Lower bound for the spin in particle decay process Is my logic right?
Suppose there is a particle $p$ that can either decay into $ \{$a spin-1 and a spin-0 particle$\}$ or  two spin-0 particles, then the lowest possible spin of $p$ is 2. This is because we need the spin to be even and large enough to accommodate the spin-1 product. 

ADDED:  $p$ is such that $p\to \pi^-+\rho^+$ and $p\to \xi+\xi$ where $\xi$ is a particle of spin 0. I want to know the lowest possible spin for $p$. 
I am assuming that conservation of total angular momentum and parity holds.
By the way, would the intrinsic parity of $p$ be negative, since I think each of $\rho^+$ and $\pi^-$ have negative intrinsic parity?
 A: If this is a homework problem then you should do two things: 1) in the future, don't ask this type of question without tagging it as a homework problem; 2) ask for a more complete description of the problem from whoever assigned it -- it hasn't been clearly stated (or restated here, perhaps). I'll only work through the first reaction, $p\to\pi^-+\rho^+$ and leave it to the reader to work out the rest.
If we assume that parity is conserved (say, by the strong interaction) and we know (from observation) that the intrinsic parities of the $\pi$ and $\rho$ are negative then, for the first reaction, $p\to\pi^-+\rho^+$, we have two possibilities: 1) The parity of $p$ is $+1$ (giving initial state parity $+1$) and the decay may occur with $\ell = 0, 2, 4, \ldots$, since the parity of the final state is $(-1)^\ell(-1)_\pi(-1)_\rho$. In this case, the lowest spin that $p$ may have is $1$ since the $\rho$ has spin-$1$ and $\pi$ has spin-$0$; 2) the parity of $p$ is $-1$. Then, by similar reasoning, the lowest spin of $p$ is $0$ since at least one unit of orbital angular momentum is needed to conserve parity.
So we see clearly that the spin and parity are strongly correlated. This is one of the ways that experimentalists can determine the parity of observed particles.
