# Two-photon scattering: colours

Is there a particular conservation principle that necessitates that the outcoming photon pair has the same frequencies as the incoming photon pair?

I'm thinking in particular of these Feynman-like diagrams. Is it only like A or can it be like B as well?

(The diagrams are $2$-dimensional and might suggest that the photons have different velocities, but that's not what I had in mind! What I do hope to convey is that the directions of the outgoing photons are different in A and B.)

If there is no such conservation principle at work, with which theory would one then be able to calculate the distribution of the amplitudes with respect to frequency-pairs?

I hope this makes sense. Otherwise please let me know.

Edit. I've added the diagrams with A', A'', and A''' below, just in case I got it wrong in A.

-
There's nothing wrong with any of the diagrams A,...,A'''. All the permutations contribute to the amplitude for photon scattering sense photons are indistinguishable bosons. – Michael Brown Feb 19 '13 at 15:42
@MichaelBrown Thanks and thanks for your insightful answer. I'm now considering if I can figure out what happens to a red, a blue, and a violet incoming photon. That seems to require a further insight. Am I on the right track in assuming that the leading diagrams for that would each involve two subsequent two-photon "loops" instead of one three-photon "loop"? But, then, doesn't the three-photon "loop" contribute anything at all? – Keep these mind Feb 19 '13 at 16:00

Energy and momentum are both conserved. Working in the centre of momentum frame the momenta of the incoming photons are equal and opposite so the total momentum is $P=p+(-p)=0$. The energies of the photons are also equal and equal to $pc$. The total energy $E=2pc$. Now let the photons scatter into two photons of energy/momentum $E_1,\ p_1$ and $E_2,\ p_2$ respectively. Since the total momentum is conserved we must have $p_1 + p_2 = 0$, so $p_2 = - p_1$. The momenta remain equal in magnitude. Further, the energies are equal: $E_1 = |p_1| c$ and $E_2 = |p_2| c = |p_1| c = E_1$. Since the energies are equal and the total energy is still $E=2pc$, we have that the energy of the final photons must be $pc$. So nothing can change except for the direction of the outgoing photons.