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 	Copyright, 1995-2017, all rights reserved, Mark Jarrell (Dept.of 
 Physics and Atronomy, Louisiana State University, LA 70803).  This 
 material may not be reproduced for profit, modified or published in any 
 form (this includes electronic redistribution) without the prior written 
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}

\title{Chapter 11: Dielectric Properties of Materials}
\author{Lindhardt}
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\pagebreak
\Large

Electromagnetic fields are essential probes of material properties 
\begin{itemize} 
	\item IR absorption 
	\item Spectroscopy 
\end{itemize}
The interaction of the field and material may be described either classically 
or Quantum mechanically.  We will first do the former. 
 
\section{Classical Dielectric Response of Materials}

     Classically, materials are characterized by their dielectric response of 
either the bound or free charge.  Both are described by Maxwells equations 
\beq
	\grad \times \CE = -\frac 1c \pde Bt, \qquad \grad \times \H = 		
	\frac{4\pi }c{\bf j} + \frac 1c\pde Dt
\eeq
and Ohm's law 
\beq
	{\bf j} = \sigma \CE\,.
\eeq 
Both effects may be combined into an effective dielectric constant $\tilde{\ep 
}$, which we will now show.  For an isotropic medium, we have 
\begin{figure}[htb]
\centerline{\includegraphics[height=2.5in,keepaspectratio,clip=true]{wave.pdf}}
\caption[]{\em{If the average excursion of the electron is small compared to 
the wavelength of the radiation $<x>\ll \lambda$, then we may ignore the
wave-vector dependence of the radiation so that $\ep(\k,\om)\approx \ep(\om)$.}}
\end{figure}
\beq
	\D(\omega ) = \ep (\omega )\CE (\omega )
\eeq
where 
\beqa
	& \CE (t) = \int d\omega e^{-i\omega t}\CE (\omega ) \qquad & \H (t) = 	
		\int d\omega e^{-i\omega t}\H (\omega ) \\
	 & \D (t) = \int d\omega e^{-i\omega t}\D (\omega ) \qquad & \B (t) = 	
		\int d\omega e^{-i\omega t}\B (\omega )
\eeqa
\beq
	\CE (\omega ) = \CE ^*(-\omega ) \Rightarrow \ \CE (t) \in \ \Re 
\eeq
Then $\grad \times \H \ = \frac{4\pi}c{\bf j} + \frac 1c\pde Dt \Rightarrow $
 
\beq
	\grad \times \int d\omega e^{-i\omega t}\H (\omega ) = \frac{4\pi }c 	
		\int d\omega e^{-i\omega t}{\bf j}(\omega ) + \frac 1c 		
		\pde{}t\int d\omega e^{-i\omega t}\D (\omega )
\eeq
\beq
	\int d\omega e^{-i\omega t}\left\{ \grad \times \H (\omega ) - 		
		\frac{4\pi }c{\bf j}(\om ) - \frac 1c(-i\om )\D (\om )\right\} 	
		\ = 0 
\eeq
\beqa
	\grad \times \H \ & = & \frac{4\pi}c{\bf j} - i\frac{\omega }c\D 	
		(\omega ) \nonumber \\
	& = &\frac{4\pi \sigma}c\CE - i\frac{\om }c\ep\CE \nonumber\\
&=& \left\{ 		
		\begin{array}{ll}
			-i\frac{\om }c\CE \left( \ep - \frac c{\om }\frac{4\pi 	
				\sigma }c\right) \ = -\frac{i\om }c\CE 		
				\tilde{\ep } \\
			\frac{4\pi }c\CE \left( \sigma - \frac c{4\pi 		
				}\frac{i\omega }c\ep \right) \ = \frac{4\pi 	
				}c\CE \tilde{\sigma }
		\end{array} \right.
\eeqa
Thus we could either define an effective conductivity $\tilde{\sigma } = \sigma 
- \frac{i\om \ep}{4\pi }$ which takes into account dielectric effects, or an 
effective dielectric constant $\tilde{\ep } = \ep + i\frac{4\pi \sigma }{\om 
}$, which accounts for conduction. 
 
\subsection{ Conditions on $\ep $ }      
 
From the reality of $\D (t)$ and $\CE (t)$, one has that $\CE (+\om ) = \CE 
^*(-\om )$ and $\D (\om ) = \D ^*(-\om )$, hence for $\D = \ep \CE $, 
\beq
	\ep (\om ) = \ep ^*(-\om )
\eeq 
Additional constraints are obtained from causality 
\beqa
	\D (t) & = & \int d\om \ep (\om )\CE (\om )e^{-i\om t} \nonumber\\
	& = & \int d\om \ep (\om )e^{-i\om t}\int \frac{dt^{\prime }}{2\pi 	
		}e^{i\om t'}\CE (t^{\prime }) \nonumber\\
	& = & \frac 1{2\pi }\int dt d\om \left( \ep (\om ) - 1 + 1\right)\CE
		(t^{\prime })e^{-i\om (t - t^{\prime })}	
\eeqa	
then we make the substitution
\beq
	\chi (\om ) \equiv \ \frac{\ep - 1}{4\pi }
\eeq
\beqa
	\D (t) = & 2\int dt d\om \left( \chi (\om ) + \frac 1{4\pi }\right) \CE	
		(t^{\prime })e^{-i\om (t - t^{\prime })} \nonumber\\
	\D (t) = & \CE (t) + 2\int dt d\om \chi (\om )\CE (t^{\prime })e^{-\om 	
		(t - t^{\prime })}
\eeqa
Define $G(t) \equiv \ 2\int d\om \chi (\om )e^{-i\om t}$, then
\beq
	\D (t) = \CE (t) + \int _{-\infty}^{\infty}dt^{\prime} 			
		G(t-t^{\prime})\CE (t^{\prime})
\eeq 
Thus, the electric displacement at time $t$ depends upon the field at other 
times; however, it {\it cannot} depend upon times $t^{\prime } > t$ by 
causality. 
 Hence 
\beq
	G(\tau ) \equiv \ 0 \qquad \tau \ < 0
\eeq
\begin{figure}
\centerline{\includegraphics[height=2.25in,keepaspectratio,clip=true]{contour.pdf}}
\caption[]{\em{If $\chi (\om )$ is analytic in the upper half
plane, then causality is assured.}}
\end{figure}

This can be enforced if $\chi (\om )$ is analytic in the upper half plane, then 
we may close 
\beqa
	G(\tau ) & = & 2\int _{-\infty}^{\infty}d\om \chi (\om )e^{-i\om \tau } 
		\nonumber\\
	& = & 2 \int\!\!\!\!\!\:\!\circ d\om \chi (\om )e^{-i\om \tau } \equiv 	
		\ 0
\eeqa
contour in the upper half plane and obtain zero for the integral since $\chi $ 
is analytic within and on the contour. 
 
\subsection{ Kramer's Kronig Relations }

One may also derive an important relation between the real and imaginary
parts of the dielectric function $\ep(\omega)$ using this analytic property.  
From the Cauchy integral formula, if $\chi $ is analytic inside and on the 
contour $C$, then
\beq
	\chi (z) = \frac 1{2\pi i}{\int\!\!\!\!\!\:\!\circ }_C \frac{\chi (\om 	
		^{\prime})}{z-\om'}d\om ^{\prime}
\eeq
Then taking the contour shown in Fig.~\ref{fig:contour2}
\begin{figure}[htb]
\centerline{\includegraphics[height=1.75in,keepaspectratio,clip=true]{contour2.pdf}}
\caption[]{\em{The contour (left) used to demonstrate the Kramer's Kronig 
Relations.  Since the contour must contain $\om$, we must deform the contour 
so that it avoids the pole $1/(\om-\om')$.}}
\label{fig:contour2}
\end{figure} 
and assuming that $\chi (\om ) \alt \ \frac 1{\om}$ for large $\om $ (in fact 
${\bf Re} \chi \ \alt \ \frac 1{\om }$ and ${\bf Im} \chi \ \alt \frac 1{\om 
^2}$) we may ignore the large semicircle.  Let $z = \om + i 0^+$, so that 
\beq
	\chi (\om + i0^+) = \frac 1{2\pi i}\int _{-\infty}^{\infty} \frac{\chi 	
		(\om ^{\prime})d\om ^{\prime }}{\om - \om ^{\prime} - i0^+}
\eeq
\beq
	2\pi i\chi (\om + i0^+) = P \int _{-\infty}^{\infty} \frac{\chi (\om 	
		^{\prime})d\om ^{\prime}}{\om - \om ^{\prime}} + i\pi \chi (\om 
		)
\eeq
\beq
	\chi (\om ) = \frac 1{i\pi }P \int _{-\infty}^{\infty}d\om ^{\prime } 	
		\frac{\chi (\om ^{\prime})}{\om - \om ^{\prime}}
\eeq
Then let $4\pi \chi (\om ) = \ep (\om ) - 1 = \ep _1 + i\ep _2 - 1$ and we get 
\beq
	\ep _1(\om ) + i\ep _2(\om ) - 1 = \frac 2{\pi }P \int 			
		_{-\infty}^{\infty} \frac{-i\ep _1(\om ^{\prime}) + \ep _2(\om 	
		^{\prime}) + 1}{\om - \om ^{\prime}}d\om ^{\prime}
\eeq	
or 
\beq
	\ep _1(\om ) - 1 = \frac 1{\pi } P \int _{-\infty}^{\infty} \frac{\ep 	
		_2(\om ^{\prime})}{\om - \om ^{\prime}}d\om ^{\prime}
\label{eq:KK12}
\eeq
\beq
	\ep _2(\om ) = -\frac 1{\pi }P \int _{-\infty}^{\infty} \frac{\ep 	
		_1(\om ^{\prime}) - 1}{\om - \om ^{\prime}}d\om ^{\prime}
\eeq
which are known as the Kramer's Kronig relations.  Many experiments 
measure $\ep _2$ and from Eq.~\ref{eq:KK12} we can calculate $\ep _1$! 
 
 
\section{ Absorption of E and M radiation }
\subsection{Transmission}
\begin{figure}[htb]
\centerline{\includegraphics[height=2.9in,keepaspectratio,clip=true]{beam.pdf}}
\caption[]{\em{In transmission experiments a laser beam is focused on a thin 
slab of some material we wish to study.}}
\label{fig:beam}
\end{figure}

Imagine that a laser beam of known frequency is normally incident upon a thin 
slab of some material (see Fig.~\ref{fig:beam}), and we are able to measure its 
transmitted intensity.

     Upon passing through a boundary, part of the beam is reflected and part 
is transmitted (see Fig.~\ref{fig:beam2}).  In the case of normal incidence, it 
is easy to calculate the related coefficients from the conditions of 
continuity of $\CE _{\perp}$ and $\H _{\perp} = \B _{\perp} (\mu = 1)$
\begin{figure}[htb]
\centerline{\includegraphics[height=2.5in,keepaspectratio,clip=true]{beam2.pdf}}
\caption[]{\em{The assumed orientation of electromagnetic fields
incident on a surface.}}
\label{fig:beam2}
\end{figure}
\beqa
	\grad \times \CE \ = -\frac 1c \pde Bt, \qquad i\k \times \CE \ = 	
		\frac{i\om }c\B \\
	|n\CE _{\perp}| = |\B_{\perp}|
\eeqa
\beq
	\left. \begin{array}{ll}
		\CE_0 + \CE^{\prime \prime} = \CE^{\prime} \\
		(\CE_0 - \CE^{\prime \prime}) = n\CE^{\prime}
	\end{array} \right\} \quad t = \frac 2{n+1} 
\eeq
In this way the other coefficients may be calculated (see Fig.~\ref{fig:beam3}). 
\begin{figure}[htb]
\centerline{\includegraphics[height=2.0in,keepaspectratio,clip=true]{beam3.pdf}}
	\caption[]{\em{Multiple events contribute the the radiation
transmitted through and reflected from a thin slab.}}
\label{fig:beam3}
\end{figure}	 
Accounting for multiple reflections the total transmitted field is 
\beq
	\CE \ = \CE _0t_1t_2e^{ik d} + \CE _0t_1r_2r_2t_2e^{3 ik d} + 	
		\cdots, \qquad \mbox{where } k = \frac{\tilde{n}\om }c
\eeq
\beq
	\CE = \frac{\CE _0t_1t_2e^{id\tilde{n}\om /c}}{1 - 			
		r_2^2e^{2id\tilde{n}\om /c}}
\eeq
If $\tilde{n} \sim \ 1$ and $d$ is small, then 
\beq
	\CE \simeq \ \CE _0e^{id\tilde{n}\om /c}
\eeq
\beq
	I \simeq \ I_0e^{id\om (\tilde{n} - \tilde{n}^*)/c} = I_0e^{-d\om 	
		2\kappa /c}
\eeq
\beq
	\kappa \ = \frac{\ep _2}{2\sqrt{\ep_1}}
\eeq
\beq
I \simeq \ I_0e^{-(\ep _2\om /c)d} \simeq \ I_0\left( 1 - \frac{\ep_2\om }cd\right)
\eeq
If the thickness $d$ is known, then the quantity 
\beq
	\frac{\om \ep _2(\om )}c = \frac{4\pi \sigma _1(\om )}c
\eeq
may be measured (the absorption coefficient). The real part of the dielectric
response, $\ep_1(\om)$ may be calculated with the Kramers Kronig relation 
\beq
	\ep _1(\om ) = 1 + \frac 1{\pi}P\int_{-\infty}^{\infty} \frac{\ep 	
		_2(\om ^{\prime})}{\om ^{\prime} - \om }d\om ^{\prime}
\eeq
According to Young Kim, this analysis works for sufficiently thin samples
in the optical regime, but typically fails in the IR where $\ep_2$ becomes
large.

\subsection{Reflectivity}

Of course, we could also have performed the experiment on a very thick slab of 
the material, and measured the reflectivity $R$ 
\begin{figure}[htb]
\centerline{\includegraphics[height=1.75in,keepaspectratio,clip=true]{reflect.pdf}}
\caption[]{\em{The coefficient of reflectivity (the ratio of the reflected to 
the incident intensities) $R/I$ depends upon both $\ep _1(\om )$ and 
$\ep _2(\om )$, making the analysis more complicated than in the transmission 
experiment.}}
\end{figure} 
However, this is a much more complicated experiment since $\frac RI$ depends
upon both $\ep _1$ and $\ep _2(\om )$, and is hence much more difficult to 
analyze [See Frederick Wooten, {\bf{Optical Properties of Solids}},
(Academic Press, San Diego, 1972)].

	To analyze these experiments, one must first measure the reflectivity,
$R(\omega)$ over the entire frequency range.  We then write
\beq
\label{eq:R_eps}
R(\omega)=r(\om)r^*(\om) = \frac{\lep n-1\rip^2 +\kappa^2}
                               {\lep n+1\rip^2 +\kappa^2}
\eeq
where $\tilde n = n + i\kappa$ and
\beq
r(\om)=\rho(\om) e^{i\th}
\eeq
so that $R(\om)=\rho(\om)^2$.
If $\rho(\om)\to 1$ and $\th\to 0$ fast enough as the frequency
increases, then we may employ the Kramers Kronig relations replacing 
$\chi$ with $\ln{r(\om)} = \ln{\rho(\om)} + i\th(\om)$, so that
\beq
\th(\om')= -\frac{2\om'}{\pi} {\rm{P}} \int_0^\infty \frac{\ln{\sqrt{R(\om)}}}{\om^2-\om'^2} d\om
\eeq

Thus, if $R(\om)$ is measured over the entire frequency range where
it is finite, then we can calculate $\th(\om)$.  This complete knowledge of 
$R$ is generally not available, and various extrapolation and fitting 
schemes are used on $R(\om)$ so that the integral above may be completed.

We may then use Eq.~\ref{eq:R_eps} above to relate $R$ and $\th$ to the
real and imaginary parts of the refractive index,
\beq
n(\om)=\frac{1-R(\om)}{1+R(\om)-2\sqrt{R(\om)}\cth(\om)}
\eeq
\beq
\kappa(\om)=\frac{2\sqrt{R(\om)}\sth(\om)}{1+R(\om)-2\sqrt{R(\om)}\cth(\om)}\,,
\eeq
and therefore the dielectric response 
$\ep(\om)=\lep  n(\om)+\kappa(\om) \rip^2$ and the optical conductivity
$\sigma(\om)=\frac{-i\om\ep}{4\pi}$.

\subsection{Model Dielectric Response}

We have seen that EM radiation is a sensitive probe of the dielectric 
properties of materials.  Absorption and reflectivity experiments allow us to 
measure some combination of $\ep_1$ or $\ep_2$, with the remainder 
reconstructed by the Kramers-Kronig relations. 
 
     In order to learn more from such measurements, we need to have detailed 
models of the materials and their corresponding dielectric properties.  For 
example, the electric field will interact with the moving charges associated 
with lattice vibrations.  At the simplest level, we can model this as the 
interaction of isolated dipoles composed of bound damped charge $e^*$
and length $s$
\beq
	p = e^*s
\eeq 
The equation of motion for this system is 
\beq
	\mu \ddot{s} = -\mu \gamma\dot{s} - \mu \om _0^2s + e^*\CE  	
\eeq
where the first term on the right-hand side is the damping force, the second 
term is the restoring force, and the third term is the external field.  
Furthermore, the polarization $P$ is
\beq
	P = \frac NVe^*s + \frac NV \al \CE
\eeq 
where $\al $ is the polarizability of the different molecules which make up the 
material.  It is to represent the polarizability of rigid bodies.  For example, 
(see Fig.~\ref{fig:sphere}) a metallic sphere has 
\begin{figure}[htb]
\centerline{\includegraphics[height=1.25in,keepaspectratio,clip=true]{sphere.pdf}}
\caption[]{\em{We may model our material as a system harmonically bound charge
and of metallic spheres with polarizablity $\alpha=a^3$.}}
\label{fig:sphere}
\end{figure}
\beq	
	\al \ = a^3
\eeq 

If we F.T. these two equations, we get 
\beq
	-\mu \om ^2s = i\om \mu rs - \mu \om _0^2s + e^*\CE (\om ) 
\eeq
\beq
	\Rightarrow \ s(\om )\left\{ \om _0^2 - \om ^2 - i\om \gamma \right\} \ 
		= \frac{e^*\CE (\om )}{\mu }
\eeq
and 
\beq
	P(\om ) = ne^*s(\om ) + n\al \CE (\om )
\eeq 
or 
\beq
	P(\om ) = \CE (\om )\left\{ \frac{ne^{*2}/\mu }{\om _0^2 - \om ^2 - 	
		i\om \gamma } + n\al \right\}
\eeq

The term in brackets is the complex electric susceptibility of the system
$\chi= \CE/P $.
\beq
	\chi (\om ) = n\al + \frac{{ne^{*2}}/{\mu}}{\om _0^2 - \om ^2 - 	
		i\om \gamma} = \frac 1{4\pi }(\ep (\om ) - 1)
\eeq  
\beq
	\ep (\om ) = 1 + 4\pi n\al + \frac{4\pi ne^{*2}/\mu }{\om _0^2 - \om 	
		^2 - i\gamma \om }, \qquad \mbox{where } \om _p^2 = \frac{4\pi 	
		ne^{*2}}{\mu }
\eeq
Or, introducing the high and zero frequency limits 
\beq
 	\ep _{\infty} = 1 + 4\pi n\al
\eeq
\beq
	\ep _0 = 1 + 4\pi n\al + \frac{\om _p^2}{\om  _0^2} = \ep _{\infty} + 	
		\frac{\om _p^2}{\om _0^2}
\eeq
\beq
	\ep (\om ) = \ep _{\infty} + \frac{\om _0^2 (\ep _0 - \ep _{\infty 	
		})}{\om _0^2 - \om ^2 - i\gamma \om}
\eeq
For our causality arguments we must have no poles in the upper complex half 
plane.  This is satisfied since $\gamma > 0$.  In addition, we need 
\beq
\chi \ = \frac 1{4\pi}(\ep (\om ) - 1) \rightarrow \ 0, \qquad \mbox{as } 
		\om \  \rightarrow \ \infty
\eeq
This means that $\ep _{\infty} = 1 + 4\pi n\al = 1$.  Of course $\al$ is 
finite.  The problem is that in making $\al \ =$ constant, we neglected the 
electron mass.  Ie., for our example of a metallic sphere, $\al \ < a^3$ for 
very high $\om $!  (See Fig.~\ref{fig:sphere2})  
\begin{figure}[htb]
\centerline{\includegraphics[height=1.75in,keepaspectratio,clip=true]{sphere2.pdf}}
\caption[]{\em{For very high $\om $, $\al < a^3$ since the electrons have a
finite mass and hence cannot respond instantaneously to changes in the field.}}
\label{fig:sphere2}
\end{figure}
due to the finite electronic mass. 
 
To analyze experiments $\ep $ is separated into real $\ep _1$ and imaginary 
parts $\ep _2$
\beq
	\ep _1(\om ) = -\frac{4\pi }{\om}\sigma _2 \ = \ep _{\infty} + 		
		\frac{(\ep_0 - \ep_{\infty})\om _0^2(\om _0^2 - \om ^2)}{
(\om_0^2 - \om ^2)^2 + \gamma ^2\om ^2}
\eeq
\beq
\ep _2(\om ) = \frac{4\pi }{\om}\sigma _1 = \frac{(\ep _0 - \ep_{\infty})
\om _0^2\gamma \om}{(\om _0^2 - \om ^2)^2 + \gamma ^2\om^2}
\eeq
\begin{figure}[htb]
\centerline{\includegraphics[height=2.75in,keepaspectratio,clip=true]{graphs.pdf}}
\caption[]{\em{Sketch of the real and imaginary parts of the dielectric 
response of the harmonically bound charge model.}}
\end{figure}
Thus a phonon mode will give a roughly Lorentzian-like line shape in
the optical conductivity $\sigma_1$ centered roughly at the phonon
frequency.  In addition, this form may be used to construct a
model reflectivity $R=|\tilde{n}-1|^2/|\tilde{n}+1|^2$, with
$\tilde{n}=\sqrt{\ep}$ which is often appended to the high end of
the reflectivity data, so that the Kramers-Kronig integrals may
be completed.


\section{ The Free-electron gas }
 
Metals have a distinct feature in their optical conductivity $\sigma
_1(\om )$ which may be emulated by the free-electron gas.  The equation of motion of the 
free-electron gas is 
\beq
	nm\ddot{s} = -\gamma \dot{s} - ne\CE 
\eeq
In steady state $\gamma \dot{s} = -ne\CE $; however 
\beq
	-ne\dot{s} = j = \sigma \CE \ = \frac{ne^2\tau }m \CE
\eeq
in the relaxation time approximation, so 
\beq
\dot{s} = -\frac{e\tau} m\CE = -\frac{ne}{\gamma }\CE \qquad \mbox{or } 	
		\gamma \ = \frac{nm}{\tau }\,.
\eeq 
Thus, the equation of motion is 
\beq
	nm\ddot{s} = -\frac{nm}{\tau }\dot{s} - ne\CE 
\eeq 
If we work in a Fourier representation, then 
\beq
	-m\om ^2s(\om ) = \frac{i\om m}{\tau} s(\om) - e\CE (\om )
\eeq
However, since $P = -ens$
\beq
	\left(m\om ^2 + \frac{i\om m}{\tau}\right) P = -ne^2\CE (\om )
\eeq
or 
\beq
	P = -\frac{ne^2/m}{\om ^2 + i\om/\tau }\CE \ = \chi \CE \ = \frac 	
		1{4\pi }(\ep - 1)\CE
\eeq
\beq
\ep \ = 1 - \frac{4\pi n^2e^2}{m\om }\frac 1{\om + i/\tau} = 1 - \frac{\om 
		_p^2}{\om }\left\{ \frac{\om - i/\tau }{\om ^2 + 1/\tau 	
		^2}\right\}
\eeq
\beq
	\ep _1 = 1 - \frac{\om_p^2}{\om }\left\{ \frac{\om }{\om ^2 + 1/\tau 	
		^2}\right\}, \qquad \ep _2 = \frac{\om _p^2}{\tau \om }\frac 	
		1{\om ^2 + 1/\tau ^2}
\eeq
However, recall that $\ep _1 = -\frac{4\pi \sigma _2}{\om }$ and $\ep _2 = 
\frac{4\pi \sigma _1}{\om }$, so that 
\beq
	\sigma _1(\om ) = \frac{\om _p^2}4 \frac{1/\tau \pi}{\om ^2 + 1/\tau 	
		^2}\,.
\eeq
\begin{figure}[htb]
\centerline{\includegraphics[height=1.75in,keepaspectratio,clip=true]{graph.pdf}}
\caption[]{\em{A sketch of the optical conductivity of a metal
at finite temperatures.  The low-frequency Drude peak is due to the coherent
transport of electrons.  The higher frequency peak is due to incoherent 
scattering of electrons from phonons or electron-electron interactions.}}
\end{figure}
 
\section{ Excitons }

     One of the most dramatic effects of the dielectric properties of 
semiconductors are excitons.  Put simply, an exciton is a hydrogenic bound 
state made up of a hole and an electron.  The Hamiltonian for such a system 
is 
\beq
	H = \frac 12m_h^*v_h^2 + \frac 12m_e^*v_e^2 - \frac{e^2}{\ep r}
\eeq
\begin{figure}[htb]
\centerline{\includegraphics[height=1.5in,keepaspectratio,clip=true]{exciton.pdf}}
\caption[]{\em{Excitons are hydrogenic bound states of an electron
and a hole.  In semiconductors with an indirect gap, such as Si, they can
be very long lived, since a phonon or defect must be involved in
the recombination process to conserve momentum.  For example, 
in ultra-pure Si ($\approx 10^{12}$ impurities per cc) the lifetime
can exceed $\tau_{Si}\approx 10^{-5}$s; whereas in direct-gap GaAs, the 
lifetime is much shorter $\tau_{GaAs}\approx 10^{-9}$s and is 
generally limited by surface states. On the right, the coordinates
of the exciton in the center of mass are shown.}}
\end{figure}
As usual, we will work in the center of mass, so 
\beq
	\frac 12m_h^*v_h^2 + \frac 12m_e^*v_e^2 = \frac 12(m_h^* + 		
		m_e^*)\dot{R}^2 + \mu \dot{r}^2
\eeq
\beq
	\bar{R} = \frac{m_e^*\bar{x}_e + m_h^*\bar{x}_h}{m_h^* + m_e^*}, \qquad 
		r = \bar{x}_e - \bar{x}_h
\eeq
The eigenenergies may be obtained from the Bohr atom solution $E_n = -\mu 
(Ze^2)^2/(2\hbar ^2n^2)$ by making the substitutions 
\beq
	Ze^2 \rightarrow \ \frac{e^2}{\ep }
\eeq
\beq
	\mu \ \rightarrow \ \mu ^* = \frac{m_h^*m_e^*}{m_h^* + m_e^*}
\eeq 
In addition there is a cm kinetic energy, let $\hbar k = P$, then 
\beq
	E_{nK} = \frac{\hbar ^2K^2}{2(m_h^* + m_e^*)} - \frac{\mu ^*e^4}{\ep 	
		^22\hbar ^2n^2} + E_g
\eeq
The binding energy is strongly reduced by the dielectric effects, since $\ep \ 
\simeq \ 10$. (it is also reduced by the effective masses, since typically $m^* 
< m$).  Thus, typically the binding energy is a small fraction of a Ryberg.  
Similarly, the size of the exciton is much larger than the hydrogen atom 
\beq
	a\simeq \frac{\ep\mu}{m_e} a_0
\eeq
This in fact is the justification for the hydrogenic approximation.
Since the orbit contains many sites the lattice may be approximated as
a continuum and thus the exciton is well approximated as a hydrogenic atom. 
\begin{figure}[htb]
\centerline{\includegraphics[height=1.25in,keepaspectratio,clip=true]{cm2.pdf}}
	\caption[]{\em{An exciton may be approximated as a hydrogenic
atom if its radius is large compared to the lattice spacing. In Si,
the radius is large due to the reduced hole and electron masses and
the enhanced dielectric constant $\ep\approx 10$.}}
\end{figure}

\edo



\section{ Longitudinal and Transverse }
 
     A solid can support both longitudinal and transverse oscillations.  (See 
Fig.~\ref{fig:waves})
\begin{figure}[htb]
\centerline{\includegraphics[height=0.75in,keepaspectratio,clip=true]{waves.pdf}}
	\caption[]{}
\label{fig:waves}
\end{figure}
\beqa
	& \mbox{Longitudinal} \qquad \qquad & \mbox{Transverse} \nonumber \\
	& P_x = P_{x0}e^{-i(\om t - qx)}& P_y = P_{y0}e^{-i(\om t - qx)} \\
	& \grad \times \P _L = 0	& \grad \times \P _T \neq \ 0 \\
	& \grad \cdot \P _L \neq \ 0	& \grad \cdot \P _T = 0 
\eeqa
Recall that the latter, curl and divergence, criteria were used to define 
transverse and longitudinal waves in E \& M (Jackson) 
 
\subsection{ Longitudinal }
 
     Longitudinal waves cannot couple to plane (transverse) and E \& M waves 
which exist in the far field of radiation.  In a medium without free charge 
(ie., an insulator) their behavior is restricted. 
\beq
	\grad \cdot \D = 4\pi\rho _{free} = 0 = \ep (\om ) \grad \cdot \CE \ = 	
		\frac{4\pi \ep (\om )}{\ep (\om ) - 1}\grad \cdot \P , 		
		\mbox{where} \grad \cdot \P \ \neq \ 0
\eeq 
\beq
	\P \ = \chi \CE \ = \frac 1{4\pi}(\ep (\om ) - 1)\CE 
\eeq
Since $\grad \cdot \P \neq \ 0$ for a longitudinal wave, it can only exist 
for $\ep (\om _L) = 0$.  Then 
\beq
	\P _L = \frac 1{4\pi }(\ep (\om ) - 1) \CE = -\frac{\CE}{4\pi}, \qquad 	
		\ep (\om _L) = 0
\eeq
so that the polarization and its field are $180^{\circ} (\pi)$ out of phase.  
As we will see, while these polarization waves cannot interact with the field, 
they can interact with electrons (this is one way that electrons lose energy). 

\subsection{ Transverse Waves }
 
Transverse waves will interact with E \& M far-field radiation, and hence must 
be solved for in the context of Maxwell's equations. 
\beq	
	\grad \times \CE \ = -\frac 1c\pde Bt, \qquad \mu \ = 1
\eeq
\beq
	\grad \times \H \ = \frac{4\pi }c\j + \frac 1c\pde {\D}t
\eeq 
Or, in terms of $\P \ = \chi \CE \ = \frac 1{4\pi }(\ep - 1)\CE , \D = \ep \CE$ 
\begin{figure}[htb]
\centerline{\includegraphics[height=2.0in,keepaspectratio,clip=true]{pspace.pdf}}
	\caption[]{}
\end{figure}
\beq
	\grad \times \P \ = \frac{\chi}c i\om \H
\eeq
\beq
	\grad \times \H \ = -\frac{i\om }c\ep \frac 1{\chi}\P
\eeq
\beq
	\Rightarrow \left\{ \begin{array}{ll}
		 iqP_{0y} = \frac{i\om }c\chi H_{0z} \\
		iqH_{0z} - \frac{i\om }c\frac{\ep }{\chi }P_{0y}
	
	\end{array} \right.
\eeq
\beq
	qP_{0y} - \frac{\om }{4\pi c}(\ep - 1) H_{0z} = 0 
\eeq
\beq
	\frac{4\pi \om}c\frac{\ep}{\ep - 1}P_{0y} - qH_{0z} = 0
\eeq
This has non trivial solutions for 
\beq
	-q^2 + \frac{4\pi \om}c \frac{\ep}{\ep -1} \frac{\om}{4\pi c}(\ep - 1) 	
		= 0
\eeq
\beq
	q^2 = \frac{\om ^2\ep}{c^2} \Rightarrow \ \om ^2 = \frac{c^2q^2}{\ep 	
		(\om )}
\eeq
These solutions of coupled electromagnetic and mechanical waves are called 
polaritons.  We can sketch their behavior, which depends upon $\ep (\om )$, 
by choosing a simple form of $\ep (\om )$ without damping.  If in the previous 
model, we choose $\gamma \ = 0$, then we get 
\beq
	\om ^2 = \frac{c^2q^2}{\ep _{\infty} + \frac{\om _0^2(\ep _0 - \ep 	
		_{\infty})}{\om _0^2 - \om ^2}}
\eeq 
\beq
	\om ^2\left\{ \ep _{\infty} + \frac{\om _0^2(\ep _0 - \ep 		
		_{\infty})}{\om _0^2 - \om ^2}\right\} \ = c^2q^2
\eeq
\beq 
 	\om ^2\ep _{\infty}(\om _0^2 - \om ^2) + \om ^2\om _0^2(\ep _0 - \ep 	
		_{\infty}) = c^2q^2(\om _0^2 - \om ^2)
\eeq
\beq
	\om ^2\ep _{\infty}\om _0^2 - \om ^4\ep _{\infty} + \om ^2\om _0^2(\ep 	
		_0 - \ep _{\infty}) - c^2q^2(\om _0^2 - \om ^2) = 0
\eeq
\beq
	-\ep _{\infty}\om ^4 + \om ^2(\ep _{\infty}\om _0^2 + \om _0^2\ep _0 - 	
		\om _0^2\ep _{\infty} + c^2q^2) - c^2q^2\om _0^2 = 0
\eeq
\beq
	\om ^2 = \frac{(\om _0^2\ep _0 + c^2q^2) \pm \sqrt{(\om _0^2\ep 	
		_0 + c^2q^2)^2 - 4\ep _{\infty}c^2q^2\om _0^2}}{2\ep _{\infty}}
\eeq
$q \rightarrow \ 0$
\beq
	\om ^2 = \frac 1{2\ep _{\infty}}\left\{\om _0^2\ep _0 + c^2q^2 \pm (\om 
		_0^2\ep _0 + c^2q^2)\left( 1 - \frac{2\ep _{\infty}c^2q^2\om 	
		_0^2}{\om _0^2\ep _0 + c^2q^2)^2}\right)\right\}
\eeq
\beq 
	\om _+^2 = \frac 1{2\ep _{\infty}}2\om _0^2\ep _0 \simeq \frac{\ep 	
		_o}{\ep _\infty}\om _0^2
\eeq
\beq
	\om _-^2 = \frac{c^2q^2\om _0^2}{\om _0^2\ep _0} = \frac{c^2q^2}{\ep 	
		_0}
\eeq
$q \rightarrow \ \infty $
\beq
	\om _+^2 = \frac{c^2q^2}{\ep _{\infty}}
\eeq
\beq
	\om _-^2 = \om _0^2
\eeq
\begin{figure}[htb]
\centerline{\includegraphics[height=2.25in,keepaspectratio,clip=true]{dispersion.pdf}}
	\caption[]{}
\end{figure} 
