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The two-vector tensor product splits into a trace singlet, symmetric traceless tensors and antisymmetric tensors. For an O(N)-invariant S-matrix, the corresponding eigenvalues are $s_0=N{S}_1+S_2+S_3$, $s_{+}=S_2+S_3$ and $s_{-}=S_2-S_3$. For $N\ge 2$, analytic unitarity requires $s_a(\theta )s_a(-\theta )=1$. With Hermitian analyticity of a two-particle S-matrix, each channel has unit modulus on the real rapidity axis.

In elastic diagonal factorized scattering, the Faddeev-Zamolodchikov algebra describes an exchange of two particle operators. Exchanging the pair twice must restore the original state. This is analytic unitarity:Keeping the reversed species in this identity avoids an unstated parity assumption. For the particular amplitudes derived below, crossing and analytic unitarity also imply $S_{\bar{A}A}=S_{A\bar{A}}$. Hermitian analyticity of a two-particle S-matrix states $S_{ab}(\theta {)}^{\ast }=S_{ba}(-{\theta }^{\ast })$ and, together with analytic unitarity, gives $|S_{ab}(\theta )|=1$ for real rapidity differences. Crossing symmetry analytically turns an incoming particle into an outgoing antiparticle, relating the two channels by $S_{A\bar{A}}(\theta )=S_{AA}(i\pi -\theta )$, with the reverse relation obtained by crossing again. The shift $i\pi$ follows from the sign reversal of the two-momentum $p(\theta )=(m\cosh \theta ,m\sinh \theta )$.

Put $u=\lambda /2$. The given amplitude is the unit-modulus hyperbolic scattering block $S_u(\theta )=\sinh [(\theta +iu)/2]/\sinh [(\theta -iu)/2]$. Its numerator and denominator interchange under $\theta \mapsto -\theta$, proving analytic unitarity. For real $u$ it also obeys Hermitian analyticity of a two-particle S-matrix. Applying crossing symmetry givesThis crossed amplitude also has unit modulus on the real axis. Thus both requested channel constraints hold.

The bound-state conclusion needs a coupling range. In the fundamental attractive range $0<u<\pi$, equivalently $0<\lambda <2\pi$, the denominator has a simple pole at $\theta =iu$ inside the physical rapidity strip. Its residue iswith positive imaginary coefficient. Under the usual one-particle interpretation of this direct-channel bound-state pole, it couples two charge-$+1$ particles to a new charge-$+2$ particle $B$. In the center-of-mass frame its constituents have analytically continued rapidities $\pm iu/2$, and their total two-momentum is $(2m\cos (u/2),0)$. Thus the relativistic bound-state mass from a rapidity pole givesIt is positive and less than the two-particle threshold $2m$. Without the coupling qualification the requested deduction is false: at $\lambda =0$ the amplitude is identically one after removing the apparent $0/0$ at the origin. A free massive complex scalar field has this diagonal amplitude, charge-$\pm 1$ particles and no isolated charge-$+2$ bound state. At $u=\pi$ the amplitude similarly becomes constant $-1$ and the apparent boundary pole cancels. Neither endpoint supplies the claimed particle.

For bound-state fusion of factorized S-matrices, represent $B$ as the residue of $A({\theta }_B+iu/2)A({\theta }_B-iu/2)$ at its bound-state separation. Move a third $A({\theta }_A)$ through both constituents using the Faddeev-Zamolodchikov algebra, then take the same residue. The bound-state normalization occurs on both sides and cancels. Consequently bootstrap fusion giveswhere $\theta ={\theta }_B-{\theta }_A$. This has analytic unitarity as a product of two shifted blocks. Its poles occur at $\theta =iu/2$ and $\theta =3iu/2$, modulo $2\pi i$; numerator zeroes occur at the corresponding negative positions, subject to cancellations at special couplings.

The nearer pole, $\theta =iu/2$, has residue $-2i\sin u$. It is a crossed-channel pole in diagonal factorized scattering, not a new direct-channel charge-$+3$ state. Indeed, the exchanged momentum has invariantusing $m_B=2m\cos (u/2)$. The exchanged particle is therefore the already present $A$, with the appropriate charge flow at the crossed vertex. Equivalently, crossing symmetry puts a positive-residue direct pole of $S_{B\bar{A}}$ at $i(\pi -u/2)$, corresponding to $B+\bar{A}⟶A$. This distinction avoids assigning an extra mass by applying the direct-channel formula to every pole.

The farther pole, $\theta =3iu/2$, lies in the physical rapidity strip only for $0<u<2\pi /3$, equivalently $0<\lambda <4\pi /3$. Its residue iswhich has positive imaginary coefficient in that range. It gives a charge-$+3$ bound state $C$. The relativistic bound-state mass from a rapidity pole now yieldsThe positive root in the admitted range isThis is the three-particle bound state from equal-mass fusion; in its own rest frame the constituent rapidities are $iu,0,-iu$. At $u=2\pi /3$ the farther apparent pole cancels because its numerator also vanishes; for $2\pi /3<u<\pi$ it lies outside the physical rapidity strip and does not require a new charge-$+3$ particle. Charge-conjugate partners carry charges $-2$ and $-3$ where the corresponding states exist. Fusion distinguishes an existing crossed-channel particle from a genuinely new direct-channel bound state, and the latter requires the stated smaller coupling range.

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.

For real $u$, this scalar S-matrix block satisfies $S_u(\theta )S_u(-\theta )=1$ and Hermitian analyticity of a two-particle S-matrix. It therefore has modulus one for real rapidity differences. For $0<u<\pi$ its pole at $\theta =iu$ has residue $2i\sin u$ and admits a direct-channel bound-state pole interpretation. Its crossed block is $\cosh [(\theta -iu)/2]/\cosh [(\theta +iu)/2]$. The endpoints $u=0$ and $u=\pi$ are removable degeneracies giving constant amplitudes, so a particle cannot be inferred there from the uncancelled denominator alone. This block specifies an amplitude; a complete theory must also obey the remaining crossing symmetry and bootstrap fusion constraints.