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- 1. SU(3)-symmetric sum rules for B ->PP decays Jordi París López MASTER THESIS SUPERVISOR: Rafel Escribano
- 2. Why do we look for Sum Rules? • Check validity of the Standard Model • Find deviations to the model ▫ Symmetry violation ▫ Sign of New Physics • Give predictions for observables 𝑖 𝐶𝑖 𝑂𝑖 = 0 𝑂𝑖 = 𝑅𝑖 − 𝐴𝑣𝑒𝑟𝑎𝑔𝑒𝑑 𝐵𝑟𝑎𝑛𝑐ℎ𝑖𝑛𝑔 𝑅𝑎𝑡𝑖𝑜𝑠 𝑅𝑖 𝐴 𝐶𝑃 𝑖 − 𝐶𝑃 𝐴𝑠𝑦𝑚𝑚𝑒𝑡𝑟𝑖𝑒𝑠
- 3. How do we obtain the Sum Rules? 𝑅𝑖 = 𝐵𝑅𝑖 + 𝐵𝑅𝑖 2 𝐴𝑖 𝐶𝑃 = 𝐵𝑅𝑖 − 𝐵𝑅𝑖 𝐵𝑅𝑖 + 𝐵𝑅𝑖 The observables are proportional to the amplitude squared: SUM RULES AMPLITUDES SQUARED AMPLITUDESEFFECTIVE HAMILTONIAN
- 4. Building the Effective Hamiltonian • Initial states: 𝐵3 = 𝐵+(𝑢 𝑏), 𝐵 𝑜(𝑑 𝑏), 𝐵𝑠 0 (𝑠 𝑏) • Final states: mesons made of (u,d,s) and ( 𝑢, 𝑑, 𝑠) 𝑃9 = − 𝜋0 2 − 𝜂0 3 − 𝜂8 6 𝜋+ 𝐾+ −𝜋− 𝜋0 2 − 𝜂0 3 − 𝜂8 6 𝐾0 −𝐾− 𝐾0 − 𝜂0 3 + 2 3 𝜂8
- 5. Building the Effective Hamiltonian • Shape of the Hamiltonian ℋ𝑒𝑓𝑓 = 𝐶 𝑖𝑗𝑘 𝐵3 (𝑖) 𝑃9 (𝑗) 𝑃9 (𝑘) ▫ 𝐶 𝑖𝑗𝑘 depends on 𝑉𝐶𝐾𝑀 • Use of SU(3) group theory to rewrite the Hamiltonian
- 6. Building the Effective Hamiltonian SU(3) decomposition ▫ 𝑞𝑞 𝑞 ⟶ 3 ⊗ 3 ⊗ 3 = 15 ⊕ 6 ⊕ 3 ⊕ 3 Tensorial definition 𝐻6 = 0 0 0 𝜆 𝑞 𝑑 0 0 𝜆 𝑞 𝑠 0 0 , 𝜆 𝑞 𝑑 0 0 0 0 0 0 −𝜆 𝑞 𝑠 𝜆 𝑞 𝑠 , −𝜆 𝑞 𝑠 0 0 0 𝜆 𝑞 𝑑 −𝜆 𝑞 𝑑 0 0 0 𝐻15 = 0 0 0 3𝜆 𝑞 𝑑 0 0 3𝜆 𝑞 𝑠 0 0 , 3𝜆 𝑞 𝑑 0 0 0 −2𝜆 𝑞 𝑑 0 0 −𝜆 𝑞 𝑠 𝜆 𝑞 𝑑 , 3𝜆 𝑞 𝑠 0 0 0 −𝜆 𝑞 𝑠 −𝜆 𝑞 𝑑 0 0 −2𝜆 𝑞 𝑠 𝐻3 = 0, 𝜆 𝑞 𝑑 , 𝜆 𝑞 𝑠 𝜆 𝑞 𝑞′ = 𝑉𝑞𝑏 ∗ 𝑉𝑞𝑞′ , q=(u,c)
- 7. Building the Effective Hamiltonian Final form: ▫ ℋ𝑒𝑓𝑓 (8) = 𝐴3 𝐵𝑖 𝐻3 𝑖 𝑃9 𝑘 𝑗 𝑃9 𝑗 𝑘 + 𝐶3 𝐵𝑖 𝑃9 𝑗 𝑖 𝑃9 𝑘 𝑗 𝐻3 𝑘 + + 𝐴6 𝐵𝑖 𝐻6 𝑘 𝑖𝑗 𝑃9 𝑗 𝑙 𝑃9 𝑙 𝑘 + 𝐶6 𝐵𝑖 𝑃9 𝑗 𝑖 𝐻6 𝑙 𝑗𝑘 𝑃9 𝑘 𝑙 + + 𝐴15 𝐵𝑖 𝐻15𝑘 𝑖𝑗 𝑃9 𝑗 𝑙 𝑃9 𝑙 𝑘 + 𝐶15 𝐵𝑖 𝑃9 𝑗 𝑖 𝐻15 𝑙 𝑗𝑘 𝑃9 𝑘 𝑙 ▫ ℋ𝑒𝑓𝑓 (1) = 𝐸3 𝐵𝑖 𝑃9 𝑗 𝑖 𝐻3 𝑗 𝑃9𝑘 𝑘 + 𝐷3 𝐵𝑖 𝐻3𝑗 𝑖 𝑃9 𝑗 𝑗 𝑃9 𝑘 𝑘 + + 𝐷6 𝐵𝑖 𝐻6 𝑘 𝑖𝑗 𝑃9 𝑗 𝑘 𝑃9 𝑙 𝑙 + 𝐷15 𝐵𝑖 𝐻15 𝑘 𝑖𝑗 𝑃9 𝑗 𝑘 𝑃9 𝑙 𝑙 ℋ𝑒𝑓𝑓 = ℋ𝑒𝑓𝑓 (8) + ℋ𝑒𝑓𝑓 (1) + ℎ. 𝑐.
- 8. Finding the Amplitudes In general the process 𝐵3 𝑖 𝑃9 𝑗 𝑃9 𝑘 appear in the Hamiltonian following Therefore: They are complex numbers, so 𝐵3 𝑖 𝑃9 𝑗 𝑃9 𝑘 𝜆 𝑞 𝑞′ ( 𝑓 𝐴3 + 𝑔 𝐴6 + ℎ 𝐴15 + ⋯ ) 𝐴𝑚𝑝𝑙𝑖𝑡𝑢𝑑𝑒 2 = |𝜆 𝑞 𝑞′ |2 · ( 𝐹 𝐴3 2 + 𝐺 𝐴6 2 + +𝐻 𝑅𝑒(𝐴3 · 𝐴6∗ ) + ⋯ ) 𝐴𝑚𝑝𝑙𝑖𝑡𝑢𝑑𝑒 ∝ 𝜆 𝑞 𝑞′ ( 𝑓 𝐴3 + 𝑔 𝐴6 + ℎ 𝐴15 + ⋯ )
- 9. Finding the Amplitudes Split processes by ∆𝑆 = 0 𝜆 𝑞 𝑑 and ∆𝑆 = 1 𝜆 𝑞 𝑠 : The processes cannot depend on their origin of SU(3). One can fix the coefficients 𝐶𝑖 by solving the ℱ = 0, 𝑒𝑡𝑐. ISSUE: 55 equations vs ≈30 processes: overdetermined system 𝑖 𝐶𝑖 𝐵𝑅𝑖 = 𝑖 𝐶𝑖( 𝐹𝑖 𝐴3 2 + 𝐺𝑖 𝐴6 2 + ⋯ ) = = 𝐴3 2 ℱ 𝑐1, … , 𝑐 𝑛 + 𝐴6 2 ℊ 𝑐1, … , 𝑐 𝑛 + ⋯ = 0 NEEDS SIMPLIFICATION
- 10. Rewriting the Hamiltonian One can use the knowledge of Feynman diagrams to rewrite the Hamiltonian into a diagramatic approach: C3,C6,C15,... terms are expressed in terms of the diagramatic “basis”
- 11. Rewriting the Hamiltonian Experimental data shows that E, A, PA, TS, CS and PS contributions are negligible in front of T, C, P, S. Amplitudes are written only as f(T,C,P,S), .e.g. Amplitudes squared will now have up to 10 terms The system of equations of the sum rules can be determined now. 𝐴𝑚𝑝𝑙𝑖𝑡𝑢𝑑𝑒 𝐵+ → 𝐾+ 𝜋0 = −𝜆 𝑞 𝑠 𝐶 + 𝑃 + 𝑇 2 𝑇 2 , 𝐶 2 , 𝑃 2 , 𝑆 2 , 𝑅𝑒 𝑇𝐶∗ , 𝑅𝑒 𝑇𝑃∗ , 𝑅𝑒 𝑇𝑆∗ , 𝑅𝑒 𝐶𝑃∗ , 𝑅𝑒 𝐶𝑆∗ , 𝑅𝑒 𝑃𝑆∗
- 12. SUM RULES Δ𝑆 = 0 Amplitudes of the processes in terms of 𝜆 𝑞 𝑑 :
- 13. SUM RULES Δ𝑆 = 0 There are 11 independent amplitudes. However, we are not considering the physical states. We’ll consider the states 𝜂 and 𝜂′ so that we can use the sum rules with tabulated data: 𝜂8 𝜂0 = − cos 𝜃 − sin 𝜃 sin 𝜃 − cos 𝜃 𝜂 𝜂′
- 14. SUM RULES Δ𝑆 = 0
- 15. SUM RULES Δ𝑆 = 0 Amplitudes of the processes in terms of 𝜆 𝑞 𝑑 :
- 16. SUM RULES Δ𝑆 = 0 The sum rules become: There are 11 constants for 10 equations: 1 coefficient must be sent to 0. 𝑖 𝐶𝑖 𝐵𝑅𝑖 = 𝑖 11 𝐶𝑖( 𝐹𝑖 𝑇 2 + 𝐺𝑖 𝐶 2 + ⋯ ) = = 𝑇 2 ℱ 𝑐1, … , 𝑐11 + 𝐶 2 ℊ 𝑐1, … , 𝑐11 + ⋯ = 0
- 17. SUM RULES Δ𝑆 = 0 Solving for every possible combination there are 3 independent sum rules. In general, they are functions of the mixing angle. Three of them are, in the limit 𝜃 = sin−1 − 1 3 :
- 18. • Build new secondary sum rules using common terms: • Check sum rules‘ zeros ⟹ 𝑆𝑈 3 symmetry check Using experimental data with the mixing angle 𝜃 = −14º for a secondary sum rule: Possibilities with the Sum Rules
- 19. • Prediction of a missing observable solving the sum rule: Using some experimental data: • Prediction of tw0 missing observables: In a hypothetical case: Leading to a region plot. Possibilities with the Sum Rules 𝑅 𝐵𝑠 0→𝜂𝐾0 = 1,1739 ± 0,1287 ⋅ 10−6
- 20. Possibilities with the Sum Rules
- 21. CONCLUSIONS • One cannot build Sum Rules considering the known processes: an approximation must be made. • The sum rules allow to give predictions to some observables up to corrections in an almost symmetric SU(3) world. • There is not enough experimental data of 𝐵 decays available to check the sum rules: in the following years sum rules should start playing relevant roles.
- 22. What now? • We must consider known SU(3) symmetry breaking contributions. ▫ The diagrammatic approach changes and so do the amplitudes. ▫ The sum rules will get corrections. ▫ Deviations from the SU(3) broken symmetry results could be regarded as signs of New Physics…