pd n, ωn, K + -, K 0 reactions at PANDA facility Yu. Rogov, First seminar of FRRC Fellows FAIR – Russia Research Center, Moscow June, , 2009
OZI rule * NN→φπ is forbidden, NN→ωπ is allowed φ production is possible only via mixing, because φ and ω are mixture of ω 0 and ω 8 : = cos 8 - sin 0 = sin 8 + cos 0 *Okubo S. // Phys. Rev V.16. P.2336
OZI rule Introducing ideal mixing angle Θ i, cosΘ i =(2/3) 1/2, sinΘ i =(1/3) 1/2, Θ i =35.5° Θ(m φ,m ω,m ω8 ), m ω8 =m(K *,ρ), Θ=39° If Z=0, then R(φ/ω)=tan 2 (Θ-Θ i )f = 4.2·10 -3
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 =3.7 0
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
Comparison with experiment R( / )=tg 2 ( - I )= 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
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
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.
Crystal Barrel, 1995, LQ hydrogen target p + p → + p + p → + R( ) = (294 97) R( ) OZI = 4.2 L=0, S=0 1 S 0 spin singlet
OBELIX, 1995, LQ, NTP, 5 mb p + p → + p + p → + R( ) = (114 10) R( ) OZI = 4.2 L=0, S=1 3 S 1 spin triplet
Different situation for annihilation from S- and P-waves R( ) = (120 12) 3 S 1 R( ) < 7.2 1 P 1 spin triplet – enhanced spin singlet – suppressed
f 2 ’(1525) / f 2 (1270) Tensor mesons: L=1, S=1, J=2 f 2 (1270) normal q q f 2 ’ (1525) s s R(f 2 ’ f 2 ) = (47 14) 3 S 1 R(f 2 ’ f 2 ) = (149 20) 1 P 1 spin triplet – suppressed spin singlet – enhanced
OBELIX, 1995, NTP, 5 mb p + p → + + + - p + p → + + + - For all events: R( ) = (5-6) R( ) OZI = 4.2 For events with M = MeV R( ) = (16-30) 10 -3
n n OBELIX, Crystal Barrel p + d → + n p + d → + n B.Pontecorvo, 1956 One-meson annihilation R( ) = (156 29) R( ) OZI = 4.2 10 -3
/ p pp ss s s ss If it were a normal quark reaction ( ) exp ~ 4 b why is it so large? ( ) was not measured
At LEAR experiemts Strong violation of the OZI rule was found in pp pp , pp ( 3 S 1 ) pd n Does it depend on –spin –orbital angular momentum –momentum transfer –isospin?
pp at PANDA Luminosity of HESR – L =2 cm -2 s -1 Cross section = 4 b at 1.4 GeV/c BR of charged mode =0.25 Registration efficiency = 0.01 N = L = 2 4 0.25 0.01 = 2 s -1 Best world statistics – 1.5 hours pp (K * K *, ) must be measured
Pontecorvo reactions Largest momentum transfer: q 2 = GeV 2 /c 2 in pd n, compared to q 2 = GeV 2 /c 2 in pp 0 Interesting physics * : two-step model pp 0, n n R( ) = (156 29) R th ( ) = (92 27) R( ) decreases with energy *Kondratyuk L.A., Sapozhnikov M.G., Phys.Lett., 1989, B220, 333. *Kondratyuk L.A. et al., Yad.Fiz., 1998, 61, 1670.
Pontecorvo reactions Reactions p + d +K 0, 0 +K 0 R=Y( K)/Y( K) = 0.92 0.15 two-step model R=0.012 because σ(KN X) > σ(KN X) Is two-step model correct?
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