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Problem 1
Problem 2
Problem 3
Massive Dirac particles in graphene: Consider a tight-binding model on a honeycomb lattice with on-site potential1 different on the sublattices A and B, .
a) Develop a general solution of this problem in terms of plane wave states, expressing the spectrum of excitations in terms of the nearest neighbor hopping amplitude and the asymmetry parameter Failed to parse (syntax error): {\displaystyle \Delta = \frac{1}{2} \left(V_A − V_B \right) } .
b) Linearize the solution found in part a) near the points and treating the asymmetry as a small perturbation, and find how the massless Dirac picture (valid for ) of low energy states of graphene is altered. That is, show that generates a finite mass of Dirac quasiparticles whose Hamiltonian is given by:
,
where .
Problem 4
Bilayer graphene: Suppose instead of a single layer of graphene one has two layers, stacked atop one another according to the Bernal stacking. This means that the B sublat- tice sites of the upper layer (layer 1) are directly above the A sublattice sites of the lower layer (layer 2). The A sublattice sites of the upper layer and the B sublattice sites of the lower layer have no “partner” atoms below/above them.
- (a) Formulate and solve the tight-binding model for this system consisting of the usual hopping t between nearest-neighbor sites in each layer, and an additional smaller hopping γ between the B1-A2 sites in different layers. Choose the zero of energy to equal that of an isolated atom. You should find 4 bands.
- (b) Two of the bands found above touch at the Fermi energy ( = 0) at the zone boundary
points ±K. Find the effective mass around these points.
- (c) Now calculate the Berry phase ( dk · A(k)) acquired by an electron encircling one of
the zone boundary points in each of the two touching bands. Use your result to argue that the bands cannot split if a small perturbation is applied provided inversion and time-reversal symmetry are preserved. Where is the inversion center for the bilayer?
- (d) Check this conclusion in a simple case: Because the B1 and A2 sites have more atoms
close by than the B2 and A1 sites, there will generally be some difference of the site energies for these orbitals. Add an energy +u to the electrons on the B1,A2 sites and −u to the electrons on the B2,A1 sites. Show that for small enough u, the two bands still touch.