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

There is a single condensate only. The additional peaks appear due to the small occupation of higher-energy states, as the increasing well depth encourages localization of particles.

See also a similar question here.

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 $$ u_{n,q}(x)=sum_ {min mathbb {Z}} tilde u_{n,q}(m) e^{i m G x}, $$ 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 $$ n(x,t)propto sum_ {min mathbb {Z}} tilde u_{n,q}(m) deltaleft(frac{m x}{hbar t}-m Gright), $$ 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.