Similarly $$ a^{\dagger}_{\alpha_6}\Phi_{3,6,10,13} = a^{\dagger}_{\alpha_6}|0001001000100100\rangle=a^{\dagger}_{\alpha_6}a_{\alpha_3}^{\dagger} a_{\alpha_6}^{\dagger} a_{\alpha_{10}}^{\dagger} a_{\alpha_{13}}^{\dagger} |0\rangle, $$ which becomes $$ -a^{\dagger}_{\alpha_4} (a_{\alpha_6}^{\dagger})^ 2 a_{\alpha_{10}}^{\dagger} a_{\alpha_{13}}^{\dagger} |0\rangle=0! $$ This gives a simple recipe:

• If one of the bits \( b_j \) is \( 1 \) and we act with a creation operator on this bit, we return a null vector
• If \( b_j=0 \), we set it to \( 1 \) and return a sign factor \( (-1)^l \), where \( l \) is the number of bits set before bit \( j \).