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Unformatted text preview: Solutions of Problem set # 7 PHYS 361 (Solid State Physics), Autumn 2009 1. Anisotropic mass (a) g ( ) = 2 Z d 3 k (2 ) 3 (  ( ~ k )) = 2 Z d 3 k (2 ) 3 (   ~ 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 ( ) = 2 Z d 3 k (2 ) 3 (   ~ 2 2 X i = s,y,z k 2 i m i ) Lets take k i m i k i then g ( ) = ( m x m y m z ) 1 / 2 2 Z d 3 k (2 ) 3 (   ~ 2 k 2 2 ) = 1 2 ~ 3 p 2(  )  M  where we used the fact that detM =  M  = ( m x m y m z ) 1 / 2 with the expression of g ( ) we get c v = 2 3 k 2 B Tg ( F ) = k 2 B T 3 ~ 3 p 2( F )  M  and m * m = g ( f ) g free ( F ) = ( F )  M  F m 3 = q (1 F )  M  m 3 / 2 (b) using p = H x , x = H p you get v a = ( ~ k a ) = 1 ~ k a + ~ 2 2 ( k k ) i M...
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 Spring '10
 Pucker
 Mass, Solid State Physics

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