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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.

Write $u_i=u^{s_i}(p_i)$ and $v_i=v^{s_i}(p_i)$. With $S_{fi}={\delta }_{fi}+i(2\pi {)}^4{\delta }^{(4)}(p_1+p_2-p_3-p_4){\mathcal{M}}_{fi}$, chooseThe relative minus sign is fixed by the canonical anticommutation relations, not by the Mandelstam variables. It follows by crossing symmetry from the identical-fermion scattering amplitude in i, or directly by the two inequivalent external contractions of $g^2(\bar{\psi }\psi {)}^2/2$. For an explicit sign check, fix incoming state order $b_1^{†}{d}_2^{†}|0⟩$ and outgoing order $b_3^{†}{d}_4^{†}|0⟩$. The direct contraction contains $(b_3^{†}{b}_1)(d_2{d}_4^{†})$, whose coefficient in the ordered product $b_3^{†}{d}_4^{†}{d}_2{b}_1$ is $-1$; the annihilation contraction $(b_3^{†}{d}_4^{†})(d_2{b}_1)$ has coefficient $+1$. The common sign can change with the external-state convention, while their relative minus sign cannot.

Let $r=m^2$. The external particles are on shell, so $s+t+u=4r$ andUse the supplied spin sums and gamma matrix trace identities. The two diagonal terms in the initial spin average areFor the interference term, the spin sum is a single Dirac trace:Terms with an odd number of Dirac gamma matrices vanish. Expanding the remaining terms givesSubstituting the Mandelstam variables reduces this toIt is real. The initial spin average therefore yieldsThe displayed answer in the source is false. At threshold, $p_i=(m,0)$, hence $s=4m^2$ and $t=u=0$. Then $\bar{v}u=\bar{u}v=0$, while the spin sums give $\overline{|\mathcal{M}{|}^2}=16g^4{m}^4$. The printed expression instead gives $20g^4{m}^4$. This is also a counterexample in the continuous threshold limit from physical nonzero incoming momenta. At $m=0$ there is a further discrepancy: the correct result is $g^4(s^2+t^2+u^2)/2$, rather than the printed $g^4(3s^2+3t^2-u^2)/2$. Changing a common phase of the scattering amplitude cannot change either comparison.

The requested $s\leftrightarrow t$ symmetry nevertheless holds for the correct answer. Crossing symmetry exchanges the two ways of pairing the four external fermion lines; the momentum-independent auxiliary-field propagator supplies no channel-dependent denominator. The relative fermionic sign can reverse the crossed scattering amplitude, but disappears in its squared modulus. With $u=4m^2-s-t$, the boxed expression is manifestly symmetric under $s\leftrightarrow t$. This is an analytic crossing relation, not a statement that positive physical $s$ and negative physical $t$ have interchangeable scattering kinematics.