Transcript More General IBA Calculations Spanning the triangle

More General IBA
Calculations
Spanning the triangle
How to use the IBA in
real life
Classifying Structure -- The Symmetry Triangle
Deformed
Sph.
Most nuclei do not exhibit the idealized symmetries but
rather lie in transitional regions. Mapping the triangle.
Mapping the Triangle with a minimum of data -exploiting an Ising-type Model –The IBA
Competition between spherical-driving
(pairing – like nucleon) and deformation-driving (esp. p-n)
interactions
H = aHsph + bHdef
Structure ~
Def.
Sph.
a/b
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
What is the
physical
meaning of 
O(6)

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 !!!
U(5)
SU(3)
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)

U(5)
SU(3)
Mapping the Entire Triangle with a minimum of data
H = ε nd
- QQ
Parameters: /ε ,  (within Q)
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 /ε varies from 0 to infinity
Spanning the Triangle
ζ
H = c [ ( 1 – ζ ) nd Qχ ·Qχ ]
4NB
0+
4+
2+
2+
1
0+
0
2.5
O(6)
ζ = 1, χ = 0
χ
0+
2γ+
4+
2+
2+
1
0+
0
0+
2.0
ζ
U(5)
SU(3)
ζ=0
ζ = 1, χ = -1.32
4+
3.33
2+
0+
1
0
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) gammaindependent limit. Describe simply with:
H = -κ Q • Q
 : 0  small as A decreases
Universal O(6) – SU(3)
Contour Plots in the CQF
O(6)

U(5)
SU(3)
SU(3)
O(6)
H = -κ Q • Q
χ = 0 O(6)
χ =  7 / 2 = - 1.32 SU(3)
( 5 χ = - 2.958 )
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
R4/2 = 2.9
A simple way to pinpoint structure.
What do we need?
• At the basic level : 2 observables (to map any point in the
symmetry triangle)
• Preferably with perpendicular trajectories in the triangle
Simplest Observable: R4/2
O(6)
 - soft
2.7
2.9
Only provides a locus of
structure
2.5
2.2
U(5)
Vibrator
3.1
3.3
SU(3)
Rotor
Contour Plots in the Triangle
O(6)
2.7
2.9
R4/2
O(6)

2

1
E (0 )
E (2 )
7
10
13
2.5
4
3.1
2.2
3.3
2.2
SU(3)
U(5)
O(6)

E ( 2 )

1
E (2 )
4
SU(3)
U(5)

2

2

1

1
B( E 2;2  0 )
B( E 2;2  2 )
7
10
13
17
O(6)
0.05
0.1
2.2
U(5)
0.01
17
SU(3)
U(5)
0.4
SU(3)
We have a problem
What we have:
What we need:
Lots of
Just one
Fortunately:

2
O(6)

E (0 )  E (2 )

1
+2.0
+2.9
E (2 )
+1.4
+0.4
+0.1
-0.1
U(5)
-1
-0.4 -2.0
-3.0
SU(3)
Mapping Structure with Simple Observables – Technique of
Orthogonal Crossing Contours
γ - soft
Vibrator
E (41 )
E (21 )
Burcu Cakirli et al.
Beta decay exp. + IBA calcs.
Rotor
E (02 )  E (22 )
E (21 )
156Er
E (02 )  E (2 )
O(6)
O(6)
2.7
2.9
R4/2
E (21 )
+2.9
+2.0
2.5
+1.0
3.1
+0.1
2.2
3.3
-0.1 -1.0
-2.0
SU(3)
U(5)
= 2.3
U(5)
-3.0
= 0.0
SU(3)
Trajectories at a Glance
O(6)
O(6)
E (02 )  E (2 )
2.7
2.9
R4/2
E (21 )
2.5
+2.9
+2.0
+1.0
3.1
+0.1
2.2
3.3
SU(3)
6
+
2
2.6
0
2.4
-2
2.2
-4
88
92
96
N
100
104
+
R4/2
4
2.8
+
8
Gd
3.0
2.0
U(5)
[ E(02) - E(2 ) ] / E(21)
U(5)
3.2
-0.1 -1.0
-2.0
-3.0
SU(3)
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?