# Hartree-Fock on a lattice: how to deal with non-diagonal charges

For a Hubbard hamiltonian, $$\hat{H} = \sum_{i,j,\sigma} t_{i j} \hat{c}^\dagger_{i \sigma} \hat{c}_{j \sigma} + \sum_i U_i \hat{n}_{i \uparrow} \hat{n}_{i \downarrow},$$ the mean-field solution can be obtained by using: $$\hat{n}_{i \uparrow} \hat{n}_{i \downarrow} \approx n_{i \uparrow} \hat{n}_{i \downarrow} + n_{i \downarrow} \hat{n}_{i \uparrow} - n_{i \uparrow} n_{i \downarrow}$$ and solving self-consistenly the resulting one-electron hamiltonian. In general, for a generic many-body term $$\hat{c}^\dagger_{i \sigma}\hat{c}^\dagger_{j \sigma^\prime}\hat{c}_{k \sigma^\prime}\hat{c}_{l \sigma},$$ the approximation is: $$\hat{c}^\dagger_{i \sigma}\hat{c}^\dagger_{j \sigma^\prime}\hat{c}_{k \sigma^\prime}\hat{c}_{l \sigma} \approx \langle \hat{c}^\dagger_{i \sigma} \hat{c}_{l \sigma} \rangle \hat{c}^\dagger_{j \sigma^\prime}\hat{c}_{k \sigma^\prime} + \langle \hat{c}^\dagger_{j \sigma^\prime}\hat{c}_{k \sigma^\prime} \rangle \hat{c}^\dagger_{i \sigma} \hat{c}_{l \sigma} - \langle \hat{c}^\dagger_{j \sigma^\prime}\hat{c}_{k \sigma^\prime} \rangle \langle \hat{c}^\dagger_{i \sigma} \hat{c}_{l \sigma} \rangle \\ -\langle \hat{c}^\dagger_{i \sigma} \hat{c}_{k \sigma^\prime} \rangle \hat{c}^\dagger_{j \sigma^\prime}\hat{c}_{l \sigma} - \langle \hat{c}^\dagger_{j \sigma^\prime} \hat{c}_{l \sigma}\rangle \hat{c}^\dagger_{i \sigma}\hat{c}_{k \sigma^\prime} + \langle \hat{c}^\dagger_{j \sigma^\prime} \hat{c}_{l \sigma} \rangle \langle \hat{c}^\dagger_{i \sigma}\hat{c}_{k \sigma^\prime} \rangle$$

(we assume the particle number is conserved, so that terms like $$\langle \hat{c}^\dagger \hat{c}^\dagger \rangle$$ are not important). Now, $$S_z$$ also being a good quantum number, if $$\sigma \neq \sigma^\prime$$ we will also ignore the terms on the second line, as done for the approximation above for Hubbard model. The Hubbard model on a lattice can be solved easily on mean-field thanks to the fact that on the self-consistency loop we only need the charges, and never non-diagonal charges (other elements of the density matrix) like $$\langle \hat{c}^\dagger_{j \sigma^\prime} \hat{c}_{l \sigma} \rangle$$.

However think now of a slightly more complicated model on a lattice. It's like the hamiltonian $$\hat{H}$$ above but now we add many-body interactions between neighboring sites; something like: $$\dfrac{1}{2}\sum_{i,j, \sigma,\sigma^\prime} J_{ij} \hat{n}_{i \sigma} \hat{n}_{j \sigma^\prime}.$$ Now, for the terms with $$\sigma \neq \sigma^\prime$$ we have the simple approximation $$\hat{n}_{i \uparrow} \hat{n}_{j \downarrow} \approx n_{i \uparrow} \hat{n}_{j \downarrow} + n_{j \downarrow} \hat{n}_{i \uparrow} - n_{i \uparrow} n_{j \downarrow}$$. However, for terms like $$\hat{n}_{i \sigma} \hat{n}_{j \sigma}$$, when doing the mean-field approximation we will unavoidable find terms like those of the second line for the generic equation above; that is, the term: $$-\langle \hat{c}^\dagger_{i \sigma} \hat{c}_{j \sigma} \rangle \hat{c}^\dagger_{j \sigma}\hat{c}_{i \sigma} - \langle \hat{c}^\dagger_{j \sigma} \hat{c}_{i \sigma}\rangle \hat{c}^\dagger_{i \sigma}\hat{c}_{j \sigma} + \langle \hat{c}^\dagger_{i \sigma} \hat{c}_{j \sigma} \rangle \langle \hat{c}^\dagger_{j \sigma}\hat{c}_{i \sigma} \rangle,$$ and here is the problem. We need to include somehow in the self-consistency the non-diagonal charges $$n_{ij\sigma}:=\langle \hat{c}^\dagger_{i \sigma} \hat{c}_{j \sigma} \rangle.$$

In order to do self-consistency on the charges we should express the $$n_{ij \sigma}$$ in terms of the $$n_{i \sigma},$$ but I don't think there is any analytical way of doing this even when all expected values are taken over Slater deterinants. Am I right? I know that I can express sums of non-diagonal charges, $$\sum_j n_{i j \sigma}$$ comfortably as sums of electron-hole interactions, $$\sum_{i \neq j} n_{i \sigma} \left( 1 - n_{i \sigma} \right)$$, which is convenient to write the Hartree-Fock energy of lattice models when the $$J_{ij}$$ parameters are constant. But in general, an isolated $$n_{ij \sigma}$$ seems untreatable. Two possible solutions come to mind:

1. Finding approximations for $$n_{ij \sigma}$$ in terms of $$n_{k \sigma}$$. How can this be done? Can the identity with $$\sum_{i \neq j} n_{i \sigma} \left( 1 - n_{i \sigma} \right)$$ (that is exact over Slater determinants) be used?

2. Doing self-consistency in terms of $$n_{ij \sigma}$$ instead of just the charges. I have tried this last option for things like PPP parametrization on hydrocarbons and I've encountered cases on which, even using a mixing algorithm on all the density matrix, the convergence is quite problematic. Has this be done before and how? (meaning, Lattice Hartree-Fock where self-consistency is done in terms of the full density matrix)

Once the Hartree-Fock approximation is done in the Hamiltonian, it becomes a quadratic one, i.e., it can be exactly diagonalized. In terms of the operators it means a linear transformation like $$c_{j\sigma}=\sum_kU_{jk}a_{k\sigma}\leftrightarrow a_{k\sigma}=\sum_jV_{kj}c_{j\sigma}$$ One can then express any product of operators in terms of new operators in the diagonal basis: $$n_{j\sigma}=c_{j\sigma}^\dagger c_{j\sigma}=...\text{etc.}$$ Thus, there are no conceptual difficulties, although the non-diagonal terms might make solving the equations self-consistenly trather tedious or require some numerics (like diagonalizing a matrix).
• @Qwertuy self-consistency means that you solve the closed system of equations for $\langle c_{i\sigma}^\dagger c_{j\sigma}\rangle$. These quantities can be expressed in terms of the solutions of the quadratic Hamiltonian, i.e., in terms of the coefficients $U_{jk}, V_{kj}$, which in turn depend on the quantities $\langle c_{i\sigma}^\dagger c_{j\sigma}\rangle$, since these are coefficients in the Hamiltonian. I am not sure I understand where the problem lies... perhaps, you could elaborate further in your post, how you set up the self-consistent equations. Commented May 19, 2022 at 13:26