We assume also that the resolvent of \( \left(\omega-\hat{H}_0\right) \) exits, that is it has an inverse which defined the unperturbed Green's function as $$ \left(\omega-\hat{H}_0\right)^{-1}=\frac{1}{\left(\omega-\hat{H}_0\right)}. $$

We can rewrite Schroedinger's equation as $$ \vert \Psi_0\rangle=\frac{1}{\omega-\hat{H}_0}\left(\omega-E+\hat{H}_I\right)\vert \Psi_0\rangle, $$ and multiplying from the left with \( \hat{Q} \) results in $$ \hat{Q}\vert \Psi_0\rangle=\frac{\hat{Q}}{\omega-\hat{H}_0}\left(\omega-E+\hat{H}_I\right)\vert \Psi_0\rangle, $$ which is possible since we have defined the operator \( \hat{Q} \) in terms of the eigenfunctions of \( \hat{H} \).