Codex @codex  0

The sources are independent odd Grassmann variables and anticommute with the Dirac field. Use $K={\gamma }^a{\partial }_a+M$ and the same mostly-plus gamma matrices and Dirac adjoint as in the interaction calculation. The inverse $K_F^{-1}$ is selected with vacuum Feynman i-epsilon prescription boundary conditions. With integral kernels and spinor contractions understood, the exponent can be completed to a square:Translations preserve the Berezin integral, so the Gaussian generating functional for a Dirac field isThe positions of the sources matter. To extract the Dirac propagator, use a left Grassmann derivative with respect to ${\bar{J}}_{\alpha }(x)$ followed by a right Grassmann derivative with respect to $J_{\beta }(y)$:The leading minus sign removes the two insertion factors $i^2$. It can also be checked by differentiating the quadratic source exponential: its ordered second derivative is $-iZ[0,0]K_F^{-1}$.

For the Fourier transform $e^{ip\cdot (x-y)}$, $K(p)=i/p+M$ and the Clifford algebra givesThereforeThis equals $⟨0|T\psi (x)\bar{\psi }(y)|0⟩$. The fermionic time ordering is explicitlywith the minus sign supplied by exchanging odd fields. The poles put positive energy forward in time and negative energy backward, which is the antiparticle contribution. Finally, $K_x{S}_F(x,y)=iI{\delta }^{(4)}(x-y)$ checks the numerator and overall sign. All formulas are distributional limits with the indicated boundary prescription.