1D Diatomic Chain Dispersion
Consider a 1D chain with two atoms in the unit cell
The coordinates of each atom in the cell
Rjs(t)=xj+ds+ujs(t);s=1,2where xj is the vector of the j-th cell, ds is the relative vector of the s-th atom in the cell, ujs(t) is the displacement of the s-th atom in the j-th cell from its equlibrium position.
For the atoms in the j-th cell, Newton’s law yields these euqations of motion
md2uj1dt2=K(uj2+uj−12−2uj1)Md2uj2dt2=K(uj1+uj+11−2uj2)where m and M are the masses of atom 1 and 2, respectively.
To find the solutions to the equations, we assume that all of the atoms move with the same frequency,
uj1(t)=Aq√mei(qxj−ωt)uj2(t)=Bq√Mei(qxj−ωt)Substitute this into the equations of motion, we have
(2Km−ω2−K√mM(1+e−iqa).−K√mM(1+eiqa)2KM−ω2)(AqaBq)=0The equations will have a solution when the determinant of the matrix equals zero, i.e.
(2Km−ω2)(2KM−ω2)−2K2mM(1+cos(qa))=0This equation is quadratic and the solution can be easily found out
ω2±=KMm[(m+M)±√m2+M2+2mMcos(qa)]where the subscript +/− denote the optical/acoustic mode, respectively.
The maximum frequency is found ath q=0 of the optical mode
ωmax=√2K(m+M)mM