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Italian Physical Society International School of Physics “Enrico Fermi” CP Violation Topics to be covered Lecture 1 1. Introduction • Why study CP violation? • Grand view - Sakharov's ideas 2. Symmetries • Mechanics • Electrodynamics • Quantum mechanics time reversal operator anti-unitary operator Lecture 2 • • • • • Edm Krammer’s degeneracy Particle physics How do you measure pion spin? How do you show that pion is a pseudoscalar? • G-parity • C,P,T in particle physics Lecture 3 3. K meson system • • • • τ-θ Puzzl Weak interaction CPT Mixing 4.CP violation • Asymmetry in partial widths • Role of final state interaction • Hyperon decays • Watson's theorem Lecture 4 •Regeneration •Explain Eq. 7.8 and Fig 7,.1 •Discovery of CP violation Cronin-Fitch experiment •Direct •indirect • In this section, we refrain from deriving the expression for ε and ε' etc. Lecture 5 CP violation in B decays Towards building the B factory EPR paradox Experiments at the B factory without much about the KM formalism Its goes back the basic question asked by man. Why do we exist? radius 104 cm 100Km 109 1013 104 We are here 1 billion years Supernova and heavy elements 1026 1m 100 Stars and milky way 30 Temp(℃) 10 1010 1015 Light elements 1034 Nucleosythesis of Helium protons and neutrons 43 Time(sec) 10 5 billion years 13.7 billion years 270 1028 cm Generation of light elements Black-body radiation The universe is filled with 3°K (-270℃)photons Expanding universe Measuring the speed of expansion universe Theory predicts universe and anti-universe anti-universe 1. You can see Armstrong’s foot step. The moon is not made out of anti-matter 2.No primary anti-particle in cosmic rays BESS experiment KEK/IPNS 10-1 He/He limit (95% C.L.) Smoot et al. (1975) 10-2 Evenson (1972) Smoot et al. (1975) 10-3 Antihelium/helium flux ratio Evenson (1972) Aizu et al. (1961) Badhwar et al. (1978) Golden et al. (1997) 10-4 Buffington et al. (1981) 10-5 Ormes et al. (1997) BESS-95 T. Saeki et al. (1998) BESS-93~95 10-6 10-7 J. Alcaraz et al. (1999) AMS01 M. Sasaki (2000) BESS-93~98 BESS-1993~2000 Preliminary BESS-Polar (2003, 20 days) PAMELA (2002~, 3 years) 10-8 AMS02 (2004~, 3 years) 10-9 10-1 1 10 102 Rigidity (GV) KEK/IPNS Yoshida 103 3.You see galaxy collisions but collisions of galaxy and anti-galaxy have been found Really absent? Alpha Matter Spectrometer Space Shuttle NASA program SEMINAR – EXPERIMENTS ON ANTIMATTER SEARCHES IN SPACE Battison We have to understand how CP is violated in the fundamental theory t 1016 sec r 1029 cm T 270C 200 / cm 3 1080 p , n 0 p ,n Why there is no anti-universe 三田一郎 名古屋大学大学院 理学研究科 1. Anti-baryon has to disappear. So, there has to be baryon number violation 2. Only anti-baryon has to disappear. So, there has to be CP violation 3. Created asymmetry will be washed out if the universe is in equilibrium. So, baryon has to be created out of equilibrium. t 1016 sec r 1029 cm T 270C X ® qe X ® qe N = 10 88 N = 10 88 These annihilations Produce light + 10 80 200 / cm 3 1080 p , n 0 p ,n N - N = 1088 + 1080 - 1088 N +N 1088 + 1080 + 1088 = 10- 8 strings The origin of quark masses The origin of CP violation So, let us go back to symmetries Mirror image x x Time reversal t t You can’t tell which is the original The symmetry is imbedded in your brain Newton’s Equation 2 d r F m 2 dt Show that this equation is invariant under parity Show that this equation is invariant under time reversal Maxwell Equation E 4 1 E 4 B J c t c B 0 1 B E 0 c t Parity E 4 1 E 4 B J c t c B 0 E 1 B 0 c t F ma F q( E v B) E E BB Charge conjugation E 4 1 E 4 J c t c B 0 B E 1 B 0 c t F ma F q( E v B) E E B B J J Time Reversal E 4 1 E 4 B J c t c B 0 1 B E 0 c t d d dt dt EE F ma B B F q( E v B) J J Our cells are controlled by electrodynamics. Why don’t we get younger? Symmetries and quantum mechanics Ω i [H,Ω]=0 i ψ and Ω ψ are both solutions to the Schorodinger Eq. i | ; t H | ; t t 1 i | ; t H 1 | ; t t i | ; t H | ; t t P: C: T: So, you can’t tell the difference! r r e e t t ψ ψ = ψ Ω†Ω ψ Ω†Ω=1 X x x x XP x xP x P x x P x = eid - x so we take e i d = + 1 P2 =1, P=P-1, P=P† P 1 XP X XP x PX x xP x | ; t H | ; t t 1 Pi P P | ; t PHP 1P | ; t t i P | ; t HP | ; t t i Show that: (1) If [P,H]=0 xHx = - xH- x H( x ) = H( - x) ¶ (2) ih P |y ,t = HP |y ,t x H y H (x ) (x y ) ¶t ¶ x ih P |y ,t = ò dy x H y y P |y ,t ¶t ¶ then ih y (- x ,t ) = H(x)y (- x ,t ) ¶t (3) y ± = 1 é y ±Py ù ë û is also a solition 2 Show that (1) - x y ± = x P† y ± = x Py ± = ± x y ± (2) y ± (- x ) = ± y ± ( x ) & & (3) y ± ( x ) is also solution to the Schordinger equation. p 2 2 e e2 H A p p A A e 2 2m 2mc 2mc If charge of a particle is flipped, and the external all fields are flipped, the motion is invariant. e e A A CHC 1 H x, p i T x, p T 1 x, p Ti T 1 i Time reversal and quantum mechanics |b>=O|b> <a|O†O|b>=<a|b> |a>=O|a> |<a|b>|=|<a|O†O|b>|=|<a|b>| <a|O†O|b>=<a|b>* O=UK K is a complex conjugate operator K α|a>+β|b> =α *K|a>+β*K|b> K † acts always to the left [α <a|+β<b|]K † =α* <a|K † + β* <b|K † 2 K =1 -1 K =K T=K Show that x,p =i is consistent with time reversal. Transformation of Schrodinger equation under T |ψ(t)>=H|ψ(t)> t -1 Ti T T|ψ(t)>=THT -1T|ψ(t)> t i |ψ(-t)>=H|ψ(-t)> t -i T|ψ(t)>=HT|ψ(t)> t T|ψ(t)>=|ψ(-t)> i Transformation of the wave function under T ψ(-t) =ò dx' x' x' ψ(-t) T ψ(t) =T ò dx' x' x' ψ(t) =ò dx' x' x' ψ(t) * ψ(x,-t)=ψ(x,t)* T ψ(t) =T ò dx' x' x' ψ(t) =ò dx' x' x' ψ(t) * T f (t) =T ò dy' y' y' f (t) =ò dy' y' y' f (t) * f (t) T †T ψ(t) =ò dx 'dy ' y ' f (t ) y ' x ' x' ψ(t) * f (t) T †T ψ(t) = ψ(t) f (t) 1. If T=K, from the definition of operator P, show that T-1PT=-P. 2. From the definition of J=rxP, show that T-1JT=-J Vx ' cos sin 0 Vx V ' sin cos 0 V y y V ' 0 0 0 z Vz Vx ' cos 0 sin Vx V ' 0 1 0 y Vy V ' sin 0 cos V z z 0 Vx ' 1 V ' 0 cos y V ' 0 sin z 0 Vx sin V y cos Vz around the z axis around the y axis around the x axis 0 i 0 0 0 0 0 0 i J z i 0 0 J x 0 0 i J y 0 0 0 0 0 1 0 i 0 i 0 0 TJT 1 J if T=K Rotation operator e iJ z 0 i 0 0 0 0 0 0 i J z i 0 0 J x 0 0 i J y 0 0 0 0 0 0 0 i 0 i 0 0 But in quantum mechanics first question we ask: what operator can we diagonalize together with H. [H , J 2] 0 [H , J ] 0 So we often use: [Ji , J j ] i ijk Jk 1 0 0 0 1 0 0 i 1 1 Jz 0 0 0 Jx 1 0 1 Jy i 0 2 2 0 0 1 0 1 0 0 i 0 i 0 Then we have to change T=K T exp i J y K