Counting determinants (and losing)
Learn
The size of the FCI problem is a counting exercise: distribute the α electrons among the M spatial orbitals, independently distribute the β electrons, multiply.
Run the numbers on systems you now know. H₂ in STO-3G: M = 2, one electron of each spin, so C(2,1)² = 4 determinants — a 4×4 Hamiltonian you could diagonalize on paper. Water in STO-3G: M = 7, five of each spin, C(7,5)² = 21² = 441. Water again in the modestly better 6-31G basis: M = 13, C(13,5)² = 1287² ≈ 1.66 million. Same molecule, one rung up in basis quality, four thousand times the determinants.
H₂ in the 6-31G basis has M = 4 spatial orbitals, one α electron, and one β electron. How many determinants does the FCI expansion contain? (Use N = C(M, Nα) · C(M, Nβ).)
That binomial explosion — not integrals, not iterations — is what confines FCI to benchmark duty. The art of post-Hartree-Fock chemistry is choosing which corner of that determinant space actually matters: singles and doubles (CISD, CCSD), the near-degenerate handful (CASSCF), or perturbative slices (MP2).
Why is FCI limited to small systems (roughly 10–20 electrons in modest bases)?