P Spring 2002 L16Richard Kass B mesons and CP violation CP violation has recently ( ) been observed in the decay of mesons containing a b-quark. Previous CP violation studies had always used mesons with an s-quark. Review of CP violation with kaons from Lecture 7. Long lifetime state with  L  5x10 -8 sec. Short lifetime state with  S  9x sec.  is a (small) complex number that allows for CP violation through mixing. The strong interaction eigenstates (with definite strangeness) are: If S=strangeness operator then: They are particle and anti-particle and by the CPT theorem have the same mass. Experimentally we find: The weak interaction eigenstates have definite masses and lifetimes: From experiments we find that |  |  2.3x10 -3.

P Spring 2002 L16Richard Kass Neutral Kaons and CP violation The standard model predicts that the quantities  +- and  00 should differ very slightly as a result of direct CP violation (CP violation in the amplitude). CP violation is now described by two complex parameters,  and , with  related to direct CP violation. The standard model estimates Re(  /  ) to be 4- 30x10 -4 ! Experimentally what is measured is the ratio of branching ratios: After many years of trying (starting in 1970’s) and some controversial experiments, a non-zero value of Re(  /  ) has been recently been measured (2 different experiments): Re(  /  )=17.2  1.8x10 -4 At this point, the measurement is more precise than the theoretical calculation! Calculating Re(  /  ) is presently one of the most challenging HEP theory projects.

P Spring 2002 L16Richard Kass B mesons and CP violation Searching for CP violation in the kaon system consisted of: 1)Get a beam of “pure” K 2 ’s (component with long lifetime)  have a long decay channel so the K 1 component decays away. 2)Look for K 2 decays that have the wrong CP:  expect CP= -1: K 2  3  look for CP= +1: K 2  2  Can we use the same technique to study CP violation with B mesons? NO! The lifetimes of the neutral B weak eigenstates are  equal so there is no way to separate the two components by allowing one of them to decay away. The kaon  difference is due to the limited phase space (m K -m 3  ) available for K  3 . There is no such limitation for B-meson decay. To study CP violation with B mesons must use another “trick”: Study the time evolution of B 0 B 0 pairs and look for a measurable quantity that depends on CP violation. Look for rate differences to the same CP final state (f): R(B 0  f)  R(B 0  f)

P Spring 2002 L16Richard Kass B mesons, CP violation, and the CKM matrix CKM in terms of W couplings to charge 2/3 quarks (best for illustrating physics!) This representation uses s 12 >>s 23 >>s 13 and c 23 =c 13 =1 Here =sin  12  12, and A, ,  are all real and . The “Wolfenstein” representaton: The Wolfenstein representation is good for relating CP violation to specific decay rates. A non-zero  gives CP violation since it provides a phase in the decay amplitude. Why do we need a phase to observe CP violation? four real parameters, phase generates CP violation

P Spring 2002 L16Richard Kass Who needs a phase ? In order to have CP violation there must be: a) two amplitudes b) two phases (weak phase, strong phase) c) only one phase changes sign under CP (weak phase) A difference between the particle and anti-particles decay rate to the same CP final state is evidence of CP violation: If the decay amplitude contains a phase that changes sign under CP then: But this won’t give CP violation since: B0B0 B0B0 f Use interference of B-meson decays to same final state (f) with/without mixing. mixing CP no mixing

P Spring 2002 L16Richard Kass B Mixing, CP violation, and the CKM matrix B mixing is exactly the same process that we discussed in “strangeness oscillations” in Lecture 11. Even simpler since the lifetimes are the same for both states (B L, B S ). A B 0 can oscillate into a B 0 via a “box” diagram: d t b b d t W W B0B0 B0B0 V tb V td V tb V td provides the weak phase necessary for CP violation in B decay. d c s s d c W W K0K0 K0K0 V cs V cd W W B0B0 B0B0 t t W W K0K0 K0K0 c c

P Spring 2002 L16Richard Kass The CP Violation Triangle Since the CKM matrix is unitary we must have: Since matrix elements can be complex numbers we can picture this relationship as a triangle. V ud V * ub V cd V * cb V td V * tb          Convenient to normalize all sides to the base of the triangle (V cd V * cb = A 3 ). In the ( ,  ) plane the triangle now becomes: (0,0)          (1,0) ( ,  )   One way to test the Standard Model is to measure the 3 sides & 3 angles and see if the triangles closes!

P Spring 2002 L16Richard Kass The CP Violation Triangle V ud V * ub V cd V * cb V td V * tb          How do we relate the sides and angles to B-meson decay? 1) Sin2  : B 0  K S : get V td V * tb from B mixing, V cb from b  c, get V cd from K 0 mixing. 2) Sin2  : 3) Sin2  : easy hard

P Spring 2002 L16Richard Kass Steps to observing CP violation with B mesons Produce B-mesons pairs using the reaction e + e -  (4S)  B 0 B 0 must build an asymmetric collider Reconstruct the decay of one of the B-mesons’s into a CP eigenstate example CP= -B 0  K S and B 0  K S Reconstruct the decay of the other B-meson to determine its flavor (“tag”) use high momentum leptons: B 0  e + or  + )X and B 0  e - or  - )X flavor of CP eigenstate also determined at time of the “tag” decay. Measure the distance (L) between the two B meson decays and convert to proper time must reconstruct the position of both B decay vertices t=L/(  c) Fit the decay time difference (t) to the functional form: dN/dt  e -  |t| [1  cp (sin2  sin(  mt)] Determine sin2   cp =  1 CP of final state =  for B 0  K S CP violating phase  m=difference between B mass eigenstates  m=0.47x10 12 h/s -B 0,+B 0

P Spring 2002 L16Richard Kass Expected signature of CP violation with B mesons Decay rate is not the same for B 0 and B 0 tag.

P Spring 2002 L16Richard Kass Why do we need an asymmetric collider? The source of B mesons is the  (4S), which has J PC =1 --. The  (4S) decays to two bosons with J P =0 -. Quantum Mechanics (application of the Einstein-Rosen-Podosky Effect) tells us that for a C=- initial state (  (4S)) the rate asymmetry: N=number of events f CP = CP eigenstate (e.g. B 0  K S ) f fl = flavor state (particle or anti-particle) (e.g. B 0  e + X) However, if we measure the time dependence of A we find: Need to measure the time dependence of decays to “see” CP violation using the B’s produced at the  (4S).

P Spring 2002 L16Richard Kass Why Do We Need an Asymmetric Collider? In order to measure time we must measure distance: t=L/v. How far do B mesons travel after being produced by the Y(4S) (at rest) at a symmetric e + e - collider? At a symmetric collider we have for the B mesons from Y(4S) decay: p lab =0.3 GeV, m B =5.28 GeV Average flight distance = (  )c  B = (p/m)(468  m)=(0.3/5.28)(468  m)=(27  m) This is too small to measure!! If the beams have unequal energies then the entire system is Lorentz Boosted:  = p lab /E cm =(p high -p low )/E cm SLAC: 9 GeV+3.1 GeV  = 0.55 = 257  m KEK:8 GeV+3.5 GeV  = 0.42 = 197  m We can measure these decay distances ! Because of the boost and the small p lab the time measurement is a z measurment. symmetric CESR asymmetric SLAC, KEK z-axis  B =1.6x sec

P Spring 2002 L16Richard Kass Recent Results on Sin2  Belle Fit distributions to: