2
$\begingroup$

Consider the Kohn-Sham equation \begin{align} \left( - \frac{\hbar^2}{2m} \nabla^2 + \nu_\mathrm{eff}(\mathbf{r}) \right) \varphi_j(\mathbf{r}) &= \varepsilon_j\varphi_j(\mathbf{r}) \end{align} The external potential will be considered the interaction potential of electrons and nuclei.

Then \begin{align} \nu_{eff}=-\sum_{j}^{K} \frac{e^2 \, Z_{j}}{|\mathbf{r_1}-\mathbf{R_j}|}+\int\frac{n(\mathbf{r_2})e^2}{r_{12}}\mathbf{dr_2} +\nu_{xc} \end{align}

This equation is solved by the method of a self-consistent field: $\varphi$ are first set, then $n$ is calculated from them, then the effective functional $\nu_{eff}$ is calculated.

However, I do not understand something.

At each step of calculating the effective potential, the term denoting the interaction of nuclei with electrons is constant? $r_{12}$ is the distance between electrons. In terms of this model, no changes in the coordinates of electrons occur at each step?

$\endgroup$
1
  • 1
    $\begingroup$ Why not on the site specifically for DFT modeling? $\endgroup$ Commented Jun 26, 2020 at 22:50

1 Answer 1

1
$\begingroup$

To start it is very important to be very careful when writing the relevant coordinates. In your Kohn-Sham equation you are writing the coordinate of the non-interacting electron that you are solving for as $\mathbf{r}$. To be consistent with this, your second equation should read:

$$ v_{\mathrm{eff}}(\mathbf{r})=v_{\mathrm{ext}}(\mathbf{r})+\int\frac{n(\mathbf{r}_2)e^2}{|\mathbf{r}-\mathbf{r}_2|}d\mathbf{r}_2+v_{\mathrm{xc}}(\mathbf{r}), $$

where I replaced your expression for the external potential with $v_{\mathrm{ext}}(\mathbf{r})$, as to answer your question it is enough to look at the Hartree potential (the second term). In the second term you have two electron coordinates, $\mathbf{r}$ and $\mathbf{r}_2$. The first is the electron coordinate at which you are calculating the effective potential $v_{\mathrm{eff}}(\mathbf{r})$, and to build the full potential you need to know its value for every $\mathbf{r}$. The second coordinate is the coordinate at which you calculate the electron charge density $n(\mathbf{r}_2)$ to then build the Hartree potential. As the integral is over all space, you consider all possible values of $\mathbf{r}_2$. Therefore, both coordinates $\mathbf{r}$ and $\mathbf{r}_2$ are changing: you are considering all possible values for both.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.