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
𝑛 = 𝜏 = +1 2 =
1
0
Protons and neutrons can be regarded as
identical particles (nucleons) with different quantum numbers
𝑝 = 𝜏 = −1 2 =
0
1
In general, a nucleon state is written as 𝑁 = 𝜙1
2
𝜙
Proton-neutron mixing:
Single-particles are mixtures of protons and neutrons
EDF with an arbitrary mixing between protons and neutrons

  ( ,  )   c  , c  , 

  ' ( ,  )   c  , ' c  , 
 ,   p , n
  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.
𝜙1 (𝑟)=𝜙1 (𝑟, 𝑛)
𝜙2 (𝑟)=𝜙2 (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 
a
(n)
i ,
,n
i 


j, p

a
( p)
j ,
, p
a
(n)
i ,

 , n   ai,  , p
( p)

i=1,…,A

Extension of the density functional
E
Skyrme
[n ,  p ]
E
Skyrm e 

[0,  ]
isoscalar
Invariant under rotation in
isospin space
isovector
Perlinska et al, PRC 69 , 014316(2004)
can be written in terms of  0 ,  3
not invariant under rotation in isospin space
isoscalar
0  n   p
Standard HF
pnHF
isovector
 1   np   pn
 2   i  np  i  pn
3  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
Tz

T
Total isospin of the system
Total energy should be independent of the
orientation of T.
Ty

T
Tx

T
All the isobaric analog states should give
exactly the same energy
Check of the code
“Isobaric analog states”
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. 48Ca (Tz=4,T=4) )
Final state
Tz

isospin T
Tz


 
  T

Ty

Tx
HF state w/o p-n
mixing
p-n mixed state


iteration
Ty
Tx
By adjusting the size and titling angle of 𝜆,
we can obtain isobaric analog states
Isocranking calculation for A=48
w/o Coulomb
48Ni
Tz
48Ca


  11(6.0) MeV for T  4 (2)
48Fe
48Ti
Tx
Energies are independent
of <Tz>
No p-n mixing at |Tz|=T
The highest and lowest weight
states are standard HF states
We have confirmed that
the results do not depend on φ.
48Cr
Result for A=48 isobars
with Coulomb
Shifted semicircle method

  (   sin  , 0 ,   cos     off )
PRC 88(2013) 061301(R).
For T=4
   90

   12 . 0 MeV
 off   8 MeV
Energies are dependent
on Tz
(almost linear dependence)
No p-n mixing at |Tz|=T
   90

gives
Tˆz  0
48Ni
48Ca
T≅4states in A=40-56 isobars
T=1 triplets in A=14 isobars
14O(g.s)
14N (excited
0+)
14C(g.s)
Our framework nicely works also for
IASs in even-even A=40-56 isobars
Excited 0+, T=1 state in odd-odd 14N
Time-reversal symmetry conserved
14N:
p-n mixed , 14C,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
Tz
48Cr
Tz
𝑻=𝟎
(Tz=0)



(Tz=0)
−𝜆 𝑥 t 𝑥
Ty

Tx
48Cr
Tx
𝑻𝒙 = 𝟒, 𝑻y = 𝑻𝒛 = 𝟎,
𝑻=𝟒
(b) Starting with the highest weight state.
Tz
48Ca
(Tz=4)
𝑻𝒛 = 𝟒, 𝑻 = 𝟒
isospin
Tz
−𝜆 𝑥 t 𝑥
Ty
Tx
Ty
48Cr
Ty
Tx
(Tz=0)
(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.
x
(b) Starting with the highest weight state.
48Ca (Tz=4)

Isocranking calc.
㉘
n
𝜃=0
- 𝜆𝑧 𝑡𝑧
~11MeV
p


z
⑳
㊽

𝜃≠0
𝜆𝑧 /2
z
𝜆𝑧 /2

x
㊽
Tz



Tx
Tz
x
Tx
𝑇 = 𝑇𝑧 = 4
𝑇𝑧 = 4
𝑇𝑧 ≠ 4
Standard HF
p-n unmixed
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
z
How to determine the size of 𝜆 ?
Isobaric analog states with T=4 in A=48 nuclei
48Ni (Tz=-|T|=-4)
48Ca (Tz=|T|=4)
Tz
㉘
㉘
Tx
p
Tz
n
x
Tx


~11MeV

z
~11MeV
⑳
  180

Standard HF
n
Fermi energy
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.
  11 . 0
This choice of 𝜆 enable us to avoid the configuration change.
With Coulomb interaction
U
Coulomb
( z )
:violates isospin symmetry
The total energy is now dependent on Tz but independent of azimuthal angle 𝜑
Initial :
Tz
final :


 ˆ
  T

Tz

T




Ty
Ty

Tx
HF state w/o p-n
mixing
Tx
p-n mixed state
larger <Tz> is favored
・
・
・

With Coulomb, the s. p. Routhians
depend on the titling angle 𝜃
w/ Coulomb
w/o Coulomb
Shifted semicircle
semicircle
x



 
0


x

z
   n   p (T z  4 )
  p   n (T z   4 )
 n   p (T z   4 )
z
Coulomb gives additional isocranking freq. effectively   off  Tˆz
w/o Coulomb

  ( x ,  y  0,  z )
w/ Coulomb

  (  x ,  y  0 ,  z   off )
(MeV)
 n   p (T z  4 )
Difference of p and n
Fermi energies