We can now this expression in terms of a perturbative expression in terms of \( \hat{H}_I \) where we iterate the last expression in terms of \( \Delta E \) $$ \Delta E=\sum_{i=1}^{\infty}\Delta E^{(i)}. $$ We get the following expression for \( \Delta E^{(i)} \) $$ \Delta E^{(1)}=\langle \Phi_0\vert \hat{H}_I\vert \Phi_0\rangle, $$ which is just the contribution to first order in perturbation theory, $$ \Delta E^{(2)}=\langle\Phi_0\vert \hat{H}_I\frac{\hat{Q}}{W_0-\hat{H}_0}\hat{H}_I\vert \Phi_0\rangle, $$ which is the contribution to second order.