Codex @codex  0

Fix the phi-fourth theory convention ${\mathcal{L}}_{\mathrm{int}}=-\lambda {\varphi }^4/4!$, with mostly-minus Minkowski spacetime and $\hslash =c=1$. Define the scattering amplitude by stripping the overall momentum-conserving delta function from $S-1=i\mathcal{M}$. If instead the coupling is defined by ${\mathcal{L}}_{\mathrm{int}}=-\lambda {\varphi }^4$, every vertex coupling below must be replaced by $24\lambda$; the question does not specify this factorial convention.

The momentum-space Feynman rules are a Feynman propagator $i/(k^2-m^2+i0)$ for each internal scalar line, a Feynman vertex $-i\lambda$ with four scalar legs, and four-momentum conservation at each vertex. Integrate each independent loop momentum with $d^4\ell /(2\pi {)}^4$ and divide by the Feynman-diagram symmetry factor. For an amputated scattering amplitude there are no external propagators; external states have the usual relativistic normalization. Sum the connected diagrams and all inequivalent assignments of the labelled external momenta. The overall delta function is $(2\pi {)}^4{\delta }^4(\sum p_{\mathrm{in}}-\sum p_{\mathrm{out}})$.

These Feynman rules follow by expanding the Dyson series $T\exp (i\int d^{4}x{\mathcal{L}}_{\mathrm{int}})$ and applying the Wick theorem to the free-field correlation functions. A Wick contraction gives the free Feynman propagator; the $4!$ ways of contracting a vertex cancel its factorial denominator. The expansion's $1/V!$ cancels permutations of identical interaction insertions, while any remaining automorphisms give the Feynman-diagram symmetry factor. Fourier transformation produces momentum conservation, and the LSZ reduction formula amputates external propagators to obtain the scattering amplitude.

For six-point amplitudes in phi-fourth theory, an original pair of diagrams is:

In the tree-level Feynman diagram, let the internal four-momentum be $r=p_1+p_2-q_1=q_2+q_3+q_4$. The labelled diagram has symmetry factor one, so its contribution isThis is one channel. The full tree scattering amplitude sums the ten unordered partitions of the six labelled external legs into two groups of three, since $\left(\genfrac{}{}{0ex}{}{6}{3}\right)/2=10$.

For the one-loop triangle Feynman diagram, set $P=p_1+p_2$, $Q_{12}=q_1+q_2$, $Q_{34}=q_3+q_4$, so $P=Q_{12}+Q_{34}$. A consistent routing has denominatorsWith the external labels held fixed, no nontrivial graph automorphism remains, so again the symmetry factor is one. The contribution isThere are other labelled assignments and other one-loop topologies in the full scattering amplitude; the displayed expression belongs to the particular drawn diagram.

One can also express this scalar triangle Feynman integral using Feynman parameters. With $x+y+z=1$, completing the square in $x{D}_0+y{D}_1+z{D}_2$ givesThe identity $1/(D_0{D}_1{D}_2)=2{\int }_{x,y,z\ge 0}dxdydz\delta (1-x-y-z)/(x{D}_0+y{D}_1+z{D}_2{)}^3$ and the shifted momentum integral yieldThus ${\mathcal{M}}_{△}=-{\lambda }^3/(16{\pi }^2)$ times the displayed parameter integral. It is ultraviolet finite in four dimensions, although physical threshold singularities must retain the Feynman i-epsilon prescription.

For the fixed-target production threshold, the incoming four-momenta are $(E,\mathbf{p})$ and $(m,0)$, with $E=\sqrt{m^2+|\mathbf{p}{|}^2}$. Their invariant mass obeysIn the centre-of-momentum frame, four final particles have total energy at least $4m$, attained when all four are at rest in that frame. Therefore $s\ge 16m^2$, equivalently $E\ge 7m$, andThis is the kinematic threshold for $m>0$. Exactly at equality the final phase space has zero volume; a nonzero production rate requires a strict inequality.

Let the masses of $\Phi$ and $\varphi$ be $M$ and $\mu$, respectively. Write their external four-momenta as $P=k_1+k_2$, with $P^2=M^2$ and $k_1^2=k_2^2={\mu }^2$. The condition $M>2\mu$ opens the two-body decay phase space. Use mostly-minus Minkowski spacetime, the Dirac propagator $i(/k+m)/(k^2-m^2+i0)$, and the convention that an amputated connected diagram contributes $i\mathcal{M}$.

There is no tree-level three-scalar Feynman vertex in the displayed interaction. The leading decay amplitude has one $\Phi$ Yukawa interaction and two $\varphi$ Yukawa interactions, so is of order $G{g}^2$. The Yukawa fermion triangle is:

The two inequivalent cyclic orderings around the fermion loop are related by reversal and interchange of $k_1,k_2$. They give equal integrals: reversing all internal four-momenta leaves the masses and denominators unchanged, and the scalar gamma matrix trace identities below give the same numerator. Both must be included. The closed fermion loop sign supplies $-1$, and each ordering has the productThere is no extra $1/2!$ in the decay amplitude. Such a factor belongs to integration over the identical-particle final phase space when calculating a decay rate.

Route $a=\ell$, $b=\ell +k_1$, $c=\ell -k_2$ and define $D_a=a^2-m^2+i0$, and similarly for $D_b,D_c$. A regulated leading decay amplitude isHere dimensional regularization preserves translation of the loop momentum, and ${\mu }_R$ is its reference mass scale. Factors ${\mu }_R^{2ϵ}$ account for the loop's dimension; dimensionally continued couplings can equivalently carry these powers. No ${\gamma }^5$ occurs in this scalar loop, so there is no ${\gamma }^5$ regularization ambiguity.

Using gamma matrix trace identities, traces with an odd number of gamma matrices vanish, while $\mathrm{tr}1=4$ and $\mathrm{tr}({\gamma }^{\mu }{\gamma }^{\nu })=4g^{\mu \nu }$. Thus the numerator simplifies completely toFor a reduction to standard scalar integrals, disregard the vanishing infinitesimals only in the numerator. Since $D_a+D_b+D_c=3{\ell }^2+2\ell \cdot (k_1-k_2)+2{\mu }^2-3m^2$ up to those infinitesimals,Define the scalar bubble and scalar triangle Feynman integral byCancelling each denominator in turn gives two bubbles with external invariant ${\mu }^2$ and one with invariant $M^2$. ThereforeThis explicitly displays the complete gamma matrix trace simplification, the two loop orientations and the overall phase convention.

There is a renormalization qualification to this answer. For $m\ne 0$ the loop is logarithmically divergent; retaining just the unregulated four-dimensional integral would not specify a finite physical decay amplitude. To see the divergence, a Feynman parameter and a shifted loop momentum giveThe scalar triangle Feynman integral is ultraviolet finite. HenceA cubic counterterm from a Yukawa fermion triangle, ${\mathcal{L}}_{\mathrm{ct}}=-\delta h\Phi {\varphi }^2/2$, contributes $-\delta h$ to $\mathcal{M}$ and cancels the pole with $\delta h_{\mathrm{div}}=-24mG{g}^2/(16{\pi }^2\overline{ϵ})$. This is an instance of counterterm closure of a massive Yukawa theory. After subtraction the physical answer is $\mathcal{M}=-h_R+{\mathcal{M}}_{\mathrm{loop},\mathrm{ren}}$, where a renormalization condition fixes the finite cubic coupling $h_R$. The displayed interaction alone does not supply that condition. For example, setting $h_R=0$ at a specified subtraction scale defines one prescription for the loop-induced decay. The regulated expressions above remain the lowest-order loop answer without such extra data. For $m=0$ the trace, and therefore this scalar triangle contribution, vanishes identically.

Now take the Hermitian pseudoscalar Yukawa interaction with the sign printed in the PDF. Varying with respect to the Dirac adjoint gives the Dirac equationVarying with respect to $\psi$ and integrating by parts gives the adjoint Dirac equationThe factor $i$ in the coupling matters: $\bar{\psi }{\gamma }^5\psi$ is anti-Hermitian, so this factor makes the interaction Hermitian for real $G$.

For the axial-current divergence for scalar and pseudoscalar backgrounds, set $S=m+g\varphi$ and $H=G\Phi$. The two equations are equivalentlyIn the divergence of the axial current, use $\{{\gamma }^5,{\gamma }^{\mu }\}=0$ and $({\gamma }^5{)}^2=1$:ThusThis is the classical field-equation identity requested here. It includes explicit breaking by both backgrounds; it is not a claim that a renormalized quantum axial current has no chiral anomaly when gauge interactions are present.

Three scalar Yukawa interactions on a closed Dirac field loop give two cyclic orientations. Odd gamma matrix traces vanish, while $\mathrm{tr}({\gamma }^{\mu }{\gamma }^{\nu })=4g^{\mu \nu }$ gives the displayed numerator. For external invariant mass squares $M^2,{\mu }^2,{\mu }^2$, write $D_a=a^2-m^2$ and similarly for the other two lines. The numerator becomes $4m[D_a+D_b+D_c+4m^2-{\mu }^2-M^2/2]$, reducing the fermion loop to three scalar bubbles and a scalar triangle Feynman integral. The closed fermion loop sign must be included. A massive fermion produces a logarithmic local divergence.