Electron–positron annihilation (Quantum electrodynamics) I'm trying to learn a bit about quantum particles interactions and I've found something in Wikipedia that I find confusing. There are two different examples with different results about the same interaction: Electron and positron annihilation.
In the first example, the annihilation produces one photon:

In the second, it produces two:

Are both examples correct? Could someone explain please?
Also, I'm curious why in the first example the antiquark can just emit a gluon. Is it because there are no conservation laws referring to gluons so they can just be emitted anytime?
Thanks, you'll notice that I haven't studied physics, I'm in fact a curious artist, so please, be patient.
 A: Yes, both examples are correct. Do notice that there is another difference between these two examples, though: in the first example, the single photon occurs only inside of the diagram (it is an internal line), while in the second example the two photons are going away (they are external lines). This makes a lot of difference.
Due to momentum conservation, it is impossible for two massive particles to annihilate into a single massless particle. The reason is that we can consider the pair of massive particles in their center of mass reference frame, where the total momentum is zero. However, once they annihilate into a massless particle, special relativity demands this massless particle to be moving at the speed of light. This implies it has non-vanishing momentum. Hence, we get a contradiction.
Notice this contradiction does not happen if there are two massless particles at the end, since each of them is going to a different direction and the total momentum is still zero.
Then how can we get a pair annihilation into a single photon in the first diagram? This is a very subtle thing. That photon is said to be off-shell, or to be a virtual particle. It happens that the internal lines of Feynman diagrams do not need to obey the so-called "mass shell condition". This is a fancy name for $E^2 = p^2 c^2 + m^2 c^4$ (which is the generalization of the famous equation $E = m c^2$ for the case where particles have non-vanishing momentum). Hence, in the second diagram we have an incoming electron and a positron, and two outgoing photons. In the first diagram, the incoming electron and positron annihilate into a virtual photon, which later creates a quark-antiquark pair.

Also, I'm curious why in the first example the antiquark can just emit a gluon. Is it because there are no conservation laws referring to gluons so they can just be emitted anytime?

Pretty much. In quantum mechanics, anything that is not forbidden is mandatory (in the sense that the probability of it happening it now zero, and hence it will happen eventually). You can also get a photon emission, for example, it needs not to be a gluon. Both the photon and the gluon are massless particles, so the energy cost for emitting them is just that of their momentum. The quark emits a gluon and recoils, losing the momentum that was released in the gluon.
At last, here's a really interesting point. You mentioned

There are two different examples with different results about the same interaction: Electron and positron annihilation.

It happens that the same input might lead to different outputs. Each of them will have a different probability of happening. In practice, we don't throw individual electrons and positrons at each other, but beams of particles, and you'll end up with a lot of different possible processes going on. So in an accelerator, you'll have outgoing photons (due to the second diagram), outgoing quarks (due to the first diagram), and etc. Other processes will also take place and make everything more complicated. For example, quarks bonding into hadrons and mesons, which are more complex particles (the proton is an example of a hadron).
