I know that when matter touches antimatter, they both explode and turn into energy. However, by my understanding, atoms don't really touch; their nuclei don't physically contact each other, since electromagnetic repulsion keeps them away.

  • If this is true, then what is the inciting moment that sets off the explosion?

  • At what point do the two substances go from being separate pockets of matter and antimatter to being energy?

  • Is it when the electron clouds get close enough?

  • How does antimatter know to explode?


3 Answers 3


Annihilation is a quantum process and as such it is probabilistic. You are correct that particles do not touch in a classical sense, but each interaction has a cross section. Particle and antiparticle are attracted to each other because of opposite charge and when they get sufficiently close, they interact. In case of electrically charged particles (like electron and positron) the interaction happens via electromagnetic field which is described by this equation of motion:

$$\partial_\nu F^{\mu\nu}=e \bar\psi \gamma^\mu \psi$$

On the left side is the kinetic term for photons (derivative of the tensor of the electromagnetic field) and on the right side is the interaction term describing charges (the $\psi$ field - for example electrons) as sources of the electromagnetic field. The equation tells us that if there is something happening in the $\psi$ field, it gives rise to photons (disturbances in the electromagnetic field). Mathematically you can say that if both $\psi$ and $\bar\psi$ are nonzero somewhere (meaning that particle and antiparticle are in the same location), then the right hand size is non-zero at that location and that means the derivative (rate of change) of $F^{\mu\nu}$ is nonzero as well. So the field $F^{\mu\nu}$ starts to "move" at that location.

When you have just one electron, it cannot spontaneously turn into photon due to charge conservation. The electron is negatively charged, photon is neutral, so this process is forbidden. But if electron and positron meet (they get close enough), their charges cancel, so the electromagnetic field has an opportunity to "steal" the energy from the $\psi$ and $\bar\psi$ field.

There is a certain probability that this will happen, but if it happens, the electron and positron (disturbances in the $\psi$ and $\bar\psi$ field) decay (disappear) and a pair of new disturbances in the $F^{\mu\nu}$ field is born (photons).

This process can be graphically represented (in a simplified way).

So how does the antimatter "know"? The $\bar\psi$ field does not "know" anything. It just obeys the equations of motion and conservation laws. When particle and antiparticle meet, they disturb other fields and there is a solid chance that they will disappear and give rise to particles in other fields. The process happens automatically, even though it is probabilistic.

There are more possible outcomes: it can happen that no photon is created and the annihilation just does not occur. Each such result has a certain probability of happening and the probability can be calculated using the mathematical tools of quantum electrodynamics (Feynman diagrams). Various possible outcomes have different diagrams associated with them and all of them interfere to affect the final probability that particle and antiparticle will annihilate.

Just a side note: when people say that annihilation turns particles into energy, it is not completely correct statement. Annihilation turns particles into other particles, for example photons (but other products are also possible, depending on circumstances). Photons are not "energy" - photons are fully fledged particles which have various features, like energy, momentum, spin.

For more details read for example the Wikipedia article or here.

  • 1
    $\begingroup$ I don't think the question is about why they come close, but at what point a particle can be said to have collided with another one. I have no expertise here, but I suspect event (particle physics) might be a good start. $\endgroup$
    – Paul
    Jan 27, 2016 at 20:30
  • $\begingroup$ @Paul You are correct. I am going to update the answer. $\endgroup$
    – mpv
    Jan 28, 2016 at 8:51
  • $\begingroup$ photon would be neutral ? why the would ? some publications doubt ? $\endgroup$
    – user46925
    Feb 3, 2016 at 12:02
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    $\begingroup$ This answer is a bit misleading -- the interaction of quantum fields is completely deterministic. You seem to say this at some points in the answer, then contradict it later with "When particle and antiparticle meet, they disturb other fields and there is a solid chance that they will disappear and give rise to particles in other fields." which is not what is 'actually happening' in the framework of QFT. $\endgroup$
    – knzhou
    Feb 3, 2016 at 19:52
  • 2
    $\begingroup$ What actually happens is that a superposition of field configurations gets produced deterministically. The other fields always get disturbed. Some of the resulting configurations have a particle and antiparticle, and some have just photons. Later, we observe one of the configurations. $\endgroup$
    – knzhou
    Feb 3, 2016 at 19:53

Your matter as well as your antimatter can be described by a propagating wave packet (for example a Gaussian). The overlap of both wave functions gives you a Luminosity $\mathcal{L}$.

Now rate of annhilation events is the cross section for this annhilation process times the Luminosity

$$R=\mathcal{L} \sigma$$

So the amount of overlap of both wave functions kind of tells if they are close enough.


You seem to think that matter is exemplified by atoms, elements and even molecules. This is incorrect. Particles are matter, and antiparticles are antimatter. An example of this, are electron and anti-electron, proton and anti-proton.
Next, you need to understand that a particle and its antiparticle are oppositely charged, therefore, they "naturally" attract each other. So if they are "within range," they will attract each other until they merge into the same space. Their "merging" causes them to turn to pure energy, which then disperses in all directions, giving the appearance of an "explosion."

There are no "electron clouds."
Neither matter or antimatter know anything, much less, that they "exploded."

  • $\begingroup$ What do you mean by "There are no electron clouds?" I thought nuclei were surrounded by probability "clouds"/orbitals, like the spdf orbitals. $\endgroup$
    – Somatic
    Feb 3, 2016 at 23:53
  • $\begingroup$ wThere are no electron clouds at the particle level. $\endgroup$
    – Guill
    Apr 15, 2017 at 23:18

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