Is the photon pair generated from the electron-positron annihilation entangled?

And would they work as a source of entangled photons suitable for experiments in quantum optics?

Yes, they are definitely entangled. Their combined energy will exactly equal the combined energy of the original electron-positron pair, for example. The same is true for combined momentum and combined spin.

  • "Combined spin" should also include any orbital angular momentum if the electron and positron briefly orbited each other (positronium) before annihilating. – Andreas Blass Nov 19 at 16:08
  • That is correct-- – S. McGrew Nov 19 at 16:13
  • And from what I understand about these things (only a layman as far as physics goes), the angular momentum of the electron-positron pair will almost always be non-zero. – EvilSnack Nov 20 at 0:57

The photon-pair is definitely entangled. It is produced from an $S$-wave state of the electron-positron system and as such has orbital angular momentum $L=0$. But it's not useful for quantum optics, the photons' energy is too high for your usual mirrors to reflect. See fig 33.15 on p. 23 of this to get an idea what happens to a 511 keV photon from two-photon annihilation of an electron-positron pair at rest once it hits matter: ionization by Compton scattering dominates the interaction, that's not what you want to happen in a mirror or lense.

It's worth keeping in mind though, that the electron-positron system (Positronium) can decay to any number of photons $>1$, though numbers higher than three are very rare. An even number of photons can be produced for decays of the singlet ground state ${}^1S_0$ (where the electron and positron are in an anti-symmetric spin state $\frac{1}{\sqrt{2}}\left( \left|\uparrow \downarrow \right\rangle - \left|\downarrow \uparrow \right\rangle\right)$, "Parapositronium"). This defines the spin entanglement in the case you had in mind, and it has actually been measured and found to agree with full entanglement, see e.g. here

An uneven number can be produced from the triplet ${}^3S_0$ ("Orthopositronium"). The triplet is much longer-lived than the singlet state (roughly thousand times), but since it consists of three states and because symmetries ($CP$) prevent the positronium from going from the triplet to the singlet state, an appreciable number of electron-positron pairs decays to three photons.

In fact, also the $S$ states of higher energy levels can decay to photons directly, but usually they will decay to the ground state first, emitting photons of energies $O(10 - 100 e\mathrm{V})$ before the actual decay to the high-energy pair or to the three continuous spectrum photons.

protected by David Z Nov 20 at 9:58

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