More General IBA Calculations Spanning the triangle How to use the IBA in real life.
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Transcript of 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.
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
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?
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