Misconceptions about Virtual Particles - Comments

[A. Neumaier]

So where is the border? Is the W in a pion decay still a transition state? What about the Ws in neutral meson mixing? What about gluons in a NLO Feynman diagram?

An intermediate state in a reaction (happening in space and time) is a transition state as long as it can only be detected as a resonance (i.e., if it does not travel far enough for its trajectory to be reconstructible from its decay products.

An intermediate line in a Feynman diagram is always a virtual particle. There is no border between objects having short-living states (resonances) and objects having no state at all (virtual particles), since these kinds of objects occupy completely different worlds. It would be like asking for the border between real people and characters in a fiction movie.

 

[mfb]

An intermediate line in a Feynman diagram is always a virtual particle.

The W* in H->WW* -> ... appears as intermediate line in a Feynman diagram (unless we have different understandings of "intermediate line"), and does not appear as proper resonance in any mass plot, so why is this a transition state? Same for pion decays.
You are contradicting yourself here.

 

[DarMM]

Resonances are still states in the Hilbert Space though. In non-relativistic quantum mechanical models you can explicitly solve, or in QFTs which have been rigorously studied to the point of full analytic control of at least some of their multi-particle states, you can see that resonances are actual physically occurring states. They simple tend to "quickly" evolve into other states.

Virtual particles however don't correspond to anything in the Hilbert space, they're simply pictorial labels on terms appearing in perturbative integrals.

 

[A. Neumaier]

welcome back, DarMM; I was missing you!

 

[A. Neumaier]

The W* in H->WW* -> ... appears as intermediate line in a Feynman diagram (unless we have different understandings of "intermediate line"), and does not appear as proper resonance in any mass plot, so why is this a transition state? Same for pion decays.
You are contradicting yourself here.

Maybe I was irritated by the star, which typically denotes a transition state; I was talking in general, not about $W^*$ in particular. . Could you please give a reference to a paper where this particluar process is discussed? Then i can tell you more.

 

[mfb]

H -> WW* -> whatever? It is one of the standard Higgs decays. The experimental papers have nice collections of references: CMS, ATLAS 1, ATLAS 2

 

[samalkhaiat]

Maybe I was irritated by the star, which typically denotes a transition state; I was talking in general, not about $W^*$ in particular. . Could you please give a reference to a paper where this particluar process is discussed? Then i can tell you more.

I don’t know your purpose of creating this unnecessary hostile environment against necessary field theory concept.
QFT deals with on-shell states as well as off-shell states. Even though the off-shell states do not trigger our detectors, their existence must be accounted for in order to explain the stuff we see in the detectors. Furthermore, it is now an experimental fact that baryons contain (beside their valence quarks) a sea of virtual quarks and gluons. Even worse for you, enormous experiments confirmed that the sea quarks of the proton have more $\bar{d}$ than $\bar{u}$! Indeed, this flavour asymmetry has been measured [1-4] to be $\bar{d} - \bar{u} = 0.118 \pm 0.012$.

And, to throw more stones on your unnecessary use of language, the “meson cloud” model [5-6] is the best model we have that can explain the above mentioned proton sea quarks flavour asymmetry. The calculations can be done even with no reference to perturbation theory.


[1] Towell R. S. et al.(FNAL E866/NuSea Collaboration), Phys.Rev.D,(2001),64, 052002.
[2] Ackerstaff K. et al.(HERMES Collaboration), Phys.Rev.Lett.(1998), 81, 5519.
[3] Arneodo M. et all. (New Muon Collaboration), Phys. Rev. D,(1994), 50, R1.
[4] Baldit A. et al. (NA51 Collaboration), Phys. Rev. Lett. B,(1994), 332, 224.
[5] Garvey G.T, Peng J-C, Prog. Part. Nucl. Phys,(2001), 47, 203.
[6] Julia-Diaz B, Riska D. O, Nucl. Phys. A,(2006), 780, 175-186.

 

[DarMM]

Furthermore, it is now an experimental factthat baryons contain (beside their valence quarks) a sea of virtual quarks and gluons.

Where has this been proven? There are solvable 2D model field theories where perturbatively a certain state looks like the Lagrangian particles* plus a sea of virtual particles, but non-perturbatively is simply a state, not containing this "sea". I don't see how protons are different.

*By which I mean one-particle states of the Lagrangian fields.

 

[DarMM]

QFT deals with on-shell states as well as off-shell states. Even though the off-shell states do not trigger our detectors, their existence must be accounted for in order to explain the stuff we see in the detectors.

Yes, in the usual formalism they are necessary, but that doesn't mean they exist. For instance in the usual formalism of GR the Christoffel symbols are necessary, but that doesn't mean there are physical "Christoffel waves" or "Christoffel fields".

 

[samalkhaiat]

Where has this been proven? There are solvable 2D model field theories where perturbatively a certain state looks like the Lagrangian particles* plus a sea of virtual particles, but non-perturbatively is simply a state, not containing this "sea". I don't see how protons are different.

*By which I mean one-particle states of the Lagrangian fields.

What are you talking about? What proof has to do with experimentally confirmed fact? And, why did you need to bring nurealistic 2D models into the disscussion?
Read the paper I mentioned first, then you understand what i was talking about.

 

[DarMM]

What are you talking about? What proof has to do with experimentally confirmed fact?

I don't mean mathematically proven, I mean where has it been experimentally demonstrated, I just used "proven" colloquially.

And, why did you need to bring nurealistic 2D models into the disscussion?

To show that what a QFT looks like perturbatively does not indicate its true behaviour. If you don't like it though, why not take 4D QCD on a lattice. Here the proton emerges as simply a state, no sea of virtual gluons.

 

[samalkhaiat]

I don't mean mathematically proven, I mean where has it been experimentally demonstrated, I just used "proven" colloquially.

If you are not in the game, then just look at the following:
[1] Towell R. S. et al.(FNAL E866/NuSea Collaboration), Phys.Rev.D,(2001),64, 052002.
[2] Ackerstaff K. et al.(HERMES Collaboration), Phys.Rev.Lett.(1998), 81, 5519.
[3] Arneodo M. et all. (New Muon Collaboration), Phys. Rev. D,(1994), 50, R1.
[4] Baldit A. et al. (NA51 Collaboration), Phys. Rev. Lett. B,(1994), 332, 224.

To show that what a QFT looks like perturbatively does not indicate its true behaviour. If you don't like it though, why not take 4D QCD on a lattice. Here the proton emerges as simply a state, no sea of virtual gluons.

Lattice QCD could not account for many observed facts, because of the ambiguous treatment of fermions. Your computer can not work with Grassmann numbers.

 

[DarMM]

Lattice QCD could not account for many observed facts, because of the ambiguous treatment of fermions. Your computer can not work with Grassmann numbers.

Computers can work with Grassmann numbers, they're just slow at doing so due to how the Grassmann algebra functions.
Plus it's not relevant to the discussion, in lattice QCD the proton is just a state, it isn't composed of a sea of particles. In perturbative lattice QCD, just as in perturbative continuum QCD, the proton is valence quarks + sea of quarks. This suggests very strongly, as it remains true at arbitrary lattice spacing, that in nonperturbative continuum QCD the proton is just a state as well.

If you are not in the game, then just look at the following:
[1] Towell R. S. et al.(FNAL E866/NuSea Collaboration), Phys.Rev.D,(2001),64, 052002.
[2] Ackerstaff K. et al.(HERMES Collaboration), Phys.Rev.Lett.(1998), 81, 5519.
[3] Arneodo M. et all. (New Muon Collaboration), Phys. Rev. D,(1994), 50, R1.
[4] Baldit A. et al. (NA51 Collaboration), Phys. Rev. Lett. B,(1994), 332, 224.

I had a look at them, I don't see them confirming what you are saying. They just show that nucleons are heavier than simple quark models suggest. This doesn't mean QCD depicts the proton as a sea of virtual quarks, or that such a sea has been observed.

 

[DarMM]

welcome back, DarMM; I was missing you!

Thanks A. Neumaier, kind of you to say. I'm looking forward to getting back into the forum.

 

[samalkhaiat]

Computers can work with Grassmann numbers, they're just slow at doing so due to how the Grassmann algebra functions.

Really? I must be an illiterate then. For your information, when dealing with fermions there are 2 kinds of problem: (1) A straightforward discretization using a chiral invariant action always leads an action which when $a \to 0$ produses a spectrum with twice as many fermions as possessed by the original theory. Various lattice actions which avoid this problem have been suggested, the most popular are the Wilson and the Kogut-Susskind models. These give up explicit chiral invariance for non-zero lattice spacing, a rather worrying matter given that chiral invariance is an important approximate symmetry of nature. (2) In the path integral formulation, which underlies the whole lattice method, the “classical” fermion fields are not true commuting numbers. They are non-commuting numbers, so cannot be directly simulated on a computer. However, it is possible to formally integrate out the fermion fields and thereby transmute the problem into one of inverting Dirac operator. In practice, this means inverting a very large matrix, so that computer time becomes a serious issue. For this reason, most people replace the Dirac operator by the unit operator, which simply corresponds to eliminating all fermion-antifermion loop diagrams. This is, in the lattice-people language, referred to as the quenched approximation. So, you might as well claim that fermions don't exist because we can get rid of them in the quenched approximation! Give yourself a break for goodness sake.

I had a look at them, I don't see them confirming what you are saying.

Are you accusing me of making up a story?

They just show that nucleons are heavier than simple quark models suggest. This doesn't mean QCD depicts the proton as a sea of virtual quarks, or that such a sea has been observed.

Really, is that all? Look, I told you something and asked to READ at least one of the 4 paper.
The title of [1] : Improved measurement of the $\bar{d}/ \bar{u}$ asymmetry in the nucleon sea.
From the abstract of [1]:
From these data, the ratio of down antiquark $\bar{d}$ to up $\bar{u}$ antiquark distributions in the proton sea is determined over a wide range in Bjorken-x.These results confirm previous measurements by E866 and extend them to lower x. From these data, $\bar{d} - \bar{u}$ and $\int (\bar{d} - \bar{u}) dx$ are evaluated for 0.015< x < 0.35 .

Did you read this part? Did you ask yourself why should there be a $\bar{d}$ and a $\bar{u}$ in the proton?

 

[DarMM]

Really? I must be an illiterate then...In practice, this means inverting a very large matrix, so that computer time becomes a serious issue. For this reason, most people replace the Dirac operator by the unit operator, which simply corresponds to eliminating all fermion-antifermion loop diagrams. This is, in the lattice-people language, referred to as the quenched approximation. So, you might as well claim that fermions don't exist because we can get rid of them in the quenched approximation! Give yourself a break for goodness sake.

The quenched approximation and the reality of virtual particles aren't remotely analogous, I never claimed the quenched approximation was valid physically. The quenched approximation is a truncation of the theory. Virtual particles appear in the perturbative expansion.

Saying a narrative of the perturbative expansion is invalid, is in no way connected to saying a truncation is equivalent to the full theory.

Also I am aware of the Nielsen-Ninomiya theorem, but this doesn't prevent computers from handling Grassmann numbers, it prevents them from handling Chiral Fermions. Computers can handle Grassmann algebras as easily as they can Complex Numbers, you could code up a Grassmann class in C++, Ruby, Python, e.t.c. in minutes.

Are you accusing me of making up a story?

Really, is that all? Look, I told you something and asked to READ at least one of the 4 paper.
The title of [1] : Improved measurement of the $\bar{d}/ \bar{u}$ asymmetry in the nucleon sea.
From the abstract of [1]:
From these data, the ratio of down antiquark $\bar{d}$ to up $\bar{u}$ antiquark distributions in the proton sea is determined over a wide range in Bjorken-x.These results confirm previous measurements by E866 and extend them to lower x. From these data, $\bar{d} - \bar{u}$ and $\int (\bar{d} - \bar{u}) dx$ are evaluated for 0.015< x < 0.35 .

Did you read this part? Did you ask yourself why should there be a $\bar{d}$ and a $\bar{u}$ in the proton?

Yes, I read that part, and I know it concerns a measurement of the Gottfried Sum observable. This is simply a smeared field difference, even perturbatively it doesn't measure quark-antiquark particle number difference. It's more a difference in field expectation values. I'm not saying observables like this are nonsense, I just don't agree with your interpretation of them. A true observation of a "gluon/quark sea" would be an observation of non-zero particle flux within the proton.

 

[atyy]

(2) In the path integral formulation, which underlies the whole lattice method, the “classical” fermion fields are not true commuting numbers. They are non-commuting numbers, so cannot be directly simulated on a computer.

The path integral formulation does not underlie all lattice methods. There are also Hamiltonian lattice formulations, eg. http://journals.aps.org/prd/abstract/10.1103/PhysRevD.11.395.

 

[A. Neumaier]

QFT deals with on-shell states as well as off-shell states.

Could you please point to a book or paper where off-shell states are given a (nonrigorous but) formal definition? I have never seen any, and I have no idea what could meant by them. All computations involving states are about states created by creation operators and are therefore necessarily on-shell. (See my list of precise definitions in the companion Insight article.)

When experimentalists talk about particle concepts they frequently use concepts without a clear (or only partially understood) formal definition. Sometimes (for example constituent quarks, valence quarks, and sea quarks) these are semiempirical concepts only loosely related to the formal concepts in quantum field theory. But the latter defines the theoretical concepts - in particular the meaning of a virtual (off-shell) particle.

Thus whatever the experimentalists talk about when they talk about states related to a meson cloud or a nucleon sea or a proton sea they are not talking about (off-shell) virtual particles but about (on-shell) bare particles in a simplified description. If they nevertheless use the terminology of virtual particles to talk about sea states as virtual quarks and gluons they are doing it without a sound formal training - they are simply mixing up the concept of a bare particle and that of a virtual particle. Since this confusion makes no difference to their experiments they don't need to care about being accurate.

But trace any experimental statement about virtual particles back to actual formulas involving states and you'll see that all states used are on-shell states. Thus their talk is only due to the loose practices that seem to suggest that virtual particles are real, indirectly observable objects with (nonexistent) off-shell states.

 

[A. Neumaier]

H -> WW* -> whatever? It is one of the standard Higgs decays. The experimental papers have nice collections of references: CMS, ATLAS 1, ATLAS 2

Thanks. The first reference talks about decay, so the decay products must be on-shell - even if they are detected only indirectly though the decay products of the decay products.

A decay process happens in time with a well-defined mean lifetime (related to the inverse imaginary part of the mass), which is impossible for a virtual particle (where all masses are real). One can also see it from the approximate [Fermi's golden rule] formula for the decay rate given in the wikipedia reference: It is expressed in terms of a matrix element of the S-matrix, which makes sense only for real particles (extermal lines).

In the light of this, do you still want to uphold your earlier claim?

The W* in H->WW* -> ... appears as intermediate line in a Feynman diagram (unless we have different understandings of "intermediate line")

 

[mfb]

@samalkhaiat: No one doubts that you can describe the proton with sea quarks and gluons, then all the experimental results you quoted apply. But you do not have to choose this description.

Thanks. The first reference talks about decay, so the decay products must be on-shell - even if they are detected only indirectly though the decay products of the decay products.

The W* is not on-shell. Its experimentally reconstructed mass (well, if we could reconstruct the neutrino properly at least...) is at most 45 GeV.

In the light of this, do you still want to uphold your earlier claim?

The W* in H->WW* -> ... appears as intermediate line in a Feynman diagram (unless we have different understandings of "intermediate line")

The W* appears as line in the Feynman diagram, and its line starts and ends within the Feynman diagram. Yes of course I do.

 

[A. Neumaier]

The W* appears as line in the Feynman diagram, and its line starts and ends within the Feynman diagram.

In this case please tell me the pages in the papers you cited for a discussion of this diagram. In particular, what is the formal meaning of the star?

 

[mfb]

The star means it is off-shell.

Here is a Feynman diagram - it is so basic that I doubt the papers include it, but the papers I linked to are exclusively about this process. Pages: all.

 

[A. Neumaier]

The star means it is off-shell.

So one W is on-shell and the other off-shell?? The Feynman diagram you linked to has instead $H\to W^-W^+$ although the text says $H\to WW^*$ which looks inconsistent.

Pages: all.

OK; I'll plough through them until I find the connection to the actual formulas used - which tell the true story. Talk in words is too often too sloppy to be sure what it means. It may take a while before I can present my conclusions.

 

[mfb]

So one W is on-shell and the other off-shell??

Yes.

The Feynman diagram you linked to has instead $H\to W^-W^+$ although the tesxt says $H\to WW^*$ which looks inconsistent.

Just a different notation for the same thing. One time with charge signs and one time with charges omitted and the "off-shellness" explicitely highlighted.

OK; I'll prough through them until I find the connection to the actual formulas used - which tell the true story.

I linked to the experimental papers, the theory papers are the references in those papers.

 

[A. Neumaier]

So one W is on-shell and the other off-shell??

Yes.

But in the Feynman diagram linked to, both W are internal lines, hence off-shell!?

 

[mfb]

One W is as much off-shell as a muon (or even an uranium atom) with its finite lifetime will be. It has a larger decay width, but that is just a quantitative difference.

 

[samalkhaiat]

@samalkhaiat: ... But you do not have to choose this description.

QCD is the underline theory for any description. However, QCD is a gauge theory and, in particular gauge, it is identical to the parton model: The parton model is a picture of the nucleon in the infinite momentum frame, i.e. in a frame where the nucleon moves with almost the speed of light. This picture is particularly useful to understand what is going on in scattering experiments involving high momentum transfer, like e.g. deep-inelastic scattering with no reference to any specific field theory. The light-front gauge is particularly appealing in the infinite-momentum frame, since it relegates dynamical aspects to kinematically suppressed contributions. For this reason, the parton model picture is commonly identified with QCD in the lightfront gauge and in the infinite-momentum frame. Strictly speaking, the light-front gauge is not more “physical” than any other gauge, but turns out to be more convenient for the interpretation of high-energy scattering experiments. One could of course choose to work in a different gauge, but one has to deal with far more complicated expressions and totally unclear physical interpretation.
In short, since thenucleon internal structure is essentially probed in the infinite-momentum frame, the parton model is by far the best model we have to calculate all experimentally relevant quantities in terms of contributions which come from the valence quarks and the sea.
Although the fundamental role of a nonperturbative pion cloud surrounding the nucleon is well understood in QCD as a consequence of the spontaneously broken chiral symmetry, QCD makes no direct definite statement about the violation of the Gottfried sum rule: Based on charge conjugation symmetry it is only possible to say that the quark sea distribution in the nucleon is equal to the antiquark distribution in the antinucleon; plus, the gluon is flavour blind $g \to q\bar{q}$, so in order to explain the violation, we were led to Sullivan’s old idea that some fraction of the nucleon's anti-quark sea distribution may be associated with non-perturbative processes like the pion cloud of the nucleon. And that seems to work fine.
For those of you who are studying DIS processes, like to dive deeper into the nucleon sea and understand the Gottfried sum rule violation, the PDF below is a good review on the subject.

 

[samalkhaiat]

Could you please point to a book or paper where off-shell states are given a (nonrigorous but) formal definition?

By on/off shell “states” I ment on/off shell “particles” and that was very very clear and obvious to everybody. So, you don’t need to make a drama out of crices.

(See my list of precise definitions in the companion Insight article.)

The reason of me participating in this thread was to tell you that this thread and your “list of precise definitions” (whatever they are) are unnecessay wast of time.

When experimentalists talk about particle concepts they frequently use concepts without a clear (or only partially understood) formal definition. Sometimes (for example constituent quarks, valence quarks, and sea quarks) these are semiempirical concepts only loosely related to the formal concepts in quantum field theory. But the latter defines the theoretical concepts - in particular the meaning of a virtual (off-shell) particle.

Yeah, why don’t tell the USA and EU governments to stop funding these experiments since they are done by un-trained people. Or, maybe you should write to Weinberg and ask him improve on his understanding of QCD and chiral symmetry, because Weinberg thinks of the physical proton as particle surrounded by a cloud of virtual measons and other hadrons.

Thus whatever the experimentalists talk about when they talk about states related to a meson cloud or a nucleon sea or a proton sea they are not talking about (off-shell) virtual particles but about (on-shell) bare particls in a simplified description. If they nevertheless use the terminology of virtual particles to talk about sea states as virtual uarks and gluons they are doing it without a sound formal training - they are simply mixing up the concept of a bare particle and that of a virtual particle. Since this confusion makes no difference to their experiments they don't need to care about being accurate.

I am a theorist, so if I write $$|P \rangle_{phy} = C_{3q}|uud\rangle + C_{5q}|uudq\bar{q}\rangle + \cdots ,$$ and consider DIS process, does the meson $q\bar{q}$ behave as real or virtual particle? Have you ever calculated or heard of deep virtual Compton scattering from the virtual baryon and meson components of a dressed nucleon?
This is exactly how we explain the violation of Gottfried sum rule: If the proton’s state contains an explicit $| \pi^{+} n \rangle$ Fock state component, a DIS probe scattering from the virtual $\pi^{+}$, which contains a $\bar{d}$ quark, will automatically lead to an excess of $\bar{d}$ over $\bar{u}$ in the proton. The all theorists Thomas et al.[1] demonstrated ,many years ago, that the nucleon’s pion cloud gives rise to unique terms in the moments of $\bar{d} - \bar{u}$ that are non-analytic in the quark mass. The leading non-analytic behaviour of the excess number of $\bar{d}$ over $\bar{u}$ arises from the infra-red behaviour of chiral loops in chiral effective theories, and is model independent.

[1] A. W. Thomas, W. Melnitchouk and F. M. Steffens, Phys. Rev. Lett. 85, 2892 (2000).

 

[andrew s 1905]

Sorry to interrupt this fascinating debate but, may I ask if virtual photon are just artifacts of the mathematical procedures in QFT in what sense, if at all, do photons mediate the EM interaction?

I am not trying to challenge the point about virtual photon only understand what this then implies wrt the EM interaction?

Regards Andrew

 

[vanhees71]

Photons are states of the electromagnetic field, and the electromagnetic field "mediates" the electromagnetic interaction. I think, the whole debate is much ado about nothing or say about sloppy language in the QFT community. Any practitioner of QFT, however, understands what "virtual particles" are, namely internal lines of Feynman diagrams, representing (in the strict perturbation expansion free) propagators of fields (in vacuum QFT the socalled Feynman propagator). That's it. Case closed ;-)).

Also the debate about the "inner structure of protons" is funny. All there is, is the attempt to describe scattering with protons, and the paradigmatic example in history of the advent of QCD as the fundamental theory of the strong interactions is indeed deep inelastic scattering. Now that we have QCD that can be described by certain approximations like the one described in #132 (light-cone gauge) and #133 ("virtual pion cloud"). These are well understood approximation schemes, and of course one should not take "virtual particles" as particles.

The same applies to unstable particles. Strictly speaking they are never particles but resonances and as such appear in the description of scattering matrix elements as internal lines. Of course, sometimes you have very longlived "resonances", which you can treat as particles with a finite lifetime. That's how their width is usually calculated in perturbation theory. You draw an external line of an unstable particle and go ahead. Of course, in fact you can also intepret this calculation also as calculating the imaginary part of the self energy of this unstable particle/resonance, i.e., you dress the free propagator appropriately to describe it as an unstable particle. This applies particularly also to the $W$ boson also discussed in a very bizzarre way in this thread. It's very clear, how the $W$ and $Z$ bosons were famously discovered in the early 80ies at the SPS in proton-antiproton collisions. Of course there you also measure the decay products, what else?

 

[bhobba]

I think, the whole debate is much ado about nothing or say about sloppy language in the QFT community. Any practitioner of QFT, however, understands what "virtual particles" are, namely internal lines of Feynman diagrams, representing (in the strict perturbation expansion free) propagators of fields (in vacuum QFT the socalled Feynman propagator). That's it. Case closed ;-)).

IMHO it indeed is much ado about nothing.

In this case even the Wikipedia article get it right:
https://en.wikipedia.org/wiki/Dyson_series

I think the much more interesting issue is the series is asymptotically divergent. How can we get answers from a divergent series?

Start a new thread about it though.

Thanks
Bill

 

[andrew s 1905]

Photons are states of the electromagnetic field, and the electromagnetic field "mediates" the electromagnetic interaction. I think, the whole debate is much ado about nothing or say about sloppy language in the QFT community. Any practitioner of QFT, however, understands what "virtual particles" are, namely internal lines of Feynman diagrams, representing (in the strict perturbation expansion free) propagators of fields (in vacuum QFT the socalled Feynman propagator). That's it. Case closed ;-)).

So just to absolutely clear then photons do not mediate the EM interaction it is the EM field that does. I assume this is true of the other gauge bosons which seems to make the common description of them as force carriers wrong.

Regards Andrew

 

[mfb]

It is a matter of description.

 

[andrew s 1905]

It is a matter of description.

So if lay people like me misuse the term "virtual particle" it is wrong but physicists can call just dismiss calling particles force carriers (when they don't seem to be) just a matter of description. I would have thought it a simple matter to say if gauge bosons carry force or not.

Regards Andrew

 

[vanhees71]

So just to absolutely clear then photons do not mediate the EM interaction it is the EM field that does. I assume this is true of the other gauge bosons which seems to make the common description of them as force carriers wrong.

Regards Andrew

Sure, it's the field which "mediates" interactions. It was a brilliant insight by Faraday to describe interactions as local via fields. This was long before it became clear that the world is relativistic and that thus "action-at-a-distance models" for interactions are very unnatural and become much more complicated. In fact it was this insight, together with Maxwell's mathematical analysis of the idea in terms of his famous equations which lead to the discovery of relativity.

Of course also the "matter particles" (in the most simple version of QED usually electrons) are described by quantum fields, and if it comes to processes like compton scattering, such a matter field "mediates" the electromagnetic interaction, but also this is a pretty mute discussion. It's usually used in popular-science attempts to explain modern high-energy physics theory to lay people, where you cannot use the appropriate math to explain it right. Then Feynman diagrams are used as if the depict scatterings of particles, where other particles are interchanged (the internal propagator lines). That's legitimate to try to explain the exciting topic to lay people, but it's very misleading to have this picture in scientific discussions. So one should take the Feynman diagrams as what they really are, namely an ingenious notation to shorten otherwise much lengthier calculations of the Dyson series for S-matrix elements.