Emission and absorption of photons
Throughout our discussion we have talked about a process like the scattering of $\alpha$-particles. But that is not essential; we could have been speaking of the creation of particles, as for instance the emission of light. When the light is emitted, a photon is “created.” In such a case, we don’t need the incoming lines in Fig. 4–4; we can consider merely that there are some atoms emitting $n$ photons, as in Fig. 4–5. So our result can also be stated: The probability that an atom will emit a photon into a particular final state is increased by the factor $(n+1)$ if there are already $n$ photons in that state. People like to summarize this result by saying that the amplitude to emit a photon is increased by the factor $\sqrt{n+1}$ when there are already $n$ photons present. It is, of course, another way of saying the same thing if it is understood to mean that this amplitude is just to be squared to get the probability. It is generally true in quantum mechanics that the amplitude to get from any condition $\phi$ to any other condition $\chi$ is the complex conjugate of the amplitude to get from $\chi$ to $\phi$: \begin{equation} \label{Eq:III:4:24} \braket{\chi}{\phi}=\braket{\phi}{\chi}\cconj. \end{equation}  We will learn about this law a little later, but for the moment we will just assume it is true. We can use it to find out how photons are scattered or absorbed out of a given state. We have that the amplitude that a photon will be added to some state, say $i$, when there are already $n$ photons present is, say, \begin{equation} \label{Eq:III:4:25} \braket{n+1}{n}=\sqrt{n+1}\,a, \end{equation}  where $a=\braket{i}{a}$ is the amplitude when there are no others present. Using Eq. (4.24), the amplitude to go the other way—from $(n+1)$ photons to $n$—is \begin{equation} \label{Eq:III:4:26} \braket{n}{n+1}=\sqrt{n+1}\,a\cconj. \end{equation}  This isn’t the way people usually say it; they don’t like to think of going from $(n+1)$ to $n$, but prefer always to start with $n$ photons present. Then they say that the amplitude to absorb a photon when there are $n$ present—in other words, to go from $n$ to $(n-1)$—is \begin{equation} \label{Eq:III:4:27} \braket{n-1}{n}=\sqrt{n}\,a\cconj. \end{equation}  which is, of course, just the same as Eq. (4.26). Then they have trouble trying to remember when to use $\sqrt{n}$ or $\sqrt{n+1}$. Here’s the way to remember: The factor is always the square root of the largest number of photons present, whether it is before or after the reaction. Equations (4.25) and (4.26) show that the law is really symmetric—it only appears unsymmetric if you write it as Eq. (4.27). There are many physical consequences of these new rules; we want to describe one of them having to do with the emission of light. Suppose we imagine a situation in which photons are contained in a box—you can imagine a box with mirrors for walls. Now say that in the box we have $n$ photons, all of the same state—the same frequency, direction, and polarization—so they can’t be distinguished, and that also there is an atom in the box that can emit another photon into the same state. Then the probability that it will emit a photon is \begin{equation} \label{Eq:III:4:28} (n+1)\abs{a}^2, \end{equation}  and the probability that it will absorb a photon is \begin{equation} \label{Eq:III:4:29} n\abs{a}^2, \end{equation}  where $\abs{a}^2$ is the probability it would emit if no photons were present. We have already discussed these rules in a somewhat different way in Chapter 42 of Vol. I. Equation (4.29) says that the probability that an atom will absorb a photon and make a transition to a higher energy state is proportional to the intensity of the light shining on it. But, as Einstein first pointed out, the rate at which an atom will make a transition downward has two parts. There is the probability that it will make a spontaneous transition $\abs{a}^2$, plus the probability of an induced transition $n\abs{a}^2$, which is proportional to the intensity of the light—that is, to the number of photons present. Furthermore, as Einstein said, the coefficients of absorption and of induced emission are equal and are related to the probability of spontaneous emission. What we learn here is that if the light intensity is measured in terms of the number of photons present (instead of as the energy per unit area, and per sec), the coefficients of absorption, of induced emission, and of spontaneous emission are all equal. This is the content of the relation between the Einstein coefficients $A$ and $B$ of Chapter 42, Vol. I, Eq. (42.18).