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Let's consider an electron-positron pair with total spin equal to zero. When it annihilates it can not emit only one photon because it would have zero momentum and nonzero energy. The pair emits two photons with opposite momenta but on the momentum-energy plain it looks like the particle goes through a forbidden state (red path on the picture below).

The first question is: How is it possible? I suppose this is because of the energy-time uncertainty. The annihilation process is instant (at least looks like on Feynman diagram) and the energy of the intermediate state is not determined. Is it correct?

If we can go through any forbidden state, why doesn't the annihilation go the blue path? This is the second question.

And the third question: Why do electron-hole pairs in semiconductors always emit photons with energy equal to the band gap? Is it just because the interaction with one photon has higher probability or there is a fundamental difference?

enter image description here

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@zephyr: Thanks! This answers my third question and may be two others. Do I see it right that two-photon (red) annihilation is not the only possible but just the most probable? Is four-photon (blue) annihilation possible but of a negligible probability because of larger number of the particles participating in the interaction? –  Maksim Zholudev Jan 16 '12 at 14:57
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No, an electron doesn't go through any forbidden state, surely not along a specific path of the kind you drew in the momentum space. The annihilation event is instantaneous, the momentum and energy are always perfectly conserved, and all external particles are always exactly on-shell. –  LuboŇ° Motl Jan 16 '12 at 15:51

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