Abstract

We have found a large spin splitting due to the Rashba spin-orbit interaction on the $\text{Bi}/\text{Ge}(111)\text{\ensuremath{-}}(\sqrt{3}\ifmmode\times\else\texttimes\fi{}\sqrt{3})R30\ifmmode^\circ\else\textdegree\fi{}$ surface by using angle-resolved photoelectron spectroscopy and first-principles electronic structure calculation. A surface resonance band derived from Bi exhibits the Rashba spin splitting with a large Rashba parameter $({\ensuremath{\alpha}}_{R})$ of $1.8\text{ }\text{eV}\text{ }\text{\AA{}}$. The spin-split states have a considerable $6s\text{\ensuremath{-}}6{p}_{z}$ hybridized character, which leads to a strong perpendicular asymmetry of the charge density in close proximity of Bi nuclei. The result suggests that the magnitude of the Rashba splitting on Bi-adsorbed surfaces should depend crucially on the local-bonding geometry of Bi.

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