It is known that matter and antimatter annihilate each other when they "touch" each other. And as far as I know, the concept of "touching" as our brain gets it is not true on the atomic level since atoms never really touch each other but only get affected by different forces.

If this is true, then when does the annihilation really happen ? Does it happen when two atoms are affected by each others repulsion force for example ?

  • 2
    $\begingroup$ I am not sure what does "physically 'touch'" mean. Matter-antimatter annihilation does not have to involve atoms, like $e^+ + e^- \rightarrow 2 \gamma$. They are described by the fields. If their fields (wavepackages) come together, then the scattering (matter-antimatter annihilation) happens $\endgroup$
    – user26143
    Commented Aug 10, 2013 at 12:48
  • $\begingroup$ same charges atoms may contact in molecules... Answer this question "How can I stand on the ground? EM or/and Pauli?" before ? note that electron and positron may bounce only $\endgroup$
    – user46925
    Commented Dec 23, 2015 at 2:24
  • $\begingroup$ @igael, what do you mean by electron and positron may bounce ? $\endgroup$ Commented Dec 23, 2015 at 12:39
  • $\begingroup$ @AbanobEbrahim : bounce without annihilation $\endgroup$
    – user46925
    Commented Dec 23, 2015 at 12:52
  • $\begingroup$ I wrote above an ambigous statement "positron may bounce only" instead of "positron may bounce also" ! sorry ... $\endgroup$
    – user46925
    Commented Dec 23, 2015 at 12:53

2 Answers 2


You should not think a particle as point like. Classically, the probability of two point like particles colliding with random location and velocity is 0, that is why you said it never happens.

However, at quantum mechanical level, these particle are described by wavefunction. It means that there is spreading in its spatial location, say 0.1nm (the minimum spreading is guaranteed by Heisenberg's uncertainty principle). The probability of annihilation can therefore be calculated by the overlap of these wavefunction and their interaction. This overlap is the meaning of "touching" in some sense.

  • $\begingroup$ "You should not think a particle as point like." Apparently, not everybody agrees: "As far as we know the electron is a point particle." $\endgroup$ Commented Jul 23, 2014 at 16:44
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    $\begingroup$ It can be (and is) simultaneously true that electrons scatter like point-particles (down to $10^{-18}\,\mathrm{m}$ experimental precision) and that their bound wave-function gives significant probability to distances on order of $10^{-10}\,\mathrm{m}$. $\endgroup$ Commented Jul 23, 2014 at 19:22
  • $\begingroup$ It turns out if you think they are point-like the probability is still nonzero due to their attractive forces, and we only have to get close enough for the kinetic energy to be convertible to a particle's worth of mass for interesting things to happen (see cosmic background radiation). $\endgroup$
    – Joshua
    Commented Oct 22, 2015 at 3:41

When one is in the micro level of particles one has to stop thinking classically, i.e. with terms we have developed from our macroscopic observations.

"Touch" at the particle level can be defined as "interaction". Our feeling of touch actually does involve the electromagnetic interactions, we touch with the field of the molecules in our hand the field of the other objects, and this touching involves the exchange of force carrying virtual particles, i.e. particles which do not have definite mass but retain their quantum numbers characterizing them.

There is a precise mathematical description given graphically by the Feynman diagrams:

e+e- annihiation

The diagram translates one to one with integrals that give the probability of this interaction happening. The exchanged electron in this diagram( or positron, depending on how one reads this) is virtual, it does not have definite mass.

Then the closer the e+ and e- are to each other the larger the probability that the scattering will result to an annihilation. So annihilation happens according to the probability of interaction.There are higher order exchanges but they give much smaller probabilities.

  • $\begingroup$ Sorry for coming back to this after that long time. But I now wonder what would happen to a nucleus made of antiprotons and antineutrons. Does the same happens as with the electron ? $\endgroup$ Commented Oct 21, 2015 at 14:16
  • $\begingroup$ Experimenters have not managed to generate antiatoms, except antihydrogne, i.e. an antiproton with a positron . In a hypothetical interaction again Feynman diagrams would describe the probabilities. If an antihydrogen met a normal atom, first the positron would annihilate with an electron ,as above, and then the antiproton would eventually be captrured by its negative charge and finally fall on the nucleus and annihilate with a proton or neutron. $\endgroup$
    – anna v
    Commented Oct 21, 2015 at 14:24
  • $\begingroup$ They say here:en.wikipedia.org/wiki/Antimatter#Antihelium that they managed to create antihelium-4 nuclei. So say an antihelium atom was released into the air, I understand that the positrons will annihilate first, but then how will the nucleus annihilate ? Even if the antiprotons are captured, then how will the neutrons annihilate ? $\endgroup$ Commented Oct 21, 2015 at 14:33
  • $\begingroup$ The antihelium nucleus because they are 8000 times or so heavier than the electrons with a double charge will have orbitals very tight around the nucleus . This means that they have a high probability of overlapping the nucleus at an S state orbital.( Look up atomic orbitals) . then the antiprotons and antinetrons will annihilate with the corresponding feynman diagrams , an example here arxiv.org/PS_cache/hep-ex/ps/9708/9708025v1.fig1-18.png $\endgroup$
    – anna v
    Commented Oct 21, 2015 at 15:00
  • $\begingroup$ "This means that they have a high probability of overlapping the nucleus", by "they" you mean the electrons, right ? And if so, then why would the nucleus annihilate ? Wouldn't we need an antihelium nucleus to overlap a helium nucleus for this to happen ? $\endgroup$ Commented Oct 21, 2015 at 21:29

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