# sol03 - Solid State Theory Solution Sheet 3 FS 2011 Prof M...

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Unformatted text preview: Solid State Theory Solution Sheet 3 FS 2011 Prof. M. Sigrist Exercise 3.1 6-Orbital tight-binding model a) The Bloch-waves constitute a basis of the (quasi-1-dimensional) Hilbert space of the atom chain. The structure of the functions in x- and y-direction can be neglected in this reasoning, since it is only remnant of the structure of the lattice sites. Since the Wannier functions are the “Fourier-transforms” of the Bloch-waves, they, too, span the whole Hilbert space. Thus, we can write H = X n,n ,j,j h w n ( x,y ; z- al ) |H| w n ( x,y ; z- al ) i c † n,j c n ,j . (1) Limiting the summation to nearest neighbors yields H = X n H n + X n 6 = n H n,n , (2) H n = X j ε n c † n,j c n,j + ( t n c † n,j +1 c n,j + h . c . ) , (3) H n,n , = X j t n,n c † n,j +1 c n ,j + t n ,n c † n,j c n ,j +1 , (4) with ε n = h w n ( x,y ; z ) |H| w n ( x,y ; z ) i , (5) t n = h w n ( x,y ; z- a ) |H| w n ( x,y ; z ) i , (6) t n,n = h w n ( x,y ; z- a ) |H| w n ( x,y ; z ) i , (7) where we have used the translational invariance of the Hamiltonian along the z- direction. b) When approximating the Wannier functions by hydrogen orbitals, the diagonal ma- trix elements become the hydrogen energies. This is obviously a bad approximation even when taking Z = 11 into account, as the outer electrons of Na see a strongly screened potential ( Z eff (3 s ) ≈ 2 . 5!). From this we see, that we cannot expect quan- titatively good ab-initio results from this starting point. However, the symmetry arguments leading to the relative signs of the matrix elements hold. In order to determine these signs, we fix the phases of the other orbitals such that the s-orbital wave functions are strictly positive. We start by considering only s-orbitals ( t n,n ≡ t n ). There, t n,n = h Ψ m,s ( x,y,z- a ) |T + X j V ( x,y,z- aj ) | Ψ m ,s ( x,y,z ) i , (8) 1 with Ψ m,s ( r ) the hydrogen wave function for the ms-orbital, the Coulomb potential V ( r ) and the kinetic energy operator T . Using that T Ψ m,s ( r ) = [ ε m- V ( r )]Ψ m,s ( r ) and ε m &amp;lt; 0, V ( r ) &amp;lt; 0, we see that the remaining integral, t n,n = h Ψ m,s ( x,y,z- a ) | ε m + X j 6 =0 V ( x,y,z- aj ) | Ψ m ,s ( x,y,z ) i , (9) is negative. It is also easy to see that, so far, t n,n = t n ,n . + +- + + +- +- +- +- +- +- +- + a) b) c) d) e) f) Figure 1: Illustration of the hydrogen orbitals in the ion chain. The overlap of two s- orbitals (a) or two p x and p y orbitals (b) is always positive. The overlap between a p x or a p y orbital with a s (c) or a p z (d) orbital vanishes. The situation is more complicated in cases e) and f). Since the magnitude of the overlap is largest in between the lattice sites, the s- p z overlap is positive and the resulting matrix element negative (e). The p z- s overlap, on the other hand, is negative, as is the sign of the product of the wave functions in the important area. Similarly, the p z- p z overlap is negative and the resulting matrix element positive (f)....
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sol03 - Solid State Theory Solution Sheet 3 FS 2011 Prof M...

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