Abstract

We address the topological nature of electronic states of step edges of Bi(111) films by first-principles band structure calculations. We confirm that the dispersion of step-edge states reflects the topological nature of underlying films, which become topologically trivial at a thickness larger than eight bilayers. This result clearly conflicts with recent claims that the step-edge state at the surface of a bulk Bi(111) crystal or a sufficiently thick Bi(111) film represents nontrivial edge states of the two-dimensional topological insulator phase expected for a very thin Bi(111) film. The trivial step-edge states have a gigantic spin splitting of one-dimensional Rashba bands and substantial intermixing with electronic states of the bulk, which might be exploited further.

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