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If I have the following reaction : $\pi^- + d \rightarrow n + n$, it's said that the two neutrons are in a state with s=1 and l=1, so that $j_f =l+s$$\space \rightarrow$ $j_f =0,1,2$ . And that the spins of the pion and deuteron should be 0 and 1 respectively, while their relative orbital angular momentum ($l_i$) is 0. That means $j_i =1$, but if $j_f = 2$ doesn't that violate the conservation of angular momentum ?

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It just means that you can't have $j_f=2$. This is the reaction from which the parity of the pion was first deduced by the way!

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