1995
DOI: 10.1016/0375-9474(94)00494-8
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Detailed study of the cluster structure of light nuclei in a three-body model (IV). Large space calculation for A = 6 nuclei with realistic nuclear forces

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Cited by 97 publications
(100 citation statements)
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“…For local spin-isospin independent interactions, 3 H, 3 He, and 4 He nuclei would be completely spatially symmetric and no sign problem would exist. For more realistic local interactions, the dominant spatially-symmetric component of the wave function and the relatively large excitation energies imply that the sign problem is not very significant for three-and four-body nuclei [34].…”
Section: Sampling Of the Pathsmentioning
confidence: 99%
“…For local spin-isospin independent interactions, 3 H, 3 He, and 4 He nuclei would be completely spatially symmetric and no sign problem would exist. For more realistic local interactions, the dominant spatially-symmetric component of the wave function and the relatively large excitation energies imply that the sign problem is not very significant for three-and four-body nuclei [34].…”
Section: Sampling Of the Pathsmentioning
confidence: 99%
“…Equation (4) represents the orthogonality condition that the total wave function (2) should be orthogonal to the Pauli-forbidden states of the 3α + N system, u F 's, which are constructed from the Pauli forbidden states between two α particles and those between α particle and extra nucleon N [94][95][96][97][98]. The Pauli-forbidden states are removed by using the Pauli-blocking operator V Pauli [99] in Eq. (6),…”
Section: Structure Of 13 C In the Four-body Cluster Modelmentioning
confidence: 99%
“…(1) in that the neutron charges equal zero (Z 1 = Z 2 = 0) and the proton mass m p is substituted by the neutron one m n , although this change of mass practically does not affect the result. The Hamiltonians for both nuclei are also characterized by a small difference between the effective nuclear interaction of a neutron with an α-particle in 14 C nucleus and the effective interaction of a proton with an α-particle in 14 O one, because the distribution of protons in the α-particle has a little larger radius than the corresponding distribution of neutrons (see, e.g., work [11]).…”
Section: Statement Of the Problemmentioning
confidence: 99%