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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?