Why is the $\rho^0 \rightarrow \pi^0 + \pi^0$ decay not allowed? I have seen this question but I am not satisfied with the answers. The $J^{PC}$ of the $\rho$ and $\pi$ are $1^{--}$ and $0^{-+}$ respectively. Taking the orbital angular momentum of the final two pions state to be $1$ seems to conserve everything ($J$, $P$, $C$, etc.) but is apparently not allowed by the spin-statistics theorem. The answers I have seen say that $L=1$ for the pions gives an anti-symmetric wavefunction. Assuming this is true, I understand why this is not allowed under the spin-statistics theorem, but I don't understand why the wavefunction anti-symmetric; that is the main question.
$\begingroup$
$\endgroup$
4
-
2$\begingroup$ possible duplicate of Why is the decay of a neutral rho meson into two neutral pions forbidden? $\endgroup$– Kyle KanosCommented Dec 19, 2014 at 3:46
-
$\begingroup$ @KyleKanos as I mentioned in my question, I had seen that question but I was not satisfied with the responses. Although both questions ask why this decay is not allowed, my question is specifically about why the final wave-function is anti-symmetric; there are no answers to this on the previous question. $\endgroup$– ginnyCommented Dec 23, 2014 at 13:37
-
$\begingroup$ Not being satisfied doesn't mean you should duplicate a question. The highest voted answer on the duplicate explains that the wavefunction should be antisymmetric; the proper thing to do would be (a) leave a comment on that post asking for clarification on that point (b) offer a bounty on that post. This last one requires something like 75 rep, so seems to be not possible for now. $\endgroup$– Kyle KanosCommented Dec 23, 2014 at 13:43
-
$\begingroup$ I think my fault here then is a poorly worded title. The other question asked why the decay is forbidden whereas my question title should have been why is the orbital angular momentum of two pions in this decay 1? Though I also thought it was better practise to ask a new question than to dig up an old question, my apologies for that. P.S. one needs 50 reputation to comment [on someone elses post], which I also do not have. $\endgroup$– ginnyCommented Dec 23, 2014 at 14:52
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
The spatial wavefunction is $Y_L^m(\theta,\phi)$. When exchanging the two particles, the spatial wavefunction becomes to $Y_L^m(\pi-\theta,\pi+\phi)$. Mathematically, we have $Y_L^m(\pi-\theta,\pi+\phi)=(-1)^L Y_L^m(\theta,\phi)$. If $L$ is odd, the spatial part is antisymmetric, otherwise symmetric.
BTW, you may post this question as a comment of the original link.
