Michael Zawodzki Phonon dispersion relation of a ... - lamp.tugraz.at

Subject: Solid State Physics (Intro.) Case: Student project Name: Michael ZawodzkiPhonon dispersion relation of a bcc lattice:Equation of motion in x -direction is:d 2 u x lmndt 2 = C √3 m[(u x l+1m+1n+1 − ux lmn ) + (ux l−1m+1n+1 − ux lmn ) + (ux l−1m−1n+1 − ux lmn )++(u x l+1m−1n+1 − ux lmn ) + (ux l+1m+1n−1 − ux lmn ) + (ux l−1m+1n−1 − ux lmn ) + (ux l−1m−1n−1 − ux lmn )++(u x l+1m−1n−1 − ux lmn ) + (uy l+1m+1n+1 − uy lmn ) − (uy l−1m+1n+1 − uy lmn ) − (uy l−1m−1n+1 − uy lmn )++(u y l+1m−1n+1 − uy lmn ) − (uy l+1m+1n−1 − uy lmn ) + (uy l−1m+1n−1 − uy lmn ) − (uy l−1m−1n−1 − uy lmn )++(u y l+1m−1n−1 − uy lmn ) + (uz l+1m+1n+1 − uz lmn ) − (uz l−1m+1n+1 − uz lmn ) − (uz l−1m−1n+1 − uz lmn )++(u z l+1m−1n+1 − uz lmn ) − (uz l+1m+1n−1 − uz lmn ) + (uz l−1m+1n−1 − uz lmn ) + (uz l−1m−1n−1 − uz lmn )−−(u z l+1m−1n−1 − uz lmn )]Forwas used.u x lmnu x lmn = ux −→ ke i(∑ 3i=1−→ ai−→ k − ωt)Primitve vectors for bcc are:We obtain:⎡A = a ⎣2−1 1 11 −1 11 1 −1⎤⎦u x lmn = ux −→ e i(l−→ k ·−→ a −→ 1 +m k ·−→ a −→ 2 +n k ·−→ a 3 ) = u x (−l+m+n)kxai( +−→ e (l−m+n)k ya+ (l+m−n)kza )2 2 2k k2,5102,081,56(C/m) 1/2 H P L 0 21,040,50,0Figure 1: Phonon dispersion of a bcc lattice.gamma - H - P - gamma - NPlot is over the symmetry lines as follows:On the next page solutions of the equation of motion are summed up in a array.1

04 − cos( a 2 (kx + ky + kz)) − cos( a 2 (3kx − ky − kz)) −cos( a 2 (kx + ky + kz)) + cos( a 2 (3kx − ky − kz)) −cos( a 2 (kx + ky + kz)) + cos( a (3kx − ky − kz))2−cos( a 2 (−kx + 3ky − kz)) + cos( a 2 (−kx − ky + 3kz)) +cos( a 2 (−kx + 3ky − kz)) + cos( a 2 (−kx − ky + 3kz)) −cos( a 2 (−kx + 3ky − kz)) − cos( a (−kx − ky + 3kz)) 2 ⎤−cos( a 2 (−kx + 3ky − kz)) − cos( a (−kx − ky + 3kz)) − √ 2 3C +cos(2 (−kx + 3ky − kz)) − cos( a 2 (−kx − ky + 3kz)) −cos( a 2 (−kx + 3ky − kz)) + cos( a (−kx − ky + 3kz))2 mω2 a −cos( a 2 (k x + ky + kz)) + cos( a 2 (3k x − ky − kz)) 4 − cos( a 2 (k x + ky + kz)) − cos( a 2 (3k x − ky − kz)) −cos( a 2 (k x + ky + kz)) − cos( a 2 (3k x − ky − kz))+cos( a 2 (−kx + 3ky − kz)) − cos( a 2 (−kx − ky + 3kz)) −cos( a 2 (−kx + 3ky − kz)) − cos( a (−kx − ky + 3kz)) − mω2 √ 2 a a +cos(2 (−kx + 3ky − kz)) + cos( a (−kx − ky + 3kz))24 − cos( a 2 (k x + ky + kz)) − cos( a 2 (3k x − ky − kz))−cos( a 2 (k x + ky + kz)) + cos( a 2 (3k x − ky − kz)) −cos( a 2 (k x + ky + kz)) − cos( a 2 (3k x − ky − kz)) − mω2 √⎡2