Decays possible? Parity conservation, bosons, fermions

[binbagsss]

1. Homework Statement

The question is to determine which decays are possible for:
i) ##P^0 ->\prod^+ \prod^-##
ii)##P^0 ->\prod^0 \prod^0##

Homework Equations

where ##J^p = 0^-, 1^- ## respectively for ##\prod^+, \prod^- , \prod^0## and ##P^0## respectively.

The Attempt at a Solution

For part i, the LHS has odd parity. ##P=(-1)^l##, so on the RHS we require ##l## to be odd.
Also need to conserve total angular momentum ##J=(l+s)+(l+s-1)+...+ | l-s |## *
On LHS ##J=1.##
##s=0##, so conservation gives ##l=-1## , which is consistent with an odd parity , so the decay is allowed.

part ii) We have the same J and P arguments, so I would have concluded the decay is possible.
The solution however is that is not because the RHS now has 2 identical bosons so the final wavefunction must be symmetric under the exchange of the two neutral pions. However this requires that the orbital angular momentum is even, so we have inconsistency.

So here's what I know :
If you swap 2 bosons the wave function has to be unchanged, but if you swap 2 fermions the wave function changes sign.
So , with this, I now don't see why we can't apply the argument to the decay in part i) - unless this property is only true for a system of identical particles??

More importantly, I don't follow the argument completely: The angular momentum being odd or even, i.e- as far as I can see the only way for ##l## to creep in, comes from the parity being odd or even- ##P=(-1)^l##, but parity is describing how the wave function behaves under a change from ##\vec r ## to ##\vec -r## So what has this got to do with swapping 2 bosons? The only possible argument I can think of would be along the lines of considering the particular case were one of the particles is situated at ##\vec r ## and the other at ##\vec -r## when we swap the bosons position??Thanks in advance !

 

[mfb]

unless this property is only true for a system of identical particles??

That is the point.
Swapping two non-identical particles leads to a completely different system, where no symmetry is relevant. Only swapping two identical particles has an interesting result.

What is P, by the way? ##\rho##?

 

[binbagsss]

That is the point.
Swapping two non-identical particles leads to a completely different system, where no symmetry is relevant. Only swapping two identical particles has an interesting result.

What is P, by the way? ##\rho##?

##P## is the parity.

 

[mfb]

I mean the particle that decays. I do not recognize capital P as a particle name, only p for protons but that does not make sense here.

 

[HalfLostAndSearching]

I would respond by saying that the reason for the difference in allowed decays between part i and part ii lies in the symmetry properties of the final wavefunction. In part i, the final state has odd parity and therefore the orbital angular momentum must be odd to satisfy the conservation of parity. This also means that the wavefunction must be antisymmetric under the exchange of the two particles, which is consistent with the properties of fermions. However, in part ii, the final state has even parity and therefore the orbital angular momentum must be even. This requires the wavefunction to be symmetric under the exchange of the two particles, which is not possible for identical bosons. This is because identical bosons must have a symmetric wavefunction, and swapping them would not change the wavefunction, leading to an inconsistency. Therefore, the different allowed decays in part i and part ii are due to the different symmetries required for the final state wavefunction.