Return to Revisions

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 $$\Delta(x)=\int \!d^dp \; \frac{e^{-ipx}}{m^2-p^2-i \epsilon}. \tag{3} \label{3} $$

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.

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