# 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?

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The statement of your question is incomplete. When considering two-body decays, the relevant quantum numbers (neglecting conserved internal coordinates) are the spins of the particles, $s, s_1, s_2$, where $s$ is the spin of your particle $p$, $s_{1,2}$, the spins of the final products, $\ell$, the orbital quantum number, and the parities of the three particles. You must also state whether the parity is conserved in the interaction that causes the decay. You have left out all information about the parities. $\ell$ may be taken to be zero unless req'd $\ne 0$ by the interaction causing decay. – MarkWayne Dec 10 '12 at 23:54
Thanks, @MarkWayne. All I know about $p$ is 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? – Talbott Dec 11 '12 at 19:39

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.