## Shell model calculation and the USD, USDA, and USDB interaction

Form the mean field calculation, the single particle energies are obtained. However, the residual interaction is still there that the actual state could be affected. Because the residual interaction produces the off-diagonal terms in the total Hamiltonian, and that mixed the single particle state.

The Shell Model calculation can calculate the nuclear structure from another approach. It started from a assumed nuclear Hamiltonian, with a basis of wavefunctions. The Hamiltonian is diagonalized with the basis, then the eigenstates are the solution of the wavefunctions and the nuclear structure, both ground state and excited states. The basis is usually the spherical harmonic with some radial function. Or it could be, in principle, can take from the result of mean field calculation. Thus, the Shell Model calculation attacks the problem directly with only assumption of the nuclear interaction.

However, the dimension of the basis of the shell model calculation could be very huge. In principle, it should be infinitely because of the completeness of vector space. Fro practical purpose, the dimension or the number of the basis has to be reduced, usually take a major shell. for example the p-shell, s-d shell, p-f shell. However, even thought the model space is limited, the number of basis is also huge. “for $^{28}$Si the 12-particle state with M=0 for the sum of the $j_z$ quantum numbers and $T_z=0$ for the sum of the %Latex t_z$quantum numbers has dimension 93,710 in the m-scheme” [B. A. Brown and B. H. Wildenthal, Ann. Rev. Nucl. Part. Sci. 38 (1998) 29-66]. Beside the huge dimensions and the difficult for diagonalizing the Hamiltonian, the truncation of the model space also affect the interaction. We can imagine that the effective interaction is different from the actual nuclear interaction, because some energy levels cannot be reached, for example, the short range hard core could produce very high energy excitation. Therefore, the results of the calculation in the truncated model space must be “re-normalized”. There are 4 problems in the shell model calculation: • the model space • the effective interaction • the diagonalization • the renormalization of the result The shell model can also calculate the excited state with $1\hbar \omega$ (1 major shell). This requires combination of the interactions between 2 major shell. For usage, say in the code OXBASH, user major concern is the choice of the interaction and model space. The shell model are able to calculate • The binding energy • The excitation energies • The nucleons separation energies • The configuration of each state • The magnetic dipole matrix elements • The Gamow-Teller (GT) transition • The spectroscopic factor • …… and more. The W interaction (or the USD) for the s-d shell was introduced by B.H. Wildenthal around 1990s. It is an parametric effective interaction deduced from fitting experimental energy levels for some s-d shell nuclei. Before it, there are some theoretical interactions that require “no parameter”, for example the G-matrix interaction is the in-medium nucleon-nucleon interaction. The problem for the USD interaction is the interpretation, because it is a black-box that it can reproduce most of the experimental result better than theoretical interactions, but no one know why and how. One possible way is translate the two-body matrix elements (TBME) back to the central, spin-orbit, tensor force. It found that the central and spin-orbit force are similar with the theoretical interactions, but the tensor force could be different. Also, there could be three-body force that implicitly included in the USD interaction. In 2006, B.A. Brown and W.A. Richter improved the USD interaction with the new data from the past 20 years [B.A. Brown, PRC 74, 034315(2006)]. The new USD interaction is called USDA and USDB. The difference between USDA and USDB is the fitting (something like that, I am not so sure), but basically, USDA and USDB only different by very little. Since the USDB has better fitting, we will focus on the USDB interaction. The single particle energy for the USDB is • $1d_{3/2} = 2.117$ • $2s_{1/2} = -3.2079$ • $1d_{5/2} = -3.9257$ in comparison, the single particle energies of the neutron of 17O of $2s_{1/2} = -3.27$ and $1d_{5/2} = -4.14$. Can to USD interaction predicts the new magic number N=16? Yes, in a report by O. Sorlin and M.-G. Porquet (Nuclear magic numbers: new features far from stability) They shows the effective single particle energy of oxygen and carbon using the monopole matrix elements of the USDB interaction. The new magic number N=16 can be observed. Advertisements ## Q-value and Biding energy I already talked on binding energy. And the Q-value is the mass different between nuclear reaction. In same case binding energy is same as the Q-value. Q-value is : $Q= \sum{m_i} - \sum{m_f}$ By energy conservation, it can be rewritten by kinetic energy. $Q= \sum{T_f} - \sum{T_i}$ But using energy is bit troublesome due to the frame transform. Using mass is simple, since it is a Lorentzian invariance. However, during scattering experiment. The kinetic energy is much more easy to measure. Thus the kinetic form is used more frequently in experiment. I will give the expression of K.E. in lab frame later. ## The most tightly bound nucleus which mean, it has most Binding Energy per nucleon. many will say it is 56Fe, which has 8.790MeV per nucleon. But in a research by Richard Shurtleff and Edward Derringh from Wentworth Institute of Technology say it is 62Ni on 1988, which has 8.795MeV per nucleon. so, the total different of the binding energy between 2 nucleus are about 300 keV, which is much larger then the contribution of the 2 extra electron in Ni then Fe. Thus raise up a question on the nuclear fusion process inside stars. why the end product is not 62Ni but 56Fe? there does not have any stable nucleus to bring 56Fe to 62Ni at the old star. this is the reason from them. and they concluded that, 56Fe is the end product is not due to just the binding energy, but also the environment. ## Deuteron The deuteron is the nucleus that contains 1 proton and 1 neutron. The spin and isospin of proton and neutron are the same, both are equal to half. It is the only stable state for 2 nucleons. Deuteron provides an unique place to study the inter nuclear force. The strong force is believed to be charge independent. Thus, the strong force can be more easily to study on deuteron due to the absent of other force or eliminate from the Coulomb force, which is understood very much. The mass of deuteron is 1876.1244 MeV. The binding energy is then 2.2245MeV. It was determined by the slow neutron capture of a proton. The emitted gamma ray is approximately equal to the binding energy and the deuteron mass was extracted. Deuteron has no excited state. It is because any excitation will easily to make the system break apart. When think deuteron as one of the family of NN system. Because of tensor force, which favor T=0 pn pair, thus only T=0, S=1 pn pair, which is deuteron is bounded. Any excitation will change the isospin from T=0 to T=1, which is unbound. The parity is positive from experiment. If we separate the deuteron wavefunction into 3 parts. The proton wavefunction, neutron wavefunction and the orbital wavefunction. Under the only force, the strong force in this system, proton and neutron are the same nucleon with different state. Thus, the parity are the same for proton and neutron. So, the product of these 2 wavefunction always has positive parity. The total parity then is solely given by the angular orbital. Any orbital wave function can be represented by the spherical harmonic, $Y(l,m)$. The parity transform is changing it to $Y(l,m) \rightarrow (-1)^l Y(l,m)$ So, the experimental face of positive parity fixed the angular momentum must be even. Ok, we just predicted the possible angular momentum from parity. The experimental fact on spin is 1. Since J = L + S, and the value of J can take every integer from |L-S| to L + S. and L must be even. The spin of proton and neutron is 1/2. Thus the possible S is 0 or 1 ( we are using L-S coupling scheme ). J = 1 = L + S , that tell us S must be odd to give out 1 for an even L. Thus S=1. So, the only possible L is 0 and 1. Thus, the possible state of deuteron is (L,S) = (0,1) or (2,1). Therefore, a deuteron is a mixed state, if without any further argument. Now, 2 out of 3 parts of the wave function symmetry were determined by symmetry argument. The isospin can now be fixed by the 2 fermions state must be antisymmetry. The spatial state symmetry is even by L = 0 or 2. And for the state (L , S) = ( 0, 1 ), the spin state is symmetric. Thus, the isospin must be antisymmetric. Since the algebra for isospin and spin are the same. We use T = 0 for the isospin. Thus a complete wavefunction is ( L , S , T ) = ( 0 , 1, 0 ). For the other possible state (L , S) = ( 2 , 1 ) , we can use same argument for isospin state. And for the degenerated state with Ms = +1, 0, -1. By the symmetry of the raising and lowering ladder operator, they all preserved the symmetry. Thus, the Ms = 0 state can only be the + state. So, we now have 2 possible states of deuteron. If the Hamiltonian is commute with L^2 and S^2, both L and L is a good quantum number and those states are eigen state. And the deuteron ground state must be one of them. ## Stability of a nucleus ( Liquid Drop Model ) when look at the table of the nuclear world, why there are some nucleus more stable then the other? which mean, why some will decay while some are not? OK, this basically the ultimate question that nuclear physics want to answer. so, the very fundamental reason, no one know. but in the elementary level, or by experimental fact and some assumption. we have Binding Energy to estimate or predict the stability of a nucleus. when the Binding Energy is larger then Zero, it must be unstable and will decay under conservation laws. if it is less then zero, it may be stable or not, it depends on whether it reach the bottom of energy level. Binding Energy can also be though as the energy required to break the nucleus. In liquid drop model, we imagine the nucleus is like a liquid. and nucleons inside just like liquid molecules. experiments show that nucleus is a spherical object. and it density is a constant. and the interaction range of nuclear force is short, few fm. thus, it likes a incompressible liquid drop. the radius of it is related to the mass number: $R^3 = A$ the Binding Energy ( = $\Delta M(A,Z,N)$ = mass deficit) is given by theoretical assumption and experimental fact. $\Delta M(A,Z,N) = - a_1 A + a_2 A^{\frac {2}{3} } + a_3 Z^2 / A^{ \frac{1}{3}} + a_4 (N-Z)^2 /A \pm a_5 A^{- \frac{3}{4} }$ the first 3 terms are theoretical assumption and the lat 2 terms are from experimental fact. All coefficients are given by experimental measurement. The first term is the “volume energy” by the nuclear force, which is proportional to the number of nucleons. the 2nd term is the “surface tension” from the “liquid”. we can see its dimension is area. (why this term is + ? ) it explained why smaller nucleus has less Binding energy. the 3rd term is the Coulomb potential term. the 4th term is the balance term. if the number of neutron and proton is no balance, the 5th term is the “Symmetry term“. for even-even of neutron and proton number, the nucleus is more stable, thus, we choose minus sign for it. for odd-odd combination, nucleus are more unstable, thus, plus sign for it. for other, like ood – even or even-odd combination, this term is zero. the value of the coefficients are: $a_1 \simeq 15.6 MeV$ $a_2 \simeq 16.8 MeV$ $a_3 \simeq 0.72 MeV$ $a_4 \simeq 23.3 MeV$ $a_5 \simeq 34 MeV$ The below plot is the Binding Energy per nucleon in Z against N. Lets use the liquid drop model and Binding Energy to look the β-decay. the β-decay conserved the mass number A. there are 2 β-decays. $\beta_- : n \rightarrow p + e^- + \bar{\nu_e}$ $\beta_+ : p \rightarrow n + e^+ + \nu_e$ so, the β+ decay decrease the number of proton while β– decay increase the number of proton. The below diagram show the β-decay for A = 22. we can see the 22Ne is stable, since no more β-decay can help to reach a lower energy level. ## 2p-2p decay of 8C and isospin-allowed 2p decay of the isobaric-analog state in 8B this paper reports another 2 protons decay mode in 8C. They also discover an “enhancement” at small relative energy of 2 protons. They also reported that an isobaric analog state, 8C and 8B, have same 2-protons decay, which is not known before. the 1st paragraph is a background and introduction. 2 protons decay is rare. lightest nucleus is 6Be and heaviest and well-studied is 45Fe. the decay time constant can be vary over 18 orders and the decay can be well treated by 3-body theory. the 2nd paragraph describes the decay channel of 8C and 8B. it uses the Q-value to explain why the 2-protons decay is possible. it is because the 1-proton decay has negative binding energy, thus, it require external energy to make it decay. while 2-protons decay has positive binding energy, thus, the decay will automatic happen in order to bring the nucleus into lower energy state. it also consider the isospin, since the particle decay is govt by strong nuclear force, thus the isospin must be conserved. and this forbid of 1-proton decay. it explains further on the concept of 2-protons decay and 2 1-protons decay. it argues that, in the 8C, the 2-protons decay is very short time, that is reflected on the large energy width, make the concept of 2 1-protons decay is a unmeasurable concept. however, for the 8Be, the life time is 7 zs (zepto-second $10^{-21}$), the 8Be moved 100 fm ( femto-meter$latex 10^{-15} ), and this length can be detected and separate the 4-protons emission in to 2 2-protons decay.

the 3rd paragraph explain the experiment apparatus – detector.

the 4th paragraph explains the excitation energy spectrum for 8C, 6Be.

the 5th and 6th paragraphs explain the excitation energy spectrum for the 6Be form 8C decay. since the 2 steps 2 – protons decay has 4 protons. the identification for the correct pair of the decay is important. they compare the energy spectrum for 8C , 6Be and 6Be from decay to do so.

the 7th paragraph tells that they anaylsis the system of 2-protons and the remaining daughter particle, by moving to center of mass frame ( actually is center of momentum frame ) and using Jacobi T coordinate system, to simplify the analysis. the Jacobi T coordinate is nothing but treating the 2-protons the 2 protons are on the arm of the T, and the daughter particle is on the foot of the T.