More General IBA Calculations Spanning the triangle How to use the IBA in real life

1. More General IBA Calculations Spanning the triangleHow to use the IBA in real life

2. Sph. Deformed Classifying Structure -- The Symmetry Triangle Most nuclei do not exhibit the idealized symmetries but rather lie in transitional regions. Mapping the triangle.

3. Mapping the Triangle with a minimum of data -- exploiting an Ising-type Model –The IBA Competition betweenspherical-driving (pairing – like nucleon)anddeformation-driving(esp. p-n)interactions H =aHsph +bHdef Structure ~a/b Def. Sph.

4. Relation of IBA Hamiltonian to Group Structure We will now see that this same Hamiltonian allows us to calculate the properties of a nucleus ANYWHERE in the triangle simply by choosing appropriate values of the parameters

5. O(6)  U(5) SU(3) V (γ) vs. χ H = -κ Q • Q What is the physical meaning of c If you think about zero point motion in a potential like this, it is clear that <γ> depends on c. For a flat potential the nucleus oscillates back and forth from 0 to 60 degrees so <γ> = 30 deg. For SU(3), <γ> will be small – nucleus is axially symmetric. Only minimum is at γ = 0o All γ excursions due to dynamical fluctuations in γ (γ-softness), not to rigid asymmetric shapes. This is confirmed experimentally !!!

6. O(6)  U(5) SU(3)

7. c Mapping the Entire Triangle with a minimum of data H =ε nd -  Q  Q Parameters: , c (within Q) /ε 2 parameters 2-D surface /ε Use of this form of the Hamiltonian, with T(E2) = aQ, is called the Consistent Q Formalism (or CQF). Roughly 94.68572382% of IBA calculations are done this way. Awkward, though that varies from 0 to infinity /ε

8. 0+ 4+ 2+ 2.5 1 2+ 0 0+ ζ H = c [ ( 1 – ζ ) nd - O(6) Qχ ·Qχ ] ζ = 1, χ = 0 4NB 0+ 2γ+ χ 3.33 4+ 2+ 0+ 2.0 4+ 1 2+ 2+ 1 ζ 0 0+ 0+ 0 U(5) SU(3) ζ = 0 ζ = 1, χ = -1.32 Spanning the Triangle

9. CQF along the O(6) – SU(3) leg H = -κ Q • Q Only a single parameter,  H =ε nd -  Q  Q Two parameters ε / and 

12. 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 Adecreases

13. O(6)  U(5) SU(3) Universal O(6) – SU(3) Contour Plots in the CQF H = -κ Q • Q χ = 0 O(6) χ = = - 1.32 SU(3) SU(3) O(6) ( χ = - 2.958 )

17. Now, what about more general calculations throughout the triangle • Spanning the triangle • How do we fix the IBA parameters for any given collective nucleus?

18. 164 Er, a typical deformed nucleus

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 2.7 2.9 2.5 3.1 3.3 2.2 A simple way to pinpoint structure. What do we need? Simplest Observable: R4/2 Only provides a locus of structure

21. Contour Plots in the Triangle 2.7 2.9 10 13 7 2.5 3.1 4 2.2 3.3 2.2 17 0.05 7 10 13 4 0.1 0.4 2.2 0.01 17 R4/2

22. We have a problem What we have: What we need: Lots of Just one +2.0 +2.9 +1.4 +0.4 +0.1 -1 -0.1 -0.4 -2.0 -3.0 Fortunately:

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

25. 156Er R4/2 = 0.0 = 2.3

26. Trajectories at a Glance R4/2

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?