Assume that we have at our disposal \( n \) different single-particle orbits \( \alpha_0,\alpha_2,\dots,\alpha_{n-1} \) and that we can distribute among these orbits \( N\le n \) particles.

A Slater determinant can then be coded as an integer of \( n \) bits. As an example, if we have \( n=16 \) single-particle states \( \alpha_0,\alpha_1,\dots,\alpha_{15} \) and \( N=4 \) fermions occupying the states \( \alpha_3 \), \( \alpha_6 \), \( \alpha_{10} \) and \( \alpha_{13} \) we could write this Slater determinant as $$ \Phi_{\Lambda} = a_{\alpha_3}^{\dagger} a_{\alpha_6}^{\dagger} a_{\alpha_{10}}^{\dagger} a_{\alpha_{13}}^{\dagger} |0\rangle. $$ The unoccupied single-particle states have bit value \( 0 \) while the occupied ones are represented by bit state \( 1 \). In the binary notation we would write this 16 bits long integer as $$ \begin{array}{cccccccccccccccc} {\alpha_0}&{\alpha_1}&{\alpha_2}&{\alpha_3}&{\alpha_4}&{\alpha_5}&{\alpha_6}&{\alpha_7} & {\alpha_8} &{\alpha_9} & {\alpha_{10}} &{\alpha_{11}} &{\alpha_{12}} &{\alpha_{13}} &{\alpha_{14}} & {\alpha_{15}} \\ {0} & {0} &{0} &{1} &{0} &{0} &{1} &{0} &{0} &{0} &{1} &{0} &{0} &{1} &{0} & {0} \\ \end{array} $$ which translates into the decimal number $$ 2^3+2^6+2^{10}+2^{13}=9288. $$ We can thus encode a Slater determinant as a bit pattern.