Mean-field calculation based on proton-neutron mixed energy density functionals

PRC 88(2013) 061301(R).

Koichi Sato (RIKEN Nishina Center) Collaborators: Jacek Dobaczewski (Univ. of Warsaw /Univ. of Jyvaskyla ) Takashi Nakatsukasa (RIKEN Nishina Center) Wojciech Satuła (Univ. of Warsaw )

Energy-density-functional calculation with proton-neutron mixing

superposition of protons and neutrons Isospin symmetry 1 0 0 1 Protons and neutrons can be regarded as identical particles (nucleons) with different quantum numbers In general, a nucleon state is written as 𝜙 1 𝜙 2 Proton-neutron mixing: Single-particles are mixtures of protons and neutrons EDF with an arbitrary mixing between protons and neutrons   (  ,  )  

c

  , 

c

 ,    

p

,

n

  ' (  ,  )  

c

  ,  '

c

 ,    ,   

p

,

n

Here, we consider p-n mixing at the Hartree-Fock level (w/o pairing) A first step toward nuclear DFT for proton-neutron pairing and its application Pairing between protons and neutrons (isoscalar T=0 and isovector T=1) Goodman, Adv. Nucl. Phys.11, (1979) 293.

Perlinska et al, PRC 69 , 014316(2004) p ？ n

Basic idea of p-n mixing Let’s consider two p-n mixed s.p. wave functions (spin indices omitted for simplicity) standard unmixed neutron and proton w. f.

𝜙 1 (𝑟 )= 𝜙 1 (𝑟, 𝑛 ) 𝜙 2 (𝑟 )= 𝜙 2 (𝑟, 𝑝 ) They contribute to the local density matrices as standard n and p densities p-n mixed densities

Hartree-Fock calculation including proton-neutron mixing (pnHF) Extension of the single-particle states 

i

,

n



j

,

p

     

a i

( ,

n

 )

a

(

j

,

p

 )   , ,

n p



i

  

a i

( ,

n

 )  ,

n

  

a i

( ,

p

 )  ,

p

i=1,…,A Extension of the density functional

E Skyrme

[ 

n

, 

p

]

E Skyrm e

 [  0 , ]

Invariant under rotation in isospin space

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

can be written in terms of

 0 ,  3

not invariant under rotation in isospin space

 0  

n



p

Standard HF  1  2  3 isovector  

np

   

i



np



pn i

 pnHF

pn

 

n

 

p

Energy density functionals are extended such that they are invariant under rotation in isospin space

We have developed a code for pnHF by extending an HF(B) solver

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….)

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 isospace

T z

Total isospin of the system

T

 Total energy should be independent of the orientation of T.

T y T



T x T

 “Isobaric analog states” All the isobaric analog states should give exactly the same energy Check of the code How to control the isospin direction ?

Isocranking calculation Analog with the tilted-axis cranking for high- spin states Isocranking term   : Input to control the isospin of the system HF eq. solved by iterative diagonalization of MF Hamiltonian. w/ p-n mixing and no Coulomb Initial state: HF solution w/o p-n mixing (e.g. 48 Ca (Tz=4,T=4) ) isospin

T



T z

  

T y

   

T

  iteration Final state

T z

p-n mixed state  

T y T x

HF state w/o p-n mixing

T x

By adjusting the size and titling angle of 𝜆 , we can obtain isobaric analog states

Isocranking calculation for A=48 w/o Coulomb

T z

    11(6.0) MeV for

T

 4 (2)

T x

Energies are independent of No p-n mixing at |Tz|=T The highest and lowest weight states are standard HF states We have confirmed that t he results do not depend on φ.

48 Ni 48 Fe 48 Cr 48 Ti 48 Ca

Result for A=48 isobars with Coulomb

Shifted semicircle method    (   sin   , 0 ,   cos    

off

) For T=4 PRC 88(2013) 061301(R).

  

off

 12 .

0 MeV   8 MeV Energies are dependent on Tz (almost linear dependence) No p-n mixing at |Tz|=T    90  gives

T

ˆ

z

 0    90  48 Ni 48 Ca

T

≅

4states in A=40-56 isobars T=1 triplets in A=14 isobars

14 O(g.s) 14 N (excited 0 + ) 14 C(g.s) Our framework nicely works also for IASs in even-even A=40-56 isobars Excited 0 + , T=1 state in odd-odd 14 N Time-reversal symmetry conserved 14 N: p-n mixed , 14 C,O: p-n unmixed HF (The origin of calc. BE is shifted by 3.2 MeV to correct the deficiency of SkM* functional)

Summary

We have solved the Hartree-Fock equations based on the EDF including p-n mixing Isospin of the system is controlled by isocranking model The p-n mixed single-reference EDF is capable of quantitatively describing the isobaric analog states For odd(even) A/2, odd(even)-T states can be obtained by isocranking e-e nuclei in their ground states with time-reversal symmetry.

Remarks: Augmented Lagrange method for constraining the isospin.

See PRC 88(2013) 061301(R).

Benchmark calculation with axially symmetric HFB solver: Sheikh et al., PRC, in press; arXiv:1403.2427

Backups

Assume we want to obtain the T=4 & Tz=0 IAS in 48Cr (w/o Coulomb) (a) Starting with the T=0 state

T z

48 Cr (Tz=0)

T x

  𝑻 = 𝟎   −𝜆 𝑥 t 𝑥

T z

48 Cr (Tz=0)

T y T x

𝑻 𝒙 = 𝟒, 𝑻 y = 𝑻 𝒛 = 𝟎, 𝑻 = 𝟒 (b) Starting with the highest weight state.

T z

48 Ca (Tz=4)

𝑻 𝒛 = 𝟒, 𝑻 = 𝟒 isospin −𝜆 𝑥 t 𝑥

T y T x T y T x T z

48 Cr (Tz=0)

T y

(a) Starting with the T=0 state Illustration by a simple model W. Satuła & R. Wyss, PRL 86, 4488 (2001).

Four fold degeneracy at ω=0 ( isospin & time-reversal) At each crossing freq., To get T=1,3, ・・ states, we make a 1p1h excitation In this study, we use the Hamiltonian based on the EDF with p-n mixed densities.

(b) Starting with the highest weight state.

48Ca (Tz=4) ~11MeV



p

⑳ ㉘ 

n

𝑇 = 𝑇 𝑧 = 4 Standard HF 

z

𝜆 𝑧 𝑡 𝑧 Isocranking calc.

𝜃 = 0 ㊽ 𝜆 𝑧 /2 𝜃 ≠ 0 

x

   

z

㊽ 

x

𝜆 𝑧 /2

T z T x

𝑇 𝑧 = 4 p-n unmixed 

z

   

x T z T x

𝑇 𝑧 ≠ 4 p-n mixed The size of the isocranking frequency is determined from the difference of the proton and neutron Fermi energies in the |Tz|=T states.

   11 .

0

How to determine the size of 𝜆 ?

Isobaric analog states with T=4 in A=48 nuclei

48Ni (Tz=-|T|=-4)

T z

㉘

T x



p

~11MeV



x

   

z

⑳ 

n

  180  Fermi energy Standard HF

48Ca (Tz=|T|=4)

㉘

T z



n T x



p

⑳   0  Standard HF We take the size of the isocranking frequency equal to the difference of the proton and neutron Fermi energies in the |Tz|=T states.

   This choice of 𝜆 enable us to avoid the configuration change.

11 .

0

~11MeV

With Coulomb interaction

U Coulomb

( 

z

) :violates isospin symmetry The total energy is now dependent on Tz but independent of azimuthal angle 𝜑 Initial :

T x T z

  

T y

 HF state w/o p-n mixing    

T

ˆ  final :

T z T

    

T y

p-n mixed state

T x

larger is favored ・  With Coulomb, the s. p. Routhians depend on the titling angle 𝜃

w/o Coulomb

semicircle



x

  0     

z

  

n

 

p

 

p

(

T z

 

n

(

T z

 4 )   4 ) 

x

w/ Coulomb

Shifted semicircle



n

 

p

(

T z

  4 ) 

z

Coulomb gives additional isocranking freq. effectively  

off



T

ˆ

z

w/o Coulomb

   ( 

x

, 

y

 0 , 

z

)   ( 

x

 , 

y

  0 , 

z

  

off

)   (MeV) 

n

 

p

(

T z

 4 ) Difference of p and n Fermi energies