1. J. F. Sharpey-Schafer Physics Department, University of Western Cape, Belleville, South Africa. Failures of Nuclear Models of Deformed Nuclei

2. Many thanks to all my colleagues S M Mullins R A Bark E A Lawrie J J Lawrie J Kau F Komati P Maine S H T Murray N J Ncapayi P Vymers P Papka iTL + Stellenbosch Univ. S P Bvumbi Univ. of Johannesburg S N T Majola Univ. of Cape Town JFS-S T E Madiba Univ. of Western Cape T S Dinoko D G Roux Rhodes Univ. A Minkova Univ. of Sofia J Timár ATOMKI, Debrecen B M Nyak ó iThemba LABS SOUTH AFRICAU.S.A. CANADA FRANCE L L Riedinger D L Hartley C Beausang M Almond M P Carpenter C J Chiara P E Garrett F G Kondev W D Kulp III T Lauritsen E A McCutchan M A Riley J L Wood C H Wu S Zu D Curien J Dudeck N Schunk

3. Bohr & Mottelson Vol II Page 363 !!

4. Nilsson Model + Paring + Cranking For 165 Yb 95 Quasi-particles (Neutrons) Pairing Cranking Jerry Garrett; Erice, Sept. 1986

5. Average Interaction Energy cos Θ 12 = j 1. j 2 / |j 1 |.|j 2 | 0+0+ 2+2+ 6+6+ 8+8+ 4+4+ Experiment for 90 41 Nb 49 pairing approx. J π = 0 + J π ≠ 0 + ↑↓ = 180° J π = 0 + ↑↑ = 0° J π ≠ 0 +

6. [660]1/2 + “Prolate” [505]11/2 - “Oblate” Configuration Dependent or Quadrupole Pairing; Assume Δ pp ≈ Δ oo >> Δ op 82 Neutrons Prolate Deformation =>

7. Full Unblocked Pairing Pairing Reduced by Odd Neutron 156 Dy 90 155 Dy 89 [505]11/2 - 162 Yb 92 gsb 163 Yb 93 gsb [521]3/2 - Jerry Garrett et al, Phys. Lett. B 118, 297 (1982)

8. Systematics of Energies of 0 2 + and [505]11/2 - states for Z = 60-68 and N = 86-98

9. Congruence of first & second Vacuum structures in 154 Gd & 152 Sm

10. What is the │0 2 + > Configuration ? nothing ۞ (t,p) & (p,t) │0 2 + > is 2p n - 2h n this gives J π but nothing on the orbit. ۞ Single particle transfer would give l n but does not populate │0 2 + >. NOT In { │0 2 + > + neutron }, look to see which orbit does NOT couple to │0 2 + >.

11. Canadian Journal of Physics 51 (1973) 1369 McMaster Løvhøiden, Burke & Waddington 157 Gd(p,t) 155 Gd Gsb|0 2 + > [521]3/2 - + |0 2 + >

12. [505]11/2 - 681 keV │0 2 + > K π =15/2 - =2 γ + + [505]11/2 - 155 64 Gd 91 High-K states

13. 1000 1200 1400 800 600 400 200 0 ExEx (keV) [521]3/2 - [651]3/2 + [505]11/2 - │0 2 + > BLOCKED 155 Gd 91 Seen by Schmidt et al; J. Phys. G12(1986)411 in (n,γ) (d,p) & (d,t) [642]5/2 + [402]5/2 + [532]3/2 - [400]1/2 + 681 keV K=3/2 - K=3/2 + {K=11/2 - } K γ =2 K=1/2 - =2-Ω K=1/2 + =2-Ω K=15/2 - =2+Ω 996 keV

14. Two Neutron Transfer to 154 Gd (N=90) N.B. Log 10 scale Shahabuddin et al; NP A340 (1980) 109 K π = 2 + Bandhead

15. O. P. Jolly; PhD thesis (1976), McMaster, Canada Proton Stripping 153 Eu(α,t) 154 Gd NB; Log 10 Scale K π = 2 + band Z=63 [411]1/2 + [413]5/2 + ΔK=2 + Z=50

16. Spin I ( ħ ) K=2 γ-band ground state band aligned i 13/2 band 156 Dy from the 148 Nd( 12 C,4n) 156 Dy reaction 2 nd vacuum Gammasphere Data Nov. 2008 Siyabonga Majola et al., to be published

17. 160 68 Er 92 Odd Spin γ-band in 160 Er Gammasphere data; Ollier et al., Phys. Rev. C83 (2011) 044309

18. 116 Cd( 48 Ca,6n) 156 Er 88 215 MeV Z = 68 N = 88 J M Rees, E S Paul et al Gammasphere Data, ANL Phys. Rev. C83 (2011) 044314 Wilets & JeanDavydov & Filippov

19. “Nuclear Models” Greiner & Maruhn “Nuclear Models” K = 2 Gamma Vibration Band Head Energy E gnd state = ½ ħω β + ħω γ E x (0,0,2,2) = ħω γ + ħ 2 / I

20. Fig. 7. Gamma phonon energy ħω γ calculated using equation (1) for even-even nuclei with neutron number from N = 88 to 98 and proton number Z = 60 (Nd) to 70 (Yb). The nuclear deformation decreases with increasing Z. This phonon energy is relatively stable with (Z,N) compared to the excitations energies of corresponding 0 2 + states shown in Fig.2.

21. HENCE: Independent of any Model, If there is a γ degree of freedom, then there must be a K = 2 band that tracks the intrinsic band due to the K = 2 projection of the Zero Point Motion in the γ direction. Same conclusion in “Ring & Schuck”

22. N=90

23. Mean field h = t + v independent ph excitations 1p1h, 2p2h, … adding 2-body residual interaction V res = V - v H = t + V res V res  Mixing of ph excitations Several ways to account for V res : leads to cooperative effects  shell model with configuration mixing  boson expansion methods: Quasi Boson Approx. (QBA) Tamm-Dankoff Approx. (TDA) Random Phase Approx. (RPA, QRPA) among others here Restriction to RPA Residual interaction and collectivity Goal : Study of vibrational degrees of freedom 0 +, 1 + -, 2 +, 3 -, … Slide from Fritz Dönau

24. The way the Soloviev qp-phonon model (QPM) works is as follows; 1] He postulates that collective phonons exist. He does not know how or what. V. G. Soloviev, Nucl. Phys. 69, 1-36 (1965) 2] So he defines a phonon operator;Qi Ψ = 0 (his equ 6) And collective states are given by;Qi + Ψ(his equ 7) And[Qi,Qj + ] = δij(his equ 8) 3] As he has no clue at all what the Qi are, he assumes they can be expanded in terms of 2qp wave functions. That is in terms of particle hole states and nothing more complex. 4] To do this he needs an interaction which he assumes is of the multipole-multipole type. 5] The phonon energy he then finds using the variation principle for Qi+ Ψ (his equ 10). But to do this he has to fit experimental data as his Hamiltonian has unknown constants κn, κp, κnp (his equ 3) which govern the strengths of the neutron-neutron, proton-proton and neutron-proton interactions. Later he gets fed up and puts all these κ s equal !! 6] In reference; V. G. Soloviev and N. Yu. Shirikova, Z. Phys. A301, 263-269 (1981), he concludes “that the two-phonon states cannot exist in deformed nuclei” (his abstract). This is because anti-symmetrization and the Pauli principle pushes up the two-phonon energy to 2.5 times twice the phonon energy in his model. At these energies the two-phonon states are well above the pairing gap and will get mixed to hell !!!

25. Two steps to find the RPA solution of the nuclear many body problem HΨ=EΨ : (i)Calculate the mean field potential (Hartree-Fock) v mf of H=t+V.  ground state properties (nuclear shape)  independent particle-hole (ph) excitations (ii) Consider the residual interaction V res = V – v mf but V res is restricted to coupling terms of 1particle-1hole type only.  coupling of 1p1h excitations to phonons (vibrational excitations) RPA state = Correlated 1particle-1hole state: Sketch of the RPA approach Slide from Fritz Dönau

26. RPA Examples §1. Superdeformation A~190; Takashi Nakatsukasa et al., Phys Rev C53(1996)2213 Showed that lowest excitations were Octupole Y 3,2 Vibrations §2. Deformed Rare Earths; Zawischa, Speth & Pal, Nucl Phys A311(1978)445 QRPA, Deformed Woods-Saxon single particle Basis, Zero-range density dependent residual interaction.  “…it is more reasonable to identify the high-lying K π = 0 + and 2 + giant quadrupole resonances with the classical β- and γ-vibrations.”  “This is due to the fact that the energies of the low-lying states are mainly given by the details of the single-particle structure at the Fermi surface whereas the high-lying states are of real collective nature.”  “…the microcsopic wave vector of the low-lying K π = 0 + states is predominantly of the pairing-vibrational type in agreement with the enhanced two-particle transfer cross sections.”

27. TPSM Triaxial Projected Shell Model Kenji Hara, Yang Sun, Javid Sheik ….. Uses a Triaxial Deformed Nilsson basis with q-q interaction and monopole & quadrupole pairing Mixed K states with K = 0,2,4… => Mixed K states with K = 0,2,4… Projection to good angular momentum results in bands with I = 0, 2, 4 … Hence “K=2” bands are not one phonon bands but arise from the γ degree of freedom. Phys Rev C61(2000)064323 Phys Rev C77(2008)034313 Phys Lett B688(2010)305

28. Conclusions  There are no such things as β-vibrations  Deformed nuclei are stiff in the β direction  “K=2 γ-bands” are real collective structures  These collective structures are built on every intrinsic configuration  Macroscopic theories not much use; e.g. IBM, X(5) etc  Boson expansions in p-h states will not get 0 2 +  Need γ degree of freedom  RPA and TPSM have hope of predicting “γ-bands”