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Status of K ±  p ± p 0

Status of K ±  p ± p 0. E. De Lucia. Strategy. Self-tag on one side using K - (nuclear interactions) Vertex with 2 tracks in DC on the signal side. BR(K ±  p ± p 0 ) = (21,13 ± 0.14)% D BR/BR = 6,6x10 -3. Method:.

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Status of K ±  p ± p 0

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  1. Status of K±p±p0 E. De Lucia

  2. Strategy • Self-tag on one side using K- (nuclear interactions) • Vertex with 2 tracks in DC on the signal side BR(K±p±p0) = (21,13 ± 0.14)% DBR/BR = 6,6x10-3 Method: Fitting the distribution of the momentum of the secondary track (p*) in the kaon reference frame we can extract BR(K±p±p0) • The selection efficiency is only related to DC reconstruction: • tracking efficiency • vertex efficiency

  3. mn peak vs p*(p mass) pp0 peak fit with p*(m mass) MC MC p* (MeV/c) p* (MeV/c) 2) Use the two “m-clusters” sample • hp:pp0 and mn have the same resolution function • p*(p mass) distribution for mn peak • p*(m mass) distribution for pp0 peak

  4. In order to extract the number of pp0decays? • Check the estimate of the mn background • - require a p0 in the calo and measure the efficiency • Take into account the 3-body decays contribution • - fit the distribution down to 190 MeV in order to have • 3-body decays contribution  only from kl3 channel The fit uses the MC shape for the 3-body decays contribution

  5. MC study of the fit 190 pb-1 pp0 mn

  6. p* (Kpp0 self-tag) p* (Kmn self-tag) p*(pp0) MC-FIT comparison Events/0.5 MeV

  7. p*(pp0) MC different contributions

  8. p* (Kpp0 self-tag) p* (Kmn self-tag) p*(mn) MC-FIT comparison

  9. p*(mn) MC different contributions

  10. The 2m-cluster sample: • reproduces the p*(mn) behavior • but • does not reproduce the p*(pp0) behavior at low and high p* • the fit overestimates 3-body decays and mn contribution • We have two possible strategies to measure BR(pp0): • 3s cut around the peak • require p0 in EMC and use the obtained p* shape

  11. p* (Kpp0 self-tag) p* (Kmn self-tag) • 3s cut around the pp0 peak Differences between p*(pp0)MC and p*(pp0)FIT fractional difference fit 0.02060+/- 0.0002 fractional difference fit 0.0208 +/- 0.0003 fractional difference fit_sel 0.0049 +/- 0.0001 fractional difference fit_sel 0.0048 +/- 0.0001

  12. Differences between p*(mn)MC and p*(mn)FIT p* (Kpp0 self-tag) p* (Kmn self-tag) The mn contribution under the pp0 peak fractional difference fit -0.00413 +/- 6.E-05 fractional difference fit -0.00388 +/- 5.E-05 fractional difference fit_sel -0.00055+/- 2.E-05 fractional difference fit_sel -0.00059 +/- 2.-05

  13. p* (Kpp0 self-tag) p* (Kmn self-tag) The 3-body contribution under the pp0 peak Differences between p*(3-body)MC and p*(3-body)FIT fractional difference fit -0.0288 +/- 0.0005 fractional difference fit -0.0268+/- 0.0006 fractional difference fit_sel -0.0160+/- 0.0004 fractional difference fit_sel -0.0154 +/- 0.0004

  14. p*(pp0)MC p*(pp0)TAG+p0 p* (Kpp0 self-tag) • require p0 in EMC and use the p* shape • Check that p0 requirement does not introduce distortions • evaluate p0 reconstruction efficiency (ep0) with pp0 tag sample • compare DATA to MC (can we use MC ep0 for Kl3 ?) 10 pb-1 pp0 tag sample

  15. p0 reconstruction efficiency (1) plab_secondary pstar_secondary

  16. p0 reconstruction efficiency (2) rvtx_2d rvtx_3d Work in progress but the idea seems good

  17. Exercise1: extract BR(pp0) and BR(mn) from p*DATA 190 pb-1 DATA pp0 ********* BR results with munu tag********** ntag_munu = 13079285 npipi0 = 781313 ernpipi0 = 971.094 BR(pipi0) = 0.198226 +/- 0.000442406 npipi0_fromsel = 832834 ernpipi0 = 971.094 BR(pipi0) fromsel = 0.211297 +/- 0.000524439 nmunu = 2.49387E+06 ernmunu = 1619.88 BR(munu) = 0.604953 +/- 0.000728481 mn There is a problem in the efficiencies……….

  18. Exercise2: extract BR(pp0) and BR(mn) from p*DATA To use the efficiency evaluated directly from data  require track to cluster association pp0 10 pb-1 mn

  19. ********* BR results with munu tag********** ntag_munu = 410065 npipi0 = 22945.7 ernpipi0 = 163.861 BR(pipi0) = 0.185681 +/- 0.00139935 npipi0_fromsel = 25180.7 ernpipi0 = 163.861 BR(pipi0) fromsel = 0.203767 +/- 0.00163508 nmunu = 81280.8 ernmunu = 289.429 BR(munu) = 0.628879 +/- 0.00252102 emn from2m-cluster ********* BR results with pipi0 tag********** ntag_pipi0 = 312400 npipi0 = 16983.7 ernpipi0 = 146.819 BR(pipi0) = 0.182244 +/- 0.00181808 npipi0_fromsel = 18628.2 ernpipi0 = 146.819 BR(pipi0) fromsel = 0.19989 +/- 0.00203922 nmunu = 61468.3 ernmunu = 252.179 BR(munu) = 0.627028 +/- 0.00380777 epp0 from emn and shift (eMCmn -eMCpp0 )

  20. Exercise0 : extract BR(pp0) and BR(mn) from p*MC 190 pb-1 pp0 ********* BR results with munu tag********** ntag_munu = 6530745 npipi0 = 433366 ernpipi0 = 3314.73 BR(pipi0) = 0.202514 +/- 0.00157059 npipi0_fromsel = 455826 ernpipi0 = 3314.73 BR(pipi0) fromsel = 0.21301 +/- 0.000478008 nmunu = 1443008 ernmunu = 1556.9 BR(munu) = 0.647016 +/- 0.000860994 mn

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