Decays possible? Parity conservation, bosons, fermions

AI Thread Summary
The discussion centers on the possible decays of the particle P^0 into different combinations of pions, specifically analyzing the decays P^0 -> π^+ π^- and P^0 -> π^0 π^0. For the first decay, conservation of parity and angular momentum allows the decay to occur since the required orbital angular momentum is odd, consistent with the odd parity of the initial state. In contrast, the second decay is not allowed because it involves two identical bosons, necessitating a symmetric wavefunction that requires even orbital angular momentum, leading to a contradiction. The conversation also touches on the implications of particle identity, emphasizing that only identical particles exhibit symmetry properties upon swapping. The key takeaway is that the rules governing decay processes are heavily influenced by the nature of the particles involved.
binbagsss
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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 !
 
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binbagsss said:
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##?
 
mfb said:
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.
 
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.
 
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