In spontaneous symmetry breaking, global symmetry broken by gauged subgroup?

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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.

This post imported from StackExchange Physics at 2014-05-21 15:21 (UCT), posted by SE-user Ayfel

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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.

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.

This post imported from StackExchange Physics at 2014-05-21 15:21 (UCT), posted by SE-user BebopButUnsteady
answered Jul 4, 2013 by (330 points)
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.

This post imported from StackExchange Physics at 2014-05-21 15:21 (UCT), posted by SE-user Heidar
@Heidar: How are you transforming $A$? It is is not an $SU(2)$ form. I think what you have shown is that if you rotate $\psi$ and rotate the pauli matrix then you get back the action, but that is not a symmetry, its just a vestige of the original symmetry.

This post imported from StackExchange Physics at 2014-05-21 15:21 (UCT), posted by SE-user BebopButUnsteady
Since we have a new field in the theory, $\hat A = A \sigma_z$, I made it transform under the global $SU(2)$ symmetry. The most natural thing was to use the adjoint transformation, $\hat A\rightarrow U\hat A U^\dagger$, which is what gauge fields usually transform under. Under this transformation there is clearly a global $SU(2)$ symmetry, however I'm not sure it makes sense to think of it as the original $SU(2)$ or not.

This post imported from StackExchange Physics at 2014-05-21 15:21 (UCT), posted by SE-user Heidar
By the way, under this transformation the $U(1)$ subalgebra is rotated into another $U(1)$ subalgebra. If one combines a global $SU(2)$ and a local $U(1)$, that seem corresponds to a local $U(1)$, just a rotated one. It looks consistent. However, I am not sure whether it is allowed to rotate the gauged subalgebra. If one assumes that $\hat A$ may only take values in the $U(1)$ subalgebra we gauged and not the whole $SU(2)$ Lie algebra, then my transformation law doesn't make sense.

This post imported from StackExchange Physics at 2014-05-21 15:21 (UCT), posted by SE-user Heidar
@Heidar: This is the same as taking a magnet with hamiltonian $m^2 +h\cdot m$, where $m$ is the magnetization and $h$ is a fixed external field, both vectors. It is true that if you rotate $h$ and $m$ you preserve the Hamiltonian. However this does not correspond to any symmetry, since we do not integrate over the field $h$ ( or vary $h$ in the classical case). Similarly, we do not integrate $\hat{A}$ over the full $SU(2)$ space.

This post imported from StackExchange Physics at 2014-05-21 15:21 (UCT), posted by SE-user BebopButUnsteady
@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 $x\rightarrow x+ x_0$, $V(x) \rightarrow V'(x) = V(x-x_0)$.

This post imported from StackExchange Physics at 2014-05-21 15:21 (UCT), posted by SE-user BebopButUnsteady
Yeah, there are certain subtleties associated with this. To me, however, it seems it's possible to save the construction. But I am not sure since I haven't thought too much about it, could be interesting to figure out.

This post imported from StackExchange Physics at 2014-05-21 15:21 (UCT), posted by SE-user Heidar
Sorry, it's a little too early in the morning for me here. I am not sure I understand your example, what do you mean with that you have broken translational symmetry but the potential is invariant?

This post imported from StackExchange Physics at 2014-05-21 15:21 (UCT), posted by SE-user Heidar
@Heider: The potential is not invariant. I'm saying what youre doing is the same as taking the classical single particle harmonic oscillator $V = (x-x_0)^2$ and saying the resulting theory is translationally invariant because $V$ doesn't change under the map $x\rightarrow x+a$, $x_0\rightarrow x_0 + a$. The point being you're not allowed to change the value of $x_0$ in a symmetry. Similarly you're not allowed to change the $U(1)$ subalgebra in a symmetry.

This post imported from StackExchange Physics at 2014-05-21 15:21 (UCT), posted by SE-user BebopButUnsteady
Oh yes, now I see your point. As I also said earlier, it might not make sense to rotate the subalgebra and my transformation law might therefore not make sense. I just haven't completely made it clear to myself, how to see there is something inconsistent. For example does $SU(2)$ conserved currents exist or not.

This post imported from StackExchange Physics at 2014-05-21 15:21 (UCT), posted by SE-user Heidar
@BebopButUnsteady : Does that mean that the transformation is $D_\mu \psi \rightarrow (U \partial_\mu + UA_\mu U^\dagger \sigma_z U)\Psi$

This post imported from StackExchange Physics at 2014-05-21 15:21 (UCT), posted by SE-user Trimok
@Trimok: you can write that if you want, but you're just replacing $\sigma_z$ with a different Pauli matrix. $A_\mu$ is just an $SU(2)$ scalar in the expression I wrote down. ($\sigma_z$ is an $SU(2)$ matrix)

This post imported from StackExchange Physics at 2014-05-21 15:21 (UCT), posted by SE-user BebopButUnsteady
@BebopButUnsteady :OK, so the transformation is $D_\mu \psi \rightarrow (U \partial_\mu + A_\mu \sigma_z U)\Psi$, which is different of $UD_\mu \psi$. So there is breaking of the $SU(2)$ global symmetry. Is it correct?

This post imported from StackExchange Physics at 2014-05-21 15:21 (UCT), posted by SE-user Trimok
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.

This post imported from StackExchange Physics at 2014-05-21 15:21 (UCT), posted by SE-user Heidar
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 $e^{i\phi} I$.

This post imported from StackExchange Physics at 2014-05-21 15:21 (UCT), posted by SE-user Heidar
I just noticed you mention that $\psi^1$ and $\psi^2$ are oppositely charged, I didn't notice it for some reason. So it seems that you were right all along, it doesn't work for this example.

This post imported from StackExchange Physics at 2014-05-21 15:21 (UCT), posted by SE-user Heidar
@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.

This post imported from StackExchange Physics at 2014-05-21 15:21 (UCT), posted by SE-user BebopButUnsteady
Oh, I get it, this makes sense, I was confused about how to transform the field A. But now it´s clear, thanks. I have to think about the normal subgroups but, it makes sense thinking about linear/non-linearly realized symmetries.

This post imported from StackExchange Physics at 2014-05-21 15:21 (UCT), posted by SE-user Ayfel
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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.

This post imported from StackExchange Physics at 2014-05-21 15:21 (UCT), posted by SE-user Trimok
answered Jul 4, 2013 by (955 points)
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).

This post imported from StackExchange Physics at 2014-05-21 15:21 (UCT), posted by SE-user user26661
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 ?

This post imported from StackExchange Physics at 2014-05-21 15:21 (UCT), posted by SE-user Trimok
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.

This post imported from StackExchange Physics at 2014-05-21 15:21 (UCT), posted by SE-user Trimok
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.

This post imported from StackExchange Physics at 2014-05-21 15:21 (UCT), posted by SE-user Ayfel

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