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Discrete Symmetries • Parity • Parity Violation • Charge Conjugation • CP Violation • Time Reversal • The CP T Theorem • Lepton number Physics 506A 4 - Discrete Symmetries Page 1 Parity • A parity transformation, P , inverts every spatial coordinate: P (t, x) = (t, −x) P 2 = I, and therefore the eigenvalues of P are ±1. • Ordinary vector v. P (v) = −v . • Scalar from v: s = v · v P (s) = P (v · v) = (−v) · (−v) = v · v = +s • Cross product of two vectors: a = v × w P (a) = P (v × w) = (−v) × (−w) = v × w = +a • Scalar from a and v: p = a · v P (p) = P (a · v) = (+a) · (−v) = −a · v = −p Scalar P (s) = +s Pseudoscalar P (p) = −p Vector P (v) = −v Pseudovector P (a) = +a Physics 506A 4 - Discrete Symmetries Page 2 Parity in Physical Systems • Two-body systems have parity pA pB (−1)ℓ P φ(12) = p1 p2 (−1)ℓ φ(12) • Intrinsically, particles and antiparticles have opposite parity Bound states like positronium e+ e− and mesons qq have parity of (−1)ℓ+1 . • Photons have a parity of (−1), and this underlies the ∆ℓ = ±1 selection rule in atomic transitions. • Note that parity is a multiplicative quantum number. This is true for all discrete symmetries. Continuous symmetries have additive quantum numbers. Physics 506A 4 - Discrete Symmetries Page 3 Example: uu mesons By conventions: u-quarks have spin 1/2 and + parity and u-quarks have spin 1/2 and - parity Parity of a uu meson is P = pu pu (−1)ℓ The intrinsic spin (S) of the uu meson is 0 or 1 but may have any orbital angular momentum (L) value. S L JP particle 0 0 0− π0 1 0 1− ρ0 0 1 1+ b1 (1235) See the PDG for a table of the quantum numbers of the low-mass mesons. Physics 506A 4 - Discrete Symmetries Page 4 Parity in the Standard Model • It has long been realized that par- ity is a respected symmetry of the strong and electromagnetic interac- tions. • In 1956, Yang and Lee realized that parity invariance had never been tested experimentally for weak in- teractions. Physics 506A 4 - Discrete Symmetries Page 5 Madame Wu’s Experiment Wu recorded the direction of the emitted electron from a 60 Co β-decay when the nuclear spin was aligned up and down The electron was emitted in the same direction independent of the spin. −→ Parity is not conserved in the weak interations Physics 506A 4 - Discrete Symmetries Page 6 Parity Violation in π Decay • Consider the weak decay π + → µ+ + νµ . Since the π is spin-0 and the µ and ν emerge back-to-back in the CM frame, the spins of the µ and ν must cancel. • Experiments show that every µ+ is left-handed, and therefore every νµ is also left-handed. • Similarly, in π − decay, both the µ− and ν µ always emerge right-handed. • If parity were conserved by the weak interaction, we would expect left-handed pairs and right-handed pairs with equal probability (just as we observe with π 0 → 2γ). • Assuming that neutrinos are massless, ALL neutrinos are left-handed ALL antineutrinos are right-handed Physics 506A 4 - Discrete Symmetries Page 7 Helicity • Only the z-component of angular momentum can be speciﬁed ⇒ align the z-axis with the direction of motion of a particle. • Deﬁne the helicity of a particle by: h = ms /s −→ −→ ⇒ ⇐ h = +1 h = −1 • For a spin-1/2 particle the helicity can be either +1 or -1. • Helicity is not Lorentz invariant unless the particle is massless Physics 506A 4 - Discrete Symmetries Page 8 Charge Conjugation I • The charge conjugation operator, C, converts a particle to i ts antiparticle. C |p = |p • In particular, C reverses every internal quantum number (e.g. charge, baryon/lepton number, strangeness, etc.). • C 2 = I implies that the only allowed eigenvalues of C are ±1. • Unlike parity, very few particles are C eigenstates. Only particles that are their own antiparticles (π 0 , η, γ) are C eigenstates. For example, ˛ +¸ ˛ −¸ C ˛π = ˛π C |γ = − |γ Physics 506A 4 - Discrete Symmetries Page 9 Charge Conjugation II • The photon has a C = −1 • f f bound states have C = (−1)ℓ+s • Charge conjugation is respected by both the strong and electromagnetic interactions. • Example: the π 0 (ℓ = s = 0 ⇒ C = +1) can decay into 2γ but not 3γ C |nγ = (−1)n |γ C ˛π 0 = ˛π 0 ˛ ¸ ˛ ¸ π 0 → 2γ is allowed (and observed) π 0 → 2γ is not allowed (and not observed < 3.1 × 10−8 ) Physics 506A 4 - Discrete Symmetries Page 10 G-Parity I • C-symmetry is of limited use. ⇒ Most particles are not C eigenstates • The C operator converts π + to π − . • These two particles have isospin assignments | 1 1 and | 1 − 1 . • A 180◦ isospin rotation gives | 1 1 = eiπI2 | 1 − 1 . • The charged pions are eigenstates under the G-parity, which combines C with a 180◦ isospin rotation: G = CeiπI2 • G-parity is mainly used to examine decays to pions (which have G = −1). G |nπ = (−1)n |nπ Physics 506A 4 - Discrete Symmetries Page 11 G-Parity II G-Parity of a few mesons Particle JP I G Decay ρ(770) 1− 1 +1 2π ω(783) 1− 0 -1 3π φ(1020) 1− 0 -1 3π f (1270) 2+ 0 +1 2π For example, the ρ(770) has G = 1 which means it should only decay to an even number of pions. Experimentally we ﬁnd that ρ −→ ππ 100% πππ < 1.2 × 10−4 Physics 506A 4 - Discrete Symmetries Page 12 CP Symmetry The combination of C and P (and time reversal T ) have special signiﬁcance. The violation of CP is the reason we live in a matter universe CP T is required to be conserved in Quantum Field Theory (QFT) Look at a pion decay example: • In the pion decay π + → µ+ (R) + νµ (L), the νµ is always left-handed (LH) • Under C, this becomes π − → µ− (R) + ν µ (L), but the ν µ is still LH ⇒ which does not occur in nature. • With C and P , though, we get a RH antineutrino. π − → µ− (L) + ν µ (R) ⇒ whiich is allowed in nature Physics 506A 4 - Discrete Symmetries Page 13 CP Violation in the Kaon Sector I ¯ • Consider the neutral kaons K 0 (ds) and K 0 (sd) These particles can mix via a second-order weak interaction: s d s d u W u u W W W u d s d s Physics 506A 4 - Discrete Symmetries Page 14 CP Violation in the Kaon Sector II ¯ • Both K 0 and K 0 are pseudoscalar mesons, therefore P = −1. and the K 0 and ¯ K 0 are a particle-antiparticle pair. As a result, under CP , we have ˛K = − ˛K 0 ¯ ¯ ˛ 0¸ ˛K = − ˛K 0 ˛ ¸ ˛ 0¸ ˛ ¸ CP CP `˛ 0 ¸ ˛ 0 ¸´ √ • Deﬁning |K1 = ˛K − ˛K¯ / 2 `˛ 0 ¸ ˛ 0 ¸´ √ |K2 = ˛K + ˛K¯ / 2 we have CP |K1 = + |K1 CP |K2 = − |K2 • If CP is conserved, then |K1 can only decay to 2π (CP = +1) |K2 can only decay to 3π (CP = −1). Physics 506A 4 - Discrete Symmetries Page 15 CP Violation in the Kaon Sector III • We observe the KS and the KL : KS → ππ τ = 0.9 × 10−10 s KL → πππ τ = 0.5 × 10−7 s 0 • K 0 and the K are mass eigenstates and are each others antiparticles • KS and the KL are CP eigenstates (have different masses) and are not antiparticles Physics 506A 4 - Discrete Symmetries Page 16 CP Violation in the Kaon Sector IV • Start out with a beam of K 0 (π − p → K 0 Λ). This will be a superposition of K1 and K2 (CP eigenstates): ˛ 0¸ √ ˛K = (|K1 + |K2 ) / 2 • The K1 component of the beam will decay away over a few centimeters, thereby leaving a nearly pure beam of K2 . ⇒ expect to see only 3π decays • Experimentally, we ﬁnd that about 1 in 440 decays is to 2π. ⇒ the KL has a small mixture of K1 : q |KL = (|K2 + ǫ |K1 ) / 1 + |ǫ|2 Physics 506A 4 - Discrete Symmetries Page 17 Other Tests of CP Violation • There are other CP -violating observables that have been measured in the kaon sector. For example, there is an asymmetry between the branching ratios of KL to π + + e− + νe versus π − + e+ + νe ¯ • Within the last few years, the BaBar and Belle experiments have measured CP violation in the B-meson sector. • CP violation should also be observable in the D-meson (charm) sector, though this will be a small effect that will be very difﬁcult to measure. • There have been earches in the charged lepton sector At UVic, we are studying the τ − → π − KS ν decay with the BaBar data • With the observation that neutrino has mass, it is expected that we will observe CP violation in the neutrino sector Physics 506A 4 - Discrete Symmetries Page 18 Why Measure CP Violation? • The universe contains much more matter than antimatter. ⇒ This requires CP violation. • Since the SM provides only one source of CP violation (CKM phase) The SM CP V appears to be inadequate to account for this asymmetry. • However the SM may have another source if there is CP V in the neutrino sector • CP violation is ubiquitous in theories of New Physics (ie, beyond SM). ⇒ SUSY can provide enough CP violation to be observable at low energies. Physics 506A 4 - Discrete Symmetries Page 19 Physics 506A 4 - Discrete Symmetries Page 20 Physics 506A 4 - Discrete Symmetries Page 21 Physics 506A 4 - Discrete Symmetries Page 22 CP Violation in τ decays 0 The goal is to search for direct CP violation in the decay τ − → h− KS ≥ 0π 0 ντ ` ´ There is no charge asymmetry between τ − → π − K 0 ντ and τ + → π + K 0 ν τ 0 0 Experimentally we observe the KS and KL mesons, which are linear combinations of K 0 and K 0 states. 0 0 Earlier experiments have shown that the KS and KL are not exact CP eigenstates. As a result the charge asymmetry is 0 Γ τ + → π + KS ν τ − Γ τ − → π − KS ντ 0 ` ´ ` ´ AQ = ` + 0 ´ ` 0 ´ (1) Γ τ → π + KS ν τ + Γ τ − → π − KS ντ Theory predicts AQ = (3.3 ± 0.1) × 10−3 Little can be learnt from this CP asymmetry, but this measurement provides a valuable calibration point in CP searches and any deviation from the Standard Model value would be evidence for new physics. Physics 506A 4 - Discrete Symmetries Page 23 Time Reversal Symmetry • Time reversal symmetry, as you might guess, reverses the time component: T (t, x) = T (−t, x) • Although we expect the weak interaction to violate T , direct T violation has not been deﬁnitively observed yet. • Experimentally, one tries to measure the rate of a reaction in both directions A + B → C + D but this is not so easy Physics 506A 4 - Discrete Symmetries Page 24 The CP T Theorem • The combination CP T is always conserved in any local quantum ﬁeld theory • CP T violation in essentially synonymous with a violation of Lorentz invariance • CP T symmetry mandates that particles and antiparticles must have certain identical properties, such as the same mass, lifetime, charge, and magnetic moment Physics 506A 4 - Discrete Symmetries Page 25 Experimental tests of CP T Citation: S. Eidelman et al. (Particle Data Group), Phys. Lett. B 592, 1 (2004) (URL: http://pdg.lbl.gov) CPT INVARIANCE (m W + ? m W ? ) / m average ? 0:002 0:007 (m e + ? m e ? ) / m average <8 10?9 , CL = 90% qe + + q e ? e <4 10?8 (g e + ? g e ? ) / gaverage (? 0:5 2:1) 10?12 ( + ? ? ) / average (2 8) 10?5 (g + ? g ? ) / g average (? 2:6 1:6) 10?8 (m + ? m ? ) / m average (2 5) 10?4 ( + ? ? ) / average (6 7) 10?4 (m K + ? m K ? ) / m average (? 0:6 1:8) 10?4 ( K + ? K ? ) / average (0:11 0:09)% (S = 1.2) K ! rate di erence/average (? 0:5 0:4)% K ! 0 rate di erence/average f ] (0:8 1:2)% in K 0 ? K 0 mixing real part of (2:9 2:7) 10?4 imaginary part of (0:02 0:05) 10?3 m K 0 ? m K 0 / m average g] <10?18 , CL = 90% (? K 0 ? ? K 0 )/m average (8 8) 10?18 phase di erence 00 ? +? (0:2 0:4) Re( 2 +? + 3 00 )? 2 3 1 L (? 3 35) 10?6 ACPT (K ) in D 0 ! K ? + , D 0 ! K + ? 0:008 0:008 m p ?m p /m p h] <1:0 10?8 , CL = 90% q q q ( mpp { mp )/ mpp (? 9 9) 10?11 p qp + qp e h] <1:0 10?8 , CL = 90% ( p + p) p (? 2:6 2:9) 10?3 (m n ? m n )/ m n (9 5) 10?5 (m ? m ) m (? 0:1 1:1) 10?5 (S = 1.6) ( ? )/ ? 0:001 0:009 ( + ? ?) / + (? 0:6 1:2) 10?3 ( + + ?) + 0:014 0:015 (m ? ? m + ) / m ? (1:1 2:7) 10?4 ( ? ? +) / ? 0:02 0:18 ( ? + +) / ? +0:01 0:05 (m ? ? m + ) / m ? (? 1 8) 10?5 ( ? ? +) / ? ? 0:002 0:040 HTTP://PDG.LBL.GOV Page 11 Created: 6/17/2004 19:40 Physics 506A 4 - Discrete Symmetries Page 26 Lepton Number • There are 3 lepton numbers: Le , Lµ and Lτ Le = +1 for e− and νe Le = −1 for e+ and ν e • Conserved in the EM and Weak interactions γ → e+ e− and π + → µ+ ν µ are allowed whereas µ+ → e+ γ is forbidden • BaBar (UVic group) put a new limit on the τ + → µ+ γ branching fraction (10−8 ) • Neutrino oscillations imply that lepton number is violated (at a very small level) Physics 506A 4 - Discrete Symmetries Page 27 Experimental tests of Lepton Number Citation: S. Eidelman et al. (Particle Data Group), Phys. Lett. B 592, 1 (2004) (URL: http://pdg.lbl.gov) TESTS OF NUMBER CONSERVATION LAWS LEPTON FAMILY NUMBER Lepton family number conservation means separate conservation of each of Le , L , L . ?(Z ! e )/?total i ] <1:7 10?6 , CL = 95% ?(Z ! e )/?total i ] <9:8 10?6 , CL = 95% ?(Z ! )/?total i ] <1:2 10?5 , CL = 95% limit on ? ! e ? conversion ( ? 32 S ! e ? 32 S) / <7 10?11 , CL = 90% ( ? 32 S ! 32 P ) ( ? Ti ! e ? Ti) / <4:3 10?12 , CL = 90% ( ? Ti ! capture) ( ? Pb ! e ? Pb) / <4:6 10?11 , CL = 90% ( ? Pb ! capture) limit on muonium ! antimuonium conversion Rg = <0:0030, CL = 90% GC / GF ?( ? ! e ? e )/?total j ] <1:2 10?2 , CL = 90% ?( ? ! e ? )/?total <1:2 10?11 , CL = 90% ?( ? ! e ? e + e ? )/?total <1:0 10?12 , CL = 90% ?( ? ! e ? 2 )/?total <7:2 10?11 , CL = 90% ?( ? ! e ? )/?total <2:7 10?6 , CL = 90% ?( ? ! ? )/?total <1:1 10?6 , CL = 90% ?( ? ! e ? 0 )/?total <3:7 10?6 , CL = 90% ?( ? ! ? 0 )/?total <4:0 10?6 , CL = 90% ?( ? ! e ? K 0 )/?total S <9:1 10?7 , CL = 90% ?( ? ! ? K 0 )/?total S <9:5 10?7 , CL = 90% ?( ? ! e ? )/?total <8:2 10?6 , CL = 90% ?( ? ! ? )/?total <9:6 10?6 , CL = 90% ?( ? ! e ? 0 )/?total <2:0 10?6 , CL = 90% ?( ? ! ? 0 )/?total <6:3 10?6 , CL = 90% ?( ? ! e ? K (892)0 )/?total <5:1 10?6 , CL = 90% ?( ? ! ? K (892)0 )/?total <7:5 10?6 , CL = 90% ?( ? ! e ? K (892)0 )/?total <7:4 10?6 , CL = 90% ?( ? ! ? K (892)0 )/?total <7:5 10?6 , CL = 90% ?( ? ! e ? )/?total <6:9 10?6 , CL = 90% ?( ? ! ? )/?total <7:0 10?6 , CL = 90% ?( ? ! e ? e + e ? )/?total <2:9 10?6 , CL = 90% ?( ? ! e ? + ? )/?total <1:8 10?6 , CL = 90% HTTP://PDG.LBL.GOV Page 12 Created: 6/17/2004 19:40 Physics 506A 4 - Discrete Symmetries Page 28 Physics 506A 4 - Discrete Symmetries Page 29 Physics 506A 4 - Discrete Symmetries Page 30 Physics 506A 4 - Discrete Symmetries Page 31 Physics 506A 4 - Discrete Symmetries Page 32 Summary • The 3 discrete symmetries P , C, and T are respected individually by the strong and electromagnetic forces. • Parity violation is the signature of the weak interaction. • The weak interaction also violates CP , as a result of a complex phase in the CKM matrix. • The combined symmetry of CP T is obeyed by all local quantum ﬁeld theories. Physics 506A 4 - Discrete Symmetries Page 33 The Quark Model • Sticks quarks and antiquarks together like Lego. • Makes no attempt to account for the complexities of low-energy QCD other than to hope that it treats all members of a multiplet in the same way. • With a handful of experimental inputs, can be used to obtain ballpark estimates for hadron masses and magnetic moments. Physics 506A 4 - Discrete Symmetries Page 34 Mesons • There are 9 distinct ﬂavor combinations that can be contructed from {u, d, s} u ¯¯ ¯ and {¯, d, s}. Group theoretically, 3 ⊗ 3 = 8 ⊕ 1, but we usually lump the 1 with the 8 to create a meson nonet. • Once we include spin, we get a pseudoscalar meson nonet (π, K, η) and a vector meson nonet (ρ, K ∗ , ω, φ). • The vector mesons are heavier, and this can be modeled by the meson mass formula: (S1 · S2 ) M = m1 + m2 + A m1 m2 With 4 parameters (mu = md = 310 MeV, ms = 483 MeV, A = (2mu )2 160 MeV), we can predict 18 meson masses to about 1%. Physics 506A 4 - Discrete Symmetries Page 35 Physics 506A 4 - Discrete Symmetries Page 36 Physics 506A 4 - Discrete Symmetries Page 37 Baryons • There are 27 distinct ﬂavor combinations that can be constructed from three copies of {u, d, s}. Group theoretically, 3 ⊗ 3 ⊗ 3 = 1 ⊕ 8 ⊕ 8 ⊕ 10. • One of the baryon octets contains the p, n, Σ, Λ, Ξ. The baryon decuplet contains the ∆, Σ ∗ , Ξ ∗ , Ω. The other octet and the singlet contain less familiar baryons. • As with the mesons, we can construct a baryon mass formula in terms of the effective masses of the quarks (different from those used for mesons) ’ and their relative spins. • No evidence for particles with more than 3 quarks Physics 506A 4 - Discrete Symmetries Page 38 Physics 506A 4 - Discrete Symmetries Page 39

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