Things I wish I d known when I was a graduate student...that are related to particle physics (part 2)

Transcription

1 Things I wish I d known when I was a graduate student...that are related to particle physics (part ) Jeffrey D. Richman University of California, Santa arbara TM AAR Physics Analysis School Feb. 1 13, 8

2 Outline Lecture 1 Constants, natural units, and order-of-magnitude estimates Overview of physics violation as an interference effect 3 kinds of violation Direct violation Lecture Thinking about symmetries Meson oscillations Time-dependent violation in decays Conclusions

3 Thinking about Symmetries Symmetries are fundamental to understanding the forces of nature. We characterize interactions by the symmetries they possess. In quantum mechanics, symmetries are nearly always represented by unitary transformations (U). ψ ψ = U ψ = = = ψ ψ Uψ Uψ ψ U U ψ ψ ψ [ U, H] = U is a symmetry of H If U is a symmetry, then U modifies the state vector...while preserving its norm. d d H ψ = i H U i U dt ψ ψ = dt ψ ( ) ( ) Solution to Schrodinger eq n. Also a solution to Sch. eq n.

4 Continuous Symmetry Transformations Continuous symmetry transformations can be written as a function of a real parameter θ, which can be a vector of parameters. U e G G U iθ G ( θ ) = ( where = since is unitary) I i δθ G for small Generator of the transformation: a QM observable! Example: the translation operator is U x e e + + δθ ip x / i( Pxx Pyy Pzz)/ ( ) = = Suppose: [ HU, ( x)] = for arb. translational invariance along xˆ [ H, P xˆ ] = x then momentum will be conserved (additively); see next slide

5 Conservation laws from continuous symmetry or transformations ψ a( p a) ψ c( p c) momentum ψ ( p ) ψ ( p eigenstates ) b b d d c, d H a, b = c, d U UHU U a, b = ψc, ψd UHUψa, ψb + ip x / ip x / = ψc, ψd e He ψa, ψb + i[( pc+ pd ) ( pa+ pb = ψc, ψd H ψa, ψb e )] x / ψ ψ ψ ψ ψ ψ ψ ψ p + p = p + p a b c d ψ, ψ H ψ, ψ = c d a b [ UH, ] = Momentum is additively conserved! (or else transition is not allowed)

6 Discrete Symmetry Transformations Discrete symmetry transformations cannot be written in terms of a continuous parameter. Such transformations must be performed in a single jump. C: a a charge conjugation P: r r parity p e e + C s e P e s p reversed T: t t motion reversal e T incoming state e outgoing state

7 P and C violation in polarized muon decay =spin direction = momentum direction reminder: discuss issue of orbital angular momentum e ν e ν e e ν e e + μ ν μ P ν μ μ C ν μ μ + Allowed Not Allowed Allowed ( Γ ) ( ) 1 Γ = ( Γ =Γ) 3 1 P and C are individually violated maximally in the weak interactions, but combined is a good symmetry even for most weak processes!

8 Conservation laws from discrete symmetry transformations ψ ( η ) a a ψ ( η ) b b ψ ( η ) c c ψ ( η ) d d C eigenstates for simplicity = ψ, ψ [ HC, ] ψ, ψ = ψ, ψ Hηη ηη Hψ, ψ c d a b c d c d a b a b = ( ηη ηη ) ψ, ψ H ψ, ψ c d a b c d a b η η or = ηη c d a b ψ, ψ H ψ, ψ = c d a b C is unitary & hermitian (discrete xf) The C eigenvalue is multiplicatively conserved! (or else transition is not allowed)

9 Cosmology: Sakharov s three conditions A. Sakharov (1967): How to generate an asymmetry between N(baryons) and N(anti-baryons) in the universe (assuming equal numbers initially)? 1. aryon-number-violating process. oth C and violation (particle helicities not relevant to particle populations) 3. Departure from thermal equilibrium Nbar Nanti-bar Γ( X Y ) Γ( X Y ) Δ ( ) i i i i hypothetical heavy particle We appear to owe our existence to some form of violation at work in the early universe.

10 A closer look at oscillations A single, general formalism based on time-dependent perturbation theory describes meson oscillations in K, D,, and s systems. KK D D s s PP K s d W uct,, uct,, Common formalism, but very different parameter values behavior in each case is very different! Key point: since the weak interactions induce transitions between P and P, these flavor-eigenstate particles are not eigenstates of the total Hamiltonian, and they do not have definite masses or lifetimes. (They are superpositions of states P L and P H that do. Want to calculate Δm and ΔΓ!) s d K

11 Discovery of Oscillations ARGUS experiment (1987) ϒ(4 S) D D D D D D 1 * + * 1 1 μ1ν1, 1 π1 * + * μν, 1 π 13 pb -1 ~ 11, pairs χ =.17 ±.5 d ARGUS, PL 19, 45 (1987) Time-integrated mixing rate: 1% (fig. courtesy D. MacFarlane)

12 Time-dependent oscillation measurement e e + ( 4S) ϒ = bb βγ =.55 CM frame boosted with respect to lab! = bd = bd D π D π + * + Correlated oscillations (EPR) ( Δ t = ) decay (tag) evolution now uncorr.! Δt> Δt< Δ z = βγ cδt * D π D π + * + *

13 Innermost Detector Subsystem: Silicon Vertex Tracker e beam pipe: R=.79 cm Installed SVT Modules ( mesons move.5 mm along beam direction.)

14 Measuring the oscillation frequency dn dt dn dt nomix mix 1 Γt = e + Δmd t 4τ [ 1 cos( )] 1 Γt = e Δmd t 4τ nomix [ 1 cos( )] dn dn very simple! dt nomix dt mix A mix = = cos( Δm t) dn dn + dt dt amplitude=1 How do you actually do this measurement? asic question: did the oscillate or not? Need to know this as a function of time! 1. When it was produced, was the meson a or?. When it decayed, was the meson a or a? 3. What is the time difference between production and decay? mix

15 Mixing asymmetry vs. Δt Δ m = -1.5 ps (fixed to PDG'4) D<1 due to mistags T=π/Δm τ Β =1.6 ps run out of events at long lifetimes A ( ) (, ) NoMix( t) Mix( t) N t N t () t = = NoMix( t) + Mix( t) N( t) + N(, t) mix

16 Does a mass really have units of s -1? A = cos( Δm t) mix 1. Put in c ( Δ ) mc t ET. Divide by ET since phase must be dimensionless ( Δmc ) ( Δmc ) t = dimensionless!.5 ps ( Δ mc ) = (.5 1 s ) (66 1 ev 1 s) 3 1 ev Explains why we don t worry about H and L in most analyses!

17 Common formalism for P P oscillations H = H + H w Weak interactions (perturbation) induce Strong and EM interactions (create bound states) Ψ t = a t P + b t P + c t f P () () () f () f We can recast the formalism in terms of a -dimensional vector space in which we only include P and P. 1 = P = 1 important! at () H11 H1 at () at () H i bt () = H1 H = bt () t bt () P P P P f f see my Les Houches lectures H H

18 Solving the eigenvector problem in timedependent perturbation theory Eigenvectors of unperturbed Hamiltonian; notation H P = m P H P = m P H f = E f f Unperturbed energies of these states depend on quark masses, strong interations, and EM interactions that bind quarks into mesons. P = e P iθ P = e P iθ ( ) P = P Work in arbitrary phase convention; keep unphysical phases explicit. physical results must not depend on them! H = P H P 11 H = P H P 1 1 H = P H P 11 H = P H P H = H + Hw

19 The two classes of transitions in mixing Get specific form of H in terms of matrix elements of H w. P H H11 H1 M M1 i Γ Γ1 = * * H1 H = M 1 M Γ 1 Γ H M 1 H * 1 1 Off-shell (virtual) intermediate states mass matrix P P crucial decay matrix factor! * M = M 1 1 Off-shell (virtual) intermediate states P On-shell (real) intermediate states Γ i 1 Γ On-shell (real) intermediate states = Γ i i * 1 1

20 Μ ij ij, Γ ij, and the story of the factor i Results from time-dependent perturbation theory M = mδ + i H j + P ij ij w in SM f ih f f H j w m E f w Γ = π ih f f H j δ( m E) ij w w f f Real/virtual separation comes from iε odd func of m -E f 1 m Ef ε lim = lim i + + ε ( m Ef) + iε ε ( m Ef) + ε ( m Ef) + ε 1 = P iπδ ( m Ef ) even func m E f

21 Solution to the eigenvalue problem H 1 H1 = H11 H = + H 1 I K H1 p p μ μ H = = q q 1 H H 1 1 To get eigenvalues of H, just add H 11 to μ μ ± = H ± ( H H ) i ± ± = M Γ 1/ 1/ * i * 1 1 Γ1 i 1 1 Γ1 q H M α = = p H M 1/ M = M ± ± Re( H1H1) 1/ Γ = Γ Im( H H ) Δ M = Re( H H ) ΔΓ= 4 Im( H H ) ± 1 1 1/ 1/ H, K have same eigenvectors. easy! 1/

22 Time evolution of the mass eigenstates The ratio α=q/p determines the eigenstates of H in terms of superpositions of the original flavor-eigenstates: 1 ( P ) + = P + α P 1+ α * i * q M1 Γ 1 α = i 1 ( P P α P ) p M1 1 = Γ 1+ α Since these are the eigenstates of H, their time dependence is simple! i ( ) 1 i M+ Γ+ t ( ) P+ () t = e P + α P 1+ α i ( ) 1 i M Γ t ( ) P () t = e P α P 1+ α exponential time dependence 1/

23 P flavor eigenstate Time evolution of states that are initially flavor eigenstates α P 1 1+ α P e e i M i M i Γ + + i Γ t mass eigenstate + + t mass eigenstate P () t = f () t P + α f () t P + P () t no longer a flavor eigenstate P () t = f () t P + α f () t P + 1 P () t = f () t P + f () t P α + 1 f+ () t = e e + e e 1 f () t = e e e e ( im t t / im t t /) + Γ+ Γ ( im t t / im t t /) + Γ+ Γ

24 Violation in Oscillations P M 1 P P * M 1 P Γ i 1 Γ i * 1 Amp( P P ) Amp( P P ) M 1 = transition amplitude via intermediate states that are virtual (off-shell) Γ 1 = transition amplitude via intermediate states are are real (on-shell: both P and P can decay into these!) The -i is a conserving phase factor. It doesn t change sign! M 1 and Γ 1 behave like -violating phase factors, as long as they are not relatively real.

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26 First Observation of T-violation ( LEAR)

27 Violation in Oscillations: Visualization Equivalent statement of condition for violation in mixing: M 1 and Γ 1 must not be collinear and both must be nonzero. Im( M Γ ) = M Γ sin( θm θ Γ ) * no violation in mixing violation in mixing Overall rotation is merely a non-physical phase convention.

28 τ τ Oscillations in the K K System Most striking feature of K K system: huge lifetime splitting between mass eigenstates. (This is quite different from the system, where the mass splitting is very small!) ( K ) S ( K ) L 5 ns 15.5 m.9 ns.7 cm 58 ( K ) ( ) ( ) L KS KS ΔΓ = Γ Γ Γ 1 s 1-1 ( ) ( ) L S Δ M = M K M K = (.534 ±.14) 1 s ev Major experimental implication: a neutral K beam evolves over distance into a nearly pure K L beam. ΔΓ ΔM The mass and lifetime splittings are comparable!

29 Violation in mixing: observation of K L π + π - Exploit the large lifetime difference between the two neutral K mass eigenstates. π + K K S component decays away K L decay region π Demonstrates that K L decays into both =-1 (usually) and =+1 final states K L is not a eigenstate. η + ( L π π ) ( S π π ) A K ( L π π ) ( S π π ) A K + + A K η A K Key point: K L beam is self-tagging. (Tagging = method in which we identify a particle P 1 by studying a particle P that is produced in association with particle P 1.) both are x1-3

30 Phenomenology of Oscillations Oscillations in the and KK systems have very different parameters! This is due to different CKM factors and different intermediate states in the mixing diagrams. Amp( ) Amp( ) M 1 * M 1 Γ i 1 (suppressed) Γ M 1 : Dominated by tt intermediate states; can be calculated reasonably well using input from lattice QCD Γ 1 : Small! Few on-shell intermediate states that both and can reach. (These are the states both can actually decay into.) i * 1

31 Γ 1 is small in oscillations In the neutral -meson system, the common modes that both and can decay into have small branching fractions, since These decays usually lead to different final states. There are some s exceptions, but the branching fractions are small. Examples: SM predicts b c and b c (, ) ccdd (, ) uudd Cabibbo suppressed b u is CKM suppressed Γ M 1 1 = Om ( b / mt ) 1 3 Expect violation in mixing to be (1 ). O not yet observed

32 Implications of a small Γ 1 in oscillations 1/ * i * M 1/ 1 1 * Γ 1 1 q H M α = = = = 1 p H i α = 1 M M1 1 Γ 1 Γ 1 =Γ =Γ No violation in mixing since don t have second amplitude with non-zero -conserving relative phase. Prob at at 1 1 cos( ) Prob at at 1 1 cos( ) Prob at at 1 1 cos( ) Prob at at 1 1 cos( ) ( ) Γt P t P t = = e [ + ΔMt ] ( ) Γt P t P t = = e [ ΔMt ] ( ) Γt P t P t = = e [ ΔMt ] ( ) Γt P t P t = = e [ + ΔMt ] used to have α, 1/ α

33 Time-dependent asymmetries from the interference between mixing and decay amplitudes y modifying the mixing measurement, we can observe whole new class of -violating phenomena: pick final states that both and can decay into. (Often a eigenstate, but doesn t have to be.) ( bd) no net oscillation f ( bd ) no net oscillation f net oscillation net oscillation Γ ( phys( t) f) Γ ( phys( t) f)

34 Time-dependent asymmetry measurement e e + ( 4S) ϒ = bb = = bd bd Correlated oscillations (EPR) ( ) ( cc) KS ( Δ t = ) decay (tag) Δt< Δt> evolution now uncorr.! ( ) ( cc) KS

35 Ingredients of the Asymmetry Measurement 1. Determine initial state: tag using other A ( ( t) f ) ( ( t) f ) ( ( t) f ) ( ( t) f ) Γ Δ Γ Δ ( Δt) Γ Δ +Γ Δ. Reconstruct the final state system. 3. Measure Δt dependence Different final states f provide access to different CKM elements and hence different -violating phases. Reason for time dependence: one of the amplitudes is due to mixing.

36 Yield (events) Data: tagged signal events for J/ψ K s and other η = -1 sinβ modes ( ) ( cc) K ( η = 1) S 437 events, purity=9% signal region M ES (GeV/c )

37 Computing the asymmetry for final states common to and Time evolution of tagged states ΔMt ΔMt ( t) = e e cos - i α sin Γ t imt ΔMt 1 ΔMt ( t) = e e cos - i sin α Γ t imt 1 We have used ΔΓ/Γ<<1 and set Γ Γ1 Γ M = ( M+ + M ) Goal: calculate f H t () f H t () f = eigenstate Simply project above eq ns onto this!

38 Decay amplitudes for (t) f vs. (t) f f H t f H ( ) () t Γ t f H imt ΔM t ΔM t = e e f H cos - i α sin f H Γ t imt M t 1 fc P H Δ ΔM t = e e f H cos - i sin α f H blue: mixing green: decay The key quantity in these asymmetries is: f H f H λ α = f H f H * * i M1 Γ1 i M1 Γ1

39 Calculation of the time-dependent asymmetry A f () t = () () f H t f H t () + () f H t f H t ( ) ( () t f ) () t f ( ) ( () t f () ) t f Γ Γ = Γ +Γ A ( t) = S sin( Δm t) - C cos( Δm t) f S Im( λ) 1 λ = C = 1+ λ 1+ λ 1 decay amplitude: λ = 1 S = Im ( λ), C = A () t = Im( λ) sin( Δm t) f

40 Results on sinβ from charmonium modes (cc) K S ( odd) modes J/ψ K L ( even) mode asymmetry is opposite! sinβ =.7 ±.4 (stat) ±.3 (sys) λ =.95 +/-.31 (stat) +/-.13 (sys) 7 M events (raw asymmetry shown above must be corrected for the dilution)

41 * i * 1 1 i 1 Γ1 Calculating λ f H q Af f H f M Γ λ = = M p A Factor from mixing α M M VV VV e * 1 1 * tb td * e iθ tb td i( θ + φ ) M Factor from decay assuming 1 decay amplitude: i( δ+ φd ) f H = a e η ( ) iθ i( δ φd ) f H = f e a e f f H D = η ( ) f e H i( θ + φ ) λ M D = η ( f ) e φ φ ( i ) (the unphysical phase, the strong phase, and a ALL cancel)

42 cos Δm t () t f How the magic works i( D ) ae δ + φ η ( f) cos () t Δm t D a e δ φ i( ) f Δm t D i η( f)sin a e e i( δ φ ) iφ M Δm t D i sin a e e i( δ + φ ) iφ M Amp( f ) Amp( ( t) f) In each case, the two interfering amplitudes have the same conserving phase from strong interactions, so it is irrelevant. At = Im( λ) sin Δmt ( ) ( )

43 Conclusions Measurements in the system are extraordinarily powerful probes of violation and the CKM matrix. From our observations so far, the CKM framework for quark couplings to the W-boson is confirmed, although there is still room for non-sm effects. Jim Cronin s Nobel Prize lecture:...the effect is telling us that at some tiny level there is a fundamental asymmetry between matter and antimatter, and it is telling us that at some tiny level interactions will show an asymmetry under the reversal of time. We know that improvements in detector technology and quality of accelerators will permit even more sensitive experiments in coming decades. We are hopeful then, that at some epoch, perhaps distant, this cryptic message from nature will be deciphered.

44 Jim Cronin s Nobel Prize lecture:...the effect is telling us that at some tiny level there is a fundamental asymmetry between matter and antimatter, and it is telling us that at some tiny level interactions will show an asymmetry under the reversal of time. We know that improvements in detector technology and quality of accelerators will permit even more sensitive experiments in coming decades. We are hopeful then, that at some epoch, perhaps distant, this cryptic message from nature will be deciphered.

45 ackup Slides

46 Some consequences of symmetries 1. Conserved quantum numbers = ψ [ H, G] ψ = ψ HG GH ψ b a b a = ( g g ) ψ H ψ b a b a or ψb H ψa =. Relations between amplitudes g b = g a (quantum number conserved) (no transition) = UHU H = U HU H φ ψ φ ψ φ ψ Uφ H Uψ = φ H ψ Same amplitudes for these transitions! 3. Existence of multiplets (states with same energies) [ HU, ] = Uψ HUψ = ψ Hψ

47 Testing for Violation of Symmetries 1. Non-conserved quantum numbers J P π + π = η = η η ( 1) = + 1 ( = ) P π π Violates parity (weak decay). +. roken relationships between amplitudes Γ Γ ( K + π ) ( K π + ) Violates 3. Masses of particles in multiplet not the same p mp = MeV/ c mn = MeV/ c n I-spin violation (quark masses, EM interaction)

48 Conjugate amplitudes when is conserved What is the relation between an amplitude and its conjugate? P iθ ( P) = e P P iθ ( P) = e P ( ) P = P Often, people choose a specific phase convention. I like to keep the non-physical phase explicit. A = f H P = f ( ) ( ) H ( ) ( ) P = = = [ θ( ) θ( )] [ θ( ) θ( )] [ θ( ) θ( )] i P f f ( ) H( ) P e Ae f H P e i P f i P f A = A assume [H,]= 1 if conserved

49 Probabilities vs. time: master equations ( ) 1 Γ+ t Γ t Γt Prob P at t P at t = = cos( ) 4 e + e + e ΔMt ( ) 1 Γ+ t Γ t Γt Prob P at t P at t = = α cos( ) 4 e + e e ΔMt ( ) 1 1 Γ+ t Γ t Γt Prob P at t P at t = = e e e cos( Mt) α 4 + Δ ( ) 1 Γ+ t Γ t Γt Prob P at t P at t = = e + e + e cos( ΔMt) 4 Notes: 1 ΔM M M+ Γ = Γ + +Γ To calculate, need 5 numbers: M, M, Γ, Γ, α + + ( ) ( P t P t ) ( P t P t ) α 1 Prob at at = Prob at at = ( P t P t ) ( P t P t ) T Prob at at = = Prob at at =

50 Apply the condition for violation in mixing What are the implications of violation for the state vectors? P P + p q 1 α = = p + q 1+ α Mass eigenstates not orthogonal! (H not hermitian). [ H, ] Since, expect that mass eigenstates are not simultaneously eigenstates. ( iθ ) P =+ = P + e P 1 1 ( iθ ) P = = P e P 1 1 You can verify that these are 1. eigenstates. If is violated, they are not mass eigenstates.

51 Condition for Violation in Oscillations conservation ([H,]=) therefore implies H e H = H = e iθ 1 iθ 1 1 H1 Conversely, there will be observable violation in the oscillations if * i * Amp( P P ) H M1 Γ1 1 q = = = α = 1 Amp( P P ) H i 1 M p 1 Γ1 Key point: for violation to occur in mixing, both M 1 and Γ 1 must be non-zero. violation in mixing will not occur due to interference of the amplitudes within M 1 (or Γ 1 ). This is why the formalism is so simple!

52 violation and the K S K L lifetime splitting violation is a small effect in K K oscillations. 1 KS = (1 ε) K (1 ε) K + + (1 + ε ) 1 KL = (1 ε) K (1 ε) K + (1 + ε ) 3 = ( ) + 1 α ε ε =.84 ± α KS Mostly =+1 can decay to π + π, π π faster decay rate K L Mostly =-1 decays to π π π, π + π π, πeν, πμν 3 body decays: slower decay rate

53 Experimental setup used for discovery of K L π + π - J.H. Christenson, J.W. Cronin, V.L. Fitch, and R.Turlay, Phys. Rev. Lett. 13, 138 (1964).

54 violation in K K oscillations: semileptonic decays The K has a slightly higher probability of decaying as a K L than as a K S. ( + ) ( + K ) L π ν KL π ν ( + ) ( + ) L π ν L π ν L L L L Γ Γ K K K K δ = Γ K +Γ K K K + K K 1 α = = K K = (3.7 ±.1) 1 1+ α 3 L S sd sd δ gives direct measure of non-orthogonality of mass eigenstates. α 1 δ.9967

55 Time evolution of particles initially tagged as K ( K ) α 1 δ.9967 In fig., increase δ by 1X α = Γ+ t Γ t Γt cos( ) 4 e + e + e ΔMt 1 Γ+ t Γ t Γt α e e e cos( Mt) 4 + Δ 1 1 Γ+ t Γ t Γt e + e e cos( ΔMt) 4 α asymmetry

56 N N A = N + N Measurement of the charge asymmetry in K semileptonic decays + + after long time: pure K L decay time violation is a small effect in K K oscillations, but it is possible to observe it!

57 Direct violation in K decays In the neutral K decays, does violation occur only between mixing amplitudes, or does it also occur between decay amplitudes? Compare two violating amplitudes η ( + ) L π π ( ) S π π A K ( ) L π π ( ) S π π A K + + A K η ε ε η ε ε η ε 3η + + 3η ε 1 ( η η ) 3 + ε Re = (1.67 ±.6) 1 ε 3 A K violation from mixing only (Direct violation due to tree-penguin interference.) violation from interference between direct decay amps

58 oscillation frequency in the SM G Δ m = m f m S( x ) V V d d 6π F * d η d W t td tb pert. QCD correction meson decay constant Inami-Lim function G * F Δ ms = η ( ) mξ f s d dmws xt VtsVtb ξ 6π Δm m s s V = ξ Δm m V d ts d td 1 η =.551±.7 xt = ( mt / mw) ag constant f f s d ξ = 1.16 ±.5 s oscillations are very fast. Current limit: Δm s >14.4 ps -1. s d

59 Matrix form of the operator for -state system [, H ] = ( is conserved) What does this statement imply for violation in oscillations? P = e P iθ P = e P iθ e iθ e iθ conservation is equivalent to 1 ( ) ( ) H = H iθ iθ iθ e H1 e e H 1 iθ iθ iθ e H 1 = e e H1 (Only need to look at off-diag components of H.) = H 1 H 1

60 coherent wave function at the Y(4S) The and mesons are produced in a coherent quantum state. ϒ (4 S) must be in a C= -1 state, since the Y(4S) decay is a strong interaction process and conserves C. 1 Ψ ( t1, t) = t 1 1 ; p t ; p ± t1 ; p t ; p C=± ( ( ) ( ) ( ) ( ) ) Major implications 1. The asymmetry between the time-integrated decay rates is zero! At the Y(4S), you must measure Δt to perform a useful asymmetry measurement.. The two neutral mesons oscillate coherently until one of them decays. (example of Einstein-Podolsky-Rosen paradox)

61 Calculating λ for specific final states * * VV tb td VudVub π + π λ = Im( )=sin( ) * * VV tb td VudV λ α ub ( b uud ) VV VV V V J / ψ K = (-1) S λ Im( λ)=sin( β) VV VV V V ( ) ( ) b ccs K K S (assuming only tree diagram for illustration) * * * tb td cs cb cd cs * * * tb td cs cb cd cs * * * VV tb td VV cs cb VcdVcs J / ψ KL λ= ( +1 ) Im( λ)=-sin( β) * * * VV tb td VV cs cb VcdVcs b ccs K K L ( ) ( )

62 Angles of the unitarity triangle Consider two complex numbers z 1 and z. z z = 1 1 = z e z e iθ 1 iθ z z / / z z 1 1 = e i( θ θ ) 1 = θ 1 arg z z 1 θ α = VV * td tb arg * VudVub V V β = * cd cb arg * VV td tb * VudV ub γ = arg * VcdVcb V V ud * ub γ α V V cd * cb VV td β * tb