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I have a question concerning the symmetry of the spin function in multiple identical particle systems. In the solutions of one of the quizzes, my professor said that the $s=3/2$ spin function is completely symmetric, which is why we need an antisymmetric spatial component (by a Slater matrix) to describe three fermions.

I understand why we would need the spatial component to be antisymmetric if the spin part is symmetric. But how do we know, from the value of s, if the spin part is symmetric or not?

Thank you.

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  • $\begingroup$ Do you mean by $s$ the spin quantum number or its projection? If it is the projection than the spin function is symmetric because all spins have to be aligned in this case and hence are indistinguishable (on the spin level). $\endgroup$ – NDewolf Apr 9 '18 at 8:40
  • $\begingroup$ Presumably this is for a 3-particle system? $\endgroup$ – ZeroTheHero Apr 9 '18 at 13:48
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I'll assume your system contains three spin-1/2 particles. With the notation $\vert s,m\rangle$, it is clear that $$ \vert 3/2,3/2\rangle = \vert \textstyle\frac{1}{2},\frac{1}{2}\rangle_1 \vert \textstyle\frac{1}{2},\frac{1}{2}\rangle_2 \vert \textstyle\frac{1}{2},\frac{1}{2}\rangle_3 $$ is symmetric under permutation of particle indices. The other $\vert 3/2,m\rangle$ will also be symmetric since they can be reached from $\vert 3/2,3/2\rangle$ by applying $L_-$, where $$ L_- = L^{(1)}_-+L^{(2)}_-+L^{(3)}_- $$ is also symmetric under permutation of particle labels and so cannot change the permutation symmetry of the states it acts on. (Here, $L^{(k)}_-$ is the lowering operator acting on particle $k$ alone.)

Note that if you have $5$ particles, then the $\vert 3/2,3/2\rangle$ state would NOT be symmetric, and thus the entire $J=3/2$ multiplet would not be symmetric either.

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