= Solution
Let the nearest-neighbour spacing be $d$, so the <primitive unit cell> has length $\ell=2d$. Denote the two longitudinal displacements in cell $n$ by $u_n,v_n$, with spring $K$ within each cell and spring $K'$ between cells. <Hooke's law> and <Newton's second law> give
$$
\boxed{m\ddot u_n=K(v_n-u_n)+K'(v_{n-1}-u_n),\qquad
m\ddot v_n=K(u_n-v_n)+K'(u_{n+1}-v_n).}
$$
For <normal modes> $u_n=u e^{i(qn\ell-\omega t)}$ and $v_n=v e^{i(qn\ell-\omega t)}$, the two amplitudes satisfy
$$
\begin{pmatrix}
K+K'-m\omega^2&-(K+K'e^{-iq\ell})\\
-(K+K'e^{iq\ell})&K+K'-m\omega^2
\end{pmatrix}\begin{pmatrix}u\\v\end{pmatrix}=0.
$$
Its vanishing <determinant> gives the <alternating-spring chain dispersion relation>
$$
\boxed{\omega_\pm^2(q)=\frac{K+K'\pm\sqrt{K^2+K'^2+2KK'\cos(q\ell)}}m,\qquad |q|\leq\frac\pi\ell.}
$$
The minus sign is the <acoustic phonon branch>, with $\omega_-(0)=0$ and $u=v$ at the origin. The plus sign is the <optical phonon branch>, with $u=-v$ at the origin. For positive springs the <acoustic phonon branch> rises from zero to $\sqrt{2\min(K,K')/m}$ at the <Brillouin zone> edge; the <optical phonon branch> falls from $\sqrt{2(K+K')/m}$ to $\sqrt{2\max(K,K')/m}$. Their long-wavelength limits are
$$
\omega_-(q)=\ell\sqrt{\frac{KK'}{2m(K+K')}}|q|+O(|q|^3),\qquad
\boxed{\omega_+(0)=\sqrt{\frac{2(K+K')}{m}}\simeq\sqrt{\frac{2K}{m}}\left(1+\frac{K'}{2K}\right).}
$$
Thus the leading frequency in the <strong-bond limit of the alternating-spring chain> is $\sqrt{2K/m}$, the internal stretching frequency of a pair with reduced mass $m/2$.
\Image[/past-exam-of-the-quantum-condensed-matter-course-of-the-university-of-cambridge/2018-q1a-alternating-springs.png]
{title=Acoustic and optical branches of an alternating-spring chain}
{description=Original calculated dispersions for $K'/K=0.2$, using the two-atom primitive cell. Frequencies are in units of $\sqrt{K/m}$.}
Back to article page
