= Solution
\b[One <mass> per cell.] If $u_j$ is the displacement from equilibrium, the two springs give $m\ddot u_j=\alpha(u_{j+1}+u_{j-1}-2u_j)$. Substituting $u_j=Ue^{i(qja-\omega t)}$ yields
$$
\boxed{\omega(q)=2\sqrt{\alpha/m}\,|\sin(qa/2)|}.
$$
The <first Brillouin zone> is $-\pi/a\leq q\leq\pi/a$; adding $2\pi/a$ leaves the sampled displacements unchanged. The branch is linear near zero, with <sound speed> $a\sqrt{\alpha/m}$, and reaches $2\sqrt{\alpha/m}$ at either zone edge.
\b[Two <masses> per cell.] Take a cell of length $2a$, with light displacement $u_j=Ue^{i(2jaq-\omega t)}$ and heavy displacement $v_j=Ve^{i[(2j+1)aq-\omega t]}$. The coupled equations reduce to
$$
\begin{pmatrix}2\alpha-m\omega^2&-2\alpha\cos qa\\
-2\alpha\cos qa&2\alpha-M\omega^2\end{pmatrix}
\begin{pmatrix}U\\V\end{pmatrix}=0.
$$
A nonzero displacement requires determinant zero:
$$
Mm\omega^4-2\alpha(M+m)\omega^2+4\alpha^2\sin^2qa=0.
$$
Solving this quadratic in $\omega^2$ gives
$$
\boxed{\omega_\pm^2=\frac\alpha{Mm}\left[M+m\pm\sqrt{(M+m)^2-4Mm\sin^2qa}\right]}.
$$
Now the first zone is $|q|\leq\pi/(2a)$. The minus branch is acoustic: as $q\to0$, both sublattices move together and $\omega_-\to0$. Its <sound speed> is $a\sqrt{2\alpha/(M+m)}$. The plus branch is optical: at $q=0$ the atoms move oppositely with $mU+MV=0$, giving finite <frequency> $\sqrt{2\alpha(1/m+1/M)}$. This terminology describes relative motion; optical activity itself also requires coupling to the electromagnetic field.
\b[<diatomic-chain zone-edge polarizations> and dimensionality.] At the edge, $\cos qa=0$, so the heavy and light sublattices decouple: the acoustic edge <frequency> is $\sqrt{2\alpha/M}$ and the optical edge <frequency> is $\sqrt{2\alpha/m}$. Hence
$$
\boxed{\Delta\omega=\sqrt{2\alpha/m}-\sqrt{2\alpha/M}}.
$$
The acoustic edge mode moves only the heavy sublattice; the optical edge mode only the light one. If $M=m$, the gap closes and the two branches are the zone-folded monatomic branch.
\Image[/past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2013/paper-2-chain-phonons.png]
{title=Original monatomic and diatomic chain dispersions, zone-edge gap and long-wavelength displacement patterns}
A three-dimensional crystal with two atoms per primitive cell has \b[six branches: three acoustic and three optical]. For propagation along a suitable symmetry direction these resolve into one longitudinal and two transverse branches of each type. The <wavevector> and <Brillouin zone> are three-dimensional; anisotropic <force> constants determine the dispersions, and polar crystals can have longitudinal–transverse optical splitting. A universal six-branch curve cannot be specified from <masses> alone.
Back to article page
