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Transcript
16.451 Lecture 23: Applications of the Shell Model Generic pattern of single particle states solved in a Woods-Saxon (rounded square well) potential model with appropriate spin-orbit interaction to reproduce the observed “magic number” pattern: State labels: nj where n labels the order of occurrence of a given l value, and the state labels for orbital angular momentum are: 0 s 1 p 2 d 3 f 4 g 5 h 6 i Each state can hold (2j +1) neutrons and (2j+1) protons, corresponding to 2(2j+1) distinct configurations of identical nucleons (mt, mj) to satisfy the Pauli exclusion principle 27/11/2003 1 Quantum numbers for a nucleus: 2 First of all, consider a “closed shell”, which corresponds to a completely filled single-particle state, e.g. 1s1/2, 1p3/2, etc... containing (2j+1) protons or neutrons: The total angular momentum is: J ji , all have the same j in a given shell M m i 0 each m value is different, ranging from –j to +j There is effectively only one configuration here, with total z-projection M = 0. Therefore, the total angular momentum of a closed shell must be J = 0! The total parity is: (1) (1) 2 j 1 Since (2j+1) is always even, the parity of a closed shell is always positive. Valence Nucleons: 3 For a closed shell + n nucleons, the angular momentum and parity is determined by the n “valence” nucleons, since the closed shell contributes J = 0+ : J n ji , (1) n (1) n i 1 The parity is uniquely determined, but there may be several different values of J that are consistent with angular momentum coupling rules. Residual interactions between the valence nucleons in principle determine which of the allowed J has the lowest energy – we can’t predict this a priori but can learn from experiment. “Holes” -- for a state that is almost full, it is simpler to consider angular momentum coupling for the missing nucleons than for the ones that are present: result for a closed shell 0 n i 1 ji 2 j 1 i ( n 1) ji magnitudes of the two partial sums have to be the same, and M values opposite. Total angular momentum for a collection of missing particles, i.e. holes, is the same as for that same collection of particles in a given shell model state. Magnetic Moments: 4 As in lecture 19, we can write: gJ J N J 1 ( J 1) Predictions are only simple in the limit of one valence nucleon, and then we have: valence proton: valence neutron: g s, p g s,n s N s N Simplifying the results (assignment 4), we find: j 1/ 2 j 1/ 2 with g ( j 1 / 2) g s / 2 N g j ( j 3 / 2) /( j 1) g s j / 2( j 1) N g sp 5.58, g sn 3.83, gp 1, gn 0 E (MeV) Application with one valence nucleon: 5 17O There are 8 protons and 9 neutrons, so we only need the low lying states in the shell model spectrum to understand the energy levels: valence n Ground state: full to here plus one neutron Ground state quantum numbers should be those of the valence neutron in the 1d5/2 state: J = 5/2+ Magnetic moment prediction: j = l + ½, odd neutron = neutron = -1.91 N measured value: -1.89 N excellent agreement! E (MeV) Excited states of 17O can be understood by promoting the valence neutron: 6 First excited state: J = ½ + ½ + state ground state full to here Excited states of 17O: 7 E (MeV) Next excited state: J = ½ explained by promoting a neutron from the filled 1 p1/2 level to the 1d5/2 level 0+ pair ½ - neutron hole E (MeV) Excited states of 17O 8 The 5/2- state is not so easy: to have negative parity, there must be an odd nucleon in a p state (or f state, but that is higher) J=? 5/ 2 5/ 2 J , J 0, 1, 2, 3, 4, 5 But two neutrons are required to have different values of mj by the Pauli principle. Writing out the allowed configurations only J = 0, 2, 4 are allowed! (J = 2 will work here) Two valence nucleons: (enough already!) 9 This problem is much more complicated! The inner “core” nucleons couple to J = 0+ but in general there is more than one possibility for the angular momentum coupling of the valence particles. a) (pp) or (nn) case: Z and N are both even in these cases, so we know that the ground state configuration will be 0+ no matter what shell model state they are in. Excited states will have higher angular momentum, with possibilities restricted by the Pauli principle. b) (np) case: Z and N are both odd in this case. Only 6 examples in the whole nuclear chart!!! In isolation, (np) prefers to form a bound state – the deuteron – with J = 1+. Example: 18F, Z = 9, N = 9 np pair: no restrictions on total J. possibilities are 0, 1, 2, 3, 4, 5 Ground state of 18F is 1+ i.e. deuteron quantum numbers! N.B. valence nucleons interact with each other, or all J values would be degenerate! Magnetic moments revisited: (see Krane, Fig. 5.9) 10 Single particle predictions: j 1/ 2 j 1/ 2 g ( j 1 / 2) g s / 2 N g j ( j 3 / 2) /( j 1) g s j / 2( j 1) N Data for odd-proton nuclei: (lines are calculated from the formulae) Agreement is not terrific, but the values lie within the two “Schmidt lines” j Odd neutron nuclei: 11 17O example: “perfect” What is wrong? • the single particle model is too simple – nucleons interact with each other • configurations may be mixed, i.e. linear combinations of different shell model states • magnetic moments of bound nucleons may not be the same as those of free nucleons... Title page of a research monograph, Oxford, 1990: From the preface: 12 Even the most sophisticated nuclear models are not completely successful! 13 from p. 53 Indium isotopes, Z = 49 data, for those with J = 9/2+ ground states theory, with residual interactions, and g-factors reduced by 50% compared to free nucleons! long-lived excited states with J = ½so, don’t be disappointed if your homework question didn’t work out perfectly! Collective Excitations in nuclei – a new class of models (Krane, Sec. 5.2) 14 Introduction: • Over half the known nuclei have configurations (Z,N) even, J = 0+ • Recall that an empirical pairing term is included the semi-empirical mass formula to account for their unusual stability. N.B. The pairing term is not accounted for in the shell model, which ignores all interactions between particles!!! • It costs too much energy to break a pair of nucleons and populate higher single particle states, so the excitations of even-even nuclei tend to be of a collective nature the nuclear matter distribution as a whole exhibits quantized vibrations in some cases and rotations in others, with characteristic frequency patterns. • Vibrational spectra are seen in nuclei that have an intrinsic spherical shape • Rotational excitations tend to occur in nuclei with permanent quadrupole deformations Vibrational states: 15 Model: quantized oscillations of a liquid droplet at constant density (why? repulsive short-distance behaviour of the N-N force!) Consider oscillations about a spherical equilibrium shape, with a time-dependent boundary surface expressed as a linear combination of spherical harmonic functions: R( , , t ) Ro 1 (t ) Y ( , ) expansion describes any shape at all, given appropriate coefficients. Each contribution can in principle oscillate at a different frequency.... Normal modes of the system correspond to excitations with a particular value of and , and these will occur at characteristic frequencies. Application to nuclei: 1. restriction to axial symmetry, i.e. = 0 2. vibrations are quantized: En = n ħ Illustration: time sequence of oscillating nuclear shapes 16 forbidden – density change! = 0 forbidden – CM moves! = 1 OK... = 2 = 3