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Brian Meadows, U. Cincinnati Discrete Symmetries Noether’s theorem – (para-phrased) “A symmetry in an interaction Lagrangian corresponds to a conserved quantity.”

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Brian Meadows, U. Cincinnati Conserved Quantities StrongE/MWeak 4-momentumYes ChargeYes Baryon #Yes SpinYes Lepton # (e, , ) --Yes Flavour (S,C,B,T) Yes No (CKM) (or Q = F) Iso-spinYesNo PYes No CPT, CP, CYes, Yes, Yes Yes, No, No

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Brian Meadows, U. Cincinnati Parity P Particles have “intrinsic parity” =± 1 P | > = - | > ; P |q> = +1 (q is a quark); etc.. We define parity of quarks (ie the proton) to be positive. (ie P=+1) It is usually possible to devise an experiment to measure the “relative parity” of other particles. Parity of 2-body system is therefore P = (-1) l 1 2 Example: parity of Fermion anti-Fermion pair (e.g. e + e - ): Whatever intrinsic parity the e - has, the e + is opposite (actually a requirement of the Dirac theory) So, P = (-1) ( l +1)

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Brian Meadows, U. Cincinnati Parity Violation Parity is strictly conserved in strong and electromagnetic interactions Helicity can be +1 or -1 for almost any particle. It can flip if you view particle from a different coordinate system BUT not if the particle travels at c! Real photons have both +1 and -1 helicities (not zero) Consequence of conservation of parity in e/m interactions Not so for neutrinos In + + + helicity of + is ALWAYS = -1 (“left-handed”) The neutrino is LEFT-HANDED (always!) Parity is “maximally violated” in weak interactions.

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Brian Meadows, U. Cincinnati Charge Conjugation C Operator C turns particle into anti-particle. C | + > = | - > ; C |K + > = |K - > ; C |q> = |q> ; etc. C 2 has eigenvalue 1 Therefore C=± 1 Since C reverses charges, E- and B-fields reverse under C. Therefore, the has C=-1 C is conserved in strong and E/M interactions. Since 0 2 , then C| 0 > = +| 0 > Since 0 2 , then C| 0 > = +| 0 > AND 0 cannot decay to 3 (experimentally, 0 3 / 0 2 < 3 10 -8 )

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Brian Meadows, U. Cincinnati Time Reversal T This, in effect, reverses the direction of time It does not reverse x, y or z.

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Brian Meadows, U. Cincinnati CPT and Time-Reversal There is compelling reason to believe that CPT is strictly conserved in all interactions It is difficult to define a Lagrangian that is not invariant under CPT T is an operator that reverses the time No states have obviously good quantum numbers for this, but you can define CP quantum number e.g. CP | + - > = (-1) L (why?) Even CP is not conserved e.g. K 0 observed to decay into + - (CP=+1) as well as into - + 0 (CP=-1) B 0 decays to J/psi K s, J/psi K L and + -

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Brian Meadows, U. Cincinnati CP Conservation Recall that P is not conserved in weak interactions since ’s are left-handed (and anti- ’s are right-handed). Therefore, C is not conserved in weak interactions either: + + + Makes a left-handed + (because is spin 0) C( + + + ) ( - - + ) makes a left-handed - (C only converts particle to anti-particle). BUT – the - has to be right-handed because the anti- is right-handed. However, the combined operation CP restores the situation CP( + + + ) ( - - + ) Because P reverses momenta AND helicities

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Brian Meadows, U. Cincinnati CP and the K 0 Particle The K 0 is a pseudo-scalar particle (P=-1), therefore P |K 0 > = - |K 0 > and P |K 0 > = - |K 0 > The C operator just turns K 0 into K 0 and vice-versa C |K 0 > = + |K 0 > and C |K 0 > = + |K 0 > Therefore, the combined operator CP is CP |K 0 > = - |K 0 > and CP |K 0 > = - |K 0 > Neither |K 0 > nor |K 0 > are CP eigen-states We can define odd- and even-CP eigen-states K 1 and K 2 : |K 1 > = (|K 0 > - |K 0 >) / \ / 2 CP |K 1 > = (+1) |K 1 > |K 2 > = (|K 0 > + |K 0 >) / \ / 2 CP |K 2 > = (-1) |K 2 >

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Brian Meadows, U. Cincinnati CP and K 0 -K 0 Mixing Experimentally, it is observed that there are two K 0 decay modes labeled as K L and K s : K s + - ( s = 0.893 x 10 -10 s) K L + - 0 ( L = 0.517 x 10 -7 s) The decay products of the K s have P = (-1) L = (-1) 0 = +1 For the K L the products have P = -1 It is tempting to assign K L to K 1 and K s to K 2 However, this is not exactly correct: V. Fitch and J. Cronin observed, in an experiment at Brookhaven, that about 1 in 500 times, either K s 3 or K L 2 So one defines K L =1/sqrt(1+ 2 ) (K 2 + K 1 ) where is the deviation from CP conservation

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Brian Meadows, U. Cincinnati CP and K 0 -K 0 Mixing It is possible for a K 0 to become a K 0 ! The main diagram contributing to mixing in the K 0 system: This contributes to the observation of CP violation in the K 0 K 0 system. It generates a difference in mass between K 1 and K 2 It is described by a phase in the CKM matrice. d d s s u, c, t W W K0K0 K0K0

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Brian Meadows, U. Cincinnati Strangeness Oscillations Graph shows I(K 0 ) and I(K 0 ) as function of t for m s / ~ = 0.5 Experimentally, measure hyperon production in matter (due to K 0, not K 0 ) as function of distance from source of K 0 ) m s / ~ = 0.498. This corresponds to m/m ~ 5 x 10 -15 !

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Observation of K 0 -K 0 Oscillations K 0 ->3 is only 34%, 39% of the decays are leptonic Observe the asymmetry in the leptonic sector Use the sign of lepton in decays K 0 + e - e K 0 - e + e Brian Meadows, U. Cincinnati Gjesdal et al, Phys.Lett.B52:113,1974 World Average:

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Brian Meadows, U. Cincinnati Other examples of “Mixing” Evidence now also exists for mixing in other neutral meson systems: K 0 - K 0 (ds) - observed in ~1960 B 0 - B 0 (bd) - observed in ~1992 B s - B s (bs) - observed in 2005 D 0 - D 0 (cu) - observed in April 2007 ! by BaBar and (almost simultaneously) by Belle Similar mass oscillation versus “flavor observations” are Observed with neutrinos, revealing that neutrino have mass.

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