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Unformatted text preview: Solid State Theory Solution Sheet 2 FS 11 Prof. M. Sigrist Exercise 2.1 Energy bands of almost free electrons on the fcc lattice The Bloch equation is written in Fourier space as [see Eq. (1.21) of lecture notes] ~ 2 2 m ( ~ k + ~ G ) 2 ε n, ~ k c ~ G + X ~ G V ~ G ~ G c ~ G = 0 . (1) For V ≡ 0 the dispersion along the ∆line is shown in Fig. 1 for the few lowest bands. The numbers indicate the degeneracy of the bands. The different lines stem from different 0.2 0.4 0.6 0.8 1 1 2 3 4 5 6 ~ k [ 2 π a ] ε [ ¯ h 2 G 2 2 m ] Γ X 1 1 4 4 1 4 8 1 4 2 V = 0 Figure 1: The dispersion along the ∆line for free electrons on a fcc lattice. The numbers indicate the degeneracy of the eigenstates. energy parabolas centered at different but equivalent points in reciprocal space. Figure 2 shows the part of the reciprocal lattice which is relevant for the lowest energy bands. a) For V = 0 the second energy level, E = 3 ~ 2 G 2 2 m , is 8fold degenerate. It stems from parabolas centered at the 8 points connected to Γ by the following reciprocal lattice vectors: ~ G 1 = G (1 , 1 , 1) , ~ G 2 = G ( 1 , 1 , 1) , ~ G 3 = G ( 1 , 1 , 1) , ~ G 4 = G (1 , 1 , 1) , ~ G 5 = G (1 , 1 , 1) , ~ G 6 = G ( 1 , 1 , 1) , ~ G 7 = G ( 1 , 1 , 1) , ~ G 8 = G (1 , 1 , 1) , (2) where G = 2 π a . The eigenfunctions are given by ψ j ( ~ r ) = h ~ r  ~ G j i = e i ~ G j · ~ r √ V (3) and form an 8 dimensional Hilbertspace. The representation Γ of O h on this sub space is defined as ˆ Γ( g )  ~ G j i =  g ~ G j i (4) 1 k z k x k y Γ ∆ X ~ G 4 ~ G 1 ~ G 2 ~ G 3 ~ G 5 ~ G 6 ~ G 7 ~ G 8 Figure 2: Section of the reciprocal lattice. The length of the cube is 4 π a . where g ∈ O h . It is easy to see that each element of the cubic point group simply permutes the ~ G j ’s. For example, a rotation by π/ 2 around the zaxis is represented as R z π/ 2 = 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 . The character of this transformation is χ Γ ( R z π/ 2 ) = tr( R z π/ 2 ) = 0. Clearly, in order to find all the characters of the representation Γ defined by Eq. (4) we don’t have to compute all the matrices. Instead, we simply have to know how many of the ~ G i ’s are invariant under a certain element. For this purpose, it is sufficient to consider one element of each conjugacy class. In the following, J will denote the inversion, C 3 (8) the conjugacy class of rotations by 2 π/ 3 around one of the diagonals of the cube, C 4 (6) the conjugacy class of the rotations by π/ 2, C 2 (6) the conjugacy class of rotations by π around an axis through the edges of the cube and C 2 4 (3) the conjugacy class of the rotations by π around an axis through surface of the cube. (The number in brackets denotes the number of elements in the corresponding conjugacy class.) One finds the following group character...
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This note was uploaded on 10/04/2011 for the course PHYS fs11 taught by Professor Sigrist during the Spring '11 term at Swiss Federal Institute of Technology Zurich.
 Spring '11
 Sigrist
 Energy

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