Handling tree structures — recursive SPs, nested sets, recursive CTEs
A Recursive Method to Calculate Nuclear Level Densities
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Transcript of A Recursive Method to Calculate Nuclear Level Densities
A Recursive Methodto Calculate
Nuclear Level Densities
• Models for nuclear level densities
• Level density for a harmonic oscillator potential
• Simple illustrations
• Extension to general potentials
Piet Van IsackerGANIL, France
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
• Solution: c3(A,Q) calculated recursively through
n1n2n3=0
∞
∑ kn1n2n3
σ
σ∑ =A
n1n2n3=0
∞
∑ n1 +n2 +n3( )kn1n2n3
σ
σ∑ =Et /hω −
32
≡Q
cd A,Q( ) = cd−1 ′ A , ′ Q ( )cd A− ′ A ,Q− ′ Q −A+ ′ A ( )′ A ′ Q ∑
withinitialvalues
cd A =0,Q( )=δQ0
cd A,Q( ) =0, if Q<Qdmin A( )
c0 A,Q( )=2s+1( )!
A! 2s+1−A( )!δQ0
Solution method
• We need the number of solutions of
• Rewrite as
• Introduce new unknowns
• Hence we find the recurrence relation:
n1n2n3=0
∞
∑ kn1n2n3
σ
σ∑ =A,
n1n2n3=0
∞
∑ n1 +n2 +n3( )kn1n2n3
σ
σ∑ =Q
n1n2=0
∞
∑n3=1
∞
∑ kn1n2n3
σ
σ∑ =A−A'
n1n2=0
∞
∑n3=1
∞
∑ n1 +n2 +n3( )kn1n2n3
σ
σ∑ =Q−Q'
with
n1n2=0
∞
∑ kn1n20σ
σ∑ = ′ A ,
n1n2=0
∞
∑ n1 +n2( )kn1n20σ
σ∑ = ′ Q
n1n2n3=0
∞
∑ ′ k n1n2n3
σ
σ∑ =A− ′ A
n1n2n3=0
∞
∑ n1 +n2 +n3( ) ′ k n1n2n3
σ
σ∑ =Q− ′ Q −A+ ′ A
′ k n1n2n3
σ ≡kn1n2n3+1σ
cd A,Q( ) = cd−1 ′ A , ′ Q ( )cd A− ′ A ,Q− ′ Q −A+ ′ A ( )′ A ′ Q ∑
Harmonic oscillator with spin
• Simple numerical implementation:
• 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
c3 70,240( ) =896647829312727644544457613187541
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];
Comparison with Fermi-gas estimate
• Fermi-gas estimate (Bethe; cfr Bohr & Mottelson):
• Correspondence:
ρ A,E( ) =148E
exp 2π 2g εF( )E /3
ρ A,E( )⇔ c3 A,Q=Q3min+E/hω( ) /hω
g ε( )= N +1( ) N +2( ), ε =Nhω
Leonhard Euler
• L Euler in Novi Commentarii Academiae Scientiarum Petropolitanae 3 (1753) 125: Tables for the ‘one-dimensional oscillator’ problem.
ρ A,E( ) =148E
exp 2π 2E/3
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
• This is found by substracting from c3(A,Q) those states that can be constructed by acting with the step-up operator for the centre-of-mass motion. Hence:
˜ c 3 A,Qe( ), Qe =Q−Q3min A( )
˜ c 3 A,Qe( ) =c3 A,Qe( )−12′ Q e=1
Qe
∑ ′ Q e +1( ) ′ Q e +2( )˜ c 3 A,Qe − ′ Q e( )
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
• Solution: c3(N,Z,Q) can be calculated recursively or through
n1n2n3=0
∞
∑ kn1n2n3
στ
στ∑ =Aτ A+ =N,A− =Z( )
n1n2n3=0
∞
∑ n1 +n2 +n3( )kn1n2n3
στ
στ∑ = Q
c3 N,Z,Q( )= c3 N,Q- ′ Q ( )c3 Z, ′ Q ( )′ Q
∑
Shell effects
• Fermi-gas estimate (Bethe; cfr Bohr & Mottelson):
• 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.
ρ N,Z,E( ) =9gnp
4
gn εFn( )gp εF
p( )
gnpE( )−5/4
12exp 2π2gnpE/ 3
gnp =gn εFn
( )+gp εFp
( )
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 ω1 =ω2 ≡ω12≠ω3
• Energy is determined by Q12 and Q3:
• Number of configurations c3(N,Z,Q12,Q3) from:
• Calculated recursively from:
Et = Q12 +1( )hω12 + Q3 +1
2( )hω3
c3 N,Z,Q12,Q3( ) = c2 ′ N , ′ Z , ′ Q 12( )′ N ′ Z ′ Q 12
∑
×c3 N − ′ N ,Z− ′ Z ,Q12 − ′ Q 12,Q3 −N + ′ N −Z+ ′ Z ( )
n1n2n3=0
∞
∑ kn1n2n3
στ
στ∑ =Aτ A+ =N,A− =Z( )
n1 +n2( )n1n2n3=0
∞
∑ kn1n2n3
στ
στ∑ = Q12, n3
n1n2n3=0
∞
∑ kn1n2n3
στ
στ∑ = Q3
Anisotropic harmonic oscillator
• Cumulative number of levels up to energy E:
• Example: Prolate & oblate. Normal & superdeformed.
F E( ) = ρ ′ E ( )0
E
∫ d ′ E
ω = ω122ω3
3
Anisotropic harmonic oscillator
• Example 1: 38Ar for 2=0.2.
• Example 2: 56Fe for 2=0.2.
F E( ) = ρ ′ E ( )0
E
∫ d ′ E
ω12 =ω 1+13δ( ), ω3 =ω 1−2
3δ( ), δ = 45/16πβ2
hω ≈h ω122ω3
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
• Solution: c(A,E)c(0,A,E) with c(i,A,E) calculated recursively through
n=1
∞
∑ knmm=1
Ω i
∑ =A, εnn=1
∞
∑ knmm=1
Ω i
∑ =E
c i,A,E( ) =Ω i
′ A ⎛ ⎝ ⎜ ⎞
⎠ c i +1,A− ′ A ,E −εi ′ A ( )
′ A ∑
withinitialvalues
c i,A=0,E( ) =δE0
c i,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?