On the metal-insulator-transition in vanadium dioxide

TL;DR

This paper investigates the metal-insulator transition in VO$_2$, proposing that the reduction in the dimensionality of k-vectors during the structural change causes the significant drop in electrical conductivity.

Contribution

It introduces a k-vector based explanation for the MIT in VO$_2$, linking crystal structure changes to electronic conduction properties.

Findings

Conductivity drops by four orders of magnitude at MIT.

Reduced dimensionality of k-vectors in monoclinic phase causes conductivity decrease.

Triclinic and trigonal crystals lack k-vectors and are insulators.

Abstract

Vanadium dioxide (VO$_2$) undergoes a metal-insulator transition (MIT) at 340 K with the structural change between tetragonal and monoclinic crystals as the temperature is lowered. The conductivity $\sigma$ drops at MIT by four orders of magnitude. The low-temperature monoclinic phase is known to have a lower ground-state energy. The existence of a $k$-vector ${\boldsymbol k}$ is prerequisite for the conduction since the ${\boldsymbol k}$ appears in the semiclassical equation of motion for the conduction electron (wave packet). Each wave packet is, by assumption, composed of the plane waves proceeding in the ${\boldsymbol k}$ direction perpendicular to the plane. The tetragonal (VO$_2$)$_3$ unit cells are periodic along the crystal's $x$-, $y$-, and z-axes, and hence there are three-dimensional $k$-vectors. The periodicity using the non-orthogonal bases does not legitimize the electron dynamics in solids. There are one-dimensional ${\boldsymbol k}$ along the c-axis for a monoclinic crystal. We believe this decrease in the dimensionality of the $k$-vectors is the cause of the conductivity drop. Triclinic and trigonal (rhombohedral) crystals have no $k$-vectors, and hence they must be insulators. The majority carriers in graphite are "electrons", which is shown by using an orthogonal unit cell for the hexagonal lattice.