- #1

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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

*are the wave functions.*

**f's**· K

_{+}is the zeroth order propagator for a spinor using the Dirac equation.

· γ

_{μ}δ

_{+}(s

^{2}

_{4,3})γ

_{μ}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 γ

_{μ}δ

_{+}(s

_{4,32})γ

_{μ}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*(t

_{4}) |H

_{SE}| f(t

_{4})>

Where H

_{SE}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.