pdn, ωn, K+-, K0 reactions at PANDA facility

Yu. Rogov, First seminar of FRRC Fellows

FAIR – Russia Research Center,

Moscow June, 9 - 10, 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 ω0 and ω8: = cos 8 - sin 0

= sin 8 + cos 0

*Okubo S. // Phys. Rev. 1977. V.16. P.2336

0)()(

)(2

ddCBAMuuCBAM

ssCBAMZ

)(3

10 ssdduu

)2(6

18 ssdduu

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

)cos()sin()(2

1 ii ssdduu

)tan()(

)(iCBAM

CBAM

OZI rule

• Okubo: production of ss states in the non-strange hadrons interactions is forbidden

• - production in pp (pp) interaction is possible either due to admixture of light quarks in the wave function or due to strange quarks admixture in the nucleon.

0)()(

)(2

ddCBAMuuCBAM

ssCBAMZ

32 102.4)(tan)(

)(

iCBA

CBA

)cos()sin()(2

1 ii ssdduu

= -i =3.70

OZI rule predicitions

32 104)(tan)(

)(

iCBA

CBA

• Universal, does not depend on energy or other properties of the initial state.

• Depends on the masses of the meson in the nonet

3

2

2 1016))1270((

))1525((

fCBA

fCBA

25.1)(

)(

CBA

CBA

Comparison with experiment

R(/)=tg2(-I)=4.2 10-3

Weighted average of all experimental data

N

R(/)=(3.300.34)10-3

• NN

R(/)=(12.780.34)10-

3

NN

R(/)=(14.551.92)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 targetp + p → + p + p → +

• R() = (29497) 10-3

• R()OZI = 4.2 10-3

• L=0, S=0 1S0 spin singlet

• OBELIX, 1995, LQ, NTP, 5 mbp + p → + p + p → +

• R() = (11410) 10-3

• R()OZI = 4.2 10-3

• L=0, S=1 3S1

• spin triplet

• Different situation for annihilation from S- and P-waves

• R() = (12012) 10-

3 3S1

• R() < 7.2 10-3 1 P1

• spin triplet – enhanced• spin singlet – suppressed

f2’(1525)/f2(1270)

• Tensor mesons:

L=1, S=1, J=2

f2(1270) normal qq

f2’ (1525) ss

• R(f2’f2) = (4714) 10-3 3S1

• R(f2’f2) = (14920) 10-3 1

P1

• spin triplet – suppressed• spin singlet – enhanced

• 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

nn

• OBELIX, Crystal Barrelp + d → + np + d → + nB.Pontecorvo, 1956One-meson annihilation• R() = (15629)

10-3

• R()OZI = 4.2 10-3

/

p

p

s

s

s

s

pp

• If it were a normal quark reaction

()exp ~ 4 b• why is it so large?() was not

measured

nbtg i 10)()()( 4

At LEAR experiemts

Strong violation of the OZI rule was found in

pp pp, pp (3S1)

pdn

Does it depend on – spin– orbital angular

momentum– momentum transfer– isospin?

Pontecorvo reactions

• Largest momentum transfer:q2=-0.782 GeV2/c2 inpdn, compared toq2=-0.360 GeV2/c2 inpp0

• Interesting physics*:two-step modelpp0, nnR() = (15629) 10-3

Rth() = (9227) 10-3

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+K0, 0+K0

R=Y(K)/Y(K) = 0.920.15two-step model R=0.012 because σ(KNX) > 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