= Coupled-chain equations and dispersion relation
Put type-A atoms at $2ja$ and type-B atoms at $(2j+1)a$. Their displacements $u_j,v_j$ obey <Newton's second law>,
$$
m_A\ddot u_j=\alpha(v_j+v_{j-1}-2u_j),\qquad
m_B\ddot v_j=\alpha(u_j+u_{j+1}-2v_j).
$$
Choose <normal modes> $u_j=Ue^{i(2jaq-\omega t)}$ and $v_j=Ve^{i((2j+1)aq-\omega t)}$. Substitution gives
$$
\begin{pmatrix}2\alpha-m_A\omega^2&-2\alpha\cos qa\\-2\alpha\cos qa&2\alpha-m_B\omega^2\end{pmatrix}
\begin{pmatrix}U\\V\end{pmatrix}=0.
$$
A nonzero displacement vector requires zero <determinant>:
$$
m_Am_B\omega^4-2\alpha(m_A+m_B)\omega^2+4\alpha^2\sin^2qa=0.
$$
Solving this quadratic in $\omega^2$ gives the <diatomic-chain dispersion relation with equal springs>,
$$
\boxed{\omega_\pm^2=\frac\alpha{m_Am_B}\left[(m_A+m_B)\pm\sqrt{(m_A+m_B)^2-4m_Am_B\sin^2qa}\right]}.
$$
The <primitive unit cell> has length $2a$, so the first <Brillouin zone> is $|q|\leq\pi/(2a)$.
Back to article page
