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I've been reading for pleasure about two-photon interaction experiments, and one thing that confuses me is why, for example, two photons in the visible spectrum cannot interact. Is this indeed the case, and if so, why?

Jules Randolph
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    see my answer here https://physics.stackexchange.com/questions/399846/is-light-intangible-to-other-light-and-how-does-all-the-intersecting-light-exis/399878#399878 – anna v Apr 18 '18 at 10:08

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Photons are electrically neutral, so their interaction is by vacuum polarisation. The involvement of virtual charged particles makes the interaction difficult at energies $\ll m_e c^2$, $m_e$ the electron mass. (Difficult, but not impossible; pair production, of course, would be.) Eq. (1) here, first derived in 1935, shows the cross-section is proportional to the sixth power of the photons' CM energy. (It's also proportional to $\alpha^4$, with $\alpha\approx 1/137$ the fine-structure constant, so obviously $\alpha^4$ is very small.)

J.G.
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  • After the pair production threshold , there are only two vertices with α as the virtual loop can break into real electrons ( for an electron loop) – anna v Apr 18 '18 at 10:47
  • @annav Which implies the two extra pre-threshold powers of $\alpha$ substantively suppress the low-energy case compared with the high-energy one. – J.G. Apr 18 '18 at 10:50