Electromagnetic transitions and $\beta$-decay

Electromagnetic multipole moments and transitions

$$ \Delta E=\sum_{i=0}^{\infty}\langle \Phi_0\vert \hat{H}_I\left\{\frac{\hat{Q}}{\omega-\hat{H}_0}\left(\omega-E+\hat{H}_I\right)\right\}^i\vert \Phi_0\rangle=$$ $$ \langle \Phi_0\vert \left(\hat{H}_I+\hat{H}_I\frac{\hat{Q}}{E-\hat{H}_0}\hat{H}_I+ \hat{H}_I\frac{\hat{Q}}{E-\hat{H}_0}\hat{H}_I\frac{\hat{Q}}{E-\hat{H}_0}\hat{H}_I+\dots\right)\vert \Phi_0\rangle. $$ This expression depends however on the exact energy $ E $ and is again not very convenient from a practical point of view. It can obviously be solved iteratively, by starting with a guess for $ E $ and then solve till some kind of self-consistency criterion has been reached.

Actually, the above expression is nothing but a rewrite again of the full Schr\"odinger equation.

Defining $ e=E-\hat{H}_0 $ and recalling that $ \hat{H}_0 $ commutes with $ \hat{Q} $ by construction and that $ \hat{Q} $ is an idempotent operator $ \hat{Q}^2=\hat{Q} $. Using this equation in the above expansion for $ \Delta E $ we can write the denominator $$ \hat{Q}\frac{1}{\hat{e}-\hat{Q}\hat{H}_I\hat{Q}}= $$ $$ \hat{Q}\left[\frac{1}{\hat{e}}+\frac{1}{\hat{e}}\hat{Q}\hat{H}_I\hat{Q} \frac{1}{\hat{e}}+\frac{1}{\hat{e}}\hat{Q}\hat{H}_I\hat{Q} \frac{1}{\hat{e}}\hat{Q}\hat{H}_I\hat{Q}\frac{1}{\hat{e}}+\dots\right]\hat{Q}. $$