Spin and orbital states inLa1.5Sr0.5CoO4studied by electronic structure calculations

Abstract

Electronic structure of the layered perovskite ${\text{La}}_{1.5}{\text{Sr}}_{0.5}{\text{CoO}}_{4}$ with a checkerboard ${\text{Co}}^{2+}/{\text{Co}}^{3+}$ charge order is studied, using the local-spin-density approximation plus Hubbard $U$ calculations including also the spin-orbit coupling and multiplet effect. Our results show that the ${\text{Co}}^{2+}$ ion is in a high spin state (HS, ${t}_{2g}^{5}{e}_{g}^{2}$) and ${\text{Co}}^{3+}$ low spin state (LS, ${t}_{2g}^{6}$). Due to a small ${\text{Co}}^{2+}\text{ }{t}_{2g}$ crystal field splitting, the spin-orbit interaction produces an orbital moment of $0.26{\ensuremath{\mu}}_{B}$ and accounts for the observed easy in-plane magnetism. Moreover, we find that the ${\text{Co}}^{3+}$ intermediate spin state (IS, ${t}_{2g}^{5}{e}_{g}^{1}$) has a multiplet splitting of several tenths of eV and the lowest-lying one is still higher than the LS ground state by 120 meV, and that the ${\text{Co}}^{3+}$ HS state $({t}_{2g}^{4}{e}_{g}^{2})$ is more unstable by 310 meV. Either the IS or HS ${\text{Co}}^{3+}$ ions would give rise to a wrong magnetic order and anisotropy.