# Would the existence of more than 16 quark flavors make QCD deconfinning?

Consider the QCD beta function. Its expansion in powers of the coupling is

$$\beta(\mu)=-(\beta_0a(\mu)+\beta_1a^2(\mu)+\ldots)$$

where $a=\alpha/4\pi$. For simplicity let's neglect everything but the one loop term $\beta_0$. This term is given by

$$\beta_0=11-\frac{2}{3}N_f$$

where $N_f$ is the number of fermion flavors. Notice that if the number of quark flavors is higher than $16$ the beta coefficient changes sign.

While performing the running of the coupling constant, when some energy thresholds are crossed we must increase the number of active fermions up to a total of 6. Now, for the fun of it, assume that more quark flavors are lurking in even higher energies. If this were to be true, and we reached a number higher than $16$, the sign of $\beta_0$ would change and the coupling would become large at high energies deconfining QCD. Is this picture right or I am I assuming something I shouldn't?

In fact, a strong coupling regime in the IR sector, which is provided by the RG flow for $N_{f} < 16$, is a necessary condition for confinement: without strong coupling, you won't get out from the perturbative regime, so you always may treat quarks as free in/out-states in most cases. Bound states, of course, may exist, but their existence is possible only due to changes in the effective constant of the expansion. Think of the electron-proton system which forms a hydrogen atom in QED.