Single Quark at Rest: The Mystery of Dark Matter?

[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.

 

[Demystifier]

Yes, but there can still be interaction.

Yes, but not much different from electromagnetic (EM) interactions. Since we can make walls made of protons and electrons that shield from EM interactions, it seems plausible that something similar might be possible for colored Yang-Mills interactions.

 

[Demystifier]

QED has many physically relevant superselection sectors, including charged ones.

But charged ones cannot be realized experimentally, right? For example, in ion collision experiments, the ions themselves are produced by starting from neutral atoms.

 

[A. Neumaier]

But charged ones cannot be realized experimentally, right? For example, in ion collision experiments, the ions themselves are produced by starting from neutral atoms.

Of course they can. Though they appear in pairs (or highr multiples). But absorbing one charged particle prepares the other oppositely charged particle for further experimentation. This is a standard case of collapse.

 

[malawi_glenn]

Yes, but not much different from electromagnetic (EM) interactions. Since we can make walls made of protons and electrons that shield from EM interactions, it seems plausible that something similar might be possible for colored Yang-Mills interactions.

There will still be color exchange with the wall/container/thing.

Maybe if you could build solid objects with WIMPs or similar :)

 

[vanhees71]

But "collapse" must not be taken too literally, particularly not in relativistic local QFT, where a collapse in the literal sense can never happen, because there cannot be faster-than-light effects of a local measurement (and all measurements we do are local).