Decay of spin-1 particle into two 1/2 particles If particle of spin 0 decays in rest frame decays in 2 particles the angular distribution will be uniform. How it will change for particle with spin 1 decaying in pair of $e^{-} e^{+}$ for example?
 A: It will change a lot, in general, and will depend on the chirality properties of the interaction. Take a look at this.
A quick example, of the type you should have had drilled into you in your introductory HEP physics course: consider the decay of J/ψ, spin 1, in its rest frame. Assume the spin of the ψ is pointing upwards. The photon EM interaction producing an $e^+e^-$ decay pair is vectorlike: it preserves chirality, so the two decay products flying off back-to-back must have opposite chiralities: either L & R, or else R & L. So, on their common axis, their total helicity, will be +1 or -1 , never, never 0.
(Their masses are so small, compared to the kinetic energy involved, that chirality and helicity are very-very close to each other, and treated as identical, for all practical purposes, here.)
This, then, means that the $e^+e^-$ decay pair loves to be  emitted along an axis in the ψ frame where its spin points up, the lepton or antilepton in the direction of this axis being R (helicity +1/2). The angular distribution of that lepton will be $|d^1_{1,1}|^2$, so $((1+\cos\theta)/2)^2$. So emission at
π/2 will be 1/4 of that; and at π, in the opposite direction, impossible!
More dramatically, if you chose the spin 0 state of ψ on that axis, the R fermion distribution on that axis will perforce be 0, since its distribution will be
$|d^1_{01}|^2=\sin^2 \theta /2$, its maximum probability occurring at right angles to this direction; and, of course, also vanishing back-to-back.
I hope you now appreciate the presence of the Wigner d rotation matrices on the most useful page of the PDG booklet, the one with Clebsches, Spherical Harmonics, and d-functions--pure gold.
