Single Quark at Rest: The Mystery of Dark Matter?

[MatthewKM]

TL;DR Summary

Hypothetically, what is the fate of a single Quark at rest?

I find it fascinating that all nucleons are three quark configurations. It is proof that each quark needs two others to remain stable. The hypothesis that early in the big bang there was a uniform foam of quarks and coalescence of threes formed protons begs the question: what percentage of quarks we not so lucky and escaped the foam as single quarks. If it were possible to imagine a single quark at rest in a complete or near complete vacuum, what would that quark do? Would it collapse to a finite point or particle? Would it dissipate? Could it's collapse be to dark matter? Could it's dissolution be to dark energy?

 

[PeroK]

Summary: Hypothetically, what is the fate of a single Quark at rest?

I find it fascinating that all nucleons are three quark configurations. It is proof that each quark needs two others to remain stable.

It's part of a more general hypothesis called colour confinement:

https://en.wikipedia.org/wiki/Color_confinement

The hypothesis that early in the big bang there was a uniform foam of quarks and coalescence of threes formed protons begs the question: what percentage of quarks we not so lucky and escaped the foam as single quarks.

None, according to QCD (Quantum Chromodynamics).

If it were possible to imagine a single quark at rest in a complete or near complete vacuum, what would that quark do?

It isn't possible to imagine a single isolated quark within QCD. The quark field in the early universe had a lot of quarks!

Would it collapse to a finite point or particle? Would it dissipate? Could it's collapse be to dark matter? Could it's dissolution be to dark energy?

None of the above.

 

[PeterDonis]

It is proof that each quark needs two others to remain stable.

No, it isn't. Mesons have only two quarks.

Finding the stable quark states requires, as @PeroK has pointed out, looking at the implications of color confinement.

what percentage of quarks we not so lucky and escaped the foam as single quarks.

None. Single free quarks are not possible. That is one of the implications of color confinement.

If it were possible to imagine a single quark at rest in a complete or near complete vacuum

Since it isn't, such questions are meaningless. You can't ask the laws of physics what would happen in a situation that is prohibited by the laws of physics.

 

[Demystifier]

Since it isn't, such questions are meaningless. You can't ask the laws of physics what would happen in a situation that is prohibited by the laws of physics.

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

It is not much different from the fact that $e^{ikx}$ is not a Hamiltonian eigenstate of the harmonic oscillator, but in principle it can be an initial state in the (rigged) Hilbert space and one can study how such initial state evolves with time under harmonic oscillator Hamiltonian.

Of course, such an initial state is probably impossible to prepare in the laboratory, so the question is not very interesting from the practical phenomenological point of view. But theoretically, I think the question makes sense.

 

[Demystifier]

It isn't possible to imagine a single isolated quark within QCD.

Nobody required it to be isolated.

 

[PeroK]

Nobody required it to be isolated.

If it were possible to imagine a single quark at rest in a complete or near complete vacuum,

 

[Demystifier]

@PeroK his words "in vacuum" can be interpreted in different ways. I interpret it as having a state with exactly one real particle (the quark). Mathematically, such a state exists in the QCD Hilbert space.

 

[PeroK]

@PeroK his words "in vacuum" can be interpreted in different ways. I interpret it as having a state with exactly one real particle (the quark). Mathematically, such a state exists in the QCD Hilbert space.

So, how does a quark evolve?

 

[Demystifier]

So, how does a quark evolve?

I don't know, the calculation of that is certainly not simple, and since practical experiment is impossible it's very likely that nobody computed it.

Of course, any initial state evolves according to
$$|\psi(t)\rangle = e^{-iHt}|\psi(0)\rangle$$
but evaluating this for the full QCD Hamiltonian $H$ when $|\psi(0)\rangle$ is not the Hamiltonian eigenstate is anything but simple.

 

[PeroK]

I don't know, the calculation of that is certainly not simple, and since practical experiment is impossible it's very likely that nobody computed it.

I don't see what the possibilities are. It can't decay, can it? It can't collapse to dark matter or dark energy, as asked by the OP. Why would it be any different from a single electron? Apart from the fact that the state appears to be disallowed by the requirement for colour confinement.

And, it seems to me, that if you drop the requirement for colour confinement, then QCD would give very different predictions.

 

[Demystifier]

And, it seems to me, that if you drop the requirement for colour confinement, then QCD would give very different predictions.

The confinement is a criterion for stable states, not for any states.

 

[Demystifier]

It can't decay, can it?

It can, provided that energy is conserved. Note that energy is defined by the full interacting Hamiltonian (not by the free one), so the single quark is not an energy eigenstate and its average energy is rather large.

 

[PeroK]

Why? It's not a Hamiltonian eigenstate so some kind of decay seems unavoidable.

So, decay is unavoidable, but it has nothing to decay into. There are no smaller charged particles to conserve charge, for example. The way out of the conundrum is to impose colour confinement and disallow the initial state. The point about QCD is that the components of the theory appear to come as part of a package - unlike QED. The constituents of QCD (quarks, gluons and colour charge) cannot be directly detected. They are only implied by the mathematical model.

You are perhaps putting a Bohmian interpretation on this, that if they are mathematically part of the theory then they must have an independent existence. The orthodox position, IMO, is that unless you have a way to relax colour confinement the quarks and gluons cannot be studied in isolation. Quarks and gluons have no independent existence outside the strictures of QCD.

 

[Demystifier]

There are no smaller charged particles to conserve charge, for example.

By smaller, you probably mean lighter. But note that a single quark in full QCD is not an energy eigenstate, it is a superposition of states with different energies and its average energy is large. Therefore states with smaller energy certainly exist.

 

[Demystifier]

You are perhaps putting a Bohmian interpretation on this, that if they exist as part of the theory then they must have an independent existence.

I see nothing particularly "Bohmian" about such a view.

 

[Demystifier]

The constituents of QCD (quarks, gluons and colour charge) cannot be directly detected.

That's not an issue here. The issue is whether such states exist in the Hilbert space. A good analogy is a Schrodinger cat. It cannot be directly detected, but it is a legitimate state in the Hilbert space. Such a state cannot be prepared in practice, but it's not impossible in principle. Even orthodox QM accepts that a Schrodinger cat state is possible in principle, despite the fact that in practice it decays (decoheres) very fast to a cat which is either dead or alive.

 

[vanhees71]

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

It's not observable within QCD, because it's a gauge-dependent quantity. Only "colorless states" can be observable. Correspondingly we observe never single quarks or gluons but only the asymptotically free hadron states (mostly mesons, bound states of a quark and an antiquark, and baryons, bound states of three quarks).

 

[Demystifier]

It's not observable within QCD, because it's a gauge-dependent quantity. Only "colorless states" can be observable.

It's like saying that in a Poincare invariant theory only zero momentum state can be observable, because only zero momentum is Poincare invariant.

 

[vanhees71]

Poincare symmetry is a true global symmetry and has nothing to do with what's observable and what not.

A local gauge symmetry is not a true symmetry but indicates an redundance of description, i.e., a gauge-dependent quantity is not completely defined by the physical situation you want to describe and thus cannot directly describe an observable.

 

[Demystifier]

It's not observable within QCD, because it's a gauge-dependent quantity. Only "colorless states" can be observable. Correspondingly we observe never single quarks or gluons but only the asymptotically free hadron states (mostly mesons, bound states of a quark and an antiquark, and baryons, bound states of three quarks).

Instead of QCD, consider a Yang-Mills theory which is not confining and asymptotically free, e.g. weak SU(2) or something like that. Instead of color it has weak isospin, or something like that. In such theories, the states don't need to have zero isospin. This illustrates that color neutrality in QCD is not a simple consequence of symmetry. Instead, it is a complicated non-perturbative effect of interactions, a rigorous explanation of which is worth a Millenium prize. Moreover, it is only a conjecture, not a mathematically proved fact.

Furthermore, the top quark decays weakly so fast that it does not have enough time to form a hadron, in this sense the top quark behaves as a single particle. https://en.wikipedia.org/wiki/Top_quark#Decay

 

[vanhees71]

The top quark thus can't be observed as an asymptotic free state either!

Then again local gauge invariance does not describe a symmetry of physics but a redundancy in the description of a physical situation. Although it's unfortunately the slang in particle physics there's thus also no spontaneous breaking of a local gauge symmetry (Elitzur's theorem) but the Higgs mechanism instead, i.e., the "would-be Goldstone modes" are "eaten up" by the appropriate gauge fields making the corresponding vector bosons massive. As in any gauge theory also in electroweak theory gauge-dependent quantities are not representing observables.

Of course, you are right in saying that the local gauge symmetry of QCD does not prove confinement, but confinement is how the non-observability of gauge-dependent quantities is "realized" in this case. The latter paradigm holds of course also for the electroweak sector, which is non-confining.

 

[malawi_glenn]

https://www.physicsforums.com/insights/how-to-avoid-breaking-physics-with-your-what-if-question/
always a good read

 

[Demystifier]

Of course, you are right in saying that the local gauge symmetry of QCD does not prove confinement, but confinement is how the non-observability of gauge-dependent quantities is "realized" in this case. The latter paradigm holds of course also for the electroweak sector, which is non-confining.

How is non-observability of gauge-dependent quantities realized in non-confining theories?

 

[Demystifier]

https://www.physicsforums.com/insights/how-to-avoid-breaking-physics-with-your-what-if-question/
always a good read

In this case, I believe that the question belongs to the type 3, while all others here seem to think that it belongs to the type 1.

 

[PeroK]

In this case, I believe that the question belongs to the type 3, while all others here seem to think that it belongs to the type 1.

Let me draw an analogy. You could consider electrons without the PEP and study the non-chemistry that results. But, that's not part of the theory of QM. QM needs the PEP to constrain the states of multi-electron systems. Eventually, of course, the PEP can be justified by QFT. But, that's not the point.

Likewise, you could relax QCD and study free quarks. I don't believe that's part of the theory - for the same reason. QCD needs colour confinement in the same way that QM needs the PEP.

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

A single quark is no more legitimate than two electrons in a non anti-symmetric state.

 

[Demystifier]

A single quark is no more legitimate than two electrons in a non anti-symmetric state.

I don't think it's a fair analogy.

For example, wikipedia https://en.wikipedia.org/wiki/Color_confinement starts with the following claim: "In quantum chromodynamics (QCD), color confinement, often simply called confinement, is the phenomenon that color-charged particles (such as quarks and gluons) cannot be isolated, and therefore cannot be directly observed in normal conditions below the Hagedorn temperature of approximately 2 terakelvin (corresponding to energies of approximately 130–140 MeV per particle)."

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.

In fact, here is a simple proof that color charged states must exist. 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. Hence the Hilbert space of the free theory is unitary equivalent to the Hilbert space of interacting theory. But the free theory obviously contains color-charged states, so it follows that the interacting theory also must contain color-charged states. (The color charge is conserved by both free and interacting QCD, so the color charge of a state cannot change by turning on the interaction.) Of course, this proof says nothing about energy/mass of such color-charged states. Their energy/mass can be very large, which can explain why we cannot easily produce them.

Another, more physical, argument for their existence is this. Since QCD is asymptotically free, if follows that color charged states is possible for a soup of quarks at a very large temperature. Now consider what happens if this initially hot color charged state starts to cool down ...

 

[vanhees71]

How is non-observability of gauge-dependent quantities realized in non-confining theories?

It's self-evident that a quantity that is not completely determined by the given physical situation cannot be an observable. Also already within QED (and of course also in classical Maxwell theory) only gauge-invariant quantities are observable. There's simply no measurement device that can measure a gauge-dependent quantity.

 

[gentzen]

A single quark is no more legitimate than two electrons in a non anti-symmetric state.

I don't think it's a fair analogy.

I think this is the crucial question whether this analogy is appropriate or not. But it is not a question of "fair". The expectation is that a local gauge symmetry must be present, and that a single quark would imply the absence of that local gauge symmetry. I don't know who is right in this case.

The argument of vanhees71 seems to be that "observability" (or "physical relevance") of a single quark would constitute a breaking of that local symmetry. So the expected symmetry would be absent, and that would be analogous to "two electrons in a non anti-symmetric state".

 

[PeroK]

"In quantum chromodynamics (QCD), color confinement, often simply called confinement, is the phenomenon that color-charged particles (such as quarks and gluons) cannot be isolated, and therefore cannot be directly observed in normal conditions below the Hagedorn temperature of approximately 2 terakelvin (corresponding to energies of approximately 130–140 MeV per particle)."

At those energies, you do not have a well-defined quark number of $1$. Isn't that the point? When you collide protons at high energy it should be physically possible to get a free quark. But, because of the energy involved to split the proton there is always available energy to create enough quarks so that they group into mesons or baryons. Without colour confinement, just by luck you'd get a free quark every now and then.

So, yes, you can have a soup of ultra-high-energy quarks that do not stabilise into mesons and baryons. But, you cannot have just the one.

It's not as hard-and-fast as the PEP, perhaps, but without (low energy) colour confinement the theory falls apart.

 

[Demystifier]

It's self-evident that a quantity that is not completely determined by the given physical situation cannot be an observable. Also already within QED (and of course also in classical Maxwell theory) only gauge-invariant quantities are observable. There's simply no measurement device that can measure a gauge-dependent quantity.

Yes, but my point is this. In QED we gauge-invariant states with non-zero electric charge. In SU(2) weak theory we have gauge-invariant states with non-zero weak isospin "charge". By analogy, in QCD there must exist gauge-invariant states with non-zero color charge. Indeed, QCD does not say that such states don't exist, it only says that they are unstable at "normal" conditions. At high temperatures, where QCD is asymptotically free and the confinement is absent, such color charged states become stable.

Another useful analogy is this. 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.

 

[PeroK]

Indeed, QCD does not say that such states don't exist, it only says that they are unstable at "normal" conditions.

My understanding is that these states are not part of the theory. Not that they exist but decay too fast for current experiments. This is why I made reference to you as a "Bohmian" early. If you take QCD as a realist theory, where quarks exist with hidden classical trajectories, then there must be free quarks for at least a short time. But, if you take QCD as an abstract mathematical theory, then these intermediate states are not part of the theory.

 

[vanhees71]

In QED the current density $\hat{j}^{\mu}(x)=q \bar{\psi} \gamma^{\mu} \psi$ is a gauge-invariant quantity. As well it fulfills the microcausality condition. Thus electric charge (as well as electric-charge-current densities) are observable (in this case in principles as well as in Nature of course).

The "color-current density" is $\hat{j}^{a,\mu}(x) = q \bar{\psi} \gamma^{\mu} \hat{T}^a \psi$ is obviously not gauge invariant and as such cannot be an observable, no matter whether the theory is confining or not.

A somewhat funny example from electroweak theory are neutrinos. On the one hand the only asymptotic free and thus observable states from a very fundamental point of view (Poincare symmetry) are the mass eigenstates. These, however cannot be produced, because according to the weak interaction only flavor eigenstates can, but these are not gauge invariant.

There's of course no problem, because what's really observed is a reaction of asymptotic free particles creating a neutrino, which then at the detector is absorbed again and what's observed are some other asymptotic free particles. In reality we never observe "free neutrinos" but only that at one place one is created and at another place one is absorbed with a specific flavor, and the flavor in the initial state needs not be the same in the final state, which we call "neutrino oscillations". All the confusing debates about energy-momentum conservation and all that in neutrino oscillations are immediately solved, when thinking in this way about, what's "really observed" when we say we "observe neutrino oscillations".

 

[Demystifier]

At those energies, you do not have a well-defined quark number of $1$. Isn't that the point? When you collide protons at high energy it should not be physically impossible to get a free quark. But, because of the energy involved to split the proton there is always available energy to create enough quarks so that they group into mesons or baryons. Without colour confinement, just by luck you'd get a free quark every now and then.

So, yes, you can have a soup of ultra-high-energy quarks that do not stabilise into mesons and baryons. But, you cannot have just the one.

It's not as hard-and-fast as the PEP, perhaps, but without (low energy) colour confinement the theory falls apart.

Fine, but suppose 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?

 

[PeroK]

Fine, but suppose 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?

Quark number is not that well-defined.

 

[vanhees71]

That can never happen, because quarks or any colored object cannot be prepared in the first place. In the case of QCD it's manifest as confinement. Though we admittedly don't fully understand this non-perturbative property (but at least from a lattice-QCD point of view there's utmost strong theoretical evidence for confinement and from a experimental point of view anyway), it's clear from confinement that we can only use color-neutral objects to initiate the creation of strongly interacting matter, as we do in heavy-ion collisions, e.g., at the LHC. You start with color-neutral Pb nuclei (gauge-invariant bound states of quarks and gluons if you wish) and all you can create is a soup of "quarks and gluons" that all together are color neutral too. There's a lot of evidence that indeed for a very short time (a few 10 fm/c) a rapidly expanding fireball is created, where the relevant "thermodynamic" degrees of freedom are quarks and gluons (but rather in the sense of in-medium quasi-particles, but even the quasi-particle picture is somewhat questionable, because they most probably have a rather broad mass spectrum), but all that can be observed are the colorless hadrons (and of course some leptons and photons), and one must deduce the properties of the hot and dense quark-gluon-plasma state from the measurable patterns of hadrons (particle abundancies, $p_T$- and rapidity-spectra, and all that).