Hello, Reading Richard Feynman’s book “Quantum Electrodynamics” (Edited by Advanced Book Classics), I read that the electron’s self-energy is infinite and that has been a trouble for QED during 20 years. Feynman proposed a solution based on a cut-off, but that’s not fully satisfactory and I think the question remains still opened. The Feynman Diagram that describes self-energy is: The integral used in the space-time dominion is: Where · ΔE is the electron's self-energy · V and T is the volume and the period of time and them both are infinite. · The f's are the wave functions. · K+ is the zeroth order propagator for a spinor using the Dirac equation. · γμδ+(s24,3)γμ is the potential caused by point '3' over point '4'. The integration is done over time and space in both points '3' and '4'. We can say that '3' emits a virtual photon and '4' absorbs it. With this data, although we have to divide by 'V' and 'T' in order to obtain the self-energy, a divergent self-energy is obtained. However, I think it's not correct to set the potential as γμδ+(s4,32)γμ since this expression is not taking into account the charge density of '3', |f(3)|2. I think, another way to get the self-energy could be obtained by quantum mechanical means simply by: ΔE = <f*(t4) |HSE| f(t4)> Where HSE is the self-energy hamiltonian, that hamiltonian should be: With this approach, the self-energy is no longer infinite, in fact for a plane-wave (dispersed over the infinite) it's zero. Can anybody tell me where I am missed? Since for me looks clear that the self-energy should not be infinite. Thanks a lot!!! Sergio Prats NOTE: in self-interactions the initial and the final states must be the same, so this Hamiltonian must not have the same effect in the vawe-function that a normal Hamiltonian.