K→πlν semileptonic form factors from two-flavor lattice QCD

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RECERCAT Home > Universitat de Barcelona > Física Quàntica i Astrofísica > Articles publicats en revistes (Física Quàntica i Astrofísica) > View document

To access the full text documents, please follow this link: http://hdl.handle.net/2445/131711

dc.contributor	Universitat de Barcelona
dc.contributor.author	Lubicz, V.
dc.contributor.author	Mescia, Federico
dc.contributor.author	Simula, S.
dc.contributor.author	Tarantino, C.
dc.contributor.author	ETM Collaboration
dc.date	2019-04-08T08:35:41Z
dc.date	2019-04-08T08:35:41Z
dc.date	2009-12-30
dc.date	2019-04-08T08:35:41Z
dc.identifier.citation	1550-7998
dc.identifier.citation	592726
dc.identifier.uri	http://hdl.handle.net/2445/131711
dc.format	5 p.
dc.format	application/pdf
dc.language.iso	eng
dc.publisher	American Physical Society
dc.relation	Reproducció del document publicat a: https://doi.org/10.1103/PhysRevD.80.111502
dc.relation	Physical Review D, 2009, vol. 80, num. 11, p. 111502(R)
dc.relation	https://doi.org/10.1103/PhysRevD.80.111502
dc.rights	(c) American Physical Society, 2009
dc.rights	info:eu-repo/semantics/openAccess
dc.subject	Física de partícules
dc.subject	Experiments
dc.subject	Particle physics
dc.subject	Experiments
dc.title	K→πlν semileptonic form factors from two-flavor lattice QCD
dc.type	info:eu-repo/semantics/article
dc.type	info:eu-repo/semantics/publishedVersion
dc.description.abstract	We present new lattice results of the K → π ℓ ν semileptonic form factors obtained from simulations with two flavors of dynamical twisted-mass fermions, using pion masses as light as 260 MeV. Our main result is f + ( 0 ) = 0.9560 ( 84 ) , which, combined with the latest experimental data for K ℓ 3 decays, leads to \| V u s \| = 0.2267 ( 5 ) exp ( 20 ) f + ( 0 ) . Using the PDG(2008) determinations of \| V u d \| and \| V u b \| our result implies for the unitarity relation \| V u d \| 2 + \| V u s \| 2 + \| V u b \| 2 = 1.0004 ( 15 ) . For the O ( p 6 ) term of the chiral expansion of f + ( 0 ) we get Δ f ≡ f + ( 0 ) − 1 − f 2 = − 0.0214 ( 84 ) .

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