Response at Finite Frequency: the Plasma Frequency
The equation of motion
Drude's model handles an oscillating field just as easily as a static one. Write
the average momentum \(\vec{p} = m\vec{v}_{d}\) and note that collisions damp it on
the timescale \(\tau\):
\begin{equation}
m\frac{\mathrm{d}\vec{v}}{\mathrm{d}t} = -e\vec{E} - \frac{m\vec{v}}{\tau} .
\end{equation}
With the field switched off this gives \(v(t) = v(0)e^{-t/\tau}\): the drift decays
exponentially, falling to \(1/e\) of its initial value in one relaxation time. That
is what \(\tau\) means operationally.
For a driving field \(\vec{E} = \vec{E}_{0}e^{-i\omega t}\), look for a steady
response \(\vec{v} = \vec{v}_{0}e^{-i\omega t}\). Substituting,
\begin{equation}
-i\omega m\vec{v}_{0} = -e\vec{E}_{0} - \frac{m\vec{v}_{0}}{\tau}
\quad\Longrightarrow\quad
\vec{v}_{0} = \frac{-e\tau\vec{E}_{0}}{m\left(1-i\omega\tau\right)},
\end{equation}
and since \(\vec{J} = -ne\vec{v}\),
\begin{equation}
\boxed{\;\sigma(\omega) = \frac{\sigma_{0}}{1-i\omega\tau}
= \frac{\sigma_{0}}{1+\omega^{2}\tau^{2}}
+ i\,\frac{\omega\tau\,\sigma_{0}}{1+\omega^{2}\tau^{2}},
\qquad \sigma_{0} = \frac{ne^{2}\tau}{m} . \;}
\end{equation}
Sign conventions in the AC response
The relation above goes with the convention
\(\vec{E}\propto e^{-i\omega t}\). If you write \(\vec{E}\propto e^{+i\omega t}\)
instead, the result is \(\sigma(\omega) = \sigma_{0}/(1+i\omega\tau)\). The two
are complex conjugates and describe identical physics, but mixing them —
quoting \(\sigma_{0}/(1-i\omega\tau)\) alongside \(e^{+i\omega t}\) — puts the sign
of the reactive part backwards, turning inductive behaviour into capacitive.
State the convention, then keep it.
Two limits matter.
Low frequency, \(\omega\tau\ll1\). Then \(\sigma\to\sigma_{0}\), real: the
metal responds in phase with the field and dissipates energy. Since
\(\tau\sim\text{10⁻¹⁴ s}\), this covers everything up to the far infrared.
High frequency, \(\omega\tau\gg1\). Then
\begin{equation}
\sigma(\omega) \to \frac{\sigma_{0}}{-i\omega\tau} = \frac{ine^{2}}{m\omega},
\end{equation}
purely imaginary. The current is 90° out of phase with the field and
no energy is absorbed: the electrons simply oscillate, too rapidly for collisions
to intervene. The metal behaves as a lossless, inertial medium.
The plasma frequency
That inertial response has a dramatic optical consequence. The dielectric
function of a medium with conductivity \(\sigma\) is
\(\epsilon(\omega) = 1 + i\sigma(\omega)/\epsilon_{0}\omega\), so in the
collisionless limit
\begin{equation}
\epsilon(\omega) = 1 - \frac{ne^{2}}{\epsilon_{0}m\omega^{2}}
= \boxed{\;1 - \frac{\omega_{p}^{2}}{\omega^{2}},
\qquad \omega_{p} = \sqrt{\frac{ne^{2}}{\epsilon_{0}m}} \;}
\end{equation}
where \(\omega_{p}\) is the
plasma frequency — the natural frequency at
which the whole electron gas oscillates against the ion background. Its quantum
is the plasmon of the table in Chapter 4.
Why metals are shiny, and where they stop being shiny
For \(\omega < \omega_{p}\), \(\epsilon < 0\), the refractive index is imaginary, and
an incident wave cannot propagate: it is totally reflected. For
\(\omega > \omega_{p}\), \(\epsilon > 0\) and the metal becomes transparent.
Since \(\omega_{p}\) for a typical metal corresponds to a photon energy of
\(5\)–15 eV, the crossover lies in the ultraviolet. Visible light
is below it and is reflected — hence metallic lustre — while the alkali
metals are genuinely transparent in the ultraviolet, one of the early
confirmations of the free-electron picture.
