Nuclear Structure Theory

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Nuclear Structure Theory Introduction to nuclear structure A. Pastore 1Department of Physics, University of York, Heslington, York, YO10 5DD, UK August 8, 2017 Introduction [C. Diget, A.P, et al Physics Education 52 (2), 024001 (2017)] [YouTube: Binding Block Channel] A.Pastore August 8, 2017 2 / 59 Introduction A.Pastore August 8, 2017 3 / 59 Degrees of freedom A.Pastore August 8, 2017 4 / 59 Binding energy How much energy there is in a nucleus? A.Pastore August 8, 2017 5 / 59 Liquid drop-model Let's make a simple model of the nucleus: Bethe-Weiszs¨acker mass formula (total energy) Z2 (N − Z)2 E = a A − a A2=3 − a − a + δ(A; Z) B V S C A1=3 asym A To obtain energy per nucleon we do EB=A Physically motivated terms Phenomenological model based on empiric observation. [ http://amdc.in2p3.fr/web/dz.html] A.Pastore August 8, 2017 6 / 59 Compare theory/experiment We observe peaks at N=8,20,28,50,82,... why? A.Pastore August 8, 2017 7 / 59 Intro to quantum mechanics! The time-dependent Schro¨edingerequations reads @ H^ Ψ(r; t) = i~ Ψ(r; t); @t p^2 @ + V^ (r; t) Ψ(r; t) = i~ Ψ(r; t): 2m @t If H^ is time independent than the time evolution and the coordinate evolution are separable. H^ (r)Ψ(r) = EΨ(r); with H defined as p^2 + V (r) Ψ(r) = EΨ(r): 2m Exercise Consider a time-independent H and derive time-indpendent Schro¨edingerequation! A.Pastore August 8, 2017 8 / 59 Example: free particle in 1 dimension One dimensional case r ! x (simplicity) ^ p^2 V (r) = 0 ) H = 2m H^ = E −~2 d2 (x) = E (x) 2m dx2 (x) = eikx only E > 0 is allowed and p 2mE k = ~ A.Pastore August 8, 2017 9 / 59 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, 2017 10 / 59 Infinite square well in 1D 0 0 < x < a V (x) = 1 x < 0; x > a Exercise: Verify the following solutions r2 nπx (x) = sin n = 1; 2;::: n a a n2π2~2 E = n 2ma2 Comments Spectrum discrete (localisation) Role of boundary conditions! A.Pastore August 8, 2017 11 / 59 Infinite square well in 1D Let's take a = 10 fm ~2=2m = 20:75 MeVfm−2 to describe a nucleus 200 0.4 150 0.2 100 (x) 0 Ψ MeV -0.2 50 -0.4 0 0 2 4 6 8 10 0 2 4 6 8 10 n x [fm] Basis The solution of the infinite square well form a complete basis! Z dx n(x)φn0 (x) = δnn0 A.Pastore August 8, 2017 12 / 59 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 px2 + y2 + z2 < a V (x; y; z) = 1 px2 + y2 + z2 > a A.Pastore August 8, 2017 13 / 59 Infinite square well in 3D Given the symmetry of the system (spherical) we can factorise the solution as (r; θ; φ) = Rnl(r)Ylm(θ; φ) You can demonstrate (exercise) that we reduce to the following equation (in 1D!) d2R 2 dR l(l + 1) nl + n;l + k2 − R = 0 dr2 r dr r2 n;l 2 2mE with k = ~2 . With simple re-scaling (z=kr) we reduce to Bessel equation. The quantisation come from imposing the solution to be zero at the edge! Bessel functionas are a basis! A.Pastore August 8, 2017 14 / 59 Solving Schro¨edingerequation on a basis We can consider 1D case (to make it simple ! l=0.) 20 Continuum 0 -20 Classic Forbidden Classic Forbidden V(x) [MeV] -40 -60 -30 -20 -10 0 10 20 30 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 ^ hα,α0 = hBasisαjHjBasisα0 i X n n(x) = Cα (Basisα) α Exercise: try to run the code... check the convergence of eigenstates as a function of basis states A.Pastore August 8, 2017 15 / 59 Solving the equation A.Pastore August 8, 2017 16 / 59 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 2 1981 Elastic scattering of 80-180MeV protons and the proton-nucleus optical potential A.Pastore August 8, 2017 17 / 59 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, 2017 18 / 59 How to build a realistic potential? The Woods-Saxon potential Courtesy: V. Som´a A.Pastore August 8, 2017 19 / 59 Spin-orbit! We have to consider a coupling between ~s and ~l Vso(r)~s · ~l Numerical code provided. A.Pastore August 8, 2017 20 / 59 A.Pastore August 8, 2017 21 / 59 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, 2017 22 / 59 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 Heff Diagonalise Heff in the restricted space A.Pastore August 8, 2017 23 / 59 Shell model How to build Heff ?? 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 Heff to data. Successful in reproducing data, not possible to use for extrapolations A.Pastore August 8, 2017 24 / 59 Nuclear Hamiltonian The general nuclear Hamiltonian reads ~2 − X 2 X X H = r + vij + Vijk + ··· + n-body terms 2m i i i≤j i≤j≤k where vij is the 2-body Nucleon-Nucleon interaction (NN) and Vijk is the 3-body one. X p vij = vp(rij )Oij p=1;n p=1;8 Oij = 1; τiτj ; σiσj ; (τiτj )(σiσj );Sij ;Sij (τiτj ); L · S; L · Sτiτj A.Pastore August 8, 2017 25 / 59 Coulomb The hamiltonian for 1 atom (fixed position) reads (in natural units ~ = me = "0 = 1 n n n n Xe r2 Xe 1 Xe Xe 1 H = − i − Z + 2 ri rij i=1 i=1 i=1 j>i We anticipate here that our goal is to find a procedure so that ne ne ne X e 1 X X ee H = h + v¯ i 2 i=1 i=1 j>i e ee where hi is a single-electron Hamiltonian of the electron i and v¯ is a residual interaction that is difficult to treat. A.Pastore August 8, 2017 26 / 59 Hartree-Fock equations ^ HF PA We write nuclear Hamiltonian as H = i h(i). We assume independent particle motion and we get a set of equations h(i)φk(i) = "kφk(i) with i = (r; στ) The total w.f. of the nucleus is a Slater determinant (we drop spin/isospin for brevity) 1 φ1(r1) : : : φ1(rN ) φ (r ) : : : φ (r ) Φ(r1; : : : ; rN ) = p 2:::::::::1 2 N A! φN (r1) : : : φN (rN ) Pauli principle! Remember that for a system of 2 particles the Slater determinant reads 1 Φ(r1; r2) = p [φ1(r1)φ2(r2) − φ1(r2)φ2(r1)] 2 A.Pastore August 8, 2017 27 / 59 Hartree-Fock equations The Hartree Fock equations read (derivation on request) " ~2 Z # Z 2 X ∗ X ∗ − r + dyφi (y)v(x; y)φi(y) φb(x) − dyφi (x)v(x; y)φi(y)φb(y) = "bφb(x) 2m i 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, 2017 28 / 59 Hartree-Fock energy The energy of the system is EHF = hΦjH^ HF jΦi or A A HF X 1 X E = t + v¯ ii 2 ij;ij i=1 i;j=1 A A X 1 X = " − v¯ ii 2 ij;ij i=1 i;j=1 HF states are fully occupied or empty (1 or 0) HF states are filled from low to high energy Koopman's theorem HF HF E [N + 1] − E [N] ≈ "N+1 A.Pastore August 8, 2017 29 / 59 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 0 0 0 0 hr1r2jV jr1r2i = V (r1; r2; r1; r2) Hermiticity, V^ + = 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.
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