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A Recursive Method
to Calculate
Nuclear Level Densities
Piet Van Isacker
GANIL, France
• Models for nuclear level densities
• Level density for a harmonic oscillator potential
• Simple illustrations
• Extension to general potentials
Models for nuclear level densities
• « An Attempt to Calculate the Number of Energy
Levels of a Heavy Nucleus » (Bethe 1936): Statistical
analysis of Fermi gas of independent particles.
• Numerous extensions: eg back shift.
• « Theory of Nuclear Level Density » (Bloch 1953);
« Influence of Shell Structure on the Level Density of a
Highly Excited Nucleus » (Rosenzweig 1957): ‘Exact’
counting methods in single-particle shell model.
• Numerous extensions (Zuker, Paar, Pezer,... ).
• « Nuclear Level Densities and Partition Functions
with Interactions » (French & Kota 1983): Effects of
residual interaction via spectral distribution method.
• « [] Level Densities [] in Monte Carlo Shell Model »
(Nakada & Alhassid 1997); « Estimating the Nuclear
Level Density with the Monte Carlo Shell Model »
(Ormand 1997): ‘Exact’ shell-model calculations.
Level density in a harmonic oscillator
• Question: How many (antisymmetric) states with an
energy Et exist for A particles in an isotropic HO?
• Answer: Given by the number of solutions of
k
n1n2n3 A
n 1 n 2 n 3 0
n
1
n 1 n 2 n 3 0
n2 n3 kn 1 n 2 n3
3
Et / Q
2
• Solution: c3(A,Q) calculated recursively through
cd A,Q c d 1 A, Qcd A A,Q Q A A
A Q
with initial values
cd A 0,Q Q 0
cd A,Q 0, if
c0 A,Q
Q Qdmin A
2s 1!
Q0
A! 2s 1 A !
Solution method
• We need the number of solutions of
kn1n2n3 A,
n 1 n 2 n 3 0
n
n
n
k
1 2 3 nnn
1 2 3
n1 n 2 n 3 0
Q
• Rewrite as
k
n 1 n 2 0 n 3 1
n 1n 2 n 3
n
n 1 n 2 0 n 3 1
1
A A'
n2 n3 kn1 n2 n 3 Q Q'
with
k
n 1 n 2 0
n1n2 0
A,
n
n 1 n 2 0
n
k
Q
2
n
1
1 n2 0
• Introduce new unknowns k n1 n 2 n 3 kn 1 n 2 n 3 1
k
n1n2n3 A A
n 1 n 2 n 3 0
n
n
n
k
1 2 3 nnn
1 2 3
n 1 n 2 n 3 0
Q Q A A
• Hence we find the recurrence relation:
cd A,Q c d 1A, Qcd A A,Q Q A A
A Q
Harmonic oscillator with spin
• Simple numerical implementation:
spin=1/2; deg=2*spin+1;
c[d_,aa_,qq_]:=c[d,aa,qq]=
Sum[c[d,aa-aap,qq-qqp-aa+aap]*c[d-1,aap,qqp],
{aap,0,aa},{qqp,qqmin[d-1,aap],qq-aa+aap-qqmin[d,aa-aap]}];
c[d_,aa_,qq_]:=Binomial[deg,aa]/; d==0 && qq==0;
c[d_,aa_,qq_]:=1/; aa==0 && qq==0;
c[d_,aa_,qq_]:=0/; aa==0 && qq!=0;
c[d_,aa_,qq_]:=0/; qq<qqmin[d,aa];
• c3(A,Q) can be calculated to very high excitation.
• Example: The number of independent Slater
determinants for A=70 (s=1/2) particles at an
excitation energy of 30 hw is
c 3 70,240 896647829312727644544457613187541
Comparison with Fermi-gas estimate
• Fermi-gas estimate (Bethe; cfr Bohr & Mottelson):
A,E
1
2
exp 2 g F E / 3
48E
• Correspondence:
A,E c3 A,Q Q3mi n E / /
g N 1N 2, N
Leonhard Euler
• L Euler in Novi Commentarii
Academiae Scientiarum
Petropolitanae 3 (1753) 125:
Tables for the ‘one-dimensional
oscillator’ problem.
A,E
1
2
exp 2 E / 3
48E
Enumeration of spurious states
• Only states that are in the ground configuration with
respect to the centre-of-mass excitation are of interest.
• c3(A,Q) includes all solutions. Let us denote the
physical solutions as
c˜ 3 A,Qe , Qe Q Q3mi n A
• This is found by substracting from c3(A,Q) those
states that can be constructed by acting with the stepup operator for the centre-of-mass motion. Hence:
c˜ 3 A,Qe c3 A,Qe
Qe
1
2 Qe 1Qe 2c˜3 A,Qe Qe
Qe 1
Harmonic oscillator with isospin
• Question: How many states with an energy E exist
for N neutrons and Z protons in a HO?
• Answer: Given by the number of solutions of
k
A
n1n2n3
n 1 n 2 n 3 0
A
N, A Z
n
n
n
k
1 2 3 nnn
1 2 3
n 1 n 2 n 3 0
= Q
• Solution: c3(N,Z,Q) can be calculated recursively or
through
c 3N,Z,Q c 3N,Q- Qc3 Z,Q
Q
Shell effects
• Fermi-gas estimate (Bethe; cfr Bohr & Mottelson):
N, Z,E
4
np
9g
gn nF gp Fp
g E
5 / 4
np
12
exp 2 gnp E / 3
2
gnp gn Fn gp Fp
• The quantity c3(N,Z,Q) can be evaluated for closed
as well as open shells => effects of shell structure on
level densities.
• Example: Comparison of 16O and 28Si.
Anisotropic harmonic oscillator
• So far: independent particles in a spherical HO =>
interaction effects (eg deformation) are not included.
• The analysis can be repeated for an anisotropic HO
with different frequencies w1, w2 and w3.
• Example: Axial symmetry with 1 2 12 3
• Energy is determined by Q12 and Q3:
Et Q12 1 12 Q3 2 3
1
• Number of configurations c3(N,Z,Q12,Q3) from:
k
A
n1n2n3
n 1 n 2 n 3 0
n1 n2 k n n n
n 1 n 2 n 3 0
1 2 3
A
N, A Z
= Q12 ,
n
k
3 n1n2n3 = Q3
n 1 n2 n 3 0
• Calculated recursively from:
c 3 N,Z,Q12 ,Q3
c N , Z , Q
N Z Q12
2
12
c 3 N N ,Z Z,Q12 Q
12,Q3 N N Z Z
Anisotropic harmonic oscillator
• Cumulative number of levels up to energy E:
E
FE E dE
0
• Example: Prolate & oblate. Normal & superdeformed.
3 1223
Anisotropic harmonic oscillator
• Example 1: 38Ar for 2=0.2.
• Example 2: 56Fe for 2=0.2.
E
FE E dE
0
12 1 13 , 3 1 23 , 45/ 16 2
3
122 3 41A 1/ 3 MeV
Extension to general potentials
• Assume single-particle levels with energies n and
degeneracies n with n=1,2,…
• Question: How many A-particle states with energy E?
• Answer: Given by the number c(A,E) of solutions of
i
k
n 1 m 1
nm
A,
i
k
n 1
n
m 1
nm
E
• Solution: c(A,E)c(0,A,E) with c(i,A,E) calculated
recursively through
i
ci, A,E
ci 1, A A, E i A
A A
with initial values
ci, A 0,E E 0
ci, A,E 0, if
E HF energy
Conclusions
• Versatile approach to compute level densities of
particles in a harmonic oscillator potential which
includes spin, isospin, deformation... (but without
residual interactions).
• Extension to a general potential [cfr. (micro)canonical
partition function for Fermi systems, S.Pratt, PRL 84
(2000) 4255].
Perspectives (general potential)
• Systematic use in combination with Hartree-Fock
calculations (eg for astrophysics).
• Spurious fraction of states can be estimated.
• Effects of the continuum can be included.
• Inclusion of interaction effects?