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Here is the problem. The parity of the negative pion was deduced from a reaction of pion with deuteron which results in two neutrons. The deuteron has spin equal to one, pion has zero spin, both had zero orbital angular momentum. The resulting protons can have spin either $0$ or $1$, from the requirement of antisymetric wavefunction, spin 0 corresponds to orbital momentums of 0,2,4,... and spin $1$ corresponds to angular momentums of 1,3,...

The total angular momentum of deuteron and pion is $J=1$, which has to be conserved. Now they supposedly deduced, that the orbital angular momentum of the two neutrons must be $L=1$ and spin $S=1$.

My question is, how can this be, since $J=L+S=1+1=2$ which is not $1$, so the total angular momentum should not be conserved and yet wikipedia and my textbook says it is?!

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    $\begingroup$ have you realized that angular momentum is a vector? $\endgroup$
    – anna v
    Oct 25, 2016 at 18:44
  • $\begingroup$ Total angular momentum quantum number doesnt have to be conserved in a reaction. Only the spin angular momentum quantum number has to be conserved $\endgroup$ Oct 25, 2016 at 18:48
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    $\begingroup$ @PrasadMani I do not understand your comment. the letters in his formula should have an arrow, the addition is vector addition; as is true also for momenta. $\endgroup$
    – anna v
    Oct 25, 2016 at 18:57
  • $\begingroup$ Actually it was directed at OP. What i said was only the spin quantum number '$s$' is conserved in a reaction, not the total ang. momentum quantum number '$j$'. My feeling was he got confused trying to conserve $j$ $\endgroup$ Oct 25, 2016 at 19:01
  • $\begingroup$ @PrasadMani ?? j =1 before and after the reaction. Since when does j fail to be conserved? the J =2 statement is flat wrong. $\endgroup$ May 17, 2017 at 16:44

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$J$ can take values $|L-S|$ to $|L+S|$, so \begin{equation} \begin{aligned} |1-1|&=0\\ |1-1|+1&=1\\ |1-1|+2&=2 \end{aligned} \end{equation}

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