part the same [itex]v[/itex] we defined above? The potential

[tex]

U=\frac{1}{2}\mu^2 \phi_a \phi_a +\frac{1}{4}\lambda (\phi_a \phi_a )^2

[/tex]

has a "wine-bottle" shape, also called the "mexican-hat" shape, if [itex] \mu^2 <0[/itex]. It's clear that for [itex]\mu^2 >0[/itex] the ground state is unique, but for [itex]\mu^2 <0[/itex] the ground state is infinitely degenerate. The equation [itex]\partial U/\partial \phi^k =0 [/itex] for the minima, [itex](\mu^2 +\lambda \phi^2)=0[/itex] has the solution

[tex]

\phi_{g}^a =v\delta_{a,0},\quad v^2=-\frac{\mu^2}{\lambda}

[/tex]

or any O(2) rotation of this vector. To force the system into a definite ground state we add a symmetry-breaking term to the action

[tex]

\Delta S=\int dx \epsilon \phi^0, \quad \epsilon >0

[/tex]

The constant [itex]\epsilon[/itex] has the dimensions of mass cubed. The equation for the stationary points now reads

[tex]

(\mu^2 +\lambda \phi^2 )\phi^a =\epsilon \delta_{a,0}

[/tex]

with the symmetry breaking (the above) the ground state has [itex]\phi_{g}^a[/itex] pointing in the [itex]a=0[/itex] direction,

[tex]

\phi_{g}^a=v\delta_{a,0},\quad (\mu^2+\lambda v^2)v=\epsilon

[/tex]

Now we go on to expand around [itex]\phi_{g}^a[/itex], and the results in the book are

with [itex]\pi=\phi^1[/itex] and [itex]\phi^0=v+\sigma[/itex] [tex]

(-\partial^2 +m_{\sigma}^2 )\sigma =0,\quad (-\partial^2 +m_{\pi}^2)\pi=0

[/tex]

and

Now the reason I asked the above question is, substituting the first definition of [itex]v^2[/itex] into these mass definitions, [itex]m_{\pi}[/itex] equals zero... [tex]

m_{\sigma}^2=\mu^2 +3\lambda v^2=2\lambda v^2+\epsilon /v

[/tex]

[tex]

m_{\pi}^2=\mu^2+\lambda v^2=\epsilon /v

[/tex]

So then, what is [itex]v^2[/itex] in the symmetry breaking equations, why use the same letter IF it's different, and not say so?

Thanks,

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# O(2) symmetry breaking trouble

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