Pion decaying to two neutrons demonstrates odd parity

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The discussion centers on the decay of a pion into two neutrons, specifically addressing the parity of the final state. The participants clarify that the reaction involves a pion interacting with a deuteron rather than a direct decay. The allowed final states for two identical fermions, as per Fermi-Dirac statistics, include 1s0, 3p0,1,2, 1d2, and 3f2,3,4. The confusion arises around the antisymmetry of the 1s0 state, which is clarified to not be a viable reaction pathway.

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Gene Naden
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Does anyone have a reference to a good explanation of this experiment. I am looking at https://quantummechanics.ucsd.edu/ph130a/130_notes/node323.html

I am unable to comprehend the reasoning by which it determines the parity of the two neutrons in the final state. Particularly when it says the requirement of F-D statistics restricts the spin and angular momentum of the final state.
 
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That is not a pion decaying to two neutrons. It is the reaction of a pion hitting a deuteron.
 
Yes, I gave the wrong reaction. Thank you for correcting me.

The reference says the allowed two-neutron final states are, because they are identical fermions,
1s0
3p0,1,2
1d2
3f2,3,4

I looked up the spectroscopic notation. It says 1s0 is a singlet state, L=1, with J=0. So how can it be antisymmetric?
 
It does not react to 1s0. Only 3p1 is possible, as explained in the website you linked.
 

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