Sign of the hopping integral in tight binding model The Hamiltonian of tight binding model reads $H=-|t|\sum\limits_{<i,j>}c_i^{\dagger}c_j+h.c.$, why is there a negative sign in the hopping term?
 A: It doesn't have to be negative. That depends on the type of band you want to model. 
You can easily diagonalize the tight-binding model by going to momentum space. In one dimension, the resulting dispersion relation is then
$$\epsilon(k) = -2t\cos(k a)$$
where $a$ is the lattice constant. 
This is a band with a minimum at $k = 0$. If you had $+|t|$ in your tight binding Hamiltonian, you'd get a band with a maximum at $k = 0$. Both exist in real systems. If you have a minimum at $k = 0$ you call it an electron-like band and if it has a maximum at $k = 0$ you call it a hole-like band.
A: The hopping term is given by
$$
t_{ij}=\int\limits_{\mathbf{r}}d\mathbf{r}\phi_i^*\left(\mathbf{r}\right)\left[-\frac{\hbar^2}{2m}\nabla^2+U(\mathbf{r})\right]\phi_j\left(\mathbf{r}\right)
$$
where $i$ and $j$ are the sites whose hopping you want to find, $\phi_{i,j}(\mathbf{r})=\phi(\mathbf{r}-\mathbf{r}_{i,j})$ ($\mathbf{r}_i$ being coordinate of $i$-th site) are the atomic orbitals, and $U(\mathbf{r})$ is the potential of the crystal lattice. So the sign depends on your choice for $U(\mathbf{r})$. If $U(\mathbf{r})$ is taken as negative (Coulomb potential) you are more likely to end up with a negative $t$ which can be a nuisance, so people just redefine it as $-|t|$.
