Past exam of the condensed matter physics major option course of the University of Oxford / 2016 / 7

For $N$ ion pairs per unit volume, a relative displacement $x$ produces ionic polarization $P=Nqx$. Solving the driven Lorentz oscillator model with the source's $e^{-i\omega t}$ convention givesThe constant ${ϵ}_{\infty }$ includes the fast electronic response: infinity here means well above the phonon band but below electronic resonances. Writing $\Delta ϵ={ϵ}_s-{ϵ}_{\infty }$ makes the numerator $\Delta ϵ{\omega }_{\mathrm{TO}}^2$. In an undamped longitudinal mode the bulk displacement field vanishes without an external free charge, so a nonzero electric field requires ${ϵ}_r({\omega }_{\mathrm{LO}})=0$. ThereforeThis is the Lyddane-Sachs-Teller relation for the single polar mode.

At normal incidence from vacuum the Fresnel equations giveBelow the transverse resonance the dielectric response increases, raising reflectivity. Between the transverse and longitudinal frequencies the undamped dielectric function is negative, giving an evanescent transmitted wave and nearly total reflection: the Reststrahlen band. Above the longitudinal edge it becomes small and positive, passing through ${ϵ}_r\simeq 1$ at the reflectivity minimum and eventually approaching its electronic background.

Reading the original graph to its limited precision gives $\hslash {\omega }_{\mathrm{TO}}\simeq 30\mathrm{meV}$ and $\hslash {\omega }_{\mathrm{LO}}\simeq 37\mathrm{meV}$, hence ${\omega }_{\mathrm{TO}}\simeq 4.6\times 10^{13}{\mathrm{s}}^{-1}$ and ${\omega }_{\mathrm{LO}}\simeq 5.6\times 10^{13}{\mathrm{s}}^{-1}$. Extrapolating the low-energy reflectivity to roughly $R(0)\simeq 0.25$ givesValues around $9$–$10$ and $5.5$–$6.5$ are appropriate to this graphical estimate. Neither the dip near $39\mathrm{meV}$ nor the finite-energy right endpoint represents the high-frequency asymptote; treating that endpoint as $R(\infty )$ would underestimate ${ϵ}_{\infty }$.

At $\omega ={\omega }_{\mathrm{TO}}$, ${ϵ}_r={ϵ}_{\infty }+iA$ with $A=\Delta ϵ{\omega }_{\mathrm{TO}}/\gamma$. Weak damping gives $A\gg {ϵ}_{\infty }$, so for the complex refractive index $n+i\kappa$,Squaring and substituting $A$ provesUse the reflectivity at the estimated transverse resonance, not automatically the plateau maximum. A representative reading $R_{\mathrm{TO}}\simeq 0.9$, with $\Delta ϵ\simeq 3.1$, gives $\gamma /{\omega }_{\mathrm{TO}}\simeq 0.0043$, $\gamma \simeq 2.0\times 10^{11}{\mathrm{s}}^{-1}$ andGraphical uncertainty in the transverse edge makes this only a few-picosecond estimate: readings $R_{\mathrm{TO}}\simeq 0.85$–$0.93$ give approximately $2$–$11\mathrm{ps}$. The displacement amplitude decays as $e^{-\gamma t/2}$ and has lifetime $2/\gamma$; phonon occupation/energy decays as $e^{-\gamma t}$ and has lifetime $1/\gamma$. Anharmonic forces allow an optical phonon to decay into two lower-energy phonons when energy and crystal momentum are conserved. Spontaneous decay can persist at low temperature; defects and electron–phonon scattering can add further linewidth.