Download - Maciej Zalewski, UW
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,