= Solution
Fix the <phi-fourth theory> convention $\mathcal L_{\mathrm{int}}=-\lambda\phi^4/4!$, with mostly-minus <Minkowski spacetime> and $\hbar=c=1$. 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 $\mathcal L_{\mathrm{int}}=-\lambda\phi^4$, 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> $T\exp(i\int d^4x\,\mathcal L_{\mathrm{int}})$ 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:
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/iii/paper-48-scattering-diagrams.png]
{title=Tree and one-loop triangle contributions to two-to-four scalar scattering}
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 is
$$
i\mathcal M_{\mathrm{tree}}=(-i\lambda)^2\frac{i}{r^2-m^2+i0},\qquad
\boxed{\mathcal M_{\mathrm{tree}}=-\frac{\lambda^2}{r^2-m^2+i0}.}
$$
This 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 $\binom63/2=10$.
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 denominators
$$
D_0=\ell^2-m^2+i0,\qquad D_1=(\ell-P)^2-m^2+i0,\qquad D_2=(\ell-Q_{34})^2-m^2+i0.
$$
With the external labels held fixed, no nontrivial graph automorphism remains, so again the <symmetry factor> is one. The contribution is
$$
\boxed{i\mathcal M_{\triangle}=(-i\lambda)^3\int\frac{d^4\ell}{(2\pi)^4}\frac{i^3}{D_0D_1D_2}
=\lambda^3\int\frac{d^4\ell}{(2\pi)^4}\frac1{D_0D_1D_2}.}
$$
There 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 $xD_0+yD_1+zD_2$ gives
$$
\Delta=m^2-xyP^2-xzQ_{34}^2-yzQ_{12}^2.
$$
The identity $1/(D_0D_1D_2)=2\int_{x,y,z\ge0}dx\,dy\,dz\,\delta(1-x-y-z)/(xD_0+yD_1+zD_2)^3$ and the shifted momentum integral yield
$$
\int\frac{d^4\ell}{(2\pi)^4}\frac1{D_0D_1D_2}
=-\frac{i}{16\pi^2}\int_{x,y,z\ge0}\frac{dx\,dy\,dz\,\delta(1-x-y-z)}{\Delta-i0}.
$$
Thus $\mathcal M_{\triangle}=-\lambda^3/(16\pi^2)$ 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 $(m,\mathbf0)$, with $E=\sqrt{m^2+|\mathbf p|^2}$. Their <invariant mass> obeys
$$
s=(p_1+p_2)^2=2m^2+2mE.
$$
In 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 $s\ge16m^2$, equivalently $E\ge7m$, and
$$
\boxed{|\mathbf p|\ge4\sqrt3\,m.}
$$
This 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.
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