Question 3 of 7: Diatomic-Chain Phonon Dispersion — Optical and Acoustical Branches

Question 3: Diatomic-Chain Phonon Dispersion — Optical and Acoustical Branches (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Diatomic linear chain, masses $M_1 > M_2$, nearest-neighbour force constant $C$, characteristic equation (3) above. Figure P3b: optical branch runs from $\omega_1$ at $K=0$ down to $\omega_2$ at $K=\pi/a$; acoustical branch runs from $0$ at $K=0$ up to $\omega_3$ at $K=\pi/a$.

Find. (a) $\omega_1,\omega_2,\omega_3$ in terms of $C,M_1,M_2$; (b) the physical origin of the optical/acoustical labels.

Approach. Solve the quadratic (3) for $\omega^2$ at the two symmetry points $K=0$ and $K=\pi/a$ where $\cos Ka$ takes the simple values $1$ and $-1$; part (b) is answered from the relative-phase pattern of $u,v$ in each branch, read from Figure P3c.

1. Part (a), zone centre $K=0$. With $\cos(0)=1$, eq. (3) reduces to $M_1M_2\omega^4-2C(M_1+M_2)\omega^2=0$, i.e. $\omega^2\big[M_1M_2\omega^2-2C(M_1+M_2)\big]=0$. The root $\omega^2=0$ is the acoustical branch (rigid translation, no restoring force); the non-zero root is the optical branch's top: $$\boxed{\omega_1=\sqrt{2C\left(\dfrac{1}{M_1}+\dfrac{1}{M_2}\right)}}$$
2. Part (a), zone boundary $K=\pi/a$. With $\cos(\pi)=-1$, eq. (3) becomes $M_1M_2\omega^4-2C(M_1+M_2)\omega^2+4C^2=0$. Solving this quadratic in $\omega^2$ by the quadratic formula and simplifying the discriminant, $\sqrt{(M_1+M_2)^2-4M_1M_2}=\sqrt{(M_1-M_2)^2}=M_1-M_2$ (since $M_1 > M_2$): $$\omega^2=\frac{C(M_1+M_2)\pm C(M_1-M_2)}{M_1M_2}\ \Longrightarrow\ \omega^2=\frac{2C}{M_2}\ \text{ or }\ \frac{2C}{M_1}$$ Since $M_2 < M_1$, the root $2C/M_2$ is the larger of the two and belongs to the (upper) optical branch, while $2C/M_1$ belongs to the (lower) acoustical branch, consistent with Figure P3b: $$\boxed{\omega_2=\sqrt{2C/M_2}\ \text{(optical)},\quad \omega_3=\sqrt{2C/M_1}\ \text{(acoustical)}}$$
3. Part (b) — why “optical” and “acoustical.” In the acoustical mode (long-wavelength limit), $u\approx v$: the two atoms of the basis move essentially in phase, exactly like the compressive wave motion of ordinary sound — hence the name, and $\omega\to0$ as $K\to0$ just as for a sound wave. In the optical mode, the amplitude ratio at $K=0$ is $u/v=-M_2/M_1$ (from eq. 1 with $\omega=\omega_1$): the two sublattices move out of phase, oscillating against one another about a stationary centre of mass. If the two atoms carry opposite charges (Figure P3c), this out-of-phase motion creates an oscillating electric dipole moment that couples directly to a transverse electromagnetic wave of the same frequency — i.e. the mode can be driven by (and radiates/absorbs) infrared light, which is the origin of the term “optical.”