Two points to note 1 The bandwidth is proportional to the hopping term 2 It is

Two points to note 1 the bandwidth is proportional to

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Two points to note 1. The bandwidth is proportional to the hopping term. 2. It is surprisingly easy to generalize the result to 2 and 3 dimensions, because all we need to do is sum over the nearest neighbours! If they are symmetrically located then the bandwidth would simply be 2 zt where z is the number of nearest neighbours or the co-ordination number of the lattice. PROBLEM : Calculate the group velocity of a particle at the bottom of the band and at the corner ( k = ± π/a ). Show that there is a point of inflection (where the second derivative changes sign) somewhere between k = 0 and k = ± π/a . How does the Bloch function look? Here’s a plot of how the functions look. The k = 0 wavefunction is shown for reference, because that has the maximum resemblance with the ”atomic” wavefunctions. Here we assumed that the atomic wavefunction is a gaussian. See Fig. 5.2.
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5.1. DIATOMIC MOLECULE AND LINEAR CHAIN OF ATOMS 85 Figure 5.2: The dots are the atomic sites, the k = 0 wavefunction shows what the atomic states are like. The other two show what the linear combination of those wavefunctions, as given by Bloch’s theorem , would look like. See eqn. 5.13.
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86 CHAPTER 5. TIGHT BINDING OR LINEAR COMBINATION OF ATOMIC ORBITALS (LCAO) Generalising to 2 and 3 dimensions: with 1 oribital per site The generalisation is easy. To handle 2 and 3d lattices we need to write the wavefunction as | ψ k ) = 1 N summationdisplay R e i k . R | φ R ) (5.26) ( r | ψ k ) = 1 N summationdisplay R e i k . R φ ( r R ) (5.27) Where the sum runs over all direct lattice vectors R and | φ R ) is the atomic state centered at R . While taking the expectation value of energy we will group the series of terms into three and ignore the interaction between sites which are not nearest neighbours or next-nearest-neighbours: H = T + V 1 + V 2 + V 3 + ... + V N (5.28) E ( k ) = ( ψ k | H | ψ k ) (5.29) = 1 N summationdisplay R = R ( φ R | H | φ R ) + 1 N summationdisplay R , R nearest neigh- bours e i k . ( R R ) ( φ R | H | φ R ) + a0 a0 a0 a0 a0 a0 a0 a0 a64 a64 a64 a64 a64 a64 a64 a64 1 N summationdisplay R , R further than nearest neigh- bours ... (5.30) E 0 + summationdisplay nearest neigh- bours e i k . R t R (5.31) Since all sites are identical, it is sufficient to sum over the nearest neighbours of the site at R = 0. PROBLEM : Consider a 2-d rectangular lattice with sides a and b . 1. Show that following eqn. 5.31 the bandstructure would be of the form E ( k x ,k y ) = E 0 2 t 1 cos( ak x ) 2 t 2 cos( bk y ) (5.32) 2. What is the reciprocal lattice? Draw the first Brillouin zone. 3. Plot the constant energy contours, assuming t 1 > t 2 > 0 and a < b . Why is this physically reasonable? 4. Plot some constant energy contours. How do the contours look for small k ? How do the shapes change at slightly larger k ? Do all constant energy contours close within the first Brillouin zone? Similarly for 3d lattices like BCC with 8 nearest neighbours and FCC with 12 neighbours can be summed up.
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