When combining N spin 's' states, is it always true that each multiplet has even or odd symmetry?(adsbygoogle = window.adsbygoogle || []).push({});

I know that's the case for N=2 and s=1/2 or 1. For s=1/2, the triplet is symmetric and the singlet is antisymmetric. For s=1, the pentlet is symmetric, the triplet antisymmetric, and the singlet symmetric. But what about different s's and different N's?

Also, I'm a bit confused about terminology. If you have a 2-neutron bound state whose total spin is zero, that means it's antisymmetric in the spins, so it must be symmetric in orbital angular momentum, i.e., [tex]Y^{m1}_{l1}(x_1)Y^{m2}_{l2}(x_2) +Y^{m1}_{l1}(x_2)Y^{m2}_{l2}(x_1)[/tex]. Under parity, due to a property of spherical harmonics, the orbital part changes by just a factor $$(-1)^{l_1+l_2}$$.

However, I'm trying to make sense of the statement that if you have a 2-neutron bound state whose total spin is zero, that means it's antisymmetric in the spins, but must have an even total orbital angular momentum. Is that true? $$l_1+l_2$$ could be odd, but the total L could be even by taking the multiplet $$l_1+l_2-1$$ rather than highest multiplet as your total.

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# Angular momentum of identical particles

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