Abstract

The spins of the low-spin Ir(4+) (S = 1/2, d(5)) ions at the octahedral sites of the oxides Sr3NiIrO6, Sr2IrO4, and Na2IrO3 exhibit preferred orientations with respect to their IrO6 octahedra. We evaluated the magnetic anisotropies of these S = 1/2 ions on the basis of density functional theory (DFT) calculations including spin-orbit coupling (SOC), and probed their origin by performing perturbation theory analyses with SOC as perturbation within the LS coupling scheme. The observed spin orientations of Sr3NiIrO6 and Sr2IrO4 are correctly predicted by DFT calculations, and are accounted for by the perturbation theory analysis. As for the spin orientation of Na2IrO3, both experimental studies and DFT calculations have not been unequivocal. Our analysis reveals that the Ir(4+) spin orientation of Na2IrO3 should have nonzero components along the c- and a-axis directions. The spin orientations determined by DFT calculations are sensitive to the accuracy of the crystal structures employed, which is explained by perturbation theory analyses when interactions between adjacent Ir(4+) ions are taken into consideration. There are indications implying that the 5d electrons of Na2IrO3 are less strongly localized compared with those of Sr3NiIrO6 and Sr2IrO4. This implication was confirmed by showing that the Madelung potentials of the Ir(4+) ions are less negative in Na2IrO3 than in Sr3NiIrO6 and Sr2IrO4. Most transition-metal S = 1/2 ions do have magnetic anisotropies because the SOC induces interactions among their crystal-field split d-states, and the associated mixing of the states modifies only the orbital parts of the states. This finding cannot be mimicked by a spin Hamiltonian because this model Hamiltonian lacks the orbital degree of freedom, thereby leading to the spin-half syndrome. The spin-orbital entanglement for the 5d spin-half ions Ir(4+) is not as strong as has been assumed.

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