# SU(3) symmetry of QCD

1. Mar 28, 2007

### touqra

I find it awkward that quarks are in fundamental representation of SU(3) while gluons are in adjoint representation of SU(3). Is there a reason as to why this is the case? Why aren't they in the same representation or in the current specific representation?

2. Mar 29, 2007

### marlon

Why is this awkward ? Good question ! Basically you are asking why there are 8 gluons and not 9.

Quarks interact with each other through gluons. Indeed, colour charges (ie quarks) interact via the exchange of colour. Thus, a gluon is able to change the quark's colour quantum number.

Now, if gluons were NOT in the adjoint representation, but any other fundamental SU(3) representation, there would be 9 of them because these representations are 9 dimensional. So why are there 8 gluons ?

One gluon is special : if you make a linear combination of (red-antired + blue-antiblue + green-antigreen)/sqrt(3), you just made a gluon that CANNOT change the colour of a quark. Do you see why ? This gluon is a singlet state and does not respect the definition of a gluon (ie a force carrier that can change the colour of quarks). So, in total we have not 9 but 8 gluons ! To describe these 8 gluons, we need an 8 dimensional space : this is the adjoint space representation of SU (3).

Here is more : http://math.ucr.edu/home/baez/physics/ParticleAndNuclear/gluons.html

marlon

Last edited: Mar 29, 2007
3. Mar 30, 2007

### nrqed

But this only changes the question to why couldn't a gluon be a singlet? There is no theoretical reason why. After all, photons are singlet under the U(1) transformations so there is nothing a priori wrong with having a singlet gluon!!

4. Mar 31, 2007

### marlon

No it does not. A gluon can indeed be in a singlet state but the point is that this state is NOT able to be a force carrier for the strong force.

Baryons, consiting out of 3 quarks, are colour neutral. This means that they are neutral to the strong force. The three internal quarks mutually interact via this strong force through the exchange of colour. So, suppose we have a baryon with internal structure : red + green + blue. The associated gluon (which is by definition always a state-antistate) red-antired + blue-antiblue + green-antigreen MUST be a singlet state (i am not saying that it cannot be a singlet state) because it must be colour neutral. If this were not the case, ie if this gluon WAS able to change the colour of the baryon i started from, the baryons would interact with each other. This is against the colour neutral baryon evidence that we have from experiments.

marlon

5. Apr 2, 2007

### xepma

Quarks carry colour-charge, which we describe in the defining rep of $$SU(3)$$ which is usually denoted by its dimension: $$3$$.

Anti-quarks live in the conjugate rep, and therefore carry (what I call) anti-colour charge. They are described by the conjugate rep: $$3^*$$.

(Note that because we work with $$SU(3)$$ the conjugate rep and the defining rep are different representations. In SU(2) the conjugate representation of the defining rep is actually equivalent to the defining rep)

Gluons are force carriers and exchange both colour and anti-colour charge. Since they carry both charges they have to be described by the product representation: $$3\otimes 3^*$$, since it is this group which describes the internal symmetry of the gluons. But Lie Group theory tells us we can decompose this representation as a sum of irreps: $$3\otimes 3^* = 8\oplus 1$$. The 1-dimension rep we forget, since this singlet state (as explained above) doesn't exchange colour charge at all. The 8 dimensional rep is just the adjoint rep of SU[3].

Last edited: Apr 2, 2007
6. Apr 2, 2007

### mormonator_rm

If Rishon theory were true, this would not be a surprise, since the singlet gluon is equivalent to a photon in Rishon theory.

Just to give a brief background, Rishon theory is an attempt to unify the leptons, quarks, and guage bosons under one doublet of super-fundamental particles, known as Rishons. The "t" Rishon has an electric charge of -1/3, and the "v" Rishon has an electric charge of 0. Each Rishon carries a magnetic moment as well, with each Rishon having magnetic quantum numbers of either 1, 2, or -3, and anti-Rishons carrying magnetic quantum numbers of either -1, -2, or 3. While the "t" and its antiparticle are massive, the "v" and its antiparticle are supposedly massless. However, the binding potential between two "t" Rishons counters the mass of one of them, making the "ttv" and "tvv" states (up and down quarks) nearly equal in mass. Heavier generations of quarks and leptons are achieved by adding "ttbar" groups to the three-Rishon bound states.

In Rishon theory, the gluons are "t v tbar vbar" in content, with allowed non-zero magnetic quantum number combinations, while the photon is simply "t v tbar vbar" with zero for a total magnetic quantum number. In essence, a colorless gluon...

7. Apr 2, 2007

### touqra

Antiquarks must live in conjugate representation because they carry anticolors. How about QED U(1) with its antifermions or the SU(2) in electroweak? I mean like are other forces the same, i.e. particles in defining rep, antiparticles in conjugate rep?

8. Apr 2, 2007

### touqra

Antiquarks must live in conjugate representation because they carry anticolors. How about QED U(1) with its antifermions or the SU(2) in electroweak? I mean like are other forces the same, i.e. particles in defining rep, antiparticles in conjugate rep?

Also, do we just throw away the singlet after the direct product? But then, what about the closure property of a group? Wouldn't it be violated? or maybe violate any mathematical stuffs?

9. Apr 3, 2007

### xepma

Yea, in the other case it's the same deal. But here's the "fun" part: in the case of SU(2) and U(1) the defining representation is actually the same representation as it's conjugate! For SU(2) you simply have $$2=2^*$$

No, this is all taken care off. I'll come back to it later, I got a class now ;)

10. Apr 3, 2007

### xepma

OK, to answer this question you first have to realise that the group ($$3\otimes 3^*$$) acts on some target space, which in this case is 9 dimensional (logically) and can be viewed as an internal strucuture of the particle (just a label of the wave funtion in some sense).

Now, when a group element acts on this target space it maps the space on to itself. This can easily be viewed as some matrix which acts on a vector space. The fact that you can view a group element as a matrix, is called a (matrix) representation of the group, because it ofcourse obeys the group structure.

In the case of Lie Groups what you quite often will find is that the target space can actually be subdivided in to smaller, subspaces, which are actually target spaces of their own. In some sense, when a group element acts on the whole target space, it actually maps the smaller subspaces on to itself: they do not mix. Such a smaller subspace actually forms a representation on its own.

But at some point you cannot subdivide a target space anymore. The representation is now called an irrep. There is a general theory which says that any (product) representation of a group can uniquely be "build up" by irreps of that same group.

In our case we have some 9-dimensional representation of SU(3), which is the product rep: $$3\otimes 3^*$$. This representation is decomposable into two irreps of SU(3): $$1$$ and $$8$$. This means that if a group element of SU(3) acts on this 9 dimensional space, then it actually maps all elements of the 8 dimensional subspace onto that same space, and the same for the 1 dimensional subspace. The 1 and 8 dimensinional subspace do not mix!

So stuff like enclosure is perfectly taken care off.

But why can we throw away the 1 dimensional subspace again? Well, from a mathematical point of view it's because this space is trivial: it's one dimensional. Since SU(3) is unitary (determinant = 1) this means that nothing happens when you act with a group element on this space. All group elements act trivially on this space, i.e. nothing happens.

11. Apr 3, 2007

### mjsd

xepma very nicely said.....
i think I know what you are getting at...here
you are not throwing away any elements of the group by neglecting the 1-rep. the 8-rep is the 8 dimensional representation of the SU(3) group... in other words, both the 1-rep and the 8-rep "contain all elements" of the group (but simply expressed in different representation space)

12. Apr 3, 2007

### George Jones

Staff Emeritus
You have to be a little more restrictive than this. This isn't true for all representations of all groups. It is true for finite-dimensional representations of compact groups (Peter-Weyl theorem) like SU(n).

13. Apr 3, 2007

### xepma

Ofcourse, ofcourse, you're totally correct.

But I'm just trying to give a general idea of how these irreps work. Naming every little detail and exception can sometimes (actually: a lot of the times) work the wrong way: it just clouds the main idea.

14. Apr 8, 2007

### samalkhaiat

Last edited: Apr 8, 2007
15. Apr 9, 2007

### samalkhaiat

16. Apr 9, 2007

### humanino

This statement is certainly incorrect. At least too vague. Here at my lab, we probe neutrons by using (virtual) photons all the time ! The neutron does not have a vanishing charge distribution.

17. Apr 9, 2007

### samalkhaiat

18. Apr 12, 2007

### mormonator_rm

19. Apr 12, 2007

### mormonator_rm

Are you referring to VMD? The photon has the same charge and quantum numbers as any hadron it could momentarily convert into, so there is essentially no difference. The photon is better for probing anything anyway since it has no rest mass.

Well, there will never be a hydrogen-like atom with a neutron instead of a proton because the cumulative charge of the neutron will not hold an electron in orbit. The neutron is still composed of smaller charged quarks, however, and therefore can still have photon interactions.

20. Apr 15, 2007

### samalkhaiat

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