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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?

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

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  • $\begingroup$ I disagree - if you look between the two equations in the paper, Jaksch actually explicitly states he is expanding out the field operators in a Wannier basis, discarding the higher modes, and that they are well localised. He's definitely doing this calculation. $\endgroup$ – user502382 Aug 20 at 23:49
  • $\begingroup$ I looked at his paper and I agree with you, I will edit my answer. $\endgroup$ – mike stone Aug 21 at 15:32
  • $\begingroup$ I can't find Jaksch's thesis, but in a quick survey of other theses dealing with optical lattices finds almost a 50/50 split between those that define J in the same way he does and those that add a minus sign (see, for example, Daley's thesis: qo.pitt.edu/downloads/daleyphdthesis.pdf ). I am inclined to agree that the latter is correct. $\endgroup$ – Rococo Aug 22 at 15:57
  • $\begingroup$ It is conventional to have the minus sign in the Hamiltonian so that the tunneling element is positive, so the possible error should be thought of as in the definition of J rather than the Hamiltonian itself. $\endgroup$ – Rococo Aug 22 at 16:04

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