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

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

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

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

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- Gabriela Popa
- Rochester Institute of Technology
- Collaborators:
- J. P. Draayer
- Louisiana State University
- J. G. Hirsch
- UNAM, Mexico
- Georgieva
- Bulgarian Academy of Science

Gabriela Popa

Introduction

What we know about the nucleus

Characteristic energy spectraTheoretical Model

Configuration space

System HamiltonianResults Conclusion and future work

Gabriela Popa

Chart of the nuclei

Z (protons)

N (number of neutrons)

Gabriela Popa

3

3

2

2

1

1

-vibration

-vibration

Gabriela Popa

8

6

6

5

6

0, 2, 4

4

4

3

2

8

0

2

2

6

4

2

0

0

spherical

deformed

E = n

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

Gabriela Popa

Experimental energy spectra of 162Dy

Gabriela Popa

N = 5

1h

h11/2

s1/2

3s

d3/2

d5/2

2d

N = 4

g7/2

1g

f9/2

p1/2

2p

f5/2

N = 3

p5/2

1f

f7/2

2s

d3/2

N = 2

s1/2

1d

d5/2

p1/2

1p

N = 1

p3/2

1s

N = 0

s1/2

l·s

l·l

Gabriela Popa

Valence space: U(Wp) U(Wn)

totalnormal unique

Wp =32Wnp =20Wup =12

Wn =44Wnn =30Wun=14

- Particle distributions: (l,m)
- protons: neutrons: p n total
- 152Nd 10 6 4 10 6 4(12,0)(18,0) (30, 0)
- 156Sm 12 6 612 6 6(12,0)(18,0) (30, 0)
- 160Gd 14 8 614 8 6(10,4)(18,4) (28, 4)
- 164Dy 16 10 616 10 6(10,4)(20,4) (30, 4)
- 168Er 18 10 818 10 8(10,4)(20,4) (30, 4)
- 172Yb 20 12 820 12 8(36,0)(12,0) (24, 0)
- 176Hf 22 14 822 14 8(8,30)(0,12) (8, 18)

Gabriela Popa

| = Ci | i

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

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

( , ) ( , ) = (, )

Gabriela Popa

Coupling proton and neutron

irreps to total (coupled) SU(3):

(lp, mp) (ln, mn)

(lp+ln , mp+mn)

+ (lp+ln- 2,mp +mn+ 1)

+ (lp+ln+1,mp+mn- 2)

+ (lp+ln- 1,mp+mn- 1)2

+ ...

S m,l (lp+ln-2m+l,mp+mn+m-2l)k

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

Gabriela Popa

(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)

(,)(,)

(,)

Gabriela Popa

Invariants Invariants

Rot(3) SU(3)

Tr(Q2) C2

Tr(Q3) C3

b2~ l2+ lm + m + 3(l + m+1)

g =tan-1 [3 m / (2 l + m+3)]

Gabriela Popa

SU(3) conserving Hamiltonian:

H = c Q•Q + aL2 + b KJ2 + asym C2 + a3C3

+ one-body and two-body terms

+ Hs.p.p + Hs.p.n+Gp HPp +Gn HPn

Rewriting

this Hamiltonian becomes ...

H = Hspp +Hspn+Gp HPp +Gn HPn+ cQ•Q

+ aL2 + b KJ2 + asymC2 + a3 C3

Gabriela Popa

.

From systematics:

= 35/A 5/3 MeV

Gp = 21/A MeV

Gn = 19/A MeV

h = 41/A 1/3 MeV

p= 0.0637 p = 0.60

n= 0.0637 n= 0.60

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

Gabriela Popa

Exp. Th.

Exp. Th.

Exp. Th.

2.5

+

8

2

+

160

8

+

Gd

6

+

+

7

4

1.5

+

6

+

2

+

5

+

0

+

4

p

+

+

=

K

0

1

3

+

2

+

8

2

p

+

= 2

K

+

0.5

6

+

4

+

2

+

0

0

p

+

=

K

0

Gabriela Popa

Coupling proton and neutron

irreps to total (coupled) SU(3):

(lp, mp) (ln, mn)

(lp+ln , mp+mn)

+ (lp+ln- 2, mp +mn+ 1)

+ (lp+ln+1, mp+mn- 2)

+ (lp+ln- 1, mp+mn- 1)2

+ ...

m,l (lp+ln-2m+l,mp+mn+m-2l)k

Scissors

Twist

Scissors + Twist

… Orientation of the p-n system is quantized

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

Gabriela Popa

Nucleus

B(M1)

mode

160 ,162

Dy

(10,4)

(18,4)

(28,8)

(

29,

6)

0.56

t

(

26,

9)

1.77

s

1

(27

;

7)

1.82

s+t

2

(27

;

7)

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+

Gabriela Popa

2.5

2

1.5

1

0.5

0

-0.5

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

+

4

+

2

+

6

+

+

+

6

4

0

+

+

4

+

4

2

+

+

+

2

2

0

+

0

+

164

0

Dy

Energy [MeV]

K = 0

+

K = 0

+

6

6

+

+

5

5

+

+

4

4

+

3

+

3

+

2

+

+

2

6

+

6

K = 2

+

+

4

e)

c)

4

+

2

+

2

+

+

0

0

g.s.

… and M1 Strengths

Gabriela Popa

3

2.5

2

1.5

1

0.5

0

Exp. Th.

Exp. Th.

Exp. Th.

Exp. Th.

+

6

168

Er

+

4

+

2

+

0

+

6

+

8

Energy [MeV]

+

8

+

4

+

8

+

6

+

6

+

7

+

+

2

7

+

+

4

4

+

+

+

6

0

6

+

2

+

2

+

+

5

5

+

0

+

0

K = 0

+

+

4

4

+

+

8

8

+

+

3

3

+

+

2

2

+

+

6

6

K = 0

a)

+

4

+

4

K = 2

+

2

+

2

+

+

0

0

g.s.

… and M1 Strengths

Gabriela Popa

Nucleus

Calculated

Experiment

Pure

S

U

(3)

Theory

160

Dy

2.48

4.24

2.32

162

Dy

3.29

4.24

2.29

164

Dy

5.63

4.36

3.05

B(M1)[N2]

Gabriela Popa

Gabriela Popa

Gabriela Popa

Gabriela Popa

Gabriela Popa

Gabriela Popa

0

2

0

gs

2

100

100

80

80

60

60

a = (36,0)(12,0)x(24,0)

b = (28,10)(12,0)x(16,10)

c = (20,20)(4,10)x(16,10)

40

40

20

20

0

0

0

2

0

gs

2

Gabriela Popa

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

- B(E2) transitions within the g.s. band well
- reproduced

- 1+ energies fall in the correct energy range
- fragmentation in the B(M1) transition probabilities correctly predicted

Gabriela Popa

- 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.

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

Gabriela Popa

- In some nuclei total strength of the M1 distribution
- is larger than the experimental value

- Consider the states with J = 1 in the configuration space

- Consider the abnormal parity levels

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

- Investigate the M1 transitions in light nuclei

- Investigate the energy spectra in super-heavy nuclei

Gabriela Popa

Thank you!

Gabriela Popa