1. 07/30/2007Lattice 2007 Sigma meson contribution in the  I=1/2 rule Takumi Doi (Univ. of Kentucky) In collaboration with T.Draper (Univ. of Kentucky) K.-F.Liu (Univ. of Kentucky) N.Mathur (JLAB) J.-B. Zhang (Zhejiang Univ.)

2. 07/30/2007Lattice 2007 Outline The long-standing puzzle in the Scalar mesons  a 0 (980), a 0 (1450), etc. The lattice QCD study for the scalar mesons Study of the spectrum Standard q-qbar meson or tetraquark ? The volume dependence study to extract the nature of the signal   as a tetraquark mesonium The role of  meson in K  2  decay

3. 0 ¯ ¯ (1) 1 ¯ + (1) 0 ++ (0)0 + ¯ (1) 1 + ¯ (1) π (137) 0 + (1/2) ρ (770) σ (600) f 0 (980) f 0 (1370) f 0 (1500) a 0 (980) a 0 (1450) a 1 (1230) K 0 * (1430) J PG (I) M (MeV) a 2 (1320) 2 + ¯ (1) f 0 (1710) K 0 * (800)

4. 07/30/2007Lattice 2007 Puzzle in scalar mesons Too many states compared to the other channels Strange meson is lighter than non-stranged mesons ?? Some have very broad width, others have narrow width   (600),  (800), f 0 (980), a 0 (980) may be tetraquark states R.L.Jaffe (1977) Diquark model, molecular model etc.

5. 07/30/2007Lattice 2007 The controversial state: σ meson σ (500): Johnson and Teller Two-pion exchange potential: Chembto, Durso, Riska; Stony Brook, Paris, … N-N potential

6. 07/30/2007Lattice 2007 M. Ablikim et al. (BES), Phys. Lett. B598, 149 (2004) M σ = 541 ± 39 MeV, Γ σ = 504 ± 84 MeV J/ψ —> ωπ + π - 

7. 07/30/2007Lattice 2007 Simulation Parameters Overlap fermion with quenched approx. Iwasaki gauge action a = 0.200(3)fm (a -1 = 1GeV) (beta=2.264) V=16 3 X28  L 3 = (3.2fm) 3 m  >= 180 MeV, m  L > 3 About 200-300 configurations Exact chiral symmetry ! N.Mathur et al.(  QCD collab.), hep-lat/0607110

8. 07/30/2007Lattice 2007 Our results shows scalar mass around 1400-1500 MeV, suggesting a 0 (1450) is a two quark state. msmsmsms Eliminate  ’  ghost state for a0 study N.Mathur et al.(  QCD collab.), hep-lat/0607110

9. 07/30/2007Lattice 2007 ππ four quark operator (I=0) ππ four quark operator (I=0) Disconnected diagrams are found to be small  neglected at this moment

10. Further study is needed to check the volume dependence of the observed states. Scattering states Scattering states (Negative scattering length) length) Scattering states Scattering states Possible BOUND state σ(600)? σ(600)?

11. 07/30/2007Lattice 2007 Volume dependence to distinguish one/two particle state Spectral weight on the lattice Normalization for the particle If the signal is one particle state: If the signal is two particle state: Physical meaning: O(1/V) corresponds to the encounter possibility of two particles

12. 07/30/2007Lattice 2007 Volume dependence of spectral weights Volume independence suggests the observed state is an one particle state W0W0W0W0 W1W1W1W1 3D-Volume 12 3 vs. 16 3 N.Mathur et al.(  QCD collab.), hep-lat/0607110

13. 07/30/2007Lattice 2007 The role of Sigma meson in Physics K  2  decay  I = 1/2 rule Sigma can enhance the I=0 channel only T.N.Pham, Phys.Rev.D33 (1986) 1499 T.Morozumi, C.S.Lim, A.I.Sanda, PRL65 (1990) 404 Usual lattice formulation K  , K  vac. + chPT No  consideration ! (However, T.Yamazaki at lat06, direct K  2  for I=2)

14. 07/30/2007Lattice 2007 Strategy 3pt correlator G(t) Almost no energy-momentum mismatch (m(K)=m(  )=500MeV) Hw K0K0   For the contribution to physical K  2 , one needs additional   2 , for which one may use experimental information Coulomb wall t

15. 07/30/2007Lattice 2007 Results The correlator is found to be very noisy Further work is necessary to extract the quantitative signal O2O2 O6O6 K Hw Wall src  Wall src K 

16. 07/30/2007Lattice 2007 Summary/Outlook We have investigated the scalar mesons with overlap fermion at the quenched level with m  > 180MeV a 0 (1450) is two quark state  ) is likely to be a tetraquark state 3D-Volume dependence study is essential Dynamical quark effect, disconnected diagram study is desirable for future study The existence of  meson may have substantial impact: The enhancement of I=0 for K  2  decay, which may solve the  I=1/2 rule Need further calculation to extract the quantitative results