Condensation of a repulsive Bose gas in the thermodynamic limit
In plain words
Prove that a gas of mutually repelling bosons at low temperature puts a finite fraction of its particles into one quantum state when the container is infinitely large at fixed density. Without interactions this is a textbook calculation; with them there is no accepted proof.
Precise statement
$N$ bosons in a box of side $L$ in $R^{3}$, $H = \operatorname{sum}_i -\operatorname{Lap}_i + \operatorname{sum}_{i<j} v(x_i - x_j)$ with $v \ge 0$ compactly supported, density $\rho = N/L^{3}$ fixed, $L \to \infty$. Prove at $T = 0$ or at small $T > 0$ that the one-body density matrix has off-diagonal long-range order, lim_{|x-y| -> infinity} gamma^(1)(x,y) = $\rho$_0 > 0. Answer: a proof for some open range of rho and $T$.
What would settle it
A proof of a nonzero condensate density in the thermodynamic limit for one continuum model.
Status in the literature
Proven in the Gross-Pitaevskii scaling (Lieb and Seiringer, 2002) and for lattice hard-core bosons at half filling (Dyson, Lieb and Simon, 1978), not for continuum gases in the thermodynamic limit; a preprint by Suto (arXiv 1710.04441, 2017) claims a proof via Feynman-Kac permutation cycles and lists no journal publication.
Related problems
- More general than Condensation on all length scales in the dilute Bose gas
- More general than Shift of the condensation temperature caused by weak repulsion