The Algebraic ApproachThe Algebraic ApproachThe Algebraic ApproachThe Algebraic Approach

1. Introduction

2. The building blocks

3. Dynamical symmetries

4. Single nucleon description

5. Critical point symmetries

6. Symmetry in n-p systems

7. Symmetry near the drip lines

Lecture 1 Lecture 2

NUCLEAR MEAN FIELD

Three ways to simplify Shell Model

Single particle motion

Describes properties in which a limited number of nucleons near the Fermi surface are involved.

Algebraic

Truncation of configuration space Dynamical symmetry

Geometrical Model

Collective motion (in phase)

Vibrations, rotations, deformations

Describes bulk properties depending in a smooth way on nucleon number

R Ij

Interacting BosonApproximation

III BBBAAA Drastic simplification of

shell model

Valence nucleons

Only certain configurations

Simple Hamiltonian-interactions

“Boson” model because it treats nucleons in pairs 2 fermions boson

154Sm 2+ states

2.89 x 1014

IBA: 26 2+ states Correctly chosen? Compare with Exp. IBA assumes that low lying collective states primarily involve excitations of pairs of fermions coupled to L = 0 (s), and L = 2 (d). [L = 1, 3 (p, f) for = - ]

IBM Basic ideas: Assume fermions couple in

pairs to bosons of spins

s boson is like a Cooper pair d boson is like a generalized pair

O+ s-boson 2+ d-boson

valence

Valence nucleons only s, d bosons

H = Hs + Hd + Hinteractions

Number of bosons fixed: N = ns + nd

= ½ # of val. protons + ½ # val. neutrons

jj

L

jjjj

L aa)(††)(

''

Why s, d bosons? Lowest state of all e-e

s nuclei is 0+

pairing, -fct 0+ First excited state in non-magic

d e-e nuclei almost always 2+

-fct gives 2+next above 0+

3- _____________1.87

Basic, attractive SD Interaction

E j j JT V j JT2 2 2 ; ;

0+

2+

4+

6+

(2J+1)+

0+ and 2+ lowest;separated from the rest.

Note key point: Bosons in IBA are pairs of fermions

in valence shell

Number of bosons for a given nucleus

is a fixed number

154

9262 Sm

N = 6 5 = N NB = 11

Compare “phonon” models we have considered.

Excitations are particle-hole excitations relative to

Fermi surface

XX XX XX XX

XX X X

X

X

X

X + + + …

Phonon number not constant

i j i ji j

phonon

Bosons counted from nearest closedshell (i.e. particles or holes).

[Eg 130Ba Z = 56 N = 74 N = 3 N= 4 ; N=7]WHY???

7/2 5/2 3/2 1/2

Pauli Principle

Consider f7/2 “shell”with 6 neutrons M

Maximum seniority = 2Maximum (d)-boson number =1

I BA Models I BA – 1 No distinction of p, n I BA – 2 Explicitly write p, n parts I BA – 3, 4 Take isospin into account p-n pairs I BFM Int. Bos. Fermion Model Odd A nuclei H = He – e + Hs.p. + Hint

I BFFM Odd – odd nuclei E’s, B(E ), B(M ), (p, t), BE’s (d, t), (d, p) s, s† , d, d† operators [ (f, p) bosons for = - states ] H = H (s, d)

core

DYNAMICAL SYMMETRY

•Describes basic states of motion available to a system - including relative motion of different constituents

G1 G2 G3 ……….

H= aC1 [G1] † bC2 [G2] † cC3 [G3] † ……

G1 G2 G3 ……….

H= aC1 [G1] † bC2 [G2] † cC3 [G3] † ……

i i i

•Dynamical symmetry breaking splits but does not mix the eigenstates

Some ‘Working Definitions’Some ‘Working Definitions’

• Have states s and d with = -2,-1,0,1,2

- 6 -dim. vector space. • Unitary transformations involving the

operators s, s†, d, d†

=> ‘rotations that form the group U(6).

• Can form 36 bilinear combinations which close on commutation,

s†s, s†d, d†s, (d

†d) (L)

(eg: [d†s,s†s] = d†s)

- these are the generators

[Analogy: Angular momentum: Jx,Jy,Jz generate rotations and form group 0(3)]

[For 0(3), use Jz,J ; J = Jx ± i Jy

Then [J+,J-] = 2Jz ; [Jz,J] = ± J]

• A Casimir operator commutes with all the generators of a group.

– Eg: C1U(6) = N; C2U(6) = N(N+5)

• Now look for subsets of generators which form a subgroup.

Eg: (d†d)(L) - 25 -U(5)

(d†d)(1), (d†d)(3) -10 -0(5)

(d†d)(1)

-3 -0(3)

• ie: U(6) U(5) 0(5) 0(3) - group chain decomposition

• Now form a Hamiltonian from the Casimir operators of the groups.

H = H = CC1U(6) 1U(6) + + CC2U(6)2U(6) CC2U(5) 2U(5) + + CC2O(5) 2O(5) ++CC2O(3)2O(3)H = H = CC1U(6) 1U(6) + + CC2U(6)2U(6) CC2U(5) 2U(5) + + CC2O(5) 2O(5) ++CC2O(3)2O(3)

All C’s commute and H is diagonal DYNAMICAL SYMMETRYDYNAMICAL SYMMETRY

Example of angular momentumExample of angular momentum

•For O(3), generators are Jz, J+ and J-

•Then [J2, Jz] = [J2, J+] = [J2, J-] = 0

• C2O(3) = J2

•Subgroup O(2) simply Jz = C1O(2)

•So H = C2O(3) + C1O(2)

•E= J(J+1) + M

O(2)

J1

J2

M= +Jz

M= -Jz

O(3)

I. “U(5)”

- Anharmonic Vibrator

II. “SU(3)”

- Axially symmetric rotor

III. “O(6)” - Gamma - unstable rotor

0(3) 0(5) U(5) U(6) J

Δn ν

dn N

0(3) SU(3) U(6) KJ ),( N

0(3) 0(5) U(6) U(6) J

Δ N

Only 3 chains from U(6)Only 3 chains from U(6)

U(5)

NeEB B2)02;2( )1(2

)02;2(

)24;2( 2

NeEB

EBB

R4/2= 2.0

SU(3)

5

)32()02;2( 2 NN

eEB B

)32(2

)52)(22(

7

10

)02;2(

)24;2(

NN

NN

EB

EB

R4/2= 3.33

O(6)

5

)4()02;2( 2 NN

eEB B

)4(

)5)(1(

7

10

)02;2(

)24;2(

NN

NN

EB

EB

R4/2= 2.5

The first O(6) nucleus ………..

Cizewski et al, Phys Rev Lett. 40, 167 (1978)

and then many more….

Transition Regions and Realistic CalculationsTransition Regions and Realistic Calculations

Q B

e T(E2) and

(2)d)(d(2)s)dd(s Q where

2L 1

a ( 2Q 2

a d

εn H

)

O(6) 0 SU(3); 27- 0;ε For

•Most nuclei do not satisfy the strict criteria of any of the 3 Dyn. Symm.

•Need numerical calculations by diagonalizing HIBA in s – d boson basis

•Can use a very simple form of the most general H

Symmetry triangle of the IBA

Gives all 3 symm. Symmetries → Analytic for E’s, B(E2)’s, etc. Transition Regions: function of ONE Parameter

– ratio of coeffs, or Extremely simple

H = εnd + a2 Q ( ) • Q ( )

ε, a2,

ε = 0

O(6)

U(5) SU(3) a2

= 7

2

a2

= 0

ε / a2

ε / a2 ε / a2

Examples

1) Well deformed nucleus – 168Er H ~ a2 Q • Q between 0 and - 7/2

Calculations with ~ - 0.4 work well.

Fix a2 from E( +12 )

2) Os isotopes O(6) → SU(3)

3) Universal Calculations: O(6) → SU(3)

4) Mapping the triangle.

O(6)

U(5) SU(3)

O(6)

U(5) SU(3)

H = ~ a2 Q • Q

: 0 to ~ - 0.4

H ~ a2 Q • Q

: 0 → - 1.32

O(6)

U(5) SU(3)

Z=38-82 2.05 < R 4/2 < 3.15 =0.03 MeV

N.V. Zamfir, R.F. Casten, Physics Letters B 341 (1994) 1-5

SummarySummary

• Algebraic approach contains aspects of both geometrical and single particle descriptions.

• Dynamical symmetries describe states of motion of system• Analytic Hamiltonian is a sum of Casimir operators of the

subgroups in the chain.• Casimir operators commute with generators of the group;

conserve a quantum number• Each Casimir lifts the degeneracy of the states without

mixing them.• Three and only three chains possible; O(6) was the

surprise.• Very simple CQF Hamiltonian describes large ranges of

low-lying structure

0

2

4

6

2

4

0

3 2

Vibrational Transitional Rotational

0

2

4

6

0

2

2

34

V(

)

V(

)

V(

)

Evolution of nuclear shape

E = nħω E = J(J+1)

?

Previously, no analytic solution to describe nuclei at the “transitional point”

Critical Point Symmetries

F. Iachello, Phys. Rev. Lett. 85, 3580 (2000); 87, 052502 (2001).

V(β)

β

Approximate potential at phase transition with infinite square well

Solve Bohr Hamiltonian with square well potential Result is analytic solution in terms of zeros of special Bessel functions

Predictions for energies and electromagnetic transition probabilities

V(

)

Spherical Vibrator

Symmetric Rotor

γ-soft

X(5)

E(5)Two solutions depending on

γ degree of freedom

E50

0

21

4 22

6 4 3 03

0

2

ξ = 1

ξ = 2τ = 0

τ = 1

R4/2 = 2.20E(02)/E(21) = 3.03

E(03)/E(21) = 3.59

X50

2

4

6

8

10

100

158

198

227

261

0

2

4

79

120

Key SignaturesE(41)/E(21) =

2.91E(02)/E(21) = 5.67

X(5) and E(5)

Searching for X(5)-like Nuclei

2.97

2.93

78

76

100

74

72

70

68

66

64

62

60

58

56Z/N 88 90 92 94 96 98 102104106108110114116

2.30 2.26 2.44 2.51 2.70 2.68 2.56 2.53 2.49 2.48 2.48

2.62 2.66 2.74 2.93 3.02 3.09 3.15 3.20 3.17 3.08 2.93 Os

2.68 2.82 2.95 3.07 3.15 3.22 3.24 3.26 3.29 3.27 3.24 3.09

2.31

2.33

2.32

2.23

2.19

2.32

2.49

2.59

2.66

2.56

2.63

2.74

3.02

3.00

2.93

2.84

2.79

3.10

3.21

3.24

3.11 3.19 3.25 3.27 3.28 3.29 3.31 3.30 3.26

3.313.12 3.23 3.27

3.29

3.31 3.31

3.31

3.31

3.31

3.29

3.283.23

3.27

3.29

3.30

3.31

3.30

3.30

3.30

3.30

3.323.293.26

3.25

3.29

3.29

3.15

2.99

P ~ 4-5

2.932.932.93

2.93

2.86

Pt

W

Hf

Yb

Er

Dy

Gd

Sm

Nd

Ce

Ba

P= NpNn

Np+Nn

Good starting point: R4/2 or P factor

β-decay studies at Yale

156DyM.A. Caprio et al., Phys. Rev. C 66, 054310 (2002).

162YbE.A.McCutchan et al., Phys. Rev. C 69, 024308 (2004).

166Hf

152SmR.F. Casten and N.V. Zamfir, Phys. Rev. Lett. 87, 052503 (2001).N.V. Zamfir et al., Phys. Rev. C 60, 054312 (1999)..

E.A.McCutchan. et al., Phys. Rev. C- submitted.

Other Yale studies: 150Nd - R.Krücken et al., Phys. Rev. Lett. 88, 232501 (2002).

Searching for E(5)-like Nuclei

Good starting point: R4/2 or P factor

134BaR.F. Casten and N.V. Zamfir, Phys. Rev. Lett. 85, 3584 (2000).

102PdN.V. Zamfir et al., Phys. Rev. C 65, 044325 (2002).

130Xe38

54

52

50

48

46

44

42

40

5452

58

56

68666462605856 807876747270

Ba

Mo

Sr

Zr

Te

Xe

Ce

Cd

Pd

Ru

Sn

1.79

1.79

1.82

1.81

1.60

1.99

1.60

2.05

2.11

2.12

2.14

2.09

1.67

2.29

1.75

2.27

1.92

1.63

1.54

2.27 2.36

2.38

2.32

2.12

1.51 2.65

3.01

2.51

2.48

2.40

2.38

1.81

3.23

3.15

2.92

2.65

2.42

2.33

1.79 1.68

2.29

2.46

2.75

3.05

3.23

2.92

2.76

2.53

2.30

1.84

2.33 2.40

2.73

2.002.09

2.56

2.38

1.85 1.87 1.88 1.86 1.80 1.71 1.63

2.39 2.38 2.33

2.58

2.96

3.06

2.89

2.50

2.07

2.83

2.48

2.09 2.04 2.01 1.942.071.99 1.72

2.042.162.242.42

2.33

2.47

2.78 2.69 2.52 2.43 2.32 2.28

2.322.382.562.692.802.93

Z/N

P~2.5

Symmetries and phases transitions in the IBMSymmetries and phases transitions in the IBM

• Challenges for neutron-rich:

– New collective modes in three fluid systems (n-skin).

– New regions of phase transition

– New examples of critical point nuclei?

– Rigid triaxiality?

D.D. Warner, Nature 420 (2002) 614