In pure QCD, we have glueballs. Pure QCD can also be thought of as QCD where the fermion masses have been sent to infinity. If we vary the fermion masses (say, for simplicity, we deal with a single flavor of fermion), then the masses of the glueball states should depend on the fermion mass, $M_\text{glue}(m_q)$. Is there any way to theoretically predict (even very crudely) what this variation would be? The basic question I have in mind is whether the masses of the glueballs are expected to increase or decrease, when you introduce fermions into the sea.

I have read through a lattice study of glueball masses, but they don't seem to include any theoretical arguments for quark-mass dependence. I would be interested in any rough heuristic available.

[I realise also that there may be some amiguity in what I mean by varying quark masses - in particular lets say $\Lambda_\text{QCD}(\overline{MS})$ is held fixed(?)]

  • $\begingroup$ Indeed, they expressly refuse to discuss dependence on fermion masses, and expressly find no such dependence on the lowest pseudo scalar mass! The reason is the situation is still confused: you'd expect the strong coupling to be weakened by the presence of fermions, but the $\eta'$ mass, on Witten-Veneziano grounds, increases with the number of fermions! $\endgroup$ Commented Feb 4, 2022 at 14:33
  • $\begingroup$ @CosmasZachos Thanks for the comment, could I ask why we expect strong coupling to be weakened in presence of fermions? $\endgroup$ Commented Feb 4, 2022 at 19:51
  • $\begingroup$ Oh, screening…. $\endgroup$ Commented Feb 4, 2022 at 19:53
  • $\begingroup$ @CosmasZachos Maybe I am slightly confused - if we agree to hold $\Lambda_\text{QCD}$ constant as we vary the quark masses, doesn't this account for the screening effect? If we don't hold $\Lambda_\text{QCD}$ constant, I'm not sure how to rigorously ask about "varying the quark masses" $\endgroup$ Commented Feb 4, 2022 at 19:56
  • $\begingroup$ Asking for rigor in a soft discussion on the contrast between Λ and glueball masses? $\endgroup$ Commented Feb 4, 2022 at 20:09

1 Answer 1


Indeed, the main issue is that you have to decide what to keep fixed as $m_q$ is varied. On the lattice it is not enough to say that $\Lambda$ is fixed, because all parameter are dimensionless (expressed in units of the lattice spacing $a$). In the real world you fix $\Lambda$ from something like $\alpha_s(m_Z)$, but this relies on the fact that we already know what we mean by $m_Z$ in units $MeV$.

In calculations that include light quarks you can fix $a$ from the proton mass (so the proton mass is not a prediction, but you can then check $\alpha_s(m_Z)$). In the pure glue theory the scale setting procedure typically involves the potential between two infinitely heavy test charges (tuned to reproduce the heavy quarkonium spectrum). The most popular choice involves the "Sommer parameter": The distance at which the dimensionless force $r_0^2F(r_0)=1.65$ is defined to be $r_0=0.5$ fm.

You can then go ahead and study $m_q$ dependence. As an example, see this paper, which claims that the scalar glueball mass increases as $m_q$ is reduced towards the physical point.

Two additional comments:

  1. In the real world (light quarks), there is an additional problem, because there are many $0^{++}$ states, and you cannot rigorously identify which one is "the glueball", and which are mostly $\bar{q}q$, with various admixtures of glue and other stuff.

  2. You can study what happens in pure glue when you add very heavy flavors, and slowly reduce their mass. I would guess that initially, this is just a QM two-level problem. There is mixing, which leads to level repulsion, which means that the glueball mass drops. I'm not certain if this argument can be made rigorous.


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