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 \( \hat{H} \).