Total-spin sector

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 a spin-one Hamiltonian $H=A{S}_z^2+B(S_x^2-S_y^2)$, Cartesian basis energies are ${\hslash }^2(A-B)$, ${\hslash }^2(A+B)$ and zero along the $x,y,z$ basis labels. The $S_x=\hslash$ state combines the latter two basis vectors equally. Their relative phase produces the displayed spin-reversal probability. Complete reversal occurs at odd multiples of $\pi /(|A+B|\hslash )$ if $A+B\ne 0$; when it vanishes those two components remain degenerate. Individual matrix signs depend on basis conventions, while the commutators and physical probability must remain consistent.

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 ${\hslash }^2$.

For $s_1=s_2=3/2$, angular momentum addition gives one copy of $j=0,1,2,3$. The identity $2{\mathbf{S}}_1\cdot {\mathbf{S}}_2={\mathbf{J}}^2-{\mathbf{S}}_1^2-{\mathbf{S}}_2^2$ gives the four eigenvalues in the title, with state degeneracy $1,3,5,7$. These exhaust the $4\times 4$ spin space. If the particles are identical fermions, an appropriate spatial exchange parity accommodates each sector; only a specified spatial restriction can remove sectors.