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Question: are they mathematically possible at all? physically?

with finite mass systems, usually the binding energy contributes to the rest-mass of the system. It would seem that even if you could bind two massless fields/particles, the coupled system would still have a finite rest mass because of the binding energy

by any chance, is this how higgs boson gives 'mass' to particles?

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3 Answers 3

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Dear lurscher, a good question. But a bound state is something that has a negative binding energy - otherwise it wouldn't be bound. If the rest mass of a bound state AB of massless objects A,B is positive, it can clearly decay to A,B with some energy without violating any conservation law. So AB surely can't be quite stable.

Metastable bound states are a different issue. I think that in some moral sense, it is true that a black hole is a bound state of gravitons, and it becomes very long-lived if it is large.

I think it is very misleading to interpret massive particles that obtain mass from the Higgs mechanism as bound states with the Higgs boson. The vacuum itself contains a Higgs condensate - but the individual particles such as the electron don't contain any Higgs excitations, so it is not really a bound state. It is the original electron that has different properties because it is moving in a different environment.

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What about "glueballs"? Bound states of massless gluons. –  David Santo Pietro May 21 '11 at 16:09
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true!!! i completely missed the sign of the binding energy.. i feel stupid. thanks for your answer. so i infer from your answer that they (bound states of massless fields) are in principle allowed, just that we don't have many cases of them in the standard model? –  lurscher May 21 '11 at 16:10
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wait, i see where you are going: any bound massless state would have total negative energy, so it would violate the strong energy conditions of QFT and produce unstable vacuums! –  lurscher May 21 '11 at 16:14
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@lurscher: you confuse energy and mass. Just because something is massless it doesn't mean it has no energy. On the contrary, it has energy proportional to its frequency. All you need to obtain a bound state is proper interaction term in your Lagrangian. Try to look up e.g. those glueballs @David mentioned. Those are (in principle) possible because of higher order Yang-Mills ($A^3, A^4$) terms in the Lagrangian. But actual properties depend on full QFT + renormalization treatment, so this is not simple. But in principle there is no problem with bound states of massless particles whatsoever. –  Marek May 21 '11 at 17:52
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@lurscher: well, my intuition tells me that those bound states will always be massive but I have no proof of this (and it actually wouldn't suprise me if some contrived mathematical model would be just able to produce bound states whose binding energy precisely compensates for the energies of the constituents). The best I can give you is that the resulting mass can be pretty much any non-negative number, so it's quite unlikely to be precisely zero :) And yes, the resulting mass is purely the effect of dynamics, because you need to know the binding energy. –  Marek May 21 '11 at 18:28

If a bound system is complicated, it may have "massless" excitations. They are quasi-particles. I think all that we observe are quasi-particles rather than free particles in an empty space.

Solitons may be considered as bound states of different "massless" harmonics in a non-linear dispersive medium.

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+1 for the soliton analogy, this is something also i heard in a nonlinear dynamics course,but i've never seen any formalism associated to it –  lurscher May 21 '11 at 18:33
    
Your question reminds me my own funny expression: "Interaction of non interacting fields". –  Vladimir Kalitvianski May 22 '11 at 0:18

In the comments here two issues are being confused.

It is different if the question is : can massless fields be bound? It is another story if the question is: can massless fields make a bound state.

Gluons are massless and are bound in a nucleon, with virtual exchanges with quarks. Photons are massless and are bound in an atom with virtual exchanges with the electrons. So the answer to the first question is, yes, massless fields can be bound.

Certainly the photons cannot form a bound state, virtual or not, because they do not carry charge with which to exchange another photon.

Gluons because they carry colour can, within an interaction,( say proton proton,) form a glueball, which would then decay into quarks and gluons with end result the mesons we know. There can be no stable glueball because of the form of QCD . It seems from lattice QCD calculations that glueballs can exist. Here is a reference that offers a glue ball candidate , the eta(1440) which decays to mesons with a 20MeV width.

So the answer is massless fields cannot form a stable bound state.

Edit: Gluons carry colour charge and thus can exchange gluons among themselves, which is how the glueball is mathematically created .

The reason it is not stable is because there are lower energy states to which it can decay, which is the same reason as for all non stable particles' decays. In order for it to be stable it would have to have a mass smaller than the combined mass of two quarks, orderof Mev. Calculations on the lattice QCD give masses that are of the order of the rho(770Mev) meson, for the lowest glue ball state. So it is a matter of the lagrangian and the constants entering the calculations. We live in a universe where there are no stable glue balls.

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What about photon BEC? You might decide you don't want to call that a bound state though... :) –  Marek May 21 '11 at 21:22
    
why cannot form it? is a global symmetry (i.e: Poincare) disallowing them? or just the fact that the total mass-energy has to be positive? –  lurscher May 21 '11 at 21:24
    
@Marec no, I would not consider BEC a bound state in the usual sense of bound in a particle form. –  anna v May 22 '11 at 6:01
    
And now that I think more about it, a glueball could decay into two photons, given a virtual quark antiquark pair generated in the ball. As photons have mass zero then a glueball will always be unstable. Even if the QCD calculations allowed a 1 Mev mass gluon, it would still decay in the end, with a width of order of kev: prd.aps.org/abstract/PRD/v25/i3/p792_1 –  anna v May 22 '11 at 15:36

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