Self-consistent low-energy meson mass spectrum

In a typical hadron mass calculation the long-range (confining) part of the interaction between quarks ($q$) is assumed to be spin-independent. Any spin dependence is then attributed to short-range one-gluon exchange. This procedure tends to give an excessively small mass difference between pseudoscalar ($P$) and vector ($V$) mesons, at least if the quark-gluon coupling is fixed by other properties of the hadron spectrum. In the present paper we introduce an approach in which a large $P-V$ mass difference arises naturally. A confining interaction does not have to be assumed a priori. The spectrum is generated by imposing duality on an infinite sum of ladder graphs without crossed quark lines; this "planar bootstrap" corresponds to the limit $N_{\mathrm{flavor}}\to \infty$ (in quantum chromodynamics we would have to take $N_{\mathrm{color}}\to \infty$ at the same time). By making a certain simple dynamical approximation we then derive an explicit infinitely rising exchange-degenerate leading Regge trajectory $\alpha \mathrm{}\left(t\right)=S_1+S_2+\mathcal{L}\left({\nu }_a\right)$ for any given equal-mass-channel hadron + hadron→hadron + hadron process; $S_1$ and $S_2$ are external spins, $\mathcal{L}\left({\nu }_a\right)\simeq -0.5+2\mathrm{}{\alpha }^{\prime }{\nu }_a$, and ${\nu }_a=s_a+\frac{(t-\Sigma {m_i}^2)}{2}$, where the $m_i$ are the external masses and ${\sqrt{s}}_a$ is the mass of an exchanged cluster $a$. By requiring the $s_a$ to be as low as possible and imposing simultaneous consistency for complete sets of meson + meson→meson + meson processes, we are able to calculate the entire natural- and unnatural-parity low-energy leading-trajectory $q\overline{q}$ mass spectrum in terms of $m_{\rho }$ and $m_{K^{*}}$ alone. We obtain ${m_{\pi }}^2\ll {m_{\rho }}^2$, a universal Regge slope ${\alpha }^{\prime }=\frac{{({m_{\rho }}^2-{m_{\pi }}^2)}^{-1\mathrm{}}}{2}$, and the usual mass formulas ${m_{\phi }}^2-{m_{K^{*}}}^2={m_{K^{*}}}^2-{m_{\rho }}^2={m_K}^2-{m_{\pi }}^2={m_{{\eta }_s}}^2-{m_K}^2$, $m_{\omega }=m_{\rho }$. In the case of ${\eta }_u=\frac{(u\overline{u}+d\overline{d})}{\sqrt{2}}$, however, we obtain ${m_{{\eta }_u}}^2=\frac{1}{3}({m_{\rho }}^2+2\mathrm{}{m_{\pi }}^2)$, which gives $m_{{\eta }_u}=0.462\mathrm{}$ GeV.