Scientific Reports (Sep 2021)

Spin structures of the ground states of four body bound systems with spin 3 cold atoms

  • Y. M. Liu,
  • C. G. Bao

DOI
https://doi.org/10.1038/s41598-021-97521-y
Journal volume & issue
Vol. 11, no. 1
pp. 1 – 8

Abstract

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Abstract We consider the case that four spin-3 atoms are confined in an optical trap. The temperature is so low that the spatial degrees of freedom have been frozen. Exact numerical and analytical solutions for the spin-states have been both obtained. Two kinds of phase-diagrams for the ground states (g.s.) have been plotted. In general, the eigen-states with the total-spin S (a good quantum number) can be expanded in terms of a few basis-states $$f_{S,i}$$ f S , i . Let $$P_{f_{S,i}}^{\lambda }$$ P f S , i λ be the probability of a pair of spins coupled to $$\lambda =0, 2, 4$$ λ = 0 , 2 , 4 , and 6 in the $$f_{S,i}$$ f S , i state. Obviously, when the strength $$g_{\lambda }$$ g λ of the $$\lambda $$ λ -channel is more negative, the basis-state with the largest $$P_{f_{S,i}}^{\lambda }$$ P f S , i λ would be more preferred by the g.s.. When two strengths are more negative, the two basis-states with the two largest probabilities would be more important components. Thus, based on the probabilities, the spin-structures (described via the basis-states) can be understood. Furthermore, all the details in the phase-diagrams, say, the critical points of transition, can also be explained. Note that, for $$f_{S,i}$$ f S , i , $$P_{f_{S,i}}^{\lambda }$$ P f S , i λ is completely determined by symmetry. Thus, symmetry plays a very important role in determining the spin-structure of the g.s..