0
$\begingroup$

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?!

$\endgroup$
  • 1
    $\begingroup$ have you realized that angular momentum is a vector? $\endgroup$ – anna v Oct 25 '16 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 $\endgroup$ – Prasad Mani Oct 25 '16 at 18:48
  • 1
    $\begingroup$ @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 '16 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 $j$ $\endgroup$ – Prasad Mani Oct 25 '16 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. $\endgroup$ – Cosmas Zachos May 17 '17 at 16:44
1
$\begingroup$

$J$ 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}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.