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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.

The conduction and valence dispersions are two cones touching at the Dirac point, with slopes $\pm \hslash v_F$. Here $\mathbf{k}$ is measured relative to one valley centre. Each cone has a density of states proportional to $|E|$, which vanishes at the touching energy. Gate-induced carriers satisfy approximately $n=C_g(V_g-V_D)/e$ per unit area. Negative and positive offsets from $V_D$ produce holes and electrons, respectively. Away from neutrality a simple charge carrier mobility model gives $\sigma \simeq e\mu |n|$, so resistance decreases on either side of a maximum near $V_D$. Disorder, electron–hole puddles, contacts and quantum transport round the peak; the cone dispersion alone fixes neither its height nor its width, nor does it require exact electron–hole symmetry in a real device.

A direct derivation of the magnetic spectrum starts from the single-valley Dirac Hamiltonian operator $H=v_F({\sigma }_x{\pi }_x+{\sigma }_y{\pi }_y)$. Introduce oscillator operators using the magnetic length ${\ell }_B=\sqrt{\hslash /(e|B|)}$. Up to exchanging sublattices or valleys,Squaring gives adjacent oscillator number operators, henceThe positive expression printed in the source describes the electron branch only; the negative-energy partners must also be included. The same square-root dependence follows from quantizing circular orbit area, with the Dirac Berry phase cancelling the usual half-integer offset.

Each spin and valley contributes $Ae|B|/h$ states in area $A$. Including two spins and two valleys gives $N_{\mathrm{LL}}=4Ae|B|/h$, or areal degeneracy $4e|B|/h$. The zero-energy level is half-filled at charge neutrality: the carrier density counts departures from that state.

For $n=10^{16}{\mathrm{m}}^{-2}$, the filling factor and fourfold graphene Hall sequence areIn a plateau, ${\rho }_{xx}\simeq 0$ and $|{\rho }_{xy}|=h/({\nu }_N{e}^2)$. The Hall sign reverses between electron and hole doping; the sketch uses its magnitude to avoid tensor-index sign conventions. Representative plateau-centre fields are $20.68$, $6.893$, $4.136$ and $2.954\mathrm{T}$ for $\nu =2,6,10,14$. Between plateaux, crossing extended states in broadened levels gives longitudinal peaks and Hall steps. Their ideal filling-centre positions have $\nu =4j$, namelyThus peaks near $10.34$, $5.17$ and $3.45\mathrm{T}$ separate the first prominent plateaux as field increases. Many weak oscillations accumulate toward zero field and are damped there. Finite-width plateaux require localized states between extended-state energies; an exactly clean Landau spectrum does not by itself prescribe plateau widths. The illustration uses an explicitly schematic broadening, not inferred experimental widths.

At $6.5\mathrm{T}$, $\nu =6.36$, close to the $\nu =6$ plateau. With $v_F=10^6\mathrm{m}{\mathrm{s}}^{-1}$, $E_1=92.5\mathrm{meV}$ and $E_2=130.8\mathrm{meV}$. The relevant adjacent gap near the chemical potential is therefore $\Delta =E_2-E_1\simeq 38.3\mathrm{meV}$, rather than the gap from the zero level. Resolving levels requires quantum-lifetime broadening $\hslash /{\tau }_q\lesssim \Delta$ and thermal broadening appreciably smaller than $\Delta$, giving the order-of-magnitude limitsThe upper temperature is an energy-scale estimate, not a guaranteed sharp-feature temperature. For oscillation amplitudes the thermal damping factor is $X/\sinh X$, $X=2{\pi }^2{k}_{B}T/(\hslash {\omega }_c^{\ast })$, with $\hslash {\omega }_c^{\ast }=\hslash eB{v}_F^2/E_F\simeq 36.7\mathrm{meV}$ and $E_F=\hslash v_F\sqrt{\pi n}$. Requiring $X\lesssim 1$ gives the more conservative scale $T\lesssim 22\mathrm{K}$ for little thermal damping. Visible weaker features may persist above that scale; their actual cutoff depends on resolution, disorder and temperature-dependent scattering. The transport lifetime and quantum lifetime also need not coincide.