= Solution
Use the $e^{-i\omega t}$ convention, so passive <photon absorption> has positive imaginary <permittivity>. For a single <Lorentz oscillator model> <resonance>, write the <photon> <energy> as $E=\hbar\omega$ and
$$
\epsilon_r(E)=\epsilon_b+\frac{S}{E_0^2-E^2-i\Gamma E},\qquad E_0=2\,\mathrm{eV},\quad S>0,\quad\Gamma>0.
$$
Here $S$ has units of <energy> squared, $\Gamma=\hbar\gamma$ is a <damping> <energy>, and $\epsilon_b$ is an approximately constant background. The corresponding real and imaginary parts are
$$
\epsilon'_r(E)=\epsilon_b+\frac{S(E_0^2-E^2)}{(E_0^2-E^2)^2+\Gamma^2E^2},\qquad \epsilon''_r(E)=\frac{S\Gamma E}{(E_0^2-E^2)^2+\Gamma^2E^2}.
$$
Thus the <real part> starts above the background, rises on the low-energy side of the <resonance>, drops rapidly through $\epsilon'_r=\epsilon_b$ at $2\,\mathrm{eV}$, and approaches the background from below at larger <energy>. The imaginary part starts at zero, has a positive <photon absorption> peak near $2\,\mathrm{eV}$, and decays above it. For weak <damping> its peak is close to the <resonance>, but not exactly fixed there for arbitrary <damping>. At asymptotically large <energy> the corrections scale as $\epsilon'_r-\epsilon_b\sim-S/E^2$ and $\epsilon''_r\sim S\Gamma/E^3$.
\Image[/past-exam-of-the-quantum-condensed-matter-course-of-the-university-of-cambridge/2013-lorentz-response.png]
{title=Illustrative Lorentz dielectric response across a 2 eV resonance}
\b[The <resonance> position is fixed at 2 eV; peak height, width and background are not specified.] The figure chooses an explicit finite <damping> and oscillator strength only to illustrate the shape over 0–5 eV. In particular, a negative real <permittivity> above <resonance> is possible for suitable strength and <damping>, but is not forced by the <resonance> <energy> alone. Setting <damping> exactly to zero would give a singular idealized response instead of a finite-width <photon absorption> peak.
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