The Lorentz permittivity of a bound-electron medium is

PHYSICS · Multi-part · Fermi University Physics · Optics & Photonics · Condensed Matter, Materials & Soft Matter · Wave propagation and light in media

What an Instrument Can See, Ch The optical constants of a medium

Problem

The Lorentz permittivity of a bound-electron medium is ϵ~(ω)=1+ωp2ω02−ω2−iγω\tilde\epsilon(\omega) = 1+\dfrac{\omega_p^2}{\omega_0^2-\omega^2-i\gamma\omega}, with ωp2=Ne2/ϵ0m\omega_p^2=\mathcal{N}e^2/\epsilon_0 m the plasma frequency fixed by the free-electron density. ((a)) Show that setting ω0=0\omega_0=0 recovers this chapter's Drude permittivity, ϵ~=1−ωp2/(ω2+iγω)\tilde\epsilon=1-\omega_p^2/(\omega^2+i\gamma\omega). ((b)) Gold's conduction electrons give ℏωp=9.0 eV\hbar\omega_p=9.0\,\mathrm{eV} and a bulk damping ℏγ=0.070 eV\hbar\gamma=0.070\,\mathrm{eV}. Using the same 633 nm633\,\mathrm{nm} photon energy as the silver calculation in this chapter, ℏω=1.959 eV\hbar\omega=1.959\,\mathrm{eV}, predict ϵ~\tilde\epsilon from the Drude model. ((c)) The measured value at 633 nm633\,\mathrm{nm} is ϵ~≈−11.7+1.2i\tilde\epsilon\approx-11.7+1.2i. By how much does the real part disagree with the Drude prediction, and why is the disagreement so much larger than the six per cent found for silver, given that gold's interband absorption edge lies near 2.4 eV2.4\,\mathrm{eV}, much closer to the photon energy than silver's edge near 4 eV4\,\mathrm{eV}?

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