- #1
jfy4
- 649
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Hi, I'm going to quote a lot of a book so that I can get some help, brace yourselves...
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:
Now we go on to expand around [itex]\phi_{g}^a[/itex], and the results in the book are
and
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,
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,