Abstract

The ground state of a system of interacting Bose particles is examined by a pair-correlation variational method based on exact eigenstates of the total particle number, without assuming macroscopic occupation of the single-particle state k = 0 (``simple'' Bose condensation). Known results are recovered in the case of repulsive interactions of weak and intermediate strengths, where the ground state does in fact exhibit simple condensation. But it is shown that in the case of weak attractive interactions the ground state exhibits ``generalized'' condensation, in which a finite fraction of particles have momenta which are macroscopically negligible, but no one single-particle state is macroscopically occupied. A weak-coupling expansion is derived for the ground-state energy in this case, the leading term of which is three times as low as that of the state one would obtain in the case of simple condensation. By the use of a pseudopotential, this result is adapted to the case of a strongly interacting system with a small negative scattering length α and fixed low density ρ, giving E0/n=−(6π ℏ2 ρ | α | /m)[1−(128/45) π−1/2(ρ | α | 3)1/2+…]. The relationships between equilibrium density, ground-state energy, and sound velocity predicted by this expression are in order-of-magnitude agreement with those obtaining for liquid He4.

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