Getting the Bose-Hubbard Hamiltonian from cold atoms In the famous paper by Dieter Jaksch, it is shown that the usual Hamiltonian for cold bosonic atoms interacting by s-wave scattering (Equation (1) in the paper):
$$ \hat{H}=\int d^3 x\hat{\psi}^\dagger(x)\left(-\frac{\hbar^2}{2m}\nabla^2+V_0(x)+V_T(x)\right)\hat{\psi}(x)+\frac{1}{2}\frac{4\pi a_s\hbar^2}{m}\int d^3x \hat{\psi}^\dagger(x)\hat{\psi}^\dagger(x)\hat{\psi}(x)\hat{\psi}(x) $$
becomes the Hubbard model Hamiltonian (Equation (2) in the paper):
$$\hat{H}=-J\sum_{\langle i,j\rangle}\hat{b}^\dagger_i\hat{b}_j+\sum_i \epsilon_i \hat{n}_i+\frac{1}{2}U\sum_i\hat{n}_i\left(\hat{n}_i-1\right)$$
under certain approximations (expanding out the field operators in a Wannier basis, throwing away higher-order modes, neglecting overlap integrals that are small, etc). In particular, we have, for adjacent sites $i$ and $j$:
$$J=\int d^3x w^*(x-x_i)\left(-\frac{\hbar^2}{2m}\nabla^2+V_0(x)\right)w(x-x_j)$$
I do not understand why there is a minus sign in front of the $J$ in the Hamiltonian. When I attempt to transform the first Hamiltonian into the second one, I get a positive $J$. All I'm doing is expanding the field operators, throwing away the small overlaps, pulling out the lattice-site creation/annihilation operators, and dubbing the energy integral of the Wannier functions that remains $J$. It's totally unclear to me how there could possibly be anything giving a sign change in this process. What am I missing?
 A: I think that you are doing too much work here! I doubt that Jaksh is  actually doing any actual calculation to go from the lattice to the continuum.  He is  simply thinking  that a postive $V(x)$ means a positive $\epsilon_i$ and a positive scattering length $a_s$ means a positive $U$, and so on. In the continuum the kinetic energy goes as $E=+k^2/2m$.  To get a positive coeffient  of $k^2$ in a tight binding  lattice model you want  a $(1-\cos(k))\sim k^2/2$. If I remember correctly, this requires a negative number before the $b^\dagger_i b_j$ hopping term.   He has just chosen his parameter $J$ to achieve this. 
Well  I looked at the paper and I agree with the OP that the Bose-Hubbard should have a $+J$ since the formula is just an evaluation of 
$$
\int \psi^\dagger(x)(-\nabla^2 +v(x))\psi(x)d^3x
$$
for $\psi(x)= \sum_i b_i w(x-x_i)$, assuming nearest neighbour overlap. The eigenvalues of the single-particle tight-binding approx 
$$
H= J\sum_{<ij>} b_i^\dagger b_j 
$$
are 
$$
J(\cos k_x+\cos k_y+\cos k_z)
$$
for a square lattice. the  eigenkets are 
$$\sum_i e^{i{\bf k}\cdot {\bf x}_i} b^\dagger_i |0\rangle.
$$ 
We would like $J$ to be negative if the minimum energy is to be at ${\bf k}=0$. I think his minus sign is just an error.
