solution_07 - Solutions of Problem set 7 PHYS 361(Solid...

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Unformatted text preview: Solutions of Problem set # 7 PHYS 361 (Solid State Physics), Autumn 2009 1. Anisotropic mass (a) g ( ε ) = 2 Z d 3 k (2 π ) 3 δ ( ε- ε ( ~ k )) = 2 Z d 3 k (2 π ) 3 δ ( ε- ε- ~ 2 2 ~ k M- 1 ~ k ) (1) Since M is symmetric, we can find a matrix U so that M- 1 can be diagonalized: k i → U ij k j , U T M- 1 U = ˜ M- 1 = diag { 1 m x , 1 m y , 1 m z } Thus g ( ε ) = 2 Z d 3 ˜ k (2 π ) 3 δ ( ε- ε- ~ 2 2 X i = s,y,z ˜ k 2 i m i ) Let’s take ˜ k i → m i k i then g ( ε ) = ( m x m y m z ) 1 / 2 2 Z d 3 k (2 π ) 3 δ ( ε- ε- ~ 2 k 2 2 ) = 1 π 2 ~ 3 p 2( ε- ε ) | M | where we used the fact that detM = | M | = ( m x m y m z ) 1 / 2 with the expression of g ( ε ) we get c v = π 2 3 k 2 B Tg ( ε F ) = k 2 B T 3 ~ 3 p 2( ε F- ε ) | M | and m * m = g ( ε f ) g free ( ε F ) = √ ( ε F- ε ) | M | √ ε F m 3 = q (1- ε ε F ) | M | m 3 / 2 (b) using ˙ p =- ∂ H ∂x , ˙ x = ∂ H ∂p you get v a = ∂ε ∂ ( ~ k a ) = 1 ~ ∂ ∂k a ε + ~ 2 2 ( k- k ) i M-...
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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.

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solution_07 - Solutions of Problem set 7 PHYS 361(Solid...

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