IBA Lecture 3
IBA Lecture 3
Mapping the entire triangleTechnique of orthogonal crossing contours (OCC)
Mapping the Entire Triangle
2 parameters
2-D surface c
H = ε nd - Q Q
Parameters: , c (within Q) /ε
/ε
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 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 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.
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?
A particularly important recent result – masses are very sensitive to the parameters of the IBA in well-deformed nuclei
Masses can take up a role like spectroscopic observables to help identify structure.
We illustrate this briefly below
Two-neutron separation energies
Sn
Ba
Sm Hf
Pb
5
7
9
11
13
15
17
19
21
23
25
52 56 60 64 68 72 76 80 84 88 92 96 100 104 108 112 116 120 124 128 132
S(2
n)
MeV
Neutron Number
Normal behavior: ~ linear segments with drops after closed shellsDiscontinuities at first order phase transitionsS2n = A + BN + S2n (Coll.)
Use any collective model to calculate the collective contributions to S2n.
Binding Energies
Which 0+ level is collective and which is a 2-quasi-particle state?
Do collective model fits, assuming one or the other 0+ state, at 1222 or 1422 keV, is the collective one. Look at calculated
contributions to separation energies. What would we expect?
Evolution of level energies in rare earth nucleiBut note:
McCutchan et al
Collective contributions to masses can vary significantly for small parameter changes in collective models, especially for large boson
numbers where the collective binding can be quite large.
B.E
(M
eV)
B.E.( , )z cS2n(Coll.) for alternate fits
to Er with N = 100 S2n(Coll.) for two calcs.
Gd – Garcia Ramos et al, 2001
IBA
Masses: a new opportunity – complementary observable to spectroscopic data in pinning down structure, especially in nuclei with large numbers of
valence nucleons. Strategies for best doing that are still being worked out. Particularly important far off stability where data will be sparse.
Sn
Ba
Sm Hf
Pb
5
7
9
11
13
15
17
19
21
23
25
52 56 60 64 68 72 76 80 84 88 92 96 100 104 108 112 116 120 124 128 132
S(2n
) MeV
Neutron Number
Cakirli et al, 2009
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
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
Trajectories at a Glance
88 92 96 100 1042.0
2.2
2.4
2.6
2.8
3.0
3.2 Gd
N
R4/2
-4
-2
0
2
4
6
8
[ E(0
+ 2) - E
(2+ )
] / E(2
+ 1)
-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
E0s,
Two nucleon transfer
Two Nucleon Transfer Reactions: A New Interpretation for Phase
Transitional Regions and Collective Nuclei
Empirical survey of (p,t) reaction strengths to 0 + states
(p, t)
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
82 86 90 94 98 102 106 110 114 118 122 126
Neutron Number
Str
on
ge
st
0+
/GS
Pt
Os
W
Hf
Yb
Er
Dy
Gd
Sm
XO
Nearly always: cross sections to excited 0+
states are a small percentage of the
ground state cross section.
In the spherical – deformed transition
region at N = 90, excited state cross sections are comparable to those of
the ground state.
The “standard interpretation” (since ca. 1960s) of 2-nucleon transfer reactions to excited 0 + states in collective nuclei
• Most nuclei: Cross sections are small because the collective components add coherently for the ground state but cancel for the orthogonal excited states. (Special (“hot”) single particle orbits can give up to ~20% of the g.s, cross section.)
• Phase transition region: Spherical and deformed states coexist and mix. Hence a reaction such as (p,t) on a deformed 156 Gd target populates both the “quasi-deformed” ground and “quasi-spherical” excited states of 154 Gd. Well-known signature of phase transitions.
Symmetry Triangle of the IBA
Sph.
Def.
Shape/phase trans.
The IBA: convenient model that spans the entire triangle of colllective structures
c
H = ε nd - Q Q Parameters: , c (within Q) /ε
z
Sph. Driving Def. Driving
H = c [
ζ ( 1 – ζ ) nd 4NB Qχ ·Qχ - ]
Competition: : 0 to infinity /ε
Span triangle with z and c
Parameters already known for many nuclei
c is an overall scale factor giving the overall energy scale. Normally, it is fit to the first 2+ state.
IBA well-suited to this: embodies wide range of collective structures and, being based on s and d bosons, naturally
contains an appropriate transfer operator for L=0 -- s-boson
• Parameters for initial, final nuclei known so calculations are parameter-free
162Hf
0
0.2
0.4
0.6
0.8
1
1.2
1 2 3 4 5 6 7 8 9 10
162Hf
164Hf
0
0.2
0.4
0.6
0.8
1
1.2
1 2 3 4 5 6 7 8 9 10
164Hf
166Hf
0
0.2
0.4
0.6
0.8
1
1.2
1 2 3 4 5 6 7 8 9 10
166Hf
168Hf
0
0.2
0.4
0.6
0.8
1
1.2
1 2 3 4 5 6 7 8 9 10
168Hf
170Hf
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1 2 3 4 5 6 7 8 9 10
170Hf
172Hf
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1 2 3 4 5 6 7 8 9 10
172Hf
Look at Hf isotopes as an example:
Exp – all excited state cross sections are small
Gd Isotopes: Undergo rapid shape transition at N=90. Excited state cross sections are comparable to g.s.
Sph.
Def.
Shape/phase trans. line
~ 105 calculations
Big
Small
So, the model works well and can be used to look at predictions for 2-nucleon transfer strengths
Expect:
Let’s see what we get !
Huh !!???
Nuclear Model Codes at YaleComputer name: Titan
Connecting to SSH: Quick connect
Host name: titan.physics.yale.eduUser name: phy664Port Number 22Password: nuclear_codes
cd phintmpico filename.in (ctrl x, yes, return)runphintm filename (w/o extension)pico filename.out (ctrl x, return)
The following slides show the IBA input files and partial output files for U(5), SU(3) and O(6)
NOTE: O(6) q. # τ similar to U(5) ν except, since Ψ (O(6)) is mixed [in U(5) basis] τ applies to each component. Hence τ is good q. # for O(6), as is ν for U(5), but N – τ = # boson pairs is conserved in U(5) but not in O(6)
Input $diag eps = 0.20, kappa = 0.00, chi =-0.00, nphmax = 6, iai = 0, iam = 6, neig = 3, mult=.t.,ell=0.0,pair=0.0,oct=0.0,ippm=1,print=.t. $ $em E2SD=1.0, E2DD=-0.00 $ SLCT 2 2+ 0+ 2 99999
Output---------------------------
L P = 0+
Basis vectors |NR> = |ND,NB,NC,LD,NF,L P> --------------------------- | 1> = | 0, 0, 0, 0, 0, 0+> | 2> = | 2, 1, 0, 0, 0, 0+> | 3> = | 3, 0, 1, 0, 0, 0+> | 4> = | 4, 2, 0, 0, 0, 0+> | 5> = | 5, 1, 1, 0, 0, 0+> | 6> = | 6, 0, 2, 0, 0, 0+> | 7> = | 6, 3, 0, 0, 0, 0+>
Energies 0.0000 0.4000 0.6000 0.8000 1.0000 1.2000 1.2000
Eigenvectors
1: 1.000 0.000 0.000 2: 0.000 1.000 0.000 3: 0.000 0.000 1.000 4: 0.000 0.000 0.000 5: 0.000 0.000 0.000 6: 0.000 0.000 0.000 7: 0.000 0.000 0.000
---------------------------L P = 1+No states---------------------------L P = 2+
Energies 0.2000 0.4000 0.6000 0.8000 0.8000 1.0000 1.0000 1.2000 1.2000---------------------------L P = 3+ Energies 0.6000 1.0000 1.2000---------------------------L P = 4+
Energies 0.4000 0.6000 0.8000 0.8000 1.0000 1.0000 1.2000 1.2000 1.2000---------------------------L P = 5+ Energies 0.8000 1.0000 1.2000 ---------------------------L P = 6+ Energies 0.6000 0.8000 1.0000 1.0000 1.2000 1.2000 1.2000--------------------------
Transitions: 2+ -> 0+ (BE2) 2+,1 -> 0+,1: 6.00000 2+,1 -> 0+,2: 2.00000 2+,1 -> 0+,3: 0.00000 2+,2 -> 0+,1: 0.00000 2+,2 -> 0+,2: 0.00000 2+,2 -> 0+,3: 2.40000 2+,3 -> 0+,1: 0.00000 2+,3 -> 0+,2: 5.60000 2+,3 -> 0+,3: 0.00000 and 0+ -> 2+ (BE2) 0+,1 -> 2+,1: 30.00000 0+,2 -> 2+,1: 10.00000 0+,3 -> 2+,1: 0.00000 0+,1 -> 2+,2: 0.00000 0+,2 -> 2+,2: 0.00000 0+,3 -> 2+,2: 12.00000 0+,1 -> 2+,3: 0.00000 0+,2 -> 2+,3: 28.00000 0+,3 -> 2+,3: 0.00000
Transitions: 4+ -> 2+ (BE2) 4+,1 -> 2+,1: 10.00000 4+,1 -> 2+,2: 0.00000 4+,1 -> 2+,3: 2.28571 4+,2 -> 2+,1: 0.00000 4+,2 -> 2+,2: 6.28571 4+,2 -> 2+,3: 0.00000 4+,3 -> 2+,1: 0.00000 4+,3 -> 2+,2: 0.00000 4+,3 -> 2+,3: 3.85714
U(5)
Basis
Energies
Pert.WaveFcts.
U(5) in U(5) basis
******************** Input file contents ******************** $diag eps = 0.00, kappa = 0.02, chi =-1.3229, nphmax = 6, iai = 0, iam = 6, neig = 5, mult=.t.,ell=0.0,pair=0.0,oct=0.0,ippm=1,print=.t. $ $em E2SD=1.0, E2DD=-2.598 $ 99999 *************************************************************---------------------------
L P = 0+
Basis vectors |NR> = |ND,NB,NC,LD,NF,L P> --------------------------- | 1> = | 0, 0, 0, 0, 0, 0+> | 2> = | 2, 1, 0, 0, 0, 0+> | 3> = | 3, 0, 1, 0, 0, 0+> | 4> = | 4, 2, 0, 0, 0, 0+> | 5> = | 5, 1, 1, 0, 0, 0+> | 6> = | 6, 0, 2, 0, 0, 0+> | 7> = | 6, 3, 0, 0, 0, 0+>
Energies 0.0000 0.6600 1.0800 1.2600 1.2600 1.5600 1.8000
Eigenvectors
1: 0.134 0.385 -0.524 -0.235 0.398 2: 0.463 0.600 -0.181 0.041 -0.069 3: -0.404 -0.204 -0.554 -0.557 -0.308 4: 0.606 -0.175 0.030 -0.375 -0.616 5: -0.422 0.456 -0.114 0.255 -0.432 6: -0.078 0.146 -0.068 0.245 -0.415 7: 0.233 -0.437 -0.606 0.606 0.057
---------------------------
L P = 1+
No states---------------------------
L P = 2+
Energies 0.0450 0.7050 0.7050 1.1250 1.1250 1.3050 1.3050 1.6050 ---------------------------
L P = 3+ Energies 0.7500 1.1700 1.6500---------------------------
L P = 4+ Energies 0.1500 0.8100 0.8100 1.2300 1.2300 1.2300 1.4100 1.4100 ---------------------------
L P = 5+ Energies 0.8850 1.3050 1.3050---------------------------
L P = 6+
Energies 0.3150 0.9750 0.9750 1.3950 1.3950 1.5750 1.5750---------------------------
Binding energy = -1.2000 , eps-eff = -0.1550
SU(3)
SU(3) wave fcts. in U(5)
basis