# Can this exist?

1. Mar 24, 2012

### gella

can there be a particle with baryon number +1 and electric charge -2 ?

2. Mar 24, 2012

### fzero

If this is a homework problem, you have to show some of your own work before you can try to help you. It's also in the wrong section, but I wouldn't have a problem helping here anyway if you showed what your thoughts are on the problem so that we know where you're getting stuck.

3. Mar 25, 2012

### francesco85

I think so; imagine a particle composed of the following quarks (d s b d ubar); if you contract, for example, d s and b in such a way that they form a color singlet and similarly d and ubar, then you have a color singlet bound state that has electric charge equal to -2 and baryon number equal to 1.

4. Mar 25, 2012

### tom.stoer

Hm, but (d s b) + (d ubar) is nothing else but a baryon-meson system and I don't see why this should be a bound state at all

5. Mar 25, 2012

### francesco85

I don't see these particles as different: I have just shown a way to contract the color indices in such a way to form a color singlet particle; in this moment, it doesn't occur to my mind a reason why this should be forbidden. (I have not an experimental proof that this bound state particle actually exists, I just guess that this state is not forbidden).

6. Mar 25, 2012

### tom.stoer

I don't say that it's forbidden, but that I can't see any reason why it should be a bound state.

Besides the different flavors which are relevant in weak interactions only this system is identical to (d d d) + (d ubar) = Δ- + π-. b/c the two particles have electric charge -1 the only reason why they should form a bound state is the strong force. But they are color singulets so they are not effected by gluons directly, only by 'residual forces' as in low-energy effective theories, mediated by pion ond vector meson exchange.

If (d s b) + (d ubar) is a bound state, what is the reason that the Δ- + π- system does not form a bound state?

7. Mar 25, 2012

### francesco85

I don't say that they must form a bound state; I just say that they can form a bound state; think to the analogy with protons and neutrons; they can exist separately but they can also form a bound state (the deuteron).

8. Mar 25, 2012

### tom.stoer

but Δ- + π- don't and I do not see any reason why they should; and experiments tell us that they can't - in contradistinction to the deuteron

9. Mar 25, 2012

### tom.stoer

I would suggest to google for pentaquarks (for which no experimental evidence has been found so far)

10. Mar 25, 2012

### francesco85

I just say that they can, not that they must. Or do you mean that there exists a principle that forbid its existence? if not, you have answered; if yes, tell me which. In the first case you just tell me that it can exist; in the second case you tell that such a particle cannot exist.
What is this principle, in your opinion? Thanks

Isn't (d s b d ubar) a pentaquark? If people have studied them, the fact that they weren't experimentally proved was not a good reason to exclude them (as it has been always done in the history of science).

PS Let me stress that if there is an experimental proof of the existence of such a particle, the question posed in this thread is meaningless (of course the answer would be yes, in that case).

Last edited: Mar 25, 2012
11. Mar 25, 2012

### tom.stoer

This is a strange argumentation.

I just say that two electrons can form a bound state, not that they must. I have no idea why they sould, but please tell me a principle which forbids this di-electron bound state.

Is this a reasonable argumentation?

I have explained why I don't see any reason that your system can form a bound state (electric repulsion, no color force, a well-known system that does not form a bound state); so please be so kind a tell me why it still should.

12. Mar 25, 2012

### francesco85

Neutrons and protons are color neutral; protons are elctrically charged; anyway they form nuclei, with lots of protons and lots of neutrons; and there is no theoretical proof that such a bound state can exist (as far as I know, QCD is not solved, in the sense that some features of the spectrum can be imposed by the breaking of SU(2)_LXSU(2)_R to SU(2)_V, but I think that it is just an assumptions, not a proof derived from the first principles of QCD). Even if the existence of bound states like the neutron and the proton (or even a nucleus) is not proved, everybody finds reasonable to assume that they can be related to QCD (I don't even know what lattice simulation can say about this problem).
You haven't told me the principle yet.
Moreover, if I was able to prove in mathematical terms, starting from the QCD, the existence of that kind of bound states, this would mean either that such a particle must exist or that, very roughly speaking, "QCD is wrong". And I would have solved the long standing problem of QCD ( thing that is clearly false! :)

Last edited: Mar 25, 2012
13. Mar 25, 2012

### tom.stoer

there is no principle that this bound state cannot exist; but there are reasons (I told you) that it is unreasonable; and there is no experimental hint that it does exist

14. Mar 25, 2012

### francesco85

Maybe I have missed your point; if the reasons you are referring to are electric repulsion and color neutrality, these are not sufficient to exclude a bound state (as I showed you with the example of the nucleus); if the reason is "a well-known system that does not form a bound state" (theoretically), well, this seems to me like a tautology (the question is "why isn't the system under consideration a bound state?"; but maybe, also in this case, I missed what you really meant; the final answer to this question, from the theoretical point of view, can be solved, in my opinion, only through the resolution of QCD); the fact that they has not been discovered yet is a still weaker motivation (in my opinion); with the same argument: since supersymmetry has not been discovered yet, it means that it does not exist.
Or are there other reasons I have missed?

If you meant that, if they existed, they would have already been discovered, well, this is a thing I didn't know and, please, can you give a reference?

ps. just as a remark: what I meant is simply that there are reasons to think the pentaquark system is reasonable, both by the impossibility to find a principle that exclude it and by the analogy with other systems.

Last edited: Mar 25, 2012
15. Mar 25, 2012

### tom.stoer

My main point is that there is no known bound state in the Δ- π- system and that there is no good reason why there should be one.

You ask for a principle why there should be no bound state; I am asking for an explanation why there should be one; let others decide whose position is more reasonable ;-)

16. Mar 25, 2012

### francesco85

I have never meant to tell other people what to think! :) (it is a 20000 light-year far from me behavior !)
I think I have understood your point of view; let me ask a final question: is it possible in your opinion to prove the existence of bound states (normal or exotic) without solving QCD? It would be an interesting problem, in my opinion.
Best,
francesco

ps as a last citation by Gell-Mann: "everything not forbidden is compulsory" :)

17. Mar 25, 2012

### tom.stoer

w/o solving QCD one can only rule out states that are forbidden by (e.g.) symmetry principles which hold even when taking full QCD into account; solving QCD in the "bound state sector" and proving or dis-proving the existence of bound states based on the dynamics rigorously is difficult - it is a so-called millennium problem http://www.claymath.org/millennium/Yang-Mills_Theory/ [Broken]

Last edited by a moderator: May 5, 2017
18. Mar 25, 2012

Staff Emeritus
He was talking about something else entirely.

There are hadronic states that do not seem to exist in nature. People have looked. As Tom points out, these don't violate any conservation law; nonetheless, they don't seem to exist. (A nucleus with only protons doesn't violate a conservation law either, but also doesn't seem to exist).

This is in that category.

19. Mar 25, 2012

### francesco85

Thanks for the pp example; i have read anyway that there is a reason why this should be forbidden, namely something related to the exclusion principle / energy considerations (correct me if I'm wrong). Is there any analogous reason why pentaquarks shouldn't be observed? Why did people study them? Can you give references? Thank you

I'm not an exegete of Gell-Mann's thoughts; see the ":)" I have used.

Last edited: Mar 26, 2012
20. Mar 25, 2012

Staff Emeritus

21. Mar 26, 2012

### francesco85

Can you be more precise please? I have found papers in which it is claimed
1) quark model does not exclude pentaquark systems
2) lattice simulations cannot exclude pentaquark systems
3) no experimental evidence (but this has already been pointed out by Tom)

Where can i find the theoretical reasons to exclude pentaquarks (analogous to the energetic/Pauli principle considerations of the pp system)?
Thanks

22. Mar 26, 2012

### tom.stoer

There is no theoretical reason to exclude pentaquarks.

The big difference to your ideas is that pentaquarks do not imply a coupling of two color-singulets but should allow for more general solutions.

23. Mar 26, 2012

### francesco85

Very clear! But I still remark that I have never said that it must be a two color singlet bound state: it can *reasonably* (for my personal taste) be a two color singlet bound state (in the sense that I have found no reasons against it and that it was the first thing that occurred to my mind in answering the very first question); of course, this does not mean that more general solutions are not allowed; I have just found a clear discussion here

"Rise and fall of pentaquarks in the qcd sum rules approach", sec.2

and references therein. There are various possibilities to mean the structure of the pentaquark system which, how it seems at a first sight, seem to analyze all the possibilities given by group theory.

Thanks for the discussion,
Francesco

Last edited: Mar 26, 2012