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Unformatted text preview: P613 HW # 2 Solutions JM Densmore 1. The Free and Independent Electron Gas in Two Dimensions (a) In a 2D system with periodic BC you can solve the SE with momen- tum k x = 2 /Ln x and k y = 2 /Ln y . The number of allowed values of k space inside a volume will be V 4 2 The volume in k space is the area of a circle with radius k f . Taking into account the number of spins for the electron n = k 2 f 2 (1) (b) 1 n = r 2 s (2) r s = 2 k f (3) (c) In 2D the density of particles in range k + dk is 2 kdk n = Z 2 kdk 4 2 f ( k ) (4) Changing to an integral over energy we find n = m h 2 Z f ( ) d (5) (6) g = m h 2 > g = < (d) Since g is constant g = 0 all the terms in the BS expansion are equal to zero. So we find n = Z f m h 2 dE + ( - f ) m h 2 (7) n = n ( - f ) m h 2 (8) = f (9) (e) From 2.67 we have n = m h 2 Z 1 e ( - ) + 1 d 1 This integral can be looked up in a table. Using the electron densityThis integral can be looked up in a table....
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- Spring '10