About applying the angular momentum conservation to $\beta$ decay or similar decays Consider the $\beta^-$ decay:
$$n^0\to p^++e^-+{\bar\nu}_e.$$ The spin angular momenta of all the particles are given by $$S_n=S_p=S_e=S_{\nu_e}=1/2.$$ Therefore, by the addition of the angular momentum, the total spin of the final state particles can be either $S_f=3/2$ or $S_f=1/2$ which is the total spin of the initial state is $S_i=1/2$ (where $S_i=S_n$ in the our case). Here, I have uniformly used uppercase letter $``S"$ to denote the spin quantum numbers.
I want to know whether the conservation of angular momentum require both $$\boxed{S_i=S_f ~~{\rm and}~~ S_{iz}=S_{fz}?}$$
Let the initial neutron is in the spin state $S_i=1/2, S_{iz}=+1/2$. If that is the case, no decay is possible in which the final state particles are produced in the states $$|S_f=3/2,S_{fz}=-3/2\rangle$$ because $S_f\neq S_i$. Same reasoning forbids, the final state particles to be produced in the spin states $$|3/2,-1/2\rangle,~~ |3/2,1/2\rangle,~~ {\rm and} ~~|3/2,3/2\rangle.$$
The only possible spin state in which the final state particles can be produced is $|S_f=1/2,S_{fz}=+1/2\rangle$ if  $|S_i=1/2, S_{iz}=+1/2\rangle$. Similarly, the only allowed total spin state in which the final state particles is $|S_f=1/2,S_{fz}=-1/2\rangle$ if  $|S_i=1/2, S_{iz}=-1/2\rangle$. And if the initial neutron is not in an eigenstate of $S_{iz}$, but say, $S_{ix}$ then there is definite probability that the final state particles will be in spin state $|S_f=1/2,S_{fz}=-1/2\rangle$ state and a definite probability to be produced in $|S_f=1/2,S_{fz}=+1/2\rangle$ state.
Please correct if there is anything wrong because this is not well explained in books.
 A: To simplfy the analysis, let's assume you are studying the reaction in the center of mass frame. Remember that while you have conservation of momentum $\vec J$, it consists of orbital anugular momentum $\vec L$ and spin anglar momentum $\vec S$, only the total is conserved: $\vec J = \vec L +\vec S$.
Since we are in the CM frame, in the initial state, $\vec L_i = 0$ and as you pointed out $ S_i = 1/2$, so $J = 1/2$. Let's chose the $z$ axis such as $(J_i)_z = 1/2$. In the final state, as you noticed $S_f = 1/2,3/2$, but you can also have $L_f \in \mathbb N$ which you forgot (since there are several particles). In addition to your solution $L=0,S=1/2$, you therefore have $3$ additional combinations (the detailed kets are found using Clebsch Gordan coefficients):

*

*$L=1,S=1/2$


*$L=1,S = 3/2$


*$L=2,S = 3/2$.
Note that unlike the $2$ particles case where  angular momentum considerations determine the full angular distribution of the outgoing particle, in theithiss case of $3$ particles it will only help restrict its form.
Hope this helps and tell me if something's not clear.
