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Niels Tuning (1) CP violation Lecture 4 N. Tuning.

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Presentation on theme: "Niels Tuning (1) CP violation Lecture 4 N. Tuning."— Presentation transcript:

1 Niels Tuning (1) CP violation Lecture 4 N. Tuning

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3 Diagonalize Yukawa matrix Y ij –Mass terms –Quarks rotate –Off diagonal terms in charged current couplings Niels Tuning (3) Recap uIuI dIdI W u d,s,b W

4 Niels Tuning (4) CKM-matrix: where are the phases? u d,s,b W Possibility 1: simply 3 ‘rotations’, and put phase on smallest: Possibility 2: parameterize according to magnitude, in O( λ):

5 This was theory, now comes experiment We already saw how the moduli |V ij | are determined Now we will work towards the measurement of the imaginary part –Parameter: η –Equivalent: angles α, β, γ. To measure this, we need the formalism of neutral meson oscillations… Niels Tuning (5)

6 Neutral Meson Oscillations (1) Start with Schrodinger equation: Find eigenvalue: Solve eigenstates: Eigenstates have diagonal Hamiltonian: mass eigenstates! (2-component state in P 0 and P 0 subspace) Niels Tuning (6)

7 Neutral Meson Oscillations (2) Two mass eigenstates Time evolution: Probability for |P 0 >  |P 0 > ! Express in M=m H +m L and Δm=m H -m L  Δm dependence

8 Meson Decays Formalism of meson oscillations: Subsequent: decay Interference P0 fP0 fP 0  P 0  f Interference(‘direct’) Decay

9 Classification of CP Violating effects 1.CP violation in decay 2.CP violation in mixing 3.CP violation in interference Niels Tuning (9)

10 Now: Im( λ f ) 1.CP violation in decay 2.CP violation in mixing 3.CP violation in interference We will investigate λ f for various final states f

11 CP violation: type 3 Niels Tuning (11)

12 Niels Tuning (12) Final state f : J/ΨK s Interference between B 0 → f CP and B 0 → B 0 → f CP –For example: B 0 → J/ΨK s and B 0 → B 0 → J/ΨK s –Lets ’ s simplify … 1)For B 0 we have: 2)Since f CP = f CP we have: 3)The amplitudes | A( B 0 →J/ψK s ) | and | A(  B 0 →J/ψK s ) | are equal: B0B0 B0B0 dd |λ f |=1

13 Relax: B 0  J/ΨK s simplifies… Niels Tuning (13) |λ f |=1 ΔΓ=0

14 Niels Tuning (14) b d d b s d d s c J/  K0K0K0K0 B0B0B0B0 b c s d d K0K0K0K0 B0B0B0B0 c d s c b d λ f for B 0 J/K 0 S

15 Niels Tuning (15) λ f for B 0 J/K 0 S Theoretically clean way to measure  Clean experimental signature Branching fraction: O(10 -4 ) “Large” compared to other CP modes! Time-dependent CP asymmetry

16 Niels Tuning (16) Remember: C and P eigenvalue C and P are good symmetries (not involving the weak interaction) –Can associate a conserved value with them (Noether Theorem) Each hadron has a conserved P and C quantum number –What are the values of the quantum numbers –Evaluate the eigenvalue of the P and C operators on each hadron P|> = p|> What values of C and P are possible for hadrons? –Symmetry operation squared gives unity so eigenvalue squared must be 1 –Possible C and P values are +1 and -1. Meaning of P quantum number –If P=1 then P|> = +1|> (wave function symmetric in space) if P=-1 then P|> = -1 |> (wave function anti-symmetric in space)

17 Niels Tuning (17) Remember: P eigenvalues for hadrons The  + meson –Quark and anti-quark composite: intrinsic P = (1)*(-1) = -1 –Orbital ground state  no extra term –P( + )=-1 The neutron –Three quark composite: intrinsic P = (1)*(1)*(1) = 1 –Orbital ground state  no extra term –P(n) = +1 The K 1 (1270) –Quark anti-quark composite: intrinsic P = (1)*(-1) = -1 –Orbital excitation with L=1  extra term (-1) 1 –P(K 1 ) = +1 Meaning: P| + > = -1| + >

18 Intermezzo: CP eigenvalue Niels Tuning (18) Remember: –P 2 = 1 (x  -x  x) –C 2 = 1 ( ψ   ψ  ψ ) –  CP 2 =1 CP | f > =  | f > Knowing this we can evaluate the effect of CP on the K 0 CP|K 0 > = -1| K 0 > CP| K 0 > = -1|K 0 > CP eigenstates: |K S > = p| K 0 > +q|K 0 > |K L > = p| K 0 > - q|K 0 > |K s > (CP=+1) →   (CP= (-1)(-1)(-1) l=0 =+1) |K L > (CP=-1) →  (CP = (-1)(-1)(-1)(-1) l=0 = -1) ( S( K )=0  L( ππ )=0 )

19 Niels Tuning (19) CP eigenvalue of final state J/K 0 S CP |J/> = +1 |J/> CP |K 0 S > = +1 |K 0 S > CP |J/K 0 S > = (-1) l |J/K 0 S > ( S( B )=0  L(J/K 0 S )=1 ) Relative minus-sign between state and CP-conjugated state: ( S(J/)=1 )

20 Niels Tuning (20) Time dependent CP violation + If final state is CP eigenstate then 2 amplitudes (w/o mixing): B 0  f and B 0  B 0  f B 0 -B 0 oscillation is periodic in time  CP violation time dependent e -iφ (i=e iπ/2  δ=90 o ) (φ=2β) Amplitude 2 Amplitude 1

21 Niels Tuning (21) Time dependent CP violation + If final state is CP eigenstate then 2 amplitudes (w/o mixing): B 0  f and B 0  B 0  f B 0 -B 0 oscillation is periodic in time  CP violation time dependent e -iφ (i=e iπ/2  δ=90 o ) (φ=2β) 1

22 Niels Tuning (22) Sum of 2 amplitudes: sensitivity to phase What do we know about the relative phases of the diagrams? B 0  f B 0  B 0  f (strong)= (weak)=0(weak)=2 (mixing)=/2 There is a phase difference of i between the B 0 andB 0 Decays are identical K 0 mixing exactly cancels V cs

23 Niels Tuning (23) Sum of 2 amplitudes: sensitivity to phase Now also look at CP-conjugate process Investigate situation at time t, such that |A 1 | = |A 2 | : Directly observable result (essentially just from counting) measure CKM phase  directly! CP + = /2+2 + = /2-2 N(B 0  f)  |A| 2  (1-cos) 2 +sin 2  = 1 -2cos+cos 2 +sin 2  = 2-2cos(/2-2)  1-sin(2) N(B 0  f)  (1+cos) 2 +sin 2  = 2+2cos(/2-2)  1+sin(2) 1-cos() sin() 1+cos() Γ( B  f)=

24 Remember! Necessary ingredients for CP violation: 1)Two (interfering) amplitudes 2)Phase difference between amplitudes –one CP conserving phase (‘strong’ phase) –one CP violating phase (‘weak’ phase) Niels Tuning (24)

25 Remember! Niels Tuning (25)

26 What do we measure? We measure quark couplings There is a complex phase in couplings! Visible when there are 2 amplitudes: Γ( B  f) = |A 1 +A 2 e i(φ+δ) | 2 Γ(B  f) = |A 1 +A 2 e i(-φ+δ) | 2 eiφeiφ e -iφ matteranti-matter Proof of existence of complex numbers… Niels Tuning (26)

27 B-system - Time-dependent CP asymmetry B 0 →J/ψK s BaBar (2002) Niels Tuning (27)

28 Niels Tuning (28) 4 Consistency with other measurements in (,) plane   Method as in Höcker et al, Eur.Phys.J.C21:225-259,2001 Prices measurement of sin(2β) agrees perfectly with other measurements and CKM model assumptions The CKM model of CP violation experimentally confirmed with high precision! 4-fold ambiguity because we measure sin(2), not  1 2 3 without sin(2)

29 β s : B s 0 J/φ : B s 0 analogue of B 0 J/K 0 S Niels Tuning (29) Replace spectator quark d  s

30 β s : B s 0 J/φ : B s 0 analogue of B 0 J/K 0 S Niels Tuning (30)

31 Remember: The “B s -triangle”: β s Niels Tuning (31) Replace d by s:

32 β s : B s 0 J/φ : B s 0 analogue of B 0 J/K 0 S Niels Tuning (32) Differences: B0B0 B0sB0s CKMV td V ts ΔΓ ~0~0.1 Final state (spin)K 0 : s=0 φ : s=1 Final state (K)K 0 mixing-

33 β s : B s 0 J/φ Niels Tuning (33) B0B0 B0sB0s CKMV td V ts ΔΓ ~0~0.1 Final state (spin)K 0 : s=0 φ : s=1 Final state (K)K 0 mixing- A║A║ A0A0 A┴A┴ l=2 l=1 l=0 3 amplitudes V ts large, oscilations fast, need good vertex detector

34 “Recent” excitement (5 March 2008) Niels Tuning (34)

35 B s  J/ψФ : B s equivalent of B  J/ψK s ! The mixing phase (V td ): φ d =2β B 0  f B 0  B 0  f Wolfenstein parametrization to O (λ 5 ): Niels Tuning (35)

36 B s  J/ψФ : B s equivalent of B  J/ψK s ! The mixing phase (V ts ): φ s =-2β s B 0  f B 0  B 0  f Wolfenstein parametrization to O (λ 5 ): BsBs BsBs s s s s - Ф Ф V ts Niels Tuning (36)

37 B s  J/ψФ : B s equivalent of B  J/ψK s ! The mixing phase (V ts ): φ s =-2β s B 0  f B 0  B 0  f BsBs BsBs s s s s - Ф Ф V ts Niels Tuning (37)

38 Break Niels Tuning (38)

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