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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. B mesons and CP violation.

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B mesons and CP violation

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  1. Richard Kass 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. B mesons and CP violation 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: Short lifetime state with tS» 9x10-11 sec. Long lifetime state with tL» 5x10-8 sec. e is a (small) complex number that allows for CP violation through mixing. From experiments we find that |e| » 2.3x10-3.

  2. Richard Kass The standard model predicts that the quantities h+- and h00 should differ very slightly as a result of direct CP violation (CP violation in the amplitude). Neutral Kaons and CP violation CP violation is now described by two complex parameters, eand e¢, with e¢ related to direct CP violation. The standard model estimates Re(e¢/e) 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(e¢/e) has been recently been measured (2 different experiments): Re(e¢/e)=17.2±1.8x10-4 At this point, the measurement is more precise than the theoretical calculation! Calculating Re(e¢/e) is presently one of the most challenging HEP theory projects.

  3. Richard Kass • Searching for CP violation in the kaon system consisted of: • Get a beam of “pure” K2’s (component with long lifetime) • Þ have a long decay channel so the K1 component decays away. • Look for K2 decays that have the wrong CP: • Þ expect CP= -1: K2®3p look for CP= +1: K2®2p • 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. B mesons and CP violation The kaon G difference is due to the limited phase space (mK-m3p) available for K®3p. 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 B0B0 pairs and look for a measurable quantity that depends on CP violation. Look for rate differences to the same CP final state (f): R(B0®f) ¹ R(B0®f)

  4. Richard Kass B mesons, CP violation, and the CKM matrix CKM in terms of W couplings to charge 2/3 quarks (best for illustrating physics!) four real parameters, phase generates CP violation The “Wolfenstein” representaton: This representation uses s12>>s23>>s13 and c23=c13 =1 Here l=sinq12 » q12, and A, r, h are all real and » 1. The Wolfenstein representation is good for relating CP violation to specific decay rates. A non-zero h gives CP violation since it provides a phase in the decay amplitude. Why do we need a phase to observe CP violation?

  5. Richard Kass A difference between the particle and anti-particles decay rate to the same CP final state is evidence of CP violation: Who needs a phase ? If the decay amplitude contains a phase that changes sign under CP then: But this won’t give CP violation since: 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) no mixing CP Use interference of B-meson decays to same final state (f) with/without mixing. B0 f B0 mixing

  6. Richard 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 (BL, BS). A B0 can oscillate into a B0 via a “box” diagram: W Vtd Vtb t d b B0 B0 t B0 B0 W t W b d W Vtd Vtb t Vtb Vtd Vtd provides the weak phase necessary for CP violation in B decay. W Vcd Vcs c d s K0 K0 W K0 K0 W c c s d Vcd Vcs c W

  7. Richard Kass Since the CKM matrix is unitary we must have: The CP Violation Triangle Since matrix elements can be complex numbers we can picture this relationship as a triangle. VtdV*tb VudV*ub f2=a f1=b f3=g VcdV*cb Convenient to normalize all sides to the base of the triangle (VcdV*cb = Al3). In the (r, h) plane the triangle now becomes: (r, h) h f2=a f3=g f1=b (0,0) (1,0) r One way to test the Standard Model is to measure the 3 sides & 3 angles and see if the triangles closes!

  8. Richard Kass How do we relate the sides and angles to B-meson decay? The CP Violation Triangle VtdV*tb VudV*ub f2=a f1=b f3=g VcdV*cb easy 1) Sin2b: B0®yKS: get VtdV*tb from B mixing, Vcb from b®c, get Vcd from K0 mixing. 2) Sin2a: 3) Sin2g: hard

  9. Richard Kass Steps to observing CP violation with B mesons Produce B-mesons pairs using the reaction e+e-®¡ (4S) ® B0B0 must build an asymmetric collider Reconstruct the decay of one of the B-mesons’s into a CP eigenstate example CP= -B0 ®yKS and B0 ®yKS Reconstruct the decay of the other B-meson to determine its flavor (“tag”) use high momentum leptons: B0 ®(e+ orm+ )X and B0 ®(e- orm- )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/(bgc) Fit the decay time difference (t) to the functional form: dN/dtµe-G|t| [1±hcp(sin2f)sin(Dmt)] Determine sin2f -B0,+B0 Dm=difference between B mass eigenstates Dm=0.47x1012h/s hcp = ±1 CP of final state = -1 for B0 ®yKS CP violating phase

  10. Richard Kass Expected signature of CP violation with B mesons Decay rate is not the same for B0 and B0 tag.

  11. Richard Kass Why do we need an asymmetric collider? The source of B mesons is the U(4S), which has JPC =1--. The U(4S) decays to two bosons with JP =0-. Quantum Mechanics (application of the Einstein-Rosen-Podosky Effect) tells us that for a C=- initial state (U(4S)) the rate asymmetry: N=number of events fCP= CP eigenstate (e.g. B0®yKS) ffl= flavor state (particle or anti-particle) (e.g. B0®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 U(4S).

  12. Richard 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: plab =0.3 GeV, mB=5.28 GeV Average flight distance <L>= (bg)ctB= (p/m)(468mm)=(0.3/5.28)(468mm)=(27mm) This is too small to measure!! tB=1.6x10-12 sec If the beams have unequal energies then the entire system is Lorentz Boosted: bg = plab /Ecm=(phigh-plow)/Ecm SLAC: 9 GeV+3.1 GeV bg = 0.55 <L>= 257mm KEK: 8 GeV+3.5 GeV bg = 0.42 <L>= 197mm We can measure these decay distances ! Because of the boost and the small plab the time measurement is a z measurment. z-axis asymmetric SLAC, KEK symmetric CESR

  13. Richard Kass Recent Results on Sin2b Fit distributions to: Belle

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