Atoms at rest; stationary states
We want now to talk a little bit about the behavior of probability amplitudes in time. We say a “little bit,” because the actual behavior in time necessarily involves the behavior in space as well. Thus, we get immediately into the most complicated possible situation if we are to do it correctly and in detail. We are always in the difficulty that we can either treat something in a logically rigorous but quite abstract way, or we can do something which is not at all rigorous but which gives us some idea of a real situation—postponing until later a more careful treatment. With regard to energy dependence, we are going to take the second course. We will make a number of statements. We will not try to be rigorous—but will just be telling you things that have been found out, to give you some feeling for the behavior of amplitudes as a function of time. As we go along, the precision of the description will increase, so don’t get nervous that we seem to be picking things out of the air. It is, of course, all out of the air—the air of experiment and of the imagination of people. But it would take us too long to go over the historical development, so we have to plunge in somewhere. We could plunge into the abstract and deduce everything—which you would not understand—or we could go through a large number of experiments to justify each statement. We choose to do something in between. An electron alone in empty space can, under certain circumstances, have a certain definite energy. For example, if it is standing still (so it has no translational motion, no momentum, or kinetic energy), it has its rest energy. A more complicated object like an atom can also have a definite energy when standing still, but it could also be internally excited to another energy level. (We will describe later the machinery of this.) We can often think of an atom in an excited state as having a definite energy, but this is really only approximately true. An atom doesn’t stay excited forever because it manages to discharge its energy by its interaction with the electromagnetic field. So there is some amplitude that a new state is generated—with the atom in a lower state, and the electromagnetic field in a higher state, of excitation. The total energy of the system is the same before and after, but the energy of the atom is reduced. So it is not precise to say an excited atom has a definite energy; but it will often be convenient and not too wrong to say that it does. [Incidentally, why does it go one way instead of the other way? Why does an atom radiate light? The answer has to do with entropy. When the energy is in the electromagnetic field, there are so many different ways it can be—so many different places where it can wander—that if we look for the equilibrium condition, we find that in the most probable situation the field is excited with a photon, and the atom is de-excited. It takes a very long time for the photon to come back and find that it can knock the atom back up again. It’s quite analogous to the classical problem: Why does an accelerating charge radiate? It isn’t that it “wants” to lose energy, because, in fact, when it radiates, the energy of the world is the same as it was before. Radiation or absorption goes in the direction of increasing entropy.] Nuclei can also exist in different energy levels, and in an approximation which disregards the electromagnetic effects, we can say that a nucleus in an excited state stays there. Although we know that it doesn’t stay there forever, it is often useful to start out with an approximation which is somewhat idealized and easier to think about. Also it is often a legitimate approximation under certain circumstances. (When we first introduced the classical laws of a falling body, we did not include friction, but there is almost never a case in which there isn’t some friction.) Then there are the subnuclear “strange particles,” which have various masses. But the heavier ones disintegrate into other light particles, so again it is not correct to say that they have a precisely definite energy. That would be true only if they lasted forever. So when we make the approximation that they have a definite energy, we are forgetting the fact that they must blow up. For the moment, then, we will intentionally forget about such processes and learn later how to take them into account. Suppose we have an atom—or an electron, or any particle—which at rest would have a definite energy $E_0$. By the energy $E_0$ we mean the mass of the whole thing times $c^2$. This mass includes any internal energy; so an excited atom has a mass which is different from the mass of the same atom in the ground state. (The ground state means the state of lowest energy.) We will call $E_0$ the “energy at rest.” For an atom at rest, the quantum mechanical amplitude to find an atom at a place is the same everywhere; it does not depend on position. This means, of course, that the probability of finding the atom anywhere is the same. But it means even more. The probability could be independent of position, and still the phase of the amplitude could vary from point to point. But for a particle at rest, the complete amplitude is identical everywhere. It does, however, depend on the time. For a particle in a state of definite energy $E_0$, the amplitude to find the particle at $(x,y,z)$ at the time $t$ is \begin{equation} \label{Eq:III:7:1} ae^{-i(E_0/\hbar)t}, \end{equation}  where $a$ is some constant. The amplitude to be at any point in space is the same for all points, but depends on time according to (7.1). We shall simply assume this rule to be true. Of course, we could also write (7.1) as \begin{equation} \label{Eq:III:7:2} ae^{-i\omega t}, \end{equation}  with \begin{equation*} \hbar\omega=E_0=Mc^2, \end{equation*}  where $M$ is the rest mass of the atomic state, or particle. There are three different ways of specifying the energy: by the frequency of an amplitude, by the energy in the classical sense, or by the inertia. They are all equivalent; they are just different ways of saying the same thing. You may be thinking that it is strange to think of a “particle” which has equal amplitudes to be found throughout all space. After all, we usually imagine a “particle” as a small object located “somewhere.”  But don’t forget the uncertainty principle. If a particle has a definite energy, it has also a definite momentum. If the uncertainty in momentum is zero, the uncertainty relation, $\Delta p\,\Delta x=\hbar$, tells us that the uncertainty in the position must be infinite, and that is just what we are saying when we say that there is the same amplitude to find the particle at all points in space. If the internal parts of an atom are in a different state with a different total energy, then the variation of the amplitude with time is different. If you don’t know in which state it is, there will be a certain amplitude to be in one state and a certain amplitude to be in another—and each of these amplitudes will have a different frequency. There will be an interference between these different components—like a beat-note—which can show up as a varying probability. Something will be “going on” inside of the atom—even though it is “at rest” in the sense that its center of mass is not drifting. However, if the atom has one definite energy, the amplitude is given by (7.1), and the absolute square of this amplitude does not depend on time. You see, then, that if a thing has a definite energy and if you ask any probability question about it, the answer is independent of time. Although the amplitudes vary with time, if the energy is definite they vary as an imaginary exponential, and the absolute value doesn’t change.  That’s why we often say that an atom in a definite energy level is in a stationary state. If you make any measurements of the things inside, you’ll find that nothing (in probability) will change in time. In order to have the probabilities change in time, we have to have the interference of two amplitudes at two different frequencies, and that means that we cannot know what the energy is. The object will have one amplitude to be in a state of one energy and another amplitude to be in a state of another energy. That’s the quantum mechanical description of something when its behavior depends on time. If we have a “condition” which is a mixture of two different states with different energies, then the amplitude for each of the two states varies with time according to Eq. (7.2), for instance, as \begin{equation} \label{Eq:III:7:3} e^{-i(E_1/\hbar)t}\quad \text{and}\quad e^{-i(E_2/\hbar)t}. \end{equation}  And if we have some combination of the two, we will have an interference. But notice that if we added a constant to both energies, it wouldn’t make any difference. If somebody else were to use a different scale of energy in which all the energies were increased (or decreased) by a constant amount—say, by the amount $A$—then the amplitudes in the two states would, from his point of view, be \begin{equation} \label{Eq:III:7:4} e^{-i(E_1+A)t/\hbar}\quad \text{and}\quad e^{-i(E_2+A)t/\hbar}. \end{equation}  All of his amplitudes would be multiplied by the same factor $e^{-i(A/\hbar)t}$, and all linear combinations, or interferences, would have the same factor. When we take the absolute squares to find the probabilities, all the answers would be the same. The choice of an origin for our energy scale makes no difference; we can measure energy from any zero we want. For relativistic purposes it is nice to measure the energy so that the rest mass is included, but for many purposes that aren’t relativistic it is often nice to subtract some standard amount from all energies that appear. For instance, in the case of an atom, it is usually convenient to subtract the energy $M_sc^2$, where $M_s$ is the mass of all the separate pieces—the nucleus and the electrons—which is, of course, different from the mass of the atom. For other problems it may be useful to subtract from all energies the amount $M_gc^2$, where $M_g$ is the mass of the whole atom in the ground state; then the energy that appears is just the excitation energy of the atom. So, sometimes we may shift our zero of energy by some very large constant, but it doesn’t make any difference, provided we shift all the energies in a particular calculation by the same constant. So much for a particle standing still.