# Urgent question on spontaneous symmetry breaking

1. Jan 22, 2006

### Kalimaa23

Hello. I've got a QFT final tomorrow, and one question is still bothering me.
Consider two lagrangians.

The first one is
$$L = \frac{1}{2} D_{\mu} \vec{\phi}.D^{\mu} \vec{\phi} + \frac{m^{2}}{2} \vec{\phi}.\vec{\phi} - \frac{\lambda}{4} \left(\vec{\phi}.\vec{\phi}\right)^{2}$$

The second one
$$L = D_{\mu} \Phi^{\dagger} D^{\mu} \Phi + m^{2} \Phi^{\dagger}\Phi - \lambda \left(\Phi^{\dagger}\Phi\right)^2$$

In the first lagrangian $$\vec{\phi}$$ is a vector with real components, in the second on it is a SU(2) doublet.

The first lagrangian obviously has a SO(3) symmetry. If we let the vacuum aquire an expectation value, the symmetry gets broken down to SO(2). The two lost degrees of freedom become Goldstone bosons, which then can give mass to two of the gauge bosons.

The second one has SU(2) symmetry. If I understand it correctly, the symmetry gets completely broken and the three gauge fields acquire mass. (like in the electroweak sector in the Standard Model, but without Weinberg mixing with the U(1) hypercharge gauge field.)

I've got two questions :

Assuming my treatment is correct, why does the first symmetry breaking give mass to two bosons, while the second one gets broken completely?

Is the first lagrangian invariant under SU(2) transformations in the vector representation? Does something need to be alter, other than going with covariant derivatives?

Does anything of this have to do with the fact that SO(3) is locally isomorphic with SU(2), but globally with SU(2) modulo 2?

Any help or insights would be grealty appreciated.

2. Jan 22, 2006

### Careful

Hi Dimitri,

**
Assuming my treatment is correct, why does the first symmetry breaking give mass to two bosons, while the second one gets broken completely?
Is the first lagrangian invariant under SU(2) transformations in the vector representation? Does something need to be alter, other than going with covariant derivatives?
Does anything of this have to do with the fact that SO(3) is locally isomorphic with SU(2), but globally with SU(2) modulo 2?
Any help or insights would be grealty appreciated. **

If I understand you correctly, you say :
(a) the SO(3) model breaks down to SO(2) - the one unbroken mode gets positive mass - the two other massless Goldstone modes give mass to two SO(3) gauge bosons
(b) in the SU(2) model, there are three real Goldstone bosons giving mass to all gauge bosons (after vacuum polarization).

And you ask why this is so. The reason is that a two dimensional complex vector space carries four real degrees of freedom while a three dimensional real vector space only 3 (of course). More concretely :
the zero's of the potential a f*f - b (f*f)^2 (a,b > 0, f a complex two vector) are determined by f = 0 and f*f = a/b which is a quadratic equation invariant under S0(4). Doing the same trick as in the SO(3) case leaves you with a residual SO(3) symmetry (three Goldstone bosons). These three Goldstone bosons give mass to three gauge bosons since the SU(2)
gets broken to 1 by vacuum polarization. As a side remark: the global SU(2) properties (being the universal double cover of SO(3)) are only relevant when dealing with fermions. The gauge degrees of freedom are coupled to the Lie algebra elements which only depend upon a neighborhood of the identity element of the group (which is the same for SO(3) and SU(2) of course).

Cheers,

Careful

Last edited: Jan 23, 2006
3. Jan 22, 2006

### Kalimaa23

Thank you Careful, that was helpful.

What about the first lagrangian? Is the vector dot product invariant under SU(2)? I would say so, but I cannot really think of an argument.

4. Jan 22, 2006

### vanesch

Staff Emeritus
The vectors are just as well 3-dimensional representations of SU(2) as the they are 3-dimensional representations of SO(3), no ?

5. Jan 22, 2006

### Kalimaa23

I see.

About the second one... does this mean that is invariant onder SO(4), not just SU(2), right?

Sorry if I appear dense, but I'm getting tired here :zzz:

6. Jan 22, 2006

### Careful

Well, the first lagrangian is not invariant under a faithful representation of SU(2), but it is under the defining representation of SO(3) though. The trick with the SU(2) gauge symmetry is that the corresponding lagrangian actually has a higher symmetry group SO(4) (which we break partially by grouping the (1,2) and (3,4) indices in complex numbers).

Last edited: Jan 22, 2006
7. Jan 22, 2006

### Kalimaa23

I'm getting confused here...
One of the questions asked for the first lagrangian was :
Make this system invariant under SU(2) transformations in the vector representation. Does this require any changes besides going over to covariant derivatives?

EDIT : So, is this statement correct : the second lagrangian is invariant under SU(2). The potential however is invariant under SO(4), choosing a vev breaks the SO(4) to SO(3). The three lost degrees of freedom become Goldstone bosons, which then give mass to the three gauge fields.

This puzzles me. Is it the SU(2) that is broken or the SO(4)

Last edited: Jan 22, 2006
8. Jan 22, 2006

### Careful

Hi, if \psi is a REAL three vector then the lagrangian is invariant under SO(3). However, using the homomorphism from SU(2) to SO(3), one can obtain a non-injective REAL representation of SU(2) on R^3. Let me recall to you that in physics we mostly speak about COMPLEX representations (for real representations, the theorem of Schur does not hold). When we speak about a theory having a particular symmetry, we mean by this that the Lagrangian is invariant under a FAITHFUL (injective) IRREDUCIBLE representation of the group. Therefore, we say that the first lagrangian is SO(3) invariant instead of SU(2) invariant.

Concerning the second lagrangian L2, suppose one would take flat partial derivatives, then L2 is invariant under global SO(4) transformations (as well as global SU(2) transformations of course). However, by plugging in the SU(2) covariant derivative, L2 is invariant under local SU(2) and global SO(4) transformations (by transforming the Lie algebra of SU(2) - embedded in the Lie algebra of SO(4) - under the adjoint representation of SO(4), that is A -> g A g^T (g in SO(4), A in su(2) embedded in so(4)).

What is broken, is the SU(2) invariance of the vacuum state by choosing a particular polarisation and by perturbing L2 (to second order) around it.

Last edited: Jan 22, 2006
9. Jan 23, 2006

### Kalimaa23

Thanks alot for the help everyone. The final went great, 19/20

10. Jan 23, 2006

### vanesch

Staff Emeritus
Great, congrats !

And then I can now ask something here. I was affraid to post it yesterday because I'm not sure myself, and I didn't want to bring in confusion before your finals.

But when I look at your second lagrangian, to me it has U(2) symmetry (and not only SU(2) symmetry), no ? And after symmetry breaking, it has U(1) symmetry ? So we seem to go from 4 real Lie parameters to 1, while in the first lagrangian, we go from SO(3) to SO(2), from 3 real Lie parameters to 1.

Now (as Lie algebras) SU(2) and O(3) are isomorphic, and SO(2) and U(1) are isomorphic, but I'm not sure about the relationship between SO(4) and U(2) off the top of my head...

11. Jan 23, 2006

### Kalimaa23

No, from what I gather the second one indeed had only SU(2) symmetry, otherwise you would extra minus signs in there. I'm not sure though. I think the whole point of the question was that in the first one a part of the symmetry remains SO(2) or U(1), same thing, and in the second one it gets completely broken.

SO(4) has 6 generators, U(2) has 4, so I they shouldn't be isomorphic, right?

12. Jan 23, 2006

### vanesch

Staff Emeritus
Well, I don't understand: The difference between SU(2) and U(2) is that you can multiply each element of SU(2) by exp(i phi) and you obtain U(2), no ? So the difference is an extra phase factor.

Well, if you look at the second lagrangian, the doublet always occurs together with its conjugate, so multiplying the doublet with a phase factor doesn't change this (because the conjugate has the opposite phase factor due to conjugation). So it looks like me that a simple phase factor (the non-S part of U(2)) is also a symmetry of the second lagrangian...

Ah, I thought one still kept the U(1) symmetry... (as above...) But that was from the top of my head. Should probably sit down and calculate.

Sure, silly me!

Last edited: Jan 23, 2006
13. Jan 23, 2006

### Careful

**
Ah, I thought one still kept the U(1) symmetry... (as above...) But that was from the top of my head. Should probably sit down and calculate.
Sure, silly me ! **

Hi Patrick,

the dimension of SU(n) = n^2 -1 and the one of U(n) = n^2 and there is indeed the difference of the U(1) factor. The Lagrangian L2 is indeed invariant under U(2), but this symmetry is *not* entirely broken, the remaining generator being : (1 - p3)/2 where p3 is the third Pauli matrix. Still you have three real Goldstone bosons (one massive boson), three massive gauge particles and one massless gauge particle (an EM gauge potential).

Cheers,

Careful