# Why is it forbidden for two photons to turn into one?

In the context of quantum field theory, why is it impossible for two photons (or other massless bosons like gluons) to collide and produce a single photon? This kind of a process is supposed to be forbidden by momentum conservation, but it was not immediately obvious to me why this is.

## 4 answers

Suppose you have two photons A and B on an x, y plane:

- Photon A is at (-20, 9) and traveling towards (20, 0).
- Photon B is at (-20, -9) and traveling towards (20, 0).
- Both photons have wavelength w.

Before collision, for one photon:

- Energy is E
- Momentum is P=Ec
- Momentum along x is Px=P*40/41

Total momentum along y cancels out.

After:

- Total momentum is 2*Px=P*80/41
- This implies energy E=Pc=c*80/41.
- However, total energy before was c*82/41.

So now we have a contradiction, because c*2/41 of energy is "disappeared". Unless you want to violate the conservation of energy, you would need to have some particle emitted that makes up for this. Most particles would introduce additional variables like charge or spin that must also be balanced, so the easiest particle would be a low-energy photon, but then it's not really a merging :). More like energy transfer from one photon to the other.

The "energy leak" comes from the opposing momentum canceling. It would not happen if the photons were moving parallel to each other, but then they couldn't collide. Or, if the new photon could move slower, but unfortunately they're locked to one speed.

I wrote this because I thought other answers were oversimplifying with the head-on case, but actually the head-on aspect is the source of the issue and the same thing is present to some extent in everything else.

However, you ask for a QFT explanation, while this is a classical one, so not sure if this is sufficient.

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Imagine two equivalent (e.g. same frequency) photons colliding with each other head-on. The linear momentum of the system is $0$ because each photon's momentum has the same magnitude but is pointing in opposite directions.

If these two photons collide and form a single photon (and nothing else), then conservation of linear momentum would mean that this photon has $0$ momentum. But a photon is, definitionally, always moving at the speed of light, and its energy is purely from its momentum (in special relativity, $E^2 = p^2c^2 + m^2c^4$ with $m=0$). For it to have zero momentum, it would have to also have zero energy which would violate conservation of energy and correspond to their not being anything.

Even for a more glancing collision, we'd either lose energy from the cancelled out parallel portions of the momentum, or we'd need to increase momentum in the perpendicular direction to make up for it which would violate conservation of linear momentum in that direction.

Thus, to satisfy conservation of energy and conservation of linear momentum simultaneously, we either need multiple particles moving at the speed of light afterwards, or the photons need to annihilate and produce a massive particle which can have zero momentum and absorb the energy as its mass (or some mixture of massive and massless particles whose net momentum is zero).

Think about how that is supposed to work. It seems you want two photons to somehow combine into a single photon and nothing else. That means the output photon must have the combined energy of the two input photons because you've provided no other place for the energy to go. And of course momentum needs to be conserved too.

However, a photon's momentum is also proportional to its energy. But how is that supposed to work when the two photons aren't traveling in the same direction? Take the extreme case of the two photons having equal energy but colliding head-on. The net momentum is 0, but the net energy twice that of each photon. You can't conserve both energy and momentum if the photons were allowed to combine.

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Olin and Derek have covered the biggest part of this: *the combination of energy and momentum conservation rules out the two-into-one combination of photons that are not co-linear*.

That might leave you optimistic about having a special case, but there is one more rule to respect: *conservation of angular momentum*.

Photons have angular momentum with magnitude $\hbar$ that is directed either parallel or anti-parallel to the linear momentum. As a result a system of two photons moving in the same direction have a total angular momentum along the direction of travel of $\pm 2\hbar$ or $0$ none of which are values that a single photons can take on. This rules out two-into-one combination of co-linear photons.

I suppose that leaves an even fainter glimmer of hope for three-into-one and higher (initially odd) combinations, but I have no idea what that would look like at tree level.

The other place to look for an escape hatch is that this discussion assumes free-space: in a medium new tricks become possible (and, indeed, frequency doubling media is a thing you can buy commercial off the shelf for some wavelengths).

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