Potential energy; energy conservation
Now we would like to discuss what happens when the energy of a particle can change. We begin by thinking of a particle which moves in a force field described by a potential. We discuss first the effect of a constant potential. Suppose that we have a large metal can which we have raised to some electrostatic potential $\phi$, as in Fig. 7–2. If there are charged objects inside the can, their potential energy will be $q\phi$, which we will call $V$, and will be absolutely independent of position. Then there can be no change in the physics inside, because the constant potential doesn’t make any difference so far as anything going on inside the can is concerned. Now there is no way we can deduce what the answer should be, so we must make a guess. The guess which works is more or less what you might expect: For the energy, we must use the sum of the potential energy $V$ and the energy $E_p$—which is itself the sum of the internal and kinetic energies. The amplitude is proportional to \begin{equation} \label{Eq:III:7:18} e^{-(i/\hbar)[(E_p+V)t-\FLPp\cdot\FLPx]}. \end{equation}  The general principle is that the coefficient of $t$, which we may call $\omega$, is always given by the total energy of the system: internal (or “mass”) energy, plus kinetic energy, plus potential energy: \begin{equation} \label{Eq:III:7:19} \hbar\omega=E_p+V. \end{equation}  Or, for nonrelativistic situations, \begin{equation} \label{Eq:III:7:20} \hbar\omega=W_{\text{int}}+\frac{p^2}{2M}+V. \end{equation}  Now what about physical phenomena inside the box? If there are several different energy states, what will we get? The amplitude for each state has the same additional factor \begin{equation*} e^{-(i/\hbar)Vt} \end{equation*}  over what it would have with $V=0$. That is just like a change in the zero of our energy scale. It produces an equal phase change in all amplitudes, but as we have seen before, this doesn’t change any of the probabilities. All the physical phenomena are the same. (We have assumed that we are talking about different states of the same charged object, so that $q\phi$ is the same for all. If an object could change its charge in going from one state to another, we would have quite another result, but conservation of charge prevents this.) So far, our assumption agrees with what we would expect for a change of energy reference level. But if it is really right, it should hold for a potential energy that is not just a constant. In general, $V$ could vary in any arbitrary way with both time and space, and the complete result for the amplitude must be given in terms of a differential equation. We don’t want to get concerned with the general case right now, but only want to get some idea about how some things happen, so we will think only of a potential that is constant in time and varies very slowly in space. Then we can make a comparison between the classical and quantum ideas. Suppose we think of the situation in Fig. 7–3, which has two boxes held at the constant potentials $\phi_1$ and $\phi_2$ and a region in between where we will assume that the potential varies smoothly from one to the other. We imagine that some particle has an amplitude to be found in any one of the regions. We also assume that the momentum is large enough so that in any small region in which there are many wavelengths, the potential is nearly constant. We would then think that in any part of the space the amplitude ought to look like (7.18) with the appropriate $V$ for that part of the space. Let’s think of a special case in which $\phi_1=0$, so that the potential energy there is zero, but in which $q\phi_2$ is negative, so that classically the particle would have more energy in the second box. Classically, it would be going faster in the second box—it would have more energy and, therefore, more momentum. Let’s see how that might come out of quantum mechanics. With our assumption, the amplitude in the first box would be proportional to \begin{equation} \label{Eq:III:7:21} e^{-(i/\hbar)[(W_{\text{int}}+p_1^2/2M+V_1)t-\FLPp_1\cdot\FLPx]}, \end{equation}  and the amplitude in the second box would be proportional to \begin{equation} \label{Eq:III:7:22} e^{-(i/\hbar)[(W_{\text{int}}+p_2^2/2M+V_2)t-\FLPp_2\cdot\FLPx]}. \end{equation}  (Let’s say that the internal energy is not being changed, but remains the same in both regions.) The question is: How do these two amplitudes match together through the region between the boxes? We are going to suppose that the potentials are all constant in time—so that nothing in the conditions varies. We will then suppose that the variations of the amplitude (that is, its phase) have the same frequency everywhere—because, so to speak, there is nothing in the “medium” that depends on time. If nothing in the space is changing, we can consider that the wave in one region “generates” subsidiary waves all over space which will all oscillate at the same frequency—just as light waves going through materials at rest do not change their frequency. If the frequencies in (7.21) and (7.22) are the same, we must have that \begin{equation} \label{Eq:III:7:23} W_{\text{int}}+\frac{p_1^2}{2M}+V_1= W_{\text{int}}+\frac{p_2^2}{2M}+V_2. \end{equation}  Both sides are just the classical total energies, so Eq. (7.23) is a statement of the conservation of energy. In other words, the classical statement of the conservation of energy is equivalent to the quantum mechanical statement that the frequencies for a particle are everywhere the same if the conditions are not changing with time. It all fits with the idea that $\hbar\omega=E$. In the special example that $V_1=0$ and $V_2$ is negative, Eq. (7.23) gives that $p_2$ is greater than $p_1$, so the wavelength of the waves is shorter in region $2$. The surfaces of equal phase are shown by the dashed lines in Fig. 7–3. We have also drawn a graph of the real part of the amplitude, which shows again how the wavelength decreases in going from region $1$ to region $2$. The group velocity of the waves, which is $p/M$, also increases in the way one would expect from the classical energy conservation, since it is just the same as Eq. (7.23). There is an interesting special case where $V_2$ gets so large that $V_2-V_1$ is greater than $p_1^2/2M$. Then $p_2^2$, which is given by \begin{equation} \label{Eq:III:7:24} p_2^2=2M\biggl[\frac{p_1^2}{2M}-V_2+V_1\biggr], \end{equation}  is negative. That means that $p_2$ is an imaginary number, say, $ip'$. Classically, we would say that the particle never gets into region $2$—it doesn’t have enough energy to climb the potential hill. Quantum mechanically, however, the amplitude is still given by Eq. (7.22); its space variation still goes as \begin{equation*} e^{(i/\hbar)\FLPp_2\cdot\FLPx}. \end{equation*}  But if $p_2$ is imaginary, the space dependence becomes a real exponential. Say that the particle was initially going in the $+x$-direction; then the amplitude would vary as \begin{equation} \label{Eq:III:7:25} e^{-p'x/\hbar}. \end{equation}  The amplitude decreases rapidly with increasing $x$. Imagine that the two regions at different potentials were very close together, so that the potential energy changed suddenly from $V_1$ to $V_2$, as shown in Fig. 7–4(a). If we plot the real part of the probability amplitude, we get the dependence shown in part (b) of the figure. The wave in the first region corresponds to a particle trying to get into the second region, but the amplitude there falls off rapidly. There is some chance that it will be observed in the second region—where it could never get classically—but the amplitude is very small except right near the boundary. The situation is very much like what we found for the total internal reflection of light. The light doesn’t normally get out, but we can observe it if we put something within a wavelength or two of the surface. You will remember that if we put a second surface close to the boundary where light was totally reflected, we could get some light transmitted into the second piece of material. The corresponding thing happens to particles in quantum mechanics. If there is a narrow region with a potential $V$, so great that the classical kinetic energy would be negative, the particle would classically never get past. But quantum mechanically, the exponentially decaying amplitude can reach across the region and give a small probability that the particle will be found on the other side where the kinetic energy is again positive. The situation is illustrated in Fig. 7–5. This effect is called the quantum mechanical “penetration of a barrier.”  The barrier penetration by a quantum mechanical amplitude gives the explanation—or description—of the $\alpha$-particle decay of a uranium nucleus. The potential energy of an $\alpha$-particle, as a function of the distance from the center, is shown in Fig. 7–6(a). If one tried to shoot an $\alpha$-particle with the energy $E$ into the nucleus, it would feel an electrostatic repulsion from the nuclear charge $z$ and would, classically, get no closer than the distance $r_1$ where its total energy is equal to the potential energy $V$. Closer in, however, the potential energy is much lower because of the strong attraction of the short-range nuclear forces. How is it then that in radioactive decay we find $\alpha$-particles which started out inside the nucleus coming out with the energy $E$? Because they start out with the energy $E$ inside the nucleus and “leak” through the potential barrier. The probability amplitude is roughly as sketched in part (b) of Fig. 7–6, although actually the exponential decay is much larger than shown. It is, in fact, quite remarkable that the mean life of an $\alpha$-particle in the uranium nucleus is as long as $4\tfrac{1}{2}$ billion years, when the natural oscillations inside the nucleus are so extremely rapid—about $10^{22}$ per sec! How can one get a number like $10^9$ years from $10^{-22}$ sec? The answer is that the exponential gives the tremendously small factor of about $e^{-45}$—which gives the very small, though definite, probability of leakage. Once the $\alpha$-particle is in the nucleus, there is almost no amplitude at all for finding it outside; however, if you take many nuclei and wait long enough, you may be lucky and find one that has come out.