For equal masses $m$ separated by $a$ and alternating spring constants $\lambda$ and $\alpha\lambda$, the primitive-cell length is $2a$ and the two branches are
$$
\omega_\pm^2(k)=\frac{\lambda}{m}
\left[1+\alpha\pm\sqrt{(1-\alpha)^2+4\alpha\cos^2(ka)}\right].
$$
The minus and plus signs give the <acoustic phonon branch> and <optical phonon branch>, respectively.

Alternating-spring chain dispersion relation

Label the two atoms in primitive cell $j$ by $u_{2j}$ and $u_{2j+1}$, with $j$ understood modulo $N$ because of the <periodic boundary condition>. Choose the spring between them to have constant $\lambda$. <Newton's second law> gives
$$
\begin{aligned}
m\ddot u_{2j}
&=\lambda(u_{2j+1}-u_{2j})
+\alpha\lambda(u_{2j-1}-u_{2j}),\\
m\ddot u_{2j+1}
&=\lambda(u_{2j}-u_{2j+1})
+\alpha\lambda(u_{2j+2}-u_{2j+1}).
\end{aligned}
$$

For a longitudinal <normal mode>, put
$$
u_{2j}=A e^{i(2jka-\omega t)},
\qquad
u_{2j+1}=B e^{i((2j+1)ka-\omega t)}.
$$
Periodicity requires $e^{2iNka}=1$, so $k=\pi r/(Na)$. Since the primitive-cell length is $2a$, wavevectors differing by $\pi/a$ describe the same mode, and the first <Brillouin zone> may be chosen as
$$
\boxed{-\frac{\pi}{2a}\leq k<\frac{\pi}{2a}.}
$$
The amplitudes obey
$$
\begin{pmatrix}
\lambda(1+\alpha)-m\omega^2&
-\lambda(e^{ika}+\alpha e^{-ika})\\
-\lambda(e^{-ika}+\alpha e^{ika})&
\lambda(1+\alpha)-m\omega^2
\end{pmatrix}
\begin{pmatrix}A\\B\end{pmatrix}=0.
$$
Setting its <determinant> to zero gives the <alternating-spring chain dispersion relation>
$$
\boxed{
\omega_\pm^2(k)=\frac{\lambda}{m}
\left[1+\alpha\pm
\sqrt{(1-\alpha)^2+4\alpha\cos^2(ka)}\right].}
$$
The lower sign is the <acoustic phonon branch> and the upper sign is the <optical phonon branch>.

At the edge of the <Brillouin zone>,
$$
\omega_\pm^2=\frac{\lambda}{m}
\left(1+\alpha\pm|1-\alpha|\right),
$$
so the frequency <band gap> is
$$
\boxed{\Delta\omega=
\sqrt{\frac{2\lambda}{m}}\,|1-\sqrt\alpha|.}
$$
It vanishes at $\alpha=1$, when the apparent two-atom primitive cell can be reduced to one atom.

Near the centre of the <Brillouin zone>, a <Taylor expansion> gives
$$
\boxed{
\omega_-(k)=
a\sqrt{\frac{2\alpha\lambda}{m(1+\alpha)}}\,|k|
+O(|k|^3),}
$$
and
$$
\boxed{
\omega_+(k)=
\sqrt{\frac{2\lambda(1+\alpha)}m}
\left[1-\frac{\alpha a^2k^2}{2(1+\alpha)^2}
+O(k^4)\right].}
$$
The <group velocity> of the acoustic branch at long wavelength is the <speed of sound>,
$$
\boxed{c_s=a\sqrt{\frac{2\alpha\lambda}{m(1+\alpha)}}.}
$$

\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ii/paper-2-alternating-spring-dispersion.png]
{title=Dispersion relation of the alternating-spring chain}
{description=The acoustic and optical branches for alpha equal to 0.35. The vertical separation at the Brillouin-zone edge is the frequency gap.}
{height=500}

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 2 / 35A / Solution

An optical phonon branch has a nonzero frequency as $k\to0$, corresponding to relative motion of degrees of freedom within one primitive cell.


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 Incoming links: Optical phonon branch