 Solution

Use physical fields proportional to the real part of $e^{-i\omega t}$. The Drude frequency response isThus the sketch is an even Lorentzian line shape, with maximum ${\sigma }_0^2$ at zero frequency, half that value at $|\omega |={\tau }^{-1}$, and an ${\omega }^{-2}$ tail. The positive-frequency half is shown in the preceding figure. Squaring the modulus differs from plotting either $|\sigma |$ or the real part of $\sigma$.

For a single carrier species, the plasma frequency and electrical conductivity satisfyThereforeThe corresponding relaxation time is $\tau \simeq 3.57\times 10^{-14}\mathrm{s}$. This is an angular frequency; the ordinary frequency is $\boxed{f_{1/2}={\omega }_{1/2}/(2\pi )\simeq 4.46\times 10^{12}\mathrm{Hz}}$. The units of electrical conductivity used in the calculation are ${\mathrm{\Omega }}^{-1}{\mathrm{m}}^{-1}$, consistently with the SI plasma-frequency relation.