Is Bose-Einstein condensate in the optical lattice a single mode condensate?

Question

I recently read about BEC loaded into the optical lattice p.200

Looking at a condensate released from a lattice after a time of flight typically on the order of a few milliseconds amounts to observing its momentum distribution. A harmonically trapped condensate has a Gaussian momentum distribution in the limit of small interactions, whereas in the Thomas-Fermi limit in which the interactions dominate over the kinetic energy contribution it has a parabolic density profile and expands self-similarly after being released. By contrast, a condensate in a periodic potential contains higher momentum contributions in multiples of 2kL, their relative weights depending on the depth of the lattice. In fact, in the tight-binding limit see Sec. IV we can consider the condensate to be split up into an array of local wave functions that expand independently after the lattice has been switched off. Eventually they all overlap and form an interference pattern that in the absence of interactions is the Fourier transform of the initial condensate.

When there is no lattice potential all particles occupy the same state, but when the lattice appears is it still a single mode condensate or multi-mode one due to appearance of additional interference peaks (particle condense in more than one state)?

Answer 1

In a lattice, forgetting about the interactions, for now, the eigenstate of the Hamiltonian are the Bloch functions $|u_{n,q}\rangle$, which have the same periodicity as the lattice (I will do everything in 1D to alleviate the notations, but it is easily generalized to other lattices). This implies that it has a Fourier series decomposition $$ u_{n,q}(x)=\sum_ {m\in \mathbb {Z}} \tilde u_{n,q}(m) e^{i m G x}, $$ with $G$ the primitive vector of the reciprocal lattice.

Assume that, as is usually the case, the lowest energy state is the state in the band $n=0$ with quasimomentum $q=0$. The BEC then forms in this state, which is macroscopically occupied. When the trap is released to perform a time-of-flight experiment, the atoms start in the state $|u_{0,0}\rangle$, but evolve with the free particle Hamiltonian. It can then be shown that under some hypothesis (no collisions, long enough times, etc.) that the density of atoms is proportional to the Fourier transform of the initial wave-function, with the wave-vector replaced by $\frac{m x}{\hbar t}$.

In the present case, this means that the density $n(x,t)$ measured after a time-of-flight of duration $t$ will be given by $$ n(x,t)\propto \sum_ {m\in \mathbb {Z}} \tilde u_{n,q}(m) \delta\left(\frac{m x}{\hbar t}-m G\right), $$ where the $\delta$ function comes from the Fourier transform of the exponentials. Thus, the density measurement gives a sum of picks corresponding to the reciprocal lattice (picks of finite width in practice due to the finite duration of the flight), with weight proportional to the Fourier coefficients of the Bloch wavefunction.