Abstract
We demonstrate by first-principles calculations that the Rashba spin-orbit splitting in the $6p$ states of a Bi adlayer on ${\text{BaTiO}}_{3}(001)$ is strongly affected by the substrate termination. For the ${\text{TiO}}_{2}$ termination the absolute splitting is very large (about $0.23\text{ }{\text{\AA{}}}^{\ensuremath{-}1}$); in contrast, the splitting becomes strongly reduced for the BaO termination (less than $0.07\text{ }{\text{\AA{}}}^{\ensuremath{-}1}$). Our findings are explained by the termination-dependent hybridization of the Bi surface states with electronic states in the substrate.
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