Showing that \( |\tilde{c}\rangle= |c'\rangle \)

We need to show that \( |\tilde{c}\rangle= |c'\rangle \). We need also to assume that the new state is not orthogonal to \( |c\rangle \), that is \( \langle c| \tilde{c}\rangle \ne 0 \). From this it follows that $$ \langle c| \tilde{c}\rangle=\langle 0| a_{i_n}\dots a_{i_1}\left(\sum_{p=i_1}^{i_n}f_{i_1p}a_{p}^{\dagger} \right) %\left(\sum_{q=i_1}^{i_n}f_{i_2 q}a_{q}^{\dagger} \right)\dots \left(\sum_{t=i_1}^{i_n}f_{i_n t}a_{t}^{\dagger} \right)\vert 0\rangle, $$ which is nothing but the determinant \( det(f_{ip}) \) which we can, using the intermediate normalization condition, normalize to one, that is $$ det(f_{ip})=1, $$ meaning that \( f \) has an inverse defined as (since we are dealing with orthogonal, and in our case unitary as well, transformations) $$ \sum_{k} f_{ik}f^{-1}_{kj} = \delta_{ij}, $$ and $$ \sum_{j} f^{-1}_{ij}f_{jk} = \delta_{ik}. $$