As I understand there are eight types of gluons (linear combinations of color/anticolor pairs with varying amplitudes) which can combine (for very short periods) to form glueballs. If there were no restrictions on which types of gluons could be combined, there would be 36 types of glueballs with 2 gluons, 120 types with 3 gluons, 330 types with 4 gluons, etc. But glueballs must be color neutral, so not all of these types are possible.

How many glueballs are possible with a given number of constituent gluons? I think this is really two questions, since it's possible that a glueball A+B and a glueball C+D would be indistinguishable despite {A, B} not being equal to {C, D}, and so there could be distinguishable vs. indistinguishable counts.

If it helps I have a graduate understanding of algebra (though mostly field theory, not group theory) albeit essentially no knowledge of physics. Since there is an explicit representation for SU(3) it may suffice to explain how to determine whether a given combination of matrices is color neutral. Maybe there are better ways.

A related question, Permissible combinations of colour states for gluons, asks where the 8 comes from; mine is essentially asking the next step after that.

  • $\begingroup$ I'm not sure, but gauge invariance might play a role. That is, we know there are 8 gluons, but their colors are totally physically equivalent. Then a red + anti-red glueball and a blue + anti-blue glueball are actually the same particle. $\endgroup$
    – knzhou
    Commented May 12, 2016 at 19:04
  • $\begingroup$ Also, since gluons are massless, I'm not sure how much sense it makes to consider glueballs with a small, finite number of gluons. Generically the number of gluons will be infinite. $\endgroup$
    – knzhou
    Commented May 12, 2016 at 19:05
  • $\begingroup$ @knzhou: Yes, that matches my understanding. If the first step is seeing which combinations of gluons produce a color-neutral (= possible) glueball, then the second would be working out how many distinct particles are formed this way. $\endgroup$
    – Charles
    Commented May 12, 2016 at 19:06
  • $\begingroup$ @knzhou: Not to put too much credence in it, but Wikipedia has "Theoretical studies of glueballs have focused on glueballs consisting of either two gluons or three gluons, by analogy to mesons and baryons that have two and three quarks respectively." which makes me think that small numbers of gluons make sense. If huge numbers are typical then I guess the combinatorial properties yield asymptotics which in turn determine something like thermodynamical entropy/temperature for large collections. $\endgroup$
    – Charles
    Commented May 12, 2016 at 19:09

1 Answer 1


1) Note that the non-relativistic gluon model of glueballs has even less justification than the non-relativistic quark model of baryons and mesons. This is because the model messes up the spin assignments: Massless gluons have only two spin states, but non-relativistic gluons have three.

2) The color quantum numbers are trivial: The product $$ [8]\times [8] = [1]+[8]+[8]+[10]+[\bar{10}]+[27] $$ only contains one color singlet.

3) In terms of spin: To get this right you have to start from a relativistic model (like the MIT bag). Then there are three states that correspond to the leading Fock state. A scalar, a pseudoscalar, and a symmetric tensor (spin 2). In terms of operators, these states correspond to $$ G^2, \;\;\;\; G\tilde{G}, \;\;\;\; T_{\mu\nu}, $$ where $T_{\mu\nu}$ is the Yang Mills stress tensor. At higher order you can get vector glueballs and glueballs with spin greater than 2.

4) From lattice calculations we know that the states described in 3) are indeed the lowest states. The ground state is a scalar, the first excited state is spin 2, and the next state is a pseudoscalar.


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