States for an electron in a one-dimensional lattice
You would, at first sight, think that a low-energy electron would have great difficulty passing through a solid crystal. The atoms are packed together with their centers only a few angstroms apart, and the effective diameter of the atom for electron scattering is roughly an angstrom or so. That is, the atoms are large, relative to their spacing, so that you would expect the mean free path between collisions to be of the order of a few angstroms—which is practically nothing. You would expect the electron to bump into one atom or another almost immediately. Nevertheless, it is a ubiquitous phenomenon of nature that if the lattice is perfect, the electrons are able to travel through the crystal smoothly and easily—almost as if they were in a vacuum. This strange fact is what lets metals conduct electricity so easily; it has also permitted the development of many practical devices. It is, for instance, what makes it possible for a transistor to imitate the radio tube. In a radio tube electrons move freely through a vacuum, while in the transistor they move freely through a crystal lattice. The machinery behind the behavior of a transistor will be described in this chapter; the next one will describe the application of these principles in various practical devices. The conduction of electrons in a crystal is one example of a very common phenomenon. Not only can electrons travel through crystals, but other “things” like atomic excitations can also travel in a similar manner. So the phenomenon which we want to discuss appears in many ways in the study of the physics of the solid state. You will remember that we have discussed many examples of two-state systems. Let’s now think of an electron which can be in either one of two positions, in each of which it is in the same kind of environment. Let’s also suppose that there is a certain amplitude to go from one position to the other, and, of course, the same amplitude to go back, just as we have discussed for the hydrogen molecular ion in Section 10–1. The laws of quantum mechanics then give the following results. There are two possible states of definite energy for the electron. Each state can be described by the amplitude for the electron to be in each of the two basic positions. In either of the definite-energy states, the magnitudes of these two amplitudes are constant in time, and the phases vary in time with the same frequency. On the other hand, if we start the electron in one position, it will later have moved to the other, and still later will swing back again to the first position. The amplitude is analogous to the motions of two coupled pendulums. Now consider a perfect crystal lattice in which we imagine that an electron can be situated in a kind of “pit” at one particular atom and with some particular energy. Suppose also that the electron has some amplitude to move into a different pit at one of the nearby atoms. It is something like the two-state system—but with an additional complication. When the electron arrives at the neighboring atom, it can afterward move on to still another position as well as return to its starting point. Now we have a situation analogous not to two coupled pendulums, but to an infinite number of pendulums all coupled together. It is something like what you see in one of those machines—made with a long row of bars mounted on a torsion wire—that is used in first-year physics to demonstrate wave propagation. If you have a harmonic oscillator which is coupled to another harmonic oscillator, and that one to another, and so on …, and if you start an irregularity in one place, the irregularity will propagate as a wave along the line. The same situation exists if you place an electron at one atom of a long chain of atoms. Usually, the simplest way of analyzing the mechanical problem is not to think in terms of what happens if a pulse is started at a definite place, but rather in terms of steady-wave solutions. There exist certain patterns of displacements which propagate through the crystal as a wave of a single, fixed frequency. Now the same thing happens with the electron—and for the same reason, because it’s described in quantum mechanics by similar equations. You must appreciate one thing, however; the amplitude for the electron to be at a place is an amplitude, not a probability. If the electron were simply leaking from one place to another, like water going through a hole, the behavior would be completely different. For example, if we had two tanks of water connected by a tube to permit some leakage from one to the other, then the levels would approach each other exponentially. But for the electron, what happens is amplitude leakage and not just a plain probability leakage. And it’s a characteristic of the imaginary term—the $i$ in the differential equations of quantum mechanics—which changes the exponential solution to an oscillatory solution. What happens then is quite different from the leakage between interconnected tanks.  We want now to analyze quantitatively the quantum mechanical situation. Imagine a one-dimensional system made of a long line of atoms as shown in Fig. 13–1(a). (A crystal is, of course, three-dimensional but the physics is very much the same; once you understand the one-dimensional case you will be able to understand what happens in three dimensions.) Next, we want to see what happens if we put a single electron on this line of atoms. Of course, in a real crystal there are already millions of electrons. But most of them (nearly all for an insulating crystal) take up positions in some pattern of motion each around its own atom—and everything is quite stationary. However, we now want to think about what happens if we put an extra electron in. We will not consider what the other ones are doing because we suppose that to change their motion involves a lot of excitation energy. We are going to add an electron as if to produce one slightly bound negative ion. In watching what the one extra electron does we are making an approximation which disregards the mechanics of the inside workings of the atoms. Of course the electron could then move to another atom, transferring the negative ion to another place. We will suppose that just as in the case of an electron jumping between two protons, the electron can jump from one atom to the neighbor on either side with a certain amplitude. Now how do we describe such a system? What will be reasonable base states? If you remember what we did when we had only two possible positions, you can guess how it will go. Suppose that in our line of atoms the spacings are all equal; and that we number the atoms in sequence, as shown in Fig. 13–1(a). One of the base states is that the electron is at atom number $6$, another base state is that the electron is at atom number $7$, or at atom number $8$, and so on. We can describe the $n$th base state by saying that the electron is at atom number $n$. Let’s say that this is the base state $\ket{n}$. Figure 13–1 shows what we mean by the three base states \begin{equation*} \ket{n-1},\:\ket{n},\:\text{and }\:\ket{n+1}. \end{equation*}  Using these base states, any state $\ket{\phi}$ of the electron in our one-dimensional crystal can be described by giving all the amplitudes $\braket{n}{\phi}$ that the state $\ket{\phi}$ is in one of the base states—which means the amplitude that it is located at one particular atom. Then we can write the state $\ket{\phi}$ as a superposition of the base states \begin{equation} \label{Eq:III:13:1} \ket{\phi}=\sum_n\ket{n}\braket{n}{\phi}. \end{equation}  Next, we are going to suppose that when the electron is at one atom, there is a certain amplitude that it will leak to the atom on either side. And we’ll take the simplest case for which it can only leak to the nearest neighbors—to get to the next-nearest neighbor, it has to go in two steps. We’ll take that the amplitudes for the electron jump from one atom to the next is $iA/\hbar$ (per unit time). For the moment we would like to write the amplitude $\braket{n}{\phi}$ to be on the $n$th atom as $C_n$. Then Eq. (13.1) will be written \begin{equation} \label{Eq:III:13:2} \ket{\phi}=\sum_n\ket{n}C_n. \end{equation}  If we knew each of the amplitudes $C_n$ at a given moment, we could take their absolute squares and get the probability that you would find the electron if you looked at atom $n$ at that time. What will the situation be at some later time? By analogy with the two-state systems we have studied, we would propose that the Hamiltonian equations for this system should be made up of equations like this: \begin{equation} \label{Eq:III:13:3} i\hbar\,\ddt{C_n(t)}{t}=E_0C_n(t)-AC_{n+1}(t)-AC_{n-1}(t). \end{equation}  The first coefficient on the right, $E_0$, is, physically, the energy the electron would have if it couldn’t leak away from one of the atoms. (It doesn’t matter what we call $E_0$; as we have seen many times, it represents really nothing but our choice of the zero of energy.)  The next term represents the amplitude per unit time that the electron is leaking into the $n$th pit from the $(n+1)$st pit; and the last term is the amplitude for leakage from the $(n-1)$st pit. As usual, we’ll assume that $A$ is a constant (independent of $t$). For a full description of the behavior of any state $\ket{\phi}$, we would have one equation like (13.3) for every one of the amplitudes $C_n$. Since we want to consider a crystal with a very large number of atoms, we’ll assume that there are an indefinitely large number of states—that the atoms go on forever in both directions. (To do the finite case, we will have to pay special attention to what happens at the ends.) If the number $N$ of our base states is indefinitely large, then also our full Hamiltonian equations are infinite in number! We’ll write down just a sample: \begin{equation} \begin{aligned} \vdots\qquad&\qquad\qquad\qquad\qquad\:\:\vdots\\ i\hbar\,\ddt{C_{n-1}}{t}&= E_0C_{n-1}-AC_{n-2}-AC_n,\\[1ex] i\hbar\,\ddt{C_n}{t}&= E_0C_n-AC_{n-1}-AC_{n+1},\\[1ex] i\hbar\,\ddt{C_{n+1}}{t}&= E_0C_{n+1}-AC_n-AC_{n+2},\\[1ex] \vdots\qquad&\qquad\qquad\qquad\qquad\:\:\vdots \end{aligned} \label{Eq:III:13:4} \end{equation} 