# Some notes on dielectric functions 1

1. Why does metal reflect light ?

A simple reason is that there are lots of free electrons in metal. The light incident on the metal is electric field which drive the free electrons to oscillate in the same frequency with the light.

We start from drude model.

According to the simplest Drude model, the motion of the electron is governed by the external electric field E, and the probability of electron scattering within unit time, denoted as $\gamma$, which is usually within infrared frequency range.

Then the equation of motion of electron is then written in semi classical way:

$m_e \frac{ d^2 x }{d t^2} = -e E - m_e \gamma \frac{d x}{d t}$, where $x$ is the position of electron. Consider electric field changing in certain frequency $\omega$: $E = E_0 e^{i \omega t}$, then we can use ansatz: $x = x_0 (\omega) e^{i \omega t }$. Insert the ansatz into equation: $x_0 = \frac{eE_0}{\omega m_e (-i\gamma+\omega)}$.

Remember the polarization density $P$ is given by: $P(\omega) = n e x(\omega)$, where $n$ is the number density of free electrons in material, so

$P(\omega) = n \cdot (-e) \cdot x_0 e^{i \omega t} = \frac{ne^2}{\omega m_e (\omega-i\gamma)} E(\omega)$

and :

$D(\omega) = E(\omega) + 4\pi P(\omega) = (1- \frac{4\pi ne^2}{\omega m_e (\omega-i\gamma)}) E(\omega)$ (Pay attention that $E$ and $P$ has the same unit!!).

Define the plasma frequency: $\omega_P^2 = \frac{4\pi ne^2}{\omega m_e}$. Then:

$D(\omega) = (1- \frac{\omega_P^2}{\omega(\omega-i\gamma)}) E(\omega)$.

Now, we have the expression for the dielectric function for free electrons:

$\epsilon(\omega) = 1 - \frac{\omega_P^2}{\omega(\omega-i\gamma)}$

(NOTE: THIS EXPRESSION IS ONLY FOR FREE ELECTRONS, THE SCATTERING PROCESSES ARE DESCRIBED BY $\gamma$.)

Here are some particular magnitude of $n$ for different types of materials:

Metal: 10^22 ~ 10^23 cm^-3 (Potassium, Sodium, Copper) –> $\omega_P$ is on the order of 10^16 Hz, which is within ultraviolet range

Semimetals: 10^17~10^20 cm^-3 ( Arsenic, Antimony, Bismuth, Graphite) –> $\omega_P$ is around 10^14 Hz ( between infrared and Vis.)

Semiconductor: 10^13~10^15 cm^-3.

The dielectric function can rewritten as:

$\epsilon(\omega) = 1 - \frac{\omega_P^2}{\omega^2+\gamma^2} - i \frac{\omega_P^2 \gamma}{\omega(\omega^2+\gamma^2)}$.

(NOTE: if I assume the time dependence is $e^{-i\omega t}$, then it becomes: $\epsilon(\omega) = 1 - \frac{\omega_P^2}{\omega^2+\gamma^2} + i \frac{\omega_P^2 \gamma}{\omega(\omega^2+\gamma^2)}$. )

Now let’s consider visible light incident on metals.

For visible light, $\omega > 10 \gamma$. In this situation, we can neglect the imaginary part of $\epsilon(\omega)$, and the real part $\epsilon_r(\omega) = 1- \frac{\omega_P^2}{\omega^2+\gamma^2}$ is negative, because the plasma frequency is in the ultraviolet range, which has higher frequency than $\omega$.

Remember $\epsilon(\omega) = ( \hat{n} )^2$, where $\hat{n} = N+iK$ is refractive index. When $\epsilon$ is almost negative real number, we have $N \sim 0$.

In this case, the reflectance: $\mathscr{R} = |(\hat{n}-1)/(\hat{n}+1|^2 = \frac{(N-1)^2+K^2}{(N+1)^2+K^2} \sim 1$, which means the light is reflected back completely.