Solution to Exercise 2

Transcription

1 Department of physics, NTNU TFY Mesoscopic Physics Spring Solution to xercise Question Apart from an adjustable constant, the nearest neighbour nn) tight binding TB) band structure for the D triangular lattice is with nn lattice vectors k x, k y ) = γ R R R j= = aˆx e ik R j + e ik R ) j, = a ˆx + a ŷ = a ˆx a ŷ Thus, ensuring that ) =, k x, k y ) = γ + γ cosk x a + cos k xa + k y a + cos k xa ) k y a = γ + γ cosk x a + cos k ) x cos ky a = γ + γ cos k xa cos k ) xa + cos ky a Here, we have used the trigonometric identities cosa + b) + cosa b) = cosacosb and cos a = cos a. With b i a j = πδ ij, one finds reciprocal lattice vectors b = π ˆx ) ŷ a b = π a ŷ so the reciprocal lattice is also triangular, with nearest neighbour distance b = b = b = π/ a. The six M points are located at the midpoints of the edges of the hexagonal BZ, one of them is: k M = b = π ŷ. a

2 The six K points are located at the corners of the BZ, one of them is: k K = π a ˆx. Using Matlab, we obtain the following dispersion relation between the Γ point k = ) and the M point, and from Γ to K:.... between Gamma and M..... between Gamma and K In both figures, the energy is given in units of γ = γ, as function of the dimensionless quantities k y a and k x a. The latter figure suggests that the band width equals γ, with min = in the Γ point k = and max = γ in the K points. A D plot of the dispersion relation over the BZ actually, over the rectangular area in k space defined by ±k K and ±k M given above, since I did not take the time to figure out how to limit the plot region to the hexagonal BZ) supports this suggestion:

3 A formal way to find the band width amounts to locating the various stationary points of k), i.e., to find where k =, and insert these k values into k). For the D gradient of to vanish, we must require both / k x = and / k y = simultaneously. Let α = k x a and β = k y a. Then Here, we have several solutions: α = sin α cos α + ) β cos = β = cos α β sin = sinα/) = and sin β/) = yields the energy minimum min = at the Γ point k x = k y =, as well as saddle points M = γ in two of the six M points, k M =, ±π/ a) cosα/) = and cos β/) = yields the remaining four saddle points M = γ in k M = ±π/a, ±π/ a) sin β/) = and cosα/) = /) cos β/) yields the six energy maxima K = γ at the K points k K = ±π/a, ) and k K = ±π/a, ±π/ a). A D plot, viewed down the axis, and with a somewhat extended region of k space, could be illustrative:

4 Question In xercise, Question, we found reciprocal lattice vectors b = π/a)ˆx ŷ) and b = π/a)ˆx+ ŷ) for the hexagonal graphene lattice. Hence, the midpoints M of the edges of the hexagonal BZ are located at k values ±b /, ±b /, and ±π/a)ˆx. The corners K of the BZ are located at ±π/ a)ŷ, ±b xˆx+/)b y ŷ) = ±π/a)ˆx+π/ a)ŷ, and ±b xˆx +/)b y ŷ) = ±π/a)ˆx π/ a)ŷ. See figure in Solution.) Hence, by making a D plot of + and in the range π/ < k x a < π/ and π/ < k y a < π/, we cover the BZ plus a little extra, / of the BZ, actually). We set = and γ = and obtain the following valence band and conduction band +, showing clearly how the VB and the CB touch at the six K points: If we imagine the various energy bands of graphene being constructed from atomic orbitals in carbon, and taking into account that each primitive cell contains two carbon atoms, each giving rise to one band pr atomic orbital, we will end up with the two bands σ and σ from the s state. Next, we have hybridized sp bands, constructed as linear combinations of the atomic s, p x, and p y states of the two carbon atoms. The sp bands are responsible for the strong C C bonds, and the corresponding wave functions are directional, pointing towards the three nearest neighbours of any given carbon atom. As discussed in the lectures, each primitive cell will contribute electronic states to each energy band. Hence, of the electrons in each primitive cell will occupy the σ and σ bands and three sp bands. We are left with electrons pr primitive cell one from each C atom), and these electrons have precisely the energy bands and + of the present problem at their disposal. It is the p z atomic orbital that is the basis for constructing these two energy bands. In chemistry, they are typically referred to as π and π bands. The notion of π bonding is probably well known from molecules like ethylene and benzene, I guess.) At zero temperature, the electrons will be in the ground state of the system, i.e., the valence

5 band will be filled and the conduction band + will be empty. Hence, graphene may be considered a semiconductor, with zero bandgap, g =. Graphene, a single layer of graphite, was first made experimentally in, using simple equipment like Scotch tape. Since then, it has been a hot research topic, both experimentally and theoretically. One important reason is the rather peculiar band structure in the vicinity of the K points. We will come back to graphene in the next xercise. The nearest neighbour tight binding solution of graphene and graphite) was published already more than sixty years ago P. R. Wallace, Phys Rev, )).