O(2) symmetry breaking trouble

In summary: Finally, the mass terms are the coefficients of the fields, so the mass for the \sigma field is m_{\sigma}^2=\mu^2+3\lambda v^2 and for the \pi field is m_{\pi}^2=\mu^2+\lambda v^2. In summary, 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. The ground state is unique for \mu^2 >0 and infinitely degenerate
  • #1
jfy4
649
3
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:
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]
My first question: Is that [itex]v[/itex] in the symmetry breaking part the same [itex]v[/itex] we defined above?

Now we go on to expand around [itex]\phi_{g}^a[/itex], and the results in the book are
[tex]
(-\partial^2 +m_{\sigma}^2 )\sigma =0,\quad (-\partial^2 +m_{\pi}^2)\pi=0
[/tex]
with [itex]\pi=\phi^1[/itex] and [itex]\phi^0=v+\sigma[/itex]
and
[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]
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...

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,
 
Physics news on Phys.org
  • #2
The vacuum expectation value, [itex]v[/itex], is defined by [itex]\phi^a_g = v \delta_{a0}[/itex]. Its value is set by the classical potential to be a root of [itex](\mu^2+ \lambda v^2)v=\epsilon[/itex]. When [itex]\epsilon =0[/itex], it takes the value [itex]v^2 = - \mu^2/\lambda[/itex], but not in general. It should be obvious why the same letter is used when viewed in this way.
 
  • #3
That helps, thanks.
 
  • #4
geez, I still can't get that second set of quoted equations. Here are my steps. I am starting with the EOM
[tex]
(-\partial^2 +\mu^2 +\lambda \phi^2)\phi^a=\epsilon \delta_{a,0}
[/tex]
and I am expanding around [itex]v[/itex] as [itex]\phi^0 =v+\sigma[/itex] and [itex]\phi^1 =\pi[/itex]. That gives me
[tex]
(-\partial^2 +\mu^2 +\lambda (\phi^0 \phi^0 + \phi^1 \phi^1))\phi^a =\epsilon \delta_{a,0}
[/tex]
[tex]
(-\partial^2 +\mu^2 +\lambda (v^2+2\sigma v +\pi^2))\phi^a =\epsilon \delta_{a,0}
[/tex]
[tex]
(-\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
[/tex]
The first equation turns to, using [itex](\mu^2+\lambda v^2)v=\epsilon [/itex]
[tex]
-\partial^2 (v+\sigma )+\mu^2 \sigma +\lambda \pi^2 v +\lambda \pi^2 \sigma +3\lambda v^2 \sigma =0
[/tex]
and the second
[tex]
-\partial^2 \pi +\mu^2 \pi +\lambda v^2 \pi +2\lambda v \sigma \pi =0
[/tex]
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
jfy4 said:
The first equation turns to, using [itex](\mu^2+\lambda v^2)v=\epsilon [/itex]
[tex]
-\partial^2 (v+\sigma )+\mu^2 \sigma +\lambda \pi^2 v +\lambda \pi^2 \sigma +3\lambda v^2 \sigma =0
[/tex]
and the second
[tex]
-\partial^2 \pi +\mu^2 \pi +\lambda v^2 \pi +2\lambda v \sigma \pi =0
[/tex]
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

[tex]
-\partial^2 \sigma+(\mu^2 +3\lambda v^2 )\sigma = - \lambda v \pi^2 -\lambda \pi^2 \sigma
[/tex]
[tex]
-\partial^2 \pi + (\mu^2 +\lambda v^2) \pi = - 2\lambda v \sigma \pi .
[/tex]

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

What is O(2) symmetry breaking trouble?

O(2) symmetry breaking trouble is a phenomenon in physics where a system that is symmetric under O(2) rotations is unable to maintain its symmetry when it reaches a certain temperature or energy level. This results in a phase transition where the system loses its rotational symmetry and adopts a more ordered state.

What causes O(2) symmetry breaking trouble?

O(2) symmetry breaking trouble is caused by thermal fluctuations or quantum fluctuations within a system. These fluctuations can disrupt the symmetrical arrangement of particles, leading to a breakdown of symmetry.

What are some examples of systems that exhibit O(2) symmetry breaking trouble?

Some examples of systems that exhibit O(2) symmetry breaking trouble include ferromagnets, liquid crystals, and superfluids. In ferromagnets, the alignment of magnetic moments breaks the O(2) rotational symmetry. In liquid crystals, the orientation of molecules causes the breakdown of O(2) symmetry. In superfluids, the collective behavior of atoms leads to a loss of rotational symmetry.

Can O(2) symmetry breaking trouble be predicted?

O(2) symmetry breaking trouble can be predicted using theoretical models and calculations. However, the exact conditions and timing of the phase transition may be difficult to determine as they can depend on various factors such as temperature, pressure, and material properties.

How is O(2) symmetry breaking trouble relevant to other scientific fields?

O(2) symmetry breaking trouble is relevant to many fields of physics, including condensed matter physics, high energy physics, and cosmology. It also has applications in materials science and engineering, as understanding and controlling symmetry breaking can lead to the development of new materials with unique properties.

Similar threads

Replies
5
Views
290
Replies
3
Views
573
  • Quantum Physics
Replies
1
Views
1K
  • Quantum Physics
Replies
1
Views
585
Replies
5
Views
1K
Replies
15
Views
2K
Replies
3
Views
745
  • Quantum Physics
Replies
9
Views
1K
Replies
1
Views
853
  • Quantum Physics
Replies
13
Views
1K
Back
Top