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

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

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. We can write $\psi (x)=\sum _{G}c_{k-G}e^{i(k-G)x},G=n(2\pi )/a$ using Bloch's theorem and similarly $V(x)=\sum _{G}V_{G}e^{-iGx}$. The net effect is that the Schrodinger equation becomes an algebraic equation in reciprocal space, with $V_{G}$ appearing as off-diagonal matrix element that couples the amplitude $c_{k+nG}$ to $c_{k+(n-1)G}$ (intuitively, the modulation due to lattice potential transfers momentum to the wavefunction). If we write the eigenvalue equation for 1-dimensional case, it looks like

${\begin{pmatrix}\ddots &\vdots &\vdots &\vdots &\vdots &\vdots \\\dots &\lambda _{k+G}-E&V_{-G}&V_{-2G}&V_{-3G}&\dots \\\dots &V_{G}&\lambda _{k}-E&V_{-G}&V_{-2G}&\dots \\\dots &V_{2G}&V_{G}&\lambda _{k-G}-E&V_{-G}&\dots \\\dots &\vdots &\vdots &\vdots &\vdots &\ddots \end{pmatrix}}{\begin{pmatrix}\vdots \\c_{k+G}\\c_{k}\\c_{k-G}\\\vdots \end{pmatrix}}=0$

The solutions to this equation, i.e. the eigenstates of the periodic Hamiltonian, are called the Bloch states. Similar to plane wave solutions for free particle Hamiltonian, the Bloch states are delocalized over the lattice sites. But for our application, where interaction term appears as a on-site term, it would be more convenient to work with a basis set with states localized at each lattice site. The wavefunction for such states, $\phi (r)$ are called Wannier functions, which is constructed as a complex linear combination of Bloch states (analogous to Fourier transform).

$\phi (r-R)=\sum _{k\in BZ}{\frac {1}{\sqrt {N}}}e^{-ikR}e^{i\phi _{n}(k)}\psi _{n,k}(r)\,$

where $R$ is any lattice vector. Wannier functions are not unique in general, but one can define a localization function and find out that there exists a unique set of maximally localized Wannier functions (MLWF) - see the review paper by Marzari and Vanderbilt (RMP 84, 1419).

We are now ready to introduce the Bose-Hubbard Hamiltonian, which sets the framework of the SF-MI transition. Assume that all the atoms are in the lowest band, and the lattice is deep so that we work in the tight-binding limit - the dynamics is described as atoms tunneling between nearest-neighboring sites.

$H=-t\sum _{\langle i,j\rangle }{\Big (}{\hat {a}}_{i}^{\dagger }{\hat {a}}_{j}+{\hat {a}}_{j}^{\dagger }{\hat {a}}_{i}{\Big )}+U_{0}\sum _{i}{\hat {a}}_{i}^{\dagger }{\hat {a}}_{i}^{\dagger }{\hat {a}}_{i}{\hat {a}}_{i}-\mu \sum _{i}{\hat {a}}_{i}^{\dagger }{\hat {a}}_{i}$

This Hamiltonian can be described from the Hamiltonian written in terms of field operator ${\hat {\psi }}(r)$ (see the notes on "Inhomogeneous Bose gases") and expanding the field operator in the Wannier function basis (${\hat {\psi }}=\sum _{i}\phi (r-R_{i}){\hat {a}}_{i}$).

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