= Scalar four-fermion scattering
{title2=$\overline{|\mathcal M|^2}=\frac{g^4}{2}\sum_{x=s,t,u}(x-4m^2)^2$}
For $\mathcal L_{\rm int}=g^2(\bar\psi\psi)^2/2$ with one species of <Dirac field>, the <tree-level Feynman diagrams> for $f(p_1)\bar f(p_2)\to f(p_3)\bar f(p_4)$ give
$$
\mathcal M=g^2\big[(\bar u_3u_1)(\bar v_2v_4)-(\bar v_2u_1)(\bar u_3v_4)\big].
$$
The <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 $g^4R/2$. Here the <Mandelstam variables> satisfy $s+t+u=4m^2$. At threshold this gives $16g^4m^4$; 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>.
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