- #1
binbagsss
- 1,259
- 11
The question is to determine which decays are possible
for:
i) ##P^0## ## ->\prod## ##^+## ## \prod## ##^-##
ii)##P^0## ## ->\prod## ##^0## ## \prod## ##^0##
where ##J^p = 0^-, 1^- ## respectively for ## \prod## ##^+##, ## \prod## ##^-## , ## \prod## ##^0## and ##P^0## respectively.
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.
(On another note, I'm not 100% on my method for determing ##s=0## for the RHS, my reasoning is as follows, please let me know if this is correct: )
By * we can attain a certain ##J## , for a fixed ##s##, with varying values of ## l ##. so ##l## varies from ##0## to at least ##J##. )
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) - this argument is based on being a boson/ fermion - nothing to do with weather the bosons are identical or not?
Secondly, I don't follow the argument completely: The angular momentum being odd or even, as far as I can see, comes from the parity being odd or even- ##P=(-1)^l##, whereas 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?Sorry it's a tad long your assitance is greatly appreciated ! Thanks in advance !