O(2) symmetry breaking trouble

[jfy4]

Hi, I'm going to quote a lot of a book so that I can get some help, brace yourselves...

First, \phi_{a} is my field with a=0,1 as internal components and my lagrangian is
<br /> 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<br />
with U=\frac{1}{2}\mu^2 \phi_a \phi_a +\frac{1}{4}\lambda (\phi_a \phi_a )^2
now in Jan Smit's book, "Introduction to Quantum Fields on a Lattice" the page has the following:

The potential
<br /> U=\frac{1}{2}\mu^2 \phi_a \phi_a +\frac{1}{4}\lambda (\phi_a \phi_a )^2 <br />
has a "wine-bottle" shape, also called the "mexican-hat" shape, if \mu^2 &lt;0. It's clear that for \mu^2 &gt;0 the ground state is unique, but for \mu^2 &lt;0 the ground state is infinitely degenerate. The equation \partial U/\partial \phi^k =0 for the minima, (\mu^2 +\lambda \phi^2)=0 has the solution
<br /> \phi_{g}^a =v\delta_{a,0},\quad v^2=-\frac{\mu^2}{\lambda}<br />
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
<br /> \Delta S=\int dx \epsilon \phi^0, \quad \epsilon &gt;0<br />
The constant \epsilon has the dimensions of mass cubed. The equation for the stationary points now reads
<br /> (\mu^2 +\lambda \phi^2 )\phi^a =\epsilon \delta_{a,0}<br />
with the symmetry breaking (the above) the ground state has \phi_{g}^a pointing in the a=0 direction,
<br /> \phi_{g}^a=v\delta_{a,0},\quad (\mu^2+\lambda v^2)v=\epsilon<br />

My first question: Is that v in the symmetry breaking part the same v we defined above?

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

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

with \pi=\phi^1 and \phi^0=v+\sigma
and

<br /> m_{\sigma}^2=\mu^2 +3\lambda v^2=2\lambda v^2+\epsilon /v<br />
<br /> m_{\pi}^2=\mu^2+\lambda v^2=\epsilon /v<br />

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

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

Thanks,

 

[fzero]

The vacuum expectation value, v, is defined by \phi^a_g = v \delta_{a0}. Its value is set by the classical potential to be a root of (\mu^2+ \lambda v^2)v=\epsilon. When \epsilon =0, it takes the value v^2 = - \mu^2/\lambda, but not in general. It should be obvious why the same letter is used when viewed in this way.

 

[jfy4]

That helps, thanks.

 

[jfy4]

geez, I still can't get that second set of quoted equations. Here are my steps. I am starting with the EOM
<br /> (-\partial^2 +\mu^2 +\lambda \phi^2)\phi^a=\epsilon \delta_{a,0}<br />
and I am expanding around v as \phi^0 =v+\sigma and \phi^1 =\pi. That gives me
<br /> (-\partial^2 +\mu^2 +\lambda (\phi^0 \phi^0 + \phi^1 \phi^1))\phi^a =\epsilon \delta_{a,0}<br />
<br /> (-\partial^2 +\mu^2 +\lambda (v^2+2\sigma v +\pi^2))\phi^a =\epsilon \delta_{a,0}<br />
<br /> (-\partial^2 +\mu^2 +\lambda (v^2+2\sigma v +\pi^2))(v+\sigma)=\epsilon ,\quad (-\partial^2 +\mu^2 +\lambda (v^2+2\sigma v +\pi^2))\pi=0<br />
The first equation turns to, using (\mu^2+\lambda v^2)v=\epsilon
<br /> -\partial^2 (v+\sigma )+\mu^2 \sigma +\lambda \pi^2 v +\lambda \pi^2 \sigma +3\lambda v^2 \sigma =0<br />
and the second
<br /> -\partial^2 \pi +\mu^2 \pi +\lambda v^2 \pi +2\lambda v \sigma \pi =0<br />
Both of these have many of the terms I need, plus a number of them that don't show up in the book. Where am I going wrong?

Thanks,

 

[fzero]

The first equation turns to, using (\mu^2+\lambda v^2)v=\epsilon
<br /> -\partial^2 (v+\sigma )+\mu^2 \sigma +\lambda \pi^2 v +\lambda \pi^2 \sigma +3\lambda v^2 \sigma =0<br />
and the second
<br /> -\partial^2 \pi +\mu^2 \pi +\lambda v^2 \pi +2\lambda v \sigma \pi =0<br />
Both of these have many of the terms I need, plus a number of them that don't show up in the book. Where am I going wrong?

Thanks,

The terms which are of higher order in the fields are interaction terms. To find the mass we want the terms in the e.o.m. that are linear , so write those as

<br /> -\partial^2 \sigma+(\mu^2 +3\lambda v^2 )\sigma = - \lambda v \pi^2 -\lambda \pi^2 \sigma <br />
<br /> -\partial^2 \pi + (\mu^2 +\lambda v^2) \pi = - 2\lambda v \sigma \pi .<br />

The terms on the RHS are just \partial V/\partial \sigma and \partial V/\partial \pi, where V(\sigma,\pi) is the part of the scalar potential without the mass terms.