8
$\begingroup$

My question is simple. Given a group $G$ broken to a subgroup $H$, gauging a possibly different subgroup Hg breaks explicitly the global symmetry $G$, generating what is known as pseudo-Goldstone bosons. Why is this?

The usual answer I get is that the gauging determines a specific direction in field space, but I really don't understand this statement, how having a subgroup $Hg$ gauged can break the global $G$ explicitly?. What I have in mind is that inside the gauge transformations $Hg$ there are included global ones as well, so this is what is confusing me.

$\endgroup$
3
$\begingroup$

How can gauging only a subgroup not break the symmetry? Isn't any group element that doesn't preserve my gauge subrgoup going to change the action.

Let's I have an $SU(2)$ symmetry, and some scalar field $\rho$ in the fundamental. If I gauge the $U(1)$ subgroup corresponding to rotations by $\sigma_z$. I have a Lagrangian

$$\mathcal{L} = |(\partial + A \sigma_z)\psi|^2 + V(|\psi|^2)$$

Doesn't this manifestly break the $SU(2)$ symmetry? By your point about global symmetries it follows that all of the point interactions must remain invariant, but the Lagrangian as a whole is not invariant.

This example could physically correspond to two fields which are oppositely charged but otherwise symmetric. The waves in the $z$ axis correspond then to charged waves like a plasmon, whereas the rotations in the $x-y$ plane are local gauge transforms. So they are clearly different.

Added

Everyone seems to be having trouble seeing that this breaks the symmetry, and wants to say that the operation $\psi \rightarrow U\psi$, $A\sigma_z \rightarrow U A\sigma_z U^{\dagger}$, where $U$ is an $SU(2)$ matrix, is a "symmetry" of the Lagrangian. Let me write out the full path integral:

$$\int\mathcal{D}A\mathcal{D}\psi\exp\{i\int d^dx \mathcal{L}\}$$ $$\mathcal{L} = (\partial_\mu A_\nu - \partial_\nu A_\mu)^2 +|(\partial_\mu + A_\mu \sigma_z)\psi|^2 + V(|\psi|^2)$$

The variable $A_\mu$ is an $SU(2)$ scalar. It is just a regular 1-form like in electromagnetism. The integration measure (which also contains whatever unimportant gauge fixing) is just a regular integral over a 1-form field. Now for example to get Noether's theorem I need to have a change of variables that leaves my path integral invariant. The map $A\sigma_z \rightarrow U A\sigma_z U^{\dagger}$ is not a change of variables - I cannot get it by changing $A$, since $A$ doesn't know anything about $SU(2)$. There is no way I can change $x\sigma_z$ to $y\sigma_x$ by making a substitution $y=f(x)$.

You could rewrite the whole thing as

$$\int\mathcal{D}\hat{A}\mathcal{D}\delta(\hat{A}_{x,y})\psi\exp\{i\int d^dx \mathcal{L}\}$$ $$\mathcal{L} = (\partial_\mu A_\nu - \partial_\nu\hat{A}_\mu)^2 +|(\partial_\mu + \hat{A}_\mu )\psi|^2 + V(|\psi|^2)$$

where $\mathcal{D}\hat{A}$ is the measure for an $SU(2)$ field. But to correspond to what I wrote you need that delta function to in the measure. Otherwise you've gauged the entire $SU(2)$ symmetry which is obviously invariant. Now you can make change of variables $A\rightarrow U\hat{A}U^\dagger$, but this doesn't leave your path integral invariant because of that huge delta function.

$\endgroup$
  • $\begingroup$ Actually it seems that it does not break the global $SU(2)$ symmetry. Let the $U(1)$ covariant derivative be $D_\mu = \partial_\mu + A_\mu\sigma_z$. Under a global $SU(2)$ transformation $\psi\rightarrow U\psi$ and $A_\mu\sigma_z \rightarrow U A_\mu\sigma_z U^\dagger$, we have the transformation $D_\mu\psi\rightarrow U D_\mu\psi$. Thus $(D_\mu\psi)^\dagger D^\mu\psi$ seem to be invariant under global $SU(2)$ and local $U(1)$ transformations, unless I've done something wrong here. $\endgroup$ – Heidar Jul 5 '13 at 7:56
  • 1
    $\begingroup$ @Heidar: Look to pick a very stupid example, the single particle lagrangian $K - V(x)$ clearly breaks translation symmetry. But it is still invariant under the map $x\rightarrow x+ x_0$, $V(x) \rightarrow V'(x) = V(x-x_0)$. $\endgroup$ – BebopButUnsteady Jul 5 '13 at 16:13
  • 1
    $\begingroup$ Having had a cup of coffee, my brain works a little bit better. I now agree that $SU(2)$ is broken in this particular case. One way to see this is that the two components $\psi = (\psi^1,\psi^2)$ have different charges under the $U(1)$. We are therefore not allowed to make a $SU(2)$ rotation $U\psi$, since that would mix $\psi^1$ and $\psi^2$ and that would not transform covariantly under the $U(1)$ part we have gauged. This is just another way to say that, we have gauged a particular $U(1)$ embedding of $SU(2)$ and may not rotate it as I did above. $\endgroup$ – Heidar Jul 5 '13 at 18:47
  • 1
    $\begingroup$ It however still seems possible to gauge a subgroup. I think it is possible if for a gauge theory $G$, we gauge a normal subgroup $H\in G$. This means that $g H g^{-1} = H$, for all $g\in G$. It is easy to show that the Lie algebra corresponding to $H$ is also preserved. Thus given the transformation laws I gave above, everything should work out. In this particular example, the $U(1)$ is clearly not a normal subgroup of $SU(2)$ and it will therefore not work. If you instead take $G=U(2)$, there is a particular $U(1)$ subgroup you can gauge and keep global $U(2)$ symmetry, thats $e^{i\phi} I$. $\endgroup$ – Heidar Jul 5 '13 at 18:53
  • 2
    $\begingroup$ @Heidar: You can definitely gauge a subgroup $H\in G$ when $G$ isomorphic to $H\times N$ without breaking anything. As for a general normal subgroup it should be fine unless there is something that happens with outer automorphisms of gauge theories. $\endgroup$ – BebopButUnsteady Jul 5 '13 at 19:44
-1
$\begingroup$

Main Reference Zee (Quantum Mechanics in a Nutshell).

1) Global symmetry

A global symmetry means that the Lagrangian is invariant by a transformation whose parameters are constant.

For a continuous global symmetry, if the symmetry of the Lagrangian is the group $G$, and if the symmetry of the vacuum is the group $H$, a subgroup of $G$, you have ($dim G-dim H$) Goldstone bosons.

For instance, take a complex scalar field $\Phi$ with a Mexican hat potential , so that the total Lagrangian density is $L = \partial \phi^\dagger \partial \phi + \mu^2 \phi^\dagger \phi - \lambda (\phi^\dagger \phi)^2$.

The group symmetry is here $G=O(2)$

Define $\phi = \rho e^{i\theta}$

Breaking the symmetry means choosing for the vacuum the minima for the potential, and a particular angle, that is :

$\rho_V = v, \theta_V = \theta_0$

The group $H$ is trivial here.

Define : $\rho = v + \chi$, where $v = \sqrt{\frac{\mu^2}{2\lambda}}$

Developping the Lagrangian, you get a term $v^2(\partial \theta)^2$, which is the dynamical part for a massless field $\theta$, so $\theta$ is our Goldstone boson (There is one because $dim G - dim H = 1 - 0 = 1 $).

So, we see, that spontaneous symmetry breaking could arise in a global continuous symmetry.

2) Local symmetry

A local symmetry means that the Lagrangian is invariant by a transformation whose parameters are functions of space-time.

"Gauging" means (continous) local symmetry. So you don't need "gauging" to have a spontaneous symmetry breaking.

With a local symmetry, some of the Goldstone Bosons are "eaten" by the Gauge field ($A_\mu$), so that these gauge fields (which are massless) become massive. In a 4d space-time dimension, a massless Gauge field has $2$ degrees of freedom, while a massive gauge field has $3$ degrees of freedom. To do that, the Gauge field has to "eat" one degree of freedom (one Goldstone boson)

3) Global symmetry as a special case of Local Symmetry

In the set of local symmetry, global symmetry is a very special case (a very special subset), where transformation parameters are constant. So, if you want, you can consider that global symmetry are "included" into local symmetry.

$\endgroup$
  • $\begingroup$ Thanks, but this however doesn´t give me any answer, I know what symmetries are, global and local, and their diferences. However is the specific case that I am talking about. For pseudogoldstone bosons, ex: Composite higgs SO(5)/SO(4) where gauging a subgroup of the remaining SO(4) (which will be the electroweak part of SM) that causes the higgs NG to get a mass, which you can see from the wavefunction renormalization (diagrams with loops from the gauge bosons). $\endgroup$ – user26661 Jul 4 '13 at 20:31
  • $\begingroup$ So, if I understand correctly (with some delay...), you have a global symmetry $G$, with spontaneous symmetry breaking with a group $H$, then you take a subgroup $Hg$ of $H$ and you gauge it. And then there would be a explicit breaking of the global symmetry G. And you wonder how could be this possible. Right ? $\endgroup$ – Trimok Jul 5 '13 at 9:48
  • $\begingroup$ It is maybe a problem of definition. Pseudo-Goldstone Bosons are associated to approximated symmetries $G$. For instance, by considering the masses of the quarks $u$ and $d$ being zero, you have an approximate symmetry $G = SU(2)_L * SU2(R)$. The spontaneous symmetry breaking gives $G = SU(2)_L * SU2(R) \rightarrow SU2(I) = H$.if the symmetry would be exact, you will have massless Goldstone Bosons, but in fact, the symmetry $G$ is not exact (because the mass of quarks $u$ and $d$ is non zero), so you have pseudo-Goldstone bosons which are light, but not massless. $G$ is not a exact symmetry. $\endgroup$ – Trimok Jul 5 '13 at 10:42
  • $\begingroup$ Sorry I didn't answer earlier. Yes this is my question. In the case you wrote(simple QCD) it is clear because the symmetry is explicity broken by the mass terms. My problem comes when the explicit symmetry breaking should come from a gauging of a subgroup. Since I don't see. In the answer below, you can see that there is a little controversy about how to properly transform the lagrangian after having a subgroup gauged, and this is what I am trying to figure out. $\endgroup$ – Ayfel Jul 5 '13 at 16:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.