From all my searching it seems like the most common explanation of virtual particles is that they are merely internal lines in Feynman diagrams and therefore just a convenient pictoral way of organising a perturbation expansion, and therefore they need not be considered 'real' in any way.

My only remaining confusion then is why the mass of virtual particles seem to always correspond to masses of real particles, when I thought the mass of virtual particles was mostly just a measure of the strength of a corresponding Yukawa potential (i.e. $m \propto \mu$ where $V \sim e^{\mu r}$, so why do these constants $\mu$ seem to always coincide with masses of real particles?

For instance Yukawa predicted the existense of $\pi$ mesons by predicting the mass of a virtual exchange particle to be between 100 and 200 $MeV/c^2$, and then later pions were detected with roughly this mass.

My question is, why would predicting a virtual particle with this mass cause someone to expect real particles with this mass, since virtual particles are just tools in perturbative expansions?

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    $\begingroup$ What do you mean by the mass of a virtual particle? It is not true that $p^\mu p_\mu = m^2$ - virtual particles are off-shell. What is true is that the propagator has a pole at the mass $m$. $\endgroup$
    – jacob1729
    Commented Nov 10, 2020 at 18:59
  • $\begingroup$ I mean eventhough in the typical propagator the virtual particle is off mass shell (in terms of $p^2 \neq m^2$, when we use a propagator of the form $\frac{1}{p^2 - m^2}$ we use a value of m (related to the Yukawa $\mu$ I think) that always seems to coincidentally be the mass of an actual particle too. $\endgroup$
    – Alex Gower
    Commented Nov 10, 2020 at 19:01
  • $\begingroup$ Matt Strassler has an interesting take on this: virtual particles are real, but they aren't actually particles. ;) profmattstrassler.com/articles-and-posts/… $\endgroup$
    – PM 2Ring
    Commented Nov 11, 2020 at 12:08
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    $\begingroup$ @FlatterMann I like what ACM says here about Matt Strassler's take on virtual particles: physics.stackexchange.com/a/275099/123208 $\endgroup$
    – PM 2Ring
    Commented Sep 30, 2022 at 14:23
  • 1
    $\begingroup$ @PM2Ring Yep, I think ACM is a lot closer to the facts there. $\endgroup$ Commented Sep 30, 2022 at 19:34

4 Answers 4


If I understand you correctly, I think there are actually two points being asked about here, and so it will be beneficial to try and separate them.

First there is a question about the relation of the parameter $m$ which appears in the Lagrangian to the masses of physical particles. And there is also a question about how the identification of a "virtual"-like structure in a scattering process should be identified with a particle of specific mass.

Let me first tackle the second question. Forgetting everything about Feynman diagrams and perturbation theory, at the end of the day we measure scattering amplitudes which are merely approximated by a sum of Feynman diagrams. It's a theorem (proven in various QFT books. I think there may also be some more down to earth discussion in Griffiths particles book, and of course the most complete proof likely appears in Weinberg QFT volume 1) that resonances in the scattering momenta (which are poles in complex momenta) always correspond to particles in our Hilbert space. Furthermore, these particles need not correspond to the "fundamental" fields we write down in our Lagrangian. For example, the pion is a composite particle, but there still exists a pion state in the Hilbert space with definite 4-momenta, and hence certain scattering processes will have resonances at the pion mass despite there being no pion field in the standard model. This relation between poles and masses is also the reason why masses of particles are sometimes referred to as the "pole-mass" of the particle, just to distinguish from the mass parameters that may appear in the Lagrangian. (As a bonus, while the real part of the pole location is related to the mass, the complex part is related to the particle lifetime).

Now, if the Feynman diagrams are a good approximation to these scattering amplitudes, it may be the case that the tree-level diagram alone is a good approximation. If the tree-level diagram has a pole, then to within the precision of that approximation, so will the scattering process and hence there will exist a particle in the theory with approximately the pole mass, up to subtleties of renormalization.

Speaking of, I will point out that my statements about pole masses are completely non-perturbative, while may statements about tree-level approximations are perturbative. When we deal with diagramatic expansions, as soon as we want to go past tree-level and involve a look diagram we must renormalize all our couplings, including the mass parameters that appear in the Lagrangian. This will necessarily shift the locations of said poles.

A complete picture of how the non-perturbative picture, diagram expansions, and renormalization all tie together can be understood via the effective action. I think it would be too much to launch into a discussion of these things here, but I will note that there is a fairly nice description in the QFT book by V. Parameswaran Nair, though it may be some effort to parse. A much quicker route (but with correspondingly less detail) would be the QFT book by Thomas Banks.

  • $\begingroup$ Thank you. Firstly, is there a name for that proved theorem connecting pole masses to particle masses? $\endgroup$
    – Alex Gower
    Commented Nov 10, 2020 at 20:17
  • $\begingroup$ Secondly, are you saying that we can only consider the resonance to be a correspond to a single virtual particle if we only consider the tree level Feynman diagrams, and that, if we include higher orders of Feynman diagrams then this single resonance peak is more complex and cannot be considered to be just a single particle? $\endgroup$
    – Alex Gower
    Commented Nov 10, 2020 at 20:18
  • $\begingroup$ @AlexGower I don't think I've ever seen a name associated to this theorem. Weinberg refers to the study of these poles as "polology" but I've never seen anyone else call it this and if I remember correctly, googling this term returns nothing. The proof is in Weinberg (chapter 10), but may be difficult to parse. Griffiths' particle physics book should contain a discussion thereof. In the middle group between readable and completely correct would likely be Peskin/Schroeder (haven't looked, so I don't know wherein it would be). $\endgroup$ Commented Nov 10, 2020 at 20:42
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    $\begingroup$ @AlexGower In general a correlator may have many resonances, each corresponding to the mass of a particle in the spectrum. If we restrict to tree-level, these poles can only come from poles in the propagators. Adding higher loop corrections will cause these poles to move around, disappear, or create new poles either from the diagrams themselves or from changes due to renormalization. Hence the poles seen in tree-level diagrams (which have the most straight-forward interpretation as :virtual" particles) are only an approximation to the true content of the theory, which goes beyond perturbation. $\endgroup$ Commented Nov 10, 2020 at 20:45
  • $\begingroup$ @RichardMyers Schwartz also uses the term "polology" $\endgroup$ Commented Nov 11, 2020 at 6:23

My only remaining confusion then is why the mass of virtual particles seem to always correspond to masses of real particles

This is not correct.


Two electrons may scatter with any invariant mass of their summed four vectors, the four vector of the virtual particle is off mass shell: the minimum invariant mass will be twice the mass of an electron, from the four vector algebra.

My question is, why would predicting a virtual particle with this mass cause someone to expect real particles with this mass, since virtual particles are just tools in perturbative expansions?

It is the the propagator that has the correct named particle mass:

The easiest way to understand that the Yukawa potential is associated with a massive field is by examining its Fourier transform.


where the integral is performed over all possible values of the 3-vector momenta $k$. In this form, and setting the scaling factor to one, $α = 1 $, the fraction $4π/(k^2+m^2)$ is seen to be the propagator or Green's function of the Klein–Gordon equation.

Reading your various comments to the other answers, may be this statement will help: The internal lines of Feynman diagrams are labeled with a particle name that is consistent with the conservation of quantum numbers at the vertices. That the propagator has the on mass shell mass is due to the mathematics of the analogous transform (which explains how the Yukawa potential requires an on mass shellmass for the propagator of the pion) shown above.

  • $\begingroup$ I understand that the virtual particles are off mass-shell in terms of the $k^2$ in the propagator, but why does the mass m there correspond to a real particle's mass? $\endgroup$
    – Alex Gower
    Commented Nov 10, 2020 at 19:22
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    $\begingroup$ @AlexGower It helps predict a quantum field's behavior. An excitation of that field on-mass-shell is called real particle, and one off-mass-shell is called virtual. The same field with that parameter, m, describes both, differently. $\endgroup$ Commented Nov 10, 2020 at 19:56

You are right in some sense but the answer is subtle. Virtual particles are a mathematical model in describing static fields, like in your example, the strong field, that is, the color force responsible for the interaction between quarks that correspond to the proton and neutron. We use this to explain what a proton really looks like (contrary to popular belief, it is not just composed of three quarks):

enter image description here


The sea of virtual particles correspond to the static field describing the color force. Yet, these virtual particles in the picture of the proton, are corresponding to the rest mass of the proton by 99%. Only 1% of the rest mass of the proton is due to the rest mass of the valence quarks.

So the answer to your question is, these virtual particles (describing the color force) are very real when it comes to their correspondence to the rest mass of the composite particle, the proton.


Although many authors talk of virtual particles, it is by no means universal. Feynman himself regarded internal lines as representing real particles, albeit particles which are not directly observed, as do I.

“In Feynman’s theory the graph corresponding to a particular matrix element is regarded, not merely as an aid to calculation, but as a picture of the physical process which gives rise to that matrix element” — Freeman Dyson 1949, Phys. Rev. 75, 486.

Calling them virtual is a philosophical statement, not a scientifically provable position. As far as science is concerned, it is just a word. I wouldn't let it bother you. You can think of it how you like. Physical predictions are not changed.

  • $\begingroup$ I'm more concerned with how the mass of the virtual particle in the propagator term seems to always be coicidentally the mass of a real particle. $\endgroup$
    – Alex Gower
    Commented Nov 10, 2020 at 19:57
  • $\begingroup$ Why would that bother you, if you think of the particle as real rather than as virtual? $\endgroup$ Commented Nov 10, 2020 at 19:59
  • $\begingroup$ Because I am currently of the viewpoint that virtual particles are just useful tools in organising perturbation expansions, so it seems very surprising that they would 'coincidentally' always have properties of real particles, and if this were just the case then I don't understand why anyone would not see them as truly real. Instead I'm sure this can't be the case, but I don't see how $\endgroup$
    – Alex Gower
    Commented Nov 10, 2020 at 20:02
  • $\begingroup$ To be honest, I don't see why anyone does not regard them as truly real either, but this bears on philosophy and psychology, not science. I have given an argument at physics.stackexchange.com/questions/587933/… which shows that particles considered virtual for one observer can become real for another. $\endgroup$ Commented Nov 10, 2020 at 20:08

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