Superfluid to Mott Insulator Transition

A condenstate in a shallow standing wave potential is a BEC, well described by a Bogoliubov approximate solution. As the potential gets deeper, though, eventually the system transitions into a state of localized atoms, with no long-range coherence, known as a Mott insulator. These physics are important in a wide range of condensed matter systems, and can be explored deeply with BECs.

The discussion of superfluid to Mott insulator (SF-MI) transition involves some basic knowledge of periodic potentials from solid state physics. We will review the concept briefly. To observe SF-MI transition, ultracold bosons are placed in an optical lattice, which is formed by a standing wave of light. A variety of optical lattice geometry can be generated by how the laser beams interfere with each other. In this article we only consider the simple (and most common) case of separable, sinusoidal potential, generated by a pair of retroreflected beams in each axis:

$V(x,y,z)=V_{0}(\sin ^{2}(kx)+\sin ^{2}(ky)+\sin ^{2}(kz))$

The Hamiltonian $-\hbar ^{2}\nabla ^{2}/2m+V$ is periodic in space (ignoring interactions for now) with period $a$, so we can use Bloch's theorem, which says that the wavefunctions of the eigenstates $|\psi \rangle$ of a periodic Hamiltonian can be written as

${\begin{array}{rcl}\psi (x)&=&e^{ikx}u_{n,k}(x)\,,{\Big (}-{\frac {\pi }{a}}\leq k\leq {\frac {\pi }{a}},n=1,2,\ldots {\Big )}\\u_{n,k}(x+a)&=&u_{n,k}(x)\end{array}}$

What this means is that if we translate the wavefunction by distance $a$, then the wavefunction can only change up to a complex phase. From a more formal point of view, because the continuous translation symmetry is lost and only discrete translation symmetry remains, momentum is only conserved modulo $\hbar K=\hbar {\frac {2\pi }{a}}$. For this reason the quantity $\hbar k$ is called quasimomentum. If the lattice potential were zero, the quasimomentum and the momentum are related by $p=\hbar k+n\hbar K$. The integer $n$ is called the band index, because if you draw the free particle dispersion $E=\hbar ^{2}k^{2}/2m$ and "folding" the parabola inside the region $-{\frac {\pi }{a}}\leq k\leq {\frac {\pi }{a}}$, you get a discrete number of energy bands. The region $-{\frac {\pi }{a}}\leq k\leq {\frac {\pi }{a}}$ in reciprocal space is called the first Brillouin zone, or simply the Brillouin zone. When the lattice potential is zero ($V_{0}\to 0$), there are points where some of the energy bands become degenerate (e.g. the first and second bands touch at the boundaries of the Brillouin zone). When the lattice potential is nonzero, the degeneracy is broken and an energy gap opens between each band.

The discussion above becomes clearer if we apply Bloch's theorem to express the Hamiltonian in reciprocal space. Applying

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