content_image/332547.jpg content_image/332546.jpg content_image/332545.jpg
and we were able to achieve the desired massless quark eigenenergy of 0.306 GeV with the following values: 

<img src='content_image/332547.jpg'>

The Long Marches did achieve the factor of eight reduction in the ground state eigenenergy. The wave functions for this massless quark solution are shown in Fig. 20. Note that ψ$_{a}$ ( x ) here begins to resemble the hydrogen-atom ground state ψ$_{a}$ ( x ) depicted by the dashed curve in Fig. 9, and it is much narrower than that shown in Fig. 3 . 

The ground state solution says nothing about the PGG claim that, for V$_{s}$ ( x ) and V$_{v}$ ( x ) having equal slopes, the spin-orbit interaction vanishes. To check this we need to find, at a minimum, the eigenenergies of the 2 p 1 2 and 2 p 3 2 excited states. For the same values of x$_{s}$ and α$_{s}$ as in Eq. (55) and after some searching for good starting values for the iterative procedure, we obtained the three n = 2 wave functions displayed in Figs. 21, 22, and 23. The wave functions here are narrower and more hydrogen-like than in the V$_{s}$ -only case (compare with Figs. 4, 5, and 6). The eigenenergies, in GeV, for these states are 

<img src='content_image/332546.jpg'>

The 2 p 3 2 state is nearly degenerate with the 2 s state, some 200 MeV higher than in Sec. II F, but this degeneracy is probably accidental. The real surprise here is the huge spin-orbit splitting between the 2 p 1 2 and 2 p 3 2 states, not what we expected from the claim made by PGG.[25] 

This difference, it turns out, is due to the presence of the Coulomb attraction term in V$_{v}$ ( x ). If we set α$_{s}$ = 0, then (again for x$_{s}$ = 0 . 2) we find the following n = 2 eigenenergies (in GeV): 

<img src='content_image/332545.jpg'>

The wave functions in this case are similar to those in Figs. 21, 22, and 23 but are somewhat broader and less peaked near the origin. These results show the expected PGG symmetry (zero spin-orbit splitting). Note, incidentally, that the 2 s state is higher in energy than the 2 p states, as in the experimental charmonium spectrum.[26] One does conclude, however, that having a significant Coulomb contribution in V$_{v}$ ( x ) will destroy the PGG symmetry. 

