---
title: "Drude model"
sort_title: "Drude model"
date: 2021-09-23
categories:
- Physics
- Electromagnetism
- Optics
layout: "concept"
---

The **Drude model**, also known as
the **Drude-Lorentz model** due to its analogy
to the [Lorentz oscillator model](/know/concept/lorentz-oscillator-model/)
classically predicts the [dielectric function](/know/concept/dielectric-function/)
and electric conductivity of a gas of free charges,
as found in metals and doped semiconductors.



## Metals

In a metal, the conduction electrons can roam freely.
When an [electromagnetic wave](/know/concept/electromagnetic-wave-equation/)
passes by, its oscillating [electric field](/know/concept/electric-field/)
$$\vb{E}(t) = \vb{E}_0 e^{- i \omega t}$$ exerts a force on those electrons,
so the displacement $$\vb{x}(t)$$ of a particle from its initial position
obeys this equation of motion:

$$\begin{aligned}
    m \dvn{2}{\vb{x}}{t}
    = q \vb{E} - \gamma m \dv{\vb{x}}{t}
\end{aligned}$$

Where $$m$$ and $$q < 0$$ are the mass and charge of the electron.
The first term is Newton's third law,
and the last term represents a damping force
slowing down the electrons at rate $$\gamma$$.

Inserting the ansatz $$\vb{x}(t) = \vb{x}_0 e^{- i \omega t}$$
and isolating for the displacement $$\vb{x}$$, we find:

$$\begin{aligned}
    \vb{x}(t)
    = \vb{x}_0 e^{- i \omega t}
    = - \frac{q \vb{E}}{m (\omega^2 + i \gamma \omega)}
\end{aligned}$$

The polarization density $$\vb{P}(t)$$ is therefore as shown below.
Note that the dipole moment vector $$\vb{p}$$ is defined
as pointing from negative to positive,
whereas the electric field $$\vb{E}$$ goes from positive to negative.
Let $$N$$ be the metal's electron density, then:

$$\begin{aligned}
    \vb{P}(t)
    = N \vb{p}(t)
    = N q \vb{x}(t)
    = - \frac{N q^2}{m (\omega^2 + i \gamma \omega)} \vb{E}(t)
\end{aligned}$$

The electric displacement field $$\vb{D}(t)$$ is then as follows,
where the parenthesized expression is the dielectric function
$$\varepsilon_r$$ of the material:

$$\begin{aligned}
    \vb{D}
    = \varepsilon_0 \vb{E} + \vb{P}
    = \varepsilon_0 \bigg( 1 - \frac{N q^2}{\varepsilon_0 m} \frac{1}{\omega^2 + i \gamma \omega} \bigg) \vb{E}
    = \varepsilon_0 \varepsilon_r \vb{E}
\end{aligned}$$

From this, we define the **plasma frequency** $$\omega_p$$
at which the conductor "resonates",
leading to so-called **plasma oscillations** of the electron density
(see also [Langmuir waves](/know/concept/langmuir-waves/)):

$$\begin{aligned}
    \varepsilon_r(\omega)
    = 1 - \frac{\omega_p^2}{\omega^2 + i \gamma \omega}
    \qquad\qquad
    \boxed{
        \omega_p
        \equiv \sqrt{\frac{N q^2}{\varepsilon_0 m}}
    }
\end{aligned}$$

Suppose that $$\gamma = 0$$,
then we can identify three distinct scenarios for $$\varepsilon_r$$ here:

*   $$\omega < \omega_p$$, so $$\varepsilon_r < 0$$,
    so the refractive index $$\sqrt{\varepsilon_r}$$ is imaginary,
    meaning high absorption and high reflectivity
    (due to the large complex index difference between media).
*   $$\omega = \omega_p$$, so $$\varepsilon = 0$$,
    allowing for self-sustained plasma oscillations.
*   $$\omega > \omega_p$$, so $$\varepsilon_r > 0$$,
    so the index $$\sqrt{\varepsilon}$$ is real and asymptotically goes to $$1$$,
    leading to high transparency and low reflectivity from air.

For most metals $$\omega_p$$ is ultraviolet,
which explains why they typically appear shiny to us.
In reality $$\gamma > 0$$, reducing the reflectivity somewhat when $$\omega < \omega_p$$.

The Drude model also lets us calculate the metal's conductivity.
We already have an expression for $$\vb{x}(t)$$,
which we differentiate to get the velocity $$\vb{v}(t)$$:

$$\begin{aligned}
    \vb{v}(t)
    = \dv{\vb{x}}{t}
    = - i \omega \vb{x}
    = \frac{i \omega q \vb{E}}{m (\omega^2 + i \gamma \omega)}
    = \frac{q \vb{E}}{m (\gamma - i \omega)}
\end{aligned}$$

Consequently the average current density $$\vb{J}(t)$$ is found to be:

$$\begin{aligned}
    \vb{J}(t)
    = N q \vb{v}(t)
    = \sigma \vb{E}(t)
\end{aligned}$$

Where $$\sigma(\omega)$$ is the **AC conductivity**,
which depends on the **DC conductivity** $$\sigma_0$$:

$$\begin{aligned}
    \boxed{
        \sigma(\omega)
        = \frac{\gamma \sigma_0}{\gamma - i \omega}
    }
    \qquad\qquad
    \boxed{
        \sigma_0
        \equiv \frac{N q^2}{\gamma m}
    }
\end{aligned}$$

Recall that $$\gamma$$ measures friction.
Specifically, Drude assumed that the electrons often collide with obstacles,
each time resetting their momentum to zero;
in that case $$\vb{v}$$ should be interpreted as the average "drift"
of many electrons in an ensemble.
The mean time between those collisions is
the **momentum scattering time** $$\tau \equiv 1 / \gamma$$, so:

$$\begin{aligned}
    \sigma(\omega)
    = \frac{\sigma_0}{1 - i \omega \tau}
    \qquad\qquad
    \sigma_0
    = \frac{N q^2 \tau}{m}
\end{aligned}$$

After defining all those quantities,
the dielectric function $$\varepsilon_r(\omega)$$ can be written as:

$$\begin{aligned}
    \boxed{
        \begin{aligned}
            \varepsilon_r(\omega)
            &= 1 - \frac{\omega_p^2}{\omega^2 + i \gamma \omega}
            \\
            &= 1 + \frac{i \sigma(\omega)}{\varepsilon_0 \omega}
        \end{aligned}
    }
\end{aligned}$$



## Doped semiconductors

Doping a semiconductor introduces
free electrons (n-type doping) or free holes (p-type doping),
which can be treated as free charge carriers moving through the material,
so the Drude model is also relevant in this case.

We must replace the carriers' true mass $$m$$ with their *effective mass* $$m^*$$
found from the material's electronic band structure.
Furthermore, semiconductors already have
a high intrinsic dielectric function $$\varepsilon_{\mathrm{int}}$$
before being doped, so the displacement field $$\vb{D}(t)$$ becomes:

$$\begin{aligned}
    \vb{D}
    = \varepsilon_0 \vb{E} + \vb{P}_{\mathrm{int}} + \vb{P}_{\mathrm{free}}
    = \varepsilon_0 \varepsilon_{\mathrm{int}} \vb{E} - \frac{N q^2}{m^* (\omega^2 + i \gamma \omega)} \vb{E}
    = \varepsilon_0 \varepsilon_r \vb{E}
\end{aligned}$$

Where $$\vb{P}_{\mathrm{int}}$$ is the intrinsic polarization before doping,
and $$\vb{P}_{\mathrm{free}}$$ is the expression we calculated above for metals.
The dielectric function $$\varepsilon_r(\omega)$$ is therefore given by:

$$\begin{aligned}
    \boxed{
        \varepsilon_r(\omega)
        = \varepsilon_{\mathrm{int}} \bigg( 1 - \frac{\omega_p^2}{\omega^2 + i \gamma \omega} \bigg)
    }
\end{aligned}$$

Where the plasma frequency $$\omega_p$$ has been redefined as follows
to include $$\varepsilon_\mathrm{int}$$:

$$\begin{aligned}
    \boxed{
        \omega_p
        \equiv \sqrt{\frac{N q^2}{\varepsilon_0 \varepsilon_{\mathrm{int}} m^*}}
    }
\end{aligned}$$

The meaning of $$\omega_p$$ is the same as for metals,
but the free carrier density $$N$$ is typically lower in this case,
so $$\omega_p$$ is usually infrared rather than ultraviolet.

Furthermore, instead of $$\varepsilon_r \to 1$$
for $$\omega \to \infty$$ like a metal,
now $$\varepsilon_r \to \varepsilon_\mathrm{int}$$.
Along the way, there is a point where $$\varepsilon_r = 1$$
and the reflectivity becomes zero. This occurs at:

$$\begin{aligned}
    \omega^2
    = \frac{\varepsilon_{\mathrm{int}}}{\varepsilon_{\mathrm{int}} - 1} \omega_p^2
\end{aligned}$$

If $$N$$ and $$\varepsilon_\mathrm{int}$$ are known,
this can be used to experimentally determine $$m^*$$
by finding which value of $$\omega_p$$ would lead to the measured zero-reflectivity point.



## References
1.  M. Fox,
    *Optical properties of solids*, 2nd edition,
    Oxford.
2.  S.H. Simon,
    *The Oxford solid state basics*,
    Oxford.