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Unformatted text preview: Solutions of Problem set # 6 PHYS 361 (Solid State Physics), Autumn 2009 1. Peierls Instability (a) The lattice constant now becomes A = 2 a , thus the smallest reciprocal lattice vector is K = 2 A = a (b) U k = 1 L V ( k ) S * k , where V ( k ) = R dxe ikx V ( x ) , S k = 1 + e ik ( a x ) andL = 2 a ; then U k = 1 2 a V ( k )[1 + e ik ( a x ) ] 1 2 a V ( k )[1 + e ika (1 + ikx )] whenxissmall ; at k = /a, U k = i 2 a 2 V ( a ) x x (c) Fermi wavevector K F = K/ 2 = 2 a (d) slove  K F U k U k  K F k = 0, you get = K F + K F k 2 q ( K F K F k 2 ) 2 +  U k  2 ; at k = /a , = K F  U k  , therefore = U k = /a   x (e) from d)  = K F + K F k 2 q ( K F K F k 2 ) 2 +  U k  2 , thus k =  K F = K F K F k 2 + q ( K F K F k 2 ) 2 +  U k  2 , then the total E e = 2 R K F K F k dk L 2 = ~ 2 L 2 ma R K F K F dk...
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This note was uploaded on 03/02/2010 for the course PHYSICS 336 taught by Professor Pucker during the Spring '10 term at King's College London.
 Spring '10
 Pucker
 Solid State Physics

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