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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?

Gugg's handiwork

(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.

enter image description here

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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
up vote 3 down vote accepted

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.

You can get the answer in any other frame by doing a Lorentz boost of this result.

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Do note that in a boosted frame, the photons will appear to have different colors, depending on their direction. – Vibert Feb 19 '13 at 14:19

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