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Is it possible to have a bound state between a scalar and a fermion? For example, a squark--anti-squark bound state, provided that the decay width is sufficiently small compared to the binding energy?

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Certainly it's possible in principle. I'm sure there are nuclei that can be described by scalars in an effective field theory of atomic physics. The resulting "atom" would be a fermion but that's ok. Whether you actually see the things or not depends on the model because of decays, as you say. Are you talking about some generic SUSY model or SUSY in the real world which, if it exists, is badly broken? – Michael Brown May 21 '13 at 0:53
How about baryons, which consists of 3 quarks and lots of gluons... Is that a satisfactory example? – Siva May 21 '13 at 2:56
@MichaelBrown : in my mind I'm thinking of a theory with broken SUSY. Suppose only one squark were light and the decay width is sufficiently small. If I pair produce these squarks at the LHC, the squark will either want to form a squark--anti-squark bound state or a squark--anti-quark bound state; I'm not sure how to determine which one occurs. – Pengpeng Xu May 21 '13 at 4:33
@Siva : I suppose the question that I'm really asking is whether if the spin of the 'valence' constituents matters when forming a bound state. In a baryon, could I replace one of the quarks with a scalar that otherwise has the same quantum numbers and is sufficiently long lived? – Pengpeng Xu May 21 '13 at 4:35
Right. My suggestion would be to look at a toy model first (say SQCD+one chiral multiplet, forget all other flavours & gauge interactions) before tackling the MSSM or whatever. It could be either-or-both. I don't know enough about the theory to tell you which has the lowest energy. If the squark is heavy enough I think you can use perturbative QCD (could be wrong here??) but you probably need to go to the lattice otherwise. But there is no selection rule saying "scalar+fermion bound states are forbidden" or anything like that. – Michael Brown May 21 '13 at 6:16

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