What are the reasons that $SU(3)$ is used for QCD?
Why wouldn't the simpler & smaller group $SO(4)$ make a better candidate?
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1$\begingroup$ You might be interested to read the answer given by Joshphysics here, which I think gives the reason for using $\mathrm{SU(3)}$. $\endgroup$– HunterCommented Jun 4, 2014 at 15:48
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1$\begingroup$ I cannot see a reason there. Unless you are implying that SU(3) is a universal cover of SO(4), which it is not... $\endgroup$– NoEscapeCommented Jun 4, 2014 at 16:06
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1$\begingroup$ No, I am implying that we should use a unitary group, rather than an orthogonal group. So then at least you would have to use $\mathrm{SU(2)} \times \mathrm{SU(2)}$ if I remember correctly. $\endgroup$– HunterCommented Jun 4, 2014 at 16:25
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1$\begingroup$ Oh ok, can you give a reference, because that would be interesting for me to read about. I was indeed under the impression that we should use unitarity to preserve the inner product on the Hilbert space. $\endgroup$– HunterCommented Jun 4, 2014 at 16:33
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1$\begingroup$ @NoEscape : Please make your question more detailed and specific if so that we know exactly what you are asking, and exactly what you are looking for in an answer. $\endgroup$– Flint72Commented Jun 4, 2014 at 16:57
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2 Answers
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It is basically an experimental result that QCD is based on $SU(3)$. For example, we know that there are 8 gluons, no more no less. Otherwise several cross sections sensible to the color factors would be off (perhaps, the most recent example is provided by the Higgs boson production cross-section via gluon fusion measured by ATLAS and CMS at CERN). A theory based on $SO(4)$ has instead 6 gauge bosons. In fact, the theory would be based on $SU(2)\times SU(2)\sim SO(4)$ and could in principle admit two gauge couplings rather than one.
And notice that these are just two examples of several others reasons why $SU(3)$ is the correct group (at least at energies $E<1$~TeV).
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2$\begingroup$ you don't agree that the number of gluons is $8$? Doesn't just that rule out your $SO(4)$? Another trivial example could be the branching ratio of the $Z$ boson to b-quark pairs that is sensitive to the number of color. $\endgroup$– TwoBsCommented Jun 4, 2014 at 16:40
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1$\begingroup$ @NoEscape : He just gave you two beyond doubt understandable examples. The strong interactions are flavour blind, and so they must be described by a compact simple unitary group. $\endgroup$– Flint72Commented Jun 4, 2014 at 16:41
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3$\begingroup$ @NoEscape : You have given no indication of your level of education, so given that you claim above that the necessarity of a unitary group is a mistake, it appears that we may assume you are at least a senior undergrad. However you then say that this answer is too vague, when anyone understanding what they are claiming, in saying that unitary groups are not necessiary, would understand the argument. Moreover, you ask for examples which have no reasonable doubt, and the recent measurement of the Higgs-like scalar boson is one such measurement. $\endgroup$– Flint72Commented Jun 4, 2014 at 16:56
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1$\begingroup$ @Flint72 I agree 100% with you in that it would be useful for us to understand the background of the OP. Just for clarification, is NoEscape correct in saying the unitarity is not a requirement. And if he is right, do you know any books/references that discuss this in more details? $\endgroup$– HunterCommented Jun 4, 2014 at 17:18
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3$\begingroup$ @NoEscape : Indeed having a PhD or lack thereof does not matter, which is one of the great things about maths and physics, namely, just because one is older it does not mean that one is correct, rather, if the maths is corret, then the maths is correct, and nobody can dispute it. Our request for clarification of your current educational standing is to make it easier for us to taloir an answer to your needs. However at this point I am more than willing to give up, since you are unwilling to help us to help you. It has been experimentally varified that there are eight gluons... $\endgroup$– Flint72Commented Jun 4, 2014 at 20:16
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I guess the simple answer is this: $SU(3)$ ist the SMALLEST group to allow for nonlinear effects (i.e. Confinement).
Side note: the fact that a model using $SU(3)$ is confirmed by experiment does not rule out other models based on other groups. But $SO(4)$ is already ruled out theoretically, because no confinement is possible.
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2$\begingroup$ Why you don't consider $SU(2)$ as having non-linear effects? It certainly contains vertices with 3 and 4 gauge bosons. In any case, why being the smallest or the minimal group in some twisted sense should be a physical requirement? For example, the matter content of the SM is all but minimal. Moreover, the fact that $SU(3)$ gives rise to confinement depends on the representations of the matter fields. If you have too many fermions in the fundamental there is no confinement even though the group is $SU(3)$. And the requirement of confinement is not theoretical, but experimental. $\endgroup$– TwoBsCommented Jun 5, 2014 at 20:17
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$\begingroup$ read 'nonlinear' as 'non-perturbative' and relax. $\endgroup$– NoEscapeCommented Jun 23, 2014 at 10:28
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$\begingroup$ $SU(2)$ may also have non-perturbative effects, such as instantons (see e.g. sphalerons en.wikipedia.org/wiki/Sphaleron). In any case, it is not enough to give the group, you need to specify the representations to claim something about e.g. confinement or asymptotic freedom. And I insist that the requirement of confinement is not theoretical, but experimental. There is nothing wrong with a non-confining gauge theory. And the request of minimality is not a physical principle, it has been violated in nature various times. $\endgroup$– TwoBsCommented Jun 23, 2014 at 12:05
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$\begingroup$ We are talking different languages... $\endgroup$– NoEscapeCommented Jun 23, 2014 at 14:46