Why can $\chi_{c}$ states not decay leptonically? I am trying to understand why the $\chi_{c}$ states of charmonium cannot decay to $l\overline{l}$ pairs. I believe it is because they have positive parity, but I'm unsure why this prevents the decay?
 A: This is a complete rewrite of my answer, kept honest by @1MegaMan1 's logically elegant comment: a tip of the hat. The lepton pair is not in a well-defined L state, so parity arguments relying on a special L are not controlling here. First for the facts.
The χ state has $J^{PC}= 0^{++}$. It comes about from $L=S=1$, so $^3P_0$. Its $C=(-)^{L+S}$. From C, it does not couple to one photon (with $C=-1$), so it can neither be produced or decay through a just one (virtual) photon state. But it does couple to two photons, γγ, and so its production from $e^+ e^-$ collisions and decay to a lepton pair are strictly allowed, but fantastically suppressed by $\alpha^2$. Indeed, the BR to two photons for the χ is $O(10^{-4})$. Getting a lepton pair as the abutment of two such virtual photons is a no-go, in practical terms.  It mostly decays by OZI-suppressed gluons to hadronic modes, instead. 
By the way, the pseudoscalar meson, $\eta_c$, with $J^{PC}= 0^{-+}$, actually $^1S_0$, has the very same C problem in coupling to pairs of leptons. Again, its decay to two photons has a BR of  $O(10^{-4})$ and I'm not sure anyone has observed/could observe  the two lepton decay mode.
