# SU(3) symmetry of QCD

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?

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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

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nrqed
Homework Helper
Gold Member
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).
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!!

But this only changes the question to why couldn't a gluon be a singlet?
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

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].

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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!!
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...

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].
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?

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].
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?

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?
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 $$2=2^*$$

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

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?

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.

mjsd
Homework Helper
xepma very nicely said.....
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?
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)

George Jones
Staff Emeritus
Gold Member
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.
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).

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).
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.

samalkhaiat
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?
Gluon field is a gauge field, gauge fields and their transformatiom laws follow from local gauge principle: "some global internal symmetry holds locally"

Let me explain this;

1) say we believe that (at the fundamental level) SU(n) is a symmetry respected by nature.

2) assuming that the building blocks of matter are fermions, then it is natural to put these fermions in the simplest representation of SU(n). This is the defining (fundamental) representation. Mathematically,this means that we describe particles by n-component object on spacetime $$q_{i}(x)$$ and that for each value of i (= 1,2,..,n) q(x) is a bispinor (Dirac field).

3) The free-field Lagrangian

$$\mathcal{L} = i\bar{q}_{i}\gamma_{\mu}\partial^{\mu}q_{i}$$

is now invariant under the global SU(n) (infinitesimal) transformations

$$\delta q_{i} = i\theta^{a}T^{a}_{ij}q_{j}$$
$$a = 1,2,...,n^{2}-1$$

where $$\theta^{a}$$ are the x-independent parameters and $$T^{a}$$ are the finite dimensional (matrix) representation of SU(n), i.e., a set of $$n^{2}-1$$ matrices satisfying the Lie algebra of SU(n)

$$[T^{a}, T^{b}] = if^{abc} T^{c}$$

The invariance under su(n) implies (by Noether theorem) the existence of conserved currents

$$J^{a}_{\mu} = i \bar{q}_{i} T^{a}_{ij}\gamma_{\mu} q_{j}$$
$$a = 1,2,...,n^{2}-1$$

This means that the fundamental particles carry $$(n^{2}-1)$$ time-independent and Lorentz scalar charges

$$Q^{a}= \int d^{3}x J^{a}_{0}$$

Notice that $$n^{2}-1$$ is the dimension of the adjoint representation of su(n) and that the currents belong to this rep. Indeed, it is easy to show that

$$\delta J^{a}_{\mu} = f^{abc}\theta^{b} J^{c}_{\mu}$$

So, if you believe that:
1) the fundamental particles interact through $$J^{a}_{\mu}$$ ,
2) the interaction lagrangian must be invariant under Lorentz and SU(n) transformations,
then you need $$n^{2}-1$$ Lorentz-vector to couple to Noether currents $$J^{a}_{\mu}$$ .

Pre-gauge theory Lagrangians had interaction term of the form

$$J^{a}_{\mu}J^{a\mu}$$

but, we now know (thanks to the local gauge principle) that the correct interaction Lagrangian has the form

$$J^{a}_{\mu}A^{a\mu}$$

4) The local gauge principle:

Allow the parameters (which play the role of local coordinates of the group manifold) to vary with x. Then demand local su(n) invariance, i.e., make the Lagrangian invariant under local su(n) transformations

$$\delta q_{i} = i \theta^{a}(x) T^{a}_{ij}q_{j}$$

It is easy to verify that $$\mathcal{L}$$ is not invariant under these transformations. To restore invariance, one must introduce a new compensating term, i.e., one should find a derivative D which transforms like q ;

$$\delta(D_{\mu}q_{i}) = i \theta^{a}(x) T^{a} (D_{\mu}q_{i})$$

and replace $$\partial$$ by $$D$$ in the matter Lagrangian.
The covariant derivative is constructed by introducing a vector (gauge) field taking values in the Lie algebra of SU(n):

$$A_{\mu} = i A_{\mu}^{a} T^{a}$$

and defining

$$D_{\mu}q = (\partial_{\mu} + A_{\mu})q$$

Thus one needs gauge fields in the number given by the number of generators of the symmetry group, i.e., $$n^{2}-1$$ .

From the above equations you can find the infinitesimal (gauge) transformations for A ;

$$\delta A^{a}_{\mu} = -\partial_{\mu}\theta^{a} + f^{abc}\theta^{b} A^{c}_{\mu}$$

You can check that these transformations do form a group:

$$[\delta_{1},\delta_{2}] A^{a}_{\mu} = \delta_{12}A^{a}_{\mu}$$

You can also see that under global transformations $$(\partial\theta=0)$$ the gauge fields transform according to the adjoint representation of the group

$$(T^{a})^{bc}= if^{abc}$$

So, Lorentz invariance requires the gauge fields to carry a spacetime index (i.e. vectors) and internal symmetry puts these vector (gauge) fields in the adjoint representation. OR, in short, quarks and gluons can not be in the same representation because Lorentz invariance forbids particles of different spin to occupy the same SU(n) multiplet.

regards

sam

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samalkhaiat
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!
The gauge principle is the reason. By gauging su(3) you get 8 vector fields (gluons) living in the adjoint representation. No singlet gluon come out from the gauge principle. So, in the contex of $$SU_{c}(3)$$ , the question is meaningless.

People who worked on the unification of the strong and EM interactions (with gauge group U(3)) suggested that the colour singlet [in 3 x 3* = 8 +1] is the photon. But this meant that this singlet would couple to proton and to neutron with the same strength. And this cannot be correct because:
1) experiments show that photon does not couple neutron.
2) the strong force does not have a component of infinite range.
3) gravity is a lot weaker than em.

regards

sam

1) experiments show that photon does not couple neutron.
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.

samalkhaiat
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.

Yes, because this photon of yours behaves as a hadron and sees the neutron structure. My statement was about the long range behaviour.For example, the absense of Hydrogin-like atom with proton replaced by neutron.

sam

The gauge principle is the reason. By gauging su(3) you get 8 vector fields (gluons) living in the adjoint representation. No singlet gluon come out from the gauge principle. So, in the contex of $$SU_{c}(3)$$ , the question is meaningless.

People who worked on the unification of the strong and EM interactions (with gauge group U(3)) suggested that the colour singlet [in 3 x 3* = 8 +1] is the photon. But this meant that this singlet would couple to proton and to neutron with the same strength. And this cannot be correct because:
1) experiments show that photon does not couple neutron.
2) the strong force does not have a component of infinite range.
3) gravity is a lot weaker than em.

regards

sam
Wait a second here.

1) I thought that I've seen that there are photoproduction experiments that involve neutrons. I'll have to look it up again in a second here.

2) Well, if you had a colorless gluon, I suppose it would not cause the same behavior as a colored gluon would when it couples to something. The EM force may simply be a colorless representation of the strong force. In Rishon theory, the strong force and electromagnetic force both arise from magnetic quantum numbers (not saying this is absolutely true, just an idea). Anyways, since the "gluon" is colorless, it will not be bound into a hadron or glueball, and therefore may escape confinement.

3) What does gravity have to do with QED and QCD at this moment? Some theorists argue that they must be kept seperate because they cannot be related. I don't particularly like this view, but then again I'm not an expert on gravity so I'll let the experts deal with that one.

Yes, because this photon of yours behaves as a hadron and sees the neutron structure.
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.

My statement was about the long range behaviour.For example, the absense of Hydrogin-like atom with proton replaced by neutron.

sam
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.

samalkhaiat
Wait a second here.

1) I thought that I've seen that there are photoproduction experiments that involve neutrons. I'll have to look it up again in a second here.

2) Well, if you had a colorless gluon, I suppose it would not cause the same behavior as a colored gluon would when it couples to something. The EM force may simply be a colorless representation of the strong force. In Rishon theory, the strong force and electromagnetic force both arise from magnetic quantum numbers (not saying this is absolutely true, just an idea). Anyways, since the "gluon" is colorless, it will not be bound into a hadron or glueball, and therefore may escape confinement.

3) What does gravity have to do with QED and QCD at this moment? Some theorists argue that they must be kept seperate because they cannot be related. I don't particularly like this view, but then again I'm not an expert on gravity so I'll let the experts deal with that one.
What I said, had nothing to do with any preon model; Rishon or non-rishon. Personally, I don't waist time on them because they are ugly and smell wrong.

Sir, I was talking about a gauge theory [based on U(3) = SU(3) X U(1)] proposed, in the past, as an electro-strong unified theory. Such theory leads to nine gauge bosons. The 9th (colour singlet) boson was identified with the photon. Also,according to this U(3)-theory, the 9th boson couples to all colour singlet baryons (including P & N) with the same strength, not (as photon does) in proportion to their charge. This means:

1) if a colour singlet gluon exists, it could be exchanged between two colour singlets (a proton & a neutron), giving rise to a long-range force with strong coupling, whereas we know that the strong force does not have such a long-range component. Indeed, the absence of a singlet gluon as well as confinement are the reasons for the very short-range nature of the strong force.
2) if a colour singlet gluon exists, and if it is identified with the photon, then nothing would prevent the existence of 1000's of baryon-electron bound states. But,in the real world, only p-e exists.
3) as for the link to gravity, you don't need an expert to figure it out for you; According to the U(3) theory, the 9th gluon (as I mentioned) interacts with all baryons with same strength. Now, since the mass of a star, planent, cucumber etc., is proportional to the number of baryons, such interaction would give an extra contribution to gravity.

So, mathematically speaking, the issue (I was addressing) is whether the symmetry of QCD is SU(3) [which needs only 8 gluons] or U(3) [which calls for 9 gluons]. In our world, experimental evidences as well as mathematical consistency, resolve the issue in favor of SU(3). This is why I said that it is meaningless to ask a questions about 9th gluon in SU(3).

regards

sam

What I said, had nothing to do with any preon model; Rishon or non-rishon. Personally, I don't waist time on them because they are ugly and smell wrong.
I don't blame you; I personally feel that Rishon models are a little too speculative, and make too few predictions. I don't know if I feel they are "ugly" per say, but they are interesting in the sense that I try to keep an open mind, and I don't think we can prove them wrong as yet.

Sir, I was talking about a gauge theory [based on U(3) = SU(3) X U(1)] proposed, in the past, as an electro-strong unified theory. Such theory leads to nine gauge bosons. The 9th (colour singlet) boson was identified with the photon. Also,according to this U(3)-theory, the 9th boson couples to all colour singlet baryons (including P & N) with the same strength, not (as photon does) in proportion to their charge. This means:

1) if a colour singlet gluon exists, it could be exchanged between two colour singlets (a proton & a neutron), giving rise to a long-range force with strong coupling, whereas we know that the strong force does not have such a long-range component. Indeed, the absence of a singlet gluon as well as confinement are the reasons for the very short-range nature of the strong force.
That entirely depends on what interpretation you have of what all fits the description of a "colour singlet". Some might be of the opinion that we can indeed include leptons in all their varieties. And why would a "colour singlet" gluon offer the same attractive potential as its "coloured" partners?

2) if a colour singlet gluon exists, and if it is identified with the photon, then nothing would prevent the existence of 1000's of baryon-electron bound states. But,in the real world, only p-e exists.
Of all the charged hadrons in existence, only the proton lasts long enough to be in a bound state with an electron for a measurable existence. The other resonances would decay before completing an orbit to be recognized as a bound state.

3) as for the link to gravity, you don't need an expert to figure it out for you; According to the U(3) theory, the 9th gluon (as I mentioned) interacts with all baryons with same strength. Now, since the mass of a star, planent, cucumber etc., is proportional to the number of baryons, such interaction would give an extra contribution to gravity.

So, mathematically speaking, the issue (I was addressing) is whether the symmetry of QCD is SU(3) [which needs only 8 gluons] or U(3) [which calls for 9 gluons]. In our world, experimental evidences as well as mathematical consistency, resolve the issue in favor of SU(3). This is why I said that it is meaningless to ask a questions about 9th gluon in SU(3).

regards

sam
I agree that in the current standard model SU(3) is clearly the only answer. As for the idea that a "9th gluon" would create a gravity contribution, I think that allegation is a little premature. There are quantum numbers to conserve, and the vector gluon would probably not contribute as you would expect. I have not looked into that myself, so I really can't tell you either. Does somebody else want to take this one?