Thank you Martin,

My question is if QCD has a property or a theorem which explains clearly what is the reason to have only baryons or mesons with color charge zero (I suppose that this is your worry with the "color algebra" that you mentioned) in the particles seen yet. It seems that there could be more combinations without preventing the quantum conservation numbers.

Dear Martin,

Thank you very much and I know that QCD quarks are confined and even with strong interaction among the gluons given also the asymptotic freedom of the quarks. The question for me is to know how is so stable the configuration of baryons and mesons when in fact you have eighteen quarks if you consider the flavor (SU(2) symmetry given 3 infinitesimal generators) and color (SU(3) symmetry with 8 infinitesimal generators). I do not know what is the reason to have only 2 or 3 quarks in the hadrons.

Dear Joachim,

Thank you very much for this information but it seems that they are all resonances or excited states no stable at all. What is the reason that they cannot be stable states as the baryons or the mesons?

Dear Daniel, in the representation modes of SU(3) as the governing symmetry of strong interaction, the emerging irreducible representations respect to hyper-charge (Y) and isospin (I3 ) coordinate,include :{3} , {6}, {8} , {10} , {27} and conjugates which represent the various multiplete i.e. baryons and mesons.Among these representation the triplet {3} and its conjugate remained as a mystery with non integer charge (due to Gellmann-Okubo formula) till the assumption of Zweig stated that these triplet are also some real multiplete called quarks ( of Mr mark !) . interestingly these {3} representation are "constituent" of all other hadrons, and  when we check  other multiplete representations, simply the possible numbers of quarks in {8} ,{10} per particle etc. are exactly 3 for baryons octet and 2 for meson. other numbers can not be included in these representations.

The answer to the question as posed is that, first of all, there isn't any mathematical or physical obstruction to the existence of multiquark bound states, with more than three quarks. It is, simply, not so easy to make them-tetraquark states have just been observed.

Multiquark states can decay into any lighter states, that are consistent with known conservation laws. So  tetraquarks can decay into two mesons-but not into a baryon and a quark. This would violate baryon number and, eventually, electric charge-unless tetraquarks carried fractional charge.

The lightest baryon  is, thus, stable-and this turns out to be the proton. Were the up quark heavier than the down quark, then the proton would be unstable and the neutron stable-and chemistry completely different.

Extensions of the Standard Model, generically, predict baryon number violation, since quarks and leptons belong to the same multiplet. An open problem is how to parametrize this appropriately, since fine tuning is necessary, in order to obtain a decay rate consistent with the stability of matter.

The relation between the electric charge of the proton and the electron is well understood. The electroweak interactions of the quarks alone are inconsistent, when quantum effects are taken into account; similarly, the electroweak interactions of the leptons alone are inconsistent. The electroweak interactions are only then consistent when quantum effects involving quarks and leptons are taken together-and the way this is done imposes constraints on the electric charges of quarks and leptons. It is this consistency relation that implies that the electric charge of the proton is opposite of that of the electron, even when all corrections are taken into account.

There is a long standing question of existence of dibaryons, proposed as long ago as in 1977, by R.Jaffe (http://en.wikipedia.org/wiki/Dibaryon). So from first principles they are allowed to exist but probably they are not stable enough to be detected or they escape detection due to other reasons (strange decay branching ratios or unexpected long/short liftime). ALICE (at LHC) is going to publish soon results on the searches of such exotic states, putting new stringent limits on their production yields. 

Well dibaryons certainly exist, namely as  the deuteron ion (whose nucleus consists of a proton and a neutron); and an example of a ``tribaryon'' would be the tritium ion.  So it will  be interesting to read what exactly ALICE will report on. 

If we agree on the "strong" interaction as the main force acting through gluon between quarks, then the "allowable"  number of quarks in members of multiplete restricted to 2 or 3 quarks, (this can be traced out in any related reference of SU(3) symmetry) therefore we should separate the " stable" and allowable hadrons, an unstable hadron may be allowable but any allowable hadron is not necessarily stable.

No, there isn't any such constraint. The only constraint, assuming quark confinement, is that the resulting configuration be a color singlet-which includes glueballs, i.e. bound states of gluons only.

Regarding stability, only baryon number conservation applies, which implies that the lightest baryon, i.e. the proton,  is stable.

@Stam   Deuterons are composed of 2 nucleons but are not dibaryons in the sense of a quark structure (H-dibaryon is a 6 quark bound system - uuddss). By analogy, like a binary star system cannot be considered as a single star. Nuclei production in high energy nucleus- nucleus collisions is another topic of great interest, studied also by ALICE. Here at work is so called a coalescence mechanism in a high density hadronic final state created after the hadronization of the quark gluon plasma. Actually one is able to go up to a helium nucleus. 

I imagine that there is some semantics here, about what is (or isn't) classified as dibaryon, so as to, apparently, exclude the, more mundane, deuteron-though i can't see how this can be done in any really useful way.

Any dibaryon is a 6 quark bound state, of course. I'm looking forward to reading the ALICE results. 

Thank you very much to everybody for the interesting contributions!

But my basic question is not resolved yet, sorry. It is well known that the baryons are not stable particles. Without involving the light quarks we could have almost the same lifetime for all of them, i.e. without involving the weak interaction. The proton is uud (one u blue and the other red) while the neutron ddu (one d red and the other green). The lifetime of the proton is practically infinite while the one of the neutron is 15 minutes if it is isolated or practically infinite if it is within a nucleus, i.e. interacting with other quarks belonging to protons or neutrons.

On the other hand, the masses of the quarks up and down are very similar which seems to mean that the stabilities need more ingredients than the pure counting of quarks. This physical scenario doesn´t prohibit the existence of more quarks for creating new fundamental heaviest particles with more than three quarks. But the reality says that this is not so simple and there are difficulties that I do not take into account. This is the key of my question.

Doesn't anybody question whether the quarks are actual particles in the first place?

Think about it, quarks are only seen during high energy collisions and only in laboratories. Their lifetime is on the order of 10^-12 second, after which they disappear and are never seen again. The fleeting observed quark mass is variable. Doesn't this sound more like debris from an accident, rather than fundamental particles?

What can we do with quarks? Do they serve any useful purpose besides occupying our time? It is as though scientists studying quarks are high paid versions of little children observing sparks from a grinding wheel.

It makes no sense that perfectly homogeneous protons having a lifetime of billions of years would be made up of variable sized particles existing on the order of 10^-12 second that completely disappear.

The reason why only singulett color states are allowed in QCD is indeed not rigorously proven. However, there is strong evidence that it is the property of the QCD vacuum. That vacuum acts like a medium that expells colour fields similarly to the ground state of a superconductor that expells magnetic fields from the superconducting material. An Attempts to analytically construct along these lines the QCD vacuum state (i.e deriving in the simplest case an analogon to the Bethe-Salpeter equation) have - to my knowledge - failed so far. The intuition to guide the calculations is obviously obscured by clouds of indices from the various symmetry groups involved.  In Lattice QCD you can observe confinement but the insight behind that is of course rather limited.

The reason why multi-quark systems are unstable might also be attributed to the role of the vacuum. A baryon is a bubble in the QCD vacuum within which a normal (QED) vacuum  is established that allows quarks and gluons to interact freely. As the QCD vacuum is by its very definition the state of lowest energy, the energy density of the bubble will be positive in relation to the surrounding space, and the outer QCD vacuum will exert inward pointing pressure on the boundary of the bubble trying to diminish the overall energy.  The mass of the bubble will inrease with its volume, i.e. with the third power of its radius. A decay into smaller bubbles will most likely have energetic advantage and be the natural decay channel for large hadrons made of multiple quarks and gluons.

There are too many issues mixed up here-that are well understood and are the topic of a course in particle physics. Some statements are wrong: the proton  is a color singlet-the u's and the down quark do not have a well defined color charge, but are in a superposition that is a singlet; similarly for the neutron. The reason the proton is stable is baryon number conservation, since it is the lightest baryon and the reason the neutron can decay is through weak interactions, that do, also, conserve baryon number (and lepton number). 

That the up and down quarks have very similar masses (which isn't a well-defined statement, however, since they are in bound states) can be understood from the fact that isospin symmetry is a very good approximation-which is, also, reflected in the fact that the pions can be understood as the ``almost Goldstone bosons'' of its spontaneous breaking. 

Dear Stam,

It seems just logic to think that if the mass of uud and ddu are almost the same, this implies that the mass of u is almost the same than d. Leaving the quantum numbers apart.

@Stam

Not sure which issues you are addressing. Weak interaction, baryon number conservation, quark mass relations - all that is correct and has influence on the stability of hadrons. The role of the vacuum for the confinement is fundamental though. 

It's not obvious, since electromagnetic (and weak)  interactions distinguish the u and the d quarks-and their electric charges differ by a factor of 2, for instance. 

Dear David Vasak,

I do not understand how the vacuum (state) can expels the colour fields, sorry. It is true that there is a so counterintuitive energy exchange as that the neutron is with a mass of 939 Mev while the W boson of the weak interaction is 80 GeV and also the gluons are very massive interacting among them in the corresponding vacuum. I cannot have an idea of how the interaction or decay in Yang-Mills is vacuum is given because the states are not possible to give as in the electronic case of Schrödinger or Dirac equations. 

Dear David Wise,

Although the quarks are a well established fundamental particles, I understand your worries. In fact it is another aspect of the same issue. I do not distinguish how the quarks can have so different properties being as hadron in nucleus or outside of it. Or how the quarks in one nucleus can move but they do not combine to obtain new kind of particles.

Perhaps you are right and we observe them just in scatterings of huge energy and this prevent us to understand what is their actual structure in cases as in the inner part of a nucleus.

Hydrogen tell us how large is the stability of isolated protons and deuterium also proves that the combination of a proton with a neutron enlarges the stability of the neutron. 

Although most of the particles are charged, but is not include the color. Thus the experimental particles are divided into two categories: no color binary (meson) and ternary(Baryon).  In other words, due to a phenomenon known as color confinement, quarks are never directly observed or found in isolation; they can be found only within hadrons, such as baryons (of which protons and neutrons are examples), and

mesons. Of course The existence of "exotic" hadrons with more valence quarks, such as tetraquarks (qqqq) and pentaquarks (qqqqq), has been conjectured but not proven.

there are hadrons with four quarks and six quarks now experimentally known... 

In principle color singlet states can be formed of 4 quarks (tetraquarks) or 4 quarks and one antiquark (pentaquarks). The experimental search of those states was an important topic few years ago. As far as I know pentaquarks have not been found, whereas the situation is a bit less clear for tetraquarks.

The LHCb Collaboration has published on the 24 April (2014) a paper which shows that the f0(500) and the f0(980) mesons, long thought to be four quark states (tetraquarks) made out of the light udud quarks (f0(500)) or susu quarks (f0(980)), are not consistent with being tetraquarks. The four quark states cannot be classified within the traditional quark model in which the strongly interacting particles (hadrons) are formed either from quark-antiquark pairs (mesons) or three quarks (baryons). They are therefore called exotic particles.

@Daniel

To understand the mechanism of confinement as a property of the vacuum you may think of the analogy of superconductivity (BSC theory, you will find more in any solid state physics textbook). In detail I described that analogy in my dissertation that you can download from this portal. It is in German, though. You may also download my work on the microscopic investigation of hadron fission in the simple quark bag model (Fission of hadrons by Spontaneous Quark-Antiquark Pair Production in Supercritical Colour Fields ). The quark bag model is a phenomenological description of the vacuum bubble picture.

Daniel,

You said:

"The lifetime of the proton is practically infinite while the one of the neutron is 15 minutes if it is isolated or practically infinite if it is within a nucleus, i.e. interacting with other quarks belonging to protons or neutrons.

On the other hand, the masses of the quarks up and down are very similar which seems to mean that the stabilities need more ingredients than the pure counting of quarks. ... But the reality says that this is not so simple and there are difficulties that I do not take into account."

So, you answered your own question.  The quark model is a rather simplistic view of hadrons and cannot explain all observed hadronic phenomena.  Quantum Chromodynamics (QCD) is the best explanation we have so far in that, so far as we can compute, it reproduces all observed hadronic phenomena.  In QCD, hadrons are described by their quantum numbers and the number of quarks is not a good quantum number.  In QCD, in principle one could calculate the fraction of time a proton spent in it's valence state of three quarks relative to the time it spent as three valence quarks plus a quark-antiquark pair and so on.  And, as you stated, these computable quantities can depend significantly on the slight differences between the masses of the quarks or their electrical charge.

Also, to echo a previous commentor, we see many stable multiquark states all the time, they are the nuclei: the deuteron, triton, helium-3, ...  A very interesting question in QCD is whether the di-neutron could actually form a bound state.  The answer appears to be quite sensitive to small differences in the quark masses.

George,

do you think that dineutron should be considered a well-defined entity? Why to produce two neutrons (three quarks each) when it could be possible other combinations if we have many protons and neutrons in a nucleus?

David Vasak,

Could you give one idea of what kind of vacuum do you refer when you mentioned superconductivity? In BCS theory what is important is the electronic bosonization. Perhaps you are thinking in the Ginzburg-Landau formalism where the spontaneous symmetry broken is made for producing goldstones. I do not know, please could you give a simple explanation because your papers are too difficult for me. 

The ``di-neutron'' is, certainly, a well-defined combination of six quarks and, as in any similar situation, it does make sense to ask, whether there could exist a configuration of 4 down quarks and 2 up quarks that has less energy than two neutrons-this is what a bound state would mean. Of course this is very challenging to compute-but, perhaps,  experiments with ultra cold neutrons could be done, cf., for example, .http://www.aps.org/units/dnp/research/harvarducn.cfm

Stam,

Till what I know, the di-neutrons are always inestables, how long is their lifetime?

Well, since they haven't been unambiguously discovered yet the answer is not known. Apparently there was an announcement a couple of years ago, cf. here 

https://agenda.infn.it/getFile.py/access?contribId=29&sessionId=12&resId=0&materialId=slides&confId=3627

but I don't know, whether it has been confirmed. 

Daniel,

As an experimental matter in the real world, this question of stability was already answered above: in the presence of electroweak interactions, there are no stable hadrons except the proton and larger (stable) nuclei.  One can discuss hypothetical stability in a world where there is only QCD or maybe QCD and QED and it is more than  just an academic exercise because it helps us understand the nature of the unstable hadronic resonances produced in experiments.

With regards to various exotic dibaryons, the theoretically interesting question is under what conditions do the dibaryons represent the ground state of the two-baryon system.  In experiment, it clearly is true for the deuteron and is ambiguous for the other cases, which suggests in the real world it's a close call.  So, what happens when we move away from real world conditions like turn off QED or vary the quark masses? It turns out that under conditions only slightly different from the real world, other dibaryons can be stable. Understanding these cases may ultimately help us understand the experimental situation better.

Here's a paper by some people doing this kind of study, but it's by no means the final answer:

http://dx.doi.org/10.1103/PhysRevD.85.054511

George,

I am not at all expert in this subject and my knowledge comes from QFT. Therefore I am quite surprised when you write sentences as: "in the presence of electroweak interactions, there are no stable hadrons except the proton and larger (stable) nuclei". For me the electroweak interaction is always present. If you have a d quark always can decay in a u quark taking into account the weak interaction and this is done quite quickly if you have only a combination of ddu without being constrained to be in a nucleus. The question for me is how there are no more stable states or quasi-stable containing more quarks than three? The infinitesimal generators of the color symmetry are 8 and the flavor 3, are they directly related with this situation for states of the Yang-Mills equations?

@Daniel

To be honest what I can say about the structure of a "superconducting" QCD vacuum is pretty vague. BCS theory describes a mechanism based on electron interaction with the ion lattice in the solid body. At low temperatures the positively charged ions vibrate slowly and can take along the electron gas. The electrons respond to the attractive force of the ions in sync. The interaction between two electrons, mediated by the ion lattice,  becomes effectively atrractive rather than repulsive as would be expected from the pure Coulomb interaction. The electron gas becomes a collective state coupled to the lattice vibrations. This is what you call the pairing mechanism or bosonisation. If a magnetic field penetrates into that environment the electron pairs start revolving around the magnetic lines and generate a magnetic field that effectively cancels out the external field within the bulk of the crystal.   The superconductor repells magnetic fields (0 magnetic permeability).

I imagine that the color algebra does not require an ion lattice in order to create singlet combinations of quarks interacting via gluons - in a vague analogy to "Cooper pairs". These combinations in order to condense into a collective QCD vacuum state must of course get rid of all quantum numbers, i.e. spin, isospin/flavor, etc. That collective vacuum state must also be randomized with respect to the Poincare group (so called Spaghetti vacuum). If color fields penetrate into the vacuum the condensate's induced dynamics would neutralize them ensuring confinement.

The question is based on a misunderstanding. The down quark *can* decay into an up quark through electroweak interactions-but the proton (a uud color singlet) can't decay to a uuu color singlet, because the uuu hadron (called the Delta++) is, in fact,  heavier. This latter baryon, in fact, decays to a proton and a charged pion. 

Multiquark bound states do exist as nuclei and, apparently, as bound states of the strong force (e.g. tetra quarks). Their stability is controlled as, in all cases, by conservation laws. So the correct question is how to understand the stability of nuclei in terms of the strong force. Color and flavor are internal quantum numbers that are independent. 

David,

Thank you for your kind answer but still I do not understand the analogy with superconductivity. In BCS the electrons close to Fermi level have an attractive interaction due to the Frölich phonons as you said for forming Cooper pairs, but this not related with the vacuum (fundamental state, i.e. lowest energy eigenvalue) in their electronic structure or in BCS.

Respect to Messner effect, i.e. the diamagnetic properties of superconductivity is also out of vacuum, in fact here it is necessary to distinguish superconductors of type I or II ,because former don't coexist with magnetic field (Abrikosov-Nielsen vortices) while latter can do it.

In any case I understand that it is quite difficult to translate one phenomenon in another in so simple form and your explanation can give me one idea your last sentence: "That collective vacuum state must also be randomized with respect to the Poincare group (so called Spaghetti vacuum). If color fields penetrate into the vacuum the condensate's induced dynamics would neutralize them ensuring confinement." 

Daniel,

The superconducting state is the state of lowest energy in the superconductor. Vacuum is in terms of field theory by definition the state of lowest energy.  It is not the "empty" space but a very complex physical state (even in QED - the "dressed" Dirac sea).

You are of course right that superconductivity is more complex but I wanted to draw a simple analogy and not go into the complications of Type I and II (vortices might have some relevance for QCD as well). Above that there are also high-temp superconductors.... :)  

Stam,

the question is that the quarks are never isolated but when a neutron (ddu) decays in a proton (uud) the transformation was due to the decay d in u plus W- and W- in electron plus neutrino. Say neutron decays in proton, electron and antineutrino. The other d quark cannot decay in the singlet uuu as you have explained very well, but it seems to me semantics to say that the new tetraquark qq(bar)q q(bar) or pentaquark qqqqq(bar) are nuclei or particles. What we should be able to predict ( this is science) is if these states are possible and how stable would be.

David Vasak,

One superconductor only have a small amount of electrons involved in the superconductivity and most of them are in states of lower energy in the solids.

They are bound states, if their energy is lower than the that of the total energy of the individual constituents. That's all. The deuteron nucleus is such an example. The *possible* states can be classified by group theory-i.e. conservation laws. 

The lifetimes, of course, depend on the dynamics. 

Stam,

Thank you and for being concrete, could you tell me if there is a pentaquark or tetraquark and what are its lifetimes? Could we get a seventhquark particle with a predicted lifetime too?

Pentaquarks haven't been discovered yet, but tetraquarks are presented here:http://arxiv.org/abs/1404.1903 

The six-quark boud state, that's the deuteron nucleus has, of course, a macroscopic lifetime in vacuum. In hot and dense nuclear matter it has been studied: http://arxiv.org/abs/nucl-th/9706021

Of course a seventh quark bound state is imaginable: take a baryon and two mesons, which immediately indicates its decay products. The calculation of lifetime is a dynamical issue. 

Stam,

A pentaquark is a baryon plus a meson and it is still without being discovered. A seventhquark can be thought as a pentaquark plus a meson too. Stam, I think that you play with particles, nuclei or number of quarks as you want for giving an answer, but I do not know if this is only a game or a physical reality. What do you think?

Once more: these are *possible* states, allowed by conservation laws. The specific states depend on the dynamics-that's why it was so hard to detect tetraquarks. Cf. the paper on tetraquarks.  

The situation is analogous to that of compounds  in chemistry: in that case the force is electromagnetism, here it's the strong force. It is hard to make compounds  with particular properrties-cf.  the pharmaceutical industry; something similar is occurring here. The situation is, of course, harder, because it's harder to control color charges than it is electric charges. 

Stam,

I understand you. Controlling the color or the the flavor of the quark would be a new kind of engineering very appealing indeed. What is very interesting for me is to know how the issue is in this direction and this is the target of my question.

Your answers are difficult to follow for me because you say, on one hand, that this is theoretically possible and without problems; but on the other hand, you say that the last decission is the experiments in ALICE. This tells me that the theory is almost a complement and the present theories of QCD are just a crude approximation to the reality. Is this true? How is actually this issue of creating new particles based in the present hadrons?

The problem is that the description of bound states of quarks and gluons is very difficult to treat theoretically, since the coupling becomes stronger, as the distance between the color charges increases. Lattice techniques are, in principle, appropriate here-in practice it's a hard problem of present-day nuclear physics research. QCD is the complete theory of the strong interactions, but it is under quantitative control at *short* distances, not at the scales, where bound states are formed. It's a non-trivial matter to realize the matching between experimentally accessible quantities (like form factors and current correlators) and theoretically computable quantities. 

Some of these states were stable when the universe was so young that they did not have time to decay.

Matts,

Thank you for your answer but could you give one idea of the physical conditions necessary to have stables such states in early universes?

Sorry, I think I gave the wrong answer. Quarks combined into hadrons when the universe was about one microsecond old, but strong interaction decay times are of the order of 10^{-22} s, so  these states were never stable.

Matts,

Why the hadrons were formed at the microseconds?

   Matts,

   Your observation is very interesting and honest, thank you. But let me to explain a little bit deeper my question. I know that the hadronic age is between 10^(-6)s to 1 second within the Inflationary Big Bang Model of the Universe, which explains that no other stable hadrons than the protons are since then. But the question for me is that if you had chosen the Inflationary epoch 10^(-32) s till 10^(-12)s (electroweak epoch) is quite complex to distinguish the states of quarks, isn´t?

Daniel,

Above the temperature 200 MeV where the QCD--hadron phase transition occurs, the

hadrons are represented by their free quark subconstituents in rhermal equilibrium. They contribute more degrees of freedom than hadrons, forming a dense medium of quark matter. This phase transition occurs about 1 microsecond after Big Bang.

When the universe cools below the phase transition the  quarks become bound in hadronic matter; the separation between quarks in the nucleons increases, and the interaction between any two quarks in a nucleon cease to interact with quarks in neighbouring nucleons.

Your question is: just how unlikely is it that more than two or three quarks get bound in the same hadron at that brief moment of the QCD--hadron phase transition. To answer it one needs a good model, but I think it is reasonable that multi-quark hadrons would be rare. Perhaps such states exist in the interior of neutron stars or quark stars. A quark star is really what you are asking for.

   Matts,

   Thank you very much.

   You have open a new possibility for looking for stable states of more than three quarks in cosmological objects. But the quark stars, in all that I know, are obtained as neutron stars very massive. Thus the gravitation is the energy involved for making more dense the quark plasma and therefore to make them free. But I think that this star would be very inestable and we came back to a similar problem of instability of the one presented by Stam in this discussion. Please, could you give us any information more about this issue?

Daniel,

Lots of neutron stars are known to exist, but what their interiors contain is not known. There could be a core of quark matter.or clumps.

The existence of quark stars is much less known. One could say that quark stars are possible and every object which is possible likely exists.

The interiors of black holes is not known either.

    Matts,

    Could be stable the quark stars, even as theoretical objects, for the moment?

Yes, they are stablebecause they are the end products of a certain collapse scenario.

The neutrons in a neutron star form a cold Fermi gas in which the quantum degeneracy pressure of the neutrons prevent the star from collapse. If the mass of the star exceeds the Tolman--Oppenheimer--Volkoff limit of 2-3 million solar masses

it may not always be stabilized against bounce. They then follow the route of further gravitational collapse to become a hypothetical quark star or a black hole.

    Matts,

   Thank you very much. Neutrons are inestable hadrons out of the atomic nucleus (around 15 min) (quarks d decay in u). In a neutron star they are also stable and as fermions the exclusion Pauli's principle avoids their collapse. What seems logic is that if you increase the gravitational force in a certain value the neutrons are so close that all the quarks are free and the star by themself could be considered as an enormous excited hadron emitting neutrinos or photons for example, and decaying again in the initial neutro stars and never in black holes if the initial mass wasn't high enough.

   In any case, all this staff is quite speculative and I do not know if could shed light on the particles formed by more than three quarks. 

The so called quark-gluon plasma is believed to be created in heavy ion collisions (http://www.if.pw.edu.pl/~kperl/HIP/1304.1452v1.pdf) but its existence is veery short (~10^-24 s) as it expands quickly and hadronizes because there no external forces to maintain it iconfined n this peculiar super hot state (T=160 MeV). In this way one reproduces state of the matter when our Universe was only microseconds old after the Big Bang. There are some speculations that under special conditions (as mentioned by Matts) such a quark gluon plasma could exist on an astronomical scale. 

     Adam,

    Thank you very much. I understand very well the hadron formation after the inflationary age of the Universe (justifying the cosmological principle of homogeneous and isotropic symmetries for the present matter of the Universe). But what seems to me difficult to justify is that there are only the present hadrons (2 or 3 quarks only), even more than the existence or not of the magnetic monopole, having so many possibilities of combinations.

    Matts introduced new objects as the quark stars (the pity is that they are not actually identified stars), but which could open a new possibility of finding. Could you give one explanation for having only combinations of 2 or 3 quarks in particles?

Daniel

As far as I know the quark-hadron transition is poorly understood, and therefore one cannot raise arguments for or against the formation of multi-quark hadrons.To me it seems likely that this transition leads so quickly to stable ordinary meson and baryon states that multi-quark hadrons have no time to form, or they are too unstable to resist decay into more stable 2- and 3-quark states. But this is just speculation.

Another speculation is that neutron stars contain clumps of quark matter in the form of multi-quark hadrons.

@Daniel

The experimental observation of he states composed of only 2 or 3 quarks was at the origin of introducing the color quantum number and the rule that the only stable configurations are colorless.  This was then formulated in a theoretical language using a group theory formalism - http://www.physics.umd.edu/courses/Phys741/xji/chapter1.pdf

Adam,

First of all thank you very much for your information of QCD. But you have touched the heart of my question, when you said that : " The experimental observation of he states composed of only 2 or 3 quarks was at the origin of introducing the color quantum number and the rule that the only stable configurations are colorless". Do you mean that more than 2 or 3 quarks are not real for making a hadron? Is there experimentally possible to play with the "quarks" for making bigger hadrons of more than 3 quarks? What is the state-of-art in this question (if this question has been of interest enough)?

Thank you in advance and sorry for coming back to this question so late.

Matts,

Thank you very much for your answer and sorry for my delay in responding. The application to cosmology is not obvious at all, but it seems that in the first moments previous to the hadron formation, around 10^(-35) seconds, i.e. in the inflationary age, it seems that the implications of having so massive particles could change the application of theorems as the of Borde-Vilenkin-Guth for singularities. I know that this far of my original question but it could help to understand a little better hypothesis as the dark matter, what do you think about?

Daniel,,

I refrain from speculating about what happened  around 10^(-35) seconds, we have next to no information about the physics from that time.

  Dear Matts,

  I am not sure to understand you, because my ignorance about the astrophysical dates is quite big, but independently of how they have born or arisen, the question is that nowadays are hadrons (baryons and mesons) only have as maximum three quarks. This number is not followed by QCD, in all that I know, and also something which can limit to this number of quarks to joint for creating an stable particle.

   Dear Arno,

   If you need one lepton as a positron to create a proton from quarks, then it seems that we are speaking about very different things. Could you explain it without going to your theory or articles? Thank you!

Dear Daniel: Less than two is forbidden because the macroscopic color charge must be white. Otherwise, the vacuum energy of a lonely quark would tend to infinity

   Dear José,

   Thank you for your answer. Less than two is one and it would be an isolated quark and it is well known in Quantum Chromodynamics that the color confinement forbids it due to the huge energy needed.

José,

There are in fact dozens of hadronic states formed by 2, 3 to 5 quarks, but they are all very unstable due to the nature of the strong force.  Only the lightest mesons and baryons are stable against strong decay, but instead they decay by the weak force. All that remains are then the protons.

   Dear Matts,

  Thank you for your answer. It is true that the proton is the lights baryon, although the GUTs theories predict a lifetime for it between 1038 or 1040 years. It is necessary to say that if it could decay then the baryonic number conservation would fail and also to say that the neutron is quite stable within a nucleus.