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Bulletin of the American Physical Society

Ground state spin of a Fermi system with random interactions

Vladimir Zelevinsky

Published Mar 8, 2007• 0 citations
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Numerical simulations show that the ground state spin of a finite many-body system with random but rotationally invariant interactions is predominantly zero. This means that it is not sufficient to explain the zero ground state spin of even-even nuclei only by pairing correlations. Simple explanations based on a statistical multiplicity, width of the level density, effective pairing, formation of condensates etc. do not work. We suggest statistical theory based on the effective mean field and geometric chaoticity of angular momentum coupling of individual particles. The results show a significant role of symmetry arguments in finite systems.
Recent developments in many-body quantum chaos show that a quantum system with strong incoherent interactions can still be described with the aid of mean quasiparticle occupation numbers as in Fermi liquid theory. We use these ideas and the geometric chaoticity of angular momentum coupling to explain the predominance of ground states with zero spin and the maximum possible spin for a system of randomly interacting fermions in a rotationally invariant mean field (an analog of the Hund rule). We show that spin ordering coexists with the chaotic features of the ground-state wave functions.
In this paper we revisit spin III ground state probabilities (denoted by P(I))P(I))P(I)) of a few integrable systems under random interactions. We obtain explicit and simple formulas of P(I)P(I)P(I) for ddd-boson systems and fermions in a j=5/2j=5/2j=5/2 shell.