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The spin-unrestricted Hartree-Fock eigenvalule equations are coupled differential equations which determine the spatial part of the electron spin orbitals.

The coupled Hartree-Fock eigenvalue equations have the following form $$ \left[ h + \sum_{j=1}^{N^\uparrow}\left( J_j^\uparrow- K_j^\uparrow\right) + \sum_{j=1}^{N^\downarrow} J_j^\downarrow \right] \varphi^\uparrow_i = \varepsilon_i^\uparrow \varphi^\uparrow_i $$ $$ \left[ h + \sum_{j=1}^{N^\downarrow}\left( J_j^\downarrow- K_j^\downarrow\right) + \sum_{j=1}^{N^\uparrow} J_j^\uparrow \right] \varphi^\downarrow_i = \varepsilon_i^\downarrow \varphi^\downarrow_i $$ where $J$ and $K$ are the Coulmb and Exchange operators, and $h$ contains the kinetic energy plus the interaction with the nuclei.

But for this equation to make sense, I have to know $N^\uparrow$ and $N^\downarrow$ in advance.

However, usual quantum chemistry codes (say, I want to solve Hartree-Fock for the Nitrogen atom)) only need the total number of electrons $N=N^\uparrow+N^\downarrow$ as an input parameter.

But how can they solve the spin-unrestricted Hartree-Fock equations without knowing $N^\uparrow$ and $N^\downarrow$? Is it dervied by Hund's rule?

I think there must be some kind of additional assumtion (like Hund's rule), because neither the exact many-electron Schrödinger equation nor the spin-unrestricted Hartree-Fock equation can determine the spin states of the electrons. I guess, for any non-relativistic theory, the spin has to be included by some emperical rules.

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You are right: the Hartree-Fock equations depend on the occupation numbers of the orbitals, which are unknown, so this is a bit of a chicken-and-egg problem. This issue appears also for usual, closed-shell restricted Hartree-Fock. In practice, the occupation numbers are guessed (or explicitly specified) at the beginning of the calculation, normally on the basis of the Aufbau principle and of Hund's rules. Quantum Chemistry programs incorporate these rules and in general do a good job at guessing the right occupation number for the electronic ground state. However, there are cases in which the get it wrong (for example if there are transition metals, which lead to a multitude of low-lying states), in which cases one must proceed by trial and error (try various occupation numbers and see which choice gives the lowest energy). There are many, many references on the Hartree-Fock method, a good one is: P Echenique and J. L. Alonso, A mathematical and computational review of Hartree–Fock SCF methods in quantum chemistry, Molecular Physics, Vol. 105, Nos. 23–24, 3057–3098, 2007. https://arxiv.org/abs/0705.0337

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