My question is very simple. What is the correct way of modelling a Coulomb interaction on a 2D lattice?

Usually for a system that is infinitely big $(N\to\infty)$ and not discrete $(a_0\to 0)$, the two particle operator is simply $$ V=\sum_{\vec{k}_1,\vec{k}_2,\vec{p}}V(\vec{p})c^{\dagger}_{\vec{k}_1+\vec{p},\uparrow}c^{\dagger}_{\vec{k}_2-\vec{p},\downarrow}c_{\vec{k}_2,\downarrow}c_{\vec{k}_1,\uparrow} $$ where I guess $$ V(\vec{p})\propto\frac{1}{\sqrt{\Vert\vec{p}\Vert^2+\kappa^2}} $$ and $\kappa$ is the screening parameter. But obviously this can't be true on a lattice where the momenta are merely crystal momenta such that $\vec{k}$ and $\vec{k}+\vec{K}$ is considered equivalent when $\vec{K}$ is a reciprocal lattice vector.

What is the correct discretized form of the Coulomb interaction on a 2D reciprocal lattice?


1 Answer 1


Not sure why the number of dimensions matters here. When dealing with a lattice we can attribute to each lattice site wave functions of its orbitals, $$ \phi_{n,j}(\mathbf{x})=\phi_{n,0}(\mathbf{x}-\mathbf{x}_j), $$ and calculate the matrix elements of the Coulomb interaction between the orbitals - just the way we would do when, e.g., studying Helium molecule. Assuming for simplicity only one orbital per site, the Coulomb interaction this takes form: $$ V=\frac{1}{2}\sum_{j_1,j_2,j_4,j_3}\sum_{\sigma_1,\sigma_2} U_{j_1j_2j_3j_4}c_{j_1,\sigma_1}^\dagger c_{j_2,\sigma_2}^\dagger c_{j_3,\sigma_2}c_{j_4,\sigma_1},\\ U_{j_1j_2j_3j_4}=\int d\mathbf{x}d\mathbf{x}' \phi_{j_1}(\mathbf{x})^*\phi_{j_2}(\mathbf{x}')^*v(\mathbf{x}-\mathbf{x}')\phi_{j_4}(\mathbf{x})\phi_{j_3}(\mathbf{x}'). $$ Remark: the usual caveat here is different order of indices for the operators and the matrix element.

This is often further simplified, neglecting the Coulomb scattering or even keeping only on-site Coulomb interaction, in which case one obtains a Hubbard model: $$ V=\frac{1}{2}\sum_j\sum_\sigma U c_{j,\sigma}^\dagger c_{j,\bar{\sigma}}^\dagger c_{j,\bar{\sigma}}c_{j,\sigma}, $$ where the interaction between the equal spins gives a trivial constant energy shift, due to the Pauli principle $$ c_{j,\sigma}^\dagger c_{j,\sigma}^\dagger c_{j,\sigma}c_{j,\sigma}=c_{j,\sigma}^\dagger c_{j,\sigma}. $$ The Hubbard model ahs been extensively studied in different numbers of dimensions - e.g., for a square or honeycomb lattice in 2D.

  • $\begingroup$ Thank you for your answer. I should have clarified my question. I am aware of the Hubbard model in real space but what I am interested in is the reciprocal space, more precisely, the 2D discrete Fourier transform of the 3D Coulomb interaction. $\endgroup$
    – Wong Harry
    Oct 26, 2021 at 16:01
  • $\begingroup$ @WongHarry Why does it pose a problem? You can take, e.g., a habbard model, diagonalize the non-interacting part of the Hamiltonian, and use this transformation in the Coulomb interaction. $\endgroup$
    – Roger V.
    Oct 26, 2021 at 16:11

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