O(2) symmetry breaking trouble

1. May 15, 2012

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
$$L=\frac{1}{2}\partial_{\mu}\phi_a \partial^\mu \phi_a +\frac{1}{2}\mu^2 \phi_a \phi_a +\frac{1}{4}\lambda (\phi_a \phi_a )^2$$
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:
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
with $\pi=\phi^1$ and $\phi^0=v+\sigma$
and
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,

2. May 15, 2012

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.

3. May 15, 2012

jfy4

That helps, thanks.

4. May 16, 2012

jfy4

geez, I still can't get that second set of quoted equations. Here are my steps. I am starting with the EOM
$$(-\partial^2 +\mu^2 +\lambda \phi^2)\phi^a=\epsilon \delta_{a,0}$$
and I am expanding around $v$ as $\phi^0 =v+\sigma$ and $\phi^1 =\pi$. That gives me
$$(-\partial^2 +\mu^2 +\lambda (\phi^0 \phi^0 + \phi^1 \phi^1))\phi^a =\epsilon \delta_{a,0}$$
$$(-\partial^2 +\mu^2 +\lambda (v^2+2\sigma v +\pi^2))\phi^a =\epsilon \delta_{a,0}$$
$$(-\partial^2 +\mu^2 +\lambda (v^2+2\sigma v +\pi^2))(v+\sigma)=\epsilon ,\quad (-\partial^2 +\mu^2 +\lambda (v^2+2\sigma v +\pi^2))\pi=0$$
The first equation turns to, using $(\mu^2+\lambda v^2)v=\epsilon$
$$-\partial^2 (v+\sigma )+\mu^2 \sigma +\lambda \pi^2 v +\lambda \pi^2 \sigma +3\lambda v^2 \sigma =0$$
and the second
$$-\partial^2 \pi +\mu^2 \pi +\lambda v^2 \pi +2\lambda v \sigma \pi =0$$
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,

5. May 16, 2012

fzero

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

$$-\partial^2 \sigma+(\mu^2 +3\lambda v^2 )\sigma = - \lambda v \pi^2 -\lambda \pi^2 \sigma$$
$$-\partial^2 \pi + (\mu^2 +\lambda v^2) \pi = - 2\lambda v \sigma \pi .$$

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.