# What happens when two photons collide with one another, head on, dead center? [duplicate]

If two photons were to collide directly, head on, and are of the same energy, what happens? Are new particles created, is energy released? Or do they just pass through one another?

Photons don't directly interact with each other, but if one photon produced an e+/e- pair then the second photon could interact with that pair.

The interaction has to conserve the energy of the two photons and conserve their momentum as well of course.

But yes they could (and most probably depending on their energy) just pass right "through" each other.

Usually absolutely nothing. Electromagnetism is linear, which means that the result of doing something with two photons is the superposition of the results of doing something with each one individually. By that reasoning, since one photon by itself just goes on its own merry way, then two photons, even if they go near each other, just go along on their respective merry ways without interacting at all.

However, the universe is not just electromagnetism! We also have charged particles, and "charged" means "interacts with photons". Therefore, photons push charged particles around. That means that photons can interact indirectly: first photon pushes particle, particle moves somewhere new, particle interacts with second photon.

Also if you have a photon with enough energy (or large enough number of photons with lower energy) then the photon(s) can actually create charged particles. Suppose it takes a photon with energy $E$ to create a charged particle. If we have two photons each with energy $E/2$, then if they get near each other we might get creation of a charged particle where this would not have happened with any one of those photons alone. It takes a lot of energy for this to happen. To create an electron you need a photon with a wavelength of

$$\lambda = \frac{c}{\nu} = \frac{c}{E/h} = \frac{c}{m c^2 / h} \approx 10^{-12} \, \text{m}$$

which is quite small: about one hundredth the diameter of an atom. Visible light is in the range of around $6 \times 10^{-7} \, \text{m}$, so you need ten thousand times smaller wavelength, meaning ten thousand times more energy, than visible light. If the photons are creating charged particles then they can interact indirectly without there being any charges floating around a-priori. That kind of makes it as if they're interacting directly because there's no obvious matter floating around mediating the interaction. In the case that the charged particles are created, mediate a photon-photon interaction, and are then destroyed, those particles are called virtual particles. This can happen even if the photons don't actually have enough energy to create the particles! Energy conservation can be violated by an amount $\Delta E$ if the interaction only lasts for time $\Delta t \approx \hbar / \Delta E$, roughly speaking.

• 10^-12 isn't ridiculous, that's a hard X-ray or gamma ray. Commented Aug 5, 2015 at 9:42

To leading order, nothing happens in any photon "collision".

To higher order there are light-light interactions that involve particle loops, but they don't (can't) depend on the geometry because we can always boost to a frame in which the pair has zero net momentum (even though you can't boost to the frame where a single photon has zero momentum).

• This is lovely! I want to emphasize the surprising punch line: every photon collision is head on, dead center. Commented Aug 5, 2015 at 3:47

Two gamma ray photons are produced by the collision of an electron and a positron. If the laws of physics are reversible at the quantum level, there is a chance that the collision of two gamma ray photons could produce a positron and an electron.

However, it's questionable that this would not violate the 2nd law of thermodynamics. It seems to me that a positron-electron system is of lower entropy than a two-photon system, simply because there are more ways to distribute the conserved energy of the system between two identical photons than between a positron and electron. The two photons could pass through each other without changing the nature of the system, whereas such a state would significantly change the positron-electron system.

Two photons moving in opposite directions ("head-on") can collide and move off in different directions (still opposite if the photons have equal energies), If they have enough energy, the photons might produce an electron-positron pair. At even higher energies, other final states are allowed by conservation of energy. The cross-sections (or probabilities) for various final states in photon-photon scattering can be calculated with high accuracy in quantum electrodynamics.