Schematic Rho Eta Plot. Bound on γ

Transcription

1 Strategies For The Determination of ffi 3 (fl) In B! DK CP David Atwood BCP4 February 19, 2001

2 Outline ffl What are we Measuring? ffl The Basic Idea ffl Methods to Extract fl: How well can we do? A Single Two Body Mode Two or More Two Body Modes A Single Three Body Mode ffl Impact of DD oscillations. ffl Conclusions

3 What are We Measuring? ffl We want to measure the CP odd phase difference between between b! ucs and b! ucs. ffl In the standard model, this phase is fl. ffl For interference to occur: Some trick must be used to match the initial and final states. There must be a strong phase difference to produce manifest CP violation. u s c s c W u γ W b b ffl CP violation implies lower bound on fl

4 Why is fl Important? ffl Confirm or refute the standard model: The Standard Model predicts that all the CKM phases and magnitudes fit consistently into an unitary matrix. If this fails, the Standard Model with three generations must be extended. ffl Observed CP violation will imply a lower bound on fl. ffl Within the unitary triangle picture a lower bound on fl will be complimentary to the information from Bs oscillations.

5 Schematic Rho Eta Plot η B s B s ρ Bound on γ

6 The Basic Idea ffl We need to dress up final and/or initial states to allow interference. ffl There are two possible ways to do this 1. B s =B s! D + s K : Here B s oscillation does the job. 2. B! D 0 =D 0 K : Interference will occur if a common D 0, D 0 final state is observed (eg. ß + ß ). (**This talk**) [Gronau, London, Wyler 91; DA Dunietz Soni 97]

7 + D s s c u K W s K u s c Ds W s s b b Bs X Bs s K u c D o u X D o u c s K u W W b u B This talk b B u

8 Common Final States ffl In principle, any hadronic state is common to D 0 and D 0. ffl Each strategy to determine fl, needs overcome the following difficulties: 1. B! D 0 K is about 100 bigger than B! D 0 K 2. It is probably not possible to measure Br(B! D 0 K ) independently. ffl In all cases, one has to worry about the possibility of D 0 D 0 oscillations.

9 Why can't we measure Br(B! D 0 K )? ffl If the D 0 decays hadronically, it will interfere quantum mechanically with the same decay mode of the D 0. (impossible) K + K + - K o D small (DCS) π π - - BIG D o - K Big small (color) - B - B ffl If the D 0 decays semi-leptonically, it is subject to a O(10 5 ) background from the direct semileptonic decay of the parent B to the same sign lepton. (difficult) B! K [D 0! e ν e + X] versus B! e ν e + X

10 Methods to Extract fl ffl In this section, I will discuss the extraction of fl by using the following types of D 0 =D 0 decay modes: 1. D 0 =D 0 to a single two body mode. 2. D 0 =D 0 to two or more two body modes. 3. D 0 =D 0 to a single three body final state.

11 D 0 =D 0 To A Single Two Body Mode. ffl If only one D 0 =D 0 decay modes is observed, there is not enough information to determine fl, however if there is large CP violation, a restrictive lower bound may be placed on sin 2 fl. ffl To enhance CP violation, it is best to consider states where D 0! X is DCS and D 0! X is CA. [DA Dunietz Soni 1997, 2000] ffl In this case, the two channels have roughly equal magnitude giving potentially large CP asymmetries.

12 ffl The free parameters of the system are: 1. fl, the total weak phase difference. 2. ο the total strong phase difference. 3. The Branching ratio a = Br(B! K D 0 ). 4. The branching ratio: b = Br(B! K D 0 ). 5. The branching Ratio c = Br(D 0! X) 6. c = Br(D 0! X) 7. The total rates d = Br(B! K [X]) and d = Br(B +! K + [X]) ffl In addition d and d are each functions of ffl, ο, a, b, c, cg so there are two equations in 3 unknowns! one parameter to nail down. ffl Thus, given a set of observations, b is a function of fl (in fact sin 2 fl) U 2 b 2 sin 2 fl 2Ub(z + 2 cos 2 fl)sin 2 fl + z 2 sin 2 fl + y 2 cos 2 fl = 0 U = (c=ac); z = (d + d)=(2ac) 1; y = (d d)=(2ac): ffl Larger CP violation results in a more restrictive lower bound on sin 2 fl and more restrictive bounds on b: sin 2 fl 1 2 (1 + z)(1 r 1 (y=(1 + z)) 2 ) (1 r z + jyj + 1) 2» u» (1 + r z jyj + 1) 2

13 Ub gamma (deg) ffl Here z = 1:5 while y = 0, 1, 1:5. ffl Note that for a given value of b, the two angular branches are the strong phase and the weak phase.

14 10 4 b(k*) (In Units of 10**-6) gamma (deg) ffl Numerical estimate for B! K Λ [D 0! X] Red is for X = K + ß Blue is for X = K s ß 0 ffl Assuming: a(k Λ ) = 6: ; b(k Λ ) = 1: fl = 90 ffi ; ο = 30 ffi for both modes. Thus d = ; μ d = 8: ff CP (K + ß ) = 0:47 ff CP (K s ß 0 ) = 0:12

15 ffl In the above example, if N B =(acceptance) = 10 8, then the 95% cl bound on fl is about 10 ffi below the ideal bound. ο ffi fl ffi fl min (95% cl) fl min

16 Possible methods to get around b ffl Use many modes and take the best lower bound (see also 3 body case to follow) ffl Use a model to estimate a range in b ffl Check an analogous decay where cross channel interference is smaller: B 0! K Λ [D 0! K + ß ]; Λ b! Λ[D 0! K + ß ] where the cross channel interference should only be ο 30%: can bound b to a possibly useful range. ffl Use a two or more of quasi-two body modes simultaneously! see next.

17 Two or More Two Body Modes ffl The data from two different modes will in general intersect in 4 points in the fl b plane. ffl Three or more modes will, in general, intersect only at the correct point the strong phases can be read off of the other branches of the curves. ffl For a sample calculation I will feed in the following randomly selected strong phases: a(k Λ ) = 6: b(k Λ )=1: Mode Br(D 0! final state) Br(D 0! final state) K + ß (2:9 ± 1:4) : K + ρ 3: : K + a 1 7: : K Λ+ ß 8: : Branching ratios in units of Mode d i d i 2 i + d i ) ff 0 ο i K + ß K s ß K + ρ K + a K s ρ K Λ+ ß

18 ffl Just two modes used: K + ß (solid) K s ß 0 (short dashes) ffl Confidence regions assuming that N B =acceptance = 10 8 : 99%; 90%; 68% b (10^-6) gamma deg

19 ffl All the modes used: ffl K + ß (solid) K s ß 0 (short dashes) K + ρ (long dashes) K + a 1 (dash-dot) K s ρ 0 (dash-dot-dot) K Λ+ ß (dash-dash-dot) ffl Confidence regions assuming that N B =acceptance = 10 8 : 99%; 90%; 68% b (10^-6) gamma deg

20 ffl Projecting the normalized likelihood distribution onto the fl axis in the cases where fl = 15 ffi ; 30 ffi ; 60 ffi and 90 ffi. ffl Confidence regions assuming that N B =acceptance = 10 8 : 99%; 90%; 68% gamma deg

21 What States Can be Used? ffl One can change D 0 to any excited D 0Λ state ffl One can change K Λ to any excited K Λ state ffl The same analysis applies if there is only one amplitude. ffl If there are multiple amplitudes (D Λ K Λ ) one can consider angular momentum analysis of the decay. [Sinha and Sinha 98] ffl Each combo must be placed on a separate fl b plot. ffl If you assume that Br(B! D 0 K Λ ) = Br(B 0! D 0 K Λ0 ) then we can also include B 0 decays on the same fl b plot. [DA in progress] ffl Likewise, given Br(Λ b! ΛK ) = Br(B! D 0 K ) then we can also include Λ b decays on the same fl b plot. ffl Phase information from a charm factory could also be used to provide additional constraints to a global fit. [Soffer 98]

22 A Single Three Body Final States ffl Many of the quasi-two body states are channels in the same three body final state, for instance: D 0! K Λ0 ß 0 ; K Λ+ ß ; K + ρ ;! K + ß ß 0 ffl Each point in the Dalitz plot may be thought as a separate mode" so in principle, such a system offers an infinitude of modes" so there is enough information in such a final state to determine both fl and b. [DA Dunietz Soni 2000] ffl To extract the amplitude, one can fit the distribution to a resonance channel model. ffl We can also construct the lower bound on sin 2 fl for each point on the Dalitz plot. ffl Generally, the lest lower bound is exactly sin 2 fl. ffl The same is true for both lower and upper bounds on b. ffl Let us define f(q) as the fraction of the dalitz plot where the bound on sin 2 fl is better than q. ffl Using the phenomenological model from E687 for the CA decay and SU(3) for the DCS decay :::

23 Figure 5a B- gamma= 90.0 xi= t/md** s/md**2 B+ gamma= 90.0 xi= t/md** s/md**2 With =acceptance B 0 s!!!! s = (p ß + p K+ ) 2 and t = (p ß0 + p ß ) 2 ffl Large partial rate asymmetry

24 Figure 5b B- gamma= 90.0 xi= t/md** s/md**2 B+ gamma= 90.0 xi= t/md** s/md**2 ffl Little partial rate asymmetry but difference in distribution

25 1 0.8 pra xi (deg) N_B xi (deg) But what do you really need? ffl Blue curve: N 3ff B PRA. to see CP violation using only ffl Red curve: N 3ff B to see CP violation using the differences in distribution.

26 f(q)** q 0.5 t/md** s/md**2 ffl f(q) is the proportion of the Dalitz plot with sin 2 fl > q. Curves for sin 2 fl = 0:25; 0:5; 0:75.

27 The Impact of DD Oscillation ffl It has been suggested that resonant effects can give a SM value of y D ο :01 while physics beyond the SM could give x D ο :01 [M Gronau 99; H Nelson 99] ffl Consider B! K [D 0! K + ß ] where the last step is DCS. If there is O(1%) chance of the D 0 oscillating to a D 0, the CA decay D 0! K + ß will be comparable and disrupt the above analysis. ffl To analyze the expected magnitude of mixing effects in this system, let us introduce an expansion parameter μ 2 = b=a. ffl Then, numerically c=c = O(μ 2 ) and if mixing is near its maximum expected value, then x D = O(μ 2 ) or y D = O(μ 2 ) and μ ο 0:1. ffl In this regime it is valid to approximate: d(fi) = (d 0 + d 1 fi)e fi ; d(fi) = (d 0 + d 1 fi)e fi ; Where fi = D t and d 0 and d 0 are the decay rates that would obtain absent mixing. ffl Using the expansion in terms of μ, d 1 =d 0 and d 1 =d 0 are O(μ) thus mixing effects are expected to be ο 10%.

28 ffl Three ways to deal with mixing are: 1. Determine the mixing parameters elsewhere and fold them into the analysis. If the mixing parameters were known, the determination of fd 0, d 0 d 1, d 1 g would give enough information to extract fl and b. This is likely to be practical only if the mixing is very large. 2. Include the possible mixing as a systematic error in your final result. [Soffer and Silva 2000] An analysis of how the error propagates through estimates that a mixing bound of x D, y D < :01 will result in a systematic error in fl of about 10 ffi

29 3. Use time dependent information to eliminate contamination due to mixing. [DA Dunietz Soni 2000] If time dependent information is available, one can reduce to the unmixed case by convoluting the data with the weighting function w = 2 fi: d 0 = Z 1 0 d(fi) w dfi d 0 = Z 1 0 d(fi) w dfi The statistical cost of this approach is: N mixing = 2(1 + ff 2 )(1 + d 1 =d 0 ) 2 N no mixing ff is the (detector time resolution) D. N mixing is the number of B's required if mixing might be present, N no mixing would be the number of B's required if mixing is known to be absent.

30 Conclusions ffl Decays of the form B! D 0 K allow the opportunity to measure fl through the interference of b! c and b! u transitions. ffl All DCS decay modes of the D 0 should be checked for CP violation. If any are found, a bound on fl may be established. ffl Data from at least three modes can be used to determine fl but as many modes as possible should be used. ffl Three body modes contain additional phase information; potentially one mode alone could be used. ffl Respectable measurements of fl may happen if N B =(accept) ο 10 8 ffl The potential for large (1%) DD μ oscillations introduces about O(10 ffi ) systematic uncertainty in fl. Time dependent information can eliminate this at the cost of a factor of ο 2 in statistic.