IBA Lecture 3. Mapping the entire triangle Technique of orthogonal crossing contours (OCC)

32
IBA Lecture 3

Transcript of IBA Lecture 3. Mapping the entire triangle Technique of orthogonal crossing contours (OCC)

Page 1: IBA Lecture 3. Mapping the entire triangle Technique of orthogonal crossing contours (OCC)

IBA Lecture 3

Page 2: IBA Lecture 3. Mapping the entire triangle Technique of orthogonal crossing contours (OCC)

Mapping the entire triangleTechnique of orthogonal crossing contours (OCC)

Page 3: IBA Lecture 3. Mapping the entire triangle Technique 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

Page 4: IBA Lecture 3. Mapping the entire triangle Technique of orthogonal crossing contours (OCC)

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

Page 5: IBA Lecture 3. Mapping the entire triangle Technique of orthogonal crossing contours (OCC)

• 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

Page 6: IBA Lecture 3. Mapping the entire triangle Technique of orthogonal crossing contours (OCC)

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

Page 7: IBA Lecture 3. Mapping the entire triangle Technique of orthogonal crossing contours (OCC)

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:

Page 8: IBA Lecture 3. Mapping the entire triangle Technique of orthogonal crossing contours (OCC)

)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.

Page 9: IBA Lecture 3. Mapping the entire triangle Technique of orthogonal crossing contours (OCC)

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?

Page 10: IBA Lecture 3. Mapping the entire triangle Technique of orthogonal crossing contours (OCC)

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

Page 11: IBA Lecture 3. Mapping the entire triangle Technique of orthogonal crossing contours (OCC)

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

Page 12: IBA Lecture 3. Mapping the entire triangle Technique of orthogonal crossing contours (OCC)

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

Page 13: IBA Lecture 3. Mapping the entire triangle Technique of orthogonal crossing contours (OCC)

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

Page 14: IBA Lecture 3. Mapping the entire triangle Technique of orthogonal crossing contours (OCC)

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

Page 15: IBA Lecture 3. Mapping the entire triangle Technique of orthogonal crossing contours (OCC)

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

Page 16: IBA Lecture 3. Mapping the entire triangle Technique of orthogonal crossing contours (OCC)

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

Page 17: IBA Lecture 3. Mapping the entire triangle Technique of orthogonal crossing contours (OCC)

E0s,

Two nucleon transfer

Page 18: IBA Lecture 3. Mapping the entire triangle Technique of orthogonal crossing contours (OCC)

Two Nucleon Transfer Reactions: A New Interpretation for Phase

Transitional Regions and Collective Nuclei

Page 19: IBA Lecture 3. Mapping the entire triangle Technique of orthogonal crossing contours (OCC)

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.

Page 20: IBA Lecture 3. Mapping the entire triangle Technique of orthogonal crossing contours (OCC)

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.

Page 21: IBA Lecture 3. Mapping the entire triangle Technique of orthogonal crossing contours (OCC)

Symmetry Triangle of the IBA

Sph.

Def.

Shape/phase trans.

Page 22: IBA Lecture 3. Mapping the entire triangle Technique of orthogonal crossing contours (OCC)

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.

Page 23: IBA Lecture 3. Mapping the entire triangle Technique of orthogonal crossing contours (OCC)

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

Page 24: IBA Lecture 3. Mapping the entire triangle Technique of orthogonal crossing contours (OCC)

Gd Isotopes: Undergo rapid shape transition at N=90. Excited state cross sections are comparable to g.s.

Page 25: IBA Lecture 3. Mapping the entire triangle Technique of orthogonal crossing contours (OCC)

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 !

Page 26: IBA Lecture 3. Mapping the entire triangle Technique of orthogonal crossing contours (OCC)

Huh !!???

Page 27: IBA Lecture 3. Mapping the entire triangle Technique of orthogonal crossing contours (OCC)
Page 28: IBA Lecture 3. Mapping the entire triangle Technique of orthogonal crossing contours (OCC)

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)

Page 29: IBA Lecture 3. Mapping the entire triangle Technique of orthogonal crossing contours (OCC)

The following slides show the IBA input files and partial output files for U(5), SU(3) and O(6)

Page 30: IBA Lecture 3. Mapping the entire triangle Technique of orthogonal crossing contours (OCC)

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)

Page 31: IBA Lecture 3. Mapping the entire triangle Technique of orthogonal crossing contours (OCC)

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

Page 32: IBA Lecture 3. Mapping the entire triangle Technique of orthogonal crossing contours (OCC)

******************** 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