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Solution to exercises

Solution to exercises. Exercise 7 The total amplitude |A+B| = √(|A| 2 + |B| 2 ) CP conjugation changes f w to – f w The magnitude of |A+B| and |A+B| is identical  no observable CP violation. CP. I. I. b). a). R. R. Solution to exercises. Exercise 7 (continued)

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Solution to exercises

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  1. Solution to exercises • Exercise 7 • The total amplitude |A+B| = √(|A|2 + |B|2) • CP conjugation changes fw to –fw • The magnitude of |A+B| and |A+B| is identical  no observable CP violation CP I I b) a) R R Niels Tuning (1)

  2. Solution to exercises • Exercise 7 (continued) • Repeat of exercise with 45o additional strong phase • Amplitude |A+B| and |A+B| now different  observable CP violation. Conclusion: you need both a CP violating and a CP invariant phase to observe CP violation CP I I B +fw fs fs R R A -fw Phase difference between A and B: φs+φw Phase difference between A and B: φs-φw φs+φw=90+45=135 Φs-φw=90-45=45 Niels Tuning (2)

  3. Solution to exercises • Exercise 8 • The decay diagram for B0 and B0 • The P operation interchanges the p+ and the p-. The C operation changes each p+ in a p- and vice versa, thus CP(p+p-) = C(p-p+) = p+p-. Therefore p+p- is an eigenstate of CP • The direct decay of the B0 proceeds to the left diagram above. The decay through mixing of the B0 proceeds through the right decay diagram above preceded by a B0/B0bar mixing diagram d d Vud Vud* u u Vub* Vub u u b b d d d d Niels Tuning (3)

  4. Solution to exercises • Exercise 8 (continued) • The direct decay has a phase –g (because it has Vub*). The decay via mixing has phase 2b (from mixing) plus a phase g (because it has Vub) adding up to 2b+g. The net phase difference is therefore 2b+g-(-g) = 2b+2g • The three angles of the unitarity triangle are a,b,g. Since in a triangle a+b+g=p by construction, you find that 2b+2g=-2a+2p=-2a Niels Tuning (4)

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