• I
kelly0303
Hello! I am a bit confused about the distinction between virtual and real particles. For example a Z boson, which has a very short lifetime, in all experiments will decay to some other stable particles (i.e. it is detected through its decay). This means that it will always appear as a propagator, and as far as I understand, the propagators are not, usually, on shell. The mass of the Z was obtained by looking for a resonance peak, and that is the quoted mass in PDG for example. But what does it mean for a Z particle to not be virtual i.e. do we even have a real Z? So my question is, especially for unstable particle, which always appear as propagators in some Feynman diagram, when is it real and when is it virtual? Thank you!

Gold Member
2022 Award

Michael Price
Hello! I am a bit confused about the distinction between virtual and real particles. For example a Z boson, which has a very short lifetime, in all experiments will decay to some other stable particles (i.e. it is detected through its decay). This means that it will always appear as a propagator, and as far as I understand, the propagators are not, usually, on shell. The mass of the Z was obtained by looking for a resonance peak, and that is the quoted mass in PDG for example. But what does it mean for a Z particle to not be virtual i.e. do we even have a real Z? So my question is, especially for unstable particle, which always appear as propagators in some Feynman diagram, when is it real and when is it virtual? Thank you!
Real particles appear in the 'in' and 'out' states, with only one vertex interacting with the rest of the diagram, and are on-shell. Virtual particles have both propagator vertices appearing inside the diagram and are off-shell.
If your Z boson survives long enough to leave the interaction volume then it would be deemed "real". But in actuality all particles are virtual, since interactions can occur anywhere.

Last edited:
kelly0303
Real particles appear in the 'in' and 'out' states, with only one vertex interacting with the rest of the diagram, and are on-shell. Virtual particles have both propagator vertices appearing inside the diagram and are off-shell.
Thank you for you reply! But this is what I am confused about. You can't have for a (say) Z boson a diagram where the Z boson comes 'in' or 'out' of the diagram. The basis for QFT (if I understood it well) is that at infinity the particles are free, then they interact, then they are free again at infinity. But a Z boson (or any unstable particle) can't come and go to infinity, as it decays very fast. So the Z boson can only appear as a propagator, otherwise it would mean that it traveled for a long time without decaying. So how does one define a real particle?

Michael Price
Thank you for you reply! But this is what I am confused about. You can't have for a (say) Z boson a diagram where the Z boson comes 'in' or 'out' of the diagram. The basis for QFT (if I understood it well) is that at infinity the particles are free, then they interact, then they are free again at infinity. But a Z boson (or any unstable particle) can't come and go to infinity, as it decays very fast. So the Z boson can only appear as a propagator, otherwise it would mean that it traveled for a long time without decaying. So how does one define a real particle?
I half answered this in an update, but let me complete it here. The interaction volume is arbitary, and the distinction between free and interacting a bit artifical. (There is no such thing as a free particle - they are just convenient fictions.). In reality all particles are virtual, and it is the so-called "real" particles that don't exist, ironically.