States of definite energy
We could study many things about an electron in a lattice, but first let’s try to find the states of definite energy. As we have seen in earlier chapters this means that we have to find a situation in which the amplitudes all change at the same frequency if they change with time at all. We look for solutions of the form \begin{equation} \label{Eq:III:13:5} C_n=a_ne^{-iEt/\hbar}. \end{equation}  The complex numbers $a_n$ tell us about the non-time-varying part of the amplitude to find the electron at the $n$th atom. If we put this trial solution into the equations of (13.4) to test them out, we get the result \begin{equation} \label{Eq:III:13:6} Ea_n=E_0a_n-Aa_{n+1}-Aa_{n-1}. \end{equation}  We have an infinite number of such equations for the infinite number of unknowns $a_n$—which is rather petrifying. All we have to do is take the determinant … but wait! Determinants are fine when there are $2$, $3$, or $4$ equations. But if there are a large number—or an infinite number—of equations, the determinants are not very convenient. We’d better just try to solve the equations directly. First, let’s label the atoms by their positions; we’ll say that the atom $n$ is at $x_n$ and the atom $(n+1)$ is at $x_{n+1}$. If the atomic spacing is $b$—as in Fig. 13–1—we will have that $x_{n+1}=x_n+b$. By choosing our origin at atom zero, we can even have it that $x_n=nb$. We can rewrite Eq. (13.5) as \begin{equation} \label{Eq:III:13:7} C_n=a(x_n)e^{-iEt/\hbar}, \end{equation}  and Eq. (13.6) would become \begin{equation} \label{Eq:III:13:8} Ea(x_n)=E_0a(x_n)-Aa(x_{n+1})-Aa(x_{n-1}). \end{equation}  Or, using the fact that $x_{n+1}=x_n+b$, we could also write \begin{equation} \label{Eq:III:13:9} Ea(x_n)=E_0a(x_n)\!-\!Aa(x_n\!+b)\!-\!Aa(x_n\!-b). \end{equation}  This equation is somewhat similar to a differential equation. It tells us that a quantity, $a(x)$, at one point, ($x_n$), is related to the same physical quantity at some neighboring points, ($x_n\pm b$). (A differential equation relates the value of a function at a point to the values at infinitesimally nearby points.) Perhaps the methods we usually use for solving differential equations will also work here; let’s try. Linear differential equations with constant coefficients can always be solved in terms of exponential functions. We can try the same thing here; let’s take as a trial solution \begin{equation} \label{Eq:III:13:10} a(x_n)=e^{ikx_n}. \end{equation}  Then Eq. (13.9) becomes \begin{equation} \label{Eq:III:13:11} Ee^{ikx_n}=E_0e^{ikx_n}-Ae^{ik(x_n+b)}-Ae^{ik(x_n-b)}. \end{equation}  We can now divide out the common factor $e^{ikx_n}$; we get \begin{equation} \label{Eq:III:13:12} E=E_0-Ae^{ikb}-Ae^{-ikb}. \end{equation}  The last two terms are just equal to $(2A\cos kb)$, so \begin{equation} \label{Eq:III:13:13} E=E_0-2A\cos kb. \end{equation}  We have found that for any choice at all for the constant $k$ there is a solution whose energy is given by this equation. There are various possible energies depending on $k$, and each $k$ corresponds to a different solution. There are an infinite number of solutions—which is not surprising, since we started out with an infinite number of base states. Let’s see what these solutions mean. For each $k$, the $a$’s are given by Eq. (13.10). The amplitudes $C_n$ are then given by \begin{equation} \label{Eq:III:13:14} C_n=e^{ikx_n}e^{-(i/\hbar)Et}, \end{equation}  where you should remember that the energy $E$ also depends on $k$ as given in Eq. (13.13). The space dependence of the amplitudes is $e^{ikx_n}$. The amplitudes oscillate as we go along from one atom to the next. We mean that, in space, the amplitude goes as a complex oscillation—the magnitude is the same at every atom, but the phase at a given time advances by the amount $(ikb)$ from one atom to the next. We can visualize what is going on by plotting a vertical line to show just the real part at each atom as we have done in Fig. 13–2. The envelope of these vertical lines (as shown by the broken-line curve) is, of course, a cosine curve. The imaginary part of $C_n$ is also an oscillating function, but is shifted $90^\circ$ in phase so that the absolute square (which is the sum of the squares of the real and imaginary parts) is the same for all the $C$’s. Thus if we pick a $k$, we get a stationary state of a particular energy $E$. And for any such state, the electron is equally likely to be found at every atom—there is no preference for one atom or the other. Only the phase is different for different atoms. Also, as time goes on the phases vary. From Eq. (13.14) the real and imaginary parts propagate along the crystal as waves—namely as the real or imaginary parts of \begin{equation} \label{Eq:III:13:15} e^{i[kx_n-(E/\hbar)t]}. \end{equation}  The wave can travel toward positive or negative $x$ depending on the sign we have picked for $k$. Notice that we have been assuming that the number $k$ that we put in our trial solution, Eq. (13.10), was a real number. We can see now why that must be so if we have an infinite line of atoms. Suppose that $k$ were an imaginary number, say $ik'$. Then the amplitudes $a_n$ would go as $e^{-k'x_n}$, which means that the amplitude would get larger and larger as we go toward large negative $x$’s—or toward large positive $x$’s if $k'$ is a negative number. This kind of solution would be O.K. if we were dealing with a line of atoms that ended, but cannot be a physical solution for an infinite chain of atoms. It would give infinite amplitudes—and, therefore, infinite probabilities—which can’t represent a real situation. Later on we will see an example in which an imaginary $k$ does make sense. The relation between the energy $E$ and the wave number $k$ as given in Eq. (13.13) is plotted in Fig. 13–3. As you can see from the figure, the energy can go from $(E_0-2A)$ at $k=0$ to $(E_0+2A)$ at $k=\pm\pi/b$. The graph is plotted for positive $A$; if $A$ were negative, the curve would simply be inverted, but the range would be the same. The significant result is that any energy is possible within a certain range or “band” of energies, but no others. According to our assumptions, if an electron in a crystal is in a stationary state, it can have no energy other than values in this band. According to Eq. (13.13), the smallest $k$’s correspond to low-energy states—$E\approx(E_0-2A)$. As $k$ increases in magnitude (toward either positive or negative values) the energy at first increases, but then reaches a maximum at $k=\pm\pi/b$, as shown in Fig. 13–3. For $k$’s larger than $\pi/b$, the energy would start to decrease again. But we do not really need to consider such values of $k$, because they do not give new states—they just repeat states we already have for smaller $k$. We can see that in the following way. Consider the lowest energy state for which $k=0$. The coefficient $a(x_n)$ is the same for all $x_n$. Now we would get the same energy for $k=2\pi/b$. But then, using Eq. (13.10), we have that \begin{equation*} a(x_n)=e^{i(2\pi/b)x_n}. \end{equation*}  However, taking $x_0$ to be at the origin, we can set $x_n=nb$; then $a(x_n)$ becomes \begin{equation*} a(x_n)=e^{i2\pi n}=1. \end{equation*}  The state described by these $a(x_n)$ is physically the same state we got for $k=0$. It does not represent a different solution. As another example, suppose that $k$ were $-\pi/4b$. The real part of $a(x_n)$ would vary as shown by curve $1$ in Fig. 13–4. If $k$ were $7\pi/4b$, the real part of $a(x_n)$ would vary as shown by curve $2$ in the figure. (The complete cosine curves don’t mean anything, of course; all that matters is their values at the points $x_n$. The curves are just to help you see how things are going.) You see that both values of $k$ give the same amplitudes at all of the $x_n$’s. The upshot is that we have all the possible solutions of our problem if we take only $k$’s in a certain limited range. We’ll pick the range between $-\pi/b$ and $+\pi/b$—the one shown in Fig. 13–3. In this range, the energy of the stationary states increases uniformly with an increase in the magnitude of $k$. One side remark about something you can play with. Suppose that the electron cannot only jump to the nearest neighbor with amplitude $iA/\hbar$, but also has the possibility to jump in one direct leap to the next nearest neighbor with some other amplitude $iB/\hbar$. You will find that the solution can again be written in the form $a_n=e^{ikx_n}$—this type of solution is universal. You will also find that the stationary states with wave number $k$ have an energy equal to $(E_0-2A\cos kb-2B\cos 2kb)$. This shows that the shape of the curve of $E$ against $k$ is not universal, but depends upon the particular assumptions of the problem. It is not always a cosine wave—it’s not even necessarily symmetrical about some horizontal line. It is true, however, that the curve always repeats itself outside of the interval from $-\pi/b$ to $\pi/b$, so you never need to worry about other values of $k$. Let’s look a little more closely at what happens for small $k$—that is, when the variations of the amplitudes from one $x_n$ to the next are quite slow. Suppose we choose our zero of energy by defining $E_0=2A$; then the minimum of the curve in Fig. 13–3 is at the zero of energy. For small enough $k$, we can write that \begin{equation*} \cos kb\approx1-k^2b^2/2, \end{equation*}  and the energy of Eq. (13.13) becomes \begin{equation} \label{Eq:III:13:16} E=Ak^2b^2. \end{equation}  We have that the energy of the state is proportional to the square of the wave number which describes the spatial variations of the amplitudes $C_n$.