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?

  • $\begingroup$ 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. $\endgroup$ Aug 21, 2015 at 12:09

1 Answer 1


Virtual particles are not real. A virtual particle is essentially defined by being associated to a propagator. It is, formally, nothing more than such a propagator. The idea of "virtual particle" doesn't even exist before you notice that you can draw pretty Feynman diagrams as a succinct representation of the way QFT amplitudes are calculated.

This is because you first derive the Feynman diagram perturbation expansion (e.g. by LSZ formalism or in the path integral formalism), then notice this expansion can formally be represented by diagrams where the external legs correspond to actual particle states and then call the internal lines "virtual particles" (because the external lines so nicely correspond to actual particles), but the formalism only says "draw an internal line of this particle type for every propagator of that particle type that appears in the integral you want to calculate".

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.

The heuristic energy-time uncertainty relation you wrote down for its "lifetime" is different - this is actually valid when you have so-called resonances, where actual intermediate bound states of the theory are formed during an interaction. Resonances are different from virtual particles in that they actually correspond to physical states of the theory, and they show up in scattering amplitudes in perfect analogy to the contribution of metastable states to non-relativistic partial wave scattering of quantum mechanics. The width of the resonance in the energy spectrum is related to its life-time by a Fourier relation, but, stricty speaking, this is not a quantum mechanical uncertainty relation like the one between position and momentum because time is not a quantum mechanical operator (see this question for some interpretations of a "time-energy uncertainty").

  • $\begingroup$ Thank you! I think I get it now. However I want to know your "opinion" about my last question: what does it mean physically when a propagator diverge? $\endgroup$
    – Crazydemon
    Aug 22, 2015 at 7:28
  • 2
    $\begingroup$ @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. $\endgroup$
    – ACuriousMind
    Aug 22, 2015 at 13:35
  • 1
    $\begingroup$ Not well known for me, I'm currently doing QED as a student. Thank you very much! $\endgroup$
    – Crazydemon
    Aug 22, 2015 at 17:33
  • $\begingroup$ @ACuriousMind is there any intuition behind why the poles of the propagator correspond to the on-mass-shell particle states of the theory? $\endgroup$
    – Alex Gower
    May 11, 2021 at 14:14

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