Scalar Propagator and Equation of motion

Question

I would like to know if the physical reason why the scalar propagator satisfies the Klein-Gordon equation is due to the fact that "pictorially" its action is to represent a "virtual" particle with spin 0 (i.e., a scalar field) that propagates from point A to point B?

I assume this is the case because, being a particle with spin zero, even if virtual, it should still obey the same dynamics as on-shell particles with spin zero. Is this assumption correct?

And does it also apply to other spin values as well as to non-free theories (in the presence of a potential and thus a modification of the equations of motion)?

Answer 1

1. The scalar propagator $$\Delta(x)=i \langle 0 |T \,\phi(x) \phi(0)|0\rangle \tag{1}$$ does not satisfy the Klein-Gordon equation $(\square +m^2) \Delta(x)=0$, but $\Delta(x)$ is a Green function of the Klein-Gordon equation, satisfying $$ (\square +m^2)\Delta(x)=\delta^{(d)}(x) \tag{2}$$ together with Feynman boundary conditions. Its Fourier representation is thus given by $$\Delta(x)=\int \!\frac{d^dp}{(2 \pi)^d} \; \frac{e^{-ipx}}{m^2-p^2-i \epsilon}. \tag{3} \label{3} $$

2. In contrast to an on-shell particle obeying the energy-momentum relation $p^2=m^2$ (corresponding to a particle in an initial or final asymptotic state), a virtual particle is off shell, i.e. $p^2 \ne 0$, in general. Thus, a virtual particle does not obey the same dynamics as a real one.

3. Spin does not affect the position of the pole of the momentum-space propagator.