 Solution

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.