I have read that the electron self-energy correction, a process in which the electron absorbs a virtual photon, changes the position of the electron (which leads to the lamb shift in Hydrogen-like atoms).

How does this process lead to a change in the electron's position? Does it occur only in Hydrogen-like atoms, or in all atoms?

  • $\begingroup$ Absorption of virtual photons is what binds an electron to a proton, yielding the hydrogen atom $\endgroup$ Apr 27, 2019 at 0:34
  • $\begingroup$ @Lewis So why does it change and "smear" the position of the electron? Is it a random process? $\endgroup$ Apr 27, 2019 at 4:09
  • $\begingroup$ What it does is s.eR $\endgroup$ Apr 27, 2019 at 13:01
  • $\begingroup$ Smear the probability function. In other words the wave function. The only difference between the self-energy and regular binding energy is the source of the virtual photon. $\endgroup$ Apr 27, 2019 at 13:06
  • $\begingroup$ @Lewis What are these sources of the virtual photon? Do electrons in all atoms experience this smearing? What parameter determines "how much" the wavefunction gets smeared? $\endgroup$ Apr 27, 2019 at 20:21

1 Answer 1


The explanation you are searching for is that the electron loses energy/momentum when "emitting the virtual photon" and therefore must be in a "lower orbit" until it reabsorbs said photon. As it is doing so all the time, it's average position is nearer to the nucleus. I must say that this is just a pretty picture for experimentalists/leymen who do not understand QFT. Virtual particles are a mathematical concept which describes the effect of what is happening on the quantum level (up to some order), but not the thing itself. In the end there is no easy way to imagine what happens b/c during an interaction the particle picture (asymptotic states) breaks down and the for us unintuitive wave nature of the "particles" takes over. The electron field kind of interacts with itself and in doing so produces higher order effects like the one you are thinking about.

  • $\begingroup$ Thanks. What determines the strength and amount of the smearing effect due to self-energy and/or binding energy? Does this happen only in hydrogen atoms, or also for all electrons in any atom? Does it happen for atoms in covalent or metallic bonds? $\endgroup$ Apr 27, 2019 at 19:53
  • $\begingroup$ @AliLavasani The self-energy is valid for any electron (or any particle w/ charge generally). But if the electron isn't bound, we can't observe it as a "lower orbit". So it happens everywhere. I think it should be observable theoretically in any bound electron, but it is a very small correction (hyperfine structure). I would imagine that for most systems (eg bigger atoms, molecules and solid state systems) our theoretical predictions are not precise enough (many body problem) for this correction to be measurable. $\endgroup$
    – paleonix
    Apr 28, 2019 at 9:29
  • $\begingroup$ Self energy is only appearing second order in alpha in perturbation theory. At all energies we can achieve, alpha is very small (1/137). At very high energies the effect of the self energy would become more important. $\endgroup$
    – paleonix
    Apr 28, 2019 at 9:34
  • $\begingroup$ Why should the electron self-energy correction and smearing in more complex atoms and metallic structures or molecules be smaller than the hydrogen atom? Many other atoms have almost the same radius and much stronger nucleus electric field. What parameter(s) determines how large this smearing effect is? $\endgroup$ Apr 28, 2019 at 9:48
  • $\begingroup$ @AliLavasani I didn't say that it is smaller, just that our calculations/models from which the not yet corrected values come are approximating more things, s.t. the uncertainty is much bigger than this correction. The hydrogen atom is a two-body problem and analytically solvable. Helium already needs some approximations to be made/numerical calculations to be solved. $\endgroup$
    – paleonix
    Apr 28, 2019 at 9:52

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