Why does a particle-antiparticle collision produce $2$ photons instead of $1$? In one of the questions that I did, it asked: What is produced when an electron and positron collides with each other? The answer is $2$ photons. Why doesn't it merge into one? After all, the photon with that amount of energy can decay back into a particle and antiparticle.
Is my understanding correct?
PS. I'm just a beginner learning about the standard model. Would be nice if the answer is simplified to a lower level. Thanks.
 A: Photons have momentum $p=\frac{E}{c}$ (despite having no mass) Thus to conserve linear momentum, multiple photons are formed, moving in different directions. Also, photons are typically formed for low energy collisions. High energy collisions can result in exotic heavy particles forming. Furthermore, the annihilation (or decay) of an electron-positron pair into a single photon can occur in the presence of a third charged particle to which the excess momentum can be transferred by a virtual photon from the electron or positron.
A: Set $c=1$ and denote the electron/positron mass by $m$. Suppose that we go to the rest frame of the collision, which we can always do when considering massive particles. Then we can write the 4-momentum of the electron as $(E, \vec{p}) =(\sqrt{m^2 + \vec{p}^2}, p_x, p_y, p_z)$ and the 4-momentum of the positron as $(\sqrt{m^2 + \vec{p}^2}, -p_x, -p_y, -p_z)$. 
At some point during the collision, the positron and electron will annihilate under the formation of one photon; however, by conservation of momentum, if nothing else happens, that photon will have 4-momentum $(E, \vec{p}) = (2 \sqrt{m^2 + \vec{p}^2},0,0,0)$. This means that it is not on-shell, as its energy and momentum do not satisfy $E = |\vec{p} |$. Therefore, something else has to happen in order for us to end up with real particles instead of virtual ones. 
Conversely, one can start with an energetic on-shell photon with $E \geq 2m$. Now we cannot go to the rest frame of the photon, because the photon is a massless particle. So suppose the photon moves in the z-direction; then it has 4-momentum $(E,0,0,E)$. If the photon then decays into a positron and an electron, one could go to the rest frame of the positron or the electron. In the rest frame of the positron (for definiteness), the positron will have four-momentum $(m, 0,0,0)$ and therefore the electron will have four-momentum $(E-m, 0,0, E)$. But now you can convince yourself that the electron is not on shell either, so again some other process has to happen to make all the outgoing particles real.
A: Your understanding is wrong.
The LEP experiments had collisions of particle antiparticle and produced at least two particles as output,not just two. Here are the published  measurements of just       charged multiplicity at LEP.
The reason they cannot go  into one photon and need at  least two photons is explained in the other answer: so as to have momentum and energy conservation at the center of mass: the electron and positron have an invariant mass in the sum of their four vectors whereas a single photon is massless: reductio ad absurdum.
They can go into one massive particle ,  as seen in this experimental plot of e+e- scattering:

but there exist no stable particles, just resonances that decay, some in several channels. The positronium is a bound state atvery low energy for a while, of e+e-, forming an atom,  before the e+e- annihilate into two photons.
