I have recently started doing DFT calculations as part of my PhD. The general approach to DFT in my department appears to be that it is regarded as a tool for doing calculations, which need not be understood in detail. While I understand this to some degree, I however still feel the need to at least understand the basic theoretical ideas behind a DFT calculation. I am therefore using this post to summarize a few of my questions about DFT starting from the basic theory.

As I understand it, the formalism of DFT rests upon the Hohenberg-Kohn theorems, stating that the ground state properties of a many-body system are unique functionals of the ground state density, which furthermore minimizes the total energy functional $E[n(\mathbf{r})]$ of the system:

$$ E[n(\mathbf{r})] = T[n(\mathbf{r})] + E_H[n(\mathbf{r})] + E_{ext}[n(\mathbf{r})] + E_{xc}[n(\mathbf{r})] $$

Here $T[n(\mathbf{r})]$ is the kinetic energy functional, $E_H[n]$ the Hartree functional and $E_{ext}[n(\mathbf{r})]$ the energy due to external potentials while $E_{xc}[n]$ is the unknown exchange-correlation energy functional, which encompasses the exchange and correlation effects of the many-body problem and must be suitably approximated by e.g. the LDA.

In practice DFT calculations are based on the Kohn-Sham approach, which makes the ansatz that there is a non-interacting system, which reproduces the density of the many-body system one wishes to describe. Consequently, instead of solving a many-body Hamiltonian, one solves a set of single-particle equations, the Kohn-Sham equations:

$$ (\nabla^2 + V^{KS})\phi_i(\mathbf{r}) = \epsilon_{i}\phi_{i}(\mathbf{r}) $$

with the Kohn-Sham potential given by:

$$ V^{KS}(\mathbf{r}) = V_{ext}(\mathbf{r}) + V_{H}(\mathbf{r}) + \frac{\delta E_{xc}}{\delta n(\mathbf{r})} $$

The density is then readily evaluated as a sum over the occupied orbitals:

$$ \rho(\mathbf{r}) = \sum_{occ}\lvert \phi_{i}(\mathbf{r}) \rvert^2 $$

My main question is essentially, if anyone can walk me through how the above theoretical ideas are used in a concrete DFT calculation. Say, e.g., I want to calculate the ground state energy of some molecule. How does a DFT calculation proceed using the equations outlined above?

My own (partial) answer: The first step is specifying the energy functional, which relies on choosing a suitable approximation for the exchange-correlation functional. My guess is then that the next step is to solve the Kohn-Sham single-particle equations. To do one needs the Kohn-Sham potential, which must be found from the functional derivative of the total energy functional. I am honestly not sure if this is how a DFT calculator proceeds, but it seems like the only way to get the Kohn-Sham potential. Having solved the KS system, one can then calculate the electron density, which should then equal the density of the interacting system. However, I speculate that the procedure may involve some sort of self-consistency such that one should in practice make an iterative procedure to find the correct interacting density.

Assuming everything is correct so far, how does one then get the ground state energy? So far we have only calculated the ground state density and the energies of the Kohn-Sham orbitals are not physical energies of the system.

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    $\begingroup$ Welcome to Chemistry! Take the tour to get familiar with this site. I'm having a hard time figuring out exactly what your question is. I'd re-format your last block to split it up since the question is buried in there. I think the answer is in your question: once you have self-consistently solved for $\rho(\mathbf{r})$ (which is the same thing as $n(\mathbf{r})$), plug back in to the energy functional, and you're done. $\endgroup$ – pentavalentcarbon Feb 16 '18 at 14:32
  • $\begingroup$ Thanks for the reply. I can understand it is a bit difficult to figure out what my question is. Essentially my main question is: How do you obtain the Kohn Sham potential to use in the Kohn-Sham equations? When I do a DFT calculation I specify the exchange correlation functional but not the KS potential. $\endgroup$ – user13514 Feb 17 '18 at 14:23
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    $\begingroup$ Just from the equations themselves, you can see that $V_{xc}[n(\mathbf{r})] = \frac{\delta E_{xc}}{\delta n(\mathbf{r})}$: the potential is the functional derivative of the energy with respect to the density. You are correct that the potential is never specified, but in this case the potential is defined by the energy functional. In practice, the Kohn-Sham equations are identical to the Hartree-Fock equations, except the (exact) exchange contribution to the Fock operator is replaced with the Kohn-Sham matrix: chemistry.stackexchange.com/a/81832/194 $\endgroup$ – pentavalentcarbon Feb 17 '18 at 14:51
  • $\begingroup$ Possible duplicate of How does one actually get the energy from the Kohn–Sham equations? $\endgroup$ – pentavalentcarbon Feb 17 '18 at 16:28