# 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?

• There might be more details in his PhD thesis. – Norbert Schuch Aug 21 '19 at 21:13
• And don't *** link to PDFs!! Even more so behind paywalls! – Norbert Schuch Aug 21 '19 at 21:14

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 $$H= J\sum_{} b_i^\dagger b_j$$ 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.