Why the Antisymmetry of Wavefunction for l = 1?

[c299792458]

Would someone please explain the following found on P. 125 of these notes http://www.hep.phys.soton.ac.uk/hepwww/staff/D.Ross/phys3002/PCCP.pdf?

>On the other hand, two $π^0$’s cannot be in an $l = 1$ state. The reason for this is that pions are bosons and so the wavefunction for two identical pions must be symmetric under interchange, whereas the wavefunction for an $l = 1$ state is antisymmetric if we interchange the two pions. This means that the decay mode $$\rho^0\to \pi^0+\pi^0$$ is forbidden.

I don't understand why the wavefunction of $l = 1$ must be antisymmetric. Perhaps I have forgotten something?

Thanks.

 

[Bill_K]

The spin part is 0 x 0 which is symmetric. And the parity of an orbital wavefunction with angular momentum ℓ is (-)^ℓ, so ℓ = 1 is antisymmetric under interchange of the particles.