= Solution
The <intrinsic parity> $\eta$ of a particle is the <parity> eigenvalue of its one-particle state at rest:
$$
\mathcal P|\mathbf0,s,m\rangle
=\eta|\mathbf0,s,m\rangle.
$$
For the particles in this question, $\eta=\pm1$. A two-particle state has <multiparticle parity> $\eta_1\eta_2(-1)^\ell$, where $\ell$ is its relative <orbital angular momentum>.
The pionic-deuterium <ground state> has atomic $\ell=0$. Since the pion has spin zero, its total angular momentum is therefore the deuteron spin, $J=1$, and its parity is $\eta_d\eta_\pi=\eta_\pi$. The two neutrons are identical <fermions>. Their antisymmetric <spin-one-half singlet state> would require even orbital $\ell$, but $S=0$ and $J=1$ would instead require $\ell=1$. Hence they must occupy the symmetric <spin-one-half triplet state>, so the orbital state is antisymmetric and $\ell$ is odd. The <addition of angular momentum> with $S=J=1$ permits $\ell=0,1,2$, leaving $\ell=1$. Thus parity conservation gives
$$
\eta_\pi
=\eta_n^2(-1)^1
=-1,
$$
so \b[the pion has intrinsic parity $\boxed{\eta_\pi=-1}$].
In $\Delta^-\to\pi^-+n$, the final total spin is $S=1/2$. Coupling it to $\ell$ to obtain $J=3/2$ gives
$$
\left|\ell-\frac12\right|
\leq\frac32
\leq\ell+\frac12,
$$
so \b[the allowed orbital angular momenta are $\boxed{\ell=1,2}$]. The observed angular probability is $\cos^2\theta=|P_1(\cos\theta)|^2$, where $P_1$ is a <Legendre polynomial>; it therefore selects the $\ell=1$ partial wave rather than $\ell=2$. The <parity selection rule for a two-body decay> now gives
$$
\eta_{\Delta^-}
=\eta_\pi\eta_n(-1)^\ell
=(-1)(+1)(-1)
=+1.
$$
Therefore \b[the delta particle has intrinsic parity $\boxed{\eta_{\Delta^-}=+1}$].
Back to article page
