Does quantum mechanics state that a photon is a positron and electron annhilating

[Agnostic]

Main Question or Discussion Point

Does QM tell us that photons are actually electrons and positrons goign through a cycle of annilation and creation?

 

[Agnostic]

can an electron and positron annhilate each other without the presence of a neutron?

 

[Integral]

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.

 

[Agnostic]

Is that to say there is a 0 probabilty of a photon going to a positron and electron?

 

[Parlyne]

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.

 

[Mentz114]

According to the path integral formulation of QM, a photon has a non-zero probability of spending time as a (virtual ?) e+/e- pair.

 

[mathman]

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.

 

[Mentz114]

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 .

 

[mathman]

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.