1. Niels Tuning (1) Particle Physics II – CP violation Lecture 1 N. Tuning

2. Niels Tuning (2) Outline 1 May –Introduction: matter and anti-matter –P, C and CP symmetries –K-system CP violation Oscillations –Cabibbo-GIM mechanism 8 May –CP violation in the Lagrangian –CKM matrix –B-system 15 May –B-factories –B  J/Psi Ks –Delta ms

3. Niels Tuning (3) Literature Slides based on the course from Wouter Verkerke. W.E. Burcham and M. Jobes, Nuclear and Particle Physics, chapters 11 and 14. Z. Ligeti, hep-ph/0302031, Introduction to Heavy Meson Decays and CP Asymmetries Y. Nir, hep-ph/0109090, CP Violation – A New Era H. Quinn, hep-ph/0111177, B Physics and CP Violation

4. Niels Tuning (4) Motivation – What is so interesting about CP violation? It’s about differences in matter and anti-matter –Why would they be different in the first place? –We see they are different: our universe is matter dominated Equal amounts of matter & anti matter (?) Matter Dominates !

5. Niels Tuning (5) Where and how do we generate the Baryon asymmetry? No definitive answer to this question yet! In 1967 A. Sacharov formulated a set of general conditions that any such mechanism has to meet 1)You need a process that violates the baryon number B: (Baryon number of matter=1, of anti-matter = -1) 2)Both C and CP symmetries should be violated 3)Conditions 1) and 2) should occur during a phase in which there is no thermal equilibrium In these lectures we will focus on 2): CP violation Apart from cosmological considerations, I will convince you that there are more interesting aspects in CP violation

6. Niels Tuning (6) What is evidence for the existence of anti-matter? Energetic photons produced in matter/anti-matter annihilation –Look at spectrum of photons in universe and look for spikes –Main problem: photons can not travel unlimited distances in the universe because of interactions with remaining cosmic background radiation and gases etc… –Conclusion: No anti-matter in 20Mpc radius. –How to look further into space? Better: Look for anti-Helium nuclei flying through space –Positrons, anti-protons can occasionally be produced in various processes, but producing anti-Helium is way too complicated by ‘regular means’: Only viable source of anti-Helium are fusion processes in ‘anti-stars’ –Presence/absence of anti-Helium says something about existence of anti-matter in distant regions of space –Large rest mass of Helium nuclei allows them to travel much further through space than photons  Conclusions of anti-Helium searches cover much larger region of space

7. Niels Tuning (7) The AMS experiment – Searching for He In essence as small particle physics experiment in space –AMS-01 brought to space through flight of Discovery shuttle –Can detect and identify many types of cosmic rays

8. Niels Tuning (8) Results of the AMS experiment Zero anti-helium found, plenty of Helium found –Rigidity of tracks is measure of particles momentum –Very high energy Helium nuclei have traveled from far  Says something about spatial reach of experiment ‘Universe with pockets of anti-matter’ hypothesis increasingly unlikely –Future AMS-02 experiment will (launch 2007) will have much increased range

9. Niels Tuning (9) Introduction – positron discovery by Anderson Result: discovery of a positively charged electron-like particle dubbed the ‘positron’ –Experimental confirmation of existence of anti-matter! Lead plate to slow down particle in chamber Incoming particle (high momentum / low curvature) Outgoing particle (low momentum / hi curvature)

10. Niels Tuning (10) Introduction – positron discovery by Anderson 4 years later Anderson confirmed this with   e + e - in lead plate using  from Thorium carbide source

11. Niels Tuning (11) Introduction – anti-neutrino: Savannah river Decisive experiment close to Savannah River nuclear reactor in South Carolina in 1956 (Nobel prize 1995) Idea: nuclear reactor provides enormous anti-neutrino flux from fission O(10 13 ) /cm 2 /sec –Try to detect inverse beta decay: + p +  n + e + (Beta decay: n  p + + e - + ) n e +  p + p +  n e + Cross over e - Invert reaction

12. Niels Tuning (12) Introduction – anti-neutrino: Savannah river How do you detect + p +  n + e + –Look for the positron through the reaction e + e -   and detect 2 photons produced simultaneously. Savannah river Detector: –Tank with 200 liters of water with 40 kg of CdCl 2 dissolved in it. –Surrounded by 110 photomultipliers for photon detection Clean signal found  direct proof of existence of neutrino –Nobel prize 1995  ν + n  p + + e - not observed –   ν ≠ ν, Lepton number must be conserved From inverse beta decay From detector material

13. Niels Tuning (13) Introduction - What about the other anti-particles? Dirac equation: for every (spin ½) particle there is an anti-particle –It took a bit longer, but more were discovered Anti-proton (1955) and anti-neutron (1955) using cyclotrons Reactions with particles and anti-particles –Q: How do you produce anti-particles anyway? –A: In pairs with particles, e.g.   e + e - But this is not the whole story as we will see later General rule: crossing symmetry –In any existing reaction you can move a particle through the arrow while turning it into an anti-particle –Example: e -   e -  (Compton scattering)   e + e - (Pair creation) Move e - to right Move  to left ( = )

14. Niels Tuning (14) Definition and discovery of C,P,CP violation

15. Niels Tuning (15) Continuous vs discrete symmetries Space, time translation & orientation symmetries are all continuous symmetries –Each symmetry operation associated with one ore more continuous parameter There are also discrete symmetries –Charge sign flip (Q  -Q) : C parity –Spatial sign flip ( x,y,z  -x,-y,-z) : P parity –Time sign flip (t  -t) : T parity Are these discrete symmetries exact symmetries that are observed by all physics in nature? –Key issue of this course

16. Niels Tuning (16) Example: People believe in symmetry… Instruction for Abel Tasman, explorer of Australia (1642): “Since many rich mines and other treasures have been found in countries north of the equator between 15 o and 40 o latitude, there is no doubt that countries alike exist south of the equator. The provinces in Peru and Chili rich of gold and silver, all positioned south of the equator, are revealing proofs hereof.”

17. Niels Tuning (17) Three Discrete Symmetries Parity, P –Parity reflects a system through the origin. Converts right-handed coordinate systems to left-handed ones. –Vectors change sign but axial vectors remain unchanged x  -x, p  -p, but L = x  p  L Charge Conjugation, C –Charge conjugation turns a particle into its anti-particle e +  e -, K -  K + Time Reversal, T –Changes, for example, the direction of motion of particles t  -t  

18. Niels Tuning (18) P-parity experiments Before 1956 physicists were convinced that the laws of nature were left-right symmetric. Strange? –A “gedanken” experiment: Consider two perfectly mirror symmetric cars: What would happen if the ignition mechanism uses, say, 60 Co  decay? “L” and “R” are fully symmetric, Each nut, bolt, molecule etc. However the engine is a black box Person “L” gets in, starts, ….. 60 km/h Person “R” gets in, starts, ….. What happens? “L” “R” Gas pedal driver Gas pedal driver

19. Niels Tuning (19) The situation in 1956 Nothing hints at the existence of any kind of Parity violating physics… –Reminder: Parity: (x,y,z)  (-x,-y,-z) –If universe is parity-symmetric, inverting all spatial coordinates would not changes laws of physics 1956: Lee and Yang publish a paper Question of Parity Conservation in Weak Interactions/ –Suggestion: Weak interaction might violate Parity symmetry. Originated from discussions at April HEP conference in Rochester, NY. Following Yang's presentation Richard Feynman brought up the question of non-conservation of parity. –Feynman himself later said, "I thought the idea (of parity violation) unlikely, but possible, and a very exciting possibility." Indeed Feynman later made a fifty dollar bet with a friend that parity would not be violated.

20. Niels Tuning (20) Parity symmetry – The situation in 1956 When the paper appeared, physicists were not immediately prompted into action. The proposition of parity non-conservation was not unequivocally denied; rather, the possibility appeared so unlikely that experimental proof did not warrant immediate attention. The physicist Freeman Dyson wrote of his reaction to the paper: "A copy of it was sent to me and I read it. I read it twice. I said, `This is very interesting,' or words to that effect. But I had not the imagination to say, `By golly, if this is true it opens up a whole new branch of physics.' And I think other physicists, with very few exceptions, at that time were as unimaginative as I."

21. Niels Tuning (21) Parity symmetry – the experiment Madame Wu –Another immigrant was now to play the next major role, Madame Chien-Shiung Wu. Arriving at Berkely in 1936 from Shanghai, Wu was one of the most ardently pursued coeds on campus. But she was also a hard worker who abhorred the marked absence of women from the American scientific establishment. She says, "... it is shameful that there are so few women in science... In China there are many, many women in physics. There is a misconception in America that women scientists are all dowdy spinsters. This is the fault of men. In Chinese society, a woman is valued for what she is, and men encourage her to accomplishments --- yet she retains eternally feminine." Idea from experiment in collaboration with Lee and Yang: Look at spin of decay products of polarized radioactive nucleus –Production mechanism involves exclusively weak interaction

22. Niels Tuning (22) Intermezzo: Spin and Parity How does the decay of a particle with spin tell you something about parity? Gedanken-experiment: decay of X  a + b –Spin: |1,1>  |½, ½ > + |½, ½> –It is important that X is maximally polarized: only then there is a single solution for the spin of the decay products. If not, e.g. |1,0>  |½, +½> + |½, -½> |1,0>  |½, -½> + |½, +½> 

23. Niels Tuning (23) Intermezzo: Spin and Parity and Helicity We introduce a new quantity: Helicity = the projection of the spin on the direction of flight of a particle H=+1 (“right-handed”) H=-1 (“left-handed”)

24. Niels Tuning (24) Intermezzo: Spin and Parity and Helicity Spin is quantized  Helicity is quantized –Possible H values for S=1/2: H=-1 and H=+1 –Most particles are linear combination of H=+1 and H=-1 states –Angular distribution for particles observed in specific helicity eigenstate: Superposition of H=+1 and H=-1 states I() RH = 1 - (v/c) cos  I() LH = 1 + (v/c) cos  H=+1 (“right-handed”) H=-1 (“left-handed”) + Constant If both helicities are produced equally in decay. If not angular distribution will not be flat

25. Niels Tuning (25) Note on Helicity Note that Helicity is not generally a Lorentz- invariant observable –Sign of particle momentum p is relative to observer. –A second observer ‘overtaking’ the particle from the lab observer perspective will see the particle moving in the opposite direction (p’ = -p)  It see the opposite Helicity Exception for massless particles: –You cannot overtake massless particles moving at speed of light –Helicity for massless particles is Lorentz-invariant intrinsic property

26. Niels Tuning (26) A realistic experiment: the Wu experiment (1956) Observe radioactive decay of Cobalt-60 nuclei –The process involved: 60 27 Co  60 28 Ni + e - + e – 60 27 Co is spin-5 and 60 28 Ni is spin4, both e- and e are spin-½ –If you start with fully polarized Co (S Z =5) the experiment is essentially the same (i.e. there is only one spin solution for the decay) |5,+5>  |4,+4> + |½,+½> + |½,+½> S=1/2 S=4

27. Niels Tuning (27) The Wu experiment – 1956 Experimental challenge: how do you obtain a sample of Co(60) where the spins are aligned in one direction –Wu’s solution: adiabatic demagnitization of Co(60) in magnetic fields at very low temperatures (~1/100 K!). Extremely challenging in 1956!

28. Niels Tuning (28) The Wu experiment – 1956 The surprising result: the counting rate is different –Electrons are preferentially emitted in direction opposite of 60 Co spin! –Careful analysis of results shows that experimental data is consistent with emission of left-handed (H=-1) electrons only at any angle!! ‘Backward’ Counting rate w.r.t unpolarized rate ‘Forward’ Counting rate w.r.t unpolarized rate 60 Co polarization decreases as function of time

29. Niels Tuning (29) The Wu experiment – 1956 Physics conclusion: –Angular distribution of electrons shows that only pairs of left- handed electrons / right-handed anti-neutrinos are emitted regardless of the emission angle –Since right-handed electrons are known to exist (for electrons H is not Lorentz-invariant anyway), this means no left-handed anti-neutrinos are produced in weak decay Parity is violated in weak processes –Not just a little bit but 100% How can you see that 60 Co violates parity symmetry? –If there is parity symmetry there should exist no measurement that can distinguish our universe from a parity-flipped universe, but we can!

30. Niels Tuning (30) Our universe vs a parity-flipped universe What happens to helicity in parity-flipped universe? –Momentum flips sign –Spin stays the same –Helicity is product and flips sign Conclusion: –Any process that produces right-handed anti-neutrinos in our universe will produce left-handed anti-neutrinos in the mirrored universe. –If left and right-handed neutrinos are not produced at the same rate the physics in the mirrored universe is different Direction of motion Orientation of spin Direction of motion Orientation of spin righthanded H=+1 lefthanded H=-1 P

31. Niels Tuning (31) Parity violation in weak decays Apply parity operation to 60 Co decay e- LH e Our universe e- RH e e- LH e e- RH e P-Flipped universe

32. Niels Tuning (32) Parity violation in weak decays Apply parity operation to 60 Co decay e- LH e Our universe (RH anti-neutrinos only) Allowed e- RH e Forbidden e- LH e Allowed e- RH e Forbidden P-Flipped universe (LH anti-neutrinos only) Preferential direction of electrons is forward Preferential direction of electrons is backward

33. Niels Tuning (33) So P is violated, what’s next? Wu’s experiment was shortly followed by another clever experiment by L. Lederman: Look at decay  +   +  –Pion has spin 0, ,  both have spin ½  spin of decay products must be oppositely aligned  Helicity of muon is same as that of neutrino. Nice feature: can also measure polarization of both neutrino ( + decay) and anti-neutrino ( - decay) Ledermans result: All neutrinos are left-handed and all anti-neutrinos are right-handed ++  

34. Niels Tuning (34) Charge conjugation symmetry Introducing C-symmetry –The C(harge) conjugation is the operation which exchanges particles and anti-particles (not just electric charge) –It is a discrete symmetry, just like P, i.e. C 2 = 1 C symmetry is broken by the weak interaction, –just like P ++   (LH) --  C OK

35. Niels Tuning (35) The Weak force and C,P parity violation What about C+P  CP symmetry? –CP symmetry is parity conjugation (x,y,z  -x,-y,z) followed by charge conjugation (X  X) ++ ++  ++  ++ Intrinsic spin PC    CP CP appears to be preserved in weak interaction!

36. Niels Tuning (36) What do we know now? C.S. Wu discovered from 60 Co decays that the weak interaction is 100% asymmetric in P-conjugation –We can distinguish our universe from a parity flipped universe by examining 60 Co decays L. Lederman et al. discovered from p+ decays that the weak interaction is 100% asymmetric in C-conjugation as well, but that CP-symmetry appears to be preserved –First important ingredient towards understanding matter/anti- matter asymmetry of the universe: weak force violates matter/anti-matter(=C) symmetry! –C violation is a required ingredient, but not enough as we will learn later Next: a precision test of CP symmetry conservation in the weak interaction

37. Niels Tuning (37) Conserved properties associated with C and P C and P are still good symmetries in any reaction 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)

38. Niels Tuning (38) Figuring out P eigenvalues for hadrons QFT rules for particle vs. anti-particles –Parity of particle and anti-particle must be opposite for fermions (spin-N+1/2) –Parity of bosons (spin N) is same for particle and anti-particle Definition of convention (i.e. arbitrary choice in def. of q vs q) –Quarks have positive parity  Anti-quarks have negative parity –e - has positive parity as well. –(Can define other way around: Notation different, physics same) Parity is a multiplicative quantum number for composites –For composite AB the parity is P(A)*P(B), Thus: –Baryons have P=1*1*1=1, anti-baryons have P=-1*-1*-1=-1 –(Anti-)mesons have P=1*-1 = -1 Excited states (with orbital angular momentum) –Get an extra factor (-1) l where l is the orbital L quantum number –Note that parity formalism is parallel to total angular momentum J=L+S formalism, it has an intrinsic component and an orbital component NB: Photon is spin-1 particle has intrinsic P of -1

39. Niels Tuning (39) Parity eigenvalues for selected 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| + >

40. Niels Tuning (40) Figuring out C eigenvalues for hadrons Only particles that are their own anti-particles are C eigenstates because C|x>  |x> = c|x> –E.g.  0,,’, 0,,, and photon C eigenvalues of quark-anti-quark pairs is determined by L and S angular momenta: C = (-1) L+S –Rule applies to all above mesons C eigenvalue of photon is -1 –Since photon is carrier of EM force, which obviously changes sign under C conjugation Example of C conservation: –Process  0   C=+1( 0 has spin 0)  (-1)*(-1) –Process  0   does not occur (and would violate C conservation)

41. Niels Tuning (41) Introduction to K 0 physics Now focusing on another little mystery in the domain of ‘strange’ mesons: what precisely is a K 0 meson? –Quark contents: K 0 =  sd, K 0 =  ds First: what is the effect of C and P conjugation on the K 0 /K 0 -bar particles? –P|K 0 > = -1|K 0 >(because  qq pair) –C|K 0 > = |K 0 >(because K 0 is anti-particle of K 0 & l+s=0) Knowing this we can evaluate the effect of CP on the K 0 –CP|K 0 > = -1|K 0 >

42. Niels Tuning (42) What are the K 0 CP Eigenstates Thus K 0 and K 0 are not CP eigenstates, –Somewhat strange since they decay through weak interaction which appears to conserve CP! Nevertheless it is possible to construct CP eigenstates as linear combinations –Remember QM: You can always construct wave functions as linear combinations of the solutions of any operator (like the Hamiltonian) –|K 1 > = 1/2(|K 0 > - |K 0 >) –|K 2 > = 1/2(|K 0 > + |K 0 >) –Proof is exercise for today Does it make sense to look at linear combinations of |K 0 > and |K 0 >? –I.e does |K 1 > represent a real particle?

43. Niels Tuning (43) K 0 /K 0 oscillations Well it might, because it turns out that the weak interaction can turn a K 0 particle into a K 0 particle! –Weird? Yes! Impossible? No! –In can be done using two consecutive weak interactions: Important implication for nature of K 0 particle –Q: If a K 0 can turn into a K 0 at any moment, how do you know what particle you’re dealing with? –A: You don’t. Any particle in the lab is always a linear combination of the two  It makes as much sense to talk about K 1 and K 2 eigenstates as it does to talk about K 0 and K 0 eigenstates K0K0 S=1 Weak force (X) Weak force S=0 S=-1 K0K0 S=-1

44. Niels Tuning (44) So what is the K 0 really? The K 0 meson is something you can only describe with quantum mechanics It is a linear combination of two particles! –You can either see it as a combination of |K 0 > and |K 0 >, eigenvalues of Strangeness (+1 vs +1) and undefined CP –Or you can see it as a combination of |K 1 > and |K 2 >, eigenvalues of CP (+1, -1), with undefined strangeness –Both representations are equivalent. Kaons are typically produced in strong interactions in a strangeness eigenstate (K 0 or K 0 ) Kaons decay through the weak interaction as eigenstates of CP (K 1 or K 2 ) –Since K 1 and K 2 don’t have definite S, S is not conserved in weak decays as we already know –If you believe a particle must have a single well defined lifetime, then K 1 and K 2 are the ‘real’ particles

45. Niels Tuning (45) So what is the K 0 really? Graphical analogy – Any object with two components can be decomposed in more than one way K0K0 Anti-K 0 K2K2 K1K1 |K>

46. Niels Tuning (46) Decays of neutral kaons Neutral kaons is the lightest strange particle  it must decay through the weak interaction If weak force conserves CP then –decay products of K 1 can only be a CP=+1 state, i.e. |K 1 > (CP=+1) →   (CP= (-1)(-1)(-1) l=0 =+1) –decay products of K 2 can only be a CP=-1 state, i.e. |K 2 > (CP=-1) →  (CP = (-1)(-1)(-1)(-1) l=0 = -1) You can use neutral kaons to precisely test that the weak force preserves CP (or not) –If you (somehow) have a pure CP=-1 K 2 state and you observe it decaying into 2 pions (with CP=+1) then you know that the weak decay violates CP… ( S( K )=0  L( ππ )=0 )

47. Niels Tuning (47) Designing a CP violation experiment How do you obtain a pure ‘beam’ of K 2 particles? –It turns out that you can do that through clever use of kinematics Exploit that decay of K into two pions is much faster than decay of K into three pions –Related to fact that energy of pions are large in 2-body decay –  1 = 0.89 x 10 -10 sec –  2 = 5.2 x 10 -8 sec (~600 times larger!) Beam of neutral Kaons automatically becomes beam of |K 2 > as all |K 1 > decay very early on… Initial K 0 beam K 1 decay early (into ) Pure K 2 beam after a while! (all decaying into πππ ) !

48. Niels Tuning (48) The Cronin & Fitch experiment Incoming K 2 beam Decay of K 2 into 3 pions If you detect two of the three pions of a K 2   decay they will generally not point along the beam line Essential idea: Look for (CP violating) K 2   decays 20 meters away from K 0 production point

49. Niels Tuning (49) The Cronin & Fitch experiment Incoming K2 beam Decay pions If K 2 decays into two pions instead of three both the reconstructed direction should be exactly along the beamline (conservation of momentum in K 2   decay) Essential idea: Look for K 2   decays 20 meters away from K 0 production point

50. Niels Tuning (50) The Cronin & Fitch experiment Incoming K 2 beam Decay pions Result: an excess of events at =0 degrees! K 2   decays (CP Violation!) Essential idea: Look for K 2   decays 20 meters away from K 0 production point K 2   decays Note scale: 99.99% of K   decays are left of plot boundary

51. Niels Tuning (51) Cronin & Fitch – Discovery of CP violation Conclusion: weak decay violates CP (as well as C and P) –But effect is tiny! (~0.05%) –Maximal (100%) violation of P symmetry easily follows from absence of right-handed neutrino, but how would you construct a physics law that violates a symmetry just a tiny little bit? Results also provides us with convention-free definition of matter vs anti-matter. –If there is no CP violation, the K 2 decays in equal amounts to  + e - e (a)  - e + e (b) –Just like CPV introduces K 2  ππ decays, it also introduces a slight asymmetry in the above decays (b) happens more often than (a) –“Positive charge is the charged carried by the lepton preferentially produced in the decay of the long-lived neutral K meson”

52. Niels Tuning (52) Kaons: K 0,K 0, K 1, K 2, K S, K L, … The kaons are produced in mass eigenstates: –| K 0 >:  sd –|  K 0 >:  ds The CP eigenstates are: –CP=+1: |K 1 > = 1/  2 (|K 0 > - |  K 0 >) –CP= -1: |K 2 > = 1/  2 (|K 0 > + |  K 0 >) The kaons decay to 2 (Short-lived) or 3 pions (Long-lived): –|K S >: predominantly CP=+1 –|K L >: predominantly CP= -1 η +- = (2.286 ± 0.014) x 10 -3 |ε| = (2.284 ± 0.014) x 10 -3

53. Niels Tuning (53) What do we know now? C and P are both violated by the weak interaction because neutrinos can only be produced in a single helicity state (LH neutrinos and RH anti-neutrinos) –Results from Wu and Lederman experiments You can associate a P quantum number to all hadrons and a C quantum number to all hadrons that are their own anti-particle –These C and P quantum numbers are conserved in any reaction not involving the weak interaction The K 0 meson is a special particle that really exists as a combination of two particle: the K 0 / K 0 combination or the K 1 / K 2 combination at your choice The Cronin and Fitch experiment shows that in addition to C and P violation the product CP is also violated, but only a tiny little bit –Origin of CP violation sofar unknown  Topic of future section

54. Niels Tuning (54) Schematic picture of selected weak decays K 0  K 0 transition –Note 1: Two W bosons required (S=2 transition) –Note 2: many vertices, but still lowest order process… d ss K0K0 K0K0 s uu W W dd

55. Niels Tuning (55) Strangeness violation in W mediated decays In 1963 N. Cabibbo made the first step to formally incorporate strangeness violation in W mediated decays 1)For the leptons, transitions only occur within a generation 2)For quarks the amount of strangeness violation can be neatly described in terms of a rotation, where  c =13.1 o Weak force transitions u d’ = dcos c + ssin c W+W+ Idea: weak interaction couples to different eigenstates than strong interaction weak eigenstates can be written as rotation of strong eigenstates

56. Niels Tuning (56) Cabibbos theory successfully correlated many decay rates Cabibbos theory successfully correlated many decay rates by counting the number of cos c and sin c terms in their decay diagram

57. Niels Tuning (57) Cabibbos theory successfully correlated many decay rates There was however one major exception which Cabibbo could not describe: K 0   +  - –Observed rate much lower than expected from Cabibbos rate correlations (expected rate  g 8 sin 2  c cos 2  c ) d ++ --  u cos c sin c W W ss

58. Niels Tuning (58) The Cabibbo-GIM mechanism Solution to K 0 decay problem in 1970 by Glashow, Iliopoulos and Maiani  postulate existence of 4 th quark –Two ‘up-type’ quarks decay into rotated ‘down-type’ states –Appealing symmetry between generations u d’=cos( c )d+sin( c )s W+W+ c s’=-sin( c )d+cos(c)s W+W+

59. Niels Tuning (59) The Cabibbo-GIM mechanism Cabibbo-GIM mechanism introduces clean formalism for quark ‘flavour’ violations in the weak interaction: The weak interaction (W boson) couples to a rotated set of down-type states Tiny problem at time of introduction: there was no evidence for a 4 th quark –Fourth ‘charm’ quark discovered in 1974  Vindication of Cabbibo/GIM mechanism Lepton sector unmixed Quark section mixed through rotation of weak w.r.t. strong eigenstates by  c

60. Niels Tuning (60) The Cabibbo-GIM mechanism How does it solve the K 0   +  - problem? –Second decay amplitude added that is almost identical to original one, but has relative minus sign  Almost fully destructive interference –Cancellation not perfect because u, c mass different d ++ --  u cos c +sin c d ++ --  c -sin c cos c ss ss

61. Niels Tuning (61) From 2 to 3 generations 2 generations: d’=0.97 d + 0.22 s (θ c =13 o ) 3 generations: d’=0.97 d + 0.22 s + 0.003 b NB: probabilities have to add up to 1: 0.97 2 +0.22 2 +0.003 2 =1 –  “Unitarity” !

62. Niels Tuning (62) What do we know about the CKM matrix? Magnitudes of elements have been measured over time –Result of a large number of measurements and calculations Magnitude of elements shown only, no information of phase

63. Niels Tuning (63) How do you measure those numbers? Magnitudes are typically determined from ratio of decay rates Example 1 – Measurement of V ud –Compare decay rates of neutron decay and muon decay –Ratio proportional to V ud 2 –|V ud | = 0.9735 ± 0.0008 –V ud of order 1

64. Niels Tuning (64) End of Lecture 1 1 May –Introduction: matter and anti-matter –P, C and CP symmetries –K-system CP violation Oscillations –Cabibbo-GIM mechanism 8 May –CP violation in the Lagrangian –CKM matrix –B-system 15 May –B-factories –B  J/Psi Ks –Delta ms

65. Niels Tuning (65) Exercises – Spin in decay: helicity 1)Decay topology: consider the process 60 27 Co  60 28 Ni+e - + ν e where the e and the ν are emitted along the polarization (z) axis. (See illustration.) We are now interested in the orientation of the spin of the electron/anti-neutrino w.r.t to the direction of flight. a)Show that there are two possibilities for decays along the 60 Co spin axis: one where the electron spin points to the same direction has the direction of motion and one where the electron spin points to the opposite direction of the direction of motion b)What are the helicities of the electron and anti-neutrino for the two emission possibilites, respectively. (Helicity is defined as: ) S=4 S=1/2 e- νe νe Ni

66. Niels Tuning (66) Exercise – Pion decay: π - → l -  ν l 2)Given a - (spin 0) in rest that decays into a  - and a  ν  (both spin ½) a)Draw the decay for both spin orientation possibilities b)What is the helicity of the anti-neutrino for both possibilities? c)Which of the two processes does not occur in nature? d)Draw the C-conjugated diagram of the  + decay that occurs in nature. Does the C-conjugated process occur in nature? Explain why (not)? e)Draw the C and P conjugated diagram of the  - decay that occurs in nature. Does this decay occur in nature? f)Can the following decay occur: π - → e -  ν e ? g)The (V-A) theory predicts that the decay rate Γ(π - → l -  ν l ) is proportional to the two-body phase space factor multiplied by (1- v/c), where v is the speed of the lepton in the pion rest frame. Derive the phase space factor (p 2 dp/dE), where p is the lepton momentum and E the total energy in the pion rest frame. h)Estimate the ratio of branching ratios: Γ(π - → e -  ν e ) /Γ(π - → μ -  ν μ )

67. Niels Tuning (67) Exercises – CP Eigenstates of the K 0 3)Consider K 1 and K 2 : |K 1 > = 1/2(|K 0 > - |  K 0 >) |K 2 > = 1/2(|K 0 > + |  K 0 >) a)Prove that K 1 and K 2 are eigenstates of the CP operation and determine the eigenvalues of K 1 and K 2. b)Given CP conservation, how do K 1 and K 2 decay? c)Why is the lifetime of K 2 so much larger? (mpi=140 MeV, mK=490MeV) In the experiment of Cronin and Fitch, a beam of almost only K 2 particles is produced from a beam K 0 particles by placing the experiment 20 meters away from the K 0 production point so that most K 1 particles will have decayed. The distribution of K 1 decay distances is an exponential distribution so even after 20 meters there will be a few K 1 left. It is important to know how many such decays there are as a K 1 decay in the detector is indistinguishable from a K 2 decaying (with CP violation) into two pions. d)Given that the decay time of the K 2 is 5.2 x 10 -8 sec, what fraction of the K 2 has not decayed yet after 20 meters (assume v  c) e)Given that the decay time of the K 1 is  1 = 0.89 x 10 -10 sec, what fraction of the K 1 has not decayed yet after 20 meters f)What is the ratio of K 1 to K 2 decays after 20 meters g)How many K 0 particles need to be produced to measure 22700 K 2 decays between 20 and 21 meters of flight with a 1% detection efficiency for particles decaying in that one-meter stretch?

68. Niels Tuning (68) Exercise – Weak decay diagrams 4) Feynman diagrams for weay decays a)Draw the diagram for π + decay. Indicate the CKM matrix element labels at each weak vertex b)Draw the diagram for  0  p e e decay Indicate the CKM matrix element labels at each weak vertex c)Draw two (topologically different) diagrams for the K 0  K 0 transition Indicate the CKM matrix element labels at each weak vertex Both diagram may have at most 4 vertices Hint: One was shown in the course material.

69. Niels Tuning (69) Solutions

70. Niels Tuning (70) Solution to exercises Exercise 1 a)The two possible configurations are shown on the right. The black arrows show the orientation of the spin. The red dashed arrows show the direction of motion b)In the left scenario H(e)=+1, H()=-1 In the right scenario H(e)=-1, H()=+1 c)See book S=4 S=1/2 e- e S=4 S=1/2 e- e

71. Niels Tuning (71) Solution to exercises Exercise 2 a)See below b)Top: H() = RH, Bottom: H()=LH c)Only the top process occurs in nature (there exist only left- handed neutrinos and right-handed anti-neutrinos) --  

72. Niels Tuning (72) Solution to exercises Exercise 2 continued d)The C-conjugated diagram of the  - decay that occurs in nature is The C conjugated process does not occur in nature, as it features a right- handed neutrino. –The CP conjugated diagram of the “- decay that occurs in nature” is This process does occur in nature, just like in the original  - decay process, because it features a left-handed neutrino. CP therefore appears to be a conserved symmetry in the weak interaction ++   ++  

73. Niels Tuning (73) Solution to exercises Exercise 2 continued f)Yes, but is surpressed! e - wants to be left handed because it is almost massless. g)See book

74. Niels Tuning (74) Solution to exercises Exercise 3 a)Start with the known CP properties of the K 0 and K 0 P|K 0 >= -|K 0 >, because quark-antiquark gives (+1)(-1)=(-1) and S,L=0 CP|K 0 > = C(-|K 0 >)= -1|K 0 > CP|K 0 > = -1 |K 0 > –To prove that |K 1 > is an eigenvalue of CP we should show that the CP operation on |K 1 > yields a constant times |K 1 >. Calculate effect of CP on |K 1 > (omitting norm factor 1/2): CP(|K 0 > - |K 0 >) = (-|K 0 > + |K 0 >) = +1(|K 0 > - K 0 >) = +1|K 1 > –Same thing for |K 2 > CP(|K 0 > + |K 0 >) = (-|K 0 > - |K 0 >) = -1(|K 0 > + K 0 >) = -1|K 2 > b)K1  pipi, K2  pipipi c)3 times pion mass is almost kaon mass: hardly any phase space!

75. Niels Tuning (75) Solution to exercises Exercise 3 continued b)Distribution of decays follows f(t)=exp(-t/) with =5.2x10 -8. To calculate the fraction that decays after 20 meters we need to calculate the fraction that decays after 20m/c = 6.6x10 -8 sec c)The primitive of f(t) is F(t) = - exp(-t/): F(0) = - = -5.2x10 8, F()=0, F(20/c)=-*exp(-(20/c)/) = -5.33x10 -10 Fraction(K 2 ) = exp(-20/c) = 27.7% d)For K 1 fraction formula is the same but with different t, Fraction(K 1 ) = exp(-749) = 10 (-749/ln(10)) =10 -325.29 = 1.9x10 -325 The ratio is 1.9x10 -325 / 27.7% = 6.9 x 10 -325. e)Fraction of decays between 20 and 21 meters is exp(-20/c)-exp(-21/c) = 0.27746 – 0.2602 = 0.017 = 1.7% The number of K 2 s to produce is therefore 22700 / 1.7% / 1% = 133 million

76. Niels Tuning (76) Solution to exercises Exercise 4 a)Feynman diagram for - decay (=u d) W-W- u d --  V ud --

77. Niels Tuning (77) Solution to exercises Exercise 4 b)Feynman diagram for  0  p e e decay ( 0 = uds) uds u u d W-W- e (RH) e-e- 00 p V us

78. Niels Tuning (78) Solution to exercises Exercise 4 b)Two Feynman diagrams for K 0 – K 0 mixing. You can make additional ones by replacing one or both of the ‘c’ quarks with ‘u’ or ‘t’ quarks d s K0K0 K0K0 d s cc d s K0K0 K0K0 d s c c V cs V * cd V cs V * cd