I am asked the following question: A kaon $K^0$ decays into two pions $\pi^0$, being this two pions particles with spin zero. Using the conservation of angular momentum and the fact that the two pions are identical particles, analyze which restrictions are imposed over the spin of the $K^0$.
This question is in an exam of an introductory course on Quantum Mechanics, so the answer should be simple and not involve anything about quarks. Here is what I did:
First of all I wrote the most general initial and final state as
Knowing that the pions are particles of spin zero, because the question specifies this, we have that $s_{\pi^0} \equiv 0$ so the final state can be reduced to
Now I consider the center of mass system. In this system we should have all the orbital angular momentums to be zero because 1) the kaon is at rest and 2) the pions are emitted in radial directions (I assume this). Thus the initial and final states become
Finally I apply the conservation of angular momentum. The final angular momentum is zero and so it must be the initial total angular momentum, thus the spin of the kaon must be zero.
I have doubts about this answer because I have never used the fact that the pions are identical particles... Do you think that what I did is correct?