# 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.

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.

• Thanks, lpz! Let me try to summarize your answer. Since $J_i=L_i+S_i=1/2$ and by conservation of angular momentum $J_f=1/2$, the possible combinations for $L_f$ and $S_f$ that produces $J_f=1/2$ are (i) $L_f=1, S_f=1/2$, (ii) $L_f=1, S_f=3/2$, (iii) $L_f=2, S_f=3/2$, and (iv) $L_f=0,S_f=1/2$. Do I get this right? Jun 12, 2022 at 19:21
• Yes that’s it. Now you just need to project your final state into these angular momentum eigenstates to extract the relevant information.
– LPZ
Jun 12, 2022 at 19:51
• You may be interested in this as well: physics.stackexchange.com/q/713386/164488 Jun 12, 2022 at 20:23