 Decomposition of three SU(2) spinors

For $Q^{\alpha \beta \gamma }$, put $S=Q^{(\alpha \beta \gamma )}$, $d_1^{\alpha }={ϵ}_{\beta \gamma }{Q}^{\alpha \beta \gamma }$, $d_2^{\alpha }={ϵ}_{\beta \gamma }{Q}^{\beta \alpha \gamma }$ and $d_3^{\alpha }={ϵ}_{\beta \gamma }{Q}^{\beta \gamma \alpha }$. With ${ϵ}_{12}=1$ and inverse ${ϵ}^{12}=-1$, the two-dimensional alternating identity gives $d_1-d_2+d_3=0$. The inverse decomposition is $Q^{\alpha \beta \gamma }=S^{\alpha \beta \gamma }+{ϵ}^{\alpha \beta }(2d_1^{\gamma }-d_2^{\gamma })/3-{ϵ}^{\alpha \gamma }(d_1^{\beta }+d_2^{\beta })/3$. The symmetric part is the spin-$3/2$ SU(2) representation, and the two independent contractions are spin-$1/2$ representations. This is the spin coupling underlying ground-state three-quark baryons, before the Pauli constraint on three-quark flavour multiplets is imposed.

// Target: quantum-field-theory.bigb