I Single Quark at Rest: The Mystery of Dark Matter?

[Demystifier]

Resonances are real as "bumps" in a cross section. What you meausure are cross sections describing the scattering of asymptotic free "in-states" to asymptotic free "out-states". It's described by a complex pole in the corresponding Green's function.

The same can be said for long-living (quasi-stable) particles, there is no sharp difference between resonances and long-living particles. Indeed, I think that's how top quark was discovered, as a short living resonance.

 

[PeterDonis]

Why?

Because once you start pulling A and B apart, the space between them is not hot quark gluon plasma, it's vacuum, and your reasoning based on everything being at high temperature and asymptotically free breaks down. (In fact, even putting a wall between A and B is enough for that reasoning to break down if the wall is not itself made of hot quark gluon plasma.)

Similarly, a single free quark at rest, which is what the OP was describing, is not part of a hot quark gluon plasma, so it's irrelevant to talk about the properties of a hot quark gluon plasma in that context.

 

[Demystifier]

Because once you start pulling A and B apart, the space between them is not hot quark gluon plasma, it's vacuum, and your reasoning based on everything being at high temperature and asymptotically free breaks down.

If that argument was right, then thermodynamics (not only of quark gluon plasma, but also of ordinary gases) would work only for systems with infinite volume, because if the volume is finite then there is an outside space so not everything is at that temperature. And yet, we know that thermodynamics works for finite systems too. In particular, quark-gluon plasma does not need to have infinite volume. Hence your argument does not convince me.

 

[PeterDonis]

If that argument was right, then thermodynamics (not only of quark gluon plasma, but also of ordinary gases) would work only for systems with infinite volume, because if the volume is finite then there is an outside space so not everything is at that temperature.

Not at all. Thermodynamics works just fine for finite systems provided you take into account what happens at their boundaries. All I am doing is pointing out what must happen at the boundaries of the quark gluon plasmas A and B once you start separating them.

quark-gluon plasma does not need to have infinite volume

I didn't say it did. I only pointed out the obvious fact that once you separate two quark-gluon plasmas, the space between is not quark-gluon plasma, so you are no longer talking about internal asymptotically free motions of quarks within a single quark-gluon plasma, you are talking about separating two quark-gluon plasmas, which requires adding energy to them, which in turn creates additional quarks to make each one color neutral individually, where before you separated them you just had one quark-gluon plasma which only needed to be color neutral overall.

 

[Demystifier]

you are talking about separating two quark-gluon plasmas, which requires adding energy to them

That's the part I don't understand. Why the separation requires adding energy?

An auxiliary question. Suppose that initially the plasma is spherical. If you want to reshape it without breaking it into two parts, does it also require energy?

 

[Demystifier]

Similarly, a single free quark at rest, which is what the OP was describing, is not part of a hot quark gluon plasma, so it's irrelevant to talk about the properties of a hot quark gluon plasma in that context.

Fair enough, but then how about my proof in #26? Would you say that the energy of such states would not only be large, but actually infinite? But it can't be infinite because the lattice is finite. This can be seen heuristically by considering two quarks: the potential energy between them increases with distance, but the distance cannot be infinite on the finite lattice.

 

[mitchell porter]

I think a relevant concept here is "color reconnection". Sorry no time to define it, but references exist.

 

[PeterDonis]

Why the separation requires adding energy?

Because that's how the strong interaction works once you get beyond distances of about the size of an atomic nucleus: increasing the distance between strongly interacting particles requires adding energy because of the strong attractive force between them. This is the flip side of asymptotic freedom: at very short distance scales, the strong interaction becomes weaker (asymptotic freedom), but at longer distance scales, the strong interaction becomes stronger.

 

[PeterDonis]

Suppose that initially the plasma is spherical. If you want to reshape it without breaking it into two parts, does it also require energy?

Heuristically, I would expect that it would because I would expect the spherical shape to be the lowest energy shape, as it is for attractive forces generally in hydrostatic equilibrium (at least if we assume negligible rotation).

 

[PeterDonis]

how about my proof in #26

Your proof there is based on the idea that if the free theory contains color charged states, the interacting theory must also contain such states. But the fact that a state exists mathematically in the Hilbert space does not guarantee that it is physically realizable.

In the free theory, the color interaction does not exist and there is no such thing as color confinement, which is a consequence of the color interaction existing. So in the free theory there indeed would be nothing to prevent what you are calling "color charged states" (though that term itself is a misnomer in the free theory since there is no color interaction in that theory) being physically realizable.

But in the interacting theory, color confinement does exist and so there is a restriction on what states are physically realizable that is not present in the free theory, despite the fact that in the finite lattice QCD case both theories are formulated on the same underlying Hilbert space.

 

[A. Neumaier]

The state he imagines (single quark state with zero momentum) is a legitimate state in the QCD Hilbert space.

No. The Hilbert space of QCD only contains colorless states. Single quarks live in a nonphysical Krein space with indefinite inner product. The Hilbert space of QCD is a subspace of this Krein space with positive definite inner product. This subspace only contains colorless states.

It does not say that color-charged particles don't exist at all. It only says that they can't be observed at small energies.

at high temperature is not the same as at high energies. Free quarks at high temperature are very complex ensembles of composite quasiparticles in a high temperature bath.

Let me consider QCD on the lattice, so that the theory is mathematically well defined by having a finite number of degrees of freedom. This guarantees that there are no UV and IR divergences and that the Haag's theorem does not apply.

Not only that, but bound states also don't exist. Neither is there a dynamics, because the theory is defined as a Euclidean QFT, not a Lorrentzian one! Lattice QCD is therefore very nonphysical. Its physical predictions require limits that invalidate all your arguments, since the limiting single quark states are not well-defined!

Since QCD is asymptotically free, if follows that color charged states is possible for a soup of quarks at a very large temperature.

How does this follow?

Can we have an isolated water molecule at low temperature? One will say no, because water molecules like to "confine" into a solid crystal. But it really means that an isolated water molecule is unstable, not that it's impossible.

This is nonsense. A single water molecule at low temperature is completely stable. It cannot crystallize for want of other water molecules...

uppose that the soup contains 1 billion quarks and 1 billion plus 1 anti-quarks. After cooling down, you get 1 billion mesons plus 1 anti-quark extra. What happens with the extra anti-quark?

In the bath there would be one extra quark to neutralize the lone antiquark! But the whole argument makes no sense because quark number is not well-defined, so you cannot count...

 

[Nugatory]

Due to interactions it is not a Hamiltonian eigenstate, but it is a state. Hence, in principle, it can be an initial state and it is legitimate to ask how such initial state would evolve with time.

Yes, and in this sense we can talk about the fate of an isolated quark…. But, as is often the case with natural language descriptions of quantum behavior, we are using the words in a way that is completely at odds with a layman’s understanding of what they mean.

 

[A. Neumaier]

in this sense we can talk about the fate of an isolated quark….

Not really. Single quark states have negative squared norm, hence no probability interpretation. Thus one cannot interpret them in any physical way.

 

[Nugatory]

Thus one cannot interpret them in any physical way.

Yes, which is why describing their forward evolution in time doesn’t correspond at all to the lay understanding of forward evolution

 

[Demystifier]

The Hilbert space of QCD only contains colorless states. Single quarks live in a nonphysical Krein space with indefinite inner product. The Hilbert space of QCD is a subspace of this Krein space with positive definite inner product. This subspace only contains colorless states.

That's interesting. Can you give a reference for more details?

 

[vanhees71]

Heuristically, I would expect that it would because I would expect the spherical shape to be the lowest energy shape, as it is for attractive forces generally in hydrostatic equilibrium (at least if we assume negligible rotation).

That's what's investigated in heavy-ion collisions. There is pretty much evidence for the fact that at the highest energies of collisions of heavy nuclei (such as at the Relativistic Heavy Ion Collider at Brookhaven National Laboratory or at the Large Hadron Collider at CERN) a socalled quark-gluon plasma is formed.

One of the most important findings is that the $p_T$ spectra of the finally detected hadrons are well-described by relativistic viscous hydrodynamics with a very small viscosity over entropy-density ratio $\eta/s$ close to the lower "quantum bound" of $1/4 \pi$. This manifests itself in the $p_T$ spectra, which follow closely a relativistic Maxwell-Jüttner distribution of a flowing medium. Further, in semicentral collisions, where the initial overlap region of the nuclei is almond shaped, you find socalled elliptic flow of the final hadrons, i.e., when looking at the angle distribution in the plane perpendicular to the beam direction you find more particles moving in direction in the plane than out of plane. This is explained by the hydrodynamical flow, because the pressure gradient along the short axis of the almond is larger than along the long axis and thus the particles are streaming more in the directon along the short axis than along the long axis, leading to a postive $v_2$ (the Fourier coefficient of the angular $p_t$-distribution with the angle measured relative to the reaction plane).

All this of course does not prove that there were really partons as the relevant degrees of freedom of the hot and dense expanding fireball. For that there's evidence from the socalled constituent-quark-number scaling of the elliptic flow, which results from a simple coalescence picture for the formation of hadrons out of the quarks and gluons within the fireball: A meson is formed by a quark and an antiquark with the total momentum given by the sum of the momenta of the quark and the antiquark, and also the $v_2$ of the quark and antiquark add up to the $v_2$ of the meson. The same holds for the formation of baryons out of three quarks (or a di-quark and a quark). This means that when plotting $v_2/N_q$ against $p_T/N_q$ the $v_2$ of mesons and baryons approximately fall on a universal line.

Last but not least, at these high-energy collisions the medium is almost net-baryon-number free, i.e., there are as many baryons and antibaryons in the fireball. This is the situation, which can be calculated by lattice QCD at finite temperature, and this calculations shows that the transition from a partonic to a hadron-resonance-gas state is a cross-over transition at a temperature of about $155 \text{MeV}$. From simple kinetic rate-equation models using phenomenological (string) models for confinement one expects that the hadron abundancies follow a thermal distribution as function of their mass, and indeed this is the case to an amazing accuracy spanning several oders of magnitude in abundancies of various hadrons and even like nuclei and anti-nuclei.

 

[A. Neumaier]

That's interesting. Can you give a reference for more details?

we already discussed this here in a long thread about QCD. It is standard material, like in QED in the Gupta-Bleuler formalism.

Kugo, T., & Ojima, I. (1979). Local covariant operator formalism of non-abelian gauge theories and quark confinement problem. Progress of Theoretical Physics Supplement, 66, 1-130.
(with 1644 citations according to Google Scholar)

 

[vanhees71]

That's of course a very good point of the operator formalism: Though it's from a practical point of view much more involved than the standard textbook treatment in terms of the Faddeev-Popov path-integral procedure, the operator formalism provides a clear distinction of what are physical states and what not within a (non-Abelian) gauge theory. It's, however, not "standard material". I know it only from this paper and a very nice textbook by Kugo on gauge theories, which is however available only in Japanese and in German, as far as I know.

 

[Demystifier]

we already discussed this here in a long thread about QCD. It is standard material, like in QED in the Gupta-Bleuler formalism.

Kugo, T., & Ojima, I. (1979). Local covariant operator formalism of non-abelian gauge theories and quark confinement problem. Progress of Theoretical Physics Supplement, 66, 1-130.
(with 1644 citations according to Google Scholar)

Very relevant paper, but it actually claims less than you said.

On page 68 it says that it is only a conjecture that colored states (with positive norm) don't exist. Indeed, it is consistent with the general understanding that QCD confinement is only a conjecture (otherwise, the Millennium problem would be already solved).

On page 69 it proves a theorem that localized colored states don't exist. That's very important and interesting, but doesn't rule out non-localized colored states.

On page 70 it "demystifies" non-localized states by discussing the case of QED, where ordinary charged particles are non-localized due to the Coulomb tail.

Now I can say something about the original problem with new glasses. Does a single quark state exist in QCD? There is a conjecture that it doesn't, but we don't know. If it exists then we know that it is not a localized state, very much like a single electron state in QED is not localized due to the Coulomb tail. Intuitively it makes sense: A quark and an anti-quark are connected by a gluon string, but a single quark, who does not have an anti-quark cousin to connect with, perhaps still can exist by forming a Coulomb-like gluon field around it.

 

[vanhees71]

What's observable on electrons in simple QED is the electromagnetic charge-current density, and this is indeed a gauge-invariant quantity, but that's not true for QCD with colour-charge-current density, which is not gauge invariant and thus in my opinion should not be an observable of the theory. The difference is of course, because QED is an Abelian and QCD a non-Abelian local gauge theory. This is, of course, a hand-waving physics argument but not a mathematical proof, which indeed should be the answer to the mentioned unsolved Millenium Problem.

 

[A. Neumaier]

On page 68 it says that it is only a conjecture that colored states (with positive norm) don't exist. Indeed, it is consistent with the general understanding that QCD confinement is only a conjecture (otherwise, the Millennium problem would be already solved).

It is a conjecture from the purely mathematical point of view, such as the conjecture that QED does or doesn't exist. from a physical point of view it is a well-established fact. Note also that the Millennium problem is neither about QCD nor about confinement, it is about the existence of QCD and the nonexistence of massless particles in QCD. So your claim is of the same level as claiming that QCD is a conjecture only...

Moreover, for a physical interpretation it is not enough that colored states of positive norm exist - they must belong to the physical Hilbert space, the kernel of a certain operator. I think there are no colored states in the physical sector.

Does a single quark state exist in QCD? There is a conjecture that it doesn't, but we don't know.

The state of the art advanced significantly since the 1979 survey paper by Kugo and Ojima. Numerical calculations from lattice QCD extrapolated to the continuum produce a negative mass square term in the quark propagator, which proves that it has no probability interpretation - propagators of physical particles must be of the Kallen-Lehmann form. So the only doubt is whether the numerical calculations are reliable. But calculations of the same sort is what gives credence to lattice QCD, so from a physical point of view, nothing more needs to be established.

 

[A. Neumaier]

It's, however, not "standard material". I know it only from this paper

Well, this paper is a much cited book-sized report! It is _the_ source for QCD on the operator level, which is what is needed if one wants to have a perturbatively valid Hilbert space.

More references:

Alkofer, R., & von Smekal, L. (2001). The infrared behavior of QCD propagators in Landau gauge. Nuclear Physics, Section A, 1(680), 133-136.

Lowdon, P. (2018). Non-perturbative constraints on the quark and ghost propagators. Nuclear Physics B, 935, 242-255.

Hayashi, Y., & Kondo, K. I. (2020). Complex poles and spectral functions of Landau gauge QCD and QCD-like theories. Physical Review D, 101(7), 074044.

The last paper discusses numerical results on the poles of the quark propagator.

 

[Demystifier]

Numerical calculations from lattice QCD extrapolated to the continuum produce a negative mass square term in the quark propagator, which proves that it has no probability interpretation - propagators of physical particles must be of the Kallen-Lehmann form.

Negative mass square term, i.e. tachyon, usually indicates that this state is unstable, not that it doesn't have a probability interpretation. Sure, such an unstable state does not have a good particle interpretation, but in field theory, a priori, physical states don't need to have a particle interpretation.

 

[Demystifier]

I think there are no colored states in the physical sector.

Fair enough, you think so as most experts in the field do. But it seems to me that there is no proof of that, even with physicists's standards of "proof". (I even have an idea that in gauge/gravity duality those hypothetical colored states could be dual to outgoing Hawking particles, but at the moment I cannot say much more about that ...)

 

[vanhees71]

Gauge/gravity dualism itself is a conjecture only too afaik.

 

[A. Neumaier]

Negative mass square term, i.e. tachyon, usually indicates that this state is unstable, not that it doesn't have a probability interpretation.

No. Tachyons have no probability interpretation.

Tachyons indicate not unstable particles but unphysical states caused by the choice of an unphysical vacuum. They have to be resolved by shifting the vacuum to the correct, physical vacuum, where tachyons are no longer present.

in field theory, a priori, physical states don't need to have a particle interpretation.

But they need to have a probability interpretation.

But it seems to me that there is no proof of that, even with physicists's standards of "proof".

Noncausal quark propagators are mathematically rigorous proof of that (Kallen-Lehmann theorem), and that the former is the case is demonstrated by physicists's standards of "proof".

 

[Demystifier]

Tachyons indicate not unstable particles but unphysical states

https://en.wikipedia.org/wiki/Tachyonic_field#Condensation

 

[Demystifier]

@A. Neumaier can you comment on the following? Something in the paper you cited [1] seems fishy to me. At page 71 it claims that Reeh-Schlieder property implies that charged particles (like a single electron) do not exist in QED, so from the fact that electrons clearly exist in the real world, it concludes that the Reeh-Schlieder property in fact does not hold in QED. But it seems wrong to me, because the Reeh-Schlieder property is a very general theorem valid for a large class of theories, including QED. Do I miss something?

[1] Kugo, T., & Ojima, I. (1979). Local covariant operator formalism of non-abelian gauge theories and quark confinement problem. Progress of Theoretical Physics Supplement, 66, 1-130

 

[vanhees71]

But the authors clearly state that one must not impose the Reeh-Schlieder property in the Abelian case, i.e., QED. I'm not very familiar with the quite subtle math of the covariant operator formalism anymore. So better let @A. Neumaier answer the mathematical details.

From a naive physicist's point of view it's pretty simple: What's observable are local, gauge invariant quantities, and in QED the electromagnetic-current operator, $j^{\mu}(x) = :q \bar{\psi}(x) \gamma^{\mu} \psi(x):$ is both local (i.e., it fulfills the microcausality condition with the Hamilton density) and gauge-invariant.

In QCD the "color currents" are $\hat{j}^{a,\mu}(x) = :g \bar{\psi}(x) \gamma^{\mu} T^a \psi(x)$, where $\psi$ is a quark field (i.e., a color triplet field transforming according to the fundamental representation of the local color-SU(3)). This is obviously not gauge invariant and thus cannot represent an observable. It transforms according to the adjoint representation of color-SU(3).

 

[Demystifier]

But the authors clearly state that one must not impose the Reeh-Schlieder property in the Abelian case, i.e., QED.

They indeed say it, but this statement seems to contradict the Reeh-Schlieder theorem. https://en.wikipedia.org/wiki/Reeh–Schlieder_theorem

 

[Demystifier]

In QCD the "color currents" are $\hat{j}^{a,\mu}(x) = :g \bar{\psi}(x) \gamma^{\mu} T^a \psi(x)$, where $\psi$ is a quark field (i.e., a color triplet field transforming according to the fundamental representation of the local color-SU(3)). This is obviously not gauge invariant and thus cannot represent an observable. It transforms according to the adjoint representation of color-SU(3).

Then why do we not have confinement in weak SU(2)? The analogous isospin current is also not gauge invariant. I guess you will say that SU(2) gauge bosons acquire mass, so gauge invariance is "broken". But then perhaps gluons also can acquire mass by some (as yet unknown) mechanism?

 

[vanhees71]

This argument has nothing to do with confinement. Also in electroweak theory, as in any gauge theory, only gauge-invariant self-adjoint local operators can represent observables.

A local gauge symmetry cannot be spontaneously broken, because it's not a symmetry in the Noetherian sense. If you try that, you get the Higgs mechanism, making the gauge bosons massive and no massless Nambu-Goldstone bosons in the physical spectrum. Unfortunately in the literature almost all authors call the Higgs mechanism "spontaneous breaking of a local gauge symmetry". That's they usual slang, you cannot get rid anymore because of (in this case in my opinion unhealthy) tradition. Of course a local gauge symmetry must never be explicitly broken (also not by an anomaly), because then the entire theory becomes inconsistent.

As in QED there is color superconductivity in QCD. In such a phase some or even all gluons get massive.

 

[vanhees71]

They indeed say it, but this statement seems to contradict the Reeh-Schlieder theorem. https://en.wikipedia.org/wiki/Reeh–Schlieder_theorem

As I'm not a axiomatic-QFT theorist, I leave this for @A. Neumaier to answer.

 

[A. Neumaier]

the Reeh-Schlieder property is a very general theorem valid for a large class of theories, including QED.

It follows from the Wightman axioms, which only describes the uncharged sector of QED, not all of QED.

perhaps gluons also can acquire mass by some (as yet unknown) mechanism?

In QCD, color symmetry is unbroken, and has to be to match experiment.

 

[Demystifier]

It follows from the Wightman axioms, which only describes the uncharged sector of QED, not all of QED.

Why do Wightman axioms not describe the charged sector? (I want a theoretical answer not depending on experimental data.)

One of the consequences of Wightman axioms is the CPT theorem. What does C stand for in this theorem, if Wightman axioms do not describe the charged sector?

 

[Demystifier]

As in QED there is color superconductivity in QCD. In such a phase some or even all gluons get massive.

Is the QCD coupling strong or weak in this phase?

 

[dextercioby]

Wightman's axioms need to be amended, if one is to treat a gauge theory such as electromagnetism. Wightman's axioms in 4D Minkowski only work for a chargeless massive scalar field. I remember seeing a whole chapter in Bogolubov et al. (1990) book on QFT regarding a rigorous construction of em. quantization in the free case.

 

[malawi_glenn]

Now going back to the "at rest" part.
What does it mean for a quantum particle to be at rest? Is it even possible?

 

[dextercioby]

"at rest" cannot be imagined as a notion with significance at microscopic/quantum level. This concept is purely a relativistic one (Newtonian, SR, GR). An electron "hits" a screen after passing through a S-G apparatus. Can we consider that the screen brings him to rest? We usually interpret that the screen's molecular structure "absorbs" it somehow.

 

[malawi_glenn]

"at rest" cannot be imagined as a notion with significance at microscopic/quantum level. This concept is purely a relativistic one (Newtonian, SR, GR). An electron "hits" a screen after passing through a S-G apparatus. Can we consider that the screen brings him to rest? We usually interpret that the screen's molecular structure "absorbs" it somehow.

Exactly. So the "free quark at rest" scenario is ill defined

Consider the following experiment. (It's a gedanken experiment, but should be possible in principle.) Consider a big container filled with a hot color neutral soup of quark-gluon plasma. Now insert a wall in the container which divides the container into two compartments, labeled A and B. The plasma in A does not need to be exactly color neutral, and similarly for B, but together they are exactly neutral. Now separate A and B at a large spatial distance from each other. Finally, after the separation, cool down A at a low temperature, so that confinement can take place. Given that A was not initially color neutral, what happens in A after the cooling?

What is the construction of this wall/container? Does it contain atomic nuclei? If yes, there will be strong/color force interactions between the wall/container and the plasma.

 

[Demystifier]

What is the construction of this wall/container? Does it contain atomic nuclei? If yes, there will be strong/color force interactions between the wall/container and the plasma.

In principle, I guess one could use Tokamak https://en.wikipedia.org/wiki/Tokamak

 

[malawi_glenn]

In principle, I guess one could use Tokamak https://en.wikipedia.org/wiki/Tokamak

Magnetic fields then, generated by some material, still non zero probablity for a quark to be near enough the material and interact?

 

[Demystifier]

Magnetic fields then, generated by some material, still non zero probablity for a quark to be near enough the material and interact?

In principle, the probability for that can be made sufficiently small.

 

[malawi_glenn]

In principle, the probability for that can be made sufficiently small.

What about the gluons? They are not affected by the magnetic field

 

[Demystifier]

What about the gluons? They are not affected by the magnetic field

Here "gluons" are not free particles, but a Coulomb-like field attached to the quarks.

 

[A. Neumaier]

Why do Wightman axioms not describe the charged sector? (I want a theoretical answer not depending on experimental data.)

Because they describe the sector of a QFT containing the vacuum, which is neutral by definition.

 

[A. Neumaier]

One of the consequences of Wightman axioms is the CPT theorem. What does C stand for in this theorem, if Wightman axioms do not describe the charged sector?

Charge conjugation is a formal operation that applies to all charges, not specifically electric charge.
https://en.wikipedia.org/wiki/C-symmetry#In_quantum_theory
In nonrelativistic theories, CPT symmetry covers electric charge. But in QED, states with different charge quantum number live in different superselection sectors, and the vacuum sector is uncharged.

But the vacuum sector contains (at least perturbatively) states made from any number of electron-positron pairs, which are neutral but on which the perturbative CPT symmetry acts nontrivially because it interchanges electrons and positrons.

 

[Demystifier]

But in QED, states with different charge quantum number live in different superselection sectors, and the vacuum sector is uncharged.

But the vacuum sector contains (at least perturbatively) states made from any number of electron-positron pairs, which are neutral but on which the perturbative CPT symmetry acts nontrivially because it interchanges electrons and positrons.

Interesting! From this point of view, would you say that the electric charged sector of the Standard Model is totally irrelevant for physics, because it can never be realized in experiment? When we "isolate" electron here, we always have a proton (or something else with a positive charge) somewhere else.

 

[malawi_glenn]

Here "gluons" are not free particles, but a Coulomb-like field attached to the quarks.

Yes, but there can still be interaction.

 

[A. Neumaier]

Now going back to the "at rest" part.
What does it mean for a quantum particle to be at rest? Is it even possible?

It means - a particle considered in its rest frame, which exists in the massive case (only).

Interesting! From this point of view, would you say that the electric charged sector of the Standard Model is totally irrelevant for physics, because it can never be realized in experiment? When we "isolate" electron here, we always have a proton (or something else with a positive charge) somewhere else.

Nonsense. QED (and the standard model) has many physically relevant superselection sectors, including charged ones. This is no contradiction with the fact that only its vacuum sector is described by Wightman's axioms.