Optical Properties of Solids over a Wide Frequency Range

Abstract

In this chapter we discuss the optical properties of solids over a wide frequency range.

Problems

19.1

Suppose that we model the interband transitions in Ge as a step function $\varepsilon _2(\omega )=\varepsilon _l$ for $E_{min}< \hbar \omega < E_{max}$ (see diagram) and $\varepsilon _2(\omega )=0$ otherwise, as shown in Fig. 19.12.

Fig. 19.12

The sample step function model used in problem 19.1 for the interband transitions in Ge

• (a) Use the Kramers–Kronig relation to find an expression for $\varepsilon _1(\omega )$ for all $\omega$. Take $\varepsilon _1=1$ in the limit $\omega =\infty$ and express your answer in terms of $E_{max}$, $E_{min}$, and $\varepsilon _l$.

• (b) For which photon energies does $\varepsilon _1(\omega )$ exhibit structure? Is your answer physically reasonable and why?

• (c) Obtain an explicit expression for $\varepsilon _1(0)$ at zero frequency, and use this result to explain why narrow gap semiconductors tend to have large dielectric constants at $\omega =0$.

• (d) Use the Kramers–Kronig relations to show the sum rule

  $$ \frac{n e^2}{m} = \frac{1}{2\pi ^2} \int _0^\infty \varepsilon _2(\omega ) \omega d\omega $$

  where n is the total carrier density of the semiconductor at a temperature T.

19.2

Using the Kramers–Kronig relation

$$ \varepsilon _1(\omega _0) - 1 = \frac{2}{\pi } \int _0^\infty \frac{\omega '\varepsilon _2(\omega ') - \omega _0\varepsilon _2(\omega _0) }{\omega '^2 - \omega _0^2} d\omega ' $$

explain why $\varepsilon _1(0)$ at $\omega _0=0$ is so large for Si [$\varepsilon _0(\mathrm{Si}) = 12$] relative to glass for which $\varepsilon _0 < 3$.

19.3

According to Johnson and Christy’s paper on the Optical Constants of Noble Metals (see Johnson et al. (1972): 4370), the equation for the dielectric permittivity in the Drude free electron theory is given by

$$\begin{aligned} \varepsilon (\omega ) = 1 - \frac{\omega _p^2}{\omega (\omega + \frac{\i }{\tau })} \end{aligned}$$

where $\varepsilon (\omega )$ is the dielectric function at frequency $(\omega )$. Here, $\omega _p$ is the plasma frequency, and $\gamma =\frac{\i }{\tau }$ is the collision frequency. Use values for gold $m^* = 0.99m_0$, and $\tau = 9.3 \times 10^{-15} s$.

• (a) Plot the real and imaginary parts of the dielectric function ($\varepsilon _1$ and $\varepsilon _2$) for Au over the wavelength range from 400-1000 nm.

• (b) Using the Kramers–Kronig relation, calculate $\varepsilon _1$ from $\varepsilon _2$ and then calculate $\varepsilon _2$ from $\varepsilon _1$. Plot these together with the original functions from (a) over the 400–1000 nm wavelength range.

19.4

Use the Kramers–Kronig relation to calculate the real part of $\varepsilon (\omega )$, given the imaginary part of $\varepsilon (\omega )$ for positive $(\omega )$ for the two cases:

• (a)
  $$\frac{\varepsilon _2(\omega )}{\varepsilon _0} = \lambda [\theta (\omega - \omega _1) - \theta (\omega - \omega _2)], \omega _2> \omega _1 > 0 $$
• (b)
  $$\frac{\varepsilon _2(\omega )}{\varepsilon _0} = \frac{\lambda \gamma \omega }{(\omega _0^2 - \omega ^2)^2 + \gamma ^2\omega ^2} $$

19.5

Show that if a linear response function, such as the linear electric susceptibility $\chi (\omega )$ or the dielectric function $(\varepsilon (\omega ) - 1)$, satisfies the following two conditions: (1) it is analytic in the upper half of the complex $\omega$-plane and (2) it approaches zero sufficiently fast as $\omega$ approaches infinity, it satisfies the Kramers–Kronig relation.

19.6

An electromagnetic wave travels inside a dielectric material with a complex index of refraction $\tilde{N}=\tilde{n}+i\tilde{k}=c/c_1$, and is reflected at the plane boundary (xy-plane). If the E vector of the incident wave is in the x-direction, the reflection coefficient for the E field is

$$ r = \frac{\tilde{n}\cos \theta _1 - \cos \theta _2}{\tilde{n}\cos \theta _1 + \cos \theta _2} $$

and if the H vector is in the x-direction, the reflection coefficient is

$$ r' = \frac{\cos \theta _1 - \tilde{n}\cos \theta _2}{\cos \theta _1 + \tilde{n}\cos \theta _2} $$

• (a) Use boundary conditions at the interface to derive the above equation for r.

• (b) Suppose that $\theta _1$ is made large enough so that total internal reflection occurs. With the aid of the above equations, find the phase angles $\phi$ and $\phi '$ of the reflected waves ${ E_r}$ and ${ E_r'}$ in each of the two cases, defining the phase angle of the incident wave to be zero at the boundary. (Note that $\cos \theta$ is imaginary under conditions of total internal reflection.) Show that

  $$ \tan \left( \frac{\phi ' - \phi }{2}\right) = \frac{\cos \theta _1 \sqrt{\sin ^2\theta _1 - (1/\tilde{n}^2)}}{\sin ^2\theta _1}. $$

• (c) For a linearly polarized incident wave, it is possible for the reflected wave to be circularly polarized. If this is possible, what must be the polarization direction of the incident wave? Write an equation that determines the required angle of incidence?

• (d) Determine the smallest value of the index of refraction for which (b) is possible, and find the corresponding angle of incidence.

19.7

• (a) Suppose that we apply a magnetic field along the (001) direction normal to the surface of a sample. Find the dependence of $\varepsilon _1$ and $\varepsilon _2$ on the magnetic field. (Hint: Use of right and left circularly polarized fields will be helpful with this problem.)

• (b) Find the dependence of the plasma frequency on magnetic field using right and left incident circularly polarized light. Sketch the result of the magnetic field on the optical reflectivity for right and left circular polarized light.