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.

  • $\begingroup$ "Combined spin" should also include any orbital angular momentum if the electron and positron briefly orbited each other (positronium) before annihilating. $\endgroup$ – Andreas Blass Nov 19 '18 at 16:08
  • $\begingroup$ That is correct-- $\endgroup$ – S. McGrew Nov 19 '18 at 16:13
  • $\begingroup$ 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. $\endgroup$ – EvilSnack Nov 20 '18 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 http://pdg.lbl.gov/2018/reviews/rpp2018-rev-passage-particles-matter.pdf 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 http://adsabs.harvard.edu/abs/2009APS..HAW.GB108S

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 '18 at 9:58

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