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Bottom Quark: Discovered in 1977

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Bottom Quark: Discovered in 1977 B CP Violation in the System An Experimental View Kevin Pitts University of Illinois February 17, 2000 B CP Violation in the System Outline 1. Introduction: Charged Current Decays the CKM Matrix B=B Mixing 2. CP violation Introduction 0 J= K the Mode S 2 3. the Measurement of sin the Fermilab Tevatron the CDF Detector 0 J= K the Sample S Flavor Tagging Results 4. Outlook: Other CP Modes 5. Summary Kevin T. Pitts 1 February 17, 2000 B CP Violation in the System Cheat Sheet Fermions (spin 1/2) generation I II III charge Leptons: e -1 e 0 free particles Quarks: c t u +2/3 s b d -1/3 q qqq bound states: q , Mesons: 0 0 B = jbd > B = jbd > + B = jbu > B = jbu > 0 0 B = jbs > = jbs > B s s 0 + 1 p K = (jds > +jsd >) = jud > S 2 0 1 p = K (jds > jsd >) = jud > L 2 0 1 p J= = jcc > = (juu > +jdd >) 2 Kevin T. Pitts 2 February 17, 2000 B CP Violation in the System Weak Decays Interactions: The only decay mechanism that changes flavor is the weak interaction. Strong Interaction: + ! K K jss >! jsu > +jsu > ! ( ) ! conserves flavor Electromagnetic interaction: 0 ! juu >! ! ( ) ! conserves flavor Weak neutral current: 0 ! Z ! bb ! conserves flavor Charged neutral current: 0 B ! D ` b ! c ! ( ) ! does not conserve flavor Kevin T. Pitts 3 February 17, 2000 B CP Violation in the System Why are B ’s so Interesting? Bottom quark: discovered in 1977. Several thousand papers since then (theory+experiment: production, decay, couplings, mixing, etc.) Two new accelerators + several new experiments have been built to study B hadrons. Why is this B east so interesting? m ' 5m it’s heavy: b p it couples most strongly to the ultramassive top ' 175m quark (m ) t p ! but it can’t decay to top! = 1:5 psec it has a long lifetime b = 450 m (c ) b 0 0 B =B it has a long mixing frequency ! fully mixes in about 4.4 lifetimes CP expected to show LARGE violating effects ' 10 30% ! LARGE 0:1% CP ! violation in the kaon system: Kevin T. Pitts 4 February 17, 2000 B CP Violation in the System The Cabibbo-Kobayashi-Maskawa Matrix Weak charged-current decay (W exchange): e- e- Vus - - W – W – - ν ν µ e s e K- – u u ν µ – 0 u π g y p J W + h:c:; L = cc cc 2 where ! ! e d L L J = ( ; ; ) u ; c ; t ) V +( s L e L L L CKM L cc b L L and ! V V V ud us ub V = V V V cd cs cb V V V td ts tb is the Cabibbo-Kobayashi-Maskawa (CKM) Mixing Matrix which rotates quark mass eigenstates into weak eigenstates. Kevin T. Pitts 5 February 17, 2000 B CP Violation in the System The Cabibbo-Kobayashi-Maskawa Matrix The Wolfenstein parameterization: 0 1 2 3 1 =2 A ( i ) 2 2 @ A V ' 1 =2 A 3 2 A (1 i ) A 1 = sin = 0:22 , sine of the Cabibbo angle, C 2 2 2 (K ! ) jV j sin = – ' us C y V V = 1 unitarity ) ) V V + V V + V V = 0 td cd ud tb cb ub The unitarity triangle Kevin T. Pitts 6 February 17, 2000 B CP Violation in the System The Unitary Triangle Goal of current and future experiments is to measure both sides and three angles in as many ways possible: overconstrain the triangle to look for new physics. Examples: 0 B B ) jV =V j and mixing are d td ts s ) jV j b ! u decays ub 0 CP ) violation in J= K S 0 + CP ! ) violation in B 0 CP ! D K ) violation in B s s Kevin T. Pitts 7 February 17, 2000 B CP Violation in the System Unitarity Relations Unitary Cabibbo-Kobayashi-Maskawa (CKM) Mixing Matrix: ! V V V ud us ub V = V V V cd cs cb V V V td ts tb y y V V = V V = 1 Unitarity ) ! ! ! V V V V V V 1 0 0 ud us ub ud cd td V V V = V V V 0 1 0 cd cs cb us cs ts V V V 0 0 1 V V V td ts tb ub cb tb V + V V + V V = 0 ) V td cd ud tb cb ub This relation is quite useful because each term is of 3 order . Kevin T. Pitts February 17, 2000 B CP Violation in the System 0 0 =B B Mixing 0 B For initial state = : 1 t 0 e P (B (t)) = (1 + cos(m t)) d 2 1 t 0 P ( B (t)) = e (1 cos(m t)) d 2 start with a B0 beam Prob(B0(t)) 0.5 – 0 Prob(B (t)) – 1/N dN/dt Prob(B0(t)+B0(t)) 0.25 0 0246810 time (nsec) 1 – – A = (N(B0)-N(B0))/(N(B0)+N(B0)) 0 Asymmetry at ∆mt=π, no B0 -1 0246810 time (nsec) 6= weak eigenstates strong eigenstates m : mass difference between the heavy and d m = 0 ) light weak eigenstates ( no mixing) d Kevin T. Pitts 8 February 17, 2000 B CP Violation in the System 0 0 =B B Mixing Vtd – – b W d – – – – B0 u,c,t u,c,t B0 W d b Vtd Vtd – – – – – b u,c,t d – B0 W W B0 u,c,t d b Vtd 0 0 B ; B : Flavor Eigenstates 0 0 B ; B : Mass Eigenstates H L m = m m d H L (' 0) = H L 0 B For initial state = : 1 1 0 t P (B (t)) = e (cosh( t) + cos(m t)) d 2 2 1 t 0 e ( 1 + cos(m t)) (t)) ' P (B d 2 0 0 B mt = ' 4:4 fully mixes to B after lifetimes Kevin T. Pitts 9 February 17, 2000 B CP Violation in the System Example Mixing Measurement Ph.D. thesis of P. Maksimovic´ PRL 80, 2057 (1998). + + 0 B ! ` D 3000 ev 0 + B ! ` D 2000 ev 0 + B ! ` D 4300 ev B top plot: charged do not mix! middle/bottom plots: N N unmixed mixed = * A N +N unmixed mixed A = 1 ) all unmixed A = 1 ) all mixed jAj < 1 due to experimental effects max Kevin T. Pitts 10 February 17, 2000 B CP Violation in the System Symmetries 1. parity inversion: P look in the mirror P (~r ) ! (~r ) ) left-handed right-handed P je >! je > L R ! 2. charge conjugation: C particle antiparticle ! particle antiparticle C jp >! jp > 3. time reversal: T run the film backwards T (t) ! (t) Symmeties are “preserved” if the operator commutes with the Hamiltonian. e.g. the electromagnetic interaction conserves all three of the above symmetries: [H ; P ] = [H ; C ] = [H ; T ] = 0 em em em Kevin T. Pitts 11 February 17, 2000 B CP Violation in the System CP Violation If CP is a valid symmetry, then an interaction for a left-handed particle must be the same as an interaction for a right-handed antiparticle " " P C If CP is violated, then some force of nature must be able to tell the difference between a left-handed quark and right-handed antiquark Note: CP violation is one of the conditions necessary for the universal matter asymmetry... some process must create (and destroy) quarks and antiquarks at unequal rates!!! Kevin T. Pitts 12 February 17, 2000 B CP Violation in the System CP Violation Since 1964, CP violation has still only been seen in the neutral kaon system. The standard model predicts that we should see it in the neutral B system as well: 0 B = jbd > 0 B = jbd > Now, we will discuss a search for CP violation in CP the neutral B system. The type of violation we are looking for is of the form: dN dN 0 0 0 0 ) 6= ) (B ! J= K (B ! J= K S S dt dt where + J= = jcc > J= ! 0 + 0 1 p (jds > +jsd >) K ! K = S S 2 0 and J= K is a CP eigenstate: S 0 0 > >= jJ= K CP jJ= K S S Kevin T. Pitts 13 February 17, 2000 B CP Violation in the System “Mixing Induced” CP Violation Two paths to the same final state: 0 0 B ! J= K direct decay ( ) S 0 0 0 B ! B ! J= K mixed decay ( ) S The mixed decay accumulates a phase relative to the direct decay. V – V cb c cb c – J/ψ –J/ψ b W c b W c 0 – 0 B V – B V cs s 0 cs s 0 KS – –KS d d d d – V V td d 0 td d 0 – – – KS t –KS b t d V s – b d V s 0 W – 0 cs 0 W 0 cs B W – B W B c – B c t W ψ t W – ψ J/ – – J/ V V c V V c d td b cb d td b cb " " 0 0 t = 0 B t = 0 produce B at produce at Kevin T. Pitts 14 February 17, 2000 B CP Violation in the System “Mixing Induced” CP Violation The “direct” decay interferes with the mixed decay. 0 ψ 0 – 0 ψ 0 B J/ KS B J/ KS – B0 B0 β β e2i e-2i 0 ! J= K 2 For B the phase arises from the S interference. 0 t = 0 produce B at : dN 0 t= 0 ( B ! J= K ) / e (1 + sin 2 sin mt) s dt 0 t = 0 produce B at : dN 0 0 t= (B ! J= K ) / e (1 sin 2 sin mt) s dt Kevin T.
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