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

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

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

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

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

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

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

 

[fzero]

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,

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.