Thanks for reply..but can you explain in detail that why T=1 states are unbounded ?

Monojit> can you explain in detail

For details you should consult f.i. Griffiths book on Particle physics. Briefly, there is a isospin dependent contribution to the energy, which lowers the energy for the T=0 state and increases the energy for the T=1 state (by an amount large enough to prevent binding).

we know that np bound state is possible..now for T=1, T3 = 0, correspond to np+pn and for T=0 we get np - pn. So I thought the problem is not with the T=1 states rather with the S =0 states. for nn state the spin state has to be singlet i. e S=0 and I read somewhere that the depth of the potential for S=0 is shallow. So is it S state dependent also ?

Monojit> we know that np bound state is possible..now for T=1

From where do you know that a T=1 np state is possible? Deuterium is T=0.

the three isospin triplet states are T=1:  nn(T3=1), np+pn(T3=0), pp(T3=-1)

  and isospin single state T=0: np - pn

So Deuterium is the anti-symmetric combination of p and n right ? and the symmetric bound state of np is not possible..is that you what trying to say ?

And it is nothing to do about the potential about the S=0 state ?

I know that deuteron is the 3S1 state but what I wanted to ask the in the last question was that why the 1S state of Deuteron do not exist ? and you are saying that it is not S=0 state but it is T=1 state which is not bounded.

This is exactly what I say. Which is just a repetition of what I learned about nuclear physics some time (40+ years)  ago.

In the old-fashioned (pre QCD) viewpoint the main binding interaction is due to exchange of pions. Which have zero spin, and hence does no lead to any spin dependent interactions. However, pions have isospin, which leads to isospin dependent interactions. (There are some additional --- weaker --- spin dependent interactions due to exchange of spin-1 rho-mesons).

Correction added: As shown in the lecture notes linked to by Stam Nicolis, pion exchange do lead to spin dependent interactions between nucleons. The whole story becomes rather complicated, and dependent on detailed values of the parameters involved.

Of course a neutron-neutron bound state is possible, thanks to the strong interactions (the Joliot-Curies had found that (n,p), (p,p) and (n,n) interactions were comparable, shortly after Chadwick had discovered the neutron, which led to the notion of isospin by Heisenberg). 

What do you mean, Stam?

There are no entries for 2N or 2He in the chart of nuclides. If I remember correctly the nucleon-nucleon isotriplet is visible as a resonance in nucleon-nucleon scattering, about 100 keV above stability threshold.

The answer was to the question, why the neutron-neutron bound state might not be possible. My answer to that is that it can be, as the result of the strong interactions. This doesn't have anything to do with nucleon-nucleon scattering, that, as its name suggest, doesn't probe the bound states of nucleons, but the scattering states. 

Indeed this is obvious, within the pion picture, since pions are spin-0 particles and, therefore, mediate attractive interactions. That it is very hard to realize in practice is another issue-but, for instance, in neutron stars, it is, precisely such an interaction that leads to the formation of a superfluid condensate.

Ultra-cold neutrons have, also, been studied experimentally: https://lpsc.in2p3.fr/Indico/getFile.py/access?contribId=6&resId=0&materialId=slides&confId=947

so it's useful to be quite precise about what exactly is the issue.

Yes, it could have been possible. In another universe, with slightly different physical parameters. But it didn't happen in our universe.

Which raises the philosophical question: How would a universe with the masses of the neutron and proton interchanged look like? 

Once more: if only strong interactions are taken into account, neutron-neutron bound states are possible. Cf. http://www.physics.umd.edu/courses/Phys741/xji/chapter6.pdf

They become disfavored, once electromagnetic and weak interactions are taken into account, which, among other effects, imply that the neutron is heavier than the proton and thus is the baryon that can decay, through weak interactions, in a way consistent with baryon number conservation.

And they do occur in our universe, in neutron stars: http://www.int.washington.edu/PROGRAMS/Schwenk.pdf

But Stam, from equations (6.19) and (6.21) in the lecture notes it follows that the potential is repulsive for the isotriplet, T=1. Hence no binding for nn. (Actually worse than I remembered).

Correction added: Further investigations show that the T=1, S=0 potential is also attractive, but slightly less so than the T=0, S=1 potential. Hence it becomes a rather complicated problem to determine whether the T=1, S=0 potential is strong enough to support bound states. Nature has solved this problem, with a negative result.

I think pairing in conventional superconductors is a different (many-body) mechanism, and actually a binding in resiprocal space, between particles on opposite sides of the Fermi surface. Then even two electrons can pair up.

The electrons pair up through their (attractive) interaction with spin-0 phonons and the neutrons pair up through their attractive interaction with spin-0 pions, as noted. They can't pair up otherwise.

Of course, in both cases, there's a competition to the pairing: for the electrons it's the repulsion, due to the electromagnetic interactions; for the neutrons it's more complicated: http://www.slac.stanford.edu/cgi-wrap/getdoc/slac-pub-4155.pdf

This reference stresses, indeed, how careful one must be in deducing the properties of the two-body interaction from the more easily accessible three-body interactions.

For a review of neutron-neutron interactions, cf. http://www.sciencedirect.com/science/article/pii/0370157389901270

that, however, focus on three-body data. It's more useful to sift the citations of the first paper in search of two-body computations.

It is obviously a more delicate question than I first thought. I did some searches, and found the statement in J. Phys. G, 23, 1619 (1997) that  "Indeed, this is the case for the two-neutron system which requires only an additional 67 keV attraction for the formation of a bound di-neutron". I am not sure if this refers to theoretical calculations or observations (of a scattering resonance?).

Anyway, I tend to believe that a stable two-neutron bound state cannot exist in free space (for a time comparable to the lifetime of the neutron, which should not be that constraining experimentally). However, I note that the existence of neutron-rich halos in some nuclei has been attributed to di-neutrons and similar multi-neutron sub-configurations. 

It is, of course, a very challenging calculation, though experiments with ultra cold neutrons could shed light on the issue. A quite recent calculation is available here:https://inspirehep.net/record/1299315/files/scoap3-fulltext.pdf, where it's stressed that the uncertainty about the neutron-neutron scattering length remains to be resolved, cf. here: http://www.int.washington.edu/talks/WorkShops/int_10_1/People/Gardestig_A/Gardestig.pdf for an overview.

Incidentally, there are, of course, strong interaction contributions that break isospin symmetry and, thus, distinguish the neutron and the proton through the fact that the down quark has a different mass from the up quark. Taking them into account is highly non-trivial. These do not lead to a finite lifetime for the neutron, however, which is due to the weak interactions. 

So the bottom line is that there isn't any theoretical reason such a bound state can't exist. While the deuteron, i.e. the (n,p) state is the most extensively studied, the other states are, simply, much less known, because much harder to study-apart from dense nuclear matter, like neutron stars.

It is interesting (and humbling) to learn that a question I thought was clearly resolved 40+ years ago actually isn't. I don't think there is anything fundamental about the issue, since it is only a delicate dynamical question, but it is very interesting anyway. Nature has of course solved the problem long ago, but if experimenters are unable to read the solution we are equally ignorant.

I would not trust any theoretical computations based on phenomenological boson exchange models, due to their own uncertain limitations. I would rather wait for very accurate ab initio lattice calculations (if I could convince the ultimate authority to live long enough for that...). 

I'm not sure, indeed, what ``Nature'' has solved-the fact that it's easier to observe (n,p) than (p,p) or (n,n) doesn't imply that (p,p) or (n,n) are forbidden. So, as long as there isn't a clear reason why they should be forbidden, they may occur and it becomes a challenging calculation. 

Lattice field theory techniques are just now starting to address nuclear physics properties quantitatively; taking into account light, dynamical, fermions is a formidable challenge-cf. http://www.epj-conferences.org/articles/epjconf/pdf/2014/03/epjconf_inpc2013_01012.pdf

and http://arxiv.org/abs/1212.1572

for a recent review (the latter shows results for the neutron-neutron potential).

Indeed, there is no actual  fundamental formulation of the nucleon nucleon interaction ( I mean not comparable to QED or even QCD). All we know is that - in our world, in free space - only np bound system occurs, not nn or pp (T.G.!!!). The later (pp) is easier to justify in terms of Coulomb repulsion than the first (nn) . Nevertheless, we just empirically  or semi-empirically (meson exchange, nonperturbative-QCD-based conjectures, low-k expansions...) formulate the strong nucleon-nucleon interaction to reproduce the experimental properties known from nucleon-nucleon scattering so that the force in T=0 channel (only present in np-system) compensates the lack of binding in the T=1 chanel (present in the three systems nn, pp and np). Clearly, not only the strong interaction is there, and the presence of the electromagnetic and weak forces (making free neutrons decay fast) certainly plays a role, but existing "first principles" formulations of the nucleon-nucleon interaction do explain the np-binding as due basically to the T=0 strong force. In a macroscopic system like a neutron star the gravitational force is what matters most.

Yet, the existence of bound borromean nuclei show that a nn-pair can exist also with the "small perturbation" introduced by the presence of a third body (say an alpha particle, a lithium core,...). Therefore it is clear that the nucleon-nucleon force is still an intriguing open problem for theorists.

In the framework of the orbital model of the structure of elementary particles, the neutron has a bonding shell formed by the orbital of an integer negative electric charge, and its core is nothing more than a proton, which in turn is formed by the orbital of an integer positive electric charge. When a neutron is joined to a proton it shares its shell, forming a two protons nucleus, corresponding to the deuteron.

For two neutrons, the shell would have two negative electrical charges. This system is very unstable due to the mutual repulsion of the two negative charges of the shell. But if a neutron is added to a deuteron the core has then three protons, and in this case the electrostatic attraction from the three protons core is strong enough to counteract the repulsion between the two negative electrical charges of the bonding shell, leading to the triton. However, if a proton is added to a deuteron the core contains then three protons wrapped by a shell formed by a single negative charge, corresponding to the helion3. In further increasing the number of nucleons, the main trend corresponds to a balance between the number of protons and neutrons.

This paradigm also provides a simple explanation of why nuclear neutrons do not decay, simply because they cease to exist by sharing their shell with adjacent protons. Interested readers can get more detailed information in the publication "Fundamentals of the orbital conception of elementary particles and of their application to the neutron and nuclear structure", in physicsessays.org, also available in arxiv.org/ftp/hep-ph/papers/0102/0102268.pdf

It is a deep intellectual error to think that, at the very starting point of matter, its elementary building blocks would have such a complex structure as the standard model gives them. The complexity of the structure of elementary particles from the standard model is highly incredible and artificial. Furthermore, the quark model has no reductive power. It appeals to 60 primary elements or key states: 36 different quarks (6 families of tri-chromatic quarks and antiquarks), and 24 different gluons (8 tri-chromatic types). Thus, it has no horizon of being unitary, moreover conceptually quite puzzling and showing a severe lack of realism. Frank Wilczek, Nobel laureate, 2004, said: "from the perspective of QCD, the foundations of nuclear physics appear distinctly unsound".

Bound or not (a resonance) the proof of existence of a n-n state woud be great. I think it could be also searched within the deuteron structure, an excited state member for the T=1 triplet. As always, experimental data is needed...

In addition to the explanation of Georges Sardin, you can find more about the neutron-neutron scattering in: Blatt and Weisskopf; Theoretical Nuclear Physics (Dover Books on Physics).

Referring to scattering of nucleons you may also have a look to the animation at:

http://perso.ya.com/sardin1/index.htm

Back to the n-n union, within the standard model framework it should exist and should be extremely strong and stable since bonded through the strong interaction mediated by the gluons without electrostatic repulsion between the two neutrons, and furthermore all sorts of neutron nuclei should also exist. The standard model is clearly defective and must appeal to all kinds of artifices in any attempt to solve its many inner incoherencies.

Higher the complexity of fundamentals and of descriptive mathematics, lower the probability of coincidence with physical reality, and farther from its reliable description. In our opinion, quarks, with their colours, fractional charges, counterfeit masses, and the eight types of gluons, are the equivalent of the epicycles, the deferents, the equant and the circular orbits of the Ptolemaic system. As long as the quark model will remain an enforced mainstream model, ahead standpoints will be inhibited. The official symptomatic omission of straightforward deductions challenging the Standard Model, is an inappropriate, counterproductive strategy.

Georges ~

How to you describe the lower-mass multiplets of elementary particles? How do you describe the decay rate of the neutral pion? How do you describe the results of deep inelastic scattering? How do you describe the charmonium and heavy-mass elementary particles? How to you describe the R -- there ratio of electron-positron annihilation to hadrons compared to  muon-antimuons?

How does the chart of nuclides look according to your model?

Kåre, if I had the answer to all your questions the model would already be fully settled, however let me point out a few points.

 All neutral particles are structured by a pair of opposite electric charges. What differentiates them is exclusively their quantum state. Of course this is not an easy task. We need again Schrödinger and Dirac mathematics to search for exact solutions. Years ago, I have proposed a basic approach for the muon, which is a most simple case, being seen as an excited state of the electron (http://perso.ya.com/sardin1/muon.pdf). In fact all elementary particles are seen as excited states of a generic fundamental state corresponding to its inner equilibrium state. All of them are departures from this fundamental state, through two different quantizations: the quantization of the structural radius, and the quantization of its vibratory state.

 In inelastic scattering, besides the scattering itself, each impinging particle may undergo a transition to a different structural quantum state.

In the case of muon-antimuon annihilation, it should be taken into account that besides their collision energy, which for simplicity can be supposed to be equal, each one releases an extra 105.7 MeV while for electron-positron annihilation each one brings only 0.51 MeV. This affects the annihilation ratio.

 About the nuclides chart it does not change it. The amend is mainly conceptual since nuclear neutrons are considered to be dissolved into their nucleus, which is a proton, and their shell, which is then shared with all protons. Nuclear neutrons do not exist, neutrons only exist as free particles, through the union of a nuclear proton and a negative charge of the nucleus shell, in order to easy their escape in a neutral form.

The n-p system is in a 3S1+3D1 state which has a bound state, the deuteron.The  n-n state is in a 1S0  state.  These assignments are a consequence of the Pauli principle and isospin conservation.  The  3S1+3D1 is more attractive due to the nucleon-nucleon tensor force, which is absent in the 1S0 state;  that extra attraction generates the n-p bound state and leaves the n-n system without a bound state.  This was worked out in the 50s by Schwinger.

The origin of the tensor force is from meson exchange (or from quark pair exchange).   The basic reason is that experiments led to this theory and nature just set it up that way.  If nature had not setup the meson coupling constants as they are, in an alternate universe with stronger attraction there could be a n-n bound state.

But we are confined to the universe as it was delivered to us.

True. Nature is set up in just one way. No bound state for the nn or pp systems. Is there a fundamental reason why the deuteron has not one single excited state? As far as I know, what we have is experimental evidence. Not more. Should this evidence change and such a state be found... our "fundamental" theories would have to be "adjusted"...

Frank all what you say is correct, it agrees with my previous statement on the T=0 channel, yet it is also true that this comes from parameter adjustments to explain what we observe (and viceversa, if parameters had different values.. The antropic principle )

In addition to Frank, a very simple pion exchange model predicts that if the strong potential of the deuteron (I=0 |pn-np> system) is -V, the potential of the nn system (as well as I=1 |np+pn> system) is +V/3. Therefore it is repulsive.

The description of the net wave function in atoms is a bit different from what we have in nuclear physics. The total wave function for two electrons is: (spatial *spin). However, for two nucleons (like two neutrons), the net wave function related as: (spatial*spin*iso-spin). The iso-spin term declares the bound state of nucleons appeared due to the nuclear force. When iso-spin is 0 (named singlet), the bound state is completely possible as we face it for deuterons. However, when iso-spin is |1| (triplet), the nucleons are not able to create a bound state. Now, for a neutron-neutron (proton-proton) system, the net iso-spin is |1|, [T3=1/2(Z-n)=2/2=1]. Thus, bound state can not  be possible.

I think that np and nn bound states are possible. See, for example ArXiv:nucl-th/0407079v1

For those who could be interested in a refreshing viewpoint let me point out two papers, one about the neutron fine-structure and the other one about the cosmic absence of the anti-proton.

https://www.researchgate.net/publication/319007116_The_neutron_fine-structure

https://www.researchgate.net/publication/323756114_The_electron_as_the_fossil_of_the_anti-proton