Taking for example the meson propagator: $$\Delta_F (x-y) = \int \frac{d^4k}{(2\pi)^4} \frac{e^{-ik(x-y)}}{k^2 - m^2 + i\epsilon}.$$ It describes a meson that propagate from a point of Minkowski space, $x$, to another one, $y$. Now, I know that an intermediate particle (like this mentioned meson) is a virtual one, so, from a physical point of view, it lives for times that satisfy: $$\Delta E \cdot \Delta t \le \hslash / 2\pi.$$ What I'm asking is:

1. where it comes out that a propagator describes virtual particles? I understand that it doesn't describe real ones because it is a solution of: $$(\square _x + m) \Delta_F (x-y) = \delta^4 (x-y).$$ Which is not the Klein-Gordon, due to the nonzero second member, but why virtual?

2. Why particles that are described by propagators are said to be "off mass-shell"? If they were on-shell, the propagator would diverge but what does this mean physically?

• Maybe Matt Strassler's article will help. A virtual particle isn't a short-lived real particle. The propagator describes field interactions. A proton and an electron "exchange field" such that hydrogen has little field left, but they don't actually throw photons back and forth. – John Duffield Aug 21 '15 at 12:09

There are no particles that are "described by propagators", or rather all particles are - it gives the probability to detect something starting at $x$ to be detected at $y$, as you said. One says that virtual particles are "off-shell" because if you want to really make the idea that those internal lines are "particles" work, they you have to observe that the momentum associated to the line is just integrated over all of momentum space, while the mass shell would restrict it to the hypersurface traced out by $k^2=m^2$. But at the end of the day, the formalism just gives you a line in a diagram that is not associated to any of the incoming or outgoing particle states. There's no basis for claiming that this line might be somehow associated to a particle state - it isn't.
• @Crazydemon: No opinion to give, it's a well-known fact (called the Källén-Lehmann spectral representation) that the poles of the propagator correspond to the $n$-particle states of the theory. In other words - the value of $p^2$ where the free propagator diverges is the mass. – ACuriousMind Aug 22 '15 at 13:35