The Bose-Hubbard-Hamiltonian reads:

$ 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?


Since you have a Hamiltonian with discrete sites, I assume that you consider a periodic lattice.

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.

  • $\begingroup$ Thank you for your answer! But shouldn't I integrate then from -Pi to Pi? Since all values of k are allowed? $\endgroup$ – QuantumMechanics Nov 8 '15 at 20:02
  • 1
    $\begingroup$ @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. $\endgroup$ – Praan Nov 8 '15 at 20:09

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