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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}

Page1 / 7

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online