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This question already has an answer here:

We know that in QFT, all particles are being created and annihilated, and because of the conservation laws, on average the number of particles remains constant.

Suppose we have an electron in an atom. Is it possible that this electron gets annihilated with a created positron, and the created electron paired with the positron takes its place? In other words, is it possible that the electron inside an atom gets destroyed and supplemented by another electron? In this case, does this mean that the position of the electron in the atom suddenly changes?

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marked as duplicate by G. Smith, John Rennie quantum-mechanics May 8 at 7:46

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ depends what you mean by position. If you mean orbital, that would violate conservation of energy. Usually that's a no. $\endgroup$ – Manu de Hanoi May 8 at 1:41
  • $\begingroup$ @Manu The electron is a point particle and has a position and a momentum (subject to HUP). The orbital is the space in which the position of the electron is more likely to be. This is what I mean by position, not the orbital. $\endgroup$ – Ali Lavasani May 8 at 3:49
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    $\begingroup$ Possible duplicate of Can vacuum fluctuations change the position of an electron? $\endgroup$ – G. Smith May 8 at 3:52
  • $\begingroup$ @G.Smith That question is mine. I want a clear yes/no answer to this question but didn't get it there. But sorry I forgot I had asked the same thing before. $\endgroup$ – Ali Lavasani May 8 at 3:55
  • $\begingroup$ I just provided a clear yes/no answer, and the answer is No. $\endgroup$ – G. Smith May 8 at 4:22
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No, this process cannot happen. There is no possible Feynman diagram for QED that corresponds to it. It is fairly easy to draw all allowed QED diagrams because there is only one possible vertex.

In previous questions, you have been interested in how QED effects perturb the electron and lead to the Lamb shift. To one-loop order, there are only two diagrams that contribute to the Lamb shift. They are shown as (a) and (b) here. The first is an electron self-energy diagram. The second is a vacuum polarization diagram with a virtual electron-positron pair.

The real electron in the atom interacts with the virtual electron-positron pair only via a virtual photon. It cannot annihilate with the virtual positron.

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  • $\begingroup$ How about pair creation by a high-energy photon? Can the electron annihilate with the created positron, leaving behind the created electron? $\endgroup$ – Ali Lavasani May 8 at 4:43
  • $\begingroup$ Yes, that should be possible, with a photon being emitted from the annihilation. The Feynman diagram for that process is easy to draw. $\endgroup$ – G. Smith May 8 at 5:13
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there are problems with the fundamental assumptions you are making in your question when you mention the QFT approach. You say: "because of the conservation laws,on average the number of particles remains constant" but this statement is actually incorrect. Consider the decay of the neutron by electroweak process: $$n \rightarrow p \, + e^-\, +\, \overline{\nu}_e$$ the "number of particles" before and after is clearly not conserved. As a matter of fact, the conservation of particles was historically one of the flaws of the early version of Quantum Mechanics: in Schrodinger equation, the probability associated to the wave function is conserved , i.e. the famous "probability current conservation" identity, which means that a particle basically can't transform into another, because that would mean its probability would go from non-zero (before) to zero (after). In the above equation, the probability to have a neutron before and after the weak interaction process is clearly not identical. QFT of course solves all this issues through the formalism of creation/destruction operators. It was not, as you certainly know, the only problem of Schrodinger equations, but just one out of many.

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