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In $V_n\otimes V_n$, interchange of the two factors commutes with SU2. On the highest-weight vector $D^r{x}_1^{n-r}{x}_2^{n-r}$ it acts by $(-1{)}^r$. Each irreducible summand in the Clebsch-Gordan decomposition has multiplicity one, so this sign is its scalar action on the entire summand. The symmetric square is the $+1$ eigenspace and the exterior square the $-1$ eigenspace. ThereforeFor instance, $S^2{V}_1=V_2$ and ${\Lambda }^2{V}_1=V_0$. The dimensions agree respectively with $(n+1)(n+2)/2$ and $n(n+1)/2$.

A boson is an identical particle whose total multiparticle quantum state is symmetric under exchange of any two particles, whereas a fermion has a total state that is antisymmetric under every exchange. The Spin-statistics theorem associates integer spin with bosons and half-integer spin with fermions.

For three distinguishable spin-one particles, the spin Hilbert space has dimensionLetbe the total angular momentum operator. Since each particle haswe haveThe Hamiltonian operator is consequentlyOn the total-spin sector with quantum number $J$, its energy eigenvalue is

The Clebsch-Gordan decomposition may be performed by first coupling particles $1$ and $2$. Their intermediate spin is $j_{12}=0,1,2$, and coupling the third spin givesThus the three spin-one angular-momentum decomposition contains total spin $J=0,1,2,3$ with multiplicities $1,3,2,1$. Multiplying each multiplicity by the multiplet dimension $2J+1$ givesThe degeneracies sum to $27$, as required.

When the particles are indistinguishable, their integer spin makes them bosons. Their common spatial wavefunction is symmetric, so their spin wavefunction must also lie in the symmetric three-spin-one subspace. Its decomposition isHence only the $J=3$ and $J=1$ levels remain, now with one copy of each multiplet:

For two equal spins, exchange acts on the coupled state $|j,m⟩$ by the Clebsch-Gordan coefficient symmetry factor $(-1{)}^{2j_1-j}$. It is therefore $-1$ when $j=2j_1-1$, making every state in that multiplet antisymmetric, and $+1$ when $j=2j_1-2$, making the next multiplet symmetric.

The Clebsch-Gordan decomposition permitsThus a total-angular-momentum singlet $j=0$ occurs exactly when $j_1=j_2$.

With ${ϵ}_{0123}=+1$, define the Pauli-Lubanski pseudovector byFor a translation $a^{\mu }$ and a Lorentz transformation $\Lambda =\exp (\omega )$, a left-handed Weyl spinor transforms asEquivalently, the active field at a fixed point uses ${\psi }^{\prime }(x)=S(\Lambda )\psi ({\Lambda }^{-1}(x-a))$.

The Clebsch-Gordan decomposition isIndeed,The symmetric spinor is the $(1,0)$ representation, while its antisymmetric part is the Lorentz scalar $(0,0)$.

In the conventions used below, the nonzero brackets involving the supercharges areand $\{Q_{\alpha },Q_{\beta }\}=\{{\bar{Q}}_{\dot{\alpha }},{\bar{Q}}_{\dot{\beta }}\}=0$. These equations, together with the Poincare algebra, are the four-dimensional $\mathcal{N}=1$ Super-Poincaré algebra without central charges.

A total-spin-$J$ sector is an irreducible subspace on which ${\mathbf{J}}^2$ has eigenvalue $J(J+1){\hslash }^2$. For two spin-one particles, Clebsch-Gordan decomposition gives sectors $J=0,1,2$ of dimensions one, three, and five.