Past exam of the symmetry and relativity course of the University of Oxford / 2017 / 2 / c

Put $\epsilon =\hslash \omega$. The total four-momentum of the perpendicular photons isAlthough each constituent is massless, the sum has $P_{\mathrm{tot}}^2=2{\epsilon }^2/c^2$. Its invariant mass is thereforeThe rest energy is the total energy in the frame where the two photon momenta cancel, not the sum of individual photon rest energies.

The centre-of-mass velocity follows from the total energy and momentum:A boost in this direction with ${\Gamma }_{\mathrm{CM}}=\sqrt{2}$ gives zero total spatial momentum and total energy $\sqrt{2}\hslash \omega$. The constituent photons remain moving at $c$ in that frame.