How can quarks exist if they are confined?

  • A
  • Thread starter DarMM
  • Start date
  • Tags
    Quarks
  • Featured
In summary, according to nonperturbative QCD, quarks and gluons don't exist and in nonperturbative QED with two spinors (e.g. proton and electron) hydrogen isn't composed of a proton and an electron. However, in some approximate sense, quarks and gluons do exist in QCD and hydrogen is composed of a proton and an electron even in full QED with protons.
  • #36
DarMM said:
how quarks can be absent from the physical Hilbert space and their associated fields ill defined as operators on the physical Hilbert space, with the obvious evidence of the multilocal nature of the proton
Nothing forbids the associated currents to be physical (gauge invariant). Only the particle interpretation must be given up.

Multilocality in the plane orthogonal to the flow direction is all that needs to be explained, I think, and currents do that.
 
Physics news on Phys.org
  • #37
DarMM said:
What I'm trying to understand in this thread is how quarks can be absent from the physical Hilbert space and their associated fields ill defined as operators on the physical Hilbert space, with the obvious evidence of the multilocal nature of the proton as mentioned by @A. Neumaier and @bhobba for Deep Inelastic scattering.
Quarks carry a color charge (of the fundamental representation) and thus are gauge dependent and thus can't be observables. The same holds for gluons which carry a color charge (of the adjoint representation) and thus also can't be observables. Only gauge-independent expectation values and gauge-invariant S-matrix element lead to a sensible definition of observables within the theory. Physically that's named "confinement", i.e., all color charges are confined in color-neutral objects which are (putatively) observable. Today these we know this are the baryons and the mesons (or both together the hadrons) and maybe some more "exotic" states like tetraquarks (maybe the XYZ states in the charm sector are such guys, but maybe they are rather "meson molecules", which is still under both theoretical and experimental debate). Then there should also be glue balls, i.e., color-neutral bound states of gluons.

Formally quarks and gluons are the quantum fields of the theory, and they can be used to build observables, e.g., ##\bar{u} \gamma_5 d##. This carries the quantum numbers of the negative pion.
 
  • #38
vanhees71 said:
Quarks carry a color charge (of the fundamental representation) and thus are gauge dependent and thus can't be observables. The same holds for gluons which carry a color charge (of the adjoint representation) and thus also can't be observables. Only gauge-independent expectation values and gauge-invariant S-matrix element lead to a sensible definition of observables within the theory. Physically that's named "confinement", i.e., all color charges are confined in color-neutral objects which are (putatively) observable. Today these we know this are the baryons and the mesons (or both together the hadrons) and maybe some more "exotic" states like tetraquarks (maybe the XYZ states in the charm sector are such guys, but maybe they are rather "meson molecules", which is still under both theoretical and experimental debate). Then there should also be glue balls, i.e., color-neutral bound states of gluons.

Formally quarks and gluons are the quantum fields of the theory, and they can be used to build observables, e.g., ##\bar{u} \gamma_5 d##. This carries the quantum numbers of the negative pion.
That's a good summary and don't take this as rude (I don't mean it that way), but I know all this. What I'm getting at is that it is odd that you can build operators for physical states out of seemingly unphysical fields acting on a state space that isn't a Hilbert Space and yet those unphysical fields seem to be echoed in certain scattering experiments. I think something like @A. Neumaier's comment is what is needed.

Basically quarks seem to unnecessary in a nonperturbative setting on one level (all states are colorless, quark states don't exist in a Hilbert space and hence one might be tempted to construct a purely hadronic Lagrangian directly on a hardonic-glueball Hilbert space), but seemingly a useful concept in other cases, e.g. Deep Inelastic Scattering.
 
  • Like
Likes dextercioby and vanhees71
  • #39
DarMM said:
and yet those unphysical fields seem to be echoed in certain scattering experiments.
vanhees71 said:
Formally quarks and gluons are the quantum fields of the theory, and they can be used to build observables, e.g., ##\bar{u} \gamma_5 d##. This carries the quantum numbers of the negative pion.
In particular, they can be used to build gauge invariant and hence in principle observable currents ##j_q=\bar{q} \gamma q## (trace over colors) for each of the six quarks ##q##. Thus while individual (colored) quarks are unphysical, the six (colorless) quark flows are physical.
 
Last edited:
  • Like
Likes dextercioby and vanhees71
  • #40
A. Neumaier said:
In particular, they can be used to build gauge invariant and hence observable currents ##j_q=\bar{q} \gamma q## (trace over colors) for each of the six quarks ##q##. Thus while individual quarks are unphysical, the six quark flows are physical.
In addition, there are physical flavor-changing currents ##j_q=\bar{q} \gamma q'## with two different quarks ##q## and ##q'##.
 
  • Like
Likes dextercioby and vanhees71
  • #41
The details of this discission are well beyond me, nevertheless I am trying to extract what you mean by "real" and "physical".

Would I be right to conclude that you mean that the entity under discussion is real and or physical if it is a coherent element of the model (rather than an artefact of the mathematical procedure used to solve it) and or that it needs to be an observable (at least in principle)?
Thanks Andrew
 
  • #42
andrew s 1905 said:
the entity under discussion is real and or physical if it is a coherent element of the model
The usage in the present thread is: A quantum field is called physical, or real (i.e., observable in principle) if it is realized as a (densely defined, distribution valued) operator on a Hilbert space with positive definite inner product.
 
Last edited:
  • Like
Likes DarMM and andrew s 1905
  • #43
I've never seen a "distribution valued operator" in my everyday life nor by any experimentalists with their refined equipment to extend our senses to the "quantum realm". I've once visited CERN, looking at the accelerator and at 2 of the main experiments (CMS and ALICE). This is what's real, but the outcome of this setup is described very well by the Standard Model (among other things indeed using distribution valued operators on a Hilbert (?) space, or however you defined the Hilbert-like space used to formulate the theory in terms of more rigorous mathematical formulations), and in my opinion that and only that is what makes the SM "realistic". It's the only sense the word "realistic" has for me as a physicist: A theory/model is realistic if and only if it describes successfully the outcome of observations in the real world. E.g., string theory is "not realistic", because it doesn't describe anything observable (yet). This may change in the future when string theorists come up with some real-world testable result, which then is checked by experimentalists with real-world equipment in the lab (or in astronomical observatories on Earth or in space).
 
  • #44
vanhees71 said:
It's the only sense the word "realistic" has for me as a physicist
But I described the sense how ''realistic'' is used in mathematical physics and in this thread. For example, $\Phi^4$ theory in 2 space-time dimensions is realistic in the above sense, while it is clearly not realistic in your sense.
vanhees71 said:
the Standard Model (among other things indeed using distribution valued operators on a Hilbert (?) space, or however you defined the Hilbert-like space used to formulate the theory in terms of more rigorous mathematical formulations)
The operator formulation of the standard model is in terms of a Krein space (topological vector space with an indefinite inner product), hence is not immediately physical in the sense of mathematical physics.
 
  • #45
I see. Well, at the end mathematical physics becomes physics when it makes predictions testable by observation. Of course, as ##\phi^4## theory in 1+1 dimensions, it can be of high value to understand the mathematics of the theory, and as such it's of course well worth studying, but mathematics has no aim at all to be "realistic" in any sense, and that makes it so useful in applications too, because it provides a concise language for all kinds of theories and models aiming at the description of "reality".
 
  • #46
vanhees71 said:
mathematics has no aim at all to be "realistic" in any sense,
But mathematics uses all sorts of everyday expressions (such as groups, fields, rings, equivalence, truth) to denote specific formal objects or relations, and pins them down by giving precise definitions. Thus these terms can be used in an unambiguous way and convey definite information. This is what makes it the right tool for precision in all sciences. This is also the way the term ''physical field'' is used in the present discussion.

However, the relation of the precisely defined terms to the loose everyday notion described by the same word may be very weak.
 
  • Like
Likes vanhees71
  • #47
Of course, I agree with that notion of math completely.
 
  • #48
@A. Neumaier has already covered this, but just to respond.
vanhees71 said:
I've never seen a "distribution valued operator" in my everyday life nor by any experimentalists with their refined equipment to extend our senses to the "quantum realm".
"Physical field" means an operator valued distribution which (after smearing) acts on the Hilbert Space. This is sort of the bare minimum needed for a field to have correlation functions and thus to be observable in some form (i.e. correspond to detector clicks). It's just a term you tend to see in non-perturbative studies of field theories.

Some of the fields one can have in QFT don't produce physical fields in this sense. An example would be the ghost fields, which don't act on the Hilbert Space (or more strictly aren't endomorphisms on it) and so aren't really connected to observables, i.e. it would be possible to completely eliminate them in an alternate way of formulating the theory.

Hence the distinction, one kind are associated with observable quantities, the other are not and can be written out of the theory.

The surprising thing (and the focus of this thread) is that nonperturbatively quark fields are the same as ghost fields in this sense. However it is confusing, because many observations (e.g. Deep Inelastic Scattering) look most natural in terms of quarks.
 
  • #49
In the pertubative theory the ghost fields cancel contributions of other unphysical field-degees of freedom. They are necessary to organize the perturbative calculation of observable predictions of the theory.
 
  • #50
vanhees71 said:
In the pertubative theory the ghost fields cancel contributions of other unphysical field-degees of freedom. They are necessary to organize the perturbative calculation of observable predictions of the theory.
Yes, perturbatively.

Nonperturbatively in some formulations, e.g. lattice gauge theory, they don't even show up.
 
  • #51
Quarks and gluons most definitely "exist" in the sense that the quantum field theory that we believe describes hadrons, QCD, has quark fields and gluon fields in it, in exactly the same way that quantum electrodynamics (QED) has electron/positron fields and photon fields in it. If we set the QCD coupling constant to zero, we get a QFT of free quarks and gluons. Though there is not yet a fully mathematically rigorous construction of QCD, there is ample evidence that we can start with a lattice version and take the zero-spacing limit, with the result being the correct QFT of hadrons. In this sense, QCD is on a much better mathematical footing than QED, which is expected to be a trivial theory that cannot survive removal of the cutoff (equivalently, taking the lattice spacing to zero).
 
  • Like
Likes vanhees71
  • #52
I don’t understand the statement that the SU(3) lattice gauge theory doesn't include quarks and gluons...

I also don’t understand which *physical Hilbert space* we’re talking about. Clearly we aren’t discussing the perturbative Hilbert space of QCD, which is the only one I know how to rigorously construct given a certain regularization scheme and which is only valid where we take a certain asymptotic limit after carefully modding out the various unphysical degrees of freedom. Still this is a valid physical sector of the as yet to be discovered full theory, so you can’t just ignore it, and this sector clearly includes Quarks and Gluons and is verified by deep inelastic scattering amongst many other experiments.

So presumably the statements in this thread are about the as yet to be defined full nonperturbative Hilbert space of the theory. But again which part? QCD exhibits different critical behavior at various phases of the theory, which is usually captured by a variety of different ad hoc methods. Most of which includes quarks and gluons at some level, but some of the resulting theories don’t necessarily agree on their overlap.

I think a lot of the ‘what are the fundamental DOF of QCD’ debates have sort of ended with the duality revolutions, where it became clear that vastly different descriptions could lead to the same underlying physical predictions
 
  • #53
Avodyne said:
QCD, has quark fields and gluon fields in it, in exactly the same way that quantum electrodynamics (QED) has electron/positron fields and photon fields in it.
But in QED, the transversal photon field is physical, while in QCD, quark fields cannot be made physical by taking the transversal part.
Haelfix said:
I don’t understand the statement that the SU(3) lattice gauge theory doesn't include quarks and gluons...
The claim by DarMM was only that it does not contain ghosts.
Haelfix said:
presumably the statements in this thread are about the as yet to be defined full nonperturbative Hilbert space of the theory. But again which part?
Yes. Most of it refers to the vacuum sector (zero density and temperature), for which Wightman's axioms are expected to hold.
Haelfix said:
I think a lot of the ‘what are the fundamental DOF of QCD’ debates have sort of ended with the duality revolutions, where it became clear that vastly different descriptions could lead to the same underlying physical predictions
But the physical Hilbert space defining the vacuum representation should still be the same, in the sense of carrying an isomorphic representation of the algebra of observables.

This is analogous to the well-known fact from representation theory of finite-dimensional Lie algebras that the same representation can be obtained in very different ways. Nevertheless, the irreducible unitary representations are uniquely determined by a few quantum numbers.
 
Last edited:
  • Like
Likes vanhees71
  • #54
Avodyne said:
Quarks and gluons most definitely "exist" in the sense that the quantum field theory that we believe describes hadrons, QCD, has quark fields and gluon fields in it, in exactly the same way that quantum electrodynamics (QED) has electron/positron fields and photon fields in it. If we set the QCD coupling constant to zero, we get a QFT of free quarks and gluons.
That's not just some simple ##g \rightarrow 0## limit though. Removing the coupling prevents it from being a gauge theory as it removes the gluon self-coupling terms, so it's a completely different (non-gauge) theory, a singular limit as such, just think of it in terms of the SU(3) fiber bundle. It's quite unlike QED, where sending ##e## to zero just decouples electrons.

Also, related to what @A. Neumaier said, some of the photon's degrees of freedom are physical, however the gluon and quark fields the entirety of their degrees of freedom map out of the Hilbert space.
 
  • #55
Haelfix said:
I don’t understand the statement that the SU(3) lattice gauge theory doesn't include quarks and gluons...
It doesn't include ghosts.

Haelfix said:
I also don’t understand which *physical Hilbert space* we’re talking about.
Same as in QED, the physical Hilbert space is a subset of the whole unphysical (Hilbert-Krein) space selected by a BRST condition or similar.

Haelfix said:
Still this is a valid physical sector of the as yet to be discovered full theory, so you can’t just ignore it, and this sector clearly includes Quarks and Gluons and is verified by deep inelastic scattering amongst many other experiments.
I don't think perturbation theory is a physical sector, it's a method of approximating calculations which is very accurate (once you don't include too many terms) within a certain range of momenta and energy, but that's not a physical sector to me.

The whole point of the thread is that it is not so clear it includes Quarks and Gluons, as they map out of the Hilbert Space and thus aren't well-defined operators on the physical Hilbert space. You can use Quark and Gluon fields, but it's not so clear it contains them any more than it contains Faddeev-Poppov ghosts
 
  • #56
DarMM said:
It doesn't include ghosts.
Sorry I apologize I misread your statement, but to be fair it was a little hard to follow b/c Lattice QCD *is* a nonperturbative approach and it manifestly does include quarks and gluons. So I'm failing to square the premise a little. How can a nonperturbative approach explicitly calculate the masses of an object quite precisely, and then the very same theory in principle somehow not include those objects? Note that this is more general than even lattice qcd... Other nonpertubative objects (like Wilson loops) can also be interpreted as extended quark anti quark pairs undergoing some sort of displacement.

DarMM said:
Same as in QED, the physical Hilbert space is a subset of the whole unphysical (Hilbert-Krein) space selected by a BRST condition or similar.
Sure but that's the physical Hilbert space of the pertubative approach. It's not really applicable in the same way in lattice qcd for instance.

DarMM said:
I don't think perturbation theory is a physical sector, it's a method of approximating calculations which is very accurate (once you don't include too many terms) within a certain range of momenta and energy, but that's not a physical sector to me.
If pertubation theory didn't map out that corner of the theory (high momentum transfer/deep UV), we wouldn't be able to say much about QCD at all and we wouldn't know anything about asymptotic freedom. Instead I view the pertubative method as the solution for the problem Avodyne mentioned above, namely the limiting behaviour of the theory when you send the QCD coupling constant to zero. So those states must clearly be included in the full Hilbert space of the theory, whatever else you may think of it.
 
Last edited:
  • #57
A. Neumaier said:
But the physical Hilbert space defining the vacuum representation should still be the same, in the sense of carrying an isomorphic representation of the algebra of observables.

This is analogous to the well-known fact from representation theory of finite-dimensional Lie algebras that the same representation can be obtained in very different ways. Nevertheless, the irreducible unitary representations are uniquely determined by a few quantum numbers.

This should definitely be true but note that in more general dualities, you can have a homomorphism of Hilbert spaces where each end carries very different quantum numbers, so the map can be quite nontrivial in general.
 
  • #58
Haelfix said:
Sorry I apologize I misread your statement, but to be fair it was a little hard to follow b/c Lattice QCD *is* a nonperturbative approach and it manifestly does include quarks and gluons. So I'm failing to square the premise a little. How can a nonperturbative approach explicitly calculate the masses of an object quite precisely, and then the very same theory in principle somehow not include those objects?
No worries about the misreading, it's a long thread even I'm not certain what I've said at times! :smile:

What objects does Lattice QCD calculate the masses of that we are saying might not exist here. The discussion here is about quarks and gluons, I don't believe lattice QCD calculates their masses, but takes them as inputs (in the quark case).

Haelfix said:
Sure but that's the physical Hilbert space of the pertubative approach. It's not really applicable in the same way in lattice qcd for instance.
No, it is also a nonperturbative statement, see for example Nakanishi's book on the Covariant Operator formalism for Gauge theories or Strocchi's book "An Introduction to Non-Perturbative Foundations of Quantum Field Theory". It works there just as well. I don't see what is explicitly perturbative about it unless I'm missing something.

Haelfix said:
If pertubation theory didn't map out that corner of the theory (high momentum transfer/deep UV), we wouldn't be able to say much about QCD at all and we wouldn't know anything about asymptotic freedom. Instead I view the pertubative method as the solution for the problem Avodyne mentioned above, namely the limiting behaviour of the theory when you send the QCD coupling constant to zero. So those states must clearly be included in the full Hilbert space of the theory, whatever else you may think of it.
First, I wasn't saying perturbation theory was invalid, just that it's not a sector. It's a method useful for calculating in a given regime of momenta, provided you don't sum too many terms.

Secondly, they simply aren't included in the Hilbert Space of the theory, since their states have negative norm. I'm not following the argument why states existing in a singular limit where the fundamental nature of the theory is lost, e.g. ##g \rightarrow 0## losing the gauge/fiber bundle structure, means those states have to be in the Hilbert space of the theory in general. From the analysis of Strocchi, Balaban and Nakinishi they don't seem to.
 
  • #59
DarMM said:
The discussion here is about quarks and gluons, I don't believe lattice QCD calculates their masses, but takes them as inputs (in the quark case).

I'm far from an expert on QCD, but my understanding is that they can go both ways. The standard way is to take Quark masses as an input and to compute the mass of some Hadron, but conversely you can take the Hadronic spectrum and compute a Quark 'mass'. Where 'mass' as you might expect is a bit of a fuzzy scheme dependent concept inside a strongly interacting composite object (it is certainly not the usual pole in the propagater, considering that there are strongly divergent infrared effects at play). In any event this is an active area of research (see the pdg section on this (p726))
http://pdg.lbl.gov/2015/download/rpp2014-Chin.Phys.C.38.090001.pdf
DarMM said:
No, it is also a nonperturbative statement, see for example Nakanishi's book on the Covariant Operator formalism for Gauge theories or Strocchi's book "An Introduction to Non-Perturbative Foundations of Quantum Field Theory". It works there just as well.

Ok, I will try to check those books out if I get the time.
Cards on the table, my cursory understanding of confinement/infrared slavery basically begins and mostly ends with this paper:

https://journals.aps.org/prd/abstract/10.1103/PhysRevD.10.2445

as well as some of Seiberg and Witten's work on the subject in the context of deformed N=2 Supersymmetry where they explicitly show the analog of the 'dual Meissner effect' which leads to the emergence of an explicit flux tube via Monopole and Dyon condensation. This is afaik, one of the only known examples of a theory where the mechanism of confinement is understood to some extent (maybe some of the experts here know of others).
Note that in this picture, it's somewhat subtle to ask question about the nature of the fundamental degrees of freedom, analogous to the situation in string theory where you can ask a similar sort of question about strings vs Dbranes. In any event, while this picture is known to not be quite right for QCD, it explains many of the desired features (Regge behavior, the linear confining potential etc).

DarMM said:
"I'm not following the argument why states existing in a singular limit where the fundamental nature of the theory is lost, e.g. g→0 losing the gauge/fiber bundle structure, means those states have to be in the Hilbert space of the theory in general. From the analysis of Strocchi, Balaban and Nakinishi they don't seem to. "

Regarding this singular limit, you are probably better served by the discussion in the other thread about asymptotic states in QCD, and the experts there can help you better than I can. In the cartoon version of the flux tube picture this limit corresponds to stretching the tube infinitely far to where it breaks, but when it breaks, it always breaks into smaller pieces of string (b/c you have to put so much energy in that you create a virtual quark-antiquark pair that then rebinds into smaller 'mesons') and thus you would never observe something other than a color singlet state (albeit one that you can make arbitrarily close to 'looking' like a free quark at high momentum transfer for the purposes of scattering experiments). Nevertheless, the quarks are still the constituent probes of that theory as well as what appears in the classical Lagrangian, and the only reason you can never see them must completely lie within the quantum dynamics of the extra QCD self interaction term.

All this to say that I completely agree that quarks as elementary particles are of a very different nature than electrons. However they are also of a different nature than virtual particles and certainly of ghost states so I don't think it's correct to imagine a final nonperturbative theory completely excising them from their description. How could that even possibly work and preserve all of the successes of the eight fold way, deep inelastic scattering and so forth..
 
  • #60
Haelfix said:
All this to say that I completely agree that quarks as elementary particles are of a very different nature than electrons. However they are also of a different nature than virtual particles and certainly of ghost states so I don't think it's correct to imagine a final nonperturbative theory completely excising them from their description. How could that even possibly work and preserve all of the successes of the eight fold way, deep inelastic scattering and so forth..
They seem to not be in the nonpertuabtive Hilbert space of QCD due to having negative norm and mapping out of the Hilbert space into the surrounding Krein space. It seems it is possible to write QCD completely in terms of hardonic-glueball operators on a hadronic-glueball Hilbert space*. Thus the confusion/point of this thread is exactly your final sentence.

*See: Taichiro Kugo, Izumi Ojima; Local Covariant Operator Formalism of Non-Abelian Gauge Theories and Quark Confinement Problem, Progress of Theoretical Physics Supplement, Volume 66, 1 February 1979, Pages 1–130

In essence if quarks are states in the Hilbert space, they can't be confined. See here as well: https://arxiv.org/abs/hep-ph/0105142
 
  • #61
Haelfix said:
you can take the Hadronic spectrum and compute a Quark 'mass'. Where 'mass' as you might expect is a bit of a fuzzy scheme dependent concept inside a strongly interacting composite object (it is certainly not the usual pole in the propagater, considering that there are strongly divergent infrared effects at play). In any event this is an active area of research (see the pdg section on this (p726))
http://pdg.lbl.gov/2015/download/rpp2014-Chin.Phys.C.38.090001.pdf
p.726, which you cite, says that the quark masses determined by lattice QCD are bare masses (i.e., parameters of the bare Lagrangians in a suitable approximation) and not physical masses (poles of the propagator). Trying to get the latter by fits to experiment produces values that violate causality: The Kallen-Lehmann decomposition of the propagator contains complex conjugate quark poles where the squared mass ##m^2## has a negative real part, while causality requires masses to be real and nonnegative. This implies that a quasi-free space based on the Fock construction but with the fitted Kallen-Lehmann decomposition leads to an indefinite inner product. Therefore it is not a Fock space in the sense in which the term is used by mathematical physicists (where a Hilbert space must result) but only an ''indefinite Fock space'' of the kind @samalhayat mentioned in a related thread.

As a consequence, the (non-free) field operators for quarks can also be defined only on a Krein space - i.e., a generaliziation of a Hilbert space obtained by replacing the definiteness condition of the inner product by the weaker nondegeneracy condition.
 
Last edited:
  • Like
Likes dextercioby
  • #62
DarMM said:
many observations (e.g. Deep Inelastic Scattering) look most natural in terms of quarks.
There is an interesting paper by Casher, Kogut, and Susskind in Phys. Rev. D19 (1974), 732-745. They discuss the exactly solvable toy model of QED_2 in 1+1 dimensional spacetime, which exhibits formal asymptotic freedom and confinement of electrons. They show that some sort of ''deep inelastic scattering of electrons'' occurs although the physical Hilbert space contains no electrons. They extend QED_2 to a model with 3 abelian quarks, leading to mesons and baryons.
 
Last edited:
  • Like
Likes DarMM
  • #63
vanhees71 said:
Indeed! In our perception of physical reality there are neither Hilbert and Fock spaces, Lie and other groups in QT, nor configuration and phase spaces, no fiberbundles, Minkowksi and pseudo-Riemannian manifolds in classical physics. These are all description of our perceptions of Nature. It is an astonishing empirical fact that we can order our perceptions (at least the "objective" ones) using these mathematical entities.
Yes. In the formulation of QM I present in the paper linked in my signature, the main object of concern is neither observable (as in standard QM) nor beable (as in the usual formulation of Bohmian mechanics), but a perceptible.
 
  • #64
DarMM said:
In what sense?
At a more fundamental level, I think all particles of the Standard Model should be thought of as quasiparticles, in the same sense in which a phonon (the quantum of sound) is a quasiparticle. See the paper linked in my signature.
 
  • #65
Demystifier said:
At a more fundamental level, I think all particles of the Standard Model should be thought of as quasiparticles, in the same sense in which a phonon (the quantum of sound) is a quasiparticle. See the paper linked in my signature.

I agree; in a strongly interacting quantum many-body system, whether it's a QFT or whether it's the Hubbard model, the excitations one sees at low energy will not have any simple relation to the "fundamental" degrees of freedom which are specified in the Hamiltonian.
 
  • Like
Likes Demystifier
  • #66
vanhees71 said:
Yes sure, there are mathematical problems with (perturbative) gauge models which are solved with some mathematical tools (Faddeev-Popov quantization is more pragmatic, while the operator approach based on BRST is also very illuminating to understand some finer aspects). No matter which mathematical sophistication is necessary, one must not forget that these are all descriptions of nature, not nature itself!
Reading over this thread and noticed I never responded to this.

Although this is true, it's not really related to what was being discussed, which was the term "physical Hilbert space". This is a purely technical term in the quantisation of gauge theories, part of the BRST framework. Although "it's not nature" is true, there's no more need for it with this term than there is for "density matrix". It's just a technical term that has the word "physical" in it.
 
<h2>1. How can quarks exist if they are confined?</h2><p>Quarks are subatomic particles that are confined within the nucleus of an atom. They are held together by the strong nuclear force, which is one of the four fundamental forces of nature. This force is strong enough to overcome the repulsive force between the positively charged protons in the nucleus, keeping the quarks confined.</p><h2>2. What is confinement in relation to quarks?</h2><p>Confinement is the phenomenon in which quarks are bound together within the nucleus of an atom. This is due to the strong nuclear force, which is responsible for holding the nucleus together. Quarks cannot exist freely on their own due to their strong interaction with this force.</p><h2>3. Why are quarks confined?</h2><p>Quarks are confined because they have a property known as color charge, which is similar to electric charge. The strong nuclear force is able to bind particles with color charge together, but it also prevents them from existing independently. This is why quarks are always found within larger particles, such as protons and neutrons.</p><h2>4. Can quarks ever be observed outside of confinement?</h2><p>No, quarks cannot be observed outside of confinement. The strong nuclear force is too strong for individual quarks to exist outside of a bound state. However, scientists have been able to indirectly study quarks through experiments and observations of their effects on other particles.</p><h2>5. How does confinement impact our understanding of the universe?</h2><p>The concept of confinement plays a crucial role in our understanding of the universe. It helps explain why quarks cannot exist independently and how they are able to form the building blocks of protons, neutrons, and other particles. Without confinement, the structure and behavior of matter as we know it would be very different.</p>

1. How can quarks exist if they are confined?

Quarks are subatomic particles that are confined within the nucleus of an atom. They are held together by the strong nuclear force, which is one of the four fundamental forces of nature. This force is strong enough to overcome the repulsive force between the positively charged protons in the nucleus, keeping the quarks confined.

2. What is confinement in relation to quarks?

Confinement is the phenomenon in which quarks are bound together within the nucleus of an atom. This is due to the strong nuclear force, which is responsible for holding the nucleus together. Quarks cannot exist freely on their own due to their strong interaction with this force.

3. Why are quarks confined?

Quarks are confined because they have a property known as color charge, which is similar to electric charge. The strong nuclear force is able to bind particles with color charge together, but it also prevents them from existing independently. This is why quarks are always found within larger particles, such as protons and neutrons.

4. Can quarks ever be observed outside of confinement?

No, quarks cannot be observed outside of confinement. The strong nuclear force is too strong for individual quarks to exist outside of a bound state. However, scientists have been able to indirectly study quarks through experiments and observations of their effects on other particles.

5. How does confinement impact our understanding of the universe?

The concept of confinement plays a crucial role in our understanding of the universe. It helps explain why quarks cannot exist independently and how they are able to form the building blocks of protons, neutrons, and other particles. Without confinement, the structure and behavior of matter as we know it would be very different.

Similar threads

  • High Energy, Nuclear, Particle Physics
Replies
4
Views
2K
  • Quantum Physics
3
Replies
75
Views
7K
  • Quantum Physics
Replies
3
Views
3K
  • Special and General Relativity
Replies
3
Views
859
  • High Energy, Nuclear, Particle Physics
Replies
1
Views
879
Replies
30
Views
2K
  • High Energy, Nuclear, Particle Physics
Replies
9
Views
2K
  • High Energy, Nuclear, Particle Physics
2
Replies
44
Views
9K
  • Beyond the Standard Models
Replies
7
Views
2K
  • High Energy, Nuclear, Particle Physics
Replies
8
Views
2K
Back
Top