There is no $κ(900)$

Details There is no $κ(900)$ There is no $κ(900)$

2000-05-21

arxiv.org/abs/hep-ph/0005208v2

Nucl.Phys.A688:823-841,2001

Physics

High Energy Physics

High Energy Physics - Phenomenology

19 pages, 5 figures Version to be published in Nuclear Physics A containing new figures and additional entries in tables to cl

Scientific paper

10.1016/S0375-9474(00)00587-X

In the I=0 sector there are more scalar mesons than can fit in one $q{\bar q}$ nonet. Consequently, many have claimed that there is in fact more than one multiplet, perhaps both $q{\bar q}$ and $qq{\bar {qq}}$. Such proposals require the existence of at least two strange isodoublets (and their antiparticles). The current PDG Tables list just one state, the $K^*_0(1430)$, while fits to data with Breit-Wigner forms and variable backgrounds can accommodate a $\kappa(900)$ too. Whether a state exists in the spectrum of hadrons is not a matter of ability to fit data along the real energy axis, but is completely specified by the number of poles in the complex energy plane. Here we perform a model-independent analytic continuation of $\pi K$ scattering results between 825 MeV and 2 GeV to determine the number and position of resonance poles. We find that there {\bf is} a $K^*_0(1430)$, but {\bf no} $\kappa(900)$. The LASS data cannot rule out the possibility of a very low mass $\kappa$ well below 825 MeV.

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