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# cha10 - PHYS 635 Solid State Physics Take home exam 1...

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PHYS 635 Solid State Physics Take home exam 1 Gregory Eremeev Fall 2004 Submitted: November 8, 2004 Problem 1:Ashcroft & Mermin, Ch.10, p.189, prob.2 a) Let’s prove β xx = β yy = β zz = β (1) β xx = - d r · Ψ * x ( r ) · Ψ x ( r ) · Δ U ( r ) = - d r · x 2 · | φ ( r ) | 2 · Δ U ( r ) = - d r · y 2 · | φ ( r ) | 2 · Δ U ( r ) = - d r · Ψ * y ( r ) · Ψ y ( r ) · Δ U ( r ) = β yy (2) Now 0 = β xx - β yy = - d r · ( x 2 - y 2 ) · | φ ( r ) | 2 · Δ U ( r ) (3) If we here make the change of variables: x = x - y ; y = x + y , which is basically rotation in space, then we will get 0 = β xx - β yy = - d r · ( x · y ) · | φ ( r ) | 2 · Δ U ( r ) = β xy = 0 (4) b) In the case of a simple cubic Bravais lattice with γ ij ( R ) negligible for all but the nearest-neighbor R let’s calculate ˜ γ xy ( k ). We should take into account 6 neighbors: R = a ( ± 1 , 0 , 0); a (0 , 0 , ± 1); a (0 , ± 1 , 0) (5) ˜ γ xy ( k ) = - e ik x a d r · xyφ ( r ) φ ( ( x - a ) 2 + y 2 + z 2 ) 1 2 Δ U ( r ) - e - ik x a d r · xyφ ( r ) φ ( ( x + a ) 2 + y 2 + z 2 ) 1 2 Δ U ( r ) - e ik y a d r · x ( y - a ) φ ( r ) φ ( x 2 + ( y - a ) 2 + z 2 ) 1 2 Δ U ( r ) - e - ik y a d r · x ( y + a ) φ ( r ) φ ( x 2 + ( y + a ) 2 + z 2 ) 1 2 Δ U ( r ) - e ik z a d r · xyφ ( r ) φ ( x 2 + y 2 + ( z - a ) 2 ) 1 2 Δ U ( r ) - e - ik z a d r · xyφ ( r ) φ ( x 2 + y 2 + ( z + a ) 2 ) 1 2 Δ U ( r ) = 0 (6) , because under rotation transformation x y, y -x each integral changes sign. c) The energies are found by setting to zero the determinant: | ( ε ( k ) - E p ) δ ij + β ij + ˜ γ ij ( k ) | = 0 (7) 1

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First we will prove that ε ( k ) - E p + β + ˜ γ xx ( k ) = ε ( k ) - lon 0 ( k ) + 4 γ 0 cos 1 2 k y a cos 1 2 k z a (8) , where ε 0 ( k ), γ 0 , γ 1 and γ 2 are those from A & M. In nearest-neighbor approximation we will have the 12
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