Weak Interactions Cabbibo Angle and Selection Rules

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1 Particle and s Cabbibo Angle and 03/22/2018 My Office Hours: Thursday 1:00-3:00 PM 212 Keen Building

2 Outline

3 Helicity Helicity: Spin quantization along direction of motion.

4 Helicity Helicity: Spin quantization along direction of motion. Helicity is just the projection of the spin onto the direction of linear momentum. The helicity of a particle is right-handed if the direction of its spin is the same as the direction of its motion and left-handed if opposite. Because the eigenvalues of spin with respect to an axis have discrete values, the eigenvalues of helicity are also discrete.

5 Helicity Helicity: Spin quantization along direction of motion. In nature, only left-handed neutrinos and right-handed antineutrinos observed!! Violates C invariance: ν e and ν e should have identical weak interactions. Violates P invariance: Parity reverses momentum while leaving spin unchanged converts ν L into ν R.

6 Helicity Helicity: Spin quantization along direction of motion. In nature, only left-handed neutrinos and right-handed antineutrinos observed!! However, observation compatible with CP invariance: CP(ν L ) = ν R

7 Helicity of Goldhaber et al. measured helicity of neutrino in 1958: Europium Samarium e Eu (J = 0) 152 Sm (J = 1) + ν e electron capture from atomic K shell Sm (J = 1) Sm (J = 0) + γ

8 Helicity of The helicity of ν e was deduced from the measured helicity of the photon by applying angular momentum conservation. The polarization of the photons was determined from their absorption in magnetized iron. Only left-handed neutrinos observed!! (in later experiments, only right-handed antineutrinos observed)

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10 The new theory makes some remarkable predictions: Heavy neutral vector boson Z 0 (+ weak reactions arising from the exchange) For example: ν α + N ν α + X Observed in 1973 neutral currents

11 Charged currents in weak decays of hadrons understood in terms of basic processes in which W ± bosons are emitted or absorped by their constituents quarks: W couples to leptons and quarks semileptonic decay: d u + e + ν e In contrast: leptonic process (W couples only to leptons) l + ν e l + ν e or e. g. τ µ + ν µ + ν τ

12 Charged currents in weak decays of hadrons understood in terms of basic processes in which W ± bosons are emitted or absorped by their constituents quarks: W couples to leptons and quarks semileptonic decay: d u + e + ν e In contrast: leptonic process (W couples only to leptons) l + ν e l + ν e or e. g. τ µ + ν µ + ν τ

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14 Families of quarks & leptons have identical weak interactions: ( ) ( ) u c and d s g W = g u d = g c s ( ) ( ) ν e ν µ e and µ Obtain the basic W ± - quark vertices by replacing ν e u e d ν µ c µ s in the basic W ± - lepton vertices. For example: µ e + ν e + ν µ d u + e + ν e e. g. n p + e + ν e

15 ( ) ( ) u c and d s g W = g u d = g ū d = g c s = g c s ( ) ( ) ν e ν µ e and µ d + ū W s + c W e + ν e W XXXXXXXXX allowed s + ū W d + c W e + ν µ W forbidden same coupling g W

16 ( ) ( ) u c and d s g W = g u d = g ū d = g c s = g c s ( ) ( ) ν e ν µ e and µ d + ū W s + c W e + ν e W XXXXXXXXX allowed s + ū W d + c W e + ν µ W forbidden Suppressed: K (s ū) µ + ν µ (s + ū W vertex).

17 Introduce quark mixing using Θ C Cabbibo angle d = d cos Θ C + s sin Θ C s = d sin Θ C + s cos Θ C and apply lepton-quark symmetry to ( ) ( ) u c d and s K s ū µ + ν µ now possible.

18 ( ) d s = ( cos θc sin θ C sin θ C cos θ C Introduce quark mixing using Θ C Cabbibo angle (Θ C 13 ) The transitions c d and s u, as compared to c s andthe d transitions u, arec therefore d and s suppressed u, as compared to byc a factor s and dof u, are sin 2 Θ C : cos 2 Θ C 1 : 20. ) ( ) d. (10.19) s Whether the state vectors d and s or the state vectors u and c are rotated, or indeed both pairs simultaneously, is a matter of convention alone. Only the difference in the rotation angles is of physical importance. Usually the vectors of the charge 1/3 quarks are rotated while those of the charge +2/3 quarksareleftuntouched. d = d cos Θ C + s sin Θ Experimentally, θ C is determined by comparing the lifetimes C and branching ratios of the s semi-leptonic and hadronic decays of various particles as shown in the sketch. This = d yields: sin Θ C + s cos Θ C sin θ C 0.22, and cos θ C (10.20) therefore suppressed by a factor of sin 2 θ C : cos 2 θ C 1:20. (10.21) ν µ ν e e u p d u ν e e p u d u u π d g µ W g u d W g. cos θ c d g u d W g.sinθ c s g. cosθ c M if g 2 n M if g 2. cos θ c Λ 0 M if g 2. cos θ c.sinθ c

19 In 1971, known particles: ν e, e, ν µ, µ & u, d, s X c quark completes lepton-quark symmetry. With third generation: Cabbibo-Kobayashi-Maskawa (CKM) Matrix d V u d V u s V u b d s = V c d V c s V c b s b V t d V t s V t b b Now, d, s, b are eigenstates of weak interaction.

20 Now, d, s, b are eigenstates of weak interaction: ( ) u ( ) c ( ) t ( ) ν e ( ) ν µ ( ) ν τ d s b e µ τ Examples: The following two decays appear similar Σ dds n udd + e + ν e Σ + uus n udd + e + + ν e However: Γ(Σ + n + e + + ν e ) Γ(Σ n + e + ν e ) <

21 Examples: The following two decays appear similar Σ dds n udd + e + ν e However: Σ + uus n udd + e + + ν e Γ(Σ + n + e + + ν e ) Γ(Σ n + e + ν e ) < Solution: Σ decay ( dds ddu ) involves single-quark transition Σ + decay ( uus ddu ) requires two transitions

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23 If S change in strangeness and Q change in total charge of quarks, then: Allowed processes: S = 0 and Q = ±1 S = Q = ±1 ( Q, S same sign) Examples (charge of u, c, t is + 2 3, charge of d, s, b is 1 3 ): u s + W + S = Q = 1 W s + c s u + W S = Q = 1 W + s + c

24 If S change in strangeness and Q change in total charge of quarks, then: Allowed processes: S = 0 and Q = ±1 S = Q = ±1 Not allowed: S = Q = ±1. ( Q, S same sign) Example: Σ + n + e + + ν e Σ n + e + ν e S = 1, Q = 1 S = 1, Q = 1 (okay)

25 If S change in strangeness and Q change in total charge of quarks, then: Allowed processes: S = 0 and Q = ±1 S = Q = ±1 ( Q, S same sign) Decay of Ω(1672) sss (S = 3) must proceed via a series of successive decays: Ω sss Ξ 0 ssu + π 0 with S = 1 (weak decay)

26 If S change in strangeness and Q change in total charge of quarks, then: Allowed processes: S = 0 and Q = ±1 S = Q = ±1 ( Q, S same sign) Decay of Ω(1672) sss (S = 3) must proceed via a series of successive decays: Ω sss Ξ 0 ssu + π 0 with S = 1 (weak decay) Ξ 0 ssu Λ sud + π 0 with S = 1 (weak decay)

27 If S change in strangeness and Q change in total charge of quarks, then: Allowed processes: S = 0 and Q = ±1 S = Q = ±1 ( Q, S same sign) Decay of Ω(1672) sss (S = 3) must proceed via a series of successive decays: Ω sss Ξ 0 ssu + π 0 with S = 1 (weak decay) Ξ 0 ssu Λ sud + π 0 with S = 1 (weak decay) Λ sud p + π with S = 1 (weak decay)