Using 2 hydrogen atom, we started from 2 atomic orbital (AO),

The atomic orbital is the ground state of hydrogen atom. From these 2 AO, we can construct linear combination of atomic orbitals (LCAO),

Notice that an electron can be in either or , that means the electron is not belong to any atom, but belong to the molecule.

The last step is adding the spin, there are 2 spins , The molecular orbital (MO) are,

The Slater determinant can be formed by any 2 of them.

We started by 1 atomic orbital, and there are and 2 atoms, then 2 LCAOs are formed, and then multiplied by 2 spin states, gives 4 molecular orbitals. From these 4 MO and 2 electrons, we can from 6 different Slater determinant.

If we have different number of AO for different atom, we have AO

The LCAO are

The total number of LCAO are all permutation of the AOs. A simple construction is that only plus or minus sign are used. Then the number of LCAO ( not orthonormal ) is

The MO are

Thus the number of Slater determinant are

For example, we have 3 atoms, there are AO, there are 16 LCAOs. Thus, we have 32 MO. With 2 electrons, there are 496 Slater determinant.

For C-H molecule, there are 6 electrons for carbon and 1 electron for hydrogen, use 6 AOs for carbon and 1 AO for hydrogen, the number of MO is 24 and 346,104 Slater determinant.

When using all Slater determinant, it is called full CI.

When only increase the AO, but limited number of Slater determinant, The energy is called Hartree limit.

A problem is to construct LCAOs for arbitrary AOs. For example, for AOs, The number of possible combination with pule or minus sign is 4. However. I cannot construct orthonormal 4 LCAOs, because the system are over constrained (16 constrains with 12 unknown of )