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Unformatted text preview: Solid State Theory Solution Sheet 4 FS 11 Prof. M. Sigrist Exercise 4.1 Onedimensional model of a semiconductor The Hamilton operator is H 1 = H + V where H = t X i c † i c i +1 + c † i +1 c i , (1) V = v X i ( 1) i c † i c i . (2) [a] Let us consider the case v = 0. We write c k = 1 √ N X j e ikj c j , c j = 1 √ N X k e ikj c k , (3) where k ∈ [ π,π ) and kN = 2 πn , n ∈ Z . The above expression is plugged into Eq. (1) and we obtain H = t X k,k 1 N X j e i [ kj k ( j +1)] + 1 N X j e i [ k ( j +1) ik j ] ! c † k c k (4) = t X k ( e ik + e ik ) c † k c k = X k ( 2 t cos k )  {z } k c † k c k . (5) The second line follows from 1 N X j e i [ kj k ( j +1)] = 1 N X j e i ( k k ) j e ik = δ k,k e ik . (6) Let us define the following oneparticle state:  φ k i = c † k  i where  i is the vacuum. It fulfills c † k c k  φ k i = c † k c k c † k  i = c † k (1 c † k c k )  i = c † k  i =  φ k i , (7) and consequently H  φ k i = k  φ k i . (8) Therefore,  φ k i is an eigenstate of the Hamilton operator. [b] Let’s consider now the case v 6 = 0. Again, the expression (3) is plugged into H 1 : H 1 = X k k c † k c k + v X k,k " 1 N X j ( 1) j e i ( k k ) j # c † k c k (9) Since 1 N X j ( 1) j e i ( k k ) j = 1 N X j e i ( k k + π ) j = δ k,k + π (10) 1 it follows that H 1 = X k k c † k c k + k + π c † k + π c k + π + v c † k c k + π + v c † k + π c k . (11) where ∑ means summation over the reduced Brillouin zone k ∈ [ π/ 2 ,π/ 2]. Note that k + π = 2 t cos( k + π ) = 2 t cos k = k . (12) Introducing ¯ c k = c k c k + π (13) the Hamilton operator is written in matrix form H 1 = X k ¯ c † k ˆ H 1 ¯ c k , (14) where ˆ H 1 = k v v k . (15) We define new operators a k and b k according to c k c k + π = u k v k v k u k a k b k = U † ¯ α k , (16) H 1 = X k ¯ α † k U † ˆ H 1 U ¯ α k . (17) We can choose U such that U † ˆ H 1 U is diagonal. The energies are obtained from the secular equation det k λ v v k λ = λ 2 2 k v 2 = 0 (18) which has the solutions λ = ± q 2 k + v 2 = ± E k . (19) Furthermore, one finds u k = v p 2 E k ( E k + k ) , v k = r E k + k 2 E k . (20) Finally, the Hamilton operator is written in the eigenbasis H 1 = X k E k a † k a k + E k b † k b k . (21) [c] The band structure of the alternating chain is shown in Fig. 1. The gap between valence and conduction band is Δ = 2 E ± π/ 2 = 2 v . The ground state for N/ 2 electrons on the chain is given by...
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 Spring '11
 Sigrist

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