With enough fermions, does QCD confinement disappear?

Question

It's been a while since I thought about QFT, but the following thing was puzzling me. I've often read that theories with a beta function that is asymptotically free at lowest order will likely suffer non-perturbative effects at low energies.

For example, Srednicki's text (page 434) notes:

[T]he gauge coupling in quantum chromodynamics gets weaker at high energies, and stronger at low energies. This has dramatic physical consequences. Perturbation theory cannot serve as a reliable guide to the low-energy physics. And indeed, in nature we do not see isolated quarks or gluons. (Quarks, in particular, have fractional electric charges and would be easy to discover.) The appropriate conclusion is that color is confined: all finite-energy states are invariant under a global $SU(3)$ transformation. This has not yet been rigorously proven, but it is the only hypothesis that is consistent with all of the available theoretical and experimental information.

However, the beta function for QCD depends on the number of fermions $n_F$ like $-(C_1 - C_2 n_F) \alpha^2$ at lowest order. From that same page, it looks like for $17$ or more fermion species (quark flavors), the beta function's leading coefficient is positive and hence the beta function is not asymptotically free at lowest order.

This motivates my question: For $17$ or more fermion species, is the corresponding theory no longer confined?

Answer 1

I believe this is heuristically the right idea: as you mention in the comments though there are some technicalities with whether asymptotic freedom and confinement are 1-to-1, and there are also some issues when you start looking at the two-loop beta function that lead to some surprises. A good paper to look at is this lattice study (Appelquist, Fleming, Neil), which investigates numerically the behaviour. I'll summarise what I believe to be the main points below:

The beta function of massless QCD has an expansion: $$\beta(g) = -\frac{1}{16 \pi^2} \left(\frac{11}{3}N_c -\frac{2}{3} N_f\right) g^3 - \frac{1}{(16\pi^2)^2} \left(\frac{34}{3} N_c^2 - \frac{1}{2} N_f \left(2 \frac{N_c^2-1}{N_c} + \frac{20}{3}N_c\right)\right)g^5 + \cdots$$ If we look at just the $g^3$ one-loop term, indeed there is a cutoff where $\beta(g) > 0$ for $N_f > 16$ and $\beta(g) < 0$ for $N_f \leq 16$. Including the two-loop $g^5$ term however introduces a zero to the beta function: So, when you go to the infrared instead of the coupling getting larger and larger, you flow instead to the conformal fixed point. We can imagine though that the exact values of $N_f$ for which this fixed point exists depends on higher loop corrections too - that's why the lattice study was done to see for which $N_f$ this conformal fixed point exists (this is the `conformal window'). In the lattice study, they end up finding that $12 \leq N_f \leq 16$ has the fixed point, and $N_f \leq 8$ has confinement.

As a footnote - the kind of amusing thing about this kind of lattice study is that (I think) you get large $N_f$ "for free". Normally, when you put fermions on the lattice, you get a problem with fermion doublers, where you have unwanted additional fermion modes that are unphysical and you need to remove via some procedure. These kinds of studies with large $N_f$ can to some extent utilise these fermion doublers to their advantage.