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?

• The parity quantum number is defined if the the state is (anti-)symmetric under reflection: that is $\psi(\vec{x}) = \pm \psi(-\vec{x})$. If parity is defined it is $\pm1$ according to the sign of the first relationship. You are probably expected to know that even orbital angular momentum states have parity +1 and odd orbital angular momentum states have parity -1. – dmckee Mar 16 '13 at 22:04
• A second import point here is that the answer has to be independent of the frame of reference, so we can freely choose to evaluate it in particle A's frame (almost always the right choice for these problems). From that it follows that A has orbital angular momentum of zero. The only non-trivial $L$ you should have in your work is that of the products. – dmckee Mar 16 '13 at 22:17
• But if we use A's frame, then there is no meaning to asking for solutions for even and odd parity A. Doesn't the problem as stated demand that we work in the lab frame? Also, am I correct in assuming that J of particle A must be an integer, because L and S of the B particles are both integers? – alexvas Mar 17 '13 at 0:54
• "there is no meaning to asking for solutions for even and odd parity A" Yes, there is. Elementary particles can have intrinsic parity. Note that parity is multiplicative (i.e. a negative intrinsic parity particle in a $l = 1$ state has positive parity overall). Also most reactions conserve parity, so the parity of A determines which final states are available to electromagnetic and strong decays (weak decays may violate parity). – dmckee Mar 17 '13 at 1:03
• I still don't see how this is relevant to the problem. Could you give an example of $S_A$, $J_A$, $L$, $S_1$, and $S_2$ that work? How does the intrinsic parity of particles translate into the total angular momentum J? – alexvas Mar 17 '13 at 1:09

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: $$\begin{array}{r|cc} & \text{spin 0} & \text{spin 1} \\ \hline \text{parity}+ & \text{decay to singlet},0^+ & \text{decay forbidden} \\ \text{parity}- & \text{decay to triplet},0^- & \text{decay to triplet},1^- \end{array}$$