Full configuration interaction theory calculations provide in principle, if we can diagonalize numerically, all states of interest. The dimensionality of the problem explodes however quickly.

The total number of Slater determinants which can be built with say \( N \) neutrons distributed among \( n \) single particle states is $$ \left (\begin{array}{c} n \\ N\end{array} \right) =\frac{n!}{(n-N)!N!}. $$

As an example, for a model space which comprises the first four major harmonic oscillator shells only, that is the \( 0s \), \( 0p \), \( 1s0d \) and \( 1p0f \) shells we have \( 40 \) single particle states for neutrons and protons. For the eight neutrons of oxygen-16 we would then have $$ \left (\begin{array}{c} 40 \\ 8\end{array} \right) =\frac{40!}{(32)!8!}\sim 8\times 10^{7}, $$ possible Slater determinants. Multiplying this with the number of proton Slater determinants we end up with approximately \( d\sim 10^{15} \) possible Slater determinants and a Hamiltonian matrix of dimension \( 10^{15}\times 10^{15} \), an intractable problem if we wish to diagonalize the Hamiltonian matrix.