Single Quark at Rest: The Mystery of Dark Matter?

[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).

 

[Demystifier]

Quark number is not that well-defined.

But the charge color is well defined. What happens after cooling down of the soup, if the initial hot soup has non-zero charge color?

 

[PeroK]

But the charge color is well defined. What happens after cooling down of the soup, if the initial hot soup has non-zero charge color?

If you could create a colour-charged soup, then QCD would be broken!

 

[Demystifier]

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.

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?

 

[Demystifier]

If you could create a colour-charged soup, then QCD would be broken!

It doesn't make sense. QCD is asymptotically free so nothing prevents colour-charged soup at high temperatures.

 

[vanhees71]

But the charge color is well defined. What happens after cooling down of the soup, if the initial hot soup has non-zero charge color?

You cannot create a non-zero color-charged soup, because you can't have a color-charged object to begin with. Asymptotic freedom only implies that at huge temperatures and/or densities the quarks and gluons move quasi freely within the soup (aka a QGP) but as a whole the soup is color-neutral.

 

[Demystifier]

You cannot create a non-zero color-charged soup, because you can't have a color-charged object to begin with.

Yes you can, see the experiment in #38.

 

[PeroK]

Yes you can, see the experiment in #38.

That experiment reminds me of the various attempts to subvert the UP. Either it would subvert QCD. Or, when you look at the fine detail, QCD would somehow ensure that both halves were colour neutral.

 

[vanhees71]

I doubt that this gedanken experiment can be done in the real world. It's of course a complicated not yet fully solved problem, how "fragmentation" works, i.e., that you always end up with color-neutral states.

 

[Demystifier]

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.

Then where do these intermediate states come from, if not from the theory?

 

[vanhees71]

I'd rather say they exist only within the theory and are not referring to anything observable. Even in non-confining theories like QED the "transient states" of interacting fields have no physical interpretation as observables.

 

[Demystifier]

I'd rather say they exist only within the theory and are not referring to anything observable. Even in non-confining theories like QED the "transient states" of interacting fields have no physical interpretation as observables.

Fine, this reduces the problem to a more elementary problem of interpretation of short living resonances. Are they real or not? Note that they are not virtual states in Feynman diagrams, so according to general postulates of QM they should be real and measurable in principle. They are not measurable by current technology, but could be measurable with future technology. In fact, there is no well defined limit how short living a particle must be to call it "resonance", the borderline between particle and resonance is not sharp.

 

[Demystifier]

That experiment reminds me of the various attempts to subvert the UP. Either it would subvert QCD. Or, when you look at the fine detail, QCD would somehow ensure that both halves were colour neutral.

I doubt that this gedanken experiment can be done in the real world. It's of course a complicated not yet fully solved problem, how "fragmentation" works, i.e., that you always end up with color-neutral states.

I don't think that color neutrality is such a fundamental principle as you think it is. I don't see any theoretical argument for that. Given that QCD is asymptotically free, I don't see why should both halves of hot quark-gluon plasma be color neutral.

 

[PeterDonis]

Now separate A and B at a large spatial distance from each other

This would require adding energy to the system which would cause the creation of additional quarks/antiquarks to make both sides color neutral. It's no different from the thought experiment of trying to pull apart a meson or baryon to obtain free quarks: you end up with additional quarks being created so you get multiple mesons or baryons instead of free quarks.

Similar remarks would apply to any attempt to prepare an initial state that was not color neutral.

 

[Demystifier]

This would require adding energy to the system which would cause the creation of additional quarks/antiquarks to make both sides color neutral.

Why? The quark gluon plasma is in the high-temperature asymptotically free phase, i.e. the particles are almost free.

 

[vanhees71]

Fine, this reduces the problem to a more elementary problem of interpretation of short living resonances. Are they real or not? Note that they are not virtual states in Feynman diagrams, so according to general postulates of QM they should be real and measurable in principle. They are not measurable by current technology, but could be measurable with future technology. In fact, there is no well defined limit how short living a particle must be to call it "resonance", the borderline between particle and resonance is not sharp.

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

 

[Demystifier]

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

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

 

[PeterDonis]

Why?

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

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

 

[Demystifier]

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

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

 

[PeterDonis]

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

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

quark-gluon plasma does not need to have infinite volume

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

 

[Demystifier]

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

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

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

 

[Demystifier]

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

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

 

[mitchell porter]

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

 

[PeterDonis]

Why the separation requires adding energy?

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

 

[PeterDonis]

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

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

 

[PeterDonis]

how about my proof in #26

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

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

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