The propagators in quantum field theory are directly related to transition amplitudes of the form $\langle{t',\vec{x}}|{t,\vec{x}}\rangle $ where $|t,\vec{x}\rangle \equiv \phi(t,\vec{x}) |0\rangle$. In these notes and here it is argued that the propagators, e.g. the Feynman propagator $$\Delta_F (x-y) = \int \frac{d^4k}{(2\pi)^4} \frac{e^{-ik(x-y)}}{k^2 - m^2 + i\epsilon}$$ that we usually consider in quantum field theory is not "the propagator for real particles".

So what's the propagator for real particles? I know the difference between real and virtual particles. My problem is that the sources linked above emphasize that the Feynman propagator is only valid for virtual particles and I want to understand why.

  • 1
    $\begingroup$ They probably just meant that this is only the free propagator. Particles in real life are interacting, so the propagator is not the same. $\endgroup$ Commented Oct 30, 2019 at 0:24
  • $\begingroup$ @doublefelix I don't think the Feynman propagator is a "free propagat". See my answer below. $\endgroup$
    – jak
    Commented Oct 30, 2019 at 8:57
  • $\begingroup$ The propagator you wrote above is only for free particles. It comes from plugging in the solution to the EOM that you get from the free lagrangian (which is no longer the solution with interaction terms) $\endgroup$ Commented Oct 30, 2019 at 9:39
  • $\begingroup$ @doublefelix I'm not sure I understand you correctly. You say the propagator is for free particles because we use the EOM and thus it describes on-shell particles? I don't think this is necessarily the case. Moreover, how then do you explain that it's not a solution of the EOM? $\endgroup$
    – jak
    Commented Oct 30, 2019 at 9:49
  • $\begingroup$ Not because we use the Equation of Motion at all, but rather because the EOM that we use comes for the Lagrangian for free particles, $\partial_\mu \phi \partial ^\mu \phi - m^2 \phi ^2$. Therefore the results we get only apply to the case that we have a free particle. And if we add an interaction term like $\phi^4$ to the lagrangian, the old solution for $\phi$ does not solve the EOM that we get from the interacting lagrangian $\endgroup$ Commented Oct 30, 2019 at 9:54

1 Answer 1


The Feynman propagator is a Green's function. This means we find $\delta(x_\mu-y_\mu)$ if we plug it into the Klein-Gordon equation. This implies that the Feynman propagator describes (a sequence of) field configuration that only exists in the presence of interactions and thus describes "virtual particles".

In contrast, the "fundamental" propagator (which also known as the Wightman function) \begin{align}D(t',\vec{x}', t,\vec{x}) \equiv \langle{{t',\vec{x}'}|{t,\vec{x}}}\rangle &= \int \frac{ \mathrm{d }k^3 }{(2\pi)^3 2\omega_{k} } {\mathrm{e }}^{i\Big( \omega_k \cdot ( t ' - t) -\vec k \cdot (\vec x ' -\vec x) \Big)} \notag \\[-1ex] &= \int \frac{ \mathrm{d }k^3 }{(2\pi)^3 2\omega_{k} } {\mathrm{e }}^{i k^\mu (x_\mu' -x_\mu) } \end{align} is indeed a solution of the free Klein-Gordon equation (i.e. the kernel). Hence, it's a (sequence of) field configuration that is realizable for free fields and thus describes how "real particles" propagate.

  • $\begingroup$ Virtual is usually a term used to describe propagators that are not subject to the on-shell condition, these appear in intermediate states in Feynman diagrams but that's the only difference. One normally calls particles that are on shell, I.e. they follow $E^2-m^2=p^2$, real. So the simple answer is, it is the Feynman propagator plus the on-shell condition, that means integrating four momenta subject to $k_0 = \sqrt{p^2+m^2}$. $\endgroup$
    – ohneVal
    Commented Oct 30, 2019 at 9:21
  • $\begingroup$ @ohneVal Thanks. If post this comment as an answer I can accept it. $\endgroup$
    – jak
    Commented Oct 30, 2019 at 9:44

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