Is the a 0(1450) a candidate for the lowest q q1³ P₀ state?

2003/02/28 by A. M. Badalian · 1 citation
Physics and Astronomy · #Contradiction #Coupling (piping) #Hadron #Nuclear physics research studies #Particle physics theoretical and experimental studies #Quantum Chromodynamics and Particle Interactions #Tensor (intrinsic definition) #Value (mathematics) #hep-ph

paper · pdf · doi:10.1134/1.1592589

published as Phys.Atom.Nucl.66:1342-1348,2003; Yad.Fiz.66:1382-1388,2003 · 13 pages

arxiv created 2003/06/30 · openalex publication_date 2003/07/01 · arxiv updated 2014/11/17 · openalex created_date 2016/06/24 · openalex updated_date 2026/08/05

Abstract

For a 0(1450), considered as the q q1³ P₀ state, “experimental” tensor splitting, c exp = −150±40 MeV, appears to be in contradiction with the conventional theory of fine structure. There is no such discrepancy if a 0(980) belongs to the q q1³ P_J multiplet. The hadronic shift of a 0(980) is shown to be greatly dependent on the value of strong coupling in spin-dependent interaction.

Citations

• Perturbative–nonperturbative interference in the static QCD interaction at small distances
• Are there short distance non-perturbative contributions to the QCD static potential?
• Nonperturbative hyperfine contribution to theb1andh1meson masses
• Determination ofαs(1 GeV)from the charmonium fine structure

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