In the ground state of an isolated Rb-87 atom, the electron (a fermion) is paired with the nucleus (a fermion) to form a composite boson. The binding energy of this pair is about 4.2 eV. When a BEC of Rb is formed, say at temperatures around 10^{-7} K as is the case in current experiments, the kinetic energy per atom is 10^{-11} eV so the pairing of two fermions within each atom is not going to be affected by collisions.

In the ground state of an isolated Rb-87 atom, the electron (a fermion) is paired with the nucleus (a fermion) to form a composite boson. The binding energy of this pair is about 4.2 eV. When a BEC of Rb is formed, say at temperatures around 10^{-7} K as is the case in current experiments, the kinetic energy per atom is 10^{-11} eV so the pairing of two fermions within each atom is not going to be affected by collisions.

So if I understand correctly, as long as interacting energy scales are lower than the bounding energies of a composite boson, it physically doesnt make sense to look at them as separate fermions? In other words, I cannot even claim to as what is happening to the individual fermions?

From a more general perspective, one would expect a composite system to be well described by, say, molecular particles (bosons) if the binding length between two constituents (fermions) is small compared to the average inter-molecular spacing. Alternatively, this may be cast in terms of (intra-molecular) binding energies much larger than the inter-molecular (kinetic) energy.

Of course, the individual constituents (fermions) are still "there", but they are in this case strongly interacting. They become relevant again if you were to lower their binding energy more and more until it became comparable to the kinetic energy.

So the basic idea is the competition between the binding energy of the constituent fermions and the relevant energy scales of the composite bosons. However, if I try to look at it from the exclusion principle perspective, when the constituent bosons form a condensate, or more theoretically are at the same energy level, then are the constituent fermions also lying on the same energy level, in an apparent violation of the exclusion principle?

If I consider the interaction between the fermions on one of the boson splits up the levels of the fermions, then it is the same for all the bosons occupying the same level. The interaction can create sub levels in the single bosonic state, but they also should be highly degenerate to the fermions? Am I correct?

You're implicitly assuming the constituent fermions were non-interacting when you refer to them as "lying on the same energy level". That single-particle picture only works for an ideal gas -- in this case, an ideal gas of bosons (so to say, the center-of-mass variables). By contrast, the relative coordinates (fermion-fermion) exhibit strong binding, but that happens on a length scale much smaller than that of the molecules (bosons) given by the kinetic energy (plus trapping potential) and can be neglected.

Rubidium BEC is probably not the best example here because, first, there is always depletion of the condensate so that not all particles are in the same (zero-momentum) state. Second, there are too many electrons in various states in a Rb atom. Third, the total angular momentum F of the atom is well-defined, but the electron and nuclear spins are not; as a result, two atoms always form a superposition of electron singlet and triplet.

A much purer situation is just two hydrogen atoms in a box, each composed of one electron and one proton. We can even align all four spins in the external magnetic field. These atoms certainly can be in the same ground state in the box. The original question was: what happens to fermions they are composed of? In particular, are protons also in the same state?

I guess, the problem is a bit confusing words "in the same state" regarding identical particles. A system of identical particles is described by a wavefunction of all their coordinates (spin and spatial) and must be properly symmetrized with respect to the permutation of particles.

I see it so that the coordinate wavefunction of two atoms occupying the same or different levels in the box is always symmetric with respect to their permutation because they are bosons. But it is antisymmetric with respect to the permutation of both protons and electrons (the spin part is symmetric). This implies that the protons are in different states.

As for the interaction strength, it is interesting to look at the BEC-BCS crossover: on one side, weakly bound bosonic molecules composed of fermionic atoms form BEC and on the other side individual fermionic atoms pair in momentum space.

Thank you. The answer was quite enlightening!!!

They certainly will because it does not matter wherether the fremions interact with each other or not within those hard spheres. The key point is the total spin of each sphere. Since the total spin is integer, the spheres are bosons and can form BEC. However, when you say that fermions do not interact within a sphere you implicitely mean direct interaction. If the fermions are identical, there is always exchange interaction, which is simply another representation of their identity and necessity of the wavefunction of two fermions be antisymmetric.

Well, I ment BEC of spheres. To have it we don't need to know what the spheres are made of. The only required thing is that the total spin of each sphere be integer. As to the correlation between fermions inside the sphere, I think, they do correlate even whithout any direct interaction. First, the total momentum of the sphere must be conserved (if it experiences only s-wave scattering by other spheres). Thus, the total spin is conserved. So, in the absence of any interaction involving spins, we will have two sorts of spheres with S=0 (singlets) and S=1 (triplets) and each sort will form independent BECs. More precisely, triplet spheres are subdivided into three states with m_S=-1, 0, +1, which are again decoupled because of momentum conservation. Moreover, triplet spheres (parallel spins) must have odd orbital momentum L to make the total wavefunction of constituent fermions antisymmetric.

You probably asked whether a sphere can be a superposition of singlet and triplet states. I guess it cannot but I have to think how to prove it.

The question you are asking relates to the question of BEC-BCS crossover, which has been extensively studied. An introduction to the topic is in my book, Solid State Physics: Essential Concepts. The basic answer is that the coherence of the wave function of the BEC is in the center-of-mass degree of freedom of the pairs, not in the relative motion of the fermions.

A possible answer to this question is given in

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