sol07 - Solid State Theory Solution Sheet 7 FS 11 Prof M...

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Solid State Theory Solution Sheet 7 FS 11 Prof. M. Sigrist Exercise 7.1 Phonons in One Dimension Given the potential energy, V = v 2 N/ 2 X i =1 ( u 2 i - u 2 i +1 ) 2 + ( u 2 i - u 2 i - 1 ) 2 , (1) and the atomic masses M and m , we can immediately write down the classical equations of motion, ¨ u 2 i = - v M (2 u 2 i - u 2 i +1 - u 2 i - 1 ) ¨ u 2 i +1 = - v m (2 u 2 i +1 - u 2 i +2 - u 2 i ) , (2) which can be simplified by using the spatial Fourier transform u 2 l = r 2 N π/ 2 X k =0 ( q 1 ,k e - i 2 kl + q * 1 ,k e i 2 kl ) , u 2 l +1 = r 2 N π/ 2 X k =0 ( q 2 ,k e - ik (2 l +1) + q * 2 ,k e ik (2 l +1) ) , (3) where k takes values in the reduced Brillouin zone, and is taken to be positive because the ± k Fourier components are not independent as u 2 i is real. In terms of the q ak , the equations of motion simplify to ¨ q 1 ,k = - 2 v M ( q 1 ,k - q 2 ,k cos k ) ¨ q 2 ,k = - 2 v m ( q 2 ,k - q 1 ,k cos k ) . (4) By taking the temporal Fourier transform, this becomes ω 2 k - 2 v M - 2 v M cos k - 2 v m cos k ω 2 k - 2 v m q 1 ,k q 2 ,k = 0 . (5) For a non-trivial solution to occur the determinant of the matrix has to vanish, which gives an equation determining ω k , ω 4 k - 2 v 1 M + 1 m ω 2 k + 4 v 2 Mm sin 2 k = 0 . (6) There are two solutions corresponding to the two phonon branches that result from the doubling of the unit cell (or folding back of the BZ), which are ω 2 k = v Mm M + m ± p M 2 + m 2 + 2 Mm cos(2 k ) (7) 1
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- 1 2 0 1 2 0 0.25 0.5 0.75 1 1.25 1.5 0 0.25 0.5 0.75 1 1.25 1.5 k Π Ω ak !!!!! v Figure 1: Acoustic and optical branch of the phonon dispersion in the reduced Brillouin zone for M/m = 10. The corresponding Lagrangian is then given by L = 2 X a =1 π/ 2 X k =0 ˙ q ak ˙ q * ak - ω 2 ak q ak q * ak , (8) and the canonical momenta are p ak = L ˙ q ak = ˙ q * ak , p * ak = L ˙ q * ak = ˙ q ak , (9) and the corresponding Hamiltonian is given by H = 2 X a =1 π/ 2 X k =0 p ak p * ak + ω 2 ak q ak q * ak (10) Using the equations of motion, ¨ q ak = - ω 2 ak q ak , (11) we can write down the general solution q ak = 1 2 ( a a,k e iωt + a a, - k e - iωt ) , (12) where a a,k and a a, - k are independent complex numbers because q ak is complex. As q ak always multiplies e - ikl when transforming back to real space, a a,k corresponds to a right- moving wave, whereas a a, - k corresponds to a left-moving one. From this we get p * ak = ˙ q ak = 2 ( a a,k e iωt - a a, - k e - iωt ) . (13) Now we can eliminate p ak and q ak from the Hamiltonian in eq. (10) by substituting the solution of the equation of motion, which leads to H = 2 X a =1 π/ 2 X k =0 ω 2 ak ( a a,k a *
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