1996
DOI: 10.1006/aphy.1996.0109
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Scissors Modes and Spin Excitations in Light Nuclei IncludingΔN=2 Excitations: Behaviour of8Be and10Be

Abstract: Shell model calculations are performed for magnetic dipole excitations in 8 Be and 10 Be, first with a quadrupole quadrupole interaction (Q } Q) and then with a realistic interaction. The calculations are performed both in a 0p space and in a large space which includes all 2 | excitations. In the 0p with Q } Q we have an analytic expression for the energies of all states. In this limit we find that in 10 Be the L=1 S=0 scissors mode with isospin T=1 is degenerate with that of T=2. By projection from an intrins… Show more

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Cited by 12 publications
(7 citation statements)
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References 29 publications
(217 reference statements)
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“…2 and Table II. For B = 0 the T = 1 and T = 2 scissors are degenerate in energy [17,14], and there are four degenerate states in all for each T . As B is made negative two T = 1 states come down in energy and two come up.…”
Section: The Fact That the 2 +mentioning
confidence: 99%
“…2 and Table II. For B = 0 the T = 1 and T = 2 scissors are degenerate in energy [17,14], and there are four degenerate states in all for each T . As B is made negative two T = 1 states come down in energy and two come up.…”
Section: The Fact That the 2 +mentioning
confidence: 99%
“…For x = 1, y = 1 one gets the best fit to the Bonn A matrix elements and this choice is used in this work. This has been discussed extensively in previous references [5,9,11].…”
Section: Results Of the Shell Model Diagonalizationsmentioning
confidence: 91%
“…Hence, besides the intruder state, we have the above two J = 0 + states as candidates for the experimental 0 + 2 state at 6.18 M eV . As noted in the previous work [5] if, in the 0p-shell model space we fit χ to get the energy of the lowest 2 + state in 10 Be to be at the experimental value of 3.368 M eV (18χ), then the two sets of triplets are at an excitation energy of 30χ which equals 5.61 M eV -not far from the experimental value. There is however a problem -in a 0p-space calculation with Q · Q, the lowest 2 + state is two-fold degenerate, corresponding to J = 2 + K = 0 and J = 2 + K = 2.…”
Section: Introductionmentioning
confidence: 96%
See 1 more Smart Citation
“…It was noted by Fayache, Sharma and Zamick [7] that the isovector scissors mode in 10 Be is at quite a high energy, E ≈ 22 MeV. One cannot obtain such a high energy for this state with an isospin-independent Q · Q interaction whose strength is chosen so that the 2 + 1 energy comes out correctly.…”
Section: Closing Remarksmentioning
confidence: 99%