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For $\Psi (p_1)\to \chi (p_2)+\varphi (k_1)+\varphi (k_2)$, the sole tree-level Feynman diagram has a Yukawa coupling interaction vertex producing an off-shell scalar field particle of four-momentum $q=p_1-p_2=k_1+k_2$, followed by its cubic scalar field interaction vertex. Its invariant decay amplitude isTwo successive Yukawa coupling interaction vertices would change the fermion species twice and leave a $\Psi$, so they cannot produce this final state. Interchanging the two identical outgoing bosons does not create another Feynman diagram; the cubic factorial is already included in the Feynman rule. The integrated decay width instead includes the identical final-state symmetry factor $1/2!$.

Put $Q=M-3m$. For $Q<0$ the decay width is zero by conservation of energy. Just above the three-particle decay threshold, all three daughters are nonrelativistic. The decay amplitude of iv remains finite: the internal scalar propagator denominator approaches $(2m{)}^2-m^2=3m^2$, and the Dirac scalar bilinear does not vanish at rest. Thus, for nonzero fixed $g$ and $\lambda$, the leading suppression comes entirely from three-body threshold phase space.

Use the spatial Dirac delta function to eliminate one daughter momentum. The remaining integral has six independent momentum coordinates and one conservation of energy constraint,Rescaling both momenta by $\sqrt{mQ}$ gives $Q^3$ from the measure and $Q^{-1}$ from the Dirac delta function. The smooth factors $1/(2E)$ do not change the power. Consequentlywith a finite positive leading coefficient when $g\lambda \ne 0$; $\Gamma =0$ for $m\ge M/3$. The $1/2!$ for the identical bosons changes only this coefficient. Equivalently, at fixed $M$ the opening is quadratic in $M/3-m$. If $g\lambda =0$, this leading tree-level Feynman diagram vanishes, so the stated leading coefficient is zero.

Use incoming fermion $p_1$ and antiparticle $p_2$, outgoing fermion $p_3$ and antiparticle $p_4$. The two tree-level Feynman diagrams are the $t$ channel, with $q_t=p_1-p_3=p_4-p_2$, and the $s$ channel, with $q_s=p_1+p_2=p_3+p_4$. They occupy the lower row of the figure. In the $t$ channel the fermion flow follows $1\to 3$ and $4\to 2$; in the $s$ channel it follows $1\to 2$ at the incoming interaction vertex and $4\to 3$ at the outgoing interaction vertex.

The two contractions again have a relative fermionic sign. With the same harmless common-phase convention as i, their sum isThe auxiliary-field propagator is the same constant $-i$ in both channels. There is no $u$ channel preserving the external fermion number assignments.

Label the incoming fermion momenta $p_1,p_2$ and outgoing fermion momenta $p_3,p_4$, with $p_1+p_2=p_3+p_4$. There are two tree-level Feynman diagrams: the $t$ channel connects $1$ to $3$ and $2$ to $4$, with exchanged momentum $q_t=p_1-p_3=p_4-p_2$; the $u$ channel connects $1$ to $4$ and $2$ to $3$, with $q_u=p_1-p_4=p_3-p_2$. Both internal lines are auxiliary-field propagators. The upper row in the figure gives these labelled connections; the outgoing labels are exchanged in its second panel to avoid a visually unnecessary crossing.

Each tree-level Feynman diagram contains two factors $ig$ and one factor $-i$, giving $i{g}^2$. The exchange of the two identical outgoing fermions gives the relative fermionic sign. Up to a common external-state phase, the corresponding scattering amplitude isThere is no annihilation channel: both incoming particles carry the same fermion number, whereas a $\sigma \bar{\psi }\psi$ interaction vertex couples one fermion to one antiparticle when both are incoming.

Set $B(x)=\bar{\psi }(x)\psi (x)$, a Grassmann-even Dirac scalar bilinear. Completing the square givesIntroduce external Grassmann sources $\eta ,\bar{\eta }$ for the Dirac field, but no source for $\sigma$. The generating functional isAt a finite regulator the change of variable ${\sigma }^{\prime }=\sigma -gB$ has unit Jacobian: it translates a bosonic integration variable by a Grassmann-even expression independent of that variable. Its oscillatory Gaussian integral can be defined with a convergence prescription and then continued. It is a constant $\mathcal{N}$ independent of $\psi ,\eta ,\bar{\eta }$. ThusThe factor $\mathcal{N}$ cancels in $Z[\eta ,\bar{\eta }]/Z[0,0]$. Every normalized Green function with only external fermion lines is a Grassmann source derivative of this same normalized generating functional, soThis is an exact Hubbard–Stratonovich transformation in reverse, rather than just an approximation based on tree-level Feynman diagrams. At the diagrammatic level the contact auxiliary-field propagator $-i{\delta }^{(4)}(x-y)$ collapses each exchanged line into the same local four-fermion interaction. The equality uses a common regulator and does not assert perturbative renormalizability of that four-dimensional four-fermion interaction.

Take $\varphi (p_1)+\bar{\varphi }(p_2)\to \varphi (p_3)+\bar{\varphi }(p_4)$ with the Mandelstam variables of i. The allowed tree-level Feynman diagram is the crossed $u$-channel $\Phi$ exchange in panel (c) of b's figure. At one end the incoming particle and outgoing antiparticle transfer charge $+2$; the other end absorbs that charge. The zero-charge $s$- and $t$-channel assignments are forbidden by the interaction vertices. ThusIn the center-of-mass frame with $|\mathbf{p}|\ll m$, $s=4m^2+O(|\mathbf{p}{|}^2)$ and $u=O(|\mathbf{p}{|}^2)$. For $M\gg m$, both scattering amplitudes approach $g^2/M^2$. The elastic two-body Lorentz-invariant phase space and incident flux agree for the two experiments. Usingfor distinguishable final particles, and an extra factor $1/2!$ when integrating the identical-particle final state over the whole sphere, givesThere is no additional division for identical incoming particles in the definition of a beam scattering cross-section. Before taking $M/m\to \infty$, the zero-momentum ratio is $\frac{1}{2}[M^2/(M^2-4m^2){]}^2$, provided the threshold is away from the heavy pole.

Near $s=M^2$, the bare $s$-channel scalar propagator makes fixed-order perturbation theory fail: repeated self-energy insertions contain powers of ${\Sigma }_{\Phi }(s)/(s-M^2)$. The $\Phi \to \varphi \varphi$ channel is open for $M>2m$, giving the leading decay widthResumming the self-energy gives a finite Breit-Wigner resonance cross section, with denominator $s-M_{\mathrm{pole}}^2+i{M}_{\mathrm{pole}}{\Gamma }_{\Phi }$ near the pole, rather than an infinite scattering cross-section. Also, $s\simeq M^2$ with $M\gg m$ lies outside the assumed nonrelativistic incoming kinematics; the ratio $1/2$ is not a resonance prediction.

Label the incoming four-momenta $p_1,p_2$ and the outgoing ones $p_3,p_4$. Define the Mandelstam variables $s=(p_1+p_2{)}^2$, $t=(p_1-p_3{)}^2$, $u=(p_1-p_4{)}^2$. Two incoming $\varphi$ particles have conserved charge $+2$, so the only tree-level Feynman diagram is an $s$-channel $\Phi$ exchange, shown in panel (b) of b's figure. Each end converts two charge-one lines into one charge-two line. There are no $t$- or $u$-channel exchanges: an incoming and an outgoing particle at such an end would transfer zero charge, for which this theory has no cubic interaction vertex.The single Feynman diagram is already symmetric under exchanging the two outgoing particles. This does not add another copy of the scattering amplitude; the identical final-state symmetry factor enters the integrated scattering cross-section instead.

Interpret the repeated species label in the Yukawa interaction as summed over $i=1,2$: each Feynman vertex is diagonal in species and equals $-ig$. Define the invariant scattering amplitude by $S-1=i(2\pi {)}^4{\delta }^{(4)}(P_f-P_i)\mathcal{M}$. There is one tree-level Feynman diagram, an $s$-channel scalar exchange. In the left panel, $j=2$ for this part; the right panel only occurs in d. Arrows show fermion-number flow, so an antifermion arrow opposes its physical momentum.

Writing $k=p_1+p_2=p_3+p_4$ and $s=k^2$, the momentum-space Feynman rules giveThe scalar carries no spin label. Average over the $2\times 2$ incoming spin states and sum the four outgoing combinations. For the real scalar Yukawa interaction, the conjugate of ${\bar{v}}_2{u}_1$ is ${\bar{u}}_1{v}_2$. The fermion spin sums then yieldThe bar denotes the initial spin average, not a Dirac adjoint. The gamma matrix trace identities giveand the second trace has the same value because $p_3+p_4=p_1+p_2$ and all external masses are $m$. ThereforeThis is the fixed-order expression away from $s=M^2$. If that pole is accessible, a physical scattering resonance requires a scalar self-energy and decay width; the tree formula alone is not finite there.

With the conventional particle labels, the allowed cubic decay channels areEach has one tree-level Feynman diagram, consisting of one cubic Feynman vertex. The final identical-leg contractions supply $2!$ and cancel the $1/2!$ in the Lagrangian density.

Taken literally with these labels, the printed $\Phi \to \varphi \varphi$ channel is forbidden: its initial Noether charge is $-2$ and its final Noether charge is $+2$. Its decay amplitude is zero, at every order preserving the global phase symmetry. If the source uses “$\Phi$” to name the heavy antiparticle instead, its intended answer is the nonzero decay amplitude $-h$ for $\bar{\Phi }\to \varphi \varphi$. The following decay width calculation concerns either allowed channel, whose widths are equal by charge conjugation.

Keep the particle labels specified under 2. There are two connected tree-level Feynman diagrams: the quartic contact interaction vertex and an $s$-channel light antiparticle exchange. The initial Noether charge is $1-2=-1$, exactly the charge of $\bar{\varphi }$. At the first cubic Feynman vertex, the incoming $\varphi$ and $\Phi$ leave the second $\varphi$ leg internal; the conjugate Feynman vertex supplies both outgoing particles.

ThusHere $s=(p_1+p_2{)}^2$ for incoming $\varphi (p_1),\Phi (p_2)$. Both terms have the same sign in the chosen S-matrix convention.

The distinction from antiparticles matters. If the paper's heavy “$\Phi$ particle” is instead intended to mean the charge-$+2$ state $\bar{\Phi }$, so that its printed decay into $\varphi \varphi$ is allowed, the exchange diagram is crossed. For that alternative physical process,This is a consistent alternative naming convention only if it is also applied to the decay in iii; one cannot use the same heavy particle label for both charge assignments.

Let $p_1,p_2$ be the incoming and $p_3,p_4$ the outgoing four-momenta, and set $s=(p_1+p_2{)}^2$, $t=(p_1-p_3{)}^2$, $u=(p_1-p_4{)}^2$, the Mandelstam variables. The only connected tree-level Feynman diagram is the $s$-channel exchange of $\bar{\Phi }$, with Noether charge $+2$. The two incoming light particles attach to ${\varphi }^2\Phi$, and the two outgoing light particles to its conjugate. There is no four-light contact Feynman vertex, nor a charge-allowed $t$ or $u$ exchange with these external labels.

Define the invariant scattering amplitude by $⟨f|S-1|i⟩=i(2\pi {)}^4{\delta }^{(4)}(P_f-P_i)\mathcal{M}$. ThenThe Feynman vertices already contain the identical-leg combinatorics. No extra factor of two multiplies this invariant scattering amplitude; the distinct identical final-state symmetry factor enters a cross section integral.

For $g=0$, the invariant scattering amplitude from the tree-level Feynman diagram in i has a pole at $s=M^2$. The intermediate heavy antiparticle can go on shell and live for a time of order its mean lifetime. Near this scattering resonance, inserting a single self-energy is no longer a uniformly small correction. Dyson resummation gives, to leading accuracy in a narrow decay width,Here $M$ is understood as the renormalized resonance mass; the real self-energy shifts the original pole position. The imaginary term removes the unphysical divergence of the fixed-order invariant scattering amplitude.

For equal incoming and outgoing light masses, the two-body momentum ratio cancels in the cross section. The angle-independent invariant scattering amplitude, integrated over the full solid angle with the identical final-state symmetry factor $1/2!$, gives $\sigma =|\mathcal{M}{|}^2/(32\pi s)$. Thus the Breit-Wigner resonance cross section isAt $s\gg M^2$ it reproduces $h^4/(32\pi s^3)$. At the resonance peak,This is also $8\pi /k_{\ast }^2$ for $k_{\ast }=M\beta /2$, the elastic identical-boson S wave partial-wave unitarity limit when this is the only open decay channel. The approximation assumes ${\Gamma }_{\mathrm{phys}}\ll M$ and a resonance sufficiently far from threshold for the width to vary slowly across it.

If one mechanically uses the unsymmetrized ${\Gamma }_{\mathrm{printed}}=2{\Gamma }_{\mathrm{phys}}$ from d, the same displayed Breit-Wigner resonance cross section would have that doubled width and predict a peak $8\pi /(M^2{\beta }^2)$. This is the literal-width substitution, rather than the consistent physical result: the given high-energy cross section already includes the identical-final-particle factor that the printed decay width integral omits.

All daughter particles are massless, so the parent mass $M$ supplies the only kinematic scale at tree level. Each vertex has two derivatives, making its scattering amplitude scale as ${\lambda }_i{M}^2$. For the four-$b$ vertex, symmetrizing over the derivative-bearing legs yields a sum of pairwise four-momentum products, which is proportional to $M^2$ by four-momentum conservation.

For $n$ massless daughters, the Lorentz invariance Lorentz-invariant phase-space measure scales as $M^{2n-4}$: each $d^{3}q/(2E)$ contributes two powers and the four-dimensional Dirac delta removes four powers. Including the parent factor $1/(2M)$ gives $\Gamma \sim |{\lambda }_i{|}^2{M}^{2n-1}$. ThusThe unspecified factors are dimensionless angular integrals and identical-particle factors. Each result has mass dimension one, as a decay width must in natural units.

Every interaction contains an even number of $b$ fields, so the exact discrete symmetry $b\mapsto -b$ forbids $\Phi \to bbb$ to every perturbative order. There is no such prohibition for $\Phi \to bb$.

The latter decay channel is absent at tree level in the displayed interactions, but contracting two $b$ legs at a $\Phi b^2(\partial b{)}^2$ vertex produces a tadpole diagram with two external $b$ legs. For example, contracting the undifferentiated pair leaves the local operator $\Phi (\partial b{)}^2$ multiplied by an integral proportional to $\int d^4\ell /({\ell }^2+i0)$. Contracting the differentiated pair instead gives a local $\Phi b^2$ contribution. Both are allowed effective field theory operators. With a ultraviolet cutoff, these local terms generically carry quadratic or quartic cutoff dependence and require matching coefficients or counterterms.

There is a relevant qualification to a claim that the one-loop decay width must be nonzero. Since $b$ is exactly massless here, the one-vertex tadpole integrals have no external scale and are scaleless integrals. They vanish in dimensional regularization. Thus the correct conclusion is allowed, and generically generated by quantum corrections or matching, not a regulator-independent nonzero one-loop prediction from ${\lambda }_b$ alone. No exact symmetry protects the two-$b$ channel; its numerical rate requires the renormalized coefficients of the allowed operators.

For ${\mathcal{L}}_{\mathrm{int}}=g^2(\bar{\psi }\psi {)}^2/2$ with one species of Dirac field, the tree-level Feynman diagrams for $f(p_1)\bar{f}(p_2)\to f(p_3)\bar{f}(p_4)$ giveThe fermionic sign is the antisymmetry of the two external contractions. The initial spin average is one quarter of the spin sum. The two diagonal contributions are $g^4(t-4m^2{)}^2$ and $g^4(s-4m^2{)}^2$; the interference Dirac trace is $R=(s-4m^2{)}^2+(t-4m^2{)}^2-(u-4m^2{)}^2$, giving the displayed result after subtracting $g^{4}R/2$. Here the Mandelstam variables satisfy $s+t+u=4m^2$. At threshold this gives $16g^4{m}^4$; for zero mass it gives $g^4(s^2+t^2+u^2)/2$. The formula describes the leading scattering amplitude of an effective interaction and is not a statement of perturbative renormalizability.

The nonlinear gauge fixing $\partial \cdot A+\lambda A^2$ produces cubic and quartic photon Feynman vertices. For physical photon polarization vectors, the two-external-leg cubic vertex reduces to a longitudinal internal momentum times the scalar product of the external polarizations. Its exchanged photon propagator cancels that momentum squared. Each of the three exchange graphs contributes $+4i{\lambda }^2({\epsilon }_i\cdot {\epsilon }_j)({\epsilon }_k\cdot {\epsilon }_l)$, cancelling the corresponding term of the quartic vertex. This vanishing physical tree-level Feynman diagram sum does not assert that the off shell four-point function vanishes.

The pseudoscalar Yukawa interaction with coefficient $-g$ has the displayed Feynman vertex. Its two-fermion tree-level Feynman diagrams are the $t$ and $u$ exchanges; antisymmetry of the final identical fermions requires their relative minus. Fermion-antifermion scattering has $t$ exchange and $s$ annihilation, again with a relative minus. With initial and final states ordered as $b_1^{†}{d}_2^{†}|0⟩$ and $b_3^{†}{d}_4^{†}|0⟩$, the normal-ordered current contributes $-d^{†}d\bar{v}{\gamma }^{5}v$ in the exchange channel but $db$ and $b^{†}{d}^{†}$ in the annihilation channel. These fermionic signs fix the overall sign as well as their interference; changing an external-state phase changes both terms together.