# In an electron-positron annihilation, in what direction are the photons released?

I read that, in an electron-positron annihilation, at least 2 photons are produced, because of the law of conservation of momentum.

my question is: in what direction are those photons released? and how accurate is this animation?

edit: now i understand that there is no specific direction for the released photons, but we can calculate the probability of them being released at a certain angle.

now is it correct that the 2 released photons must have opposite directions?

• Hi This is a classical picture, but with elementary particles, there is no definite path they take, only the probability of being found somewhere. The animation should not be taken as accurate, Regard
– user74893
Mar 29, 2015 at 11:18

It is a standard exercise in quantum electrodynamics to find the angular dependence of the differential cross section. Which more or less means how probable it is for the photons to scatter at a certain angle, given the energy of the incident particles.

So assuming the spins of the electron-positron pair are averaged, and that you don't care about the photon polarization, if you define $$\theta$$ to be the angle the outgoing photons make with the direction of the incident particles, and you work in the COM frame, you get

\begin{align*} \left| \frac{d\sigma}{d \Omega}\right|\propto \frac{E^2 + p^2 \text{cos}^2\theta + 2m_e^2}{m_e^2 + p^2 \text{cos}^2\theta} - \frac{2m_e^4}{\left( m_e^2 + p^2 \text{cos}^2\theta \right)^2} \end{align*} Where $$E$$ is the energy of the incoming electron, and $$p$$ is its momentum, and $$c\equiv 1$$.

For very energetic $$e^+ e^-$$ this becomes $$\left| \frac{d\sigma}{d \Omega}\right|\propto 2\text{csc}^2 \theta -1$$

fun fact: for slow $$e^+ e^-$$ their De Broglie size will be big as to effectively set a lower limit on the impact parameter, which will make the formation of the meta stable bound state positronium much more likely. After that, the decay will depend on the specifics of the bound state.

This may follow Einstein's equation and may appear to fit into classical picture, but this is taking place between two particles obeying Fermi Dirac statistics. Hence, it is a quantum phenomenon, and the direction of photon emission is arbitrary, as required by the fundamental assumption of quantum mechanics.

• Capital letters and full stops, please!! Mar 29, 2015 at 17:25