Klein-Gordon Feynman Green's Function

Fourier transforming converts the wave operator $\Box$ into $k^2$ . This is motivated by the action of $\Box$ on the plane wave giving the momentum dependence in the Fourier transform:

$\Box e^{ik\cdot x} = \partial^2 e^{ik \cdot x} \sim k^2 e^{ik\cdot x},$

which implies the conversion $\Box \sim k^2$ .

In the convention given, $k^2 = -m^2$ for a particle that is on-shell. Particles in QFT correspond to the poles of the Green's function, so the denominator should take the form $k^2+m^2$ .

Getting the signs right in Fourier transforming the derivative in QFT can be tricky, especially because the metric convention depends on the author. This problem uses a somewhat atypical convention for the metric than is usually taught in introductory QFT, in which $k^2 = -m^2$ instead of $k^2 = m^2$ . Note also that in field theory textbooks there is also often a factor of $i$ in the definition of the Green's function which has been omitted from this article for consistency.

In any case, the conversion $\Box \sim k^2$ , even up to sign and numerical factors, implies the inversion of the operator:

$(\Box + m^2)^{-1} \to \frac{1}{k^2+m^2}$

which is the desired Green's function. Whether it is the Feynman or retarded propagator is not particularly important for this question, since this only specifies a pole convention for integration to position space.