1. This solution reprinted from the solutions manual for the revised edition.

Transcription

1 Chapter Two 1. This solution reprinted from the solutions manual for the revised edition. 2. This solution reprinted from the solutions manual for the revised edition. 15

2 3. This solution reprinted from the solutions manual for the revised edition. 16

3 4. First, restating equations from the textbook, ν e = cos θ ν 1 sin θ ν 2 ν µ = sin θ ν 1 + cos θ ν 2 ) and E = pc (1+ m2 c 2 2p 2 Now, let the initial state ν e evolve in time to become a state α,t in the usual fashion α,t = e iht/ h ν e = cos θe ie1t/ h ν 1 sin θe ie2t/ h ν 2 ] = e [e ipct/ h im2 1 c3t/2p h cos θ ν 1 e im2 2 c3t/2p h sin θ ν 2 The probability that this state is observed to be a ν e is P (ν e ν e )= ν e α,t 2 = e im2 1 c3t/2p h cos 2 θ + e im2 2 c3t/2p h sin 2 θ 2 = cos 2 θ + e i m2 c 3t/2p h sin 2 θ 2 [ ] m = cos 4 θ + sin 4 θ + 2 cos 2 θ sin 2 2 c 3 t θ cos 2p h [ ] m = 1 sin 2 2θ sin 2 2 c 3 t 4p h Writing the nominal neutrino energy as E = pc and the flight distance L = ct we have [ ] P (ν e ν e )=1 sin 2 2θ sin 2 m 2 c 4 L 4E hc It is quite customary to ignore the factor of c 4 and agree to measure mass in units of energy, typically ev. The neutrino oscillation probability from KamLAND is plotted here: 17

4 Survival Probability Data - BG - Geo ν e Expectation based on osci. parameters determined by KamLAND L 0 /E νe (km/mev) The minimum in the oscillation probability directly gives us sin 2 2θ, that is 1 sin 2 2θ 0.4 so θ 25 The wavelength gives the mass difference parameter. We have 40 km MeV 4 hc 8π 200 MeV fm =2π = m2 m 2 where we explicitly agree to measure m 2 in ev 2. Therefore m 2 = 40π ev /10 3 = ev 2 The results from a detailed analysis by the collaboration, in Physical Review Letters 100(2008)221803, are tan 2 θ =0.56 (θ = 37 ) and m 2 = ev 2. The full analysis not only includes the fact that the source reactors are at varying distances (although clustered at a nominal distance), but also that neutrino oscillations are over three generations. 5. This solution reprinted from the solutions manual for the revised edition. 18

5 6. This solution reprinted from the solutions manual for the revised edition. Note: This is the proof of the so-called dipole sum rule. 7. This solution reprinted from the solutions manual for the revised edition. 19

6 8. This solution reprinted from the solutions manual for the revised edition. 9. This solution reprinted from the solutions manual for the revised edition. 20

7 10. This solution reprinted from the solutions manual for the revised edition. 21

8 22

9 11. This solution reprinted from the solutions manual for the revised edition. 23

10 12. This solution reprinted from the solutions manual for the revised edition. 24

11 13. This solution reprinted from the solutions manual for the revised edition. 14. This solution reprinted from the solutions manual for the revised edition. 25

12 15. This solution reprinted from the solutions manual for the revised edition. 16. This solution reprinted from the solutions manual for the revised edition. 26

13 17. This solution reprinted from the solutions manual for the revised edition. 27

14 18. This solution reprinted from the solutions manual for the revised edition. 19. This solution reprinted from the solutions manual for the revised edition. 28

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16 20. This solution reprinted from the solutions manual for the revised edition. 21. Starting with (2.5.17a), namely g(x, t) = exp( t 2 +2tx), carry out the suggested integral i.e. n=0 m=0 [ H n (x)h m (x)e x2 dx g(x, t)g(x, s)e x2 dx = = ] 1 (n!) 2 tn s m = π 1/2 e 2st (t+s)2 +2x(t+s) x 2 dx = e 2st e [x (t+s)]2 dx = π 1/2 e 2st n=0 2 n n! tn s n The sum on the right only includes terms where t and s have the same power, so the normalization integral on the left must be zero if n m. When n = m this gives [ ] 1 H n (x)h n (x)e x2 1/2 2n dx = π (n!) 2 n! or H 2 n(x)e x2 dx = π 1/2 2 n n! which is (2.5.29). In order to normalize the wave function (2.5.28), we compute ( ) mω h u n(x)u n (x)dx = c n 2 Hn 2 x e mωx2 / h dx = c n 2 h mω π1/2 2 n n! = 1 so that c n =(mω/π h) 1/4 (2 n n!) 1/2, taking c n to be real. Compare to (B.4.3). 30

17 22. This solution reprinted from the solutions manual for the revised edition. 23. This solution reprinted from the solutions manual for the revised edition. 31

18 24. This solution reprinted from the solutions manual for the revised edition. 25. This solution reprinted from the solutions manual for the revised edition. 32

19 26. This solution reprinted from the solutions manual for the revised edition. 33

20 27. Note: This was Problem 36 in Chapter Five in the Revised Edition. It was moved to this chapter because density of states is explicitly worked out now in this chapter. It seems, though, that I should have reworded the problem a bit. Refer back to the discussion in Section 2.5. The wave function is u E (x) = 1 L eik x where k x = 2π L n x and k y = 2π L n y and n x and n y are integers, with p = hk. The energy is E = p2 2m = h2 2m (k2 x + k 2 y)= 2π2 h 2 ml 2 (n2 x + n 2 y)= 2π2 h 2 ml 2 n2 so de = 4π2 h 2 ml 2 ndn The number of states with n between n and n + dn, and φ and φ + dφ, is ( ) 2 L dn = ndndφ = m dedφ 2π h so the density of states is simply m(l/2π h) 2. Remarkably, this result independent of energy. 34

21 28. This solution reprinted from the solutions manual for the revised edition. 35

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23 29. This solution reprinted from the solutions manual for the revised edition. 37

24 30. This solution reprinted from the solutions manual for the revised edition. 31. This solution reprinted from the solutions manual for the revised edition. 38

25 32. This solution reprinted from the solutions manual for the revised edition. 33. This solution reprinted from the solutions manual for the revised edition. 39

26 34. This solution reprinted from the solutions manual for the revised edition. 40

27 35. This solution reprinted from the solutions manual for the revised edition. 41

28 36. This solution reprinted from the solutions manual for the revised edition. 42

29 37. This solution reprinted from the solutions manual for the revised edition. 43

30 38. This solution reprinted from the solutions manual for the revised edition. 39. This solution reprinted from the solutions manual for the revised edition. 44

31 40.This solution reprinted from the solutions manual for the revised edition. 45

32 46