# Self-energy series expression in terms of unperturbed Green function for exited states

I would like to understand how to arrive at the series in equation (36) in this paper https://arxiv.org/abs/cond-mat/0506438, specifically $$\Sigma(E) = V+VG'_0(E)V+VG'_0(E)VG'_0(E)V$$ where $$G'_0(E)$$ is the free propagator when operating on excited states and 0 when operating on the ground state.

I'd imagine you can derive it from the Dyson equation $$\Sigma = G_0^{-1}-G^{-1},$$ but I don't understand why the Dyson equation doesn't just give $$\Sigma = V$$.

• and what is $V$? – AccidentalFourierTransform Sep 13 '19 at 22:33
• It's the interaction part of the hamiltonian that's taken as the perturbation. – Daniel Sep 14 '19 at 14:12