Hybrid Charmonium and the ρ − π Puzzle

Abstract

Using the method of QCD sum rules, we estimate the energy of the lowest hybrid charmonium state, and find it to be at the energy of theΨ′(2S) state, about 600 MeV above theJ/Ψ(1S)state. Since our solution is not consistent with a pure hybrid at this energy, we conclude that theΨ′(2S) state is probably an admixedcc¯and hybridcc¯gstate. From this conjecture, we find a possible explanation of the famousρ−πpuzzle.

Determining 1−−heavy hybrid masses via QCD sum rules

“…In former studies [16,18], people also fought the same target but with some differences from this work. In previous works, operator expansion only up to dimension four, but in our study the dimension six operators are taken into account.…”

Sigma and pion decays of mixed heavy hybrids

“…The vector component provides a separate sum rule, but the sum rule with the scalar component has less error (as determined from the method of QCD sum rules in previous calculations-see, e.g., Refs [5,4]). From Eq(7) we obtain…”

“…For the Υ(nS) states this provides an explanation for the unusual two-pion decays [6]; and for the Ψ ′ (2S) this solves the famous ρ − π problem. Prior to this, we had studied the Ψ ′ (2S), also using the method of QCD sum rules, and found [5] that it was not a pure hybrid. In the present work we estimate the decay of the Ψ ′ (2S) to the J/Ψ(1S) with a sigma or two free pions as a test of our mixed heavy quark hybrid meson theory.…”

Ψ(2S) Decay to J/Ψ(1S) + 2 π or J/Ψ(1S) + σ + 2 π

“…This solves many puzzles [4,5]. In Ref [3] it was found that the ratios of cross sections, with the σ a broad 2-pion resonance,…”