Why must strong force theory be QCD?

In summary, the strong interaction is described by a local SU(3)-gauge theory due to its empirical success and the fact that it provides the necessary features for the Gell-Mann model for hadrons. SU(2) was also proposed early on, but would not have enough gluons for asymptotic freedom. Additionally, SU(3) allows for patterns of charges in the electroweak sector that match electroweak phenomenology and avoid anomalous breaking of the electroweak gauge symmetry. There is also evidence for asymptotic freedom in QCD, making SU(3) a favorable choice. Choosing SU(2) instead would result in a very different spectrum and dynamics, which have been extensively tested and ruled out
  • #1
ndung200790
519
0
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.
 
Physics news on Phys.org
  • #2
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
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
clem said:
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.

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
tom.stoer said:
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.

[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.

I have never seen any claim that SU(2) QCD would look very different phenomenologically

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 [itex]\pm 1/2[/itex] and the flavor multiplets have very different distribution of charges.
 
Last edited:
  • #6
The reference is to SU(2) of the color degree of freedom, not flavor.
 
  • #7
fzero said:
VVVV For SU(2), the baryons are quark bilinears. The quarks would presumably have electric charge [itex]\pm 1/2[/itex] and the flavor multiplets have very different distribution of charges.

I am not sure exactly what you mean. It surely depends on the rep of the quarks, right? Let's say QCD is governed by SU(2), call is colorspin. Let's 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
fzero said:
The spectrum would be very different. In particular you wouldn't be able to produce baryons with electric charge 2 and a reasonable lifetime.

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:
  • #9
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
torus said:
I am not sure exactly what you mean. It surely depends on the rep of the quarks, right? Let's say QCD is governed by SU(2), call is colorspin. Let's 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

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

[tex]b_1 = - ( 11 - 4 N_f ) \frac{N}{3}, [/tex]

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

tom.stoer said:
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.

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
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
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
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
Haelfix said:
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!
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
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
Haelfix said:
It won't, but it just could never be confused with the strong force of the real world.
I don't get it. Please try again.

You claim that with SU(2) color symmetry
Haelfix said:
there are no quark-quark bound states
Why?

Haelfix said:
... 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.
I fully agree, but that is (nearly) irrelevant in the IR.
 
  • #17
tom.stoer said:
I fully agree, but that is (nearly) irrelevant in the IR.

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
fzero said:
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.
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
tom.stoer said:
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)

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
no problem; let's continue with Haelfix

Haelfix said:
It won't, but it just could never be confused with the strong force of the real world.
I don't get it. Please try again.

You claim that with SU(2) color symmetry
Haelfix said:
there are no quark-quark bound states
Why?
 
  • #21
Hey Tom. I am not saying that SU(2) color would not form bound states. Just that those bound states that it does form, could never be confused for the real world QCD spectrum since amongst other reasons you don't have complex representations (only real or pseudoreal) and one requires color neutrality.

So I mean mesons in QCD are 3 * 3bar = 1 + 8. The singlet state here is all important..

Whereas a diquark state is 3 * 3 = 3bar + 6.. Not color neutral!

So yea, since we observe quark-antiquark bound states, and not quark-quark bound states, it rules out SU(2)c as a symmetry of the real world.

Incidentally, its a cute question that we give to our grad students for their comp exams, and that is to calculate the weak force bound states and their lifetimes.

As for asymptotic freedom.. Well the flipside is infrared slavery. The question whether or not infrared slavery implies confinement, mass gaps and so on is a difficult question as you know since we don't quite understand the precise mechanism at play. They seem to go together in the common examples.
 
  • #22
Haelfix said:
Hey Tom. I am not saying that SU(2) color would not form bound states. Just that those bound states that it does form, could never be confused for the real world QCD spectrum since amongst other reasons you don't have complex representations (only real or pseudoreal) and one requires color neutrality.

So I mean mesons in QCD are 3 * 3bar = 1 + 8. The singlet state here is all important..

Whereas a diquark state is 3 * 3 = 3bar + 6.. Not color neutral!

So yea, since we observe quark-antiquark bound states, and not quark-quark bound states, it rules out SU(2)c as a symmetry of the real world.

I don't agree.

Both meson states and diquark = baryon states of SU(2) color are constructed as 2 * 2 = 1 + 3 where '1' is (for both mesons and baryons!) the physical color singulet. Using 3 * 3 would correspond to quarks in the adjoint rep. of SU(2) color which is of course something totally different and does by no means correspond to QCD.
 
Last edited:
  • #23
There is direct experimental evidence for quarks having 3 colors. Electron-positron collisions produce hadrons at three times the rate that one would expect for colorless quarks.

Particle Data Group - 2010 Reviews, Tables, Plots > Kinematics, Cross-Section Formulae, and Plots > Plots of cross sections and related quantities (rev.). With a center-of-mass energy more than about 1.5 to 2 GeV, one gets a very good fit with QCD calculations. The lowest-order effect is not QCD, of course, but electromagnetic: virtual photon -> quark + antiquark. The factor of 3 comes from
virtual photon -> red quark + cyan antiquark
virtual photon -> green quark + magenta antiquark
virtual photon -> blue quark + yellow antiquark

Furthermore, electron-positron collisions produce quark-antiquark pairs with the angular distribution expected of quarks being spin-1/2 particles.
http://www.sprace.org.br/slc/file.php/6/HEP-I-II/HEP_6_1.pdf [Broken]


From the hypothesis that mesons and baryons are gauge-singlet states, one can determine what gauge groups that they may follow. Mesons are quark-antiquark states, and are thus trivial, while baryons are 3-quark states, and are the interesting case. So we must find which gauge group allows 3 quarks to be in a singlet state.

We do this by finding conserved quantities or congruence classes for the simple Lie algebras. Their irreducible representations or irreps are denoted by highest-weight vectors; these are vectors of nonnegative integers. The conserved quantities are calculated from highest-weight vector w by
Q(irrep with w) = (c.w) mod c0
where c0 and the components of c are all nonnegative integers.

All weights in an irrep have a Q value equal to that of the irrep as a whole, and
Q(product of irreps with HWV's w1 and w2) = Q(irrep with w1) + Q(irrep with w2)

Going over the simple Lie algebras:
A(n) - SU(n+1) - c0 = n+1
B(n) - SO(2n+1) - c0 = 2
C(n) - Sp(2n) - c0 = 2
D(n) - SO(2n) - even n: two with c0 = 2, odd n: one with c0 = 4
G2 - none
F4 - none
E6 - c0 = 3
E7 - c0 = 2
E8 - none

For SU(2)/SO(3), this is evident in boson vs. fermion parity.

So to get a c0 that can divide 3, one can only have SU(3n) or E6. Since quarks have 3 color states, that gives us SU(3). The quarks are in the 3 irrep, or {1,0}, the antiquarks in the 3* irrep or {0,1}, and the gluon in the adjoint, 8 irrep, or {1,1}. The conserved quantity here is "triality": Q(w) = (w1 + 2*w2) mod 3. Q(quarks) = 1, Q(antiquarks) = 2, Q(gluon) = 0, Q(mesons) = 0, Q(baryons) = 0.


Furthermore, three (anti)quarks can combine to make a colorless, {0,0} singlet. This combination is antisymmetric, which resolves a puzzle with the quark model. In the baryons, spin and flavor must be combined in symmetric fashion. This is evident from the Delta++ (uuu), the Delta- (ddd) and the Omega- (sss), all spin-3/2 (spins parallel). This violates Fermi-Dirac statistics, expected from quarks having spin 1/2, but being antisymmetric in color yields an overall antisymmetry, thus preserving F-D stats.


The gauge particle of QCD has been named the gluon, and it's also been seen indirectly in electron-positron collision. An energetic quark-antiquark pair makes a two jets of hadrons, one for the quark and one for the antiquark. One of the two can radiate a gluon, and it will make an additional jet. The angular distribution of the resulting gluon can be used to find its spin, and as expected for a gauge particle, its spin is 1.
 
Last edited by a moderator:
  • #24
I offer the follow hypothesis for consideration.

The basis of the hypothesis is that the strong force (SF) can be described as a special unitary group operation where SU(3) represents a matrix multiplication of SU(2), and conversely and equally, SU(2) represents a matrix multiplication of SU(3). Using Feynman path integral approach, the strong force for any stable isotope has identical probability of being described by application of either special unitary group operation.

SU(3) = [3 * SU(2)] / 2 = SF

SU(2) = [2 * SU(3)] / 3 = SF

Any comments appreciated.
 
  • #25
No idea what you mean.

As we said a coupleof times: the qualitative picture (asymptotic freedom, confinement, chiral symmetry breaking, ...) is identical forSU(2) and SU(3); but the detailed phenomenology (bound states = hadron multiplets, spectrum = masses, charges, decay width, ...) is different.

You could use SU(2)color and SU(n)flavor to calculate DIS, spectra in lattice QCD etc. - and you'll find results differung from nature
 
  • #26
tom.stoer said:
I don't agree.

Both meson states and diquark = baryon states of SU(2) color are constructed as 2 * 2 = 1 + 3 where '1' is (for both mesons and baryons!) the physical color singulet. Using 3 * 3 would correspond to quarks in the adjoint rep. of SU(2) color which is of course something totally different and does by no means correspond to QCD.

Right but then you don't have any hope of having something resembling the quark model. I agree that you can make a color singlet state out of the fundamental of SU(2) of course.

Another issue for SU(2)c is that again b/c of the pseudo reality of the representation, the mass generation mechanism for the quarks is going to be quite unlike the real world as it is not a chiral representation and the HIggs mechanism doesn't necessarily work.

I'd have to think about it, as its somewhat subtle but I think there will be some sort of mass generation for the quarks arising from chiral symmetry breaking. Its a little upside down from the usual process (normally chiral symmetry is broken by the quark masses).
 
  • #27
Haelfix, thanks for answering, I think we perfectly agree
 

1. Why is QCD the accepted theory for the strong force?

The Quantum Chromodynamics (QCD) theory has been extensively tested and supported by numerous experiments and observations. It accurately explains the behavior and interactions of subatomic particles, particularly those affected by the strong force. It has also successfully predicted new particles and phenomena, further solidifying its validity as the accepted theory for the strong force.

2. How does QCD explain the strong force?

QCD is a quantum field theory that describes the strong force as the exchange of particles called gluons between quarks. These gluons carry the color charge, which is the property that interacts with the strong force. The theory also explains the phenomenon of confinement, where quarks cannot exist in isolation due to the strong force.

3. What makes QCD different from other theories of the strong force?

QCD is a part of the Standard Model of particle physics, which is the most comprehensive and successful theory in explaining the behavior of subatomic particles. It differs from other theories of the strong force, such as the Yukawa theory, in that it takes into account the principles of quantum mechanics and special relativity.

4. Can QCD be unified with other fundamental forces?

Currently, QCD cannot be unified with the other fundamental forces (electromagnetic, weak, and gravitational) into a single theory. However, it is a critical component of the Grand Unified Theory (GUT), which seeks to unify all the forces of nature into one unified theory.

5. Are there any limitations or challenges in QCD?

QCD has been successful in explaining the strong force, but there are still some limitations and challenges in the theory. One of the major challenges is the mathematical complexity of the theory, making it difficult to make precise calculations. Additionally, QCD does not address the phenomenon of gravity, and therefore, cannot be considered a complete theory of all the fundamental forces.

Similar threads

  • High Energy, Nuclear, Particle Physics
Replies
6
Views
1K
  • High Energy, Nuclear, Particle Physics
Replies
6
Views
2K
  • High Energy, Nuclear, Particle Physics
2
Replies
38
Views
3K
  • High Energy, Nuclear, Particle Physics
Replies
1
Views
1K
  • High Energy, Nuclear, Particle Physics
Replies
1
Views
879
  • High Energy, Nuclear, Particle Physics
Replies
2
Views
1K
  • High Energy, Nuclear, Particle Physics
Replies
27
Views
4K
  • High Energy, Nuclear, Particle Physics
Replies
4
Views
2K
  • High Energy, Nuclear, Particle Physics
Replies
8
Views
2K
  • High Energy, Nuclear, Particle Physics
Replies
2
Views
2K
Back
Top