1. Terminating states as a unique laboratory for testing nuclear energy density functional Maciej Zalewski, UW under supervision of W. Satuła Kazimierz Dolny, 30.09.2006 Outline: -fine tuning of LEDF parameters using terminating states, -time odd fields and spin-orbit strenght, -phenomenological restoring of broken rotational symmetry in I max -1 states

2. Local Energy Density Functional Theory with Skyrme force induced parameters  | H |  Slater determinat 20 parameters We may treat LDF as a starting point and adjust C parameters Skyrme force parameters are fitted to set of data (nuclear matter, masses and radii of nuclei). There is no one obvious way to obtain this parameters hence there is great number of parametrizations.

3. Examples of band terminating states in 46 Ti 24 Terminating states: -the best example of almost unperturbed single particle motion, -uniquely defined (N  Z), -configuration mixing beyond mean-field expected to be marginal, -shape-polarization effects included already at the level of the SHF, -good to test badly known time- odd fields, -seem to be ideal for fine tuning of particle-hole interaction. cranking +3/2 +1/2 -1/2 -3/2 +7/2 +5/2 +3/2 +1/2 -1/2 -3/2 -5/2 -7/2 p-h proton neutron 20 -1/2 +7/2 +5/2 +3/2 +1/2 -3/2 -5/2 -7/2 +3/2 +1/2 -1/2 -3/2 fully filled partially filled 14  00 +3  across the gap f 7/2 (n=6) f 7/2 (n=7) d 3/2 17  00 energy scale (bulk properties) spin-orbit dominates!!!  E = f 7/2 n I max E( ) E( ) - d 3/2 f 7/2 n+1 I max The idea is to calculate the difference

4. Skyrme force induced LEDF -0,5 0,0 0,5 1,0 0,70,80,91,0 SLy4 SLy5 SkO SIII SkM* SkP MSk1 SkXc g0g0 g1g1 m* „locked” by the local gauge invariance „free” i.e. not constrained by data Hence, the isoscalar Landau parameters induced by the Skyrme: are more or less „random” scales with m* 0.40 -0.19

5. Spin fields Original Skyrme force induced LEDF Landau parameters: g 0 =0.4; g 1 =0.19 iduced LEDF H. Zduńczuk, W. Satuła, R. Wyss, Phys. Rev. C58 (2005) 024305 - dicrease of ΔE exp -ΔE th and unification of isotopic and isotonic dependence

6. W0;W0;W0W0 * [MeV fm 5 ] 0 0.5 1.0 1.5 2.0 120140160180 W0W0 W0W0 * SkM* MSk1 SkP SkXc Sly.. SIII SkO  E [MeV] Correlation between spin-orbit strenght and ΔE „standard” s-o term: „scaled” s-o term: VERSUS H. Zduńczuk, W. Satuła, R. Wyss et al, Int. Jour. of Mod. Phys.A422

7. 0.0 0.2 0.4 -0.2 Landau LEDF: spin-orbit reduced by 5% 42 Ca 44 Ca 44 Sc 45 Sc 45 Ti 46 Ti 47 V 0.12 0.14 0.16 0.18 SkO SLy4 1/30-1.3 W 1 /W 0  [MeV] EE SkO 1/3 0 W 1 /W 0 ~ -1.3  E exp -  E th [MeV] (N-Z)/A Modification of spin-orbit strenght H. Zduńczuk, W. Satuła, R. Wyss, Phys. Rev. C58 (2005) 024305 Standard Skyrme s-o: W=W’ W 1 /W 0 = 1/3 Non-standard Skyrme s-o: W=-W’ W 1 /W 0 = -1 W 1 /W 0 = -1.3 Reinhard/Flocard: SkO Brown (SkXc): W’=0 W 1 /W 0 = 0 -further dicrease of ΔE exp -ΔE th by ~200keV to acceptable level,

8. Terminating states f 7/2 n cranking +3/2 +1/2 -1/2 -3/2 +7/2 +5/2 +3/2 +1/2 -1/2 -3/2 -5/2 protons neutrons 20 +7/2 +5/2 +3/2 +1/2 -1/2 -3/2 -5/2 +3/2 +1/2 -1/2 -3/2 f 7/2 19/2  00 -1  Signature change -7/2 d 3/2 17/2  I max I max -1

9. Energy difference between I max and I max -1 f 7/2 n states -3.0 -2.5 -2.0 -1.5 -0.5 47 V 43 Sc 44 Sc 45 Sc 44 Ti 46 V 45 Ti 42 Sc 46 Ti E(I max )-E(I max -1) [MeV] EXP SM SLy4 SLy4  results mean field calculations disagree with SM and Exp. data -there should be one I max -1 state, -mean-field solutions break rotational symmetry. Assuming that particles from f 7/2 shell (outside the core) play role only, we may treat |I max, I max > state as a vacuum for creation and anihilation operators â +, â Example for 43 Sc a b (spurious state)(I max -1 state)

10. Restoration of rotational symmetry – two methods Requiring λ 1 =0, we have: Method A We assume that we know a,b coefficients: Method B e 1 = E(I max )-E(ν) e 1 = E(I max )-E(π) We set zero of the enargy scale at the energy of I max state. Interaction between |π> and |ν> Energies of ‘spurious’ and I max -1 states

11. Restoration symmetry -3.0 -2.5 -2.0 -1.5 -0.5 EXP SM SLy4  47 V 43 Sc 44 Sc 45 Sc 44 Ti 46 V 45 Ti 42 Sc 46 Ti SLy4  E(I max )-E(I max -1) [MeV] -good agreement with SM and Exp. -in general – these methods works really good ! -method B seems to work slightly better, H. Zduńczuk, J. Dobaczewski, W. Satuła – see poster by H. Zduńczuk 0.50 0.55 0.60 0.65 0.70 47 V 43 Sc 44 Sc 45 Sc 44 Ti 46 V 45 Ti 42 Sc 46 Ti phenomenology from neutron from proton  -state probability in I max -1 Results of calculations with angular momentum projection support this conclusion

12. Energy difference between I max and I max -1 [f 7/2 n+1 d 3/2 -1 ] states Three possible mean-field I max -1 states: -neutron signature change in f 7/2 shell, -proton signature change in f 7/2 shell, -proton signature change in d 3/2 shell. Now there is one ‘spurious’ state and two I max -1 states +3/2 +1/2 -1/2 -3/2 +7/2 +5/2 +3/2 +1/2 -1/2 -3/2 -5/2 -7/2 12  f 7/2 neutron signature change 1/2  +3/2 +1/2 -1/2 -3/2 +7/2 +5/2 +3/2 +1/2 -1/2 -3/2 -5/2 -7/2 12  f 7/2 proton signature change 1/2  +3/2 +1/2 -1/2 -3/2 +7/2 +5/2 +3/2 +1/2 -1/2 -3/2 -5/2 -7/2 12  d 3/2 proton signature change 1/2  protons neutrons a c b

13. Symmetry restoring – two methods again We allow complex V and set it to obtain λ 1 =0 (in this case – complex congugation in Hamiltonian) Method B We assume we know a, b, c coeficients and require λ 1 =0 : Method A where: e i = E(I max )-E(i) i= ν, π, π, We set zero of the enargy scale at the energy of I max state. Energies of ‘spurious’ and I max -1 states

14. Results of restoring rotational symmetry 455keV 0 0.5 1.0 1.5 2.0 42 Ca 44 Ca 43 Sc 44 Sc 45 Sc 45 Ti 46 Ti 47 V exp SM SLy4 corrected SLy4 E(I max ) – E(I max -1) [MeV] The lowest I max -1 state 42 Ca 44 Ca 43 Sc 44 Sc 45 Sc 45 Ti 46 Ti 47 V SkO corrected SkO exp SM Sly4 -constant offset of ~450keV -details of isotopic and isotonic dependance reproduced remarkably well SKO -average value is good -discrepancies in isotonic and isotopic dependance The lowest I max -1 state

15. Summary -terminating states are excellent for testing nuclear energy density functional, -mean field solutions of I max states are in excellent agreement with experimental data, -I max -1 states cannot be reproduced by mean field! They break rotational symmetry which can be easily restored,