Addition of angular momentum

For two subsystems, the total operator is $\mathbf{J}={\mathbf{J}}_1\otimes I+I\otimes {\mathbf{J}}_2$, and $J_{\pm }=J_{1\pm }+J_{2\pm }$. The coupled basis diagonalizes ${\mathbf{J}}^2$ and $J_z$.

For a composite quantum system, the total angular momentum operator is the sum of the constituent angular momenta, with each summand acting on its own tensor factor. Its simultaneous eigenstates with $J_z$ organize the state space into total-spin sectors.

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.

Coupling three distinguishable spin-one systems giveswhere the labels denote total spin and coefficients denote multiplicities. The dimensions are respectively $1$, $3\cdot 3$, $2\cdot 5$, and $7$, summing to $27$.

The completely symmetric subspace of three spin-one systems has dimensionand decomposes into one total-spin-$3$ multiplet and one total-spin-$1$ multiplet:The highest-weight product state generates the seven-dimensional spin-$3$ multiplet; its three-dimensional invariant complement is spin $1$.

The spin-one Cartesian basis $|x⟩,|y⟩,|z⟩$ transforms as a three-dimensional vector under rotations. In its two-site tensor square, the scalar, antisymmetric, and symmetric traceless subspaces are respectively the total-spin $0$, $1$, and $2$ sectors.

For two spins of magnitude $s$, ${\mathbf{S}}_1\cdot {\mathbf{S}}_2=\frac{1}{2}({\mathbf{J}}^2-{\mathbf{S}}_1^2-{\mathbf{S}}_2^2)$. For two spin-one particles its eigenvalues in total-spin sectors $J=0,1,2$ are $-2,-1,1$ in units of ${\hslash }^2$.

The tensor product of spin-$j_1$ and spin-$j_2$ irreducible representations decomposes into one copy of each spinThe change-of-basis entries between uncoupled and coupled states are Clebsch--Gordan coefficients.

With $J=j_1+j_2$, the maximal state is the product $|J,J⟩=|j_1,j_1⟩|j_2,j_2⟩$. Lowering it once givesThe orthogonal combination is the highest-weight state $|J-1,J-1⟩$.

Two equal spins $j$ have the unique total-spin-zero stateIt is invariant under joint rotations.

For two spin-one-half particles, the singlet isEvery component of the total spin annihilates it, and ${\mathbf{S}}^2$ has eigenvalue zero.

The symmetric two-particle statesform the spin-one triplet. On this subspace, ${\mathbf{S}}^2$ has eigenvalue $1(1+1){\hslash }^2=2{\hslash }^2$.