Hexagonal Indium Double Layer on Si(111)-($\sqrt{7}\times\sqrt{3})$

TL;DR

Density-functional calculations reveal that the hexagonal In/Si(111)-($\

Contribution

The study demonstrates that the hexagonal surface is actually a double layer of indium, not a single atom-thick layer, revising previous assumptions.

Findings

The surface consists of a double layer of In atoms.

The double layer explains coexistence of phases and similar electronic structures.

The surface is not a one-atom-thick In overlayer.

Abstract

Density-functional calculations are used to verify the atomic structure of the hexagonal In/Si(111)-($\sqrt{7}\times\sqrt{3}$) surface, which has been considered to represent an ultimate two-dimensional (2D) limit of metallic In overlayers. Contrary to the prevailing assumption, this surface consists of not a single layer but a double layer of In atoms, which corresponds to a hexagonal deformation of the well-established $rectangular$ In double layer formed on Si(111)-($\sqrt{7}\times\sqrt{3}$) [Park $et$ $al$., Phys. Rev. Lett. 109, 166102 (2012)]. The same double-layer thickness accounts well for the typical coexistence of the hexagonal and rectangular phases and their similar 2D electronic structures. It is thus conclusive that, regardless of rectangular or hexagonal, the In/Si(111)-($\sqrt{7}\times\sqrt{3}$) surface does not represent an one-atom-thick In overlayer.