Post on 21-Dec-2015
More General IBA Calculations
Spanning the triangle
Transition regions away from dynamical symmetries in the
IBA
Mapping the TriangleMapping the Triangle
2 parameters
2-D surface
/ε
H = ε nd - Q Q Parameters: , (within Q)
= 0 ε = 0
/ε
Along the O(6) – SU(3)
leg
H = -κ Q • Q
Only a single parameter,
1
IBACQF Predictions for 168Er
γ
g
Along the O(6) – SU(3)
leg
H = -κ Q • Q
Only a single parameter,
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 decreases2
Along the O(6) – SU(3)
leg
H = -κ Q • Q
Only a single parameter,
““Universal” IBA Calculations Universal” IBA Calculations for the SU(3) for the SU(3) –– O(6) leg 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 NB ]
Can plot any observable as a set of contours vs. NB and χ.
3
Universal O(6) – SU(3) Universal O(6) – SU(3) Contour PlotsContour Plots
7 2/
H = -κ Q • Q
χ = 0 O(6) χ = = - 1.32 SU(3)
5( χ = - 2.958 )
Along the O(6) – SU(3)
leg
H = -κ Q • Q
Only a single parameter,
Mapping the Mapping the EntireEntire Triangle Triangle
2 parameters
2-D surface
H = ε nd - Q Q
Parameters: , (within Q)
varies from 0 to infinity: unpleasant.
What to do? Rewrite Hamiltonian slightly.
/ε
/ε
/ε
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
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?
Backups
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
U(5) SU(3)
O(6)
3.3
3.1
2.92.7
2.5
2.2
R4/2
N = 10
Lets do some together
• Pick a nucleus, any collective nucleus 152-Gd (N=10) 186-W (N=11) Data0+ 0 keV 0 keV2+ 344 1224+ 755 3966+ 1227 8090+ 615 8832+ 1109 737
R42 = 2.19 zeta ~ 0.4 3.24 zeta ~ 0.7R02 = -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
70:100:5 Alaga
Initial,
final
spins
K values
of the
initial,
final
states70:100:5