Consider the process $$\rho^0\rightarrow\pi^0+\pi^0$$ The $\rho^0$ has $J=1$ whilst the two pions have $S_{tot}=0$ and thus require $L_{tot}=1$ by conservation of angular momentum. Consequently this process is forbidden because the pions are identical bosons and must have a wavefunction that is symmetric under particle exchange and $L_{tot}=1$ does not allow this.

I am happy with this but for the last sentence. Could somebody go into how the collective orbital angular momentum of a system of two particles relates to the symmetry of the two particle wavefunction under exchange of particles? Thank you for any help :)


1 Answer 1


This is due to the property of the angular part of the wave function, namely the spherical harmonics. When you have a parity transform (inversion through the origin), spherical harmonics transform like this: $$Y_{lm}(\theta, \phi) \rightarrow (-1)^l Y_{lm}(\theta, \phi)$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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