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

[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).

 

[A. Neumaier]

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).

Well, the collapse is an effective description, projecting away the part irrelevant for the subsequent experiments. Once one measures (and absorption counts as a measurement), relativistic invariance is broken anyway...

 

[vanhees71]

Yes, but it's not something that violates causality and locality. It's an update of the description after a preparation/measurement procedure, but I think that's off-topic in this very interesting thread about what's observable in gauge theories.

I'm a bit sceptic towards the axiomatic-field-theory approach since it hasn't brought us closer to the goal of formulating relativistic (interacting) QFTs mathematically rigorously. If I understand your remarks in this thread right, the Wightman system of axioms can't even describe electron-electron scattering within its scheme since this is a state with non-zero net charge. I think then we have to admit that we don't have a mathematically satisfactory description yet and have to live with the physicists' pragmatic approach of declaring perturbative, maybe also resummed/unitarizing approaches (e.g., in the Schwinger-Dyson approach to QCD) as the best that can be done with the mathematical tools at hand. What then you can "prove" (at a physicists' level of rigor) is the gauge-independence of the S-matrix ("vacuum QFT") or also observables defined within "finite-temperature QFT" as, e.g., black-body radiation or the electromagnetic radiation from a heavy-ion collision from a fireball of locally equilibrated hot and dense medium (QGP-HRG fireball).

 

[A. Neumaier]

If I understand your remarks in this thread right, the Wightman system of axioms can't even describe electron-electron scattering within its scheme since this is a state with non-zero net charge.

The Haag-Kastler framework is more permissive in this respect. since it is not bound to the vacuum sector. In the Wightman framework you need to describe electron-electron scattering by keeping some far away positrons around (in the AQFT parlance a 'charge behind the moon'), which is a bit ugly.

In spite of these shortcomings, the Clay Millennium problem is about existence of pure Yang-Mills QFTs in the Wightman framework, where glueballs are bound states in the vacuum sector.

Things are different if one admits indefinite scalar products; then there is a well-developed and predictive perturbative algebraic theory in Krein spaces.

 

[vanhees71]

I guess this is the formalization of Gupta-Bleuler quantization for the Abelian (QED) case and "covariant operator formalism" for non-Abelian gauge theories a la Kugo, Ojima et al?

 

[Demystifier]

Also in electroweak theory, as in any gauge theory, only gauge-invariant self-adjoint local operators can represent observables.

But in electroweak theory, a physical state can have a non-zero SU(2) charge, am I right? I don't fully understand why physical states can have a non-zero SU(2) charge, but cannot have a non-zero SU(3) (color) charge; where does the difference come from, exactly?

 

[malawi_glenn]

But in electroweak theory, a physical state can have a non-zero SU(2) charge, am I right? I don't fully understand why physical states can have a non-zero SU(2) charge, but cannot have a non-zero SU(3) (color) charge; where does the difference come from, exactly?

Isn't it entirely up to if the model has asymptotic freedom or not? Quarks and SU(3) is not enough, you need to specify representation and number of states for the sign of the beta function

 

[vanhees71]

This is pretty subtle. A standard reference concerning the gauge invariance of observables is

https://doi.org/10.1016/0550-3213(81)90448-X

 

[malawi_glenn]

This is pretty subtle. A standard reference concerning the gauge invariance of observables is

https://doi.org/10.1016/0550-3213(81)90448-X

I wish I had access
...

 

[Demystifier]

This is pretty subtle.

So if I'm wrong in this thread, I believe I'm wrong in a non-trivial and interesting way.

 

[vanhees71]

That's for sure right. It's a highly non-trivial question, how to justify the standard perturbative calculuations in electroweak theory, using non-gauge invariant quantities.

 

[A. Neumaier]

I guess this is the formalization of Gupta-Bleuler quantization for the Abelian (QED) case and "covariant operator formalism" for non-Abelian gauge theories a la Kugo, Ojima et al?

yes.

 

[Demystifier]

I found an interesting analogy in condensed matter physics. For that matter I present two excerpts from the book Guidry and Sun "Symmetry, Broken Symmetry and Topology in Modern Physics".

The idea that colored states of quarks should exist in interacting QCD because they exist in free QCD is analogous to Fermi liquids, where interacting states of electrons are very similar to free electrons, related to them by adiabatic change of the interaction. The opposite idea, that such states don't exist in interacting QCD, is analogous to Cooper pairs (which are analogous to QCD mesons) the energy of which is non-analytic in the coupling constant and diverging for zero coupling.

 

[A. Neumaier]

The idea that colored states of quarks should exist in interacting QCD because they exist in free QCD is analogous to Fermi liquids, where interacting states of electrons are very similar to free electrons, related to them by adiabatic change of the interaction. The opposite idea, that such states don't exist in interacting QCD, is analogous to Cooper pairs (which are analogous to QCD mesons) the energy of which is non-analytic in the coupling constant and diverging for zero coupling.

One problem with your analogies is that they don't show that the first is not realized in QCD.

Another problem is that condensed matter theory is about quasiparticles in a nonrelativistic QFT whose vacuum is a thermal state with a structure very different from the QCD vacuum.

 

[vanhees71]

In QCD colored states are not "realized", because they don't specify a gauge-independent observable. They are simply not interpretable in any physical way.

 

[Demystifier]

In QCD colored states are not "realized", because they don't specify a gauge-independent observable. They are simply not interpretable in any physical way.

A purely theoretical question (not realized in actual nature). What if we extend the Standard Model by adding couplings between Higgs and gluons, analogous to the couplings between Higgs and weak gauge bosons? In that case gluons become massive due to the Higgs mechanism and gauge invariance becomes spontaneously broken. In that theory colored states could be realized, right? And yet the theory is still gauge invariant at a more fundamental level.

 

[vanhees71]

No, in gauge theories gauge dependent "states" don't describe any physics. It doesn't matter whether the gauge symmetry is "higgsed" or not. It also never can be spontaneously broken!

 

[Demystifier]

No, in gauge theories gauge dependent "states" don't describe any physics. It doesn't matter whether the gauge symmetry is "higgsed" or not. It also never can be spontaneously broken!

So how is one W-boson, or one electron, weak-SU(2) gauge invariant?

 

[vanhees71]

What's measured are S-matrix elements between physical states, which are gauge invariant. For a careful, pedagogical discussion of electroweak gauge theory, see

https://static.uni-graz.at/fileadmin/_Persoenliche_Webseite/maas_axel/ew2021.pdf

 

[Demystifier]

What's measured are S-matrix elements between physical states, which are gauge invariant.

That's misleading. An S-matrix element is a probability amplitude. As such, it can only be measured in an ensemble. A single measurement never gives you the S-matrix element. On the other hand, a single measurement detects a single electron or a single W-boson. It's not clear in what sense it's gauge-invariant under weak-SU(2) gauge transformations.

 

[vanhees71]

Once more:

https://doi.org/10.1016/0550-3213(81)90448-X

Anything what's observable, including cross sections (S-matrix elements), must be gauge invariant. Gauge-dependent quantities take arbitrary values and thus don't describe unique phenomena in Nature.

 

[Demystifier]

Anything what's observable, including cross sections (S-matrix elements), must be gauge invariant.

Fine, but it still doesn't explain how one electron state is gauge invariant. It must be so, but I don't see how to prove that it indeed is.

 

[vanhees71]

That's a pretty subtle question. For a thorough discussion, see the already above quoted lecture notes by Axel Maas:

https://static.uni-graz.at/fileadmin/_Persoenliche_Webseite/maas_axel/ew2021.pdf

 

[Demystifier]

That's a pretty subtle question. For a thorough discussion, see the already above quoted lecture notes by Axel Maas:

https://static.uni-graz.at/fileadmin/_Persoenliche_Webseite/maas_axel/ew2021.pdf

I guess the answer to my question is something like this. Physical states are those which are annihilated by the BRST charge. A gauge transformation of a physical state, say one-electron state, adds a zero-norm state to the initial state, so it represents the same physical state. That looks fine.

But then, the same should work for QCD in deconfined phase. The one-quark state in the deconfined phase of QCD should be gauge invariant, very much like the one-electron state is gauge invariant in electro-weak gauge theory.

You are an expert for heavy ion collisions. I think such a collision can produce a deconfined phase of QCD, so maybe a one-quark state can appear in that context?

 

[A. Neumaier]

But then, the same should work for QCD in deconfined phase. The one-quark state in the deconfined phase of QCD should be gauge invariant, very much like the one-electron state is gauge invariant in electro-weak gauge theory.

The deconfined phase of QCD is a many-particle state, hence the phrase ''The one-quark state in the deconfined phase of QCD'' is meaningless.

 

[vanhees71]

I guess the answer to my question is something like this. Physical states are those which are annihilated by the BRST charge. A gauge transformation of a physical state, say one-electron state, adds a zero-norm state to the initial state, so it represents the same physical state. That looks fine.

But then, the same should work for QCD in deconfined phase. The one-quark state in the deconfined phase of QCD should be gauge invariant, very much like the one-electron state is gauge invariant in electro-weak gauge theory.

You are an expert for heavy ion collisions. I think such a collision can produce a deconfined phase of QCD, so maybe a one-quark state can appear in that context?

That's a misconception from the 1980ies. Today it's clear that the "QGP" which can be produced in heavy-ion collisions is far from being a thermal-equilibrium ideal gas of massless quarks and gluons. One rather has a strongly coupled liquid, and still there's a kind of confinement. How to really understand this state is far from being clearly resolved.

In the vacuum it's qualitatively clear: Quarks and gluons cannot be observed as asymptotic free states due to confinement. All we observe are asymptotic free color-neutral hadrons (usually baryons and mesons but also maybe more exotic tetra or maybe penta-quark states).

The difference between QCD and QFD is that the latter is not confining (not even asymptotic free).

 

[Demystifier]

The deconfined phase of QCD is a many-particle state, hence the phrase ''The one-quark state in the deconfined phase of QCD'' is meaningless.

Not necessarily. For example, if a many-particle state is a simple tensor product of one-particle pure states, then we can talk of well defined one-particle pure states.

 

[A. Neumaier]

Not necessarily. For example, if a many-particle state is a simple tensor product of one-particle pure states, then we can talk of well defined one-particle pure states.

A different phase is a different sector of states. The Hilbert space of the deconfined phase is disjoint with the Hilbert space of the vacuum sector. There is no decomposition of the states of the deconfined phase into well defined one-particle pure states.

 

[Demystifier]

A different phase is a different sector of states. The Hilbert space of the deconfined phase is disjoint with the Hilbert space of the vacuum sector. There is no decomposition of the states of the deconfined phase into well defined one-particle pure states.

Instead of taking about phases of QCD, which apparently nobody understand very well, let us talk about phases of water. Ice is a "confined" phase, steam is a "deconfined" phase. I think we can talk about single molecules within steam.

 

[A. Neumaier]

Instead of taking about phases of QCD, which apparently nobody understand very well, let us talk about phases of water. Ice is a "confined" phase, steam is a "deconfined" phase. I think we can talk about single molecules within steam.

Yes. But this is because here particle number is conserved and the molecules are not quasiparticles. As a consequence, the phase transition doesn't generate a different sector unless the thermodynamic limit is taken, where the particle structure disappears.

The sector structure in relativistic QFT is quite different from that of nonrelativistic theories.

 

[Demystifier]

Yes. But this is because here particle number is conserved and the molecules are not quasiparticles. As a consequence, the phase transition doesn't generate a different sector unless the thermodynamic limit is taken, where the particle structure disappears.

The sector structure in relativistic QFT is quite different from that of nonrelativistic theories.

1. Since you point out that molecules are not quasiparticles, are you suggesting that some would-be analogous objects in QCD are quasiparticles? Which ones?

2. Why does particle structure disappear in the thermodynamic limit?

 

[A. Neumaier]

1. Since you point out that molecules are not quasiparticles, are you suggesting that some would-be analogous objects in QCD are quasiparticles? Which ones?

The quasiparticle structure in nonrelativistic QFTs is analogous to the sector structure of relativistic theories, obtained by redefining the vacuum of the Fock space. The renormalization in relativistic QFT does the same. There the unphysical, bare particles are the Fock particles, whereas the physical, renormalized particles are what would be called quasiparticles in a nonrelativistic setting.

2. Why does particle structure disappear in the thermodynamic limit?

Because the thermodynamic limit (where the volume and the number of particles become infinite) changes the structure of the $C^*$-algebra of observables (particle number is no longer an observable) and hence the state space. Without the thermodynamic limit one has no discontinuous phase diagrams, hence no rigorous phase transitions.

 

[Demystifier]

The quasiparticle structure in nonrelativistic QFTs is analogous to the sector structure of relativistic theories, obtained by redefining the vacuum of the Fock space. The renormalization in relativistic QFT does the same. There the unphysical, bare particles are the Fock particles, whereas the physical, renormalized particles are what would be called quasiparticles in a nonrelativistic setting.

Because the thermodynamic limit (where the volume and the number of particles become infinite) changes the structure of the $C^*$-algebra of observables (particle number is no longer an observable) and hence the state space. Without the thermodynamic limit one has no discontinuous phase diagrams, hence no rigorous phase transitions.

I see. But what if we look at all this from the the point of view of lattice QCD? The lattice is assumed to be finite, so there is no thermodynamic limit in a rigorous sense. In a regime in which it describes confinement, the bare quarks are unphysical particles, while the dressed states, like mesons and nucleons, are physical quasiparticles. But even on a finite lattice one can study phase transitions in a non-rigorous phenomenological sense. So in this sense I would expect that, at high energies/temperatures, the bare quarks become physical particles, very much like electrons in Fermi gas. Or at least, that the quarks only wear a "swimming suit", so that the physical states don't differ much from bare quarks, very much like the quasiparticles in Fermi liquids. Does it make sense to you?

 

[A. Neumaier]

very much like electrons in Fermi gas.

Electrons in a Fermi gas are quasiparticles, not particles. The vacuum structure of them is different from that of the bare vacuum, and the quasiparticle Fock space is disjoint from the bare particle Fock space.
If you treat everything on a finite lattice, the quasiparticles can be described in terms of the bare particles, but they are very complex bare multiparticle states. Thus the single lattice electron quasiparticle has essentially nothing to do with the single bare lattice electron.

Similarly, in lattice QCD, the unconfined lattice quarks of a quark-gluon plasma are in fact very complex bare multiquark states, and the confined lattice quarks of vacuum QCD are another class of very complex bare multiquark states of a very different nature (which is clearly revealed in the thermodynamic limit).

 

[Demystifier]

Electrons in a Fermi gas are quasiparticles, not particles.

Can you support it by a reference?

 

[A. Neumaier]

Can you support it by a reference?

Biss, H., Sobirey, L., Luick, N., Bohlen, M., Kinnunen, J. J., Bruun, G. M., ... & Moritz, H. (2022). Excitation spectrum and superfluid gap of an ultracold Fermi gas. Physical Review Letters, 128(10), 100401.
https://arxiv.org/abs/2105.09820

 

[A. Neumaier]

Electrons in a Fermi gas are quasiparticles, not particles.

The point is that the quasi vacuum defined by the Fermi surface depends on temperature, hence one has a different sector for each temperature. The standard vacuum corresponds to zero temperature.

 

[Demystifier]

Biss, H., Sobirey, L., Luick, N., Bohlen, M., Kinnunen, J. J., Bruun, G. M., ... & Moritz, H. (2022). Excitation spectrum and superfluid gap of an ultracold Fermi gas. Physical Review Letters, 128(10), 100401.
https://arxiv.org/abs/2105.09820

From the Introduction:
"For an interacting Fermi gas, the relevant quasiparticles are particle-hole excitations, where one particle is removed from the Fermi sea and a hole is created in its place."

That makes sense. But originally you said that electrons are quasiparticles, not that particle-hole excitations are quasiparticles.

 

[Demystifier]

The point is that the quasi vacuum defined by the Fermi surface depends on temperature, hence one has a different sector for each temperature. The standard vacuum corresponds to zero temperature.

But in real experiments we always have a finite number of electrons confined in a finite volume. Under this condition, all sectors are unitary equivalent.

 

[vanhees71]

The point is that the quasi vacuum defined by the Fermi surface depends on temperature, hence one has a different sector for each temperature. The standard vacuum corresponds to zero temperature.

In more physical terms, here the ground state is called ground state, and the vacuum is always the state with no particles present.

In a many-body system in thermal equilibrium often you can describe the system in terms of "quasi particles". These describe usually collective excitations of the ground state but look similar to "particles" in the Green's-function formalism, and different phases of a system are characterized by different kinds of "thermodynamically relevant quasi-particles".

E.g., in a Fermi liquid you can have a "normal phase". There you have a "ground state" which is filled up to the Fermi level, and quasi-particle-like excitations which may look like the "real particles" but with a different mass. Also holes in the Fermi sea can look like quasi particles with the opposite charges. Then you can have bound states of holes and particles ("excitons") etc. etc. Other quasi-particle-like excitations can be also of bosonic nature, e.g., phonons, plasmons, etc.

Phases are usually characterized by an order parameter. E.g., in the normal phase of a Fermi liquid, you have $\langle \psi \psi \rangle=0$ but if there is any (effective) attractive interaction between the Fermions at sufficiently low temperatures the ground state is no longer a Fermi sea but also Cooper pairing occurs, where $\langle \psi \psi \rangle \neq 0$, and the Cooper pairs (which are pairs of two electrons close to the Fermi surface with opposite momenta and usually also opposite spin) become quasi-particles too. A specialty here is that the non-zero order parameter describes an electrically charged state (with $q=-2e$) where the charge is the electric charge, i.e., coupling to the electromagnetic field, i.e., the symmetry under multiplication with a phase factor is a local symmetry, and thus you have no spontaneous symmetry breaking but a "hidden local gauge symmetry" aka the "Anderson-Higgs-Brout-Englert-Kibble-et-al mechanism", making the corresponding gauge field, i.e., here the photons, massive. This explains the Meisner effect.

 

[vanhees71]

From the Introduction:
"For an interacting Fermi gas, the relevant quasiparticles are particle-hole excitations, where one particle is removed from the Fermi sea and a hole is created in its place."

That makes sense. But originally you said that electrons are quasiparticles, not that particle-hole excitations are quasiparticles.

These are one kind of a whole zoo of quasi-particles :-).

 

[Demystifier]

These are one kind of a whole zoo of quasi-particles :-).

Which zoo is bigger, quasiparticles in condensed matter, or hadrons in particle physics?
Allegedly, there is a proof that this problem is undecidable.

 

[Vanadium 50]

Quasiparticles may be bigger, but hadrons are nicer.

 

[A. Neumaier]

From the Introduction:
"For an interacting Fermi gas, the relevant quasiparticles are particle-hole excitations, where one particle is removed from the Fermi sea and a hole is created in its place."

That makes sense. But originally you said that electrons are quasiparticles, not that particle-hole excitations are quasiparticles.

This is like electron-positron creation in QED. What is created is a 2-quasi-particle state in which you can identify one as the electron the other as the hole.

But in real experiments we always have a finite number of electrons confined in a finite volume. Under this condition, all sectors are unitary equivalent.

Yes, but the quasiparticles in different sectors are multiparticle states very different from the single-particle states at 0 temperature.

 

[vanhees71]

I think there are more quasiparticles in condensed matter than hadrons in particle physics. Particularly there are more exotic ones like Weyl fermions and (maybe/maybe not) Majorana fermions, magnetic monopoles, etc. etc.

On the other hand there's also many-body physics of strongly interacting matter, and there you get also quasi particles in various QCD phases. The only problem is, it's harder to be observed. We rely on heavy-ion collisions, where you produce for a few 10 fm/c a blob of strongly interacting matter with some few 10 fm in spatial extension, and all you can observe are the hadrons in the state after thermal freeze-out and the particle abundancies which are more or less fixed at chemical freeze-out, which is close to the confinement-deconfinement transition region (for the matter created in heavy-ion collisions at the highest available energies at RHIC and LHC it's matter close to 0 net-baryon density with a cross-over pseudo-critical temperature of about 150 MeV/k). Then you have also electromagnetic probes ("dileptons" and "photons"), which are irradiated during the entire fireball evolution, which gives you a space-time averaged glimp about the spectral properties of the em. current-current correlation function providing some information on the medium modifications of the light vector mesons in the hadronic phase and their "melting" into the vector-vector correlation functions of the quarks and anti-quarks, etc. etc. Here the "quas-iparticle picture" becomes problematic since usually the "relevant excitations" have a quite large spectral width, i.e., the quasi-particle formulation of kinetic theory, derived from the full QFT non-equilibrium many-body formalism (Schwinger-Keldysh time contour -> Kadanoff Baym equations in the Wigner representation -> coarse graining to a BUU equation) becomes problematic, and you need so-called "off-shell transport" descriptions.

 

[Demystifier]

In a last couple of days I was reading about a whole new line of research about IR physics, reviewed in https://arxiv.org/abs/1703.05448. In particular, it is argued that large gauge transformations can map physical states into new physically inequivalent states.

In the context of this thread, it is argued that, in the absence of confinement, quarks in a color singlet state may physically turn into a color non-singlet state. https://arxiv.org/abs/1707.08016