Are there virtual particles for every kind of particle there is?

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    $\begingroup$ Did you see this earlier answer - I think it goes a long way to answering your question. If it does not, please explain what is missing. $\endgroup$
    – Floris
    Commented Jul 30, 2015 at 17:21
  • $\begingroup$ Good. I will nominate this question for "closing as a duplicate". That means it will remain here and if people phrase a question like yours, they will be directed to the earlier question / answer. $\endgroup$
    – Floris
    Commented Jul 30, 2015 at 17:31
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    $\begingroup$ Like Bosoneando said below, don't take "virtual particles" too seriously. Because they're field quanta, not short-lived real particles that pop into existence like magic. Spntaneously, like worms from mud They're abstract things. It's like you divvy up an electromagnetic field into chunks and say each is a virtual particle. So are there virtual particles for every kind of particle there is? No, because there aren't any virtual particles really. Because they're virtual. Gluons in ordinary hadrons are virtual too. $\endgroup$ Commented Jul 30, 2015 at 21:44
  • $\begingroup$ That actually helped a lot, thanks for phrasing it that way. $\endgroup$
    – Alex
    Commented Jul 30, 2015 at 21:49

1 Answer 1


Don't take "virtual particles" too seriously.

Quantum Field Theory for a single (non-interacting) field is quite easy to solve. One of the things that can be calculated is the Feynman propagator $G_F(x,y)$, wich describes the probability amplitude for the propagation of a particle from spacetime point $x$ to spacetime point $y$. Obviouslly, the Feynman propagator will depend on the mass, four-momentum and quantum numbers of the particle.

But single, non-interacting fields are so boring. We'll be more interested in interacting fields. This is also the case that can be tested experimentally: measuring cross sections, decay widths and all that stuff. But we run in trouble: QFT for interacting fields is extremely hard!

So, we have to use a low-energy approximation: perturbation theory [Even if the energies in particle colliders are really high by ours standards, they are low enough to apply perturbation theory]. In the lagrangian you will find terms such as $$\mathcal{L}_i \sim g \psi^\alpha \phi^\beta$$ where $\psi$ and $\phi$ are two fields, $\alpha$ and $\beta$ integer exponents and $g$ a coupling constant, which quantifies the strength of the interaction. The trick to calculate any observable is to make an expansion of the time-evolution operator in powers of the coupling $g$. The math is pretty tedious, but in the end you find out that by a miraculous coincidence (in fact, Wick's theorem) all the observables depend on the Feynman propagators of the fields involved in the interaction lagrangian (with one caveat: in the propagator, the mass and four-momentum have a wrong relationship $p^\mu p_\mu \neq m^2$).

And here is where the genious of Feynman comes into play: each term (a horrendous integral) in the perturbative expansion can be represented as [a sum of] pretty diagrams. In a Feynman diagram, each external line is one of the incoming/outcoming particles, each internal line is one of the propagators due to Wick's theorem, and the vertices between lines are determined according to the interaction lagrangian (in the example above, in each vertex there are $\alpha$ lines of type $\psi$ and $\beta$ lines of type $\phi$). Feynman diagrams are a very useful tool to summarize and write the terms in the perturbative expansion. They are NOT a depiction of particles moving around, colliding, merging and branching. In this sense, the internal lines in these diagrams aren't meant to represent particles, even though they are usually called 'virtual particles'. Maybe 'internal propagator' would be less misleading.

So, tl;dr: "Virtual particles" are not a special type of particle, and there are Feynman diagrams with internal propagators corresponding to all the fields in the Lagrangian.


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