# P780.02 Spring 2003 L14Richard Kass B mesons and CP violation CP violation has recently (1999-2001) been observed in the decay of mesons containing a b-quark.

## Presentation on theme: "P780.02 Spring 2003 L14Richard Kass B mesons and CP violation CP violation has recently (1999-2001) been observed in the decay of mesons containing a b-quark."— Presentation transcript:

P780.02 Spring 2003 L14Richard Kass B mesons and CP violation CP violation has recently (1999-2001) 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  9x10 -11 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.

P780.02 Spring 2003 L14Richard 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.

P780.02 Spring 2003 L14Richard 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)

P780.02 Spring 2003 L14Richard 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

P780.02 Spring 2003 L14Richard 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-particle 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

P780.02 Spring 2003 L14Richard Kass B Mixing, CP violation, and the CKM matrix B mixing is exactly the same process that we discussed in “strangeness oscillations”. 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

P780.02 Spring 2003 L14Richard 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!

P780.02 Spring 2003 L14Richard 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 experimentally

P780.02 Spring 2003 L14Richard 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

P780.02 Spring 2003 L14Richard Kass Expected signature of CP violation with B mesons Decay rate is not the same for B 0 and B 0 tag.

P780.02 Spring 2003 L14Richard 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).

P780.02 Spring 2003 L14Richard 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 =1.6x10 -12 sec

P780.02 Spring 2003 L14Richard Kass Recent Results on Sin2  Belle Fit distributions to:

P780.02 Spring 2003 L14Richard Kass CP violation with K-mesons Vs. B-mesons observe CPV with “wrong”Look for differences in decay distributions (time) CP states (CP-  CP+).  3/1000 decays violate CPCPV is large (10-15%) for certain final states. use K L 0 , K L 0  l But the final states have small branching ratios: sin2  BRs  5x10 -4 : B 0  K s, B 0  K L sin2  BRs  5x10 -6 : B 0  +  -, B 0  0  0 K L mesons B 0 mesons can make  10 7 K L /seccan make  5B 0 /sec explores limited region of the CKM “triangles” provide a detailed check of standard model. Very mature field (1964)Field is just beginning. Experiments are planned maybe one more experimentfor next 20 years (SLAC, KEK, CERN, Fermilab)

P780.02 Spring 2003 L14Richard Kass CP Violation with Leptons If quarks can violate CP why not CP violation with leptons? The CP violation that we see in “outer space” is much larger than the CPV we see with quarks. Maybe it’s the leptons that generated the cosmic baryon/anti-baryon asymmetry! To date, no experimental evidence for CPV in lepton sector. But if neutrinos violate CP who would know? Very hard to do an experiment with neutrinos that is sensitive to CPV We know that neutrinos “mix”: v e  v  v e  v  Results from SuperK and SNO imply that neutrinos: have mass (small) neutrino lepton number not conserved (must modify standard model) Is there a “CKM”-like matrix (3x3 unitary) for neutrinos ? If neutrino mixing, why not CP violation?

Download ppt "P780.02 Spring 2003 L14Richard Kass B mesons and CP violation CP violation has recently (1999-2001) been observed in the decay of mesons containing a b-quark."

Similar presentations