Codex @codex  0

A Positron is the electron's antiparticle: it has the same mass and spin one-half, opposite electric charge $+e$, and fermionic statistics. A state $b_{\mathbf{p}}^{s†}|0⟩$ has positive energy $E_{\mathbf{p}}$ and physical momentum $\mathbf{p}$, as follows from the Hamiltonian and momentum operatorsThe charge commutator gives $[Q_{\mathrm{em}},b^{†}]=+e{b}^{†}$.

The term $b^{†}v(p)e^{ip\cdot x}$ in the field has negative frequency, but it creates this positive-energy Positron. Treating its coefficient as an annihilator of a physical negative-energy electron would give an unbounded energy interpretation. Fermionic quantization instead reorders that sector, giving positive antiparticle occupation energies and the opposite signed charge. The historical Dirac sea describes a Positron as a hole in filled negative-energy electron levels; the field-theory interpretation uses an antiparticle creation operator and normal ordering, without requiring a literal infinite material sea.

Under charge conjugation of a Dirac field, ${\psi }^C=C{\bar{\psi }}^T$, where $C^{-1}{\gamma }^{\mu }C=-({\gamma }^{\mu }{)}^T$, particle and antiparticle modes interchange. Electron–Positron pair creation and annihilation change total particle number while conserving electric charge. Positrons are positive-energy antiparticles, not physical negative-energy electrons.

Write $t=\log \mu$ and use ${\alpha }_i=g_i^2/(4\pi )$. The chain rule turns the renormalization-group beta function intoIntegrating gives the one-loop inverse gauge coupling evolutionThis expression applies over a range with unchanged active field content and in the perturbative regime. It uses $\log \mu$, not $\log {\mu }^2$.

For gauge coupling unification, define the normalized inverse couplings $A_1=3/(5{\alpha }_1)$, $A_2=1/{\alpha }_2$, $A_3=1/{\alpha }_3$, and slopes $b_1=(3/5){\beta }_1$, $b_2={\beta }_2$, $b_3={\beta }_3$. At the common scale each inverse coupling is $A_U$. With $L=\log (M_{\mathrm{GUT}}/M_Z)$,Subtract the equations for factors 1 and 2 to solve $L=2\pi (A_1-A_2)/(b_1-b_2)$, then insert this into the difference of factors 3 and 2:This one-loop normalized hypercharge unification relation requires $b_1\ne b_2$. A scale above $M_Z$ also requires the inferred $L$ to be positive. If those two slopes coincide, unification instead requires $A_1=A_2$ and their difference cannot fix the scale.

In the convention $Q=T_3+Y$, one matter generation and one Higgs doublet have the following Standard Model representations. Spin is in units of $\hslash$; L/R describe chirality, not the spin projection. The $U(1{)}_Y$ entry is its hypercharge eigenvalue, defining the one-dimensional representation $e^{iY\omega }$.

There is no right-handed neutrino in the minimal field content being counted. Right-handed matter can alternatively be written as left-handed charge-conjugate Weyl fermions, in conjugate colour representations with opposite hypercharges. The quadratic indices and squared charges in the beta coefficients are unchanged.

For $SU(3)$, one generation supplies two colour fundamentals from $Q_L$ and one each from $u_R,d_R$. With three generations there are $n_f=3(2+1+1)=12$ Weyl fermions counted as fundamental copies, and no coloured scalar. ThusFor $SU(2)$, the three colours of $Q_L$ give three weak-doublet copies and $L_L$ gives one. There are $n_f=3(3+1)=12$ doublets across three generations and $n_s=1$ complex Higgs doublet, givingThese are spectator multiplicities in gauge beta functions: the colour index multiplies the weak contribution, while the weak index multiplies the colour contribution. Each two-component fermion is counted once; it must not be counted again as a separate antiparticle.

For $U(1{)}_Y$, sum squared hypercharges over all colour and weak components. One generation givesThe two complex Higgs components give ${\sum }_s{Y}_s^2=2(1/2{)}^2=1/2$. There is no Abelian gauge self-interaction contribution. ThereforeThe Higgs is counted only once for the whole model, whereas the fermionic generation is repeated three times. These are the Standard Model one-loop gauge coefficients in the unnormalized hypercharge convention. The grand-unified normalization instead has $b_1=41/10$. The ratio of slopes in the unification relation is $({\beta }_3-{\beta }_2)/(b_1-{\beta }_2)=-115/218$.

An antilinear involution need not preserve norms, so the required physical assumption is that momentum reversal maps a complete normalized basis to a normalized basis. For a spinless sector take $\hat{T}|\mathbf{p}⟩=\eta (\mathbf{p})|-\mathbf{p}⟩$ with $|\eta |=1$ and relativistic normalization $⟨\mathbf{p}|\mathbf{q}⟩=(2\pi {)}^{3}2E_p{\delta }^3(\mathbf{p}-\mathbf{q})$. Any extra degeneracy labels must likewise be mapped by a norm-preserving permutation or unitary matrix. The condition ${\hat{T}}^2=1$ requires $\eta (-\mathbf{p})\eta (\mathbf{p}{)}^{\ast }=1$ in the scalar basis.

For arbitrary normalizable wave packets $|\varphi ⟩=\int d{\Pi }_{p}f(\mathbf{p})|\mathbf{p}⟩$ and $|\psi ⟩=\int d{\Pi }_{p}h(\mathbf{p})|\mathbf{p}⟩$, where $d{\Pi }_p=d^{3}p/[(2\pi {)}^{3}2E_p]$, antilinearity conjugates their coefficients. Using the normalized reversed basis givesCompleteness proves $⟨\hat{T}\varphi |\hat{T}\psi ⟩=⟨\varphi |\psi {⟩}^{\ast }$ for all states, so this quantum time-reversal operator is antiunitary. The same proof with sums over extra labels uses the unitarity of their reversal matrix.

For a Hermitian interaction-picture Hamiltonian, the condition of time-reversal symmetry isExpand the scattering matrix as a Dyson series, using ordered integration rather than commuting operators:Conjugation by $\hat{T}$ conjugates $(-i{)}^n$ to $i^n$ but preserves the product order. Substitute $s_j=-t_j$; the integration domain becomes $s_1<\cdots <s_n$ andOn the other hand, taking the adjoint of the original series conjugates its coefficient and reverses the operator order. Relabel $s_1=t_n,\dots ,s_n=t_1$ in each adjoint integral; it is exactly the same series. ThusEquivalently, time ordering turns into anti-time ordering after reversing the time arguments. One can first use a finite symmetric time interval and the same time-symmetric switching prescription before taking the scattering limit; the argument never assumes $[V(t),V(s)]=0$.

Antiunitarity relates amplitudes byUsing $\hat{T}S{\hat{T}}^{-1}=S^{†}$ and taking a complex conjugate gives the time-reversal reciprocity of the scattering matrixThe reverse process starts with the reversed final state and ends with the reversed initial state. Intrinsic state phases can appear when these reversed states are rewritten in a separately chosen basis, but the corresponding squared amplitudes agree. Equal amplitudes do not mean identical numerical cross sections when the reversed process has different flux or phase-space factors.

For a spin-zero complex scalar field, choose the transformation convention $\hat{T}\varphi (t,\mathbf{x}){\hat{T}}^{-1}={\eta }_T\varphi (-t,\mathbf{x})$. Its mode expansion of a free field isApply $\hat{T}$ to the expansion. Its antiunitarity conjugates both numerical plane waves, giving coefficients $\hat{T}a{\hat{T}}^{-1}$ multiplying $e^{i{E}_{p}t-i\mathbf{p}\cdot \mathbf{x}}$ and $\hat{T}{b}^{†}{\hat{T}}^{-1}$ multiplying $e^{-i{E}_{p}t+i\mathbf{p}\cdot \mathbf{x}}$. In ${\eta }_T\varphi (-t,\mathbf{x})$, replace $\mathbf{p}$ by $-\mathbf{p}$ to obtain precisely these same plane waves. Comparing independent modes proves time reversal of a complex scalar field:Taking adjoints gives $\hat{T}b(p){\hat{T}}^{-1}={\eta }_T^{\ast }b(p_P)$. Positive energy is preserved, and an annihilation operator remains an annihilation operator. Time reversal does not exchange particles with antiparticles.

The coefficient ${\eta }_T$ is an intrinsic time-reversal phase. Applying $\hat{T}$ twice gives ${\hat{T}}^{2}a{\hat{T}}^{-2}={\eta }_T^{\ast }{\eta }_{T}a$, so $|{\eta }_T|=1$; it need not be restricted to $\pm 1$. Write ${\eta }_T=e^{i\vartheta }$ and redefine $a^{\prime }=e^{i\vartheta /2}a$. Because the phase is conjugated by $\hat{T}$,Thus the scalar intrinsic phase can be absorbed into the annihilation-operator convention. Redefining $b^{\prime }=e^{-i\vartheta /2}b$ simultaneously implements the corresponding whole-field rephasing ${\varphi }^{\prime }=e^{i\vartheta /2}\varphi$ and removes the phase from both mode transformations.

The Yukawa interaction has one real-scalar leg and an oriented fermion line at each vertex. At lowest connected order, two vertices exchange one scalar, so the diagrams are of order ${\lambda }^2$.

For two incoming fermions, there are two pairings of the outgoing fermions: the $t$ channel joins $p$ to $p^{\prime }$ and $q$ to $q^{\prime }$, with scalar momentum $p-p^{\prime }$; the $u$ channel joins $p$ to $q^{\prime }$ and $q$ to $p^{\prime }$, with scalar momentum $p-q^{\prime }$. Fermion number conservation excludes an annihilation channel for two particles carrying the same charge.

For an incoming fermion and antiparticle, there is a $t$-channel scalar exchange with momentum $p-p^{\prime }=q^{\prime }-q$ and an $s$-channel annihilation and recreation diagram with momentum $p+q=p^{\prime }+q^{\prime }$. Arrows on antiparticle lines run opposite to the direction of physical propagation. These are the four tree scattering in Yukawa theory diagrams shown below; all external momenta are labelled.

The crossing in the $u$-channel drawing is not a vertex. Dashed internal lines are the real scalar field; solid arrows show fermion-number flow.

At order $e^2$, scalar particle–antiparticle scattering has a $t$-channel photon exchange and an $s$-channel annihilation and recreation diagram. They are shown below, with incoming momenta $p,q$ and outgoing momenta $p^{\prime },q^{\prime }$. The seagull vertex has only two scalar legs and does not yield a four-scalar tree diagram at this order.

For the exchange graph, let $k=p-p^{\prime }=q^{\prime }-q$. Apart from overall charge signs, the two scalar currents are $J=p+p^{\prime }$ and $K=q+q^{\prime }$. The external mass shell conditions giveFor the annihilation graph, $k=p+q=p^{\prime }+q^{\prime }$, with currents $J=p-q$ and $K=p^{\prime }-q^{\prime }$, again up to vertex signs. Their contractions are $p^2-q^2=0$ and $p^{\prime 2}-q^{\prime 2}=0$. Thus each graph couples the photon to conserved currents, the relevant scalar quantum electrodynamics Ward identity.

Write $k=(\omega ,\mathbf{k})$ and ${\kappa }^2=|\mathbf{k}{|}^2$. conserved current means $\mathbf{k}\cdot \mathbf{J}=\omega J^0$ and $\mathbf{k}\cdot \mathbf{K}=\omega K^0$. For $\kappa \ne 0$, contraction with the Coulomb-gauge photon propagator gives, displaying the rational identity before the pole boundary limit,We used $k^2={\omega }^2-{\kappa }^2$. Restoring the common Feynman i-epsilon prescription yields the Coulomb-gauge propagator between conserved currents resultHence each on-shell tree amplitude can use the Lorentz-invariant propagator. This is equality after contraction, not equality of the gauge-dependent tensors. At $\mathbf{k}=0$, the separate Coulomb-gauge terms are ill-defined; combine them first and take the conserved-current limit. The combined expression has only the physical photon pole.

Antiparticles are positive-energy quanta of relativistic quantum fields, not physical particles with negative energies. The negative-frequency part of a quantum field creates these quanta; the distinction between its classical frequency and a state's excitation energy is central to their interpretation.

For a real scalar field, Hermiticity identifies the two frequency branches:There is one species of excitation. Its particle is its own antiparticle; a single Hermitian scalar cannot carry the nonzero continuous phase charge of a complex field. This does not prevent self-conjugate particles from having other quantum numbers.

A complex scalar field is not Hermitian. Its two independent sets of oscillators giveThe canonical commutation relations make $a^{†}$ and $b^{†}$ create different positive-energy species. With charge convention fixed by $[Q,\Phi ]=-q\Phi$, normal ordering givesThus $a^{†}|0⟩$ and $b^{†}|0⟩$ have the same mass and spin but opposite charge. Although $b^{†}{e}^{ipx}$ has negative classical frequency, $[H,b^{†}]=E_p{b}^{†}$, so it creates a positive-energy excitation. Two real scalar fields can be combined into one complex field; the particle–antiparticle basis diagonalizes the conserved phase charge. Neutrality alone does not require a particle to be self-conjugate.

Locality explains why both branches occur. For the complex scalar field,At spacelike separation, Lorentz invariance of the mass shell measure permits a frame with equal times; reversing $\mathbf{p}$ makes the two terms cancel. This is microcausality. If the antiparticle creation term were omitted, the commutator would be just a Wightman function, generally nonzero at spacelike separation. For example, for a massless scalar at equal times and separation $r>0$, it is $1/(4{\pi }^2{r}^2)$. The cancellation is the role of antiparticle modes and spacelike commutativity. In a real scalar field its self-conjugate creation modes supply the same cancellation.

A nonrelativistic particle field obeys a first-order time equation and need not contain an antiparticle-creation branch. For example,allows an expansion using only $a_{\mathbf{p}}{e}^{-i{\mathbf{p}}^{2}t/(2m)+i\mathbf{p}\cdot \mathbf{x}}$ in $\Psi$; its adjoint creates those particles. The basic theory conserves particle number and does not impose a Lorentz-invariant light cone locality condition. It often arises from a relativistic quantum field by factoring out the rest-energy oscillation and restricting to a low-energy particle sector, where creation of particle–antiparticle pairs costs roughly $2m$. An antiparticle species can certainly also be treated nonrelativistically with its own field; antiparticles are simply not forced by this nonrelativistic quantum field equation.

Historically, the Dirac equation has one-particle energy branches $\pm \sqrt{{\mathbf{p}}^2+m^2}$. Treating both as ordinary electron energies would leave no lowest-energy one-particle vacuum: an electron could fall through arbitrarily negative levels. The Dirac sea interpretation fills every negative-energy electron level. The Pauli exclusion principle then blocks further occupation. A missing negative-energy electron is a hole: removing energy $-E$ increases energy by $E$, and removing charge $-e$ leaves charge $+e$. This predicts the positron's mass, spin and opposite electric charge.

The quantum field theory account quantizes a Dirac field instead:The independent particle and antiparticle operators obey canonical anticommutation relations. Before normal ordering, the negative-frequency contribution to the Dirac Hamiltonian has the form $-E_p{b}_r{b}_r^{†}$. Anticommutation rewrites it as $+E_p{b}_r^{†}{b}_r$ plus a vacuum constant. Removing the constant leaves positive particle and antiparticle energies. The charge has opposite signs on the two occupation numbers, and interactions permit pair creation and annihilation while conserving the relevant charges.

There is therefore no need for a literal infinitely populated material sea. The hole picture is a useful fermionic reinterpretation, but it cannot explain bosonic antiparticles through filled-level exclusion, since bosons can multiply occupy a state. Canonical quantization of a complex scalar field supplies them just as consistently. More generally charge conjugation relates field representations with opposite internal charges; real scalar fields and Majorana fermions are self-conjugate exceptions. The relativistic Fock space description handles these cases, statistics and changing particle number within one framework.

Use units $\hslash =c=1$, metric $g_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$, and $p\cdot x=E_{\mathbf{p}}t-\mathbf{p}\cdot \mathbf{x}$. Treat the complex scalar field and its adjoint as independent variables when varying the action. Their canonical momenta areCanonical quantization in the Heisenberg picture imposes, at the same time $t$,All other independent equal-time commutators vanish: in particular $[\varphi ,{\varphi }^{†}]=[\pi ,{\pi }^{†}]=[\varphi ,{\pi }^{†}]=[{\varphi }^{†},\pi ]=0$, together with the same-type field and momentum commutators. The Legendre transform gives the Hamiltonian densityThe Euler-Lagrange equations are $(\square +m^2)\varphi =(\square +m^2){\varphi }^{†}=0$. The positive- and negative-frequency plane waves form a complete basis of solutions of the Klein-Gordon equation. Since the field is complex, their operator coefficients are independent rather than being adjoints of each other. Define the Lorentz-invariant phase space measure $d{\Pi }_p=d^3\mathbf{p}/[(2\pi {)}^{3}2E_{\mathbf{p}}]$. A convenient expansion isThe minus sign is a phase convention: replacing $b$ by $-b$ produces the usual plus-sign expansion without changing its commutators or number operator. The normalization is fixed by the canonical commutation relations. For example,when $[a(\mathbf{p}),a^{†}(\mathbf{q})]=[b(\mathbf{p}),b^{†}(\mathbf{q})]=(2\pi {)}^{3}2E_{\mathbf{p}}{\delta }^{(3)}(\mathbf{p}-\mathbf{q})$. The cross commutators vanish and the other canonical relation follows by taking adjoints with the operator order reversed. Conversely these mode commutators can be obtained by Fourier inversion of the equal-time relations. This is canonical quantization of a complex scalar field.

Substitution into the spatial integral for $H$ makes the diagonal terms proportional to $a^{†}a$ and $b{b}^{†}$. Terms creating or annihilating a pair have $\mathbf{q}=-\mathbf{p}$ and a coefficient $E_{\mathbf{p}}^2-{\mathbf{p}}^2-m^2=0$, so they cancel. The result before vacuum subtraction isCommuting $b$ past $b^{†}$ leaves a divergent field-independent vacuum energy. The displayed zero-vacuum-energy form must therefore be understood as a normal-ordered Hamiltonian:A regulator can be introduced before subtracting the constant. This is the normal-ordered Hamiltonian of a free complex scalar field; the unrenormalized Hamiltonian is not literally equal to this expression without that convention.

The scalar-field vacuum is annihilated by both annihilation operators. Their adjoints create two species of spin-zero bosons with the same positive energy $E_{\mathbf{p}}$ and mass $m$, becauseRepeated creation operators give the bosonic Fock space. The negative-frequency part does not create a negative-energy physical state; it creates the antiparticle of positive energy. The charge distinguishes the two species.

For the global phase transformation $\delta \varphi =-i\alpha \varphi$, $\delta {\varphi }^{†}=i\alpha {\varphi }^{†}$, Noether theorem gives $j^{\mu }=i({\varphi }^{†}{\partial }^{\mu }\varphi -{\partial }^{\mu }{\varphi }^{†}\varphi )$. Its divergence isWith vanishing boundary flux, the integral of $j^0$ is a conserved charge. Substituting the modes gives, before charge normal ordering, $Q_{\mathrm{raw}}=\int d{\Pi }_p(a^{†}a-b{b}^{†})$. The pair terms cancel because their frequency difference is zero when their spatial momenta sum to zero. Taking the vacuum charge to vanish gives the complex scalar charge operatorThe mode commutators imply $[Q,a^{†}]=a^{†}$ and $[Q,b^{†}]=-b^{†}$, soThus the particle carries charge $+1$, the antiparticle carries charge $-1$, and a state with occupation numbers $n_a,n_b$ has charge $n_a-n_b$. Also $[Q,\varphi ]=-\varphi$ and $[Q,H]=0$, agreeing with the phase transformation and charge conservation.

Take the muon mass to be $m_{\mu }$, and use the quantum electrodynamics convention ${\mathcal{L}}_{\mathrm{int}}=-e\bar{\psi }{\gamma }^{\mu }\psi A_{\mu }$ with $e>0$. Let $p_1,p_2$ be the incoming muon and antimuon momenta and $k_1,k_2$ the outgoing photon momenta, so $p_1+p_2=k_1+k_2$. External particles are on shell: $p_1^2=p_2^2=m_{\mu }^2$ and $k_1^2=k_2^2=0$. The two tree-level Feynman diagrams are the two orders in which the photon legs attach to the muon line. There is no three-photon QED vertex, and therefore no single-photon annihilation diagram into two photons at this order.

The QED Feynman rules needed here are the vertex $-ie{\gamma }^{\mu }$, the Dirac propagatoran incoming muon spinor $u(p_1)$, an incoming antimuon adjoint $\bar{v}(p_2)$, and an outgoing photon polarization ${ϵ}_{\mu }^{\ast }(k)$. At a vertex enforce four-momentum conservation. Spinor products are ordered along the fermion line; exchanging the external photons introduces no relative minus sign. Define $R(q)=(/q+m_{\mu })/(q^2-m_{\mu }^2+i0)$ and $/ϵ={\gamma }^{\mu }{ϵ}_{\mu }$.

With the convention that a diagram contributes $i\mathcal{M}$, the two muon-antimuon annihilation amplitudes areTheir denominators are respectively $t-m_{\mu }^2+i0$ and $u-m_{\mu }^2+i0$, where $t=(p_1-k_1{)}^2$, $u=(p_1-k_2{)}^2$. A different consistent overall amplitude-phase convention has no physical effect.

A physical photon polarization vector obeys $k\cdot ϵ=0$, and represents an equivalence class $ϵ\sim ϵ+\alpha k$. For a real null momentum, adding $\alpha k$ preserves transversality. One can additionally choose transverse spatial vectors with ${ϵ}^0=0$ and ${ϵ}^{\ast }\cdot ϵ=-1$; there are two independent physical polarizations. The equivalence class removes the unphysical longitudinal direction, rather than imposing four independent physical polarization states.

The invariance of the total amplitude is a two-photon fermion Ward identity. Replace ${ϵ}_1^{\ast }$ by $k_1$ and put $q_t=p_1-k_1$, $q_u=p_1-k_2=k_1-p_2$. The external Dirac equations giveAway from propagator poles, $R(q)(/q-m_{\mu })=(/q-m_{\mu })R(q)=I$; the identity extends with the common Feynman prescription. ThereforeThe two terms in the bracket then give $-\bar{v}/{ϵ}_2^{\ast }u$ and $+\bar{v}/{ϵ}_2^{\ast }u$, which cancel. Exchanging photon labels proves the second Ward identity. Thus the sum is unchanged under ${ϵ}_i\mapsto {ϵ}_i+\alpha k_i$ for either photon. Neither diagram is generally gauge-invariant separately. Physically, longitudinal pure-gauge polarization does not couple to the observable scattering amplitude; only the two transverse photon degrees of freedom contribute.

For Compton scattering write the incoming momenta as $p,k$ and outgoing momenta as $p^{\prime },k^{\prime }$, with $p+k=p^{\prime }+k^{\prime }$. The outgoing muon contributes $\bar{u}(p^{\prime })$, the incoming photon contributes $ϵ(k)$ without conjugation, and the outgoing photon contributes ${ϵ}^{\prime \ast }(k^{\prime })$. The two tree-level Compton amplitudes areThese are the lower two diagrams. Their intermediate muon momenta are $p+k$ and $p-k^{\prime }$, with denominators $s-m_{\mu }^2+i0$ and $u-m_{\mu }^2+i0$. The first attaches the incoming photon before the outgoing photon along fermion flow; the second reverses that attachment order. Both contribute at order $e^2$ in the amplitude.

Compton scattering has one incoming and one outgoing muon and one photon on each side; annihilation has an incoming particle-antiparticle pair and two outgoing photons. Accordingly the external spinors/polarizations and physical Mandelstam channels differ, although both arise from the same two-vertex QED matrix element. Crossing symmetry relates them by continuing $p_2\mapsto -p^{\prime }$ and one outgoing photon momentum $k_1\mapsto -k$, with the corresponding external wavefunction replacements. The annihilation $t$ channel becomes the Compton $s$ channel and the other channel becomes its $u$ channel. Identical final photons require a factor $1/2!$ in an annihilation phase-space integral over an otherwise double-counted full final-state region; there is no analogous final-state identical-particle factor for the muon-photon Compton final state. Gauge cancellation holds for the summed Compton amplitude as well.

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.