I just want to know how does an electron felt the presence of a positron before they are converted into energy? Also how does the electron tell if it is positron or proton if this makes any difference?

  • 1
    $\begingroup$ I didn't know electrons have feelings. Might as well be more cautious, not to hurt their feelings... $\endgroup$
    – DanielC
    Feb 28, 2018 at 8:03
  • $\begingroup$ The question is posed in nontechnical language, which makes it unclear whether there is any well-defined sense in which it can be answered. Depending on how the question is construed, the answer might be basically "Bohr radius" of positronium, which is on the order of 0.1 nm. $\endgroup$
    – user4552
    Mar 1, 2018 at 1:11

3 Answers 3


In QED, the basic annihilation process is described by a $\bar{\psi }{{A}^{\mu }}{{\gamma }_{\mu }}\psi $ interaction term, in which the electron $(\psi )$, positron $(\bar{\psi })$ , and photon $(A)$ field operators act at exactly the same point. However, there are small, higher-order corrections to this term, which arguably make it non-local. (This is fancy technical jargon for John Rennie’s caveat, when interactions become significant.)

This is the process that plays in ${{e}^{-}}{{e}^{+}}\to {{\mu }^{-}}{{\mu }^{+}}$ . Some different happens in ${{e}^{-}}{{e}^{+}}\to \gamma \gamma $, where the incoming electron and positron are annihilated at two different points.

The idea that two particles must have zero separation to annihilate may seem surprising in classical terms, but not in quantum mechanics, where particles of definite momentum have undefined positions.


I don't think there is a simple answer to this.

The problem is that in quantum electrodynamics the particles we call electrons and positrons are the states described by the quantum field in the limit of negligible interactions i.e. when the particle is too far from any other particles for any significant interaction to occur. In this limit the particles are described by the free field states i.e. Fock states.

The problem is that when interactions become significant the states of the quantum field are perturbed away from the free field states. We can calculate this perturbation, using (unsurprisingly) perturbation theory, to calculate scattering probabilities, but we don't actually know what the states are. So in the annihilation process it isn't really the case that there is one electron and one positron present because the state of the quantum field cannot simply be described as an electron state and a positron state. If we insist on trying to describe it using the free field states then we have to conclude there are other particles present as well i.e. the virtual particles.

The point of all this is that during the annihilation process the electron-positron separation isn't well defined so it doesn't make sense to ask how close the two get before they are annihilated.


As a greasy handed experimenter I have a simple answer to this. Look through all the high-energy, exclusive data you have available on $e^+ + e^- \to 2\gamma$ (or possibly $e^+ + e^- \to \mu^+ + \mu^-$), pick the event with the largest squared four-momentum transfer $Q^2 = -t$ (or invariant mass $s$), and use that to characterize a length scale $$ L = \frac{hc}{\sqrt{-t}} \quad\text{or}\quad \frac{hc}{\sqrt{s}}\;,$$ which you declare to be the best experimental answer to date.

I assume the winning event is to be found somewhere in the LEP II datasets, but I have no idea what the answer might be.

However, there is no theoretical upper limit on those experiment observables and therefore no theoretical lower limit on the distance (short of the Planck scale). Finding such a lower limit would mean either a discovery that the charged leptons are not fundamental or that you've hit the scale of a more fundamental underlying theory (string theory, super-symmetry, etc.)


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