Why can't a spin-1 particle decay into two identical spin-0 particles?

Question

I've got this far:

Suppose a spin-1 particle with $j$,$l$,$s$ decays into a system of two identical spin-0 particles with $J$,$L$,$S$.

The RHS must have total spin $S=0$, so $J=L$ which must be even since the particles are Bosons and their states must be symmetric in their exchange.

Then by conservation and addition of angular momenta, the LHS must have $L=J=j=\left\{\begin{array} \.l-1,l,l+1 &\text{ if }l\neq0 \\ 1 &\text{ if }l=0 \end{array}\right.$.

Mistakes are likely.

Two possible ways to proceed (but I can't motivate either):

1. Can I choose to take $l=0$ for some reason?
2. I know (I think) that a system of two identical Bosons has to have $l$ even, because of symmetry of their wavefunctions under exchange of the two particles. Does the value of $l$ for a single Boson also have to be even? If so, can I then somehow rule out the $j=l\neq 0$ case and get a contradiction?

Answer 1

Suppose that your decay does not violate parity, and let us take the example of strong interaction process involving mesons.

Looking at this wiki paragraph and array, you will understand that each meson has a total angular moment, and a parity, noted $J^P$, and there are several possibilities for $S$ and $L$, for a given $J^P$ ($J^P$ is a characteristic of the particle)

Let us precise the example with the decay of the neutral rho meson $\rho^0$ into 2 pions.

The rho meson has $J^P=1^-$, so we see in the wiki array, that $S=1$ and $L=0$ is a possibility, while $S=1$ and $L=2$ is an other possibility. Each of the pion has $J^P=0^-$, and we see that the only possibility is $S=0$ and $L=0$, for each pion.

We decide to prepare a rho meson in the state $S=1,L=0$

A key point is that, even if each pion itself has $L=0$, the system of 2 pions may have a relative orbital momentum $L_r$, so that the total angular momentum of the 2 pions system (remembering that $S_{TOT}=0$) is then $J=L_r$. By conservation of the total angular momentum, this gives $L_r=1$

An other point is that, in the strong interaction, the parity is conserved, so we have :

$(-1) = (-1)(-1)(-1)^{L_r}$, which says that $L_r$ is odd, which is not in contradiction with the fact that $L_r=1$

But, in fact, the decay $\rho^0 \to \pi^+, \pi ^-$ is possible, while the decay $\rho^0 \to \pi^o, \pi^0$ is not possible. Why ?

There is one more constraint, in the case where the $2$ final particles are identical, the relative orbital momentum $L_r$ must corresponds to the statistical nature of the particles, so it has to be even for bosons, and odd for fermions.

Now, pions are bosons, so, in the decay $\rho^0 \to \pi^o, \pi^0$, $L_r$ must be even, which is in contradiction with the above conclusions ($L_r=1$). So, this decay is not allowed.

The decay $\rho^0 \to \pi^+, \pi ^-$ is allowed, because the particles are different, so a odd relative orbital momentum $L_r=1$ is allowed (to be rigourous, you have also to check other conservation laws, like charge, C-Parity, G-Parity, and so on (see mesons)

Answer 2

You know that s=1 since it was given in the question, so you know that L = J = j = l+1. Moreover, L must be even as you stated so l must be odd. If you could make the assumption that l must be even then you would have a contradiction, but I don't know if you can. Did you get any further with this?