pdn, ωn, K +  - , K 0  reactions at PANDA facility

1. pdn, ωn, K+-, K0  reactions at PANDA facility Yu. Rogov, First seminar of FRRCFellows FAIR – Russia Research Center, Moscow June, 9 - 10, 2009

2. OZI rule* • NN→φπ is forbidden, NN→ωπ is allowed • φ production is possible only via mixing, because φ and ω are mixture of ω0 and ω8: = cos8 - sin0 = sin8 + cos0 *Okubo S. // Phys. Rev. 1977. V.16. P.2336

3. OZI rule • Introducing ideal mixing angle Θi, cosΘi=(2/3)1/2, sinΘi=(1/3)1/2, Θi=35.5° • Θ(mφ,mω,mω8), mω8=m(K*,ρ), Θ=39° • If Z=0, then • R(φ/ω)=tan2(Θ-Θi)f = 4.2·10-3

4. OZI rule • Okubo: production of ss states in the non-strange hadrons interactions is forbidden •  - production in pp (pp) interaction is possible either due to admixture of light quarks in the  wave function or due to strange quarks admixture in the nucleon.  = -i =3.70

5. OZI rule predicitions • Universal, does not depend on energy or other properties of the initial state. • Depends on the masses of the meson in the nonet

6. Comparison with experiment • R(/)=tg2(-I)=4.2 10-3 • Weighted average of all experimental data • N • R(/)=(3.300.34)10-3 • NN • R(/)=(12.780.34)10-3 • NN • R(/)=(14.551.92)10-3

7. The OZI rule is always correct, its violation is only apparent Violation indicates on non-trivial physics: • Strange degrees of freedom in the nucleon • Role of gluon degrees of freedom

8. About hydrogen target • Proton annihilation at rest • Slow antiproton capture on an orbit of ppbar atom with large principal quantum number n~30. • Low pressure gas (~mbar): cascade to lower level, annihilation from P levels with n=2 • Liquid H: Stark mixing between various angular momentum states, annihilation from large n and L=0, i.e. S states.

9.  • Crystal Barrel, 1995, LQ hydrogen target • p + p →  +  • p + p →  +  • R() = (29497)  10-3 • R()OZI = 4.2  10-3 • L=0, S=0  1S0 spin singlet

10.  • OBELIX, 1995, LQ, NTP, 5 mb • p + p →  +  • p + p →  +  • R() = (11410)  10-3 • R()OZI = 4.2  10-3 • L=0, S=1  3S1 • spin triplet

11.  • Different situation for annihilation from S- and P-waves • R() = (12012)  10-3 3S1 • R() < 7.2 10-3 1 P1 • spin triplet – enhanced • spin singlet – suppressed

12. f2’(1525)/f2(1270) • Tensor mesons: • L=1, S=1, J=2 • f2(1270)  normal qq • f2’(1525)  ss • R(f2’f2) = (4714)  10-3 3S1 • R(f2’f2) = (14920)  10-3 1 P1 • spin triplet – suppressed • spin singlet – enhanced

13.  • OBELIX, 1995, NTP, 5 mb • p + p →  + + + - • p + p →  + + + - For all events: • R() = (5-6)  10-3 • R()OZI = 4.2  10-3 For events with M=300-500 MeV • R() = (16-30)  10-3

14. nn • OBELIX, Crystal Barrel • p + d →  + n • p + d →  + n B.Pontecorvo, 1956 One-meson annihilation • R() = (15629)  10-3 • R()OZI = 4.2  10-3

15. / • If it were a normal quark reaction • ()exp~ 4 b • why is it so large? • () was not measured s p s p s s

16. Strong violation of the OZI rule was found in • pp • pp, • pp (3S1) • pdn • Does it depend on • spin • orbital angular momentum • momentum transfer • isospin? At LEAR experiemts

17. Pontecorvo reactions • Largest momentum transfer:q2=-0.782 GeV2/c2 inpdn, compared toq2=-0.360 GeV2/c2 inpp0 • Interesting physics*:two-step modelpp0, nnR() = (15629)  10-3Rth() = (9227)  10-3R() decreases with energy *Kondratyuk L.A., Sapozhnikov M.G., Phys.Lett., 1989, B220, 333. *Kondratyuk L.A. et al., Yad.Fiz., 1998, 61, 1670.

18. Pontecorvo reactions • Reactions p + d+K0, 0+K0R=Y(K)/Y(K) = 0.920.15two-step model R=0.012 because σ(KNX) > σ(KN X) • Is two-step model correct?

19. Pontecorvo reactions at PANDA • Energy region never measured before • High statistics expected • Noone really knows why OZI rule is violated in such selective manner • Assuming that OZI rule is precisely correct, we have a hint to non-trivial physics