# Symmetry breaking: what is the subgroup left?

1. May 18, 2010

### krishna mohan

Hi....

I have studied the standard model and know that spontaneous symmetry breaking by a vev breaks SU(2)xU(1) to a U(1). How do we know to what group a vev will break the original group? I have heard of Dynkin diagrams. Are they only for continuous groups? Is there any other method for discrete groups?

Any reference given will be appreciated.

2. May 18, 2010

### ansgar

it depends on how the vacuum which breaks the symmetry looks. If the vev for SM would be something different than (0, v)^T then there would be another symmetry group left for instance.

3. May 18, 2010

### bapowell

When the scalar field that breaks the symmetry takes on a vev, count the field dimensions orthogonal to the vev (the number of 0's in (0,v)^T in the above post). These are the goldstone modes. They correspond to the broken generators of the original symmetry group. Whatever 'good' generators you're left with define the subgroup.

I believe Dynkin diagrams arise only in the context of Lie groups, which are continuous.

4. May 19, 2010

### krishna mohan

In the case of the Standard Model, I think that a single vev (v1, v2)^T will always break in to the same subgroup, the U(1). This is because such a vev can always be taken to the form
(0,v)^T by a basis transformation. And the subgroup into which the original group is broken into should not depend on the basis chosen....

But my question remains....In the case of the vev in the Standard Model, it was easy to see which generators leave it invariant. But is there a systematic procedure to solve the problem in more general cases?

5. May 26, 2010

### DJsTeLF

I have just covered this in a MSci course on the Standard Model (SM) although, given that my exam is some weeks away, I would described my understanding of it as novice at best and it may therefore take me a few days to research answers to any questions that might arise. That said, here goes:

The most general case I can see would be the SSB (spontaneous symmetry breaking) of non-Abelian gauge symmetries, as is the case in the GWS-model, that is, the electroweak sector of the SM. I will try to outline the essence of the procedure for a general field psi that in this case is a complex doublet of fields (ie 4 in total) but could just as well be something else. (For brevity I will avoid mention of the chirality considerations that, due to observed CP-violation in weak interaction, led GS&W to propose the full version of their model.)

- look for the ground state of the Lagrangian that breaks (partially) the non-Abelian gauge symmetry SU(2) x U(1). Because of Lorentz invariance, that we wish to preserve (if aiming for a physically-useful QFT at high energy), only scalar quantities are allowed to acquire non-trivial vacuum expectation values (vev), that is only <0|psi|0> /= 0. (/= meaning not equal to).

- one then is free to choose to align the vev with the 'down' direction (for example) of the local SU(2) internal symmetry (often called "weak isospin" in this case).

- the resulting 3 fields (call them theta 1,2,3 to avoid confusion) can then be 'gauged' away (aka 'swallowed by the longitudinal components of the SU(2)vector bosons') by performing appropriate local (gauge) SU(2) rotations on the fields.

- the resulting mass spectrum (ie. which fields have acquired mass and which have not) of the broken phase may then be obtained by substituting the resulting expression for the psi fields back in the the Lagrangian and focussing only on terms quadratic in the various particle excitations.

The spectrum in this case is that a non-trivial mixing of the fields occurs as a results of the SSB. It results in the W1 and W2 fields having a positive definite mass term whereas the W3 and B^mu fields have a non-diagonal mass term, ie. they are 'mixed'. They can be unmixed into the massive Z0 and massless photon fields you will be familiar with from particle physics by performing an appropriate rotation in field space. This results in what is known as the 'weak mixing or Weinberg angle' which appears in the coupling the the photon field, ie. proportional to the electron charge.

When all is said and done the group symmetries of the fields have undergone a change from SU(2)xU(1) -> U_em(1) where the latter is the unbroken Abelian group of 'electromagnetism' (not identical to the original B^mu due the the non-trivial mixing).

6. May 26, 2010

### arivero

It is important to keep in mind that a broken lagrangian is still different from its low energy limit.

U(1)_em is the subgroup left, but it is only the low energy particles, E << M, the ones which seem to believe to live in a world which a U(1) lagrangian.

On the other side, the particles with E >> M believe to live in a world with a chiral SU(2)xU(1) lagrangian.

The particles in the middle do not believe in gauge field lagrangians.

For Kaluza Klein "gravity lagrangians" the situation is peculiar: E<<M live in a 5 dimensional space time; E>>M live in a 7 dimensional space time, but nobody knows what dimension the E ~ M particles believe to live.

7. May 26, 2010

### bapowell

Arivero,

That sounds surprising. I've never heard of symmetry restoration occurring for large momentum modes. Can you please post a reference?

8. May 27, 2010

### krishna mohan

I seem to have seen that somewhere...that when you are far away from the origin in field space, you can disregard the fact that the minimum is not exactly at zero....but I don't remember where...

I think the statement I made above may not be clear and might not be even accurate...but that is what comes to my mind now...

Could we have the reference please?

9. May 27, 2010

### bapowell

I'm not sure why the high momentum modes of other fields in the theory should see a restored symmetry. For example, the W/Z bosons are massive whether or not their energies are above the electroweak symmetry breaking scale, E > M.

Reference! Reference! Reference!

10. May 27, 2010

### arivero

Hmm??? This is physics, lads, no mathematics :-D Of course that the W and Z keep being massive, the point is that the decay rate of a particle, due to the broken gauge interaction, approaches to the decay rate calculated with the unbroken interaction. This should be the interpretation of a particle physicist. A theoretical physicist tells things as "... the GUT scale where ... the full GUT symmetry is getting restored"

11. May 27, 2010

### arivero

Thinking again, there is something murky here. At m_q<<M_W, quark decay rate goes as m^5, for m_q >> M_W, it goes as m^3, so there is a transition, as I told. But still there is a mass scale, because the decay width itself has units of eV.

Perhaps decay rate is not the adequate object to ilustrate the point. In a truly unbroken symmetry, all the members of a gauge multiplet have equal mass.

12. May 27, 2010

### bapowell

OK. Yes, I was not thinking in terms of decay rates, and that's a very interesting result. I'm guessing this is essentially renormalization group flow -- at scales large relative to the transition, the relevant fields become essentially massless.

13. May 28, 2010

### arivero

Yes but the point you raised implies to ask if there is a transition, at all, in GUT schemes. For ferromagnets and Curie Point, what happens is that above Curie temperature the potential is cuadratic, below curie temperature the potential is mexican-hat quartic, and in the transition it has flat second, third and four derivatives. But here, we consider M_W fixed and just go to higher and higher energies. Is it the same thing to say M_W --> 0 that M_W << E? The problem is more complex because we have two scales: Fermi constant and W mass.

14. May 29, 2010

### fermi

This thread has developed into a new direction, which is just fine, but I feel that the original question was never properly answered. Quickly reminding again what it was:
The answer is the following: given a compact Lie group, how it gets broken down to a smaller subgroup by the vev depends on the group itself as well as the representation of the scalar field. Two different scalar fields under the same group break it differently. I am not sure if this is worked out in detail for the Exceptional Lie groups. But for all other compact simple Lie groups the answer is known at least for the scalar field representations of which dimension is not too large. I refer you to a simple, intuitive yet elegant analysis by Ling-Fong Li from an old 1974 paper: Physical Review D, page 1723, March 1974. Later other groups and representations not analyzed by Li were also categorized in terms of symmetry breaking patterns.

15. May 31, 2010

### krishna mohan

And for discrete groups?

16. Jun 2, 2010

### xepma

Higgs symmetry breaking from a gauge group G to a discrete subgroup H in two dimensions can lead to very funky physics. The idea is that the Higgs mechanisms turns all the gauge bosons (the mediating particle) into very heavy particles. This essentially means the interactions is ultra-short ranged (just like the weak interaction).

However, what can still remain is an Aharanov-Bohm effect between two particles that couple with each other through the gauge force. The Aharanov-Bohm effect doesn't care about the mass of the gauge bosons, since it is not mediated by these bosons. But in two dimensions it turns out that this effect does become a type of force, namely a topological force. This is very, very cool, as the force can essentially be interpret as a weird kind of statistics between the particles that are gauge invariant under the residual gauge group H. Such particles are called anyons.

My story probably doesnt make too much sense, but you can read more about it here http://arxiv.org/abs/hep-th/9511201 :-)