No, but an electron and a positron can annihilate to create a photon. this is not a statement about ALL photons.
Electron positron annihilation not have to happen in the presence of a neutron.
Is that to say there is a 0 probabilty of a photon going to a positron and electron?
A single photon cannot spontaneously become an electron and a positron, as it would be unable to simultaneously conserve energy and momentum. For the same reason, an electron and a positron cannot annihilate into a single photon.
However, two colliding photons (if they have sufficient energy) can form and electron and a positron; and, an electron and positron can annihilate to form two photons.
According to the path integral formulation of QM, a photon has a non-zero probability of spending time as a (virtual ?) e+/e- pair.
In addition to the 2 photon into electron-positron pair, it is possible for a single photon (of enough energy >1.22 Mev) to form a pair when in the presence of a nucleus to get the required energy and momentum balance. This reaction is quite common in nuclear reactors and in lead shielding.
Mathman, with the single photon near a nucleus, does the electric field of the nucleus play a part ? Presumably a 1.22 Mev photon is a gamma-ray type .
The electric field plays a role, since (I presume) the electromagnetic force is the only force involved. As far as the photon energy, yes it is a gamma ray. Gamma rays from nuclear reactors can be as high as 10 Mev (maybe even higher).
P.S. Error (by me) - threshold for pair production is 1.022 Mev, not 1.22.