Hatree-Fock theory is foundational to many subsequent electronic structure theories. It is an independent particle model or mean filed theory. Consider we have two non-interacting electrons. In that case, the Hamiltonian would be separable, and the total wavefunction $\Psi(\textbf{r}_1, \textbf{r}_2)$ would be product of the individual wave function. Now if we consider two electrons are forming a single system, then there are two issues. (1) We can no longer ignore the electron-electron interaction. (2) The wavefunction describing fermions must be antisymmetric with respect to the interchange of any set of space-spin coordinates. A simple Hartree product fails to satisfy that condition:
$$ \Psi_{HP}(\textbf{r}_1, \textbf{r}_2, \cdots, \textbf{r}_N) = \phi_1(\textbf{r}_1) \phi_2(\textbf{r}_2) \cdots \phi_N(\textbf{r}_N) $$
In order to satisfy the antisymmetry condition, for our two electron system we can formulate a total wavefunction of the form:
$$ \Psi(\textbf{r}_1, \textbf{r}_2) = \frac{1}{\sqrt{2}} [\chi_1(\textbf{r}_1) \chi_2(\textbf{r}_2) - \chi_1(\textbf{r}_2)\chi_2(\textbf{r}_1)] $$
Slater determinant
The above equation can be written as:
$$ \Psi(\textbf{r}_1, \textbf{r}_2) = \frac{1}{\sqrt{2}} \begin{vmatrix} \chi_1(\textbf{r}_1) & \chi_2(\textbf{r}_1) \ \chi_1(\textbf{r}_2) & \chi_2(\textbf{r}_2) \end{vmatrix} $$
Now what happens if we have more than two electrons? We can generalize the above determinant form to $N$ electrons:
$$ \Psi = \frac{1}{\sqrt{N!}} \begin{vmatrix} \chi_1(\textbf{r}_1) & \chi_2(\textbf{r}_1) & \cdots & \chi_N(\textbf{r}_1) \ \chi_1(\textbf{r}_2) & \chi_2(\textbf{r}_2) & \cdots & \chi_N(\textbf{r}_2) \ \vdots & \vdots & \ddots & \vdots \ \chi_1(\textbf{r}_N) & \chi_2(\textbf{r}_N) & \cdots & \chi_N(\textbf{r}_N) \end{vmatrix} $$
The above antisymmetrized product can describe electrons that move independently of each other while they experience an average (mean-field) Coulomb force.