c q b c q V 2 � � q b O V 2 c q b c q V 2 � q � q b O V 2 c q V 2 ma 4 π

# C q b c q v 2 ? ? q b o v 2 c q b c q v 2 ? q ? q b o

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c q + b = c q V 2( ǫ ǫ 0 q + b ) + O ( V 2 ) , c q + b = c q V 2( ǫ 0 q ǫ 0 q + b ) + O ( V 2 ) , = c q V 2 ma 4 π planckover2pi1 2 ( q + π a ) + O ( V 2 ); c q + b = c q V 2 4 planckover2pi1 2 ( q 2 π 2 a 2 ) parenleftBig 1 qa π parenrightBig + O ( V 2 ) . (1.b.18) The story changes, however, for c q b . It is not hard to jump a bit in the calculation and see c q b = c q V 2( ǫ ǫ 0 q b ) + O ( V 2 ) . (1.b.19) Now, from our calculation of the eigenenergies at the Bragg plane we know that ǫ π/a ǫ 0 π/a b = ± V 2 V 2 ma 16 π planckover2pi1 2 ( q + π a ) + O ( V 3 ) , (1.b.20) so we see c q b = c q V 2 parenleftbigg ± V 2 V 2 ma 16 π planckover2pi1 2 ( q + π a ) parenrightbigg + O ( V 2 ) , = c q 1 ± parenleftbigg 1 V 2 ma 8 π planckover2pi1 2 ( q + π a ) parenrightbigg + O ( V 2 ) , c q b = ± c q parenleftBigg 1 + Vm 8 planckover2pi1 2 ( q 2 π 2 a 2 ) parenleftBig qa π 1 parenrightBig parenrightBigg + O ( V 2 ) . (1.b.21) Putting all this together, we see ψ ± q = π a ( r ) = c q e iqr braceleftBigg 1 ± e ibr + Vm 4 planckover2pi1 2 ( q 2 π 2 a 2 ) parenleftBig 1 aq π parenrightBig parenleftbigg e ibr e ibr 2 parenrightbigg bracerightBigg , so that ψ + ( r ) c q e iqr braceleftBigg 2 e i br 2 cos parenleftBig πr a parenrightBig + i Vm 2 planckover2pi1 2 ( q 2 π 2 a 2 ) parenleftBig 1 aq π parenrightBig bracketleftbigg sin parenleftbigg 2 πr a parenrightbigg i e ibr 4 bracketrightbigg bracerightBigg ; (1.b.22) and ψ ( r ) c q e iqr braceleftBigg 2 ie i br 2 sin parenleftBig πr a parenrightBig + Vm 2 planckover2pi1 2 ( q 2 π 2 a 2 ) parenleftBig 1 aq π parenrightBig bracketleftbigg cos parenleftbigg 2 πr a parenrightbigg e ibr 4 bracketrightbigg bracerightBigg . (1.b.23) ´ oπǫρ ´ ǫδǫιπoι tieaccentlowercase ησαι
6 JACOB LEWIS BOURJAILY -0.4 -0.2 0 0.2 0.4 -0.4 -0.2 0 0.2 0.4 -2 -1 0 1 2 0 4 Figure 2. The first Brillouin zone dispersion for a tight-binding model on a two- dimensional square lattice. Problem 2: Tight-Binding Model on a Square Lattice Consider a tight-binding model on a square, two-dimensional square lattice (lattice spacing a ) with on-site energy ǫ 0 and nearest-neighbour hopping matrix element t : H = summationdisplay r braceleftBig ǫ 0 | r )( r | + t bracketleftBig | r )( r + a ˆ x | + | r )( r a ˆ x | + | r )( r + a ˆ y | + | r )( r a ˆ y | bracketrightBigbracerightBig . a) We are to obtain the dispersion relation for this model. Just for the sake of clearing up notation, our Bravais lattice here will be generated by vectora 1 = a (1 , 0) and vectora 2 = a (0 , 1) which has the associated reciprocal lattice generated by vector b 1 = 2 π a (1 , 0) and vector b 2 = 2 π a (0 , 1). We will write all momenta in terms of the reciprocal lattice, so vector q = q 1 vector b 1 + q 2 vector b 2 . Using Bloch’s theorem it is quite easy to see that the Hamiltonian of this system is given by = braceleftbig ǫ 0 + t ( e ivector q · vectora 1 + e ivector q · vectora 1 + e ivector q · vectora 2 + e ivector q · vectora 2 )bracerightbig ψ, (2.a.1) = braceleftbig ǫ 0 + t ( e i 2 πq 1 + e i 2 πq 1 + e i 2 πq 2 + e i 2 πq 2 )bracerightbig ψ, (2.a.2) = braceleftbig ǫ 0 + 2 t

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• Fall '06
• JacobBourjaily
• Physics, Work, Condensed matter physics, Reciprocal lattice, Brillouin zone, Jacob Lewis Bourjaily, ´δǫι πoιη σαι