Post History

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.