 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 $E=\hslash \omega$ andHere $S$ has units of energy squared, $\Gamma =\hslash \gamma$ is a damping energy, and ${ϵ}_b$ is an approximately constant background. The corresponding real and imaginary parts areThus the real part starts above the background, rises on the low-energy side of the resonance, drops rapidly through ${ϵ}_r^{\prime }={ϵ}_b$ 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 ${ϵ}_r^{\prime }-{ϵ}_b\sim -S/E^2$ and ${ϵ}_r^{\prime \prime }\sim S\Gamma /E^3$.

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.