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I just recently began to study quantum chemistry and need some clarification for the construction of Kohn-Sham orbitals.
Consider a system of several atoms. Let there be $N$ electrons in this system. We can reduce the description of such a system to the $\psi_i$ single-electron wave functions describing the orbital, where the index $i$ takes a value from $1$ to $N$. Now consider the same system in the framework of the density functional theory. The description converts to molecular wave-functions $\varphi_j$. Molecular orbitals are obtained from the Kohn-Sham equation: \begin{align} \left( - \frac{\hbar^2}{2m} \nabla_j^2 + \nu_\mathrm{eff}(\mathbf{r}) \right) \varphi_j(\mathbf{r}) &= \varepsilon_j\varphi_j(\mathbf{r}), & j &= 1,\dots, K. \end{align}

How is it determined how many molecular orbitals are now? $K=N$? How do the new molecular orbitals make up the overall wave function of the system?

I am a little confused by the fact that atomic orbitals form molecular ones according to very complex principles and it is unclear how this is taken into account.

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    $\begingroup$ Yes K = N. K-S functions are single-electron functions so you need one for each electron. $\endgroup$
    – Andrew
    Commented Jun 7, 2020 at 20:27
  • $\begingroup$ Thanks for the answer. If $\varphi_i$ is already molecular orbitals, it turns out they don’t have a strict binding to the electron of a particular atom? Moreover, the molecular orbital is a linear combination of atomic orbitals (if work within the framework of MO LCAO) $\endgroup$
    – Disciple
    Commented Jun 8, 2020 at 13:16
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    $\begingroup$ I'm not sure what you mean by "strict binding to the electron of a particular atom". In a molecule, electrons aren't associated with specific atoms, nor even with specific orbitals, as all the electrons on the molecule are interchangeable. We use the concept of orbitals to create a mathematical description of the net electron density on the molecule. The mathematics of Kohn-Sham orbitals are such that each one represents the contribution of one electron. $\endgroup$
    – Andrew
    Commented Jun 8, 2020 at 15:46

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