Department of physics, NTNU TFY4340 Mesoscopic Physics Spring 2010

Exercise 2

Assistance: Tuesday February 2.

Question 1

Find the nearest–neighbour tight–binding band structure E(k) for the 2D triangular lattice (nearest neighbour distance a). Assume one atomic orbital pr atom and choose the energy of this orbital so that E(k = 0) = 0. Overlap integrals are δ_{R,0} and transfer integrals are γ for (the six) nearest neighbours. Plot E(k) along lines from the Γ–point to the edge(s) and corner(s) of the (hexagonal) 1BZ, the socalled M– and K–point(s), respectively. What is the band width in this model? Hint: Remember that the reciprocal lattice (b_{1}, b_{2}) can be constructed from b_{i} · a_{j} = 2πδ_{ij}, see Exercise 1. Use Matlab, gnuplot, or any other favorite program for plotting.

Question 2

A similar approach as in the previous question, to the hexagonal graphene lattice, yields the two bands

E^{±}(k) = E_{0} ∓ γ\sqrt{1 + 4 cos \frac{3k_{x}a}{2} cos \frac{\sqrt{3}k_{y}a}{2} + 4 cos^{2} \frac{\sqrt{3}k_{y}a}{2}}

(γ < 0), due to the two–atomic primitive unit cell. (Carbon atoms at the origin and in ax̂; primitive lattice vectors a_{1} = (3a/2)x̂ − (\sqrt{3}a/2)ŷ and a_{2} = (3a/2)x̂ + (\sqrt{3}a/2)ŷ). Show that the valence band E^{−} and the conduction band E^{+} touch at the K–point(s) (i.e. the corners of the 1BZ), hence, the band gap is E_{g} = 0. Investigate the band structure more closely by playing around with your favorite software for 2D and 3D plotting.