Nuclear forces
We have seen that the system of a hydrogen atom and a proton has an energy of interaction due to the exchange of the single electron which varies at large separations $R$ as \begin{equation} \label{Eq:III:10:11} \frac{e^{-\alpha R}}{R}, \end{equation}  with $\alpha=\sqrt{2mW_H}/\hbar$. (One usually says that there is an exchange of a “virtual” electron when—as here—the electron has to jump across a space where it would have a negative energy. More specifically, a “virtual exchange” means that the phenomenon involves a quantum mechanical interference between an exchanged state and a nonexchanged state.) Now we might ask the following question: Could it be that forces between other kinds of particles have an analogous origin? What about, for example, the nuclear force between a neutron and a proton, or between two protons? In an attempt to explain the nature of nuclear forces, Yukawa proposed that the force between two nucleons is due to a similar exchange effect—only, in this case, due to the virtual exchange, not of an electron, but of a new particle, which he called a “meson.” Today, we would identify Yukawa’s meson with the $\pi$-meson (or “pion”) produced in high-energy collisions of protons or other particles. Let’s see, as an example, what kind of a force we would expect from the exchange of a positive pion ($\pi^+$) of mass $m_\pi$ between a proton and a neutron. Just as a hydrogen atom H$^0$ can go into a proton p$^+$ by giving up an electron e$^-$, \begin{equation} \label{Eq:III:10:12} \text{H}^0\to\text{p}^++\text{e}^-, \end{equation}  a proton p$^+$ can go into a neutron n$^0$ by giving up a $\pi^+$ meson: \begin{equation} \label{Eq:III:10:13} \text{p}^+\to\text{n}^0+\pi^+. \end{equation}  So if we have a proton at $a$ and a neutron at $b$ separated by the distance $R$, the proton can become a neutron by emitting a $\pi^+$, which is then absorbed by the neutron at $b$, turning it into a proton. There is an energy of interaction of the two-nucleon (plus pion) system which depends on the amplitude $A$ for the pion exchange—just as we found for the electron exchange in the $\text{H}_2^+$ ion. In the process (10.12), the energy of the H$^0$ atom is less than that of the proton by $W_H$ (calculating nonrelativistically, and omitting the rest energy $mc^2$ of the electron), so the electron has a negative kinetic energy—or imaginary momentum—as in Eq. (10.9). In the nuclear process (10.13), the proton and neutron have almost equal masses, so the $\pi^+$ will have zero total energy. The relation between the total energy $E$ and the momentum $p$ for a pion of mass $m_\pi$ is \begin{equation*} E^2=p^2c^2+m_\pi^2c^4. \end{equation*}  Since $E$ is zero (or at least negligible in comparison with $m_\pi$), the momentum is again imaginary: \begin{equation*} p=im_\pi c. \end{equation*}  Using the same arguments we gave for the amplitude that a bound electron would penetrate the barrier in the space between two protons, we get for the nuclear case an exchange amplitude $A$ which should—for large $R$—go as \begin{equation} \label{Eq:III:10:14} \frac{e^{-(m_\pi c/\hbar)R}}{R}. \end{equation}  The interaction energy is proportional to $A$, and so varies in the same way. We get an energy variation in the form of the so-called Yukawa potential between two nucleons. Incidentally, we obtained this same formula earlier directly from the differential equation for the motion of a pion in free space [see Chapter 28, Vol. II, Eq. (28.18)]. We can, following the same line of argument, discuss the interaction between two protons (or between two neutrons) which results from the exchange of a neutral pion ($\pi^0$). The basic process is now \begin{equation} \label{Eq:III:10:15} \text{p}^+\to\text{p}^++\pi^0. \end{equation}  A proton can emit a virtual $\pi^0$, but then it remains still a proton. If we have two protons, proton No. $1$ can emit a virtual $\pi^0$ which is absorbed by proton No. $2$. At the end, we still have two protons. This is somewhat different from the $\text{H}_2^+$ ion. There the H$^0$ went into a different condition—the proton—after emitting the electron. Now we are assuming that a proton can emit a $\pi^0$ without changing its character. Such processes are, in fact, observed in high-energy collisions. The process is analogous to the way that an electron emits a photon and ends up still an electron: \begin{equation} \label{Eq:III:10:16} \text{e}\to\text{e}+\text{photon}. \end{equation}  We do not “see” the photons inside the electrons before they are emitted or after they are absorbed, and their emission does not change the “nature” of the electron. Going back to the two protons, there is an interaction energy which arises from the amplitude $A$ that one proton emits a neutral pion which travels across (with imaginary momentum) to the other proton and is absorbed there. This amplitude is again proportional to (10.14), with $m_\pi$ the mass of the neutral pion. All the same arguments give an equal interaction energy for two neutrons. Since the nuclear forces (disregarding electrical effects) between neutron and proton, between proton and proton, between neutron and neutron are the same, we conclude that the masses of the charged and neutral pions should be the same. Experimentally, the masses are indeed very nearly equal, and the small difference is about what one would expect from electric self-energy corrections (see Chapter 28, Vol. II). There are other kinds of particles—like K-mesons—which can be exchanged between two nucleons. It is also possible for two pions to be exchanged at the same time. But all of these other exchanged “objects” have a rest mass $m_x$ higher than the pion mass $m_\pi$, and lead to terms in the exchange amplitude which vary as \begin{equation*} \frac{e^{-(m_xc/\hbar)R}}{R}. \end{equation*}  These terms die out faster with increasing $R$ than the one-meson term. No one knows, today, how to calculate these higher-mass terms, but for large enough values of $R$ only the one-pion term survives. And, indeed, those experiments which involve nuclear interactions only at large distances do show that the interaction energy is as predicted from the one-pion exchange theory. In the classical theory of electricity and magnetism, the Coulomb electrostatic interaction and the radiation of light by an accelerating charge are closely related—both come out of the Maxwell equations. We have seen in the quantum theory that light can be represented as the quantum excitations of the harmonic oscillations of the classical electromagnetic fields in a box. Alternatively, the quantum theory can be set up by describing light in terms of particles—photons—which obey Bose statistics. We emphasized in Section 4–5 that the two alternative points of view always give identical predictions. Can the second point of view be carried through completely to include all electromagnetic effects? In particular, if we want to describe the electromagnetic field purely in terms of Bose particles—that is, in terms of photons—what is the Coulomb force due to? From the “particle” point of view the Coulomb interaction between two electrons comes from the exchange of a virtual photon. One electron emits a photon—as in reaction (10.16)—which goes over to the second electron, where it is absorbed in the reverse of the same reaction. The interaction energy is again given by a formula like (10.14), but now with $m_\pi$ replaced by the rest mass of the photon—which is zero. So the virtual exchange of a photon between two electrons gives an interaction energy that varies simply inversely as $R$, the distance between the two electrons—just the normal Coulomb potential energy! In the “particle” theory of electromagnetism, the process of a virtual photon exchange gives rise to all the phenomena of electrostatics.