# Why must strong force theory be QCD?

1. Jan 17, 2012

### ndung200790

Please teach me this:
Why must strong interaction theory be QCD(SU(3)) theory?Because asymptotic freedom of strong force is the characteristic of any non-Abelian gauges theories.E.g why must it not be SU(2)?
Thank you very much for your kind helping.

2. Jan 17, 2012

### vanhees71

There's no a-priori reason why the strong interaction is described with a local SU(3)-gauge theory. Any asymptotically free gauge theory would do.

Thus, you need empirical input. One is that in Gell-Mann's model for hadrons, each quark flavor should appear in three different colors (and antiquarks the corresponding anticolors), and SU(3) provides this feature in its fundamental representation and the complex conjugate of this representation. Thus, it's a good ansatz to start with a color SU(3)-gauge model.

Taken the standard model as a whole, three colors are also good since they allow for patterns of charges in the electroweak sector matching electroweak phenomenology and at the same time allow to avoid the anomalous breaking of the electroweak (chiral) gauge symmetry. This works perfectly since the quarks have their known -1/3, 2/3 electric charges and come in three colors. However, this is not the only possible pattern of charge-quantum numbers which conspires such that this anomaly is avoided.

So the strongest reason for choosing SU(3) as the gauge group, i.e., QCD for the strong interaction is its empirical success.

3. Jan 17, 2012

### clem

SU(3) has eight gluons, leading to asymptotic freedom. SU(2), which was also proposed early on, would have only three gluons, not enough for asymptotic freedom. There is some evidence for asymptotic freedom in the ln(Q^2) breaking of scaling in deep inelastic scattering, so SU(3) is favored.

4. Jan 17, 2012

### tom.stoer

I guess you mean the coefficient (11 Ncolor - 2 Nflavor) in the QCD beta function with Ncolor = 3 and Nflavor = 6. Note that this would allow for small Ncolor or larger Nflavor w/o spoilng asymptiotic freedom. I have never seen any claim that SU(2) QCD would look very different phenomenologically

5. Jan 17, 2012

### fzero

[STRIKE]It's actually the other way around because[/STRIKE] the beta function has an extra minus sign: -(11 Ncolor - 2 Nflavor) .

Edit: Nevermind, the minus sign doesn't contradict what you wrote.

The spectrum would be very different. In particular you wouldn't be able to produce baryons with electric charge 2 and a reasonable lifetime.

VVVV For SU(2), the baryons are quark bilinears. The quarks would presumably have electric charge $\pm 1/2$ and the flavor multiplets have very different distribution of charges.

Last edited: Jan 17, 2012
6. Jan 17, 2012

### clem

The reference is to SU(2) of the color degree of freedom, not flavor.

7. Jan 17, 2012

### torus

I am not sure exactly what you mean. It surely depends on the rep of the quarks, right? Lets say QCD is governed by SU(2), call is colorspin. Lets stick to a 3-d rep for the quarks, i.e. put quarks in colorspin-1 reps. Then you can form mesons (couple two colorspins to 0) and baryons (there is a singlet in 1*1*1). So you might as well asign the same electric charge to these quarks as usual.

This model will of course go awry somewhere in the multiplet structure.

regards,
torus

8. Jan 18, 2012

### tom.stoer

I am not sure whether SU(2) color with unchanged weak and flavor structure would be anomaly free. But let's try

The nucleon in the fundamental rep. of SU(2) isospin would be |ud> with s=0 and el. charge q= e/3.
The Delta's in the adjoint rep. of SU(2) isospin would be |uu>, |dd> and |ud> with s=1 and q= 4/3, -2/3 and 0, respectively.
The pseudo-scalar mesons wouldn't change, neither their quantum numbers nor their masses (OK, tiny changes).

Nevertheless there would be color-neutrality of physical states, confinement, chiral symmetry breaking, asymptotic freedom, ..., so the overall picture would not change so much.

Last edited: Jan 18, 2012
9. Jan 18, 2012

### humanino

I do not mean to deflect the discussion. The behavior of unbroken SU(2) is an interesting question.

I would like to mention that in addition to spectral properties, there are detailed dynamic prediction for scaling violation patterns. One can look at jet production, transverse momentum dependencies, bremsstrahlung partons vs multiplicities, multiplicity flows between jets, inclusive hadron distributions within jets, all forms of event shapes and have been tested when predictions are possible. They rule out any other possibility but SU(3).

10. Jan 18, 2012

### fzero

Adjoint matter changes the beta function significantly. For SU(N) you get (ignoring any other matter fields)

$$b_1 = - ( 11 - 4 N_f ) \frac{N}{3},$$

so the theory is only asymptotically free with 1 or 2 adjoint flavors.

Well you end up with fractionally charged states here, but anomaly cancellation is an important point. I was only addressing the possibility of producing realistic light states. It's obvious that you can't do it with SU(2), but you can obviously have some types of mesons and baryons, just not anything that you could mistake for what is observed.

11. Jan 18, 2012

### tom.stoer

Let me summarize: any SU(N) gauge theory with fermions in the fund. color rep. and appropriate flavor structure would show the same qualitative properties as QCD, but the detailed properties (spectrum, charges, DIS, ...) would change significantly; so experimental quantitative tests rule out anything else but SU(3).

12. Jan 18, 2012

### clem

Three quarks can be in an antisymmetric state of either SU(2)color or SU(3)color.
Then all baryon properties will be the same in either color group.
There would be no difference in the flavor or spin properties.
I think some people are still confusing SU(2)color with SU(2)flavor.

13. Jan 18, 2012

### Haelfix

Its pretty easy to see why you can't have SU(2) color as a gauge group for the strong force as it doesn't admit complex representations. There are no quark-quark bound states!

Less obvious is why you couldn't have a bigger group, like SU(6). There you really do need to work on analyzing hadronic scattering cross sections in order to pin down the structure constants. Historically, electron-positron to hadron and pi meson to gamma-gamma were mostly sufficient.

So its not quite right to say any nonabelian group would work, there are general theoretical constraints. For instance one also needs a completely antisymmetric color singlet baryonic state as well, and that constrains the allowed structure significantly.

14. Jan 18, 2012

### tom.stoer

Yes, there is only one fund. rep., i.e. in SU(2) you have 2* = 2, whereas in SU(N) with N>2 you have N* ≠ N. But why would that rule out confinement? or SU(2) baryons? Aren't there lattice gauge calculations for SU(2) baryons?

15. Jan 18, 2012

### Haelfix

It won't, but it just could never be confused with the strong force of the real world.

More generally, the only thing that I can think off that will spoil asymptotic freedom is as Fzero has mentioned, the fact that the beta functions of certain gauge groups might not allow 6 species of quarks.

16. Jan 18, 2012

### tom.stoer

I don't get it. Please try again.

You claim that with SU(2) color symmetry
Why?

I fully agree, but that is (nearly) irrelevant in the IR.

17. Jan 18, 2012

### fzero

Asymptotic freedom in the UV implies confinement in the IR (and vice versa). Since there is no new physics over the relevant energy scales, you can't have one without the other.

18. Jan 18, 2012

### tom.stoer

Why? How should IR and UV be related? Afaik there are gauge groups like G2 with different center showing no confinement on the lattice.

Asymptotic fredom can be derived from perturbative calculations of beta functions, whereas for confinement we do not have a proof yet. (the perturbative calculation of beta functions breaks down in the IR; you can't deduce confinement from this breakdown, only the breakdown itself)

19. Jan 18, 2012

### fzero

Thanks, I was a bit hasty to say that without qualifications. I believe that it is true for SU(N) gauge theories, but it is lore, not a proven fact. The breakdown of perturbation theory on its own doesn't do anything to remove the singularity that appears in the running coupling. I wasn't really thinking about finite temperature or Higgs phases that can substantially modify the physics.

20. Jan 18, 2012

### tom.stoer

no problem; let's continue with Haelfix

I don't get it. Please try again.

You claim that with SU(2) color symmetry
Why?

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