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Discrete Symmetries

Discrete Symmetries. Parity. The operation that reverses the spatial coordinates is called the parity operation P : Its eigenvalue is P P= § 1 since the eigenvalue of P 2 is 1: Examples:. Parity. Particles have “intrinsic parity” = § 1

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Discrete Symmetries

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  1. Discrete Symmetries Brian Meadows, U. Cincinnati.

  2. Parity • The operation that reverses the spatial coordinates is called the parity operation P: Its eigenvalue isP • P=§1 since the eigenvalue of P2 is 1: • Examples: Brian Meadows, U. Cincinnati

  3. Parity • Particles have “intrinsic parity” =§ 1 P |> = - |> ; P |q> = +1 (q is a quark); etc.. • We define parity of quarks (ie the proton) to be +. (ieP=+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) Brian Meadows, U. Cincinnati

  4. The Effect of Parity Brian Meadows, U. Cincinnati

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

  6. Charge Conjugation • Operator C turns particle into anti-particle. • C |+> = |-> ; C |K+> = |K-> ; C |q> = |q> ; etc. • C can only be a good quantum number for neutrals • C2 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) Brian Meadows, U. Cincinnati

  7. Charge Conjugation for Charged Particles • For charged particles, it is convenient to define a related operator G = C ei I2 • The exponential rotates a state about the I2 axis by  • This flips the sign of I3 making charge change sign • Like P, this introduces a factor (-1)I • Then G|+> = C ei I2|+> = C(-1)I|-> = -|+> • Therefore, the G-parity of the  is -1. • Like C, G-parity is conserved in strong and e/m interactions • Example: What is the G-parity of the  meson? • The  is known to decay strongly to  • Therefore its G-parity is +1 Brian Meadows, U. Cincinnati

  8. 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. K0 observed to decay into +- (CP=+1) as well as into -+0 (CP=-1) • B0 decays to J/psi Ks, J/psi KL and +- Brian Meadows, U. Cincinnati

  9. The Effect of T Brian Meadows, U. Cincinnati

  10. 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 Brian Meadows, U. Cincinnati

  11. CP and the K0 Particle • The K0 is a pseudo-scalar particle, therefore P |K0> = - |K0> and P |K0> = - |K0> • The C operator just turns K0 into K0 and vice-versa C |K0> = + |K0> and C |K0> = + |K0> • Therefore, the combined operator CP is CP |K0> = - |K0> and CP |K0> = - |K0> • So we can define odd- and even-CP eigen-states K1 and K2: |K1> = (|K0> - |K0>) / \/2  CP |K1> = (-1) |K1> |K2> = (|K0> + |K0>) / \/2  CP |K1> = (+1) |K1> Brian Meadows, U. Cincinnati

  12. CP and K0-K0 Mixing • Experimentally, it is observed that there are two K0 decay modes labeled as KL and Ks: Ks +-(s= 0.893 x 10-10 s) KL  +-0(L= 0.517 x 10-7 s) • The decay products of the Ks have P = (-1)L = (-1)0 = +1 • For the KL the products have P = -1 • It is tempting to assign KL to K1 and Ks to K2 However, this is not exactly correct: V. Fitch and J. Cronin observed, in an experiment at Brookhaven, that about 1 in 500 times, either Ks 3por KL  2p Brian Meadows, U. Cincinnati

  13. CP and K0-K0 Mixing • Rather than assigning KL to K1 and Ks to K2we define: • Fitch and Cronin’s (and subsequent) experiments lead to Brian Meadows, U. Cincinnati

  14. W d s K0 K0 u, c, t u, c, t W d s CP and K0-K0 Mixing • It is possible for a K0 to become a K0 ! • The main diagram contributing to mixing in the K0 system: • This contributes to the observation of CP violation in the K0K0 system. • It generates a difference in mass between K1 and K2 Brian Meadows, U. Cincinnati

  15. K0-K0 Mixing • The Ks and KL wave-functions have time-dependence related to their complex masses M+½i~ / |Ks.L> ´s,L(t)=K,L(0)exp [- i(Ms,L /~+ ½ i/s,L)t] • If KL = K1 and Ks = K2 K0 = [L(t) + s(t)] / \/2 • At time t the intensity of a beam of K0 will be IK0(t) = |K0|2 = ¼ [e -t /s + e -t /L + 2 e- ½ (1/ s +1/ L)t cos (m t/~)] where  = +1 for K0 and -1 for K0(m ´ML-Ms) • This is referred to as “K0-K0 mixing” Brian Meadows, U. Cincinnati

  16. Strangeness Oscillations • Graph shows I(K0) and I(K0) as function of t for D mts/ ~ = 0.5 • Experimentally, measure hyperon production in matter (due to K^0, not K^0) as function of distance from source of K0) • D m ts/ ~ = 0.498. • This corresponds to D m/m ¼ 5 x 10-15 ! Brian Meadows, U. Cincinnati

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