The Large-$N$ Limit with Vanishing Leading Order Condensate for Zero Pion Mass

Abstract

It is conventionally assumed that the negative mass squared term in the linear sigma model version of the pion Lagrangian is $M^2 \sim \Lambda_{\rm QCD}^2$ in powers of $N_c$. We consider the case where $M^2 \sim \Lambda^2_{\rm QCD}/N_c$ so that to leading order in $N_c$ this symmetry breaking term vanishes. We present some arguments why this might be plausible. One might think that such a radical assumption would contradict lattice Monte Carlo data on QCD as function of $N_c$. We show that the linear sigma model gives a fair description of the data of DeGrand and Liu both for $N_c = 3$, and for variable $N_c$. The values of quark masses considered by DeGrand and Liu, and by Bali et. al. turn out to be too large to resolve the case we consider from that of the conventional large $N_c$ limit. We argue that for quark masses $m_{q} \ll \Lambda_{\rm QCD}/N_c^{3/2}$, both the baryon mass and nucleon size scale as $\sqrt{N_c}$. For $m_{q} \gg \Lambda_{\rm QCD}/N_c^{3/2}$ the conventional large-$N_c$ counting holds. The physical values of quark masses for QCD correspond to the small quark mass limit. We find pion-nucleon coupling strengths are reduced to order ${\cal O}(1)$ rather than ${\cal O}(N_c)$. Under the assumption that in the large $N_c$ limit the sigma meson mass is larger than that of the omega, and that the omega-nucleon coupling constant is larger than that of the sigma, we argue that the nucleon-nucleon large range potential is weakly attractive and admits an interaction energy of order $\Lambda_{\rm QCD}/N_c^{5/2} \sim 10$ MeV. With these assumptions on coupling and masses, there is no strong long range attractive channel for nucleon-nucleon interactions, so that nuclear matter at densities much smaller than that where nucleons strongly interact is a weakly interacting configuration of nucleons with strongly interacting localized cores.