Abstract

To understand the role that degeneracy, hybridization, and nesting play in the magnetic and pairing properties of multiorbital Hubbard models we here study numerically two types of two-orbital models, both with holelike and electron-like Fermi surfaces (FS's) that are related by nesting vectors $(\ensuremath{\pi},0)$ and $(0,\ensuremath{\pi})$. In one case the bands that determine the FS's arise from strongly hybridized degenerate ${d}_{xz}$ and ${d}_{yz}$ orbitals, while in the other the two bands are determined by nondegenerate and nonhybridized $s$-like orbitals. Using a variety of techniques, in the weak-coupling regime it is shown that only the model with hybridized bands develops metallic magnetic order, while the other model exhibits an ordered excitonic orbital-transverse spin state that is insulating and does not have a local magnetization. However, both models display similar insulating magnetic stripe ordering in the strong-coupling limit. These results indicate that nesting is a necessary but not sufficient condition for the development of ordered states with finite local magnetization in multiorbital Hubbard systems; the additional ingredient appears to be that the nested portions of the bands need to have the same orbital flavor. This condition can be achieved via strong hybridization of the orbitals in weak coupling or via the FS reconstruction induced by the Coulomb interactions in the strong-coupling regime. This effect also affects the pairing symmetry as demonstrated by the study of the dominant pairing channels for the two models.

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