Why the Antisymmetry of Wavefunction for l = 1?

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SUMMARY

The discussion centers on the antisymmetry of the wavefunction for two identical pions in an l = 1 state, as outlined in the provided notes. Pions, being bosons, require a symmetric wavefunction under particle interchange, while the wavefunction for an l = 1 state is inherently antisymmetric. This contradiction leads to the conclusion that the decay mode \(\rho^0 \to \pi^0 + \pi^0\) is forbidden due to the incompatible symmetry requirements. The spin part of the wavefunction is symmetric, but the orbital part, characterized by angular momentum ℓ, results in an overall antisymmetric wavefunction for ℓ = 1.

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

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