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Prob6_solution

Prob6_solution - Solutions of Problem set 6 PHYS 361(Solid...

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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 π A = π a (b) U k = 1 L ˜ V ( k ) S * k , where ˜ V ( k ) = R dxe - ikx V ( x ) , S k = 1 + e ik ( a - δx ) and L = 2 a ; then U k = 1 2 a ˜ V ( k )[1 + e - ik ( a - δx ) ] 1 2 a ˜ V ( k )[1 + e - ika (1 + ikδx )] when δx is small ; at k = π/a, U k = - i π 2 a 2 ˜ V ( π a ) δx δx (c) Fermi wavevector K F = K/ 2 = π 2 a (d) slove ε - ε 0 K F U k - U - k ε - ε 0 K F - k = 0, you get ε ± = ε 0 K F + ε 0 K F - k 2 ± q ( ε 0 K F - ε 0 K F - k 2 ) 2 + | U k | 2 ; at k = π/a , ε = ε 0 K F - | U k | , therefore Δ ε = -| U k = π/a | ∝ - δx (e) from d) ε - = ε 0 K F + ε 0 K F - k 2 - q ( ε 0 K F - ε 0 K F - k 2 ) 2 + | U k | 2 , thus Δ ε k = ε - - ε 0 K F = - ε 0 K F - ε 0 K F - k 2 + q ( ε 0 K F - ε 0 K F - k 2 ) 2 + | U k | 2 , then the total Δ E e = 2 R K F - K F Δ ε k dk L 2 π = - ~ 2 L 2 ma R K F - K F dk ( k - π 2 a ) + p ( k - π 2 a ) 2 + C 2 , where C = 2 ma ~ 2 π U k ; Δ E e = - ~ 2 L 2 ma - 2 k 2 F + k F p 4 k 2 F + C 2 + 1 2 C 2 ln C - 2 k F + 4 k 2 F + C 2 . When C is small ( U k is small), Δ E e can be simplified to Δ E e = - ~ 2 L 8 ma C 2 h 1 -

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Prob6_solution - Solutions of Problem set 6 PHYS 361(Solid...

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