# Why do DFT calculations output molecular orbitals?

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?

• Do you know how this works in time independent DFT? I'm worried you are confusing the two. Apr 11, 2022 at 9:11
• 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. Apr 11, 2022 at 9:11
• @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? Apr 11, 2022 at 16:11
• 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. Apr 11, 2022 at 16:34
• Oh I see what you mean. I was pretty sure that distinction wasn't necessary but included it just in case. Apr 11, 2022 at 18:17

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.