Electromagnetic Breakup of Three- and Four-Body Nuclei

The Lorentz Integral Transform (Lit) Method

“…It is evident that a smaller σ I leads to a longer range LIT functionΨ. As discussed in [27], for σ I = 0.1 MeV even R max =300 fm is not completely sufficient, since only the resonance itself is resolved, but not the cross section at higher energies.…”

“…The reason for such a LIT case study with a resonance lies in the results of a previous LIT calculation for the (e,e') longitudinal and transverse form factors of 4 He [28], where a resonance in the Coulomb monopole transition was obtained, but its width could not be determined. In the case study it is shown that for a proper resolution of a resonant structure it is very important to take into account the LIT functionΨ up to rather large distances [27]. This has been checked (i) by solving (6) imposing at a two-nucleon distance r = R max an asymptotic boundary condition which leads to a strong fall-off ofΨ( 3 P 1 ) and…”

“…In the already mentioned review article of the LIT method [16], such a case has been investigated using the AV14 NN potential [25]. The L Further LIT calculations for the deuteron total photoabsorption cross sections have already been discussed in [26,27]. For the aim of the present discussion, i.e.…”

“…the way of working of the LIT method, [27] is particularly interesting. It is a case study for a fictitious np system with a low-lying and narrow resonance in the 3 P 1 nucleon-nucleon partial wave (obtained by an additional attractive term, for details see [27]). The results of a conventional calculation with the fictitious np system for the 3 P 1 phase shifts and the "deuteron MeV.…”

“…However, in a general case the resonance width is not known beforehand and the question arises what should be the proper value for σ I in such a case. As pointed out in [27] one has to proceed as follows. One performs a LIT calculation with a given σ I and compares the result to a LIT with a δ-peak in the response function or cross section.…”