Scattering from imperfections in the lattice
We want now to consider the case of a single electron in a crystal which is not perfect. Our earlier analysis says that perfect crystals have perfect conductivity—that electrons can go slipping through the crystal, as in a vacuum, without friction. One of the most important things that can stop an electron from going on forever is an imperfection or irregularity in the crystal. As an example, suppose that somewhere in the crystal there is a missing atom; or suppose that someone put one wrong atom at one of the atomic sites so that things there are different than at the other atomic sites. Say the energy $E_0$ or the amplitude $A$ could be different. How would we describe what happens then?  To be specific, we will return to the one-dimensional case and we will assume that atom number “zero” is an “impurity” atom and has a different value of $E_0$ than any of the other atoms. Let’s call this energy $(E_0+F)$. What happens? When an electron arrives at atom “zero” there is some probability that the electron is scattered backwards. If a wave packet is moving along and it reaches a place where things are a little bit different, some of it will continue onward and some of it will bounce back. It’s quite difficult to analyze such a situation using a wave packet, because everything varies in time. It is much easier to work with steady-state solutions. So we will work with stationary states, which we will find can be made up of continuous waves which have transmitted and reflected parts. In three dimensions we would call the reflected part the scattered wave, since it would spread out in various directions. We start out with a set of equations which are just like the ones in Eq. (13.6) except that the equation for $n=0$ is different from all the rest. The five equations for $n=-2$, $-1$, $0$, $+1$, and $+2$ look like this: \begin{equation} \begin{aligned} \vdots\quad&\qquad\qquad\qquad\quad\vdots\\ Ea_{-2}&=E_0a_{-2}-Aa_{-1}-Aa_{-3},\\[2pt] Ea_{-1}&=E_0a_{-1}-Aa_0-Aa_{-2},\\[2pt] Ea_0&=(E_0+F)a_0-Aa_1-Aa_{-1},\\[2pt] Ea_1&=E_0a_1-Aa_2-Aa_0,\\[2pt] Ea_2&=E_0a_2-Aa_3-Aa_1,\\ \vdots\quad&\qquad\qquad\qquad\quad\vdots \end{aligned} \label{Eq:III:13:28} \end{equation}  There are, of course, all the other equations for $\abs{n}$ is greater than 2. They will look just like Eq. (13.6). For the general case, we really ought to use a different $A$ for the amplitude that the electron jumps to or from atom “zero,” but the main features of what goes on will come out of a simplified example in which all the $A$’s are equal. Equation (13.10) would still work as a solution for all of the equations except the one for atom “zero”—it isn’t right for that one equation. We need a different solution which we can cook up in the following way. Equation (13.10) represents a wave going in the positive $x$-direction. A wave going in the negative $x$-direction would have been an equally good solution. It would be written \begin{equation*} a(x_n)=e^{-ikx_n}. \end{equation*}  The most general solution we could have taken for Eq. (13.6) would be a combination of a forward and a backward wave, namely \begin{equation} \label{Eq:III:13:29} a_n=\alpha e^{ikx_n}+\beta e^{-ikx_n}. \end{equation}  This solution represents a complex wave of amplitude $\alpha$ moving in the $+x$-direction and a wave of amplitude $\beta$ moving in the $-x$-direction. Now take a look at the set of equations for our new problem—the ones in (13.28) together with those for all the other atoms. The equations involving $a_n$’s with $n\leq-1$ are all satisfied by Eq. (13.29), with the condition that $k$ is related to $E$ and the lattice spacing $b$ by \begin{equation} \label{Eq:III:13:30} E=E_0-2A\cos kb. \end{equation}  The physical meaning is an “incident” wave of amplitude $\alpha$ approaching atom “zero” (the “scatterer”) from the left, and a “scattered” or “reflected” wave of amplitude $\beta$ going back toward the left. We do not lose any generality if we set the amplitude $\alpha$ of the incident wave equal to $1$. Then the amplitude $\beta$ is, in general, a complex number. We can say all the same things about the solutions of $a_n$ for $n\geq1$. The coefficients could be different, so we would have for them \begin{equation} \label{Eq:III:13:31} a_n=\gamma e^{ikx_n}+\delta e^{-ikx_n},\quad \text{for}\quad n\geq1. \end{equation}  Here, $\gamma$ is the amplitude of a wave going to the right and $\delta$ a wave coming from the right. We want to consider the physical situation in which a wave is originally started only from the left, and there is only a “transmitted” wave that comes out beyond the scatterer—or impurity atom. We will try for a solution in which $\delta=0$. We can, certainly, satisfy all of the equations for the $a_n$ except for the middle three in Eq. (13.28) by the following trial solutions. \begin{align} a_n\:(\text{for $n<0$}) &=e^{ikx_n}+\beta e^{-ikx_n},\notag\\[1ex] \label{Eq:III:13:32} a_n\:(\text{for $n>0$}) &=\gamma e^{ikx_n}. \end{align} The situation we are talking about is illustrated in Fig. 13–6. By using the formulas in Eq. (13.32) for $a_{-1}$ and $a_{+1}$, the three middle equations of Eq. (13.28) will allow us to solve for $a_0$ and also for the two coefficients $\beta$ and $\gamma$. So we have found a complete solution. Setting $x_n=nb$, we have to solve the three equations  \begin{equation} \begin{gathered} (E-E_0)\{e^{ik(-b)}+\beta e^{-ik(-b)}\}= -A\{a_0+e^{ik(-2b)}+\beta e^{-ik(-2b)}\},\\[1ex] (E-E_0-F)a_0= -A\{\gamma e^{ikb}+e^{ik(-b)}+\beta e^{-ik(-b)}\},\\[1ex] (E-E_0)\gamma e^{ikb}= -A\{\gamma e^{ik(2b)}+a_0\}. \end{gathered} \label{Eq:III:13:33} \end{equation} 
 \begin{equation} \begin{aligned} (E\!-\!E_0)\{e^{ik(-b)}\!&+\beta e^{-ik(-b)}\}=\\ &-A\{a_0\!+e^{ik(-2b)}\!+\beta e^{-ik(-2b)}\},\\[1ex] (E\!-\!E_0\!-\!F)a_0=\!&-\!A\{\gamma e^{ikb}\!+e^{ik(-b)}\!+\beta e^{-ik(-b)}\},\\[1ex] (E\!-\!E_0)\gamma e^{ikb}=\!&-\!A\{\gamma e^{ik(2b)}\!+a_0\}. \end{aligned} \label{Eq:III:13:33} \end{equation} 
 Remember that $E$ is given in terms of $k$ by Eq. (13.30). If you substitute this value for $E$ into the equations, and remember that $\cos x=\tfrac{1}{2}(e^{ix}+e^{-ix})$, you get from the first equation that \begin{equation} \label{Eq:III:13:34} a_0=1+\beta, \end{equation}  and from the third equation that \begin{equation} \label{Eq:III:13:35} a_0=\gamma. \end{equation}  These are consistent only if \begin{equation} \label{Eq:III:13:36} \gamma=1+\beta. \end{equation} This equation says that the transmitted wave ($\gamma$) is just the original incident wave ($1$) with an added wave ($\beta$) equal to the reflected wave. This is not always true, but happens to be so for a scattering at one atom only. If there were a clump of impurity atoms, the amount added to the forward wave would not necessarily be the same as the reflected wave. We can get the amplitude $\beta$ of the reflected wave from the middle equation of Eq. (13.33); we find that \begin{equation} \label{Eq:III:13:37} \beta=\frac{-F}{F-2iA\sin kb}. \end{equation}  We have the complete solution for the lattice with one unusual atom. You may be wondering how the transmitted wave can be “more” than the incident wave as it appears in Eq. (13.34). Remember, though, that $\beta$ and $\gamma$ are complex numbers and that the number of particles (or rather, the probability of finding a particle) in a wave is proportional to the absolute square of the amplitude. In fact, there will be “conservation of electrons” only if \begin{equation} \label{Eq:III:13:38} \abs{\beta}^2+\abs{\gamma}^2=1. \end{equation}  You can show that this is true for our solution.