1. More General IBA Calculations Spanning the triangle

2. Transition regions away from dynamical symmetries in the IBA

3. Mapping the Triangle Mapping the Triangle 2 parameters 2-D surface   /ε H = ε n d -  Q  Q Parameters: ,  (within Q)  = 0 ε = 0  /ε

4. Along the O(6) – SU(3) leg H = -κ Q Q Only a single parameter, 

5. 1

7. Along the O(6) – SU(3) leg H = -κ Q Q Only a single parameter, 

8. Os isotopes from A = 186 to 192: Structure varies from a moderately gamma soft rotor to close to the O(6) gamma- independent limit. Describe simply with: H = -κ Q Q  : 0  small as A decreases 2

9. Along the O(6) – SU(3) leg H = -κ Q Q Only a single parameter, 

10. “Universal” IBA Calculations for the SU(3) – O(6) leg H = - κ Q Q κ is just energy scale factor Ψ ’ s, B(E2) ’ s independent of κ Results depend only on χ [ and, of course, vary with N B ] Can plot any observable as a set of contours vs. N B and χ. 3

11. Universal O(6) – SU(3) Contour Plots H = -κ Q Q χ = 0 O(6) χ = = - 1.32 SU(3) ( χ = - 2.958 )

16. Along the O(6) – SU(3) leg H = -κ Q Q Only a single parameter, 

17. Mapping the Entire Triangle Mapping the Entire Triangle 2 parameters 2-D surface  H = ε n d -  Q  Q Parameters: ,  (within Q) varies from 0 to infinity: unpleasant.  What to do? Rewrite Hamiltonian slightly.  /ε

18. Spanning the Triangle H = c [ ζ ( 1 – ζ ) n d 4N B Q χ ·Q χ - ] ζ χ U(5) 0+0+ 2+2+ 0+0+ 2+2+ 4+4+ 0 2.0 1 ζ = 0 O(6) 0+0+ 2+2+ 0+0+ 2+2+ 4+4+ 0 2.5 1 ζ = 1, χ = 0 SU(3) 2γ+2γ+ 0+0+ 2+2+ 4+4+ 3.33 1 0+0+ 0 ζ = 1, χ = -1.32

19. H has two parameters. A given observable can only specify one of them. What does this imply? An observable gives a contour of constant values within the triangle = 2.9 R4/2

20. At the basic level : 2 observables (to map any point in the symmetry triangle) Preferably with perpendicular trajectories in the triangle A simple way to pinpoint structure. What do we need? Simplest Observable: R 4/2 Only provides a locus of structure 3.3 3.1 2.9 2.7 2.5 2.2

21. Contour Plots in the Triangle 3.3 3.1 2.9 2.7 2.5 2.2 R 4/2 2.2 4 7 13 10 17 2.2 4 7 10 13 17 0.1 0.05 0.01 0.4

22. We have a problem What we have: Lots of What we need: Just one +2.9 +2.0 +1.4 +0.4 +0.1 -0.1 -0.4 -2.0-3.0 Fortunately:

23. VibratorRotor γ - soft Mapping Structure with Simple Observables – Technique of Orthogonal Crossing Contours Burcu Cakirli et al. Beta decay exp. + IBA calcs.

25. R 4/2 = 2.3 = 0.0 156 Er

26. Trajectories at a Glance R 4/2

27. Evolution of Structure Complementarity of macroscopic and microscopic approaches. Why do certain nuclei exhibit specific symmetries? Why these evolutionary trajectories? What will happen far from stability in regions of proton-neutron asymmetry and/or weak binding?

28. Backups

29. +2.9 +2.0 +1.4 +0.4 +0.1 -0.1 -0.4 -2.0-3.0 3.3 3.1 2.9 2.7 2.5 2.2 R 4/2 N = 10

30. Lets do some together Pick a nucleus, any collective nucleus 152-Gd (N=10) 186-W (N=11) Data 0+ 0 keV 0 keV 2+ 344 122 4+ 755 396 6+ 1227 809 0+ 615 883 2+ 1109 737 R42 = 2.19 zeta ~ 0.4 3.24 zeta ~ 0.7 R02 = -1.43 chi ~ =-1.32 +1.2 chi ~ -0.7 For N = 10 and kappa = 0.02 Epsilson = 4 x 0.02 x 10 [ (1 – zeta)/zeta] eps = 0.8 x [0.6 /0.4] ~ 1.2 0.8 x [0.3/0.7] ~ 0.33 STARTING POINTS – NEED TO FINE TUNE At the end, need to normalize energies to first J = 2 state. For now just look at energy ratios

31. 70:100:5 Alaga

32. Initial, final spins K values of the initial, final states 70:100:5