I currently have a problem where a decay calculation $J^P=(1/2)^+$ decays into a meson $0^-$ and a nucleon $(1/2)^+$. While $J = L + S$ is conserved, it still appears that parity isn't. $L=0$ in this case. Is this consistent with $N(1440)\rightarrow \pi + N$ where $N$ is any nucleon?
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What is your evidence that this decay has orbital angular momentum $L=0$? I'm prepared to disbelieve your question and state that in the decay $N\to N\pi$ the orbital angular momentum is $p$-wave, not $s$-wave, for exactly the reason you state in your question: strong decays conserve parity.
Remember that angular momentum addition gives $\vec J = \vec L + \vec S$, not necessarily $J=L+S$.
The Particle Data Group doesn't assign an orbital angular momentum to $N\to N\pi$, which is their usual approach when only one value of $L$ is allowed. However the following decay branches do have specific orbital angular momenta called out: \begin{align} N(1440)1/2^+ &\to \Delta(1232)3/2^+ + \pi(140)0^- & \text{$p$-wave only} \\ N(1520)3/2^- &\to \Delta(1232)3/2^+ + \pi(140)0^- & \text{$s$- and $d$-wave} \\ N(1535)1/2^- &\to \Delta(1232)3/2^+ + \pi(140)0^- & \text{$d$-wave only} \\ N(1680)5/2^+ &\to \Delta(1232)3/2^+ + \pi(140)0^- & \text{$p$- and $f$-wave} \end{align}
Notice that only the parity-conserving values of $L$ are allowed, and that the $d$- and $f$-wave decays can't be explained unless you use vector addition of angular momentum.
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$\begingroup$ Thanks! Turns out I actually just had confusion with how the orbital angular momentum works. Got it sorted out now though! $\endgroup$ Commented Sep 19, 2016 at 17:27