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Insertion of STC into TRT at the Department of Physics, Oxford
Credit: CERN

David Bacher

Graduate Student

Sub department

  • Particle Physics

Research groups

  • LHCb
david.bacher@physics.ox.ac.uk
Telephone: (2)73382
Denys Wilkinson Building, room 653
  • About
  • Publications

Observation of the decays $$ {B}_{(s)}^0 $$ → Ds1(2536)∓K±

Journal of High Energy Physics Springer 2023:10 (2023) 106

Authors:

R Aaij, ASW Abdelmotteleb, C Abellan Beteta, F Abudinén, T Ackernley, B Adeva, M Adinolfi, P Adlarson, H Afsharnia, C Agapopoulou, CA Aidala, Z Ajaltouni, S Akar, K Akiba, P Albicocco, J Albrecht, F Alessio, M Alexander, A Alfonso Albero, Z Aliouche, P Alvarez Cartelle, R Amalric, S Amato, JL Amey, Y Amhis, L An, L Anderlini, M Andersson, A Andreianov, P Andreola, M Andreotti, D Andreou, D Ao, F Archilli, A Artamonov, M Artuso, E Aslanides, M Atzeni, B Audurier, D Bacher, I Bachiller Perea, S Bachmann, M Bachmayer, JJ Back, A Bailly-reyre, P Baladron Rodriguez, V Balagura, W Baldini, J Baptista de Souza Leite, M Barbetti

Abstract:

A bstract This paper reports the observation of the decays $$ {B}_{(s)}^0 $$ B s 0 → D s 1 (2536) ∓ K ± using proton-proton collision data collected by the LHCb experiment, corresponding to an integrated luminosity of 9 fb − 1 . The branching fractions of these decays are measured relative to the normalisation channel B 0 → $$ \overline{D} $$ D ¯ 0 K + K − . The D s 1 (2536) − meson is reconstructed in the $$ \overline{D} $$ D ¯ * (2007) 0 K − decay channel and the products of branching fractions are measured to be $$ {\displaystyle \begin{array}{c}\mathcal{B}\left({B}_s^0\to {D}_{s1}{(2536)}^{\mp }{K}^{\pm}\right)\times \mathcal{B}\left({D}_{s1}{(2536)}^{-}\to {\overline{D}}^{\ast }{(2007)}^0{K}^{-}\right)\\ {}=\left(2.49\pm 0.11\pm 0.12\pm 0.25\pm 0.06\right)\times {10}^{-5},\\ {}\mathcal{B}\left({B}^0\to {D}_{s1}{(2536)}^{\mp }{K}^{\pm}\right)\times \mathcal{B}\left({D}_{s1}{(2536)}^{-}\to {\overline{D}}^{\ast }{(2007)}^0{K}^{-}\right)\\ {}=\left(0.510\pm 0.021\pm 0.036\pm 0.050\right)\times {10}^{-5}.\end{array}} $$ B B s 0 → D s 1 2536 ∓ K ± × B D s 1 2536 − → D ¯ ∗ 2007 0 K − = 2.49 ± 0.11 ± 0.12 ± 0.25 ± 0.06 × 10 − 5 , B B 0 → D s 1 2536 ∓ K ± × B D s 1 2536 − → D ¯ ∗ 2007 0 K − .
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