Physics of CP Violation (III)

Transcript

1 Physics of CP Violation (III) Special Lecture at Tsinghua Univiersity, Beijing, China March 2011 Tatsuya Nakada École Polytechnique Fédéale de Lausanne (EPFL)

2 Experimental observation of CP violation in the decay amplitudes From: Reε 0 CP in oscillations From: Imη 0 + From: Reη 0 + CP in interplay between decay and oscillations CP in decay amplitudes is inconclusive... dedicated experiments to look for this... Comparing η + and η 00.

3 η 2 + = A ( K L π + π ) 2 ( ) 2 = N S A K S π + π η 2 00 = A ( K L π 0 π 0 ) 2 A K S π 0 π 0 + N L + ( ) 2 = N S 00 N L 00 ( ) ( ) = ε + ε N K L π + π N K S π + π ( ) ( ) = ε 2 ε N K L π 0 π 0 N K S π 0 π 0 η 00 2 η + 2 =1 6Re ε ε = N 00 + S N L N 00 + L N S N( K L π 0 π 0 )N K S π + π N( K S π 0 π 0 )N K L π + π ( ) ( )

4 NA48

5 KTeV

6 Regeneration σ Kn, σ Kp > σ Kn, σ Kp K 0 = (ds) K 0 = (ds) p = (uud), n = (udd) material = p and n K L K L and K S K 0 K 0 K 0 α K 0 = 1 α ( K 0 + K 0 )+ 1+α ( K 0 K 0 ) 2 2 = 1 α K S + 1+α K L 2 2 α = σ K N σ KN 1

7 Measure π + π and π 0 π 0 at the same time: N 00 S = NS, NL = NL NA31, NA48 K L is regenerated from K S : E731, KTeV N 00 L = rns, NL = rns No normalization is required, but efficiencies, acceptances etc. have to be corrected

8 Effort over 30 years! ± η + η 00 ε ε Note; 3 Re = η + /η 00 1, i.e. Re 0 ε ε

9 Kaon interferometer e + e (@1GeV) virtual γ φ(1020) K K (charged or neutral) all due to electromagnetic interactions φ(1020) K K C = 1 P = 1 K P K K C K K K KK is a quantum superposition of K K K K C = 1 P = 1

10 Or... φ(1020) K K s = 1 K L = 1 KK L=1 = dω Y 1m ( e )K 1 m= 1 Projection to a momentum state 1 Y 1m m= 1 ( e )K e K K e K e e +Y 1m ( e )K e { } 1 = Y 1m e m= 1 ( ){ K } e K e K e K e K e

11 K 0 K 0 For neutral kaons, they oscillates, but... KK L=1 t t = 0 K 0 K 0 K 0 () = dω Y 1m e K 0 K 0 K 0 1 m= 1 ( )K () K () e t e t at time t K 0 K 0 K 0 K 0 K 0 K 0 K 0 K 0 = 0 = 0

12 Only the allowed oscillations are t = 0 sometime later... K 0 K 0 K 0 K 0 K 0 K 0 K 0 K 0 K 0 K 0 K 0 K 0 One kaon seems to know what the other does!! Also K 0 K 0 K 0 K 0 = K S K L K L K S but no K S K S or K L K L

13 K 0 K 0 L=1 t () K S p e iλ S t K L p e iλ L t K L p e iλ L t K S p e iλ S t It cannot decay into a same final state, e.g. π + π at the same time. ( π + π ( ) π + π ) H p p W K0 K 0 + L=1() t A S p A + S p ( η + η + ) = 0 It can decay into π + π and π 0 π 0 at the same time only if η + η 00. ( π + π ( ) π 0 π 0 ) H p p W K0 K 0 + L=1() t A S p A 00 S p ( η + η 00)

14 KLOE experiment at DAFNE storage ring

15 K S π + π, K L π + π CP violating decays!! An ideal way to produce K S beam 1) Identify K L decay with the decay time. 2) opposite side is K S.

16 To summarise CP violation can be generated through three interference mechanisms: Interfering processes CP phase CP phase Dispersive (M 12 ) and Weak interaction Dispersion vs absorptive (Γ 12 ) paths phases absorption (i) in P-P oscillations CP violation in the oscillation Different decay Weak interaction Strong interaction amplitudes phases phases CP violation in the decay amplitudes Decay with and without Weak interaction Time evolution oscillations phases functions CP violation in the interplay between the decay and oscillation All three mechanisms seen in the neutral kaon system: Re ε, Re ε, Im η +, Im η 00

17 Standard Model and CP violation Strong interaction Electromagnetic interaction Weak interaction gluons photons neutral current charged current: W ± Up type quark spinor field Q = 2/3 U = u c t Down type quark spinor field Q = 1/3 D = d s b example d L W V cd coupling there are 3 3 = 9 V s c L

18 u L, c L, t L u L, c L, t L W V + ud, V us, V ub, Vud, V us, W + d L, s L, b L d L, s L, b L L V ij U i γ μ (1 γ 5 ) D j W μ + V ij D i γ μ (1 γ 5 ) U j W μ CP conjugation L CP V ij D i γ μ (1 γ 5 ) U j W μ + V ij U i γ μ (1 γ 5 ) D j W μ If V ij = V ij L = L CP : i.e. CP conservation

19 Let us look at now: V ij U i γ μ (1 γ 5 ) D j One family V 1 free phase 1 free modula = V e iφ V e iφ u γ μ (1 γ 5 ) d V u γ μ (1 γ 5 ) d Changing u quark phase: u u e iφ Unitarity: V V = VV = E (one constraint) V 2 = 1 0 free phase 0 free modula NO CP

20 Two families V = V ud V cd V us V cs 4 free phase 4 free moduli (or rotation angles) 1) The phase of V ij can be absorbed by adjusting the phase differences between i- and j- quark 4 quarks = 3 phase differences 4 3 = 1 phase left 2) Unitarity V V = VV = E: four constraints: 1 off-diagonal constraint for the phase 1 1 = 0 phase left = three constraint for the rest 4 3 = 1 rotation angle left V is real, i.e. no CP.

21 Explicit demonstration V ud e iφ ud u γ μ (1 γ 5 ) d + V us e iφ us u γ μ (1 γ 5 ) s + V cd e iφ cd c γ μ (1 γ 5 ) d + V cs e iφ cs c γ μ (1 γ 5 ) s u u e iφ ud V ud u γ μ (1 γ 5 ) d + V us e i(φ us φ ud ) u γ μ (1 γ 5 ) s + V cd e iφ cd c γ μ (1 γ 5 ) d + V cs e iφ cs c γ μ (1 γ 5 ) s s s e i(φ us φ ud ), c c e i(φ cs φ us + φ ud ) V ud u γ μ (1 γ 5 ) d + V us u γ μ (1 γ 5 ) s + V cd e iδ c γ μ (1 γ 5 ) d + V cs c γ μ (1 γ 5 ) s Out of four quark, three quark phases can be adjusted: 4 free phase 1 free phase

22 Unitarity: V V = VV = E (4 constraints) V ud * V cd * V us * V cs * V ud V us V cd V cs * * V ud V us + V cd V cs = 0 V ud V us + V cd V cs e iδ = 0 δ = π :0 free phase V ud V cd V us V cs = 0 V ud 2 + V cd 2 = 1, V us 2 + V cs 2 = 1 1 free modula or rotation angle V 11 = cos θ, V 22 = cos θ, V 12 = sin θ, V 21 = sin θ V = cos θ sin θ sin θ cos θ One rotation angle without phase: NO CP (Cabibbo angle)

23 Three families V ud V cd V us V ub V cs V cb V td V ts V tb 9 free phase 9 free moduli (or rotation angles) Out of six quark, five quark phases can be adjusted: 9 free phase 4 free phase Out of nine unitarity constraints, three are for the phases 4 free phase 1 free phase Three rotation angles with one phase: * * 0 * 1 * 0 * 0 * 0 * 1 The rest (six) are for the rotation angles 9 free rotation angles 3 free rotation angles CP can be generated

24 CKM matrix and Wolfenstein s parameters V ud V us V ub V CKM = V cd V cs V cb V td V ts V tb 1 λ2 2 λ ia 2 λ 5 η Aλ 3 1 ˆ ρ iη ˆ λ 1 λ2 ( ) Aλ 2 iaλ 4 η 1 2 Aλ 3 ( ρ iη) Aλ 2 ˆ ρ = ρ 1 λ2, η ˆ = ρ 1 η2 2 2 A ~ 1, λ ~ 0.22, ρ 0 but η 0???

25 Electroweak theory with 3 families can naturally accommodate CP violation in the charged current induced interactions through the complex Cabibbo-Kobayashi-Maskawa quark mixing matrix V, with 4 parameters.

26 Standard Model for the kaon system final initial M 12 short range interactions K 0 d _ s d _ s u, c, t W W W + u, c, t _ _ _ u, c, t u, c, t W + + QCD corrections s _ d s _ d K 0 long range 2π, 3π interactions K 0 K 0 2π, 3π hadronic resonances Large uncertainties in the theoretical calculations of M 12

27 Short distance part ΔS=2 H effective = G 2 2 Fm W λ 16π 2 i λ j K 0 M V A K 0 B ij + K 0 M S P K 0 C ij i,j= u,c,t ( ) G F : Fermi constant m W : W mass quark operators CKM elements M V A = d γ μ ( 1 γ 5 )s ( )s [ ] [ ] d γ μ 1 γ 5 [ ][ d ( 1 γ 5 )s] M S P = d ( 1 γ 5 )s λ i = V is V id integrating out W B ij C ij Real part M 12, Imaginary part Γ 12 C ij = 0 for m s /m c <<1

28 K 0 M V A K 0 = K 0 d γ μ ( 1 γ 5 )s [ ( )s]k 0 [ ] d γ μ 1 γ 5 = 8 3 B K K0 d γ μ ( 1 γ 5 )s 0 0 d γ μ ( 1 γ 5 )s K 0 vacuum insertion K 0 d _ s W+quark states s _ d K 0 8B K 3 d K B parameters have to be obtained from QCD calculations. _ s s _ d K 0 K 0 d γ μ ( 1 γ 5 )s 0 = K 0 ( CP) ( CP)d γ μ ( 1 γ 5 )scp ( ) ( CP)0 = K 0 ( CP)d γ μ ( 1 γ 5 )scp ( ) 0 e iθ CP = K 0 s γ μ ( 1 γ 5 )d 0 e iθ CP

29 K 0 M V A K 0 = 8 3 B K 0 d γ μ ( 1 γ 5 )s K 0 2 e iθ CP = 4B K f 2 Km K _ 3 ν 0 e e iθ CP m K : Kaon mass f K : decay constant s s 0 K 0 K = f _ _ K / 2 d u M 12 = G 2 F 12π f KB K m K m W λ 2 c η 1 S 0 ( x c )+ λ 2 t η 2 S 0 ( x t )+ λ c λ t η 3 S 0 x c, x t x c =(m c /m W ) 2, x t =(m t /m W ) 2 ( ) ( ) [ ] e i π θ CP η 1 = 1.38 ± 0.20 η 2 = 0.57 ± 0.01 η 3 = 0.47 ± 0.04 QCD corrections If no CP CKM is real arg M 12 = θ CP + π (if B K > 0)

30 λ t2 = V ts 2 V td 2 λ c2 = V cs 2 V cd 2 A 4 λ 10 λ 2 S 0 (x t ) 2.5 S 0 (x c ) λ c λ t = V cs V cd V ts V td A 2 λ 6 S 0 (x c, x t ) total contributions the biggest contribution to M 12 is from the charm loop Standard Model Γ 12 for the kaon system K 0 2π, 3π, 2πγ, 2πe + e, etc. K 0 Theoretical calculation on Γ 12 very difficult.

31 However, M 12. Γ 12 are measured experimentally; arg M 12 can be determined from the short distance interactions. arg M 12 = Im M 12 M 12 = 2ImM 12 m S m L theoretical short distance calculation experimental value argγ 12 = θ CP in the CKM phase convention ε = M 12 Γ 12 Δ M Γ 4 M Γ 1+ i 2 M Γ 12 Reε = ΔmΔΓ 4Δm 2 + ΔΓ 2 arg M 12 Standard Model

32 Im λ t 2 = Im V ts 2 V td 2 Im λ c 2 = Im V cs 2 V cd 2 Im λ c λ t = Im V cs V cd V ts V td 2A 4 λ 10 ( 1 ˆ ρ )ˆ η ( 1 ˆ ρ )ˆ η 2A 2 λ 6 η η 2A 2 λ 6 η ˆ η ˆ S 0 (x t ) 2.5 S 0 (x c ) S 0 (x c, x t ) ( 1 ˆ ρ )ˆ η η η ˆ The t-t and t-c box diagrams dominate Im M 12

33 The ε in the Standard Model u W + W + s d d s u, c, t u, d u γ u, d d d d d u, c, t s d u, c, t s d W + u, d u, d W + u, d γ Z 0 u, d d d d d I = 0 and I = 2 V ud V us V ud V us V cd V cs V td V ts tree real γ and Z 0 penguin complex W + s d u, d u, c, t g u, d d d I = 0 V ud V us V cd V cs V td V ts Phase difference between I = 0 and I = 2 weak amplitudes g penguin complex

34 QCD and non-perturbative effects very very complicated calculations!! Current Standard Model prediction ε +9 Re = (10 ± 2 ± 6) 10 4 ε 6 (Pich 04) Errors are due to - renormalization scale - m s - short-long distance QCD matching 1) The Standard Model predictions are compatible with the measurement. 2) Hadronic uncertainties in the theoretical predictions are too large to make a precision test.

35 Other interesting kaon decays: K + _ π + νν Z 0 ν W + ν s d u, c, t u u ν ν e,μ,τ W W s d u, c, t u u V td V ts V cd V cs A K πυυ = π + υ υ H eff K + H eff = G F α 2 2π sin 2 θ W ( λ c X l NL + λ t Xx ( t )) s γ μ ( 1 γ 5 )d l=e,μ,τ ( ( )υ L ) ( ) υ L γ μ 1 γ 5 Re( λ l c X NL ) , Im( λ l c X NL ) η ( ) ( 1 ˆ ρ ), Im( λ t Xx ( t )) η ˆ Re λ t Xx ( t ) real part imaginary part neglected

36 Use _ K + π 0 e + ν (data) for the hadronic matrix element. _ _ _ d s u s π + u K + = 2 π 0 u K + s γ μ ( 1 γ 5 )d isospin relation s γ μ ( 1 γ 5 )d Br prediction with a relatively small theoretical uncertainty. Current Standard Model predictions: Br(K + π + νν) = (0.80 ± 0.11) (isospin breaking taken into account) BNL787, 1995 data: (4.2 ) PRL based on one event More data taken, no new candidate BNL787, : (1.5 ) PRL Further data taken, one more candidate BNL787, : (1.57 ) PRL

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38 25 K + π + π K + μ + ν 10 5 one event P (MeV/c)

39 E787(95-98)/E949(02) combined results (1.47 ) PRL Standard Model prediction (8.8 ± 1.2) signal region The data is compatible with the Standard Model prediction. Could become a way to extract (ρ, η) from the kaon decays. K + π + π 0

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41 CERN NA62 under construction expect ~100 K + π + νν events

42 K L 0 π 0 νν CP violating decay CP(π 0 νν) = +1 Z 0 ν W + ν s d u, c, t d d ν ν e,μ,τ W W s d u, c, t d d π 0 υ υ H eff K 0 π 0 υ υ H eff K 0 a = π 0 υ υ( CPT ) ( CPT )H W ( CPT ) ( CPT )K 0 = a e i ( θ CP θ T ) CPT symmetry No hadronic final state interactions.

43 As for the K + decay, H eff = G F α 2 2π sin 2 θ W ( λ c X l NL + λ t Xx ( t )) s γ μ ( 1 γ 5 )d l=e,μ,τ ( ( )υ L ) ( ) υ L γ μ 1 γ 5 Re( λ l c X NL ) , Im( λ l c X NL ) η ( ) ( 1 ˆ ρ ), Im( λ t Xx ( t )) η ˆ Re λ t Xx ( t ) neglected - real part imaginary part - imaginary part is given by λ t

44 π 0 ν ν H W K L = 1 [ 2 a πυ υ ( 1 2ε)e iφ Γ a ] πυ υ = a πυ υ 2 = a πυ υ 2 [ 1 ( 1 2ε)e i ( 2φ πυ υ +θ T θ CP +φ Γ )] [ 1 ( 1 2ε)e i ( 2φ πυ υ 2φ 0 )] Δφ πυ υ = φ πυ υ φ 0 = a πυ υ [ 2 1 cos2δφ πυ υ + isin2δφ πυ υ 2εe i2δφ ] πυ υ = 2a πυ υ sin 2 Δφ πυ υ + isinδφ πυ υ cosδφ πυ υ + o ε () [ ] = 2ia πυ υ sinδφ πυ υ [ cosδφ πυ υ isinδφ πυ υ + o() ε ] 2ia πυ υ e iφ πυ υ sinδφ πυ υ e iδφ πυ υ = 2ia πυ υ sinδφ πυ υ eiφ 0 π 0 ν ν H W K L = 2 a πυ υ sinδφ πυ υ 2 Im a πυ υ (CKM phase convention) CP due to the interplay between decays and oscillations much bigger than CP in the oscillations different from 2π

45 Imaginary part can come only from s t d. _ hadronic part: K 0 π 0 ν ν = K + π 0 e + ν _ d _ s _ u _ s π 0 d K 0 = π 0 u K + s γ μ ( 1 γ 5 )d s γ μ ( 1 γ 5 )d Br = Br τ L 3α 2 K L π 0 υυ K + π 0 e + υ τ + V 2 us 2π 2 sin 4 θ W * Im( V td V ts )Xx t ( 3.0 ± 0.6) [ ( )] 2 Only one term simpler than K + decays more reliable calculation Br(K 0 π 0 ν ν) < % CL (KTeV, PLB 99) future experiment planned but a tough experiment π 0 γγ π 0 eeγ

46 JPARC E14: few K 0 π 0 νν if SM

47 Note: π 0 ν ν H eff K S = a πυ υ 2 = a πυ υ 2 [ 1+ ( 1 2ε)e i ( 2φ πυ υ +θ T θ CP +φ Γ )] [ 1+ ( 1 2ε)e i ( 2φ πυ υ 2φ 0 )] = a πυ υ [ 2 1+ cos2δφ πυ υ isin2δφ πυ υ 2εe i2δφ ] πυ υ = 2a πυ υ cos 2 Δφ πυ υ isinδφ πυ υ cosδφ πυ υ + o ε () [ ] 2a πυ υ cosδφ πυ υ [ cosδφ πυ υ isinδφ πυ υ ] = 2 a πυ υ cosδφ πυ υ e iφ 0 η π 0 ν ν π 0 ν ν H eff K L π 0 ν ν H eff K S = i tan Δφ πυ υ O 1 () η + ε = O( 10 3 )

48 Theoretical accuracy of the Standard Model predictions in the kaon sector will be limited to >10% (may be) except K 0 _ π 0 νν which will be experimentally challenging!