Strong correlation effects in a two-dimensional Bose gas with quartic dispersion

Motivated by the fundamental question of the fate of interacting bosons in flat bands, we consider a two-dimensional Bose gas at zero temperature with an underlying quartic single-particle dispersion in one spatial direction. This type of band structure can be realized using the NIST scheme of spin-orbit coupling [Y.-J. Lin, K. Jiménez-Garcia, and I. B. Spielman, Nature (London) 471, 83 (2011)], in the regime where the lower-band dispersion has the form ${\varepsilon }_{\mathit{k}}\sim k_x^4/4+k_y^2+...$, or using the shaken lattice scheme of Parker et al. [C. V. Parker, L.-C. Ha, and C. Chin, Nat. Phys. 9, 769 (2013)]. We numerically compare the ground-state energies of the mean-field Bose-Einstein condensate (BEC) and various trial wave functions, where bosons avoid each other at short distances. We discover that, at low densities, several types of strongly correlated states have an energy per particle ($\epsilon$), which scales with density ($n$) as $\epsilon \sim n^{4/3}$, in contrast to $\epsilon \sim n$ for the weakly interacting Bose gas. These competing states include a Wigner crystal, quasicondensates described in terms of properly symmetrized fermionic states, and variational wave functions of Jastrow type. We find that one of the latter has the lowest energy among the states we consider. This Jastrow-type state has a strongly reduced, but finite, condensate fraction, and true off-diagonal long-range order, which suggests that the ground state of interacting bosons with quartic dispersion is a strongly correlated condensate reminiscent of superfluid helium-4. Our results show that even for weakly interacting bosons in higher dimensions, one can explore the crossover from a weakly coupled BEC to a strongly correlated condensate by simply tuning the single-particle dispersion or density.

Figure 1

We show the correlation function $\rho \left(\mathit{r}\right)$ for two different states: (a) composite-fermion state ${\psi }_{B,\mathrm{cf}}^{\mathrm{sq}}$ (red dashed-dotted line) and (b) Jastrow wave function ${\psi }_{\mathrm{J}}$ (black line). Here, we set $y=0$ and concentrate at the dependence on $x$. ${\psi }_{B,\mathrm{cf}}^{\mathrm{sq}}$ wave function has algebraically decaying correlations, i.e., it has a quasi-long-range order, while ${\psi }_{\mathrm{J}}$ has a true long-range order (see text for details).