Anisotropic radiation from excited nuclear states I come from particle physics background, but I want to understand something related to nuclear physics. I don't have any training in nuclear physics, but please feel free to share any equations.
If you had asked me how a radiation from a decay looked like in the center of mass frame, I would have answered "isotropic of course!" until very recently. Why would the nature prefer a specific direction for the emission if the kinematics is not the limiting factor? I realized that some decays are intrinsically anisotropic.  To give an example you can check the following quotes about the excited $^{8}Be^*$ decay from this article:

It is known that the 18.15 MeV transition has a very large
(8:1) forward–backward anisotropy [16, 17], which is caused by the interference of the E1 amplitude due to the direct capture process and the M1 amplitude of the 441 keV and 1030 keV resonances.

Another example could be the figure below from this article which shows angular distributions of gamma radiations from the $^{7}Li(p,\gamma)^{8}Be$ reaction for different proton energies:



I did my research, but couldn't understand the reason for this behaviour. Can anyone explain this anisotropy in simple terms? It is even better if you can also provide other anisotropic events as examples. Maybe you can also suggest some lecture notes for a quick introduction to the topic. Thanks.
 A: Nuclear scattering and excitations can be messy to understand, but the existence of anisotropy in these nuclear reactions is no more mysterious than the anisotropy in a clean particle physics process such as $e^+ e^- \rightarrow J/\psi \rightarrow \mu^+ \mu^-$, where the anisotropic angular distribution of the muons is$$\frac{d\sigma}{d\cos\theta}\propto 1+\cos^2\theta$$
The "orientation" of any intermediate state and hence the angular distribution of any final-state particles is not random but is instead constrained by angular-momentum conservation.
As @Jon_Custer notes in a comment, the initial state in any scattering process is not isotropic – the incident beam defines an axis. There is no requirement that the final state be isotropic relative to that axis, and the angular distribution relative to that axis depends on how much different angular-momentum states contribute.
For example, Feynman III 18–5 "Measuring a nuclear spin" discusses how the angular distributions in $^{12}\mathrm{C}+^{12}\!\mathrm{C}\rightarrow ^{16}\!\mathrm{O} + \alpha_1 + \alpha_2$ can be use to determine the spin of intermediate $^{20}\mathrm{Ne}^∗$ excited states.
More generally in both nuclear and particle physics, partial wave analysis uses the angular distribution to determine the spin-parity of any resonant intermediate states, such as nuclear, mesonic, or baryonic excitations.
The $^{7}\mathrm{Li}(p,\gamma)^{8}\mathrm{Be}$ scattering shown in the question proceeds through low angular momentum non-resonant proton scattering and p-wave ($\ell=1$) production of $\,\mathrm{J^P}=1^+$ $\,^{8}\mathrm{Be}^*$  excitations, as discussed by the article cited in the question
and also
Liberman.
There is mixing of intermediate states and interference between different channels, and the strength of the various resonant and non-resonant contributions depends on the proton energy, so the angular distributions change with energy in ways that are not trivial to understand.
