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I am confused of whether or not the expected electromagnetic field generated by the point-like electric charge of the electron distributed smoothly across space as a probability distribution creates the presence of an effective field mass.

I know that the probability current of the electron is $\langle \gamma^{0k} \rangle$, which is conserved. Multiplying the probability current of the electron by its total charge $q$ gives the charge current density $J^k = q\langle \gamma^{0k} \rangle$. From the current I can calculate the expected four-potential field as \begin{align} A^k(r,t) = \int \dfrac{J^k(r',t-c/r')}{|r-r'|} d^3 r' . \end{align}

From $A^k$ I can calculate the expected electromagnetic fields as $F^{jk} = \partial ^j A^k - \partial^k A^j$. Finally I can calculate the expected electromagnetic energy as \begin{align} U_{\text{eff}} = \dfrac{1}{8\pi} \int F^{jk}F^{jk} d^3 r . \end{align} Applying Einstein's relation energy is proportional to mass, I get the following effect field mass of the electron as \begin{align} M_{\text{eff}} = \dfrac{U_{\text{eff}}}{c^2} . \end{align}

Is $M_{\text{eff}}$ a real observable? I was unable to find an analytic solution for $M_{\text{eff}}$, however I did compute $M_{\text{eff}}$ for Gaussian distributed probability functions for the electron with varying standard deviation (spacial localization), and I noticed that for standard deviations of the order of $10^{-10}$ meters (lattice size), $M_{\text{eff}}$ was very small in comparison to the electron's rest mass. When the standard deviation was $10^{-15}$ meters (nucleus size), the $M_{\text{eff}}$ was comparable in size to the rest mass of the electron.

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If the electron is point-like, the Poynting energy expression is infinite and thus it is useless for calculation of the energy of the electron in the Einstein sense $E=mc^2$, which is finite.

If the electron is cloud-like, the Poynting energy expression is finite and may give part of electron's energy; not all of it, since some non-electromagnetic forces have to hold the electron together and the total energy will depend on these as well.

According to common interpretation of some fine experiments, electrons are known to be smaller than $10^{-18}~$m today and are possibly points.

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i know the energy required to excite the electron would be enormous. I'm not sure but it could be around the plank mass. I'm not exactly sure how that factors into what forces are holding it together. As for what kind of field mass it generates inside the cloud i would think maybe its of order of a single photon of around the wavelength matching the size of the cloud.

i imagine that you can think of the field around an electron as positron electron pairs. i don't know how the dirac equation would behave. i imagine you would be dealing with quantized photons. i don't know you could think of this problem as some thing like an atom. The uncertainty principle applies to the hole atom. If for example the atom tunnels through a barrier it take all its parts with it. It doesn't leave behind electrons. So if you are describing what is going on around in the atom you don't have to consider the two different possibilities. So i'm thinking all you need to calculate the field mass is to pretend the electron is a fixed point in space.

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