Past exam of the physics A course of the University of Cambridge / 2017 / Paper 2 / Section C / C10 / Solution

A phonon is one quantum of a crystal's vibrational normal modes. In the harmonic approximation, each mode is a quantum harmonic oscillator with energy levels $(n+1/2)\hslash \omega$. A phonon carries excitation energy $\hslash \omega$ and crystal momentum $\hslash q$ modulo a reciprocal lattice vector. It is a collective lattice excitation, not one atom travelling along the chain. Phonons obey Bose-Einstein statistics; their number need not be conserved in thermal equilibrium.

Put type-A atoms at $2ja$ and type-B atoms at $(2j+1)a$. Their displacements $u_j,v_j$ obey Newton's second law,Choose normal modes $u_j=U{e}^{i(2jaq-\omega t)}$ and $v_j=V{e}^{i((2j+1)aq-\omega t)}$. Substitution givesA nonzero displacement vector requires zero determinant:Solving this quadratic in ${\omega }^2$ gives the diatomic-chain dispersion relation with equal springs,The primitive unit cell has length $2a$, so the first Brillouin zone is $|q|\le \pi /(2a)$.

The minus solution is the acoustic phonon branch. As $q\to 0$, ${\omega }_{-}$ tends to zero and $U=V$: adjacent atoms move together in a long-wavelength compression. The plus solution is the optical phonon branch, with nonzero zone-centre frequency and $m_{A}U+m_{B}V=0$: the two sublattices move against one another with fixed cell centre of mass. For oppositely charged ions, this relative displacement can couple to an electric field, motivating the optical name.

At the zone boundary the frequencies are $\sqrt{2\alpha /m_{\mathrm{heavy}}}$ and $\sqrt{2\alpha /m_{\mathrm{light}}}$. Different masses produce a gap; equal masses make the two branches meet, recovering a folded monatomic dispersion.

Put $S=m_A+m_B$ and use $\sin qa\simeq qa$. Expanding the square root givesThus, to leading nonzero orders,The absolute value ensures a nonnegative angular frequency for either direction. The speed of sound is the initial acoustic slope $v_s=a\sqrt{2\alpha /(m_A+m_B)}$.

The vertical axis of the original figure is angular frequency, in $10^{13}\mathrm{rad}{\mathrm{s}}^{-1}$, not ordinary frequency. Its zone-centre optical intercept is approximately ${\omega }_{+}(0)=4.8\times 10^{13}{\mathrm{s}}^{-1}$. Convert the relative atomic masses using $m_u=1.66054\times 10^{-27}\mathrm{kg}$:The long-wavelength inference of diatomic-chain parameters then givesNo factor $2\pi$ is needed because the graph already plots $\omega$.

Reading near the origin, an acoustic point is approximately $(q,\omega )=(1.9\times 10^9{\mathrm{m}}^{-1},0.9\times 10^{13}{\mathrm{s}}^{-1})$. The initial slope therefore estimatesCombining the slope with the inferred spring constant givesThis $a$ is the model's adjacent Na-plane to Cl-plane spacing; its cell length is $2a$. The slope is a graphical estimate, so values around $4.5$–$5.0\mathrm{km}{\mathrm{s}}^{-1}$ and spacings near $0.2\mathrm{nm}$ are consistent with the reading precision. A real ionic crystal has interactions beyond the scalar nearest-neighbour model; an intercept-and-slope fit need not reproduce the entire measured dispersion or independently measured structural spacing exactly.