Question 3 of 7: Diatomic Linear Chain — Optical and Acoustical Phonon Branches
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — 17-Phys-A6: Solid State Physics — May 2019 (3 hours; closed book; useful equations, physical constants and a marking scheme are annexed to the paper). Any FIVE of the SEVEN questions constitute a complete exam paper (the first five as they appear in the answer book are marked); all seven are solved below as a complete study resource.
Reference texts: C. Kittel, Introduction to Solid State Physics, 8th ed. (Ch. 1–3, 4, 6, 8, 14, 18); N. W. Ashcroft & N. D. Mermin, Solid State Physics, 1st ed. (Ch. 2, 4–7, 22, 28, 31–32).
Question 3: Diatomic Linear Chain — Optical and Acoustical Phonon Branches (20 marks)
Given. Equal atom mass $M$ on every site; alternating force constants $C_1=C$ (weak) and $C_2=10C$ (strong); equations (1)–(2) above for the two-sublattice traveling-wave ansatz; the dispersion is required at the zone centre $K=0$ and zone boundary $K=\pi/a$ (Figure P3b shows three band-edge frequencies $\omega_1,\omega_2,\omega_3$).
Find. (a) $\omega_1=\omega_{opt}(K=0)$, $\omega_2=\omega_{opt}(K=\pi/a)$, $\omega_3=\omega_{ac}(K=\pi/a)$ (the acoustic branch is $\omega=0$ at $K=0$, not separately labelled); (b) the physical origin of the optical/acoustical names.
Fig. 3 — Optical (upper) and acoustical (lower) branches for $C_2/C_1=10$, in units of $\sqrt{C/M}$: $\omega_1=\sqrt{22C/M}$ at $K=0$, $\omega_2=\sqrt{20C/M}$ and $\omega_3=\sqrt{2C/M}$ at $K=\pi/a$.
Approach. Set the determinant of eqs. (1)–(2) to zero to get $\omega^2(K)$ in closed form, then evaluate at $K=0$ and $K=\pi/a$.
Part (a) — determinant condition. Writing $D=(C_1+C_2)-\omega^2M$ and $Z=C_1+C_2e^{-iKa}$ (so the second equation's bracket is $Z^{*}$), the coefficient determinant is $D^2-|Z|^2=0$, i.e. $D=\pm|Z|$ with $|Z|^2=C_1^2+C_2^2+2C_1C_2\cos Ka$. Hence
$$\begin{aligned}
\Delta(K)&=\sqrt{C_1^2+C_2^2+2C_1C_2\cos Ka}\\
\omega^2(K)&=\frac{(C_1+C_2)\mp\Delta(K)}{M}
\end{aligned}$$
the minus sign giving the lower (acoustical) branch, the plus sign the upper (optical) branch (this is the standard reduction; the sign choice is fixed by requiring $\omega_{ac}(K{=}0)=0$). At $K=0$ ($\cos Ka=1$): $\sqrt{(C_1+C_2)^2}=C_1+C_2$, so $\omega_{ac}=0$ and
$$\omega_1=\omega_{opt}(0)=\sqrt{\frac{2(C_1+C_2)}{M}}=\sqrt{\frac{2(C+10C)}{M}}=\sqrt{\frac{22C}{M}}$$
At $K=\pi/a$ ($\cos Ka=-1$): $\sqrt{(C_1-C_2)^2}=|C_1-C_2|=C_2-C_1$ (since $C_2>C_1$), so
$$\begin{aligned}
\omega_2&=\omega_{opt}(\pi/a)=\sqrt{\frac{2C_2}{M}}=\sqrt{\frac{20C}{M}}=2\sqrt{\frac{5C}{M}}\\
\omega_3&=\omega_{ac}(\pi/a)=\sqrt{\frac{2C_1}{M}}=\sqrt{\frac{2C}{M}}
\end{aligned}$$
$$\boxed{\begin{aligned}
\omega_1&=\sqrt{22C/M}\\
\omega_2&=2\sqrt{5C/M}\\
\omega_3&=\sqrt{2C/M}
\end{aligned}}$$
Note the gap $\omega_2>\omega_3$ at the zone boundary is a direct consequence of $C_1\ne C_2$; if the two force constants were equal the two branches would meet.
Part (b) — why ‘optical’ and ‘acoustical’. In the acoustical branch, at long wavelength ($K\to0$) the two atoms in each unit cell move in phase, with the same amplitude and direction — the cell moves as a rigid unit, exactly like the compressions of an ordinary sound wave, and $\omega\to0$ as $K\to0$ (no restoring force for a uniform rigid translation of the whole crystal). In the optical branch, the two atoms move out of phase (opposite directions) even as $K\to0$, so the unit cell develops an oscillating dipole moment if the two atoms carry opposite effective charge — this is precisely the motion that couples to (and can be driven or detected by) infrared light, which is why this branch is called ‘optical’. It also has a nonzero frequency $\omega_1\ne0$ at $K=0$ because the two atoms are still oscillating against each other even at infinite wavelength.
