Light scattering
Let’s use these results to solve a somewhat more complicated problem—but also one which is somewhat more real. We suppose that the same atoms are sitting in their ground state ($j=0$), and scatter an incoming beam of light. Let’s say that the light is going initially in the $+z$-direction, so that we have photons coming up to the atom from the $-z$-direction, as shown in Fig. 18–4(a). We can consider the scattering of light as a two-step process: The photon is absorbed, and then is re-emitted. If we start with a RHC photon as in Fig. 18–4(a), and angular momentum is conserved, the atom will be in an $m=+1$ state after the absorption—as shown in Fig. 18–4(b). We call the amplitude for this process $c$. The atom can then emit a RHC photon in the direction $\theta$—as in Fig. 18–4(c). The total amplitude that a RHC photon is scattered in the direction $\theta$ is just $c$ times (18.1). Let’s call this scattering amplitude $\bracket{R'}{S}{R}$; we have \begin{equation} \label{Eq:III:18:3} \bracket{R'}{S}{R}=\frac{ac}{2}\,(1+\cos\theta). \end{equation}  There is also an amplitude that a RHC photon will be absorbed and that a LHC photon will be emitted. The product of the two amplitudes is the amplitude $\bracket{L'}{S}{R}$ that a RHC photon is scattered as a LHC photon. Using (18.2), we have \begin{equation} \label{Eq:III:18:4} \bracket{L'}{S}{R}=-\frac{ac}{2}\,(1-\cos\theta). \end{equation}  Now let’s ask about what happens if a LHC photon comes in. When it is absorbed, the atom will go into an $m=-1$ state. By the same kind of arguments we used in the preceding section, we can show that this amplitude must be $-c$. The amplitude that an atom in the $m=-1$ state will emit a RHC photon at the angle $\theta$ is $a$ times the amplitude $\bracket{+}{R_y(\theta)}{-}$, which is $\tfrac{1}{2}(1-\cos\theta)$. So we have \begin{equation} \label{Eq:III:18:5} \bracket{R'}{S}{L}=-\frac{ac}{2}\,(1-\cos\theta). \end{equation} Finally, the amplitude for a LHC photon to be scattered as a LHC photon is \begin{equation} \label{Eq:III:18:6} \bracket{L'}{S}{L}=\frac{ac}{2}\,(1+\cos\theta). \end{equation} (There are two minus signs which cancel.)  If we make a measurement of the scattered intensity for any given combination of circular polarizations it will be proportional to the square of one of our four amplitudes. For instance, with an incoming beam of RHC light the intensity of the RHC light in the scattered radiation will vary as $(1+\cos\theta)^2$.  That’s all very well, but suppose we start out with linearly polarized light. What then? If we have $x$-polarized light, it can be represented as a superposition of RHC and LHC light. We write (see Section 11-4) \begin{equation} \label{Eq:III:18:7} \ket{x}=\frac{1}{\sqrt{2}}\,(\ket{R}+\ket{L}). \end{equation} Or, if we have $y$-polarized light, we would have \begin{equation} \label{Eq:III:18:8} \ket{y}=-\frac{i}{\sqrt{2}}\,(\ket{R}-\ket{L}). \end{equation} Now what do you want to know? Do you want the amplitude that an $x$-polarized photon will scatter into a RHC photon at the angle $\theta$? You can get it by the usual rule for combining amplitudes. First, multiply (18.7) by $\bra{R'}\,S$ to get \begin{equation} \label{Eq:III:18:9} \bracket{R'}{S}{x}=\!\frac{1}{\sqrt{2}}\, (\bracket{R'}{S}{R}\!+\!\bracket{R'}{S}{L}), \end{equation} and then use (18.3) and (18.5) for the two amplitudes. You get \begin{equation} \label{Eq:III:18:10} \bracket{R'}{S}{x}=\frac{ac}{\sqrt{2}}\cos\theta. \end{equation} If you wanted the amplitude that an $x$-photon would scatter into a LHC photon, you would get \begin{equation} \label{Eq:III:18:11} \bracket{L'}{S}{x}=\frac{ac}{\sqrt{2}}\cos\theta. \end{equation}  Finally, suppose you wanted to know the amplitude that an $x$-polarized photon will scatter while keeping its $x$-polarization. What you want is $\bracket{x'}{S}{x}$. This can be written as  \begin{equation} \label{Eq:III:18:12} \bracket{x'}{S}{x}=\braket{x'}{R'}\bracket{R'}{S}{x}+ \braket{x'}{L'}\bracket{L'}{S}{x}. \end{equation} 
 \begin{align} \bracket{x'}{S}{x}=\,&\braket{x'}{R'}\bracket{R'}{S}{x}\notag\\[.5ex] \label{Eq:III:18:12} +\,&\braket{x'}{L'}\bracket{L'}{S}{x}. \end{align}  If you then use the relations  \begin{align} \label{Eq:III:18:13} \ket{R'}&=\frac{1}{\sqrt{2}}\,(\ket{x'}+i\,\ket{y'}),\\[1ex] \label{Eq:III:18:14} \ket{L'}&=\frac{1}{\sqrt{2}}\,(\ket{x'}-i\,\ket{y'}) \end{align}  it follows that  \begin{align} \label{Eq:III:18:15} \braket{x'}{R'}&=\frac{1}{\sqrt{2}},\\[1ex] \label{Eq:III:18:16} \braket{x'}{L'}&=\frac{1}{\sqrt{2}}. \end{align}  So you get that  \begin{equation} \label{Eq:III:18:17} \bracket{x'}{S}{x}=ac\cos\theta. \end{equation}  The answer is that a beam of $x$-polarized light will be scattered at the direction $\theta$ (in the $xz$-plane) with an intensity proportional to $\cos^2\theta$. If you ask about $y$-polarized light, you find that \begin{equation} \label{Eq:III:18:18} \bracket{y'}{S}{x}=0. \end{equation} So the scattered light is completely polarized in the $x$-direction.  Now we notice something interesting. The results (18.17) and (18.18) correspond exactly to the classical theory of light scattering we gave in Vol. I, Section 32-5, where we imagined that the electron was bound to the atom by a linear restoring force—so that it acted like a classical oscillator. Perhaps you are thinking: “It’s so much easier in the classical theory; if it gives the right answer why bother with the quantum theory?” For one thing, we have considered so far only the special—though common—case of an atom with a $j=1$ excited state and a $j=0$ ground state. If the excited state had spin two, you would get a different result. Also, there is no reason why the model of an electron attached to a spring and driven by an oscillating electric field should work for a single photon. But we have found that it does in fact work, and that the polarization and intensities come out right. So in a certain sense we are bringing the whole course around to the real truth. Whereas we have, in Vol. I, done the theory of the index of refraction, and of light scattering, by the classical theory, we have now shown that the quantum theory gives the same result for the most common case. In effect we have now done the polarization of sky light, for instance, by quantum mechanical arguments, which is the only truly legitimate way.  It should be, of course, that all the classical theories which work are supported ultimately by legitimate quantum arguments. Naturally, those things which we have spent a great deal of time in explaining to you were selected from just those parts of classical physics which still maintain validity in quantum mechanics. You’ll notice that we did not discuss in great detail any model of the atom which has electrons going around in orbits. That’s because such a model doesn’t give results which agree with the quantum mechanics. But the electron on a spring—which is not, in a sense, at all the way an atom “looks”—does work, and so we used that model for the theory of the index of refraction. 