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For one Grassmann variable, define the Berezin integral by $\int d\theta 1=0$ and $\int d\theta \theta =1$, extending linearly. Since ${\theta }^2=0$, these rules determine it on every polynomial. With several Grassmann variables, integrate from the rightmost differential first. Equivalently, $\int d{\theta }_n\cdots d{\theta }_1$ extracts the coefficient of ${\theta }_1\cdots {\theta }_n$ in that order. Repeated indices give zero, while permuting distinct generators gives the fermionic sign of that permutation. Hencewhere $\epsilon$ is the Levi-Civita symbol.

For the doubled variables, fix the paired measure orientation byThe exponential in the Grassmann Gaussian integral has a finite Taylor expansion. Only its term of degree $2n$ survives integration. Each barred and unbarred index must occur exactly once; the $n!$ orders of the even bilinears cancel the Taylor factorial. Reordering the unbarred indices gives the determinant permutation sign. ThereforeThis identity holds even for a singular matrix. The normalized expectations below require $B$ invertible.

Put $C=B^{-1}$. Differentiating with respect to $B_{ji}$ inserts $-{\bar{\theta }}_j{\theta }_i={\theta }_i{\bar{\theta }}_j$. The derivative of a determinant then gives the ordered Grassmann Gaussian contractionsDifferentiate once more, using $\partial C_{ij}/\partial B_{lk}=-C_{il}{C}_{kj}$. The result isThe minus sign is the interchange sign in the fermionic Wick theorem. In particular $⟨\bar{\theta }M\theta ⟩=-\mathrm{tr}(MC)$. Reordering each barred-unbarred pair in the product and contracting the indices givesThe covariance of Grassmann bilinears is the connected second term. Its Feynman diagram is one oriented fermion loop through both insertions; the first term has two separate fermion loops, one at each insertion. Each closed fermion loop contributes a minus sign, so two loops give a plus and one gives a minus.

For the Dirac field, write $D=\gamma \cdot \partial +m$ and use signature $(-,+,\dots ,+)$, consistent with the denominator in the question. A finite regulator turns the free fermionic path integral into the same Gaussian problem with $B=iD$, because $e^{iS}=e^{-i\bar{\psi }D\psi }$. The time-ordered product covariance is consequently $(i{D}_F{)}^{-1}=-i{D}_F^{-1}$, where the subscript specifies the Feynman i-epsilon prescription.

With the Fourier transform convention $\psi (x)=\int d^{d}p(2\pi {)}^{-d}{e}^{ip\cdot x}\tilde{\psi }(p)$, the derivative kernel is $i\gamma \cdot p+m$. The gamma matrices obeysince their anticommutator makes $(\gamma \cdot p{)}^2=p^{2}I$. Thus the mostly-plus Dirac propagator with a real derivative kernel isFor $p^2=-p_0^2+{\mathbf{p}}^2$, the denominator is $p_0^2-{\mathbf{p}}^2-m^2+i0$. Its positive-energy pole is below and negative-energy pole above the real axis, giving time ordering. Here $S_F$ is defined without the leading factor $i$, as in the question; conventions that include that factor in the name Dirac propagator give the same covariance.