For a BEC, you want atoms to be in the same quantum state, not necessarily at the same position.
For a BEC, the temperature is low enough so that the de Broglie wavelength $\lambda_{\mathrm{dB}} \propto 1/\sqrt{T}$ is larger than the interatomic spacing $\propto n^{-1/3}$, $n$ being the density. This means that the wave nature of the atoms is large enough for it to be felt by other atoms, in other words atoms "see" each other even without exactly sitting on top of each other. This is just to further justify the claim that you don't need atoms at the same position. Actually, if you had a perfect box potential of side $L$, and you reached BEC, then the atoms will macroscopically occupy the ground state $ |\Psi|^2 \propto \sin^2(x/L)$ which is very much extended. If you let $L\rightarrow \infty$, the atomic distribution becomes flat. So, again, very much atoms not at the same positions.
Ok, so now interactions and collapse.
First of all, BEC is a non-interacting effect. It is not driven by a competition of interaction terms, but solely by Bose-Einstein statistics. It is experimentally interesting that BEC seems to exist also in interacting systems, though there is no general theoretical proof. By BEC in an interacting system I mean macroscopic occupation of the ground state + Off-Diagonal Long-Range Order (ODLRO) — so not all superfluids are BECs. Let me also point out that you need interactions to reach a BEC as you need to reach thermal equilibrium.
The interaction strength among weakly interacting Bose-condensed bosons is quantified by a $g n$ term in the Hamiltonian, where $g$ is $4\pi\hbar^2 a/m$ (Gross-Pitaevski equation). You can make this interaction attractive with $a<0$ and repulsive with $a>0$, where $a$ is the scattering length and it is given by $a(B) = a_0 f(B)$, where $a_0$ is the background scattering length in the presence on no external magnetic field $B$ ($f$ is some function).
The pressure of a weakly interacting Bose-condensed gas is (at $T=0$):
$$ P = -\frac{\partial E}{\partial V} = \frac{1}{2}gn^2.$$
Because $n^2$ is always positive, the condition for stability (i.e. not to collapse) is $P>0$ and hence $g>0 \Rightarrow a>0$ i.e. a repulsive system. With a positive pressure, the gas expands until it hits a wall (e.g. the confining potential). But if $P<0$ then the system is intrinsically unstable and collapses.
Rb-87 is "easy" because its background scattering length is positive and therefore trivially allows for a stable BEC. K-39, on the other hand, has a negative background scattering length so its "BEC" would collapse (and eventually explode). But its scattering length can be made repulsive by the use of a Feshbach resonance (applying a field $B$ to change $a$) so that it can undergo BEC.
Such a theory would be an effective field theory. It is often the case that different degrees of freedom are separated by large differences in energy. For example when solving the wave equation for a hydrogen atom we can, to a good approximation, treat the nucleus as a point charge and ignore its internal structure. We need only worry about the nuclear structure at energies above a GeV or so when the non-zero size of the nucleus becomes important.
In the context of quantum field theories an example is the effective theory describing the strong nuclear force as a force between hadrons due to the exchange of virtual pions (there is an interesting review here). Neither the hadrons nor the pions are fundamental particles, but as long as we use the theory only at energies well below a GeV it works fine.
In principle I guess we could construct an effective field theory to describe the interactions between atoms in a BEC, but it isn't obvious that this would be a useful thing to do. QFT becomes important when the energies are high enough to make particle production and annihilation possible, but BECs can only be created at very, very low energies i.e. temperatures only a hair's breadth from absolute zero. In this regime it's hard to see what a QFT would add to regular quantum mechanics.
Best Answer
First, let me say I'm not sure what is meant by a BEC with $T\gt 0$. Condensation is a finite temperature phenomenon, which occurs due to the presence of pair-wise interactions (generally attractive, but pairing can happen even for repulsive potential) in a many-body system. For instance, in a superconductor below some critical temperature $T_c$, electrons with opposite momenta and spin (s-wave pairing) pair up to form a bound state called a Cooper pair.
The ground state of the unpaired electron gas for $T \gt T_c$ is characterized by the Fermi energy $E_F$. After condensation, the many-body system has a new ground state at energy $E_{bcs} = E_F - \Delta $, where $\Delta \sim k_B T_c$ is the binding energy of a Cooper pair. $\Delta$ is also known as the gap.
For $T\lt T_c$ all the electrons are not paired up due to thermal fluctuations. However, the number of unpaired electrons as a fraction of the total number of electrons (the condensate fraction) goes as $N_{free}/N_{all} = 1- (T/T_c)^\alpha$, where $\alpha\gt 0$. The number of free electrons drops rapidly as $T$ is decreased below $T_c$. In lab setups, BEC's generally undergo some form of evaporative cooling to get rid of particles with energies greater than $\Delta$. At this point the condensate can be treated as a gas of interacting (quasi)particles (cooper pairs) with an approximate hard-core repulsion.
So the gas, before and after condensate formation, is always at finite temperature! This is reflected, for instance, in the dependance of the condensate fraction on $T$ as mentioned above.
The mean-field solutions for low-energy excitations of the condensate are given by the Gross-Pitaevskii equation(GPE):
$$ \left( - \frac{\hbar^2}{2m}\frac{\partial^2}{\partial r^2} + V(r) + \frac{4\pi\hbar^2 a_s}{m} |\psi(r)|^2 \right) \psi(r) = \mu \psi(r) $$
where $a_s$ is the scattering length for the hard-core boson interaction, with $a \lt 0$ for an attractive interaction and $a \gt 0$ for a repulsive interaction.
Presumably one should be able to construct a canonical ensemble with solutions $\psi(k)$ ($k$ being a momentum label of the above equation), but this is by no means obvious because of the non-linearity represented by the $|\psi(r)|^2$ term. Here a "zero temperature" state would correspond to a perfect BEC with no inhomogeneities, i.e. the vacuum solution of the GPE. However, the entire system is at some finite temperature $T \lt T_c$ as noted above. The resulting thermal fluctuations will manifest in the form of inhomogeneities in the condensate, the exact form of which will be determined by the solutions of the GPE.
Of course, the GPE's regime of validity is that of dilute bose gases ($l_p \gg a_s$ - the average interparticle separation $l_p$ is much greater than the scattering length). For strong coupling I do not know of any similar analytical formalism. If I had to take a wild guess I'd say that the strong-coupling regime could be made analytically tractable by mapping it to a dual gravitational system, but that's another story altogether.
As $l_p$ approaches $a_s$ from above, the GPE breaks down and it will have singular solutions for any given $T$ and these are likely the singularities that you are referring to.
Reference: The single best reference I can suggest is Fetter and Walecka's book on many-body physics. I'm sure you can find more compact sources with a little effort. But generally the brief explanations leave one wanting for a comprehensive approach such as the one F&W provides.
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