Mean-field calculation including protonneutron mixing in atomic nuclei―toward proton-neutron pairing―
Koichi Sato (RIKEN Nishina Center)
Collaborators:
Takashi Nakatsukasa (RIKEN)
Wojciech Satuła (Univ. of Warsaw )Jacek Dobaczewski (Univ. of Warsaw /Univ. of Jyvaskyla )Symposium on ‘Quarks to Universe in Computational Science’ (QUCS2012), Dec. 13-16, 2012

“Nuclear DFT (density functional theory) with proton-neutron pairingand its application”(e.g. GT transition in r-process)Proton-neutron (p-n) pairing:

Goodman, Adv. Nucl. Phys.11, (1979) 293.

Pairing between protons and neutrons (isoscalar T=0 and isovector T=1)
Related (?) physics:
Wigner energy, β decays , interplay between T = 0 and T = 1 states in N = Z nuclei,
isospin mixing and mirror symmetry breaking, α decay and α clustering, moments of inertia,
deformation properties, etc.
Perlinska et al, PRC 69 , 014316(2004)

For a review on p-n pairing, see A. Afanasjev, arXiv:1205.2134

pProton-neutron mixing:n？

Quasiparticles are mixtures of protons and neutrons
Density functionals with an arbitrary mixing between protons and neutronsρτ (α , β ) = Ψ cα+ ,τ cβ ,τ Ψ

τ = p, nρττ ' (α , β ) = Ψ cα+ ,τ cβ ,τ ' Ψ

τ ,τ ′ = p, n

As a first step, we consider p-n mixing at the Hartree-Fock level without pairing.

Hartree-Fock calculation including proton-neutron mixing
Extension of the single particle states
ψ i ,n = ∑ ai(,nα) α , nα

ψ j, p = ∑ a
α( p)
j ,α

ψ i = ∑ ai(,nα) α , n + ∑ ai(,pβ) β , pα, pα

i=1,…,A

Extension of the density functional

ESkyrme[ρn , ρ p ]β

E Skyrme′ [ ρ 0 , ρ ]isoscalar

Invariant under rotation in
isospin space

isovector
Perlinska et al, PRC 69 , 014316(2004)

can be written in terms ofρ0 , ρ3

not invariant under rotation in isospin spaceρ0 = ρn + ρ p

ρ1 = ρ np + ρ pn
ρ 2 = −iρ np + iρ pn
ρ3 = ρ n − ρ p

Energy density functional is extended
such that it becomes invariant under rotation in isospin space

Hartree-Fock calculation including proton-neutron mixingExtension of the single particle states
ψ i ,n = ∑ ai(,nα) α , nα

ψ j, p = ∑ aα( p)j ,αψ i = ∑ ai(,nα) α , n + ∑ ai(,pβ) β , pα, pα

i=1,…,A

Extension of the density functional

ESkyrmeβ[ρn , ρ p ]
E Skyrme′ [ ρ 0 , ρ ]isoscalar

Invariant under rotation inisospace

isovector
Perlinska et al, PRC 69 , 014316(2004)

p-n mixed Hartree-Fock Hamiltonian :hn , h pn-nn-pp-np-p

hmixed =Off-diagonal blocks are zeroin the normal HF w/o p-n mixing

It is a hard task to develop a code from scratch …
Use of a existing HFB code

HFODD(1997-)
http://www.fuw.edu.pl/~dobaczew/hfodd/hfodd.html
J. Dobaczewski, J. Dudek, Comp. Phys. Comm 102 (1997) 166.
J. Dobaczewski, J. Dudek, Comp. Phys. Comm. 102 (1997) 183.
J. Dobaczewski, J. Dudek, Comp. Phys. Comm. 131 (2000) 164.
J. Dobaczewski, P. Olbratowski, Comp. Phys. Comm. 158 (2004) 158.
J. Dobaczewski, P. Olbratowski, Comp. Phys. Comm. 167 (2005) 214.
J. Dobaczewski, et al., Comp. Phys. Comm. 180 (2009) 2391.
J. Dobaczewski, et al., Comp. Phys. Comm. 183 (2012) 166.

••
•
•
•

Skyrme energy density functional
Hartree-Fock or Hartree-Fock-Bogoliubov
No spatial & time-reversal symmetry restriction
Harmonic-oscillator basis
Multi-function (constrained HFB, cranking, angular mom.
projection, isospin projection, finite temperature….)

Road map
HFODD (Skyrme-HF(B) code)

Proton-neutron mixing at HF level

Test calc. for A=14 and 48 nuclei

Isospin projection with p-n mixing
to remove spurious isospin mixing
Proton-neutron pairing (T=1 & T=0)

QRPA with FAM, Isospin projection w/ p-n pairing, etc.
e.g. calc. of GT strengths of n-rich nuclei

Test calculations for p-n mixing
EDF with p-n mixing is correctly implemented?
w/o Coulomb force (and w/ equal proton and neutron masses):invariant under rotation in isospin spaceTz

T

Total isospin of the system

Rotational invariance in isospace

TTx
TTy

Total energy should be independent of theorientation of T.
Check of the code

How to control the isospin direction ?

Isocranking calculation Hˆ ′ = Hˆ − λ ⋅ T Isocranking termλ : Input frequency to control the isospin of the system
HF eq. solved by iterative diagonalization of MF Hamiltonian.
w/ p-n mixing and no Coulomb
Determine s. t. proton and neutron
Fermi energies become equal

Initial state: HF solution w/o p-nmixing (e.g. 14C(Tz=1,T~1))

Final state
TzTz

θ
ϕTx

 − λ ⋅T

λTy

HF state w/o p-nmixing

λiterationTx

Typ-n mixed stateCalculation for A=14 isobars• w/ p-n mixing and no Coulomb• w/ p-n mixing and Coulomb

Isobaric symmetryE
T=20

14O(Tz= -1)

14N(Tz= 0)

12

14C(Tz= +1)

Bohr & Mottelson, “Nuclear Structure” vol.1

Tz

(Initial : 14C)

Without Coulomb interaction

14C(θ , ϕ ) = (90 ,0)(θ , ϕ ) = (45 ,45 )single-particle energy:

single-particle energy:

single-particle energy:

NO)

ENERGY

NO)

<T>

NO)

ENERGY

<T>

1)
2)
3)
4)
5)
6)
7)
8)
9)
10)
11)
12)
13)
14)

-38.4191001)-38.419-38.4191002)-38.419-32.707-1003)-32.707-32.707-1004)-32.707-22.021005)-22.02-22.021006)-22.02-21.9461007)-21.946
thesame-21.946
100s. p energies8)-21.946-15.947-1009)-15.947-15.947-10010)-15.947-15.902-10011)-15.902-15.902-10012)-15.902-15.181
10013)-15.181-15.18110014)-15.181

0000
000
00
0
0
0001)
2)
3)
4)
5)
6)
7)
8)
9)
10)
11)
12)
13)
14)

-38.419-38.41932.707-32.707-22.02-22.02-21.946-21.946-15.947-15.947-15.902-15.902-15.181-15.181

70.7
70.7
-70.7
-70.7
70.7
70.7
70.7
70.7
-70.7
-70.7
-70.7
-70.7
70.7
70.7

<T>

ENERGYTˆz = 1

Total energy: -114.611699

Tˆz = 02n

Total energy: -114.611699

Tx

the same
total energiesTz

T = 1 τ z × 200
245

100: pure n
Ty

Ty
Tx

45

Tˆz = 0.707

Tz
Ty

Tz

p50% neutron &50% protonTz= 5.5・・・・・・

Total energy: -114.611699

・・・(θ , ϕ ) = (0,0)

λTx

45



-100: pure pWith Coulomb interaction

U Coulomb (τ z ):violates isospin symmetry

The total energy is now dependent on Tz
but independent of Tx and Ty

final state:
Tz

Initial state:Tz ̂− λ ⋅T



θ
Tλ

λ

λTy

Ty

ϕTx

Tx
HF state w/o p-nmixing

p-n mixed stateWith Coulomb, larger <Tz> is favored

Tz

With Coulomb interaction

θ = 0

θ = 4514C

θTx



ϕ =0

λ = 11.0MeV

θ = 180

14O

single-particle energy:

Single-particle energy:

single-particle energy:NO)
1)
2)
3)
4)
5)
6)
7)
8)
9)
10)
11)
12)
13)
14)

NO)
1)
2)
3)
4)
5)
6)
7)
8)
9)
10)
11)
12)
13)
14)

NO)
1)
2)
3)
4)
5)
6)
7)
8)
9)
10)
11)
12)
13)
14)

<T>
82.0
82.0
-82.0
-82.0
80.7
80.7
80.7
80.7
80.1
80.1
-80.6
-80.6
-80.7
-80.7

Total energy: -106.16

Tz

ENERGY
-34.362-34.362-32.527-32.527-18.41-18.41-18.324-18.324-15.868-15.868-15.821-15.821-11.868-11.868

Total energy: -100.70

Tz<Tz>=1<Tz>=0.802

Tx<T>-100-100
100
100-100-100-100-100
100
100
100
100-100-100・・・

Total energy: -106.68

ENERGY-37.980-37.980-29.880-29.880-21.666-21.666-21.587-21.587-14.942-14.942-13.469-13.469-13.419-13.419

・・・

・・・

ENERGY <T>-38.346
100-38.346
100-29.620-100-29.620-100-22.000
100-22.000
100-21.922
100-21.922
100-15.259
100-15.259
100-13.227-100-13.227-100-13.178-100-13.178-100

Tx

Total and s. p. energies depend on <Tz>

Total isospin and λ are not parallel for p-n mixed statesNo p-n mixing for θ = 0 and θ = 180
(Neither Coulomb nor isocranking term contains Tx and Ty. )Tz<Tz>=-1Tx

Tz dep. of the total energy and comparison with data

14OTz
14N (excited 0+ state)

Tx

The origin oｆ exp.
BE shifted

Tz
Tx

14C

Tz

Tx

Tz θTxexp.: binding energy(+excitation energy for 14N)
calc.: calculated forevery 15o of θ from 0 to 180o

Result for A=48 isobars

Initial : 48Cr(Tz=0, T~0)
48Ni



λ = 11.0



λ = 6.048Cr

λ = 0.0

48Ca

48Cr

<Tz>

<Tz>
With Coulomb

Without Coulomb
Without Coulomb

Increase the size of isofrequency λ to make high-isospin states
With Coulomb
Tune the size of lambda depending on the tilting angle θ
Diabatic blocking with isospin to avoid oscillatory (ping-pong) divergence
J. Dobaczewski, J. Dudek, Comp. Phys. Comm. 131 (2000) 164.

Constrained HF with Augmented Lagrange method ˆˆ   2
E′ = E − λ ⋅ T + CT − T A. Staszczak, Eur. Phys. J. A 46, 85–90 (2010)

Excitation energies for 48Cr(Tz=0)

constraints Tz=0 and Tx=0,2,4,6,…

--> nuclear symmetry energy

preliminary

Sum of squared residuals :T ≅0

a*Tx*(Tx+1) fit : 24.2
b*Tx2 fit: 6.1

Tx

Tx

a=1.20956 (MeV)
b=1.38078 (MeV)Tˆx − Tx

2Tz

Tˆz − Tz

0Energy
Not yet43.6E-050<1.0E-08-390.06

69.8E-04

0

5.7E-06-359.768

9.4E-040

3.1E-06-320.89Summary
We have developed a code for the DFT calculations with✔ proton-neutron mixing at the Hartree-Fock level✔ diabatic blocking method using isospin✔ constraints on isospin with augmented Lagrange method
and performed test calculations for A=14 & 48 systems

Ongoing
Isospin projection to remove spurious isospin mixing
W. Satuła et al., PRC 81, 054310 (2010).

FutureProton-neutron pairing (T=0 & T=1; non-rotating & rotating)Charge exchange reaction in neutron-rich nuclei (QRPA calculation)
A very wide applicability expected