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 Imax-1 states

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.

Examples of band terminating states in 46Ti24

Terminating states:-the best example of almost unperturbed single particle motion,-uniquely defined (NZ),-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

protonneutron

20

-1/2

+7/2

+5/2

+3/2+1/2

-3/2

-5/2

-7/2

+3/2+1/2

-1/2

-3/2fully filled

partially filled

14

0

+3acrossthe gap

f7/2

(n=6)f7/2

(n=7)d3/2

-1

17

0

energy scale(bulk properties)

spin-orbitdominates!!!

E = f7/2

nImax

E( )

E( ) -

d3/2 f7/2

n+1Imax

-1The idea is to calculate the difference

Skyrme force induced LEDF

-0,5

0,0

0,5

1,0

0,7 0,8 0,9 1,0

SLy4SLy4

SLy5SLy5

SkOSkOSIIISIII

SkM*SkM*

SkPSkP

MSk1MSk1

SkXcSkXc

g0

g1

m*

„locked” by thelocal gaugeinvariance

„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

Spin fields

Original Skyrme force induced LEDF

Landau parameters:g0=0.4; g1=0.19

iduced LEDF

0

0.5

1.0

1.5

2.0

2.5

(a) Skyrme LEDF:

42Ca 44Ca44Sc 45Sc45Ti 46Ti47V

(N-Z)/A

(b) Landau LEDF:

42Ca 44Ca44Sc 45Sc45Ti 46Ti47V

(N-Z)/A

SkM*

SkXc

SLy4

SkP

SkO

SLy5

SIII

Eex

p-

Eth

[M

eV]

H. Zduńczuk, W. Satuła, R. Wyss, Phys. Rev. C58 (2005) 024305

- dicrease of ΔEexp-ΔEth and unification of isotopic and isotonic dependence

W0; W0* [MeV fm5]

0

0.5

1.0

1.5

2.0

120 140 160 180

W0

W0* SkM*

MSk1

SkP

SkXcSly.. 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

0**

0 Wm

mW

0.0

0.2

0.4

-0.2

Landau LEDF:

spin-orbit reduced by 5%

42Ca 44Ca44Sc 45Sc45Ti 46Ti47V

0.12

0.14

0.16

0.18

SkO

SLy4

1/3 0 -1 -1.3W1/W0

[

MeV

]ESkO

1/30

W1/W0 ~ -1.3

-1E

exp-

Eth

[M

eV]

(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’ W1/W0 = 1/3

Non-standard Skyrme s-o:

W=-W’ W1/W0 = -1

W1/W0 = -1.3

Reinhard/Flocard:

SkO

Brown (SkXc):

W’=0 W1/W0 = 0

-further dicrease of ΔEexp-ΔEth by ~200keV to acceptable level,

Terminating states f7/2n

cranking

+3/2

+1/2

-1/2

-3/2

+7/2

+5/2

+3/2

+1/2

-1/2

-3/2

-5/2

protonsneutrons

20 +7/2

+5/2

+3/2

+1/2

-1/2

-3/2

-5/2

+3/2

+1/2

-1/2

-3/2

f7/2

19/2

0

-1Signaturechange

-7/2 -7/2

d3/2

17/2

Imax Imax-1

Energy difference between Imax and Imax-1 f7/2n states

-3.0

-2.5

-2.0

-1.5

-1.0

-0.5

47V

43Sc44Sc

45Sc44Ti

46V45Ti42Sc 46Ti

E(I

max

)-E

(Im

ax-1

) [M

eV]

EXPSMSLy4 SLy4

results mean field calculations disagree with SM and Exp. data

-there should be one Imax-1 state,-mean-field solutions break rotational symmetry.

Assuming that particles from f7/2 shell (outside the core) play role only , we

may treat |Imax, Imax> state as a vacuum for creation and anihilation operators â+, â

maxmax2/72/5maxmax2/52/3maxmax ;ˆˆ7;ˆˆ321;ˆ IIaaIIaaIII

Example for 43Sc a b

baII 1, maxmax abII 1,1 maxmax(spurious state) (Imax-1 state)

Restoration of rotational symmetry – two methods

21221 eeeeV Requiring λ1=0, we have:

Method A

We assume that we know a,b coefficients: abr

22

21

1 1 r

ere

2

12

22 1 r

ere

221

1

)(

r

eerV

Method B

b

a

b

a

eV

Ve2,1

2

1

e1= E(Imax)-E(ν)

e1= E(Imax)-E(π)

We set zero of the enargy scale at the energy of Imax state.

Interaction between |π> and |ν>

Energies of ‘spurious’ and Imax-1 states

Restoration symmetry

-3.0

-2.5

-2.0

-1.5

-1.0

-0.5

EXPSMSLy4

47V

43Sc44Sc

45Sc44Ti

46V45Ti42Sc 46Ti

SLy4

E(I

max

)-E

(Im

ax-1

) [M

eV]

-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

47V

43Sc44Sc

45Sc44Ti

46V45Ti42Sc 46Ti

phenomenology

from neutronfrom proton

-state probability in Imax-1Results of calculations with angular momentum projection support this conclusion

Energy difference between Imax and Imax-1 [f7/2n+1 d3/2

-1] states

Three possible mean-field Imax-1 states:

-neutron signature change in f7/2 shell,

-proton signature change in f7/2 shell,

-proton signature change in d3/2 shell.

Now there is one ‘spurious’ state and two Imax-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

12f7/2

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

12f7/2 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

12d3/2 proton signature change

1/2

protonsneutrons

maxmax2/12/3

maxmax2/72/5

maxmax2/52/3

maxmaxmaxmax

;ˆˆ3

;ˆˆ7

;ˆˆ12

1;;ˆ

IIaa

IIaa

IIaa

IIIII

a

c

b

Symmetry restoring – two methods again

jiij eeV We allow complex V and set it to obtain λ1=0 (in this case – complex

congugation in Hamiltonian)

Method B

ab

cebeaeV

2

23

22

21

12

ac

cebeaeV

2

23

22

21

13

bc

cebeaeV

2

23

22

21

23

We assume we know a, b, c coeficients and require λ1=0 :

Method A

Z42

1 23,2

where:

223

213

212323121

321

VVVeeeeeeZ

eee

c

b

a

c

b

a

eVV

VeV

VVe

i

32313

23212

13121

ei= E(Imax)-E(i)

i= ν, π, π,

We set zero of the enargy scale at the energy of Imax state.

Energies of ‘spurious’ and Imax-1 states

Results of restoring rotational symmetry

455keV

0

0.5

1.0

1.5

2.0

42Ca 44Ca 43Sc 44Sc 45Sc 45Ti 46Ti 47V

expSM

SLy4corrected SLy4

E(I

max

) –

E(I

max

-1)

[MeV

]

The lowest Imax-1 state

42Ca 44Ca 43Sc 44Sc 45Sc 45Ti 46Ti 47V

SkOcorrected SkO

expSM

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 Imax-1 state

Summary

- terminating states are excellent for testing nuclear energy density functional,

- mean field solutions of Imax states are in excellent agreement with experimental data,

- Imax-1 states cannot be reproduced by mean field! They break rotational symmetry which can be easily restored,