More General IBA Calculations

Spanning the triangle

How to use the IBA in real life

Classifying Structure -- The Symmetry TriangleClassifying Structure -- The Symmetry Triangle

Most nuclei do not exhibit the idealized symmetries but rather lie in transitional regions. Mapping the triangle.

Sph.

Deformed

Mapping the Triangle with a minimum of data -- Mapping the Triangle with a minimum of data -- exploiting an Ising-type Model –The IBAexploiting an Ising-type Model –The IBA

Competition between spherical-driving (pairing – like nucleon) and deformation-driving (esp. p-n)

interactions

H = aHsph + bHdef Structure ~ a/b

Sph.

Def.

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

V (γ) vs. χ H = -κ Q • Q

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 !!!

O(6)

SU(3)U(5)

What is the physical

meaning of If you think about zero point motion in a potential like

this, it is clear that

<γ> depends on 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.

O(6)

SU(3)U(5)

Mapping the Mapping the EntireEntire Triangle with a minimum of data Triangle with a minimum of data

2 parameters

2-D surface

H = ε nd - Q Q Parameters: , (within Q)

Awkward, though that varies from 0 to infinity

/ε

/ε

/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.

Spanning the Triangle

H = c [

ζ ( 1 – ζ ) nd

4NB Qχ ·Qχ - ]

ζ

χ

U(5)0+

2+ 0+

2+

4+

0

2.01

ζ = 0

O(6)

0+

2+

0+

2+

4+

0

2.51

ζ = 1, χ = 0

SU(3)

2γ+

0+

2+

4+ 3.33

10+ 0

ζ = 1, χ = -1.32

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

IBACQF Predictions for 168Er

γ

g

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

Universal O(6) – SU(3) Universal O(6) – SU(3) Contour Plots in the CQFContour Plots in the CQF

7 2/

H = -κ Q • Q

χ = 0 O(6) χ = = - 1.32 SU(3)

5( χ = - 2.958 )

O(6)

SU(3)U(5)

SU(3) O(6)

Now, what about more general calculations throughout the triangle

• Spanning the triangle• How do we fix the IBA parameters for any

given collective nucleus?

164 Er, a typical deformed nucleus

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

= 2.9R4/2

• 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: R4/2

Only provides a locus of structure

Vibrator Rotor

- soft

U(5) SU(3)

O(6)

3.3

3.1

2.92.7

2.5

2.2

Contour Plots in the TriangleContour Plots in the Triangle

U(5) SU(3)

O(6)

3.3

3.1

2.92.7

2.5

2.2

R4/2

SU(3)U(5)

O(6)

2.2

4

7

1310

17

2.2

4

7

1013

17

SU(3)U(5)

O(6)

SU(3)U(5)

O(6)

0.1

0.05

0.010.4

)2(

)2(

1

E

E

)2(

)0(

1

2

E

E

)22;2(

)02;2(

12

12

EB

EB

We have a problemWe have a problemWhat we have:

Lots of

What we need:

Just one

U(5) SU(3)

O(6)

+2.9+2.0

+1.4+0.4

+0.1

-0.1

-0.4

-1

-2.0 -3.0

)2(

)2()0(

1

2

E

EE

Fortunately:

)2(

)2()0(

1

22

E

EE)2(

)4(

1

1

E

EVibrator Rotor

γ - soft

Mapping Structure with Simple Observables – Technique of Orthogonal Crossing Contours

Burcu Cakirli et al.Beta decay exp. + IBA calcs.

SU(3)U(5)

O(6)

3.3

3.1

2.92.7

2.5

2.2

-3.0

-1.0-2.0

-0.1

+0.1

+1.0

+2.0

+2.9

U(5) SU(3)

O(6)

R4/2

)2(

)2()0(

1

2

E

EE

= 2.3 = 0.0

156Er

Trajectories at a Glance

-3.0

-1.0-2.0

-0.1

+0.1

+1.0

+2.0

+2.9

U(5) SU(3)

O(6)

SU(3)U(5)

O(6)

3.3

3.1

2.92.7

2.5

2.2

R4/2 )2(

)2()0(

1

2

E

EE

Evolution of StructureEvolution 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?