npsc-2003gabriela popa microscopic interpretation of the excited k = 0 +, 2 + bands of deformed...

NPSC-2003 Gabriela Popa

Microscopic interpretation of the excited K = 0+, 2+ bands

of deformed nuclei

Gabriela PopaRochester Institute of Technology

Collaborators:J. P. Draayer

Louisiana State University

J. G. HirschUNAM, Mexico

A. GeorgievaBulgarian Academy of Science


Outline

IntroductionIntroductionWhat we know about the nucleusWhat we know about the nucleusCharacteristic energy spectraCharacteristic energy spectraTheoretical Model Theoretical Model Configuration spaceConfiguration spaceSystem HamiltonianSystem HamiltonianResults Results Conclusion and future workConclusion and future work

Nuclear Physics Summer School June 15-27, 2003 Gabriela Popa

Chart of the nuclei

N (number of neutrons)

Z (protons)


Nuclear vibrations

-vibration

3

1

2

3

-vibration1

2

0, 2, 4

00246

8

6

8

20246

23456

spherical deformed

E = n E = I(I+1)/(2)

Schematic level schemes of spherical and deformed nuclei


Experimental energy levels

0

0.5

1

1.5

2

0

0.5

1

1.5

2

10 12 14 16 18 20 22

152Nd156Sm160Gd164Dy 168Er172Yb176Hf

2

21

02

0g.s.

Number of Pairs of Particles

Nuclear Physics Summer School June 15-27, 2003 Gabriela Popa

ExperimentalExperimental energy spectra of 162Dy

-0.5

0

0.5

1

1.5

2

2.5

3

162Dy

g.s.

K = 2

K = 0K = 0

2+4+

6+

8+

0+

2+4+6+

0+

4+

7+

2+

4+

6+

8+

3+

5+

7+

6+

5+1+

1+

1+

2+4+6+8+

0+

K = 4Ene

rgy

[MeV

]


Single particle energy levels

1s

1p

2s

1g

N = 3

N = 2

N = 1

N = 0

N = 4

1d

l·l l·s

d5/2

d3/2s1/2

1f2p

3s2d

1hN = 5

s1/2

p1/2p3/2

f7/2f5/2

p5/2

f9/2p1/2

d3/2d5/2

h11/2s1/2

g7/2


Valence space: U() U()

total normal unique

=32 n =20 u

=12 =44 n

=30 u=14

Particle distributions: protons: neutrons: total 152Nd 10 6 4 10 6 4 (12,0) (18,0) (30, 0)156Sm 12 6 6 12 6 6 (12,0) (18,0) (30, 0) 160Gd 14 8 6 14 8 6 (10,4) (18,4) (28, 4) 164Dy 16 10 6 16 10 6 (10,4) (20,4) (30, 4) 168Er 18 10 8 18 10 8 (10,4) (20,4) (30, 4) 172Yb 20 12 8 20 12 8 (36,0) (12,0) (24, 0) 176Hf 22 14 8 22 14 8 (8,30) (0,12) (8, 18)

Particle distribution


Wave Function

| = Ci | i

|i = |{; } (,)KL S; JM

( = , ) = n [f ] ( , ), S

( , ) ( , ) = (, )


Coupling proton and neutron irreps to total (coupled) SU(3):

+++- ++++-+-+-

m,l +-2m+l++m-2l)k

with the multiplicity denoted by k = k(m,l)

Direct Product Coupling


(10,4) (18,4) (28,8) (29,6) (30,4) (31,2) (32,0) (26,9) (27,7)(10,4) (20,0) (30,4)(10,4) (16,5) (26,9) (27,7)(10,4) (17,3) (27,7)(12,0) (18,4) (30,4)(12,0) (20,0) (32,0)(8,5) (18,4) (26,9) (27,7)(9,3) (18,4) (27,7)(7,7) (18,4) (25,11) (26,9) (27,7)(7,7) (20,0) (27,7)(4,10) (18,4) (22,14)

(,)(,) (,)

21st SU(3) irreps corresponding to the highest

C2 values were used in 160Gd


Tricks

Invariants InvariantsRot(3) SU(3)Tr(Q2) C2

Tr(Q3) C3 ~+ + + + =tan+


SU(3) conserving Hamiltonian:

H = Q•Q + a L2 + b KJ2 + asym C2 + a3 C3

this Hamiltonian becomes ...

+ one-body and two-body terms

+ Hs.p.Hs.p.

G HPG HP

Rewriting

H = Hsp

+Hsp+G HP

+G HP+Q•Q

+ a L2 + b KJ2 + asym C2 + a3 C3

.

System Hamiltonian


Parameters of the Pseudo-SU(3) Hamiltonian

Fit to experiment(fine tuning): a, b, asym and a3

= 0.0637 = 0.60

= 0.0637 = 0.60

From systematics: = 35/A 5/3 MeVG= 21/A MeVG= 19/A MeVh = 41/A 1/3 MeV


0

0.5

1

1.5

2

2.5

K0+

K+

K0

2

+

160Gd

0+

6+

0+

2+

2+

2+

4+

4+

8+

6+

3+

8+

7+

6+

4+

5+

8+

Exp. Th. Exp. Th. Exp. Th.

Energy spectrum for 160Gd


Coupling proton and neutron irreps to total (coupled) SU(3):

+++- ++++-+-+-

m,l +-2m+l++m-2l)k

Twist

Scissors

Scissors + Twist

… Orientation of the - system is quantized with the multiplicity denoted by k = k(m,l)

Protons

Neutrons

Direct Product Coupling


Nucleus B(M1) mode

160 ,162Dy (10,4) (18,4) (28,8) ( 29, 6) 0.56 t( 26, 9) 1.77 s(27; 7)1 1.82 s+t(27; 7)2 0.083 t+s

164 Dy (10,4) (20,4) (30,8) (31,6) 0.56 t(28,9) 1.83 s(29,7) 1.88 s+t(29,7) 0.09 t+s

() () () ()1+

M1 Transition Strengths [ 2N]

in the Pure Symmetry Limit of the

Pseudo SU(3) Model


0

0.2

0.4

0.6

0.8

1

1.2

1.4

2.2 3.0 3.7 4.5

Th.Exp.

164Dy

B(M

1,

0+ -

> 1+)

[n2 ]

Energy [MeV]

f)

-0.5

0

0.5

1

1.5

2

2.5

164 Dy

e)K = 2

K = 0 K = 0

2+4+

6+

0+2+4+

6+

0+

2+4+6+

0+

2+4+

0+

2+4+

0+

2+4+6+

0+

2+

4+6+

3+5+

2+4+6+

3+5+

g.s.

c)

Ene

rgy

[MeV

]Exp. Th. Exp. Th. Exp. Th. Exp. Th.

Energy levels of 164Dy

… and M1 Strengths


0

0.2

0.4

0.6

0.8

1

2.0 2.5 3.0 3.5 4.0 4.5

Th.168Er

B(M

1,

0+ -

> 1+)

[n2 ]

Energy [MeV]

b)

0

0.5

1

1.5

2

2.5

3

168 Er

K = 0

K = 0K = 2

0 +2 +

8 +

6 +

4 +

0 +2 +

8 +

6 +

4 +

0 +2 +

8 +

6 +

4 +

0 +2 +4 +

2 +

8 +

6 +

4 +

7 +

3 +5 +

2 +

8 +

6 +

4 +

7 +

3 +5 +

0 +2 +6 +

6 +

4 +0 +2 +

6+

4 +

g.s.

Ene

rgy

[MeV

]

Exp. Th. Exp. Th.Exp. Th. Exp. Th.

a)

Energy Levels of 168Er

… and M1 Strengths


Nucleus

ExperimentCalculated

Pure S U (3) Theory

160 Dy 2.48 4.24 2.32162 Dy 3.29 4.24 2.29164 Dy 5.63 4.36 3.05

B(M1)[N2]

Total B(M1) strength (N2)


First excited K=0+ and K=2+ states

0

0.5

1

1.5

2

0

0.5

1

1.5

2

10 12 14 16 18 20 22

152Nd 156Sm 160Gd 164Dy 168Er 172Yb 176Hf

2

21

02

0g.s.

Number of pairs of particles


0

10

20

30

40

50

60

70

80

0

10

20

30

40

50

60

70

80

0gs

2 02

0gs

2 02

156Sm

(30,0)(12,0)x(18,0)(26,6)(12,0)x(14,2)(26,2)(8,2)x(18,0)(24,6)(12,0)x(12,6)

SU

(3)

con

tent

[%

]


0

10

20

30

40

50

60

70

0

10

20

30

40

50

60

70

0gs

2 02

0gs

2 02

160Gd

a = (28,8)(10,4)x(18,4)b = (30,4)(10,4)x(18,4)c = (30,4)(10,4)x(20,4)d = (30,4)(12,0)x(18,4)e = (32,0)(10,4)x(18,4)f = (32,0)(12,0)x(20,0)


0

20

40

60

80

100

120

0

20

40

60

80

100

120

0gs

2 02

0gs

2 02

164Dy

a = (30,8)(10,4)x(20,4)b = (32,4)(10,4)x(20,4)c = (32,4)(12,0)x(20,4) d = (32,4)(10,4)x(20,4)e = (34,0)(10,4)x(20,4)f = (34,0)(12,0)x(22,0)g = (24,14)(10,4)x(14,10)


0

20

40

60

80

100

0

20

40

60

80

100

0gs

2 02

0gs

2 02

168Er

a = (30,8)(10,4)x(20,4)b = (32,4)(10,4)x(20,4)c = (32,4)(12,0)x(20,4)d = (32,4)(10,4)x(22,0)e = (34,0)(10,4)x(20,4)f = (24,14)(10,4)x(14,10)


0

20

40

60

80

100

0

20

40

60

80

100

0gs

2 02

0gs

2 02

a = (36,0)(12,0)x(24,0)b = (28,10)(12,0)x(16,10)c = (20,20)(4,10)x(16,10)

SU(3) content in 172Yb


Conclusions

•B(E2) transitions within the g.s. band wellreproduced

ground state, , first and second excited K = 0+ bands well described by a few representations

•calculated results in good agreement with the low-energy spectra

•1+ energies fall in the correct energy range

•fragmentation in the B(M1) transition probabilities correctly predicted


Conclusions

A proper description of collective properties of the first excited K = 0+ and K=2+ states must take into account the mixing of different SU(3)-irreps, which is driven by the Hamiltonian.

A microscopic interpretation of the relative position of the collective band, as well as that of the levels within the band, follows from an evaluation of the primary SU(3) content of the collective states. The latter are closely linked to nuclear deformation.

If the leading configuration is triaxial (nonzero ), the ground and bands belong to the same SU(3) irrep; if the leading SU(3) configuration is axial (=0), the K =0 and bands come from the same SU(3) irrep.


Future work

• In some nuclei total strength of the M1 distributionis larger than the experimental value

•Investigate the energy spectra in super-heavy nuclei•Investigate the M1 transitions in light nuclei

• There are new experiments that determine the inter-band B(E2) transitions• Improve the model to calculate these transitions

• Consider the abnormal parity levels• Consider the states with J = 1 in the configuration space

npsc-2003gabriela popa microscopic interpretation of the excited k = 0 +, 2 + bands of deformed...

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