Particle Physics 2002/2003Part of the “Particle and Astroparticle Physics” Master’s Curriculum VI. Discrete Symmetries and CP violation Particle Physics I I.Introduction, history & overview (2) II.Concepts (5): Units (h=c=1) Relativistic kinematics Cross section, lifetime, decay width, … Symmetries (quark model, …) III.Quantum Electro Dynamics: QED (7) Spin 0 electrodynamics (Klein-Gordon) Spin ½ electrodynamics (Dirac) Experimental highlights: “g-2”, e  e , … Particle Physics II IV.Quantum Chromo Dynamics: QCD (4) Colour concept and partons High q 2 strong interaction Structure functions Experimental highlights:  s, e  p, … V.Quantum Flavour Dynamics: QFD (6) Low q 2 weak interaction High q 2 weak interaction Experimental highlights: LEP VI.Origin of matter? (6) Strange particles GIM (why does the charm exist?) K 0 -K 0, oscillations, CP violation B 0 -B 0 oscillations Current CP violation experiment VII.Origin of mass? (2) Symmetry breaking Higgs particle: in e  e  and in pp File on “paling”: graven/ED_MASTER/master2003.ppt

Strange Particles Isospin “Strangeness” E  threshold (GeV) In general: more difficult (i.e. higher threshold) to make S=-1 particles then S=+1 when target is either p or n, and beam consists of  + or  - because for S=+1, need to get an antiquark from somewhere… Quite useful: can make ‘pure’ K 0 or K + sample by running below threshold Note: long lifetime!

Observations: 1.High production cross-section 2.Long lifetime Conclusion: must always be produced in pairs! Details:create a new quantum number, “strangeness“ which is conserved by the production process hence pair production by strong force however, the decay must violate “strangeness” if only weak force is “strangeness violating” then it is responsible for the decay process hence (relatively) long lifetime… “V particle”: particles that are produced in pairs and thus leaves a ‘v’ trial in a bubble chamber picture

Strange Particles m K ~ 494 MeV/c 2 1.No strange particles lighter than Kaons exist Decay must violate “strangeness” 2.Strong force conserves “strangeness” Decay is a pure weak interaction Isospin “Strangeness” Semi-leptonic decays: particle and anti-particle are distinct from one another! “  Q=  S rule” Hadronic and leptonic decays: particle and anti-particle behave the same

Weak decays: K - vs.  -

Universality of weak interactions Weak doublets:

K 0 decays: A problem?  Z 0 boson will now couple to uu and d’d’... This generates a “FCNC”, (Flavour Changing Neutral Current)… need to do more. Or, to put it the other way around: The absence of FCNC requires V to be unitary the reason there are no FCNC is that V C is unitary:

K 0 decays: enter the charm To (almost) cancel this diagram, lets introduce another up-type quark, and have it interact through a W with the orthogonal combination of (d,s) This new ’c’ quark causes an additional diagram that (almost) cancels the one above… If m u  m c, then the cancellation would be complete! This is called “GIM suppression” leads to a prediction of the charm mass of 1.5—2 GeV, prior to the discovery of J/ 

In general: the weak eigenstates are not the mass eigenstates! If all quarks were the same mass, this could not happen as we could take any linear combination of quarks as the mass eigenstates… And as long as V is unitary, there will be no FCNC! note: can (and will) extend this to 3 families later Q: what happens if V C =1 (i.e.  C =0)? A: the s quark (and thus all S  0 particles) would be stable!!! Q: how many independent parameters does V have when there are 2 generations (i.e. is  C all there is?). How about 3 generations? A: 2* (2*2-1)=1; 2* (2*3-1)=4

Intermezzo: Discovery of the J/  By studying the decay of strange particles, the existence of the charm and its properties (eg. mass, weak couplings) were predicted before its discovery – but nobody believed it! Brookhaven: J SLAC:  Sam Ting and Burt Richter got the 1976 Nobel prize for their discovery

Back to K 0 decays… Phys. Rev. 97, 1387 (1955) Known: 1.K 0 can decay to  +  - Hypothesized: 1.K 0 has a distinct anti-particle K 0 Claims: 1.K 0 (K 0 ) is a “particle mixture” with two distinct lifetimes 2.Each lifetime has its own set of decay modes 3.No more than 50% of K 0 (K 0 ) will decay to  +  -

K 0 and CP symmetry Known decay: Assuming CP symmetry, this should be possible as well: Assuming the reverse reaction is allowed, particle can “mix” into anti-particle, and vice-versa… How does this system evolve in time? (ignore decays for the time being) Mixing causes tiny off-diagonal element: With completely different eigenstates!

K 0 decay and CP: K 1 and K 2 Phys Rev 103,1901 (1956) Only the CP even state (K 1 ) can decay into 2 pions (which are CP even) The odd K 2 state will decay into 3 particles instead ( , ,  e,…). There is a huge difference between K 0  and K 0   in phasespace (~600x!). So the CP even state will decay much faster K 1 and K 2 are their own antiparticle, but one is CP even, the other CP odd: CP: +1 -1

2 vs. 3 particle phase space

More on time evolution Tag K 0 and K 0 decay by semileptonic decay (remember the  S=  Q rule?) K 1 decays K 2 decays

Testing CP conservation Easy to create a pure K 2 beam: just “wait” until the K 1 component has decayed… If CP conserved, should not see the decay to 2 pions in this K 2 beam This is exactely what was tested by Cronin & Fitch in 1957…  Main background:  +  -  0 K2+-K2+- Effect is tiny: about 2/1000 … and for this experiment they got the Nobel price in 1980…

Interference K L and K S are no longer orthogonal:

CPLEAR, Phys.Rep. 374(2003) T violation in mixing t=0 t CP Note: 1.This measurement allows one to make an ABSOLUTE distinction between matter and anti-matter 2.Don’t need to know the specific value of decay amplitudes; only need:

(2x)2 ways to decay… t=0 t Amplitude CP

3 ways to break CP CP violation in decay CP violation in mixing CP violation in the interference between mixing and decay

3 Ways to break CP

The Final Result… If   =0: only “K S ” like decays If    0: not only “K L ” like decays, down by |   | 2, but alsointerference contribution, down by |   | The interference term has a sign difference for K 0 and K 0 bar! A(K S )     (t) A(K L ) K 0 (t=0) If CP were conserved, K L wouldn’t decay to    , and there would be no interference… A(K S )     (t) -A(K L ) K 0 (t=0) CP

CPLEAR Use the strangeness conservation of the strong interactions to perform Tagged K 0 and K 0 production: At t=0, events with a  K + are a pure K 0 bar sample  K - are a pure K 0 sample

CPLEAR

Results: CP in Interference CPLEAR, PLB 1999 (K 0 -K 0 )/(K 0 +K 0 ) decaytime /  S Approx equal K S      and K L     rate: Maximal interference! Mainly K S     decays Mainly K L    decays K 0 bar K0K0 Note: rates are normalized to each other in the range (,) Interference maximal:

Details…

CP violation: when? Pf Pbar Introduce A, Abar into the picture CP violation seems to occurs in interference What kinds of interference can we have? Mixing Decay Mixing vs. Decay

Basic Equations: Neutral Meson Mixing If At t=0, only a(t) and b(t) are non-zero We are only interested in a(t) and b(t), and not c i (t) t is large compared to the strong-interaction scale Then one can make an approximation (Wigner-Weisskopf) which considerably simplifies things: In general, want to know the time evolution of:

Basic Equations: Neutral Meson Mixing  is not Hermitian: otherwise mesons would only oscillate, and never decay… instead: Virtual Intermediate States Real Intermediate States

Basic Equations: Neutral Meson Mixing  is not Hermitian: otherwise mesons would only oscillate, and never decay… instead: M describes oscillations,  decays

Phase conventions Because of the requirement of phase independence, R has only 7 (physical) parameters CPT invariance: T invariance: CP invariance: Requiring CPT reduces this to X parameters (SHOW!)

Solving the master equations

Computing Dm and DG in K0 mixing Okun p88? Cahn-Goldhaber, chapter 15 Or: why is kaon mixing so different from B mixing And why is D mixing different again??? Actually, why is the B lifetime so large? as expected, the D lifetime is much less than the K0S one Show that mixing vanishes if all quark masses are equal

Solution (CP violating case) Nobel Lecture Val Fitch

Intermezzo: KS regeneration i.e. why the helium bag? Or: another way to measure dm

Enter the B meson… Long lifetime! -> Vcb must be tiny! Third generation -> VCKM It mixes! -> top must be VERY heavy

And then there were 3… The CKM matrix Unexpected long B lifetime! => Vcb must be small! Eg. B+ to mu+ not observed, only limits

Mixing of neutral mesons

D0 mixing vs. Bd mixing Mixing dominated by Vtd Not yet observed! Experimental limit goes here How do we measure mixing?? Compute Vtd from the measured dm values Must have heavy (>100 GeV) top! As VtdVtb**2 isn’t very large (0.2**6)

Bd mixing vs. Bs mixing Mixing Dominated by VtdMixing Dominated by Vts Lifetime difference dominated by Vcd: tiny Lifetime difference dominated by Vcs: Expect 10-20% Not yet observed! Experimental limit goes here Some other effects of O(30%) lead to the SM expectation of ~18

CP violation in mixing Why small? Experimental results Kaons, Bd mesons

Is P a good symmetry? Parity violation observed in 60 Co experiment in Magnetic field e-e-  Parity transformation e-e-  Observed I() = 1 +  (v/c) cos  with  = Co JJ   More electrons emitted opposite the J direction. Parity violation!  1950 ‘60‘70‘80‘ Parity violation C. Yang and T. Lee, 1956 C. S. Wu, 1957 Sketch and photograph of apparatus used to study beta decay in polarized cobalt-60 nuclei. The specimen, a cerium magnesium nitrate crystal containing a thin surface layer of radioactive cobalt-60, was supported in a cerium magnesium nitrate housing within an evacuated glass vessel (lower half of photograph). An anthracene crystal about 2 cm above the cobalt-60 source served as a scintillation counter for beta-ray detection. Lucite rod (upper half of photograph) transmitted flashes from the counter to a photomultiplier (not shown). Magnet on either side of the specimen was used to cool it to approximately K by adiabatic demagnetization. Inductance coil is part of a magnetic thermometer for determining specimen temperature. B

Intermezzo: CP

Exercise: Show CP |pi+pi-> = + | pi+pi-> CP |pi+pi-pi0> = - |pi+pi-pi0>

More details about Mixing: regeneration Q: why does eg. K* 0 not mix? It has the same quark content… A: it decays to K + pi -, a strong decay – it just isn’t stable enough! A2: it is a vector particle… How do we make sure KL -> p+p- is really the same final state As KS->pipi ? maybe we’re missing a particle that takes away very little momentum? NOTE: beta decay spectrum was ‘solved’ by introducing a new particle (the neutrino) Let them interfere!

Trace Theorems