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For particles in the vector representation of the orthogonal group, the two-body S-matrix is a linear combination of the identity $I$, permutation $P$ and trace contraction $K$: $S=S_{2}I+S_{3}P+S_{1}K$. Here $K(e_i\otimes e_j)={\delta }_{ij}{\sum }_a{e}_a\otimes e_a$, $P^2=I$, $PK=KP=K$ and $K^2=NK$. The orthogonal-invariant scattering channels diagonalize these three operators simultaneously.

Interpret the paired indices $(i,k)$ as incoming and $(j,l)$ as outgoing. For self-conjugate particles in the real vector representation of the orthogonal group, crossing the second incoming particle with the second outgoing one givesThe continuation $\theta \mapsto i\pi -\theta$ exchanges the trace and permutation tensor structures and leaves the identity structure fixed. Therefore crossing symmetry imposesThese are analytic relations; they do not determine the three functions completely.

For unitarity, act on the two-species tensor product with the O(N)-invariant S-matrix $S=S_{2}I+S_{3}P+S_{1}K$. The invariant operators obey $P^2=I$, $PK=KP=K$ and $K^2=NK$. Writing $S_a^{-}=S_a(-\theta )$ and multiplying $S(\theta )S(-\theta )=I$ givesThe three invariant operators are independent for $N\ge 2$. Equivalently, the orthogonal-invariant scattering channels have amplitudesand each satisfies $s_a(\theta )s_a(-\theta )=1$. The trace singlet, symmetric traceless and antisymmetric channels have dimensions $1$, $N(N+1)/2-1$ and $N(N-1)/2$. For $N=2$ these are $1,2,1$. With Hermitian analyticity of a two-particle S-matrix, $S_a(-\theta )=S_a(\theta {)}^{\ast }$ on the real axis, so each channel has unit modulus. This expresses conservation of scattering probability.

The Faddeev-Zamolodchikov algebra orders particle operators by rapidity. Its associativity requires that a product of three operators be independent of parentheses and, in particular, that both sequences of adjacent exchanges give the same final ordered species word with the same coefficient. This is the Faddeev-Zamolodchikov associativity constraint, or spectral Yang-Baxter equation. The exchanges do not use the ordinary creation/annihilation normal ordering convention.

Take ${\theta }_1>{\theta }_2>{\theta }_3$, with $\theta ={\theta }_1-{\theta }_2$, ${\theta }^{\prime }={\theta }_2-{\theta }_3$, and abbreviateFor $N=2$, start with $A_1({\theta }_1)A_1({\theta }_2)A_2({\theta }_3)$ and compare the coefficient of $A_1({\theta }_3)A_2({\theta }_2)A_1({\theta }_1)$. First exchange positions $1,2$, then $2,3$, then $1,2$. The initial equal-index exchange givesFrom the first term, the specified final word is reached through $b_2{c}_3$; from the second term, the intermediate equal-index exchange uses $b_1$ to produce species $1$, followed by $c_2$. Thus this exchange route has coefficientFor the other route, first exchange positions $2,3$, then $1,2$, then $2,3$. The first exchange is between different indices and gives $c_2{A}_1({\theta }_1)A_2({\theta }_3)A_1({\theta }_2)+c_3{A}_1({\theta }_1)A_1({\theta }_3)A_2({\theta }_2)$. The first term reaches the target through $b_3{a}_3$, and the second reaches it through $(b_1+b_2+b_3)a_2$. ThereforeEquate these two coefficients, cancel the common term $a_2{b}_2{c}_3$, and rearrange:Restoring the three arguments gives the required scalar Yang-Baxter equation. The derivation starts in the ordered physical region $\theta ,{\theta }^{\prime }>0$; the identity extends to other values by the same scattering analytic continuation, wherever its factors are defined. Associativity must hold coefficient by coefficient for every species word; this displayed relation is one necessary component.