Feynman propagator

The Feynman propagator is a vacuum time-ordered two-point function, with the conventional overall factors stated. For a free real scalar field, $⟨0|T\varphi (x)\varphi (y)|0⟩$ has momentum-space form $i/(p^2-m^2+i0)$ in the usual Minkowski convention.

Time ordering places the later field to the left, and the pole prescription selects the vacuum boundary condition. The propagator is the Green function used for internal lines in perturbative amplitudes; it is not a directly measured probability for a particle to follow a particular path. It can be nonzero at spacelike separation without enabling superluminal signalling. Causal response is instead organized by the retarded Green function and local commutators, so these different two-point functions must not be interchanged.

With $p\cdot x=Et-\mathbf{p}\cdot \mathbf{x}$, placing an annihilation operator on $e^{+ipx}$ rather than $e^{-ipx}$ makes its vacuum time-ordered contraction equal to $D_F^{\ast }$, with the opposite pole prescription. For positive time separation, the zero-spatial-momentum factor is $e^{+imt}/(2m)$ rather than $e^{-imt}/(2m)$. Replacing the integration variable $p$ by $-p$ does not change the sign of $i0$ and cannot fix the physical positive-energy boundary condition.

With ${\Delta }_F=⟨0|T\{\varphi (x)\varphi (y)\}|0⟩$, for a mode of positive energy $E>0$ the Feynman i-epsilon prescription puts poles $+E-i0$ and $-E+i0$ in opposite half-planes. The energy contour integral closes clockwise below for positive time separation and counterclockwise above for negative separation. The residue theorem gives $e^{-iE|t|}/(2E)$ in either case. Thus the four-dimensional Fourier integral has numerator $i$ and denominator $p^2-m^2+i0$. The integral is understood as a distribution limit, not an ordinary absolutely convergent integral; changing the definition to $i{\Delta }_F$ would change the displayed normalization.

At fixed spatial momentum with mode energy $E>0$, $g_E(t)=e^{-iE|t|}/(2E)$ is continuous, with derivatives $-i/2$ and $+i/2$ at zero from opposite sides. Hence its second distributional derivative has a $-i\delta (t)$ term, while away from zero $g_E^{\prime \prime }+E^2{g}_E=0$. Therefore $({\partial }_t^2+E^2)g_E=-i\delta (t)$ and, after the spatial Fourier transform, $(\square +m^2){\Delta }_F=-i{\delta }^4$. This checks the propagator numerator, contour signs and equal-time canonical commutation relations independently of residue calculation.

The Feynman $iϵ$ prescription displaces poles or tilts an integration contour by an infinitesimal imaginary part. It selects vacuum boundary conditions and damps oscillatory integrals at infinite time.