= Solution
Take nearest-neighbour separation $d$, a common <spring constant> $K$, and one scalar displacement per atom. For a monatomic chain, the <harmonic approximation near a potential minimum>, <Newton's second law> and a <normal mode> give
$$
m\ddot u_j=K(u_{j+1}+u_{j-1}-2u_j),\qquad
u_j=u_0e^{i(kjd-\omega t)},\qquad
\boxed{\omega=2\sqrt{K/m}\,|\sin(kd/2)|}.
$$
The <monatomic-chain dispersion relation> has one <acoustic phonon branch>, with $\omega\simeq d\sqrt{K/m}|k|$ near the origin. The first <Brillouin zone> ends at $\boxed{k=\pm\pi/d}$, where $\omega=2\sqrt{K/m}$ and the <group velocity> vanishes.
For alternating <masses> $m,M$, choose a two-atom cell of length $2d$. The coupled <normal modes> give the <diatomic-chain dispersion relation with equal springs>,
$$
\boxed{\omega_\pm^2=K\left(\frac1m+\frac1M\right)\pm K\sqrt{\left(\frac1m+\frac1M\right)^2-\frac4{mM}\sin^2(kd)}}.
$$
The first <Brillouin zone> now ends at $\boxed{k=\pm\pi/(2d)}$. The minus branch is the <acoustic phonon branch>; its two atoms move together at long <wavelength>. The plus branch is the <optical phonon branch>; at $k=0$ the atoms move oppositely with a stationary centre of <mass> and $\omega_+^2=2K(1/m+1/M)$. For $M>m$ the edge frequencies are $\sqrt{2K/M}$ and $\sqrt{2K/m}$, with a gap between them.
If $M=m$, the two branches reduce to $2\sqrt{K/m}|\sin(kd/2)|$ and $2\sqrt{K/m}|\cos(kd/2)|$, touching at $\sqrt{2K/m}$. This is <zone folding in an equal-mass diatomic chain>: identical sites actually have a <primitive unit cell> of length $d$. A doubled cell folds the monatomic branch but opens no gap. Equal masses alone do not specify all possible spring models; the sketches assume equal nearest-neighbour <spring constants>.
\Image[/past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2018/paper-2-a4-chain-dispersion.png]
{title=Monatomic and two-atom-chain dispersion relations, with explicitly chosen cell lengths and zone edges}
Back to article page
