Electric dipole radiation
In the last chapter we developed the idea of the conservation of angular momentum in quantum mechanics, and showed how it might be used to predict the angular distribution of the proton from the disintegration of the $\Lambda$-particle. We want now to give you a number of other, similar, illustrations of the consequences of momentum conservation in atomic systems. Our first example is the radiation of light from an atom. The conservation of angular momentum (among other things) will determine the polarization and angular distribution of the emitted photons.  Suppose we have an atom which is in an excited state of definite angular momentum—say with a spin of one—and it makes a transition to a state of angular momentum zero at a lower energy, emitting a photon. The problem is to figure out the angular distribution and polarization of the photons. (This problem is almost exactly the same as the $\Lambda^0$ disintegration, except that we have spin-one instead of spin one-half particles.) Since the upper state of the atom is spin one, there are three possibilities for its $z$-component of angular momentum. The value of $m$ could be $+1$, or $0$, or $-1$. We will take $m=+1$ for our example. Once you see how it goes, you can work out the other cases. We suppose that the atom is sitting with its angular momentum along the $+z$-axis—as in Fig. 18–1(a)—and ask with what amplitude it will emit right circularly polarized light upward along the $z$-axis, so that the atom ends up with zero angular momentum—as shown in part (b) of the figure. Well, we don’t know the answer to that. But we do know that right circularly polarized light has one unit of angular momentum about its direction of propagation. So after the photon is emitted, the situation would have to be as shown in Fig. 18–1(b)—the atom is left with zero angular momentum about the $z$-axis, since we have assumed an atom whose lower state is spin zero. We will let $a$ stand for the amplitude for such an event. More precisely, we let $a$ be the amplitude to emit a photon into a certain small solid angle $\Delta\Omega$, centered on the $z$-axis, during a time $dt$. Notice that the amplitude to emit a LHC photon in the same direction is zero. The net angular momentum about the $z$-axis would be $-1$ for such a photon and zero for the atom for a total of $-1$, which would not conserve angular momentum.  Similarly, if the spin of the atom is initially “down” ($-1$ along the $z$-axis), it can emit only a LHC polarized photon in the direction of the $+z$-axis, as shown in Fig. 18–2. We will let $b$ stand for the amplitude for this event—meaning again the amplitude that the photon goes into a certain solid angle $\Delta\Omega$. On the other hand, if the atom is in the $m=0$ state, it cannot emit a photon in the $+z$-direction at all, because a photon can have only the angular momentum $+1$ or $-1$ along its direction of motion.  Next, we can show that $b$ is related to $a$. Suppose we perform an inversion of the situation in Fig. 18–1, which means that we should imagine what the system would look like if we were to move each part of the system to an equivalent point on the opposite side of the origin. This does not mean that we should reflect the angular momentum vectors, because they are artificial. We should, rather, invert the actual character of the motion that would correspond to such an angular momentum. In Fig. 18–3(a) and (b) we show what the process of Fig. 18–1 looks like before and after an inversion with respect to the center of the atom. Notice that the sense of rotation of the atom is unchanged.1 In the inverted system of Fig. 18–3(b) we have an atom with $m=+1$ emitting a LHC photon downward.  If we now rotate the system of Fig. 18–3(b) by $180^\circ$ about the $x$- or $y$-axis, it becomes identical to Fig. 18–2. The combination of the inversion and rotation turns the second process into the first. Using Table 17–2, we see that a rotation of $180^\circ$ about the $y$-axis just throws an $m=-1$ state into an $m=+1$ state, so the amplitude $b$ must be equal to the amplitude $a$ except for a possible sign change due to the inversion. The sign change in the inversion will depend on the parities of the initial and final state of the atom.  In atomic processes, parity is conserved, so the parity of the whole system must be the same before and after the photon emission. What happens will depend on whether the parities of the initial and final states of the atom are even or odd—the angular distribution of the radiation will be different for different cases. We will take the common case of odd parity for the initial state and even parity for the final state; it will give what is called “electric dipole radiation.” (If the initial and final states have the same parity we say there is “magnetic dipole radiation,” which has the character of the radiation from an oscillating current in a loop.) If the parity of the initial state is odd, its amplitude reverses its sign in the inversion which takes the system from (a) to (b) of Fig. 18–3. The final state of the atom has even parity, so its amplitude doesn’t change sign. If the reaction is going to conserve parity, the amplitude $b$ must be equal to $a$ in magnitude but of the opposite sign.  We conclude that if the amplitude is $a$ that an $m=+1$ state will emit a photon upward, then for the assumed parities of the initial and final states the amplitude that an $m=-1$ state will emit a LHC photon upward is $-a$.2
 We have all we need to know to find the amplitude for a photon to be emitted at any angle $\theta$ with respect to the $z$-axis. Suppose we have an atom originally polarized with $m=+1$. We can resolve this state into $+1$, $0$, and $-1$ states with respect to a new $z'$-axis in the direction of the photon emission. The amplitudes for these three states are just the ones given in the lower half of Table 17–2. The amplitude that a RHC photon is emitted in the direction $\theta$ is then $a$ times the amplitude to have $m=+1$ in that direction, namely, \begin{equation} \label{Eq:III:18:1} a\bracket{+}{R_y(\theta)}{+}=\frac{a}{2}\,(1+\cos\theta). \end{equation} The amplitude that a LHC photon is emitted in the same direction is $-a$ times the amplitude to have $m=-1$ in the new direction. Using Table 17–2, it is \begin{equation} \label{Eq:III:18:2} -a\bracket{-}{R_y(\theta)}{+}=-\frac{a}{2}\,(1-\cos\theta). \end{equation} If you are interested in other polarizations you can find out the amplitude for them from the superposition of these two amplitudes. To get the intensity of any component as a function of angle, you must, of course, take the absolute square of the amplitudes. 