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I initially assumed a quark and its anti quark would annihilate into 2 photons (like electrons with positrons) and this does seem to be the case at least sometimes (e.g. For $\pi^0$s).

However in our particle physics course we also learnt about an example where a charm and anticharm quark annihilate into a virtual gluon (initially).

My question is why don't they just always annihilate into photons, and what determines whether they decay into photons or gluons or something else?

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  • $\begingroup$ @AccidentalFourierTransform That seems more like an answer than a comment; please consider re-posting in the appropriate place. $\endgroup$ – rob Apr 6 at 14:39
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In extension to the comment by AccidentalFourierTransform, I give a, maybe oversimplified, picture of how to interpret these different "particle reaction channels". I like to think of these different Feynman graphs in the following way.

Considering simple non relativistic single particle quantum mechanics, you already know that there is no definite trajectory you can assign to a particle, but every possible trajectory has a certain probability. So if you want to know how a particle propagates from an initial point to a final point in space you can only assign a probability to each path. It is more or less the same for Feynman diagrams. The outer lines correspond to the initial and final state of your system. Now you can choose different "trajectories" so to say. A simple example would be that an electron and a positron "annihilate" to a virtual photon which then "decays" into an electron and a positron again. The same process could happen where there is no virtual photon but a virtual Z boson. These two processes are two different "trajectories" from one initial state to another, but to describe sensible physics you have to take into account all possible "trajectories"/"particle reactions".

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My question is why don't they just always annihilate into photons, and what determines whether they decay into photons or gluons or something else?

I think it is important to add to the chosen answer this crucial difference between electromagnetic and strong vertices, it is not just trajectories, it is mainly the coupling constants.

The difference in the probability of seeing photons in a quark antiquark interaction or seeing gluons is due to the coupling constants difference, 1/137 for electromagnetic interactions, 1 for strong. Generally strong wins in probabilities.

As for the pi0 , have a look how it decays, it is the different coupling constants that control the probabilities, see the e+ on e- annihilation. ( Charge pions go through the weak interaction, neutral through the electromagnetic,that is why their lifetime is so short with respect to charged, see the weak coupling. In the case of charge pion the masses of W and Z also depress probabilities due to the propagators)

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As others have said, nothing in Quantum Mechanics is definate or certain, it's all a case of probabilities. This is described in Heisenburg's Uncertainty Principle. Schroedinger's Equation may be used to calculate the actual probability of each specific outcome/event. You can see The Uncertainty Principle in action for real in an 'A' Level physics lab, with an electron diffraction tube. You use a low voltage element to "boil-off" thermionic electrons. Then accellerate them up through a 5,000 volt anode and on through a carbon slide. The electrons then hit the end of the tube and you see light and dark fringes. The bright fringes indicate where many of the electrons hit the tube, the dark spaces where they do not hit, and the fringes get less and less bright as you move away from the centre, or line of least deviation as a result of the electrons interacting with the carbon atoms' electric fields. This shows that the electrons' trajectories are not certain. The fringes appear to show the electron particles acting to form a 'wave' pattern, as in 'wave-particle duality'. The thing is, the wave fringes show a probability wave for where the electron particles impact. If you added-up all the impacts, it would come to 100% on the electrons. Hope this helps.

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    $\begingroup$ What's there not to like? The Uncertainty Principle is just like magic. $\endgroup$ – user291781 Apr 7 at 10:16

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