The 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 1Lecture 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 I j Interacting Boson Approximation

Basic, attractive SD Interaction (2J+1) and 2 + lowest; separated from the rest.

Bosons counted from nearest closed shell (i.e. particles or holes). [Eg 130 Ba Z = 56 N = 74 N  = 3 N = 4 ; N=7] WHY???  7/2  5/2  3/2  1/2 Pauli Principle Consider f 7/2 “shell” with 6 neutrons M Maximum seniority = 2 Maximum (d)-boson number =1

DYNAMICAL SYMMETRY Describes basic states of motion available to a system - including relative motion of different constituents G 1  G 2  G 3  ………. H= aC 1 [G 1 ] † bC 2 [G 2 ] † cC 3 [G 3 ] † …… i ii i Dynamical symmetry breaking splits but does not mix the eigenstates

Some ‘Working Definitions’ Have states s and d  with  = - 2, - 1,0,1, 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: J x,J y,J z generate rotations and form group 0(3)] [For 0(3), use J z,J  ; J  = J x ± i J y Then [J +,J - ] = 2J z ; [J z,J  ] = ± J  ]

A Casimir operator commutes with all the generators of a group. –Eg:C 1U(6) = N;C 2U(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) (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 =  C 1U(6) +  C 2U(6)  C 2U(5) +  C 2O(5) +  C 2O(3) All C’s commute and H is diagonal DYNAMICAL SYMMETRY

Example of angular momentum For O(3), generators are J z, J + and J - Then [J 2, J z ] = [J 2, J + ] = [J 2, J - ] = 0 C 2O(3) = J 2 Subgroup O(2) simply J z = C 1O(2) So H =  C 2O(3) +  C 1O(2) E=  J(J+1) +  M O(2) J1J1 J2J2 M= +J z M= -J z O(3) 

I. “U(5)” - Anharmonic Vibrator II. “SU(3)” - Axially symmetric rotor III. “O(6)” - Gamma - unstable rotor Only 3 chains from U(6)

U(5) R 4/2 = 2.0

SU(3) R 4/2 = 3.33

O(6) R 4/2 = 2.5

The first O(6) nucleus ……….. Cizewski et al, Phys Rev Lett. 40, 167 (1978)

and then many more….

Transition Regions and Realistic Calculations Most nuclei do not satisfy the strict criteria of any of the 3 Dyn. Symm. Need numerical calculations by diagonalizing H IBA in s – d boson basis Can use a very simple form of the most general H

Z= < R 4/2 < 3.15  =0.03 MeV N.V. Zamfir, R.F. Casten, Physics Letters B 341 (1994) 1-5

Summary 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

VibrationalTransitionalRotational Evolution of nuclear shape E = nħωE = 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, (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 Spherical Vibrator Symmetric Rotor γ -soft X(5) E(5) Two solutions depending on γ degree of freedom

ξ = 1 ξ = 2 τ = 0 τ = 1 R 4/2 = 2.20 E(0 2 )/E(2 1 ) = 3.03 E(0 3 )/E(2 1 ) = 3.59 Key Signatures E(4 1 )/E(2 1 ) = 2.91 E(0 2 )/E(2 1 ) = 5.67 X(5) and E(5)

Searching for X(5)-like Nuclei P= N p N n N p +N n Good starting point: R 4/2 or P factor β-decay studies at Yale 156 Dy M.A. Caprio et al., Phys. Rev. C 66, (2002). 162 Yb E.A.McCutchan et al., Phys. Rev. C 69, (2004). 166 Hf 152 Sm R.F. Casten and N.V. Zamfir, Phys. Rev. Lett. 87, (2001). N.V. Zamfir et al., Phys. Rev. C 60, (1999).. E.A.McCutchan. et al., Phys. Rev. C- submitted. Other Yale studies: 150 Nd - R.Krücken et al., Phys. Rev. Lett. 88, (2002).

Searching for E(5)-like Nuclei Good starting point: R 4/2 or P factor 134 Ba R.F. Casten and N.V. Zamfir, Phys. Rev. Lett. 85, 3584 (2000). 102 Pd N.V. Zamfir et al., Phys. Rev. C 65, (2002). 130 Xe Ba Mo Sr Zr Te Xe Ce Cd Pd Ru Sn Z/N P~2.5

Symmetries 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