# What is the total angular momentum for non “bound” states?

My question has to do with the computation of the parity of a state. For example, if we want to compute the intrinsic parity of the $\pi_0$ meson, we do the following : $P(u\overline{u}) = P(u)P(\overline{u})(-1)^l = -1$. Since $P(u) = -P(\overline{u})$ and $l = 0$ for the $\pi_0$ meson.

Say now we have the decay $\pi_0 \rightarrow \nu_e + \overline{\nu}_e$. If I want to compute the intrinsic parity of the final state, do I have to take into account an eventual $l$ due to the "joint" angular momentum of the system $\nu_e + \overline{\nu}_e$ ?

In other words, do I have $P(\nu_e+\overline{\nu}_e)= P(\nu_e)P(\overline{\nu_e})(-1)^l$ or $P(\nu_e)P(\overline{\nu_e})$.

To extend, what about in general for a decay $a \rightarrow b+c$. Can $b$ and $c$ always have a relative angular momentum $l$? If not, under which condition can they have one? (I know that in $\rho_0 \rightarrow \pi^+ + \pi^-$ we have to take into account a total $l$).

Note that decays to states with large $\ell$ are suppressed compared to states with small $\ell$. A hand-waving way to think about this is to imagine that the initial and final wavefunctions must overlap and recall the radial hydrogen wavefunctions, which go like $(r/a)^\ell$ near the origin; the length scale $a$ for a decay is set by the wavelengths of the particles in the final states. I vaguely recall that "allowed," "first-forbidden," "second-forbidden," etc. beta-decays are grouped by the orbital angular momentum which must be carried by the decay products.
Your proposed $\pi\to\nu\bar\nu$ is a thorny example: the pion hasn't any spin, but a $\nu\bar\nu$ traveling back-to-back must have total spin $\hbar$. If that decay happens (current upper limit on the branching ratio is $10^{-6}$) it must have $\ell=1$ to conserve angular momentum. But parity isn't conserved in weak decays anyway. You might have more luck thinking about $\pi\to\gamma\gamma$, which is the dominant decay mode for the $\pi^0$.
• Suppose that helicity and chirality are equal for neutrinos. We cannot have l=0 in the decay, since we would need a right handed neutrino, so that their spins sum to 0. So it right in this case to say that the two neutrinos can only be produced in $l=1$ (or higher) state ? Otherwise, thank you for your answer it clears things up well. – Frotaur Jan 13 '17 at 17:50
• Close -- the final state must have exactly $\ell=1$. There's no way to combine $s=1$ and $\ell=2$ (or higher) to make zero angular momentum, either. – rob Jan 13 '17 at 17:53
• Ah yes thank you, indeed I had overlooked the fact that we must conserve both $J_z$ and $J$. – Frotaur Jan 13 '17 at 18:05
• Sorry to come back on this, but what about initial state particles ? Say I want to compute the parity of $e^+e^-$ in the $e^+e^-\rightarrow \nu \overline{\nu}$ vertex. Can I have a non-zero l in the initial state ? My guess would be yes again, but i'd like to get confirmation. – Frotaur Jan 15 '17 at 1:19