sol05 - Solid State Theory Solution Sheet 5 FS 11 Prof. M....

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Unformatted text preview: Solid State Theory Solution Sheet 5 FS 11 Prof. M. Sigrist Exercise 5.1 Graphene a) We want to calculate the π-energy bands of graphene using the tight-binding method. Those bonds are due to electrons in the 2 p z orbitals. Graphene has two inequivalent carbon atoms per unit cell, which we call A and B . The primitive lattice vectors a i join atoms on the same sublattice ( A or B ), so that the reference points of the unit cells (the A-atoms in figure 1) form a triangular lattice. Setting the lattice constant a to unity they are given by a 1 = √ 3 2 , 1 2 ! a 2 = √ 3 2 ,- 1 2 ! . (1) The reciprocal lattice vectors c i can be found as usually by demanding that they satisfy c i · a j = 2 πδ ij . (2) This leads to c 1 = 4 π √ 3 1 2 , √ 3 2 ! c 2 = 4 π √ 3 1 2 ,- √ 3 2 ! (3) With the reciprocal lattice vectors the Brillouin zone can be constructed and is seen to be a regular hexagon (figure 1). Note that after identifying points that differ by a reciprocal lattice vector only two of the corners are inequivalent. Next we find the general form of the tight-binding model for the bands originating in the p z-orbitals when only nearest neighbor hopping is taken into account. The p z-orbitals are symmetric under the rotations of the plane so that the hopping matrix elements do not depend on the direction. Hence we have only one hopping parameter, which we denote by t . Furthermore, every A-atom has neighbors in B only. The onsite terms have to be the same on both sublattices, so that they can be absorbed into the chemical potential (we work at fixed filling with two electrons per unit cell). 1 a 1 a 2 x y A B c c Γ M K k x k y b 1 3 2 Figure 1: Left: Primitive lattice vectors for the honeycomb lattice. The vectors join points on one triangular sublattice. Right: Hexagonal Brillouin zone constructed from the reciprocal lattice vectors c 1 and c 2 . To describe the hopping terms we define the vectors b 1 = 1 √ 3 , , b 2 =- 1 2 √ 3 , 1 2 , and b 3 =- 1 2 √ 3 ,- 1 2 (4) that point from an A atom to its three nearest neighbors. Denoting the postion of the i-th atom on sublattice A by R a,i , the Hamiltonian is given by H = t X i 3 X j =1 c † ( R a,i ) c ( R a,i + b j ) + h.c. . (5) Now we use the Fourier transform given on the exercise sheet,...
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sol05 - Solid State Theory Solution Sheet 5 FS 11 Prof. M....

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