# sol04 - Solid State Theory Solution Sheet 4 FS 11 Prof M...

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Unformatted text preview: Solid State Theory Solution Sheet 4 FS 11 Prof. M. Sigrist Exercise 4.1 One-dimensional 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 one-particle 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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sol04 - Solid State Theory Solution Sheet 4 FS 11 Prof M...

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