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For a real auxiliary field with quadratic Lagrangian density $-a{\sigma }^2/2$, $a\ne 0$, the Feynman propagator is the inverse quadratic kernel multiplied by $i$: $D_{\sigma }(p)=-i/a$. It is independent of momentum; in position space it is the contact distribution $-i{\delta }^{(d)}(x-y)/a$. Two interaction vertices $ig$ joined by this auxiliary-field propagator give $i{g}^2/a$. Completing the square in $-a{\sigma }^2/2+g\sigma B$ produces $g^2{B}^2/(2a)$, proving the equivalence to a local interaction by a Gaussian integral.

The auxiliary field $\sigma$ has quadratic kernel $-1$ and no kinetic term. Inverting that kernel, rather than introducing an artificial scalar mass, gives the momentum space Feynman rulesThe auxiliary-field propagator is independent of momentum; in position space it is $-i{\delta }^{(4)}(x-y)$. At each interaction vertex attach $(2\pi {)}^4{\delta }^{(4)}(\sum p)$ for momentum conservation. Integrate independent loop momenta with $d^{4}p/(2\pi {)}^4$, respect fermion flow, include the closed fermion loop sign $-1$, and use the appropriate symmetry factor. Incoming and outgoing fermions carry $u_s(p)$ and ${\bar{u}}_s(p)$; incoming and outgoing antiparticles carry ${\bar{v}}_s(p)$ and $v_s(p)$. Exchanging external fermions introduces the fermionic sign. Here $/p={\gamma }^{\mu }{p}_{\mu }$ is Feynman slash notation.

For a dimensional check, $[\psi ]=3/2$, $[\sigma ]=2$, $[g]=-1$ in mass dimensions. Eliminating $\sigma$ consequently produces a four-fermion interaction of negative coupling mass dimension, not a propagating scalar degree of freedom.

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.

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.

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.