My understanding is that DFT finds the electron density which minimizes some energy functional. How does it make the connection from this optimized density to molecular orbitals?

  • $\begingroup$ Do you know how this works in time independent DFT? I'm worried you are confusing the two. $\endgroup$
    – Ian Bush
    Apr 11, 2022 at 9:11
  • 3
    $\begingroup$ IIRC, most implementations of DFT and TD-DFT use Kohn-Sham orbitals, and the density is expressed in terms of those orbitals. So, the program always deals with orbitals, the density is calculated from the orbitals. $\endgroup$
    – S R Maiti
    Apr 11, 2022 at 9:11
  • $\begingroup$ @IanBush No, I wasn’t aware there was a qualitative difference in how the connection was made in TD vs TI. Based on S R Maiti’s answer though I’m under the impression that there isn’t. What did you have in mind? $\endgroup$ Apr 11, 2022 at 16:11
  • $\begingroup$ As long as you know there is no difference that is fine - I was just concerned you keep mentioning the time dependent part when that has been totally irrelevant for your last two questions. $\endgroup$
    – Ian Bush
    Apr 11, 2022 at 16:34
  • $\begingroup$ Oh I see what you mean. I was pretty sure that distinction wasn't necessary but included it just in case. $\endgroup$ Apr 11, 2022 at 18:17

1 Answer 1


The short answer is they do not. We can identify two distinct ways of doing DFT calculations, OF-DFT (orbital-free DFT) and KS-DFT (Kohn-Sham DFT).

Let us start with the OF-DFT formalism, as it is the formalism that is more in the 'spirit' of DFT.

In OF-DFT the energy is given as:

$$E\left[\rho\right]=\int_{\Omega}\phi_{\mathrm{ext}}\left(\boldsymbol{r}\right)\rho\left(\boldsymbol{r}\right)\mathrm{d}\boldsymbol{r} + \frac{1}{2}\int_{\Omega}\frac{\rho\left(\boldsymbol{r}_{1}\right)\rho\left(\boldsymbol{r}_{2}\right)}{\left|\boldsymbol{r}_{1}-\boldsymbol{r}_{2}\right|}\mathrm{d}\boldsymbol{r}_{2}\mathrm{d}\boldsymbol{r}_{1} + T\left[\rho\right]+E_{\mathrm{x}}\left[\rho\right]+E_{\mathrm{c}}\left[\rho\right] $$

The first term is the external potential, the second term is the Coloumb interaction of the electronic density with the electronic density (electron-electron repulsion including self-repulsion), the term is the kinetic energy functional, and the last two terms are the exchange- and correlation-contributions.

As can be seen in the above equation, there is no dependency on orbitals anywhere.

Now let us examine the KS-DFT formalism:

$$ E\left[\rho_{\Phi^{\mathrm{KS}}}\right] = \int_{\Omega}\phi_{\mathrm{ext}}\left(\boldsymbol{r}\right)\rho_{\Phi^{\mathrm{KS}}}\left(\boldsymbol{r}\right)\mathrm{d}\boldsymbol{r} + \frac{1}{2}\int_{\Omega}\frac{\rho_{\Phi^{\mathrm{KS}}}\left(\boldsymbol{r}_{1}\right)\rho_{\Phi^{\mathrm{KS}}}\left(\boldsymbol{r}_{2}\right)}{\left|\boldsymbol{r}_{1}-\boldsymbol{r}_{2}\right|}\mathrm{d}\boldsymbol{r}_{2}\mathrm{d}\boldsymbol{r}_{1}+ \left< {\Phi^{\mathrm{KS}}} \left| \hat{T} \right| {\Phi^{\mathrm{KS}}} \right> + E_\mathrm{x}\left[ \rho_{\Phi^{\mathrm{KS}}} \right] + E_\mathrm{c}\left[ \rho_{\Phi^{\mathrm{KS}}} \right] $$

We can see that in the KS-DFT formalism all of the terms, except for the kinetic energy, are the same as in OF-DFT formlism. Note, that the density here is contructed from the KS-orbitals.

In the KS-DFT formalism we get the kinetric energy contribution by appliying the kinetic energy functional on the KS-orbitals, whereas, in the OF-DFT formlism the kinetic energy is a functional of the denisity.

Now to address your question directly. In (KS)-DFT we do not construct the orbitals from the optimized denisty. Instead we contruct the denisty from the orbitals. I.e. the density is:

$$ \rho_{\Phi^{\mathrm{KS}}} = \sum_{i=1}^N \left|\phi^\mathrm{KS}_i\right| $$

However, this raises the question, why do we bother KS-orbtals? Here, the unsatisfying answer is; Because we do not know any accurate kinetic energy functionals.


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