Itinerant Ferromagnetism in the Multiorbital Hubbard Model: A Dynamical Mean-Field Study

In order to resolve the long-standing issue of how itinerant ferromagnetism is affected by lattice structure and Hund’s coupling, we compare various three-dimensional lattice structures in the single- and multiorbital Hubbard models with the dynamical mean-field theory with an improved quantum Monte Carlo algorithm that preserves the spin-SU(2) symmetry. The result indicates that both the lattice structure and the $d$-orbital degeneracy are essential for the ferromagnetism in the parameter region representing a transition metal. Specifically, (a) Hund’s coupling, despite the common belief, is important, which is here identified to come from particle-hole scatterings, and (b) the ferromagnetism is a correlation effect (outside the Stoner picture) as indicated from the band-filling dependence.

Figure 1

The density of states for noninteracting electrons on the fcc lattice. Red figures represent the band filling (per spin), while black ones the corresponding total band filling for a two-orbital system.

Figure 2

Inverse spin susceptibility $\chi (\mathbf{0},0{)}^{-1}$ against temperature for the two-orbital Hubbard model on the fcc lattice for $U=4$ at $n=1.5$ (a) and $n=0.4$ (b). We switch on $U^{\prime }$ (blue) and then $J$ (orange and red) preserving the rotational symmetry ($U=U^{\prime }+2J$). The solid lines are guide to the eye, and the dashed lines extrapolations.

Figure 3

Spin susceptibility against the band filling $n$ for the two-orbital Hubbard model on the fcc lattice at $T=0.1$ for various values of $U$, with $J=0.5$ and the relation $U^{\prime }=U-2J$ fixed.

Figure 4

Wave number dependence of the susceptibility on the fcc lattice for various values of $n=1.25–2.0$ for $U=4$, $U^{\prime }=3$, $J=0.5$, $T=0.1$.

Figure 5

Local spin moment (black) versus $J$ in the two-orbital Hubbard model on the fcc lattice for (a) $n=1.5$ and (b) $n=0.75$ with $T=0.1$, $U=4$ and the relation $U^{\prime }=U-2J$ fixed. The corresponding susceptibility $\chi (\mathbf{0},0)$ is plotted in blue.