(De)localization index

I find the method used in this paper to be quite interesting. With a bit of research, I can now understand and reproduce some of their results.

First let’s rememeber the exchange density matrix:

\begin{aligned} \rho_x(r_1,r_2) &= |\rho(r_1,r_2)|^2 = |\sum_i \psi_i^*(r_1) \psi_i (r_2) |^2\\ & = \sum_i \psi_i^* (r_1) \psi_i (r_2) \sum_j \psi_j^*(r_2) \psi_j (r_1)\\ & = \sum_{ij} \psi_i^* (r_1) \psi_j^*(r_2) \psi_i (r_2) \psi_j(r_1) \end{aligned}

which is the numerator of the exchange integral. Previously, we have demonstrated that the probability of finding electron $1$ at $r_1$ and electron $2$ at $r_2$ is:

P(r_1,r_2) = \rho(r_1)\rho(r_2) - |\rho(r_1,r_2)|^2 = \rho(r_1)\rho(r_2) - \rho_x(r_1,r_2).

Immediately we know that:

\rho_x(r_1,r_2) = \rho(r_1)\rho(r_2) - P(r_1,r_2) ,

so that if one could obtain $\rho(r_1)\rho(r_2)$ and $P(r_1,r_2)$, $\rho_x(r_1,r_2)$ can be obtained.

If we integrate over the entire space for both $r_1$ and $r_2$, we get:

\begin{aligned} \int_{\R^3} \int_{\R^3} \rho_x(r_1, r_2) dr_1 dr_2 &= \int_{\R^3} \int_{\R^3} \sum^{occ}_{ij} \psi_i^* (r_1) \psi_j^*(r_2) \psi_i (r_2) \psi_j(r_1) dr_1 dr_2 \\ &= \sum^{occ}_{i=1} \sum_{i=j} \int_{\R^3} \psi_i^* (r_1) \psi_j(r_1) dr_1 \int_{\R^3} \psi_j^* (r_2) \psi_i(r_2) dr_2 \\ &+ \sum^{occ}_{i=1} \sum_{i \neq j} \int_{\R^3} \psi_i^* (r_1) \psi_j(r_1) dr_1 \int_{\R^3} \psi_j^* (r_2) \psi_i(r_2) dr_2 \\ &= \sum^{occ}_{i=1} \sum_{i=j} \int_{\R^3} \psi_i^* (r_1) \psi_j(r_1) dr_1 \int_{\R^3} \psi_j^* (r_2) \psi_i(r_2) dr_2 + 0 \\ &= N , \end{aligned}

where we’ve used the fact that $\psi$ is orthonormalized and summing to $occ$ gives back the number of electron $N$.

However, if we break up the integration using the Bader basin (each basin being associated with an atom) of the charge density, we can get two different types of “indcies”:

Delocalization index: measures the number of electron pairs that are being shared between quantum atoms $\Omega_{\mathrm{A}}$ and $\Omega_{\mathrm{B}}$.

\delta(\mathrm{A}, \mathrm{B})=\int_{\Omega_{\mathrm{A}}} \int_{\Omega_{\mathrm{B}}} \rho_{\mathrm{x}}\left(\mathbf{r}_1, \mathbf{r}_2\right) \mathrm{d} \mathbf{r}_1 \mathrm{~d} \mathbf{r}_2

Localization index: measures the electron population of each atom basin $\Omega_{\mathrm{A}}$.

\lambda(\mathrm{A})=\int_{\Omega_{\mathrm{A}}} \int_{\Omega_{\mathrm{A}}} \rho_{\mathrm{x}}\left(\mathbf{r}_1, \mathbf{r}_2\right) \mathrm{d} \mathbf{r}_1 \mathrm{~d} \mathbf{r}_2

and we can write:

\int_{\R^3} \int_{\R^3} \rho_x(r_1, r_2) dr_1 dr_2 = N = \sum_A \lambda(\mathrm{A}) + \sum_A \sum_{B \neq A} \delta(\mathrm{A}, \mathrm{B})

and the basin associated charge can be recovered by:

N(\Omega) = \lambda(\mathrm{A}) + \frac{1}{2} \sum_{B \neq A} \delta(\mathrm{A}, \mathrm{B})

Examples

Download examples.

I’ve prepared to examples using VASP and vaspwfc to generate DI and LI for diamond and MgO. Fist, we run DFT code and then do the Bader partition of the unit cell to obtain atomic basins. After that, we read the DFT wavefunction and calculate the exchange density, and then integrate over the basin.

For Diamond, there are two basin. The calculated LI and DI are:

Calculated DI : 0.95314
Calculated LI (basin 1) : 1.04748
Calculated LI (basin 2) : 1.04748

For MgO, there are two basin. The calculated LI and DI are:

Calculated DI : 0.17690
Calculated LI (basin 1) : 3.63374
Calculated LI (basin 2) : 0.01378

These results are comparable with the one obtained from the critic2 program with quantum espresso, as referenced in this paper.

However, I wasn’t able to exactly reproduce the results of this paper. Might have something to do with the PAW transformation or some discrepancies in the definition of DI and LI, I’ll put it aside for now.