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Part II CP Violation in the SM

Part II CP Violation in the SM. C hris P arkes. Outline. THEORETICAL CONCEPTS Introductory concepts Matter and antimatter Symmetries and conservation laws Discrete symmetries P, C and T CP Violation in the Standard Model Kaons and discovery of CP violation Mixing in neutral mesons

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Part II CP Violation in the SM

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  1. Part II CP Violation in the SM Chris Parkes

  2. Outline • THEORETICAL CONCEPTS • Introductory concepts • Matter and antimatter • Symmetries and conservation laws • Discrete symmetries P, C and T • CP Violation in the Standard Model • Kaons and discovery of CP violation • Mixing in neutral mesons • Cabibbo theory and GIM mechanism • The CKM matrix and the Unitarity Triangle • Types of CP violation

  3. Kaons and discovery of CP violation

  4. Whatabout the product CP? Weak interactions experimentally proven to: • Violate P : Wu et al. experiment, 1956 • Violate C : Lederman et al., 1956 (just think about the pion decay below and non-existence of right-handed neutrinos) • But is C+P  CP symmetry conserved or violated?  +  Intrinsic spin C P  + + Initially CP appears to be preservedin weakinteractions …!  +  CP

  5. Kaon mesons: in two isospin doublets Part of pseudo-scalar JP=0- mesons octet with p, h Introducing kaons K+ = us Ko = ds I3=+1/2 I3=-1/2 Ko= ds K- = us S=+1 S=-1 • Kaon production: (pion beam hitting a target) Ko : - + p  o + Ko But from baryon number conservation: Ko : + + p  K+ + Ko + p Or Ko : - + p  o + Ko + n +n Requires higher energy S 0 0 -1 +1 S 0 0 +1 -1 0 Muchhigher S 0 0 +1 -1 0 0

  6. What precisely is a K0 meson? Now we know the quark contents: K0=sd,K0 =sd First: what is the effect of C and P on the K0 andK0particles? (because l=0q qbarpair) (because l=0q qbarpair) effect of CP: Bottom line: the flavoureigenstatesK0 andK0are notCP eigenstates Neutral kaons (1/2)

  7. Nevertheless it is possible to construct CP eigenstates as linear combinations Can always be done in quantum mechanics, to construct CP eigenstates |K1> = 1/2(|K0> +|K0>) |K2> = 1/2(|K0> -|K0>) Then: CP |K1> = +1 |K1> CP |K2> = -1 |K2> Does it make sense to look at these linear combinations? i.e. do these represent real particles? Predictions were: The K1 must decay to 2 pions given CP conservation of the weak interactions This 2 pion neutral kaon decay was the decay observed and therefore known The same arguments predict that K2 must decay to 3 pions History tells us it made sense! The K2 = KL (“K-long”) was discovered in 1956 after being predicted (difference between K2 and KL to be discussed later) Neutral kaons (2/2)

  8. How do you obtain a pure ‘beam’ of K2 particles? It turns out that you can do that through clever use of kinematics Exploit that decay of neutral K (K1) into two pions is much faster than decay of neutral K (K2) into three pions Mass K0 =498 MeV, Mass π0,π+/- =135 / 140 MeV Therefore K2 must have a longer lifetime thank K1 since small decay phase space t1= ~0.9 x 10-10 sec t2= ~5.2 x 10-8 sec (~600 times larger!) Beam of neutral kaons automatically becomes beam of |K2>as all |K1> decay very early on… Looking closer at KL decays Pure K2 beam after a while!(all decaying into πππ) ! K1 decay early (into pp) Initial K0beam

  9. The Cronin & Fitch experiment (1/3) Essential idea: Look for (CP violating) K2 pp decays 20 meters away from K0 production point π0 Decay of K2 into 3 pions Incoming K2 beam J.H. Christenson, J.W. Cronin, V.L. Fitch, R. Turley PRL 13,138 (1964) π+ Vector sum of p(π-),p(π+) π- If you detect two of the three pionsof a K2 ppp decay they will generallynot point along the beam line

  10. The Cronin & Fitch experiment (2/3) Essential idea: Look for (CP violating) K2 pp decays 20 meters away from K0 production point Decaying pions Incoming K2 beam J.H. Christenson et al., PRL 13,138 (1964) If K2 decays into two pions instead ofthree both the reconstructed directionshould be exactly along the beamline(conservation of momentum in K2 pp decay)

  11. The Cronin & Fitch experiment (3/3) K2 ppdecays(CP Violation!) Weak interactions violate CP Effect is tiny, ~0.05% ! K2 ppp decays K2 p+p-+X p+- = pp+ + pp- q = angle between pK2 and p+- If X = 0, p+- = pK2: cos q = 1 If X 0, p+-pK2: cos q 1 Note scale: 99.99% of K ppp decaysare left of plot boundary Result: an excess of events at Q=0 degrees!

  12. Almost but not quite!

  13. with |ε| <<1

  14. Key Points So Far • K0, K0 are not CP eigenstates – need to make linear combination • Short lived and long-lived Kaon states • CP Violated (a tiny bit) in Kaon decays • Describe this through Ks, KL as mixture of K0 K0

  15. Mixing in neutral mesons HEALTH WARNING : We are about to change notation P1,P2 are like Ks, KL (rather than K1,K2)

  16. Particle can transform into its own anti-particle • neutral meson states Po, Po • P could be Ko, Do, Bo, or Bso Kaon oscillations s d W- _ _ u, c, t u, c, t s W+ - d K0 K0 d s u, c, t _ _ W+ W- u, c, t s d • So say at t=0, pure Ko, • later a superposition of states

  17. Here for general derivation we have labelled states 1,2

  18. neutral meson states Po, Po P could be Ko, Do, Bo, or Bso with internal quantum number F Such that F=0 strong/EM interactions but F0 for weak interactions obeys time-dependent Schrödinger equation M, : hermitian 2x2 matrices, mass matrix and decay matrix mass/lifetime particle = antiparticle Solution of form No Mixing – Simplest Case

  19. neutral meson states Po, Po P could be Ko, Do, Bo, or Bso with internal quantum number F Such that F=0 strong/EM interactions but F0 for weak interactions obeys time-dependent Schrödinger equation M, : hermitian 2x2 matrices, mass matrix and decay matrix H11=H22 from CPT invariance (mass/lifetime particle = antiparticle) Time evolution of neutral mesons mixed states (1/4) H is the total hamiltonian: EM+strong+weak

  20. Solve Schrödinger for the eigenstates of H: of the form with complex parameters p and q satisfying Time evolution of the eigenstates: Time evolution of neutral mesons mixed states (2/4) Compare with Ks, KL as mixtures of K0, K0 If equal mixtures, like K1 K2

  21. Time evolution of neutral mesons mixed states (3/4) • Some facts and definitions: • Characteristic equation • Eigenvector equation: e.g.

  22. Time evolution of neutral mesons mixed states (4/4) • Evolution of weak/flavour eigenstates: • Time evolution of mixing probabilities: decay terms Interference term i.e. if start with P0, what is probability that after time t that have state P0 ? Parameter x determines “speed” of oscillations compared to the lifetime

  23. Hints: for proving probabilities Starting point Turn this around, gives Time evolution Use these to find

  24. Δmd = 0.507 ± 0.004 ps−1 xd = 0.770 ± 0.008 Δms= 17.719 ± 0.043 ps−1 xs = 26.63 ± 0.18 Lifetimes very different (factor 600) x = 0.00419 ± 0.00211

  25. Key Points So Far • K0, K0 are not CP eigenstates – need to make linear combination • Short lived and long-lived Kaon states • CP Violated (a tiny bit) in Kaon decays • Describe this through Ks, KL as mixture of K0 K0 • Neutral mesons oscillate from particle to anti-particle • Can describe neutral meson oscillations through mixture of P0 P0 • Mass differences and width determine the rates of oscillations • Very different for different mesons (Bs,B,D,K)

  26. Cabibbo theory and GIM mechanism

  27. In 1963 N. Cabibbo made the first step to formally incorporate strangeness violation in weak decays For the leptons, transitions only occur within a generation For quarks the amount of strangeness violation can be neatly described in terms of a rotation, where qc=13.1o Cabibbo rotation and angle (1/3) Weakforcetransitions u Idea: weak interaction couples to different eigenstates than strong interactionweak eigenstates can be writtenas rotation of strongeigenstates W+ d’ = dcosqc + ssinqc

  28. Cabibbo’s theory successfully correlated many decay rates by counting the number ofcosqcandsinqcterms in their decay diagram: Cabibbo rotation and angle (2/3) E.g.

  29. There was however one major exception which Cabibbocould not describe: K0 m+m-(branching ratio ~7.10-9) Observed rate much lower than expected from Cabibbo’s ratecorrelations (expected rate  g8 sin2qc cos2qc) Cabibbo rotation and angle (3/3) s d cosqc sinqc u W W nm m- m+

  30. The GIM mechanism (1/2) • In 1970 Glashow, Iliopoulos and Maiani publish a model forweak interactions with a lepton-hadron symmetry • The weak interaction couples to a rotated set of down-type quarks: the up-type quarks weakly decay to “rotated” down-type quarks • The Cabibbo-GIM model postulates the existence of a 4th quark : the charm (c) quark ! … discovered experimentally in 1974: J/Y  cc state 2D rotation matrix Leptonsectorunmixed Quark section mixed throughrotation of weak w.r.t. strong eigenstates by qc

  31. The GIM mechanism (2/2) • There is also an interesting symmetry between quark generations: u c W+ W+ d’=cos(qc)d+sin(qc)s s’=-sin(qc)d+cos(qc)s Cabibbo mixing matrix The d quark as seen by the W, the weakeigenstate d’, is not the same as the masseigenstate (the d)

  32. GIM suppression • The model also explains the smallness • of the K0 m+m-decay See alsoBsm+m-discussion later s d s d cosqc sinqc -sinqc cosqc u c W W W W nm nm m- m+ m- m+ expected rate  (g4sinqccosqc- g4 sinqccosqc)2 The cancellation is not perfect – these are only the vertex factors – as the masses of c and u are different

  33. The CKM matrix and the Unitarity Triangle

  34. Recall: Since H = H(Vij), complex Vij would generate [T,H]  0  CP violation only if: How to incorporate CP violation in the SM? • How does CP conjugation (or, equivalently, T conjugation)act on the Hamiltonian H ? Simple exercise: hence “anti-unitary” T (and CP) operation corresponds to complex conjugation ! CP conservation is: (up to unphysical phase) =

  35. Brilliant idea from Kobayashi and Maskawa(Prog. Theor. Phys. 49, 652(1973) ) Try and extend number of families (based on GIM ideas).E.g. with 3: … as mass and flavour eigenstates need not be the same (rotated) In other words this matrix relates the weak states to the physical states The CKM matrix (1/2) Kobayashi Maskawa ud’ c s’ t b’ Imagine a new doublet of quarks 3D rotation matrix 2D rotation matrix

  36. Standard Model weak charged current Feynman diagram amplitude proportional to VijUiDj U (D) are up (down) type quark vectors Vijis the quark mixing matrix, the CKM matrix for 3 families this is a 3x3 matrix u c t d s b U = D = The CKM matrix (2/2) Can estimate relative probabilities of transitions from factors of |Vij |2

  37. As the CKM matrix elements are connected to probabilities of transition, the matrix has to be unitary: CKM matrix – number of parameters (1/2) Values of elements:a purely experimental matter In general, for N generations, N2 constraints Sum of probabilities must add to 1 e.g. t must decay to either b, s, or d so • Freedom to change phase of quark fields 2N-1 phases are irrelevant (choose i and j, i≠j) • Rotation matrix has N(N-1)/2 angles

  38. CKM matrix – number of parameters (2/2) • NxN complex element matrix: 2N2 parameters Total - unitarity constraints - phase freedom: ‘free’ parameters (rotations +phases) Number of phases • Example for N = 1 generation: • 2 unknowns – modulus and phase: • unitarity determines |V| = 1 • the phase is arbitrary (non-physical) no phase, no CPV

  39. CKM matrix – number of parameters (2/2) • NxN complex element matrix: 2N2 parameters Total - unitarity constraints - phase freedom: ‘free’ parameters (rotations +phases) Number of phases • Example for N = 2 generations: • 8 unknowns – 4 moduli and 4 phases • unitarity gives 4 constraints : • for 4 quarks, we can adjust 3 relative phases only one parameter, a rotation (= Cabibbo angle) left: no phase  no CPV

  40. CKM matrix – number of parameters (2/2) • NxN complex element matrix: 2N2 parameters Total - unitarity constraints - phase freedom: ‘free’ parameters (rotations +phases) Number of phases • Example for N = 3 generations: • 18 unknowns – 9 moduli and 9 phases • unitarity gives 9 constraints • for 6 quarks, we can adjust 5 relative phases 4 unknown parameters left: 3 rotation (Euler) angles and 1 phase  CPV ! In requiring CP violation with this structureof weak interactions K&M predicteda 3rd family of quarks!

  41. CKM matrix – Particle Data Group (PDG) parameterization 3D rotation matrix form Define: Cij= cos ij Sij=sin ij • 3 angles 12, 23, 13 phase  VCKM = R23 x R13 x R12 C12 S12 0 -S12 C12 0 0 0 1 • 0 0 • 0 C23 S23 • 0 -S23 C23 R23 = R12 = C13 0 S13 e-i 0 1 0 -S13e-i 0 C13 R13 =

  42. CKM matrix - Wolfenstein parameters Introduced in 1983: • 3 angles  = S12 , A = S23/S212 ,  = S13cos/ S13S23 • 1 phase  = S13sin/ S12S23 A ~ 1, ~ 0.22, ≠ 0 but  ≠ 0 ??? VCKM(3) terms in up to 3 CKMterms in4,5 Note:smallest couplings are complex (CP-violation)

  43. CKM matrix - Wolfenstein parameters Introduced in 1983: • 3 angles  = S12 , A = S23/S212 ,  = S13cos/ S13S23 • 1 phase  = S13sin/ S12S23 A ~ 1, ~ 0.22, ≠ 0 but  ≠ 0 ??? VCKM(3) terms in up to 3 CKMterms in4,5 Note:smallest couplings are complex (CP-violation)

  44. CKM matrix - Wolfenstein parameters Introduced in 1983: • 3 angles  = S12 , A = S23/S212 ,  = S13cos/ S13S23 • 1 phase  = S13sin/ S12S23 A ~ 1, ~ 0.22, ≠ 0 but  ≠ 0 ??? VCKM(3) terms in up to 3 CKMterms in4,5 Note:smallest couplings are complex (CP-violation)

  45. CKM matrix - hierarchy Charge: +2/3 Charge: 1/3 ~ 0.22 top bottom strange charm flavour-changing transitions by weak charged current (boldness indicates transition probability  |Vij|) up down

  46. CKM – Unitarity Triangle • Three complex numbers, which sum to zero • Divide by so that the middle element is 1 (and real) • Plot as vectors on an Argand diagram • If all numbers real – triangle has no area – No CP violation • Hence, get a triangle • ‘Unitarity’ or ‘CKM triangle’ • Triangle if SM is correct. • Otherwise triangle will not close, • Angles won’t add to 180o Imaginary Real

  47. Unitarity conditions and triangles : no phase info. Plot on Argand diagram: 6 triangles in complex plane db: sb: ds: ut: ct: uc:

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