Introduction to nuclear structure

Transcription

1 Introduction to nuclear structure A. Pastore 1 Department of Physics, University of York, Heslington, York, YO10 5DD, UK August 8, 2017

2 Introduction [C. Diget, A.P, et al Physics Education 52 (2), (2017)] [YouTube: Binding Block Channel] A.Pastore August 8, / 59

3 Introduction A.Pastore August 8, / 59

4 Degrees of freedom A.Pastore August 8, / 59

5 Binding energy How much energy there is in a nucleus? A.Pastore August 8, / 59

6 Liquid drop-model Let s make a simple model of the nucleus: Bethe-Weiszsäcker mass formula (total energy) E B = a V A a S A 2/3 Z 2 a C A 1/3 a (N Z) 2 asym + δ(a, Z) A To obtain energy per nucleon we do E B /A Physically motivated terms Phenomenological model based on empiric observation. [ A.Pastore August 8, / 59

7 Compare theory/experiment We observe peaks at N=8,20,28,50,82,... why? A.Pastore August 8, / 59

8 Intro to quantum mechanics! The time-dependent Schroëdinger equations reads ĤΨ(r, t) = i Ψ(r, t), t [ ] ˆp 2 2m + ˆV (r, t) Ψ(r, t) = i Ψ(r, t). t If Ĥ is time independent than the time evolution and the coordinate evolution are separable. Ĥ(r)Ψ(r) = EΨ(r), with H defined as [ ] ˆp 2 2m + V (r) Ψ(r) = EΨ(r). Exercise Consider a time-independent H and derive time-indpendent Schroëdinger equation! A.Pastore August 8, / 59

9 Example: free particle in 1 dimension One dimensional case r x (simplicity) V (r) = 0 Ĥ = ˆp2 2m Ĥψ = Eψ only E > 0 is allowed and 2 2m d 2 ψ(x) = Eψ(x) dx2 ψ(x) = e ikx k = 2mE A.Pastore August 8, / 59

10 Fermi gas Consider a system of several free Fermions. Pauli principle two or more identical fermions (particles with half-integer spin) cannot occupy the same quantum state within a quantum system simultaneously Quantum numbers k, σ, τ A.Pastore August 8, / 59

11 Infinite square well in 1D { 0 0 < x < a V (x) = x < 0, x > a Exercise: Verify the following solutions 2 ( nπx ) ψ n (x) = a sin a E n = n2 π 2 2 2ma 2 n = 1, 2,... Comments Spectrum discrete (localisation) Role of boundary conditions! A.Pastore August 8, / 59

12 Infinite square well in 1D Let s take a = 10 fm 2 /2m = MeVfm 2 to describe a nucleus MeV n Ψ(x) x [fm] Basis The solution of the infinite square well form a complete basis! dxψ n(x)φ n (x) = δ nn A.Pastore August 8, / 59

13 Infinite square well in 3D In physics it is important to consider symmetries of the system Cartesian Spherical Cylindrical x x = r sin θ cos φ x = ρ cos φ y y = r sin θ sin φ x = ρ sin φ z z = r cos θ z=z According to the symmetry of the Hamiltonian we can use the most adapted coordinate system. { 0 x2 + y V (x, y, z) = 2 + z 2 < a x 2 + y 2 + z 2 > a A.Pastore August 8, / 59

14 Infinite square well in 3D Given the symmetry of the system (spherical) we can factorise the solution as d 2 R nl dr 2 ψ(r, θ, φ) = R nl (r)y lm (θ, φ) You can demonstrate (exercise) that we reduce to the following equation (in 1D!) ( ) + k 2 R n,l = r dr n,l dr l(l + 1) r 2 with k 2 = 2mE. With simple re-scaling (z=kr) we reduce to Bessel equation. The quantisation come 2 from imposing the solution to be zero at the edge! Bessel functionas are a basis! A.Pastore August 8, / 59

15 Solving Schroëdinger equation on a basis We can consider 1D case (to make it simple l=0.) 20 0 Continuum V(x) [MeV] -20 Classic Forbidden Classic Forbidden x [fm] A nucleus is more reasonably represented via finite-well (approximation). How to solve the Schroedinger equation? We can project on a basis and solve eigenvalue problem h α,α = Basis α Ĥ Basis α ψ n(x) = α C n α(basis α) Exercise: try to run the code... check the convergence of eigenstates as a function of basis states A.Pastore August 8, / 59

16 Solving the equation A.Pastore August 8, / 59

17 How to build a realistic potential? We want to use QM We use experimental data to adjust the potential Ingredient 1: independent particle motion A. Nadasen et al. Phys. Rev C Elastic scattering of MeV protons and the proton-nucleus optical potential A.Pastore August 8, / 59

18 How to build a realistic potential? Ingredient 2: density profile Hint: we need a potential that looks like the density profile A.Pastore August 8, / 59

19 How to build a realistic potential? The Woods-Saxon potential Courtesy: V. Somá A.Pastore August 8, / 59

20 Spin-orbit! We have to consider a coupling between s and l Numerical code provided. V so (r) s l A.Pastore August 8, / 59

21 A.Pastore August 8, / 59

22 Single particle evolution with mass Single particle states move with mass (crossings) and deformations. Possible change in magic numbers away from stability? A.Pastore August 8, / 59

23 Shell model Build an effective 1-body potential Tune parameters to obtain magic numbers Define an active Hilbert space (valence space) Build valence-space Hamiltonian H eff Diagonalise H eff in the restricted space A.Pastore August 8, / 59

24 Shell model How to build H eff?? Ab-initio: use projection methods to go from full to restricted Hilbert space. Universal and systematic but very time-consuming, advanced many-body methods are required Phenomenological: fit H eff to data. Successful in reproducing data, not possible to use for extrapolations A.Pastore August 8, / 59

25 Nuclear Hamiltonian The general nuclear Hamiltonian reads H = 2 2m 2 i + v ij + V ijk + + n-body terms i i j i j k where v ij is the 2-body Nucleon-Nucleon interaction (NN) and V ijk is the 3-body one. v ij = v p(r ij )O p ij p=1,n O p=1,8 ij = 1, τ i τ j, σ i σ j, (τ i τ j )(σ i σ j ), S ij, S ij (τ i τ j ), L S, L Sτ i τ j A.Pastore August 8, / 59

26 Coulomb The hamiltonian for 1 atom (fixed position) reads (in natural units = m e = ε 0 = 1 n e H = i=1 2 ne i 2 Z i=1 1 n e + r i n e 1 r ij i=1 j>i We anticipate here that our goal is to find a procedure so that n e H = h e i + 1 n e n e 2 i=1 i=1 j>i where h e i is a single-electron Hamiltonian of the electron i and v ee is a residual interaction that is difficult to treat. v ee A.Pastore August 8, / 59

27 Hartree-Fock equations We write nuclear Hamiltonian as ĤHF = A i h(i). We assume independent particle motion and we get a set of equations h(i)φ k (i) = ε k φ k (i) with i = (r, στ) The total w.f. of the nucleus is a Slater determinant (we drop spin/isospin for brevity) Φ(r 1,..., r N ) = 1 φ 1(r 1)... φ 1(r N ) φ 2(r... 1) φ 2(r... N ) A! φ N (r 1)... φ N (r N ) Pauli principle! Remember that for a system of 2 particles the Slater determinant reads Φ(r 1, r 2) = 1 2 [φ 1(r 1)φ 2(r 2) φ 1(r 2)φ 2(r 1)] A.Pastore August 8, / 59

28 Hartree-Fock equations The Hartree Fock equations read (derivation on request) [ 2 2m 2 + i ] dyφ i (y)v(x, y)φ i(y) φ b (x) dyφ i (x)v(x, y)φ i(y)φ b (y) = ε b φ b (x) i To solve this equation we have to set up a self-consistent procedure. 1 choose a set of single-particle states that are supposed to not be too far from the solution. 2 The HF hamiltonian is computed 3 New single-particle states are found 4 Iterate until convergence A.Pastore August 8, / 59

29 Hartree-Fock energy The energy of the system is E HF = Φ ĤHF Φ or E HF = = A t ii i=1 A ε ii 1 2 i=1 A i,j=1 A i,j=1 v ij,ij v ij,ij HF states are fully occupied or empty (1 or 0) HF states are filled from low to high energy Koopman s theorem E HF [N + 1] E HF [N] ε N+1 A.Pastore August 8, / 59

30 What about the interaction? Warning We can not use bare NN interaction in HF method! Independent particle motion not compatible with hard cores! Need extra correlation beyond Hartree-Fock to have a converged result r 1r 2 V r 1r 2 = V (r 1, r 2, r 1, r 2) Hermiticity, ˆV + = V, to have real eigenvalues. Invariance under the exchange of coordinates, V (1, 2) = V (2, 1), so that the interaction does not change the exchange symmetry of the wavefunction. Translational invariance and Rotational invariance, the system behaves equally if you change coordinates. Galilean invariance, in the case of non relativistic systems the potential is not change if the system moves at constant velocity. Space reflection, there is no parity violation in the strong interaction. Time reversal, equation of motion must not depend on the time direction. A.Pastore August 8, / 59

31 Example 1: Skyrme Skyrme interaction was proposed already in the 50 ṽ Skyrme (r 12 ) = t 0 (1 + x 0 P σ )δ(r 1 [ r 2 ) ] t 1(1 + x 1 P σ ) δ(r 1 r 2 )k 2 + k 2 δ(r 1 r 2 ) Momentum Dependent +t 2 (1 + x 2 P σ )k δ(r 1 r 2 ) k Momentum Dependent t 3(1 + x 3 P σ )ρ α (R)δ(r 1 r 2 ) Density Dependent +iw 0 (σ 1 + σ 2 )k δ(r 1 r 2 ) k Spin Orbit with k the relative momentum operator k = 1 ( 1 2). 2i ρ α (R) is the density dependent term, usually with 1/6 α 2/3 and 2R = r 1 + r 2. Parameters The parameters t i=0,1,2,3, x i=0,1,2,3 W 0 are free parameters adjusted on nuclear data A.Pastore August 8, / 59

32 Example II: Gogny Gogny interaction was proposed in the 80s ṽ Gogny (r 12 ) = i (W ib i P σ H i P τ M i P σp τ ) exp r2 12 /µ2 i Central term +t 3 (1 + x 3 P σ )ρ α (R)δ(r 1 r 2 ) Density Dependent +iw 0 (σ 1 + σ 2 )k δ(r 1 r 2 ) k Spin Orbit Notice Almost all interaction contain a zero-range spin-orbit term and zero-range density dependent! Parameters The parameters H i, W i, B i, M i, t 3, x 3 W 0 are free parameters adjusted on nuclear data A.Pastore August 8, / 59

33 Infinite nuclear matter We consider a medium infinite in size No Coulomb non-polarised (we can remove) with isospin asymmetry Y = ρn ρp ρ n+ρ p A.Pastore August 8, / 59

34 Hartree-Fock calculation Consider an interaction of the type V = t 0(1 + x 0P σ)δ(r 1 r 2) + 1 ( 6 t3(1 + x3pσ)ρ r1 + r ) α 2 δ(r1 r 2) 2 this is the simplest form of the Skyrme interaction. We calculate the HF energy for special case Y=0 (equal number of protons and neutrons). In infinite matter the w.f. reads the system is uniform! φ(k) = 1 V exp ikr χ 1 2 σ χ 1 2 τ E A = E kin ij V (1 P xp σp τ ) ij Remember that the system is degenerate (2 in spin and 2 isospin) and ij P 2 σ/τ = 1 P x = 1 Exercise: calculate the energy of this interaction. A.Pastore August 8, / 59

35 Hartree-Fock calculation The explicit calculation lead to E A = 3 2 kf 2 5 2m + 3 t3 t0ρ ρ1+α We can plot E/A as a function of the density of the system. Experimental data: ρ sat = 0.16 fm 3 and E/A = 16 MeV (remember a V Drop!) We can calculate t 0, t 3 as of Liquid t 0 = MeVfm 3 t 3 = MeVfm 6 we take α = 1 so we can link the two body density dependent to a contact three-body. Exercise: do the calculations!!! A.Pastore August 8, / 59

36 Hartree-Fock calculation With such a simple model we can then plot the EoS in symmetric matter E/A [MeV] ρ [fm -3 ] Symmetric Neutron matter We use EoS to calculate the mass and the radius of a NS via e Tolman-Oppenheimer-Volkoff (TOV) equations for the total pressure P and the enclosed mass m dp (r) dr dm(r) dr = Gm(r)ε(r) r 2 = 4πr 2 ε(r), [ ( 1 + P (r)ε(r) ) ( )] [ c πr3 P (r) ε(r)c 2 1 2Gm(r) ] 1 rc 2, where G is the gravitational constant and ε(r) is the total energy density of the system [We need to include mass contribution!!]. A.Pastore August 8, / 59

37 More realistic interactions Energy per particle in SNM (panel a) and PNM (panel b) for some effective interactions at HF level. symbols refer to ab-initio results. E/A [MeV] BHF M3Y-P2 M3Y-P3 M3Y-P4 M3Y-P5 M3Y-P6 M3Y-P7 a) E/N [MeV] M3Y-P2 M3Y-P3 M3Y-P4 M3Y-P5 M3Y-P6 M3Y-P7 b) ρ [fm -3 ] ρ [fm -3 ] Infinite matter Excellent medium to test isovector properties (pure neutron matter) against more sophisticated many-body methods! A.Pastore August 8, / 59

38 Hartree-Fock calculation M/M Sun Neutrons only Radius [10 km] We consider only neutrons. β-equilibrium should be imposed!! Possible extra degrees of freedom at high density n, p Λ, Σ,.... Using astrophysical constraints to constraint nuclear physics (??) A.Pastore August 8, / 59

39 Mass-radius relation The shaded region enclosed by a full line is obtained from quiescent low-mass X ray binary mass and radius observations. The upper limit on NS mass is indicated by a grey line. Taken from R. Sellahewa, A. Rios; Phys. Rev. C 90, (2014) A.Pastore August 8, / 59

40 What is a neutron star? A.Pastore August 8, / 59

41 What is a neutron star? A.Pastore August 8, / 59

42 What is a neutron star? A.Pastore August 8, / 59

43 Finite nuclei: example in 208 Pb We can use HF formalism to treat double-magic nuclei (what happens in open shell???) A.Pastore August 8, / 59

44 Finite nuclei: example in 208 Pb HF potential The potential looks similar to WS We have an effective mass the interaction dresses nucleons. We need to consider Coulomb for protons A.Pastore August 8, / 59

45 Finite nuclei: example in 208 Pb Skyrme model m is energy independent in Skyrme model. Finite-range (or higher order Skyrme) models have an explicit momentum (energy) dependence A.Pastore August 8, / 59

46 Finite nuclei: example in 208 Pb A.Pastore August 8, / 59

47 How to build an effective interaction? A.Pastore August 8, / 59

48 How to build an effective interaction? A.Pastore August 8, / 59

49 How to build an effective interaction? Global performance of total binding energy E and 2-n separation energy S 2n A.Pastore August 8, / 59

50 What is missing? A.Pastore August 8, / 59

51 Superfluidity A.Pastore August 8, / 59

52 Superfluidity A.Pastore August 8, / 59

53 Superfluidity A.Pastore August 8, / 59

54 Superfluidity Bardeen-Cooper-Schrieffer (BCS) theory J. Bardeen, L. N. Cooper, and J. R. Schrieffer Phys. Rev. 108, 1175 (1957) A.Pastore August 8, / 59

55 Superfluidity Pairing favours Cooper pairs. The system gains energy by forming pairs even if they are made up with nucleons on higher lying states BCS model can be seen as an extension of Hartree-Fock What s the origin of pairing? Phonon exchange vs bare interaction? Proton/neutron pairing? Surface/Volume effects? A.Pastore August 8, / 59

56 BCS equations We consider HF ground state as starting point. We freeze s.p. scheme We solve number and gap equation We consider T = 1 pairing (only) and J = L = 0 (maximum overlap) We obtain new occupation probabilities v 2 impact on density (iterate?) k k = m>0 U mv mv k km m A = k V 2 k Occupation factor V 2 (U 2 ) stands for how occupied (empty) is a state. V 2 < 1 around ε F nuclei. Uk 2 + Vk 2 = 1 Remember HF V 2 = 1, 0! for open-shell A.Pastore August 8, / 59

57 BCS equations V 2 k = 1 [ 1 ε ] k λ 2 Eqp k E k qp = (ε k λ) k k A.Pastore August 8, / 59

58 Pairing correlations We define the experimental gap as exp odd = 1 2 [E b(n + 1) 2E b (N) + E b (N 1)] exp even = 1 [ exp ] (N 1) + exp odd odd 2 (N + 1) A.Pastore August 8, / 59

59 Pairing correlations Pairing correlations play crucial role in determining 2-neutron drip-line S 2n = E(N + 2) E(N) A.Pastore August 8, / 59