B to \eta(') K branching ratio puzzle and OZI violation contribution

Slide 1: B η(,)K Branching Ratio Puzzle Jen-Feng Hsu IoP, Academia Sinica 24. Nov. 2007 Dept. of Physics, NCU

Slide 2: The Players B ± : ub , u b Fermilab Photo Archive B 0 : db K ± : us , u s K 0 : ds ⎧ 1 η⎫ ⎪ linear combination of bases ⎨ ηq = (uu + dd ) ⎬ 2 η ′⎭ ⎪η = ss ⎩ s ⎛η ⎞ ⎛ cos φ − sin φ ⎞⎛η q ⎞ ⎜ ⎟=⎜ ⎜η ' ⎟ ⎜ sin φ ⎟⎜ ⎟ ⎝ ⎠ ⎝ cos φ ⎟⎜η s ⎟ ⎠⎝ ⎠ B η(,)K Branching Ratio Puzzle 2/12

Slide 3: B Decays into η(,) and K K+ s ¯ u ¯ b Z0 u ¯ B+ g ηq u u Feynman rules (diagrams) give Amplitude Probability α |Amplitude|2 Γ( B → η (,) K ) Branching ratio Br(B η(,) K): Γ( B → anything ) relative probability B η(,)K Branching Ratio Puzzle 3/12

Slide 4: Underestimated Branching Ratios QCD theories that “predict” branching ratios: NF, PQCD, QCDF, SCET,… Puzzle: Branching ratios were underestimated. Experimental Data: B η(,)K Branching Ratio Puzzle 4/12

Slide 5: A Test of Parameter A. G. Akeroyd, C. H. Chen and C. Q. Geng, Phys.Rev.D75:054003, 2007 mqq= mqq(masses, decay const., mixing angle) B η(,)K Branching Ratio Puzzle 5/12

Slide 6: Proposing a Mechanism H.-n. Li and Y. Y. Charng proposed to investigate the Okubo-Zweig-Iizuka (OZI) suppressed contribution Griffiths, Introduction to Elementary Particles, p. 167 B η(,)K Branching Ratio Puzzle 6/12

Slide 7: Proposing a Mechanism H.-n. Li and Y. Y. Charng proposed to investigate the Okubo-Zweig-Iizuka (OZI) suppressed contribution OZI-suppressed Griffiths, Introduction to Elementary Particles, p. 167 B η(,)K Branching Ratio Puzzle 7/12

Slide 8: OZI in B η(,)K ¯ b u ¯ B+ ? ηs u u B η(,)K Branching Ratio Puzzle 8/12

Slide 9: OZI in B η(,)K ¯ b u ¯ ¯ u s ¯ B+ ? ηs u u u s OZI-suppressed B η(,)K Branching Ratio Puzzle 9/12

Slide 10: Major OZI-suppressed Contribution µ ¶ Ã fsq ! 2 † m2 η 0 1 fss Mqs = U (φ) U (φ) fqs 0 m2 0 η fqq 1 B η(,)K Branching Ratio Puzzle 10/12

Slide 11: Major OZI-suppressed Contribution ¯ b u ¯ B+ g u u Z Z Bη 8 Fq+ q (q 2 ) = √ πm2 CF dx1 dx2 b1 db1 b2 db2 φB (x1 , b1 ) B 2 (∙ µ ¶ ¸ A 2 × (1 + x2 ρ)φq (x2 ) + rq − 1 − 2x2 φq (x2 ) + rq (1 − 2x2 )φq (x2 ) E(t(1) )h(x1 , x2 , b1 , b2 ) T P ρ ) +2rq φP E(t(2 )h(x2 , x1 , b2 , b1 ) q m2qq rq = 2mq mB B η(,)K Branching Ratio Puzzle 11/12

Slide 12: Result Without OZI With OZI mqq mqq B η(,)K Branching Ratio Puzzle 12/12