Question:
Particle A, whose spin $\mathbf{J}$ is less than 2, decays into two identical spin-1/2 particles of type B.
What are the allowed values of the orbital angular momentum $\mathbf{L}$, the combined spin $\mathbf{S} = \mathbf{s}_1+\mathbf{s}_2$ (where $\mathbf{s}_1$ and $\mathbf{s}_2$ are the spin vectors of the B particles), and the total angular momentum $\mathbf{J}$ of the decay products? Give the results from two cases: first if particle A has odd parity and then if A has even parity.
My thoughts:
Particle A can be spin-1/2, spin-1, or spin-3/2. Since $\mathbf{J}<2$, we see that there are four possibilities for A:
$$
\begin{align*}
&(1): \;\;\mathbf{S}_A = 1/2 \quad\quad \mathbf{L}_A = 0 \quad\Rightarrow\quad \mathbf{J} = 1/2 \\
&(2):\;\;\mathbf{S}_A = 1/2 \quad\quad \mathbf{L}_A = 1 \quad\Rightarrow\quad \mathbf{J} = 3/2 \\
&(3):\;\;\mathbf{S}_A = 1 \quad\quad\quad \mathbf{L}_A = 0 \quad\Rightarrow\quad \mathbf{J} = 1 \\
&(4):\;\;\mathbf{S}_A = 3/2 \quad\quad \mathbf{L}_A = 0 \quad\Rightarrow\quad \mathbf{J} = 3/2 \\
\end{align*}
$$
The total spin of the B particles can be either $1$ or $0$, and each particle can individually have an orbital angular momentum, along with the angular momentum of the particles as a system. With this thought, cases 1,2, and 4 are impossible because the orbital angular momentum of the B particles is an integer, as is their total spin (and therefore their total angular momentum too). Thus we find that only case 3 is allowed, so the total angular momentum of the B particles is $1$ and their orbital angular momentum is $0$ (so $\mathbf{J}=1$).
I have a strong feeling that this is incorrect, because the question asks for the cases when A has odd parity and even parity (what does that even mean?!) so I suspect there should be more than one possible answer. Where did I go wrong?
Best Answer
Hmmm, an old question without a satisfactory answer. I'll have a go.
The spins of the two $B$ may combine as \begin{align} \text{singlet}\quad|s_1s_2\rangle &= \frac{\left|\uparrow\downarrow\right> - \left|\uparrow\downarrow\right\rangle}{\sqrt2}, & \text{or triplet}\quad|s_1s_2\rangle &= \frac{\left|\uparrow\downarrow\right> + \left|\uparrow\downarrow\right\rangle}{\sqrt2}. \end{align}
Since the two $B$ have spin-1/2, they obey Fermi-Dirac statistics and their total wavefunction must be antisymmetric under exchange. Therefore the antisymmetric spin singlet can only be paired with $L=0,2,\cdots$, which have even parity; the symmetric spin triplet must be paired with $L=1,3,\cdots$, so that the parity of the orbital angular momentum wavefunction makes the entire wavefunction change sign if the two $B$ are interchanged.
The intrinsic parity of $B$ doesn't contribute to the overall parity of the final state, since there are two of them; if the $B$ have negative parity, the overall intrinsic parity of the pair is still positive. Thus the allowed final states are \begin{align} \text{positive parity}&: & \text{spin singlet (antisymmetric)} && L&=0 \\ \text{negative parity}&: & \text{spin triplet (symmetric)} && L&=1 \end{align} I've omitted the spin singlet combined with $L=2$, since we do not have that much angular momentum. Likewise there is no combination of the spin triplet and $L=3$ which can be produced from a $J=2$ initial state.
If $A$ has definite parity and parity is conserved in the decay, only one of these possibilities is allowed. In fact, the way that angular momentum combines (with the possibility of the spin triplet plus the $L=1$ wavefunction adding to a $J^P=0^-$ final state) means that the parity has more to do with the allowed final spin state than $A$'s spin does: $$ \begin{array}{r|cc} & \text{spin 0} & \text{spin 1} \\ \hline \text{parity}+ & \text{decay to singlet},0^+ & \text{decay forbidden} \\ \text{parity}- & \text{decay to triplet},0^- & \text{decay to triplet},1^- \end{array} $$