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Masses (Binding energies) and the IBA

Masses (Binding energies) and the IBA. Extra structure-dependent binding: energy depression of the lowest collective state.

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Masses (Binding energies) and the IBA

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  1. Masses (Binding energies) and the IBA Extra structure-dependent binding: energy depression of the lowest collective state

  2. The dominant features of separation energies are: the linear behavior along chains of isotopes, the generally parallel slopes for different Z’s and the huge drops after magic numbers.Specifically structure-dependent effects are superposed on this secular behavior. With recent capabilities to measure masses precisely, we can now access the physics that these non-linearities point to. How can we calculate them theoretically?

  3. Classifying Structure -- The Symmetry Triangle Def. Sph. 1st order Ph. Tr

  4. S (2n) MeV Neutron Number

  5. Binding energies and separation energies in the semi-empirical mass formula BE(A, Z) = avA – asA2/3– acZ (Z – 1)A-1/3– aA(A – 2Z)2 A-1 BE’s: linear and quadratic in N, Z or A. Hence, separation energies are linear. 4 3 2 3 S2n 2 (av - aA) - asA-1/3 +acZ (Z - 1) A-4/3 + 8aA Z 2 A (A – 2) Reflects shell filling (asymmetry term dominates).

  6. The only practical model to calculate a wide variety of collective nuclear observables is the IBA. Virtually all previous efforts with the IBA have focused on collectivity and its manifestations in energy levels and electromagnetic moments and transition rates.However, the IBA can be extended to incorporate mass predictions, and the opportunities this seems to provide are quite exciting. Where do masses come into the IBA and how do we need to modify the usual IBA calculations to provide masses? (Note: I use masses and binding energies interchangeably.)

  7. Let’s review group chains and degeneracy-breaking. Consider a Hamiltonian that is a function ONLY of: s†s + d†d That is: H = a(s†s + d†d) = a (ns + nd ) = aN Note that such a term does NOT normally appear in the IBA – because we deal usually with only a single nucleus. The H above can be written in terms of Casimir operators of U(6). We’ll see now what that means. In H, the energies depend ONLY on the total number of bosons, that is, on the total number of valence nucleons. ALL the states with a given N are degenerate. These states are a “representation” of the group U(6) with the quantum number N. U(6) has OTHER representations, corresponding to OTHER values of N, but THOSE states are in DIFFERENT NUCLEI (numbers of valence nucleons).

  8. Now, add a term to this Hamiltonian: H’ = H + b d†d = aN + b nd Now the energies depend not only on N but also on nd States of a given nd are now degenerate. They are “representations” of the group U(5). States with different nd are not degenerate

  9. H’ = aN + b d†d = a N + b nd N + 2 2a Here we are interested in masses, or binding energies. So, we are comparing the total binding of different nuclei. So, we are interested in the relative energies of the different representations of U(6) N + 1 a 2 2b 1 b N 0 0 0 nd E U(6) U(5) H’ = aN + b d†d

  10. Masses: Expanding the Concept of a Dynamical Symmetry N -1 N N +1 (N)

  11. Above, I discussed what happens with cases where the Hamiltonian corresponds to a specific dynamical symmetry but that was just to illustrate the idea. The same basic concepts work for calculations anywhere in the triangle. One needs to compare total binding energies of two different nuclei and this will give separation energies.Where does the binding from collective effects enter in the IBA? How does the IBA produce extra binding? What is the BASIC physics ?

  12. Recall: IBA Hamiltonian Most general IBA Hamiltonian in terms with up to four boson operators (given N) These terms CHANGE the numbers of s and d bosons: MIX basis states of the model Crucial for structure Crucial for masses This lowering of the lowest state (ground state) represents extra BINDING, changing the mass. We can calculate this with the IBA

  13. Separation energies in the IBA S2n (N) = BE (N) – BE (N – 1) Since we are now dealing with two different nuclei, we need to take account of the energy differences of different representations of U(6) or, more generally, of the ground state energies for two adjacent even-even nuclei. That is, we need to incorporate NEW terms in the IBA Hamiltonian corresponding to the linear and quadratic Casimir operators of U(6). This sounds fancy but isn’t. They simply give terms in the BE that are linear and quadratic in boson number and therefore reflect the same physics as the semi-empirical mass formula. BUT: This linear dependence does NOT include correlations or collective effects !! This is precisely where the IBA, as a model of nuclear STRUCTURE, can contribute. It gives a way to calculate those effects. It turns out, as I discovered only yesterday, that this has some very significant consequences way beyond the obvious ones !!!

  14. Separation energies in the IBA So, the energies in the IBA will be given by those two linear and quadratic terms PLUS the contribution from the collective correlations E′ = E0 + A’N + B’/2 N (N – 1) + BE(IBA) S2n (N) = (A’ – B’/2) + B’N + [BEIBA (N) – BEIBA (N – 1)] S2n (N) = A + BN + [BEIBA (N) – BEIBA (N – 1)] S2n (N) = A + BN + S2n(IBA) (N)

  15. So, how do we calculate structure in the IBA? Recall:

  16. Binding energies and separation energies in the IBA for the dynamical symmetries

  17. Recall Spectrum of U(5) No mixing of states of given number of d-bosons What is spectrum? Equally spaced levels defined by number of d bosons 3 2 1 0 nd 6+, 4+, 3+, 2+, 0+ 4+, 2+, 0+ 2+ 0+ S2(Nv) = Av + BvNvU(5) NO collective correlations, hence separation energies are purely linear

  18. Recall: IBA Hamiltonian Most general IBA Hamiltonian in terms with up to four boson operators (given N) These terms CHANGE the numbers of s and d bosons: MIX basis states of the model

  19. Separation energies in the IBA for the dynamical symmetries S2(Nv) = Av + BvNvU(5) NO collective correlations. Linear S2(Nv) = Av + BvNv +8𝜅(N +Nv) + 10𝜅SU(3) Linear dependence PLUS LINEAR collective correlations if kappa is CONSTANT O(6) Similar to SU(3) -- I don’t have the formula at hand. It is also linear if kappa is constant

  20. So, if a series of nuclei are described by a dynamical symmetry AND the coefficients on the Hamiltonian are CONSTANT, the separation energies are linear.This seldom happens.If the coefficients of the symmetry vary or, more commonly, if the structure of the nuclei varies, then there will be non-linear contributions to the separation energies

  21. How do we calculate BE’s with the IBA for nuclei away from the dynamical symmetriesin practice? • First, establish the parameters for the IBA calculation – epsilon and chi, using the OCC technique described earlier, with, possibly, some fine tuning ( use Kappa = 0.02 here). • For masses, we need to worry, however, about the ABSOLUTE energies. So, we cannot just get relative energies but need to fit the overall scale of the IBA calculations to the excitation energies, especially E(2). • Then, we use: • We write this: • We then fit A and B to the separation energy data corrected for collective contributions, add back the IBA part . See if the deviations from linearityare reproduced. S2n (N) = A + BN + S2n(IBA) (N) S2n (N)exp - S2n(IBA) (N) = A + BN

  22. Examples of calculations of BE’s or S2n values in the IBA

  23. IBA with constant parameters

  24. Realistic IBA calculations for Gd and Os – the collective contribution from the IBA NORMALIZED

  25. S (2n) MeV Neutron Number

  26. Now, here’s the shocker:

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