Bose-Hubbard in momentum space

$H=-t\sum_{<i,j>} c_i^\dagger c_j+\frac{1}{2}U\sum c_i^\dagger c_i^\dagger c_i c_i -\mu\sum c_i^\dagger c_i$

I can use a FT to get from space to momentum repsresentation:

$c_i=\sum a_k e^{-ikr}\\ c_i^\dagger=\sum a_k^\dagger e^{ikr}$

My question is, over which values runs k now? Minus Infinity to Infinity?

For a periodic one-dimensional chain with unit lattice constant, all independent solutions, labelled by $k$, can be chosen to lie in the first Brillouin zone $k \in [-\pi,\pi]$ of the 1D $k$-space. This is because the only values of the wave function that matter are at the sites of the lattice. The states with $k$ and $k+2\pi$ represent the same state. Topologically, the first Brillouin zone (BZ) is a torus.
Your sum should therefore run over all $k$ in the first BZ: $\sum_{k \in BZ} = \frac{N}{2\pi} \int_{-\pi}^\pi dk$ with $N$ the number of sites.
For higher dimensional periodic systems, the Brillouin zone has a more complicated geometry depending on the crystal symmetry. In the simplest case, for a square lattice in 2D, it just a square centred on the origin of the 2D $\vec k$-space. So in this case, your sum should go over all $\vec k$ of this square.
• @QuantumMechanics The amount of $k$ points is given by the number of unit cells $N$. For very large $N$, $k$ is almost continuous and $\sum_k = \frac{N}{2\pi} \int_{-\pi}^{\pi} dk$. I edited my answer. – Praan Nov 8 '15 at 20:09