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I was wondering for the three-site Bose Hubbard model how we can get the equations of motion out of the following hamiltonian:

$$ \hat{H} = \hbar \chi \sum_{j} \hat{\alpha}_{j}^{\dagger} \hat{\alpha}_{j}^{\dagger} \hat{\alpha}_{j} \hat{\alpha}_{j} - \hbar J \sum_{\left( j k \right)} \hat{\alpha}_{j}^{\dagger} \hat{\alpha}_{k}$$

I know that the equations of motion can be obtained by replacing the bosonic operators with complex numbers in the Heisenberg equations of motion and assuming that all moments factorise... The solution is said to be:

$$ \frac{d \alpha_{j}}{d t} = - 2 i \chi \left| \alpha_{j} \right|^{2} \alpha_{j} + i J \sum_{k = j \pm 1} \alpha_{k}$$

where the $\alpha_j$ are classical variables representing atomic amplitudes.

Furthermore, I was wondering for the three-sit Bose Hubbard model, if it is true that for j $\geq 4$, $\alpha_j =0$? Or how do I have to interpret the Sum?

Thank you very much for the help!!

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Just solve this coupled set of diffrential equations: \begin{equation} \left\{ \begin{array}{ll} \frac{\partial}{\partial t}\psi_1= -\frac{Ji}{2}(-\psi_2 + \frac{2UN}{J}|\psi_1|^2\psi_1) \\\\ \frac{\partial}{\partial t}\psi_2= -\frac{Ji}{2}(-\psi_1 - \psi_3 + \frac{2UN}{J}|\psi_2|^2\psi_2) \\\\ \frac{\partial}{\partial t}\psi_3= -\frac{Ji}{2}(-\psi_1 + \frac{2UN}{J}|\psi_3|^2\psi_3) \end{array} \right. \end{equation} Mark that U,N and J are constants contained within the system.

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