Measuring a nuclear spin
We would like to show you one example of the application of the coefficients we have just described. It has to do with a recent, interesting experiment which you will now be able to understand. Some physicists wanted to find out the spin of a certain excited state of the Ne$^{20}$ nucleus. To do this, they bombarded a carbon target with a beam of accelerated carbon ions, and produced the desired excited state of Ne$^{20}$—called Ne$^{20*}$—in the reaction \begin{equation*} \text{C}^{12}+\text{C}^{12}\to \text{Ne}^{20*}+\alpha_1, \end{equation*} where $\alpha_1$ is the $\alpha$-particle, or He$^4$. Several of the excited states of Ne$^{20}$ produced this way are unstable and disintegrate in the reaction \begin{equation*} \text{Ne}^{20*}\to\text{O}^{16}+\alpha_2. \end{equation*} So experimentally there are two $\alpha$-particles which come out of the reaction. We call them $\alpha_1$ and $\alpha_2$; since they come off with different energies, they can be distinguished from each other. Also, by picking a particular energy for $\alpha_1$ we can pick out any particular excited state of the Ne$^{20}$.  The experiment was set up as shown in Fig. 18–9. A beam of $16$-MeV carbon ions was directed onto a thin foil of carbon. The first $\alpha$-particle was counted in a silicon diffused junction detector marked $\alpha_1$—set to accept $\alpha$-particles of the proper energy moving in the forward direction (with respect to the incident C$^{12}$ beam). The second $\alpha$-particle was picked up in the counter $\alpha_2$ at the angle $\theta$ with respect to $\alpha_1$. The counting rate of coincidence signals from $\alpha_1$ and $\alpha_2$ were measured as a function of the angle $\theta$.  The idea of the experiment is the following. First, you need to know that the spins of C$^{12}$, O$^{16}$, and the $\alpha$-particle are all zero. If we call the direction of motion of the initial C$^{12}$ the $+z$-direction, then we know that the Ne$^{20*}$ must have zero angular momentum about the $z$-axis. None of the other particles has any spin; the C$^{12}$ arrives along the $z$-axis and the $\alpha_1$ leaves along the $z$-axis so they can’t have any angular momentum about it. So whatever the spin $j$ of the Ne$^{20*}$ is, we know that it is in the state $\ket{j,0}$. Now what will happen when the Ne$^{20*}$ disintegrates into an O$^{16}$ and the second $\alpha$-particle?  Well, the $\alpha$-particle is picked up in the counter $\alpha_2$ and to conserve momentum the O$^{16}$ must go off in the opposite direction.8
About the new axis through $\alpha_2$, there can be no component of angular momentum. The final state has zero angular momentum about the new axis, so the Ne$^{20*}$ can disintegrate this way only if it has some amplitude to have $m'$ equal to zero, where $m'$ is the quantum number of the component of angular momentum about the new axis. In fact, the probability of observing $\alpha_2$ at the angle $\theta$ is just the square of the amplitude (or matrix element) \begin{equation} \label{Eq:III:18:41} \bracket{j,0}{R_y(\theta)}{j,0}. \end{equation}  To find the spin of the Ne$^{20*}$ state in question, the intensity of the second $\alpha$-particle was plotted as a function of angle and compared with the theoretical curves for various values of $j$. As we said in the last section, the amplitudes $\bracket{j,0}{R_y(\theta)}{j,0}$ are just the functions $P_j(\cos\theta)$. So the possible angular distributions are curves of $[P_j(\cos\theta)]^2$. The experimental results are shown in Fig. 18–10 for two of the excited states. You can see that the angular distribution for the $5.80$-MeV state fits very well the curve for $[P_1(\cos\theta)]^2$, and so it must be a spin-one state. The data for the $5.63$-MeV state, on the other hand, are quite different; they fit the curve $[P_3(\cos\theta)]^2$. The state has a spin of $3$.  From this experiment we have been able to find out the angular momentum of two of the excited states of Ne$^{20*}$. This information can then be used for trying to understand what the configuration of protons and neutrons is inside this nucleus—one more piece of information about the mysterious nuclear forces. 