Scientific Reports (Sep 2022)

Spin-textures of medium-body boson systems with trapped spin-f cold atoms

  • Y. Z. He,
  • C. G. Bao,
  • Z. B. Li

DOI
https://doi.org/10.1038/s41598-022-19184-7
Journal volume & issue
Vol. 12, no. 1
pp. 1 – 10

Abstract

Read online

Abstract The spin-textures of bound medium-body systems with spin- $$\mathfrak {f}$$ f atoms ( $$\mathfrak {f}\ge 3$$ f ≥ 3 ) have been studied. The Hamiltonian is assumed to be dominated by the two-body interaction favoring parallel spins. The system with particle number $$N=8$$ N = 8 and $$\mathfrak {f}=3$$ f = 3 is first chosen, and the Hamiltonian is exactly diagonalized by using Fock-states as basis-states, thereby all the eigenenergies and eigenstates are obtained and a detailed analysis is made. Then the cases with $$N=13$$ N = 13 and $$\mathfrak {f}=4$$ f = 4 are further studied. Since the total spin S is conserved, the eigenstates having the same S form an S-group. Let the lowest (highest) energy state of an S-group be called a bottom-state (top-state). We found that all the bottom-states are bipartite product states with constituent states describing fully polarized subsystems containing $$N_1$$ N 1 and $$N_2$$ N 2 ( $$\le N_1$$ ≤ N 1 ) particles, respectively. For two bottom-states different in $$N_2$$ N 2 , the one with a larger $$N_2$$ N 2 is higher. For two having the same $$N_2$$ N 2 , the one with a smaller S is higher. Whereas all the top-states are found to be essentially a product state of the pairs, in each pair the two spins are coupled to $$\lambda$$ λ if the strength of the $$\lambda$$ λ -channel is more repulsive than the others. For the states belonging to an S-group, the higher one would contain more pieces. As the energy goes up, larger pieces (those containing more than two particles) will disappear.