My novice understanding to the background of this question: I have heard of annihilation turning matter into "pure energy". This is typically illustrated with an electron/positron interacting to form some gamma photons. However, quark/antiquark interactions can form more types of particles (not just photons). Furthermore, I imagine that smashing any two particles (e.g. quark/electron/neutrino/etc) would produce some set of particles that are not the original.

Question: Is there any important distinction between "two particles interacting and forming other particles" vs "two particles interacting and annihilating into other particles"? Does the word "annihilation" have any physical meaning beyond "particles interacting, where one happens to be matter and the other happens to be antimatter"?


We call annihilation interactions of particles where all the quantum numbers inputed to the interaction add up to zero, so just energy remains that can have an output of a number of new particles , the sum of quantum numbers of these particles will be also zero. It is the quantum number information that is "annihilated".

  • $\begingroup$ Thanks for the clear answer. Followup: Is there a reference for me to understand "all the quantum numbers"? Searches online seem to have different opinions about what a "quantum number" is. $\endgroup$
    – perpetual
    Apr 8 '21 at 11:57
  • $\begingroup$ In the standard model of particle physics, there is a table where the basic consituents of matter, the elementary particles, are shown,with all their quantum numbers, the antiparticles are an equal table. All interactions are basically interactions of these particles, so when two with equal and opposite quantum numbers interact, the result is only energy.en.wikipedia.org/wiki/Standard_Model $\endgroup$
    – anna v
    Apr 8 '21 at 14:02
  • $\begingroup$ In this review indico.cern.ch/event/447008/contributions/1953687/attachments/… page 40 $\endgroup$
    – anna v
    Apr 8 '21 at 14:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.