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Why does a photon colliding with an atomic nucleus cause pair production?

...I asked why a photon colliding with a atomic nucleus can become an electron and a positron. The answer that I thought was most illuminating explained that a photon spends some of its travel time as a particle-antiparticle pair of an electron and a positron. If it strikes the nucleus at the right time, this pair will be separated. It was explained that this is because 'Quantum Electrodynamics allows it'.

Why does Quantum Electrodynamics allow a photon to exist temporarily as an electron and a positron?

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I think there's a misconception in the question - it gives the impression of a situation where a propagating photon somehow objectively oscillates between being a photon and being an electron positron pair. That's not an accurate picture of the process of photon propagation. In fact there is no picture of the propagation - all QFT gives you is the probability amplitude to go from A in one polarization state to B in another. This amplitude gets corrected by some suggestive pictures which show the photon splitting and recombining, but these pictures are just tools to aid the calculation. –  twistor59 Mar 11 '13 at 21:06
    
John, maybe it will help clarify the question if you edit in some words about why you think QED shouldn't allow this to happen. –  David Z Mar 12 '13 at 1:03

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If the photon energy $E$ is sufficient to produce a pair ($E>2mc^2$), then a pair can be produced in a collision. It is due to equivalence of mass and energy. The same energy can be "represented" differently - as photons, particles, etc. Kind of conversion of one form of energy into another.

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I understand it's "allowed" but not where the rules come from that allow it... I'm missing something fundamental about why the photon would do this and not just stay a photon... –  John Mar 11 '13 at 20:36
    
@John, this is simply the observed behavior of quantum mechanics. It even has a name: "the totalitarian rule" (totalitarian because everything that is not forbidden is compulsory...). There is a level at which physics explains why (in terms of more fundamental physics), but at it root physics is descriptive not prescriptive and simply says "this is how things are observed to be". –  dmckee Mar 11 '13 at 20:46
    
@John: A photon can stay a photon too. There is no obligation to convert into a pair. They say there are different channels of scattering with different issues including staying a photon as one of possible issues. –  Vladimir Kalitvianski Mar 11 '13 at 21:20
    
Isn't it in fact that "everything that's not forbidden by symmetry does happen"? :) –  Lagerbaer Apr 12 '13 at 2:46
    
@Lagerbaer: Yes, a collision may have many different channels (issues) of occurring. –  Vladimir Kalitvianski Apr 12 '13 at 7:21

It is all hidden in the QED Lagrangian:

One can answer this question in a simple way in terms of the QED Lagrangian, at the electron-field interaction part:

$L=\bar\psi(\partial_\mu\gamma^\mu-m_e+eA_\mu\gamma^\mu)\psi$

This tells us that interactions of the form:

$e^++e^-\rightarrow\gamma$..................(1)

$\gamma\rightarrow e^++e^-$..................(2)

are allowed. This then means that since (1) and (2) are allowed at tree level, at higher order QED the interaction (at one loop level)

$\gamma\rightarrow e^++e^-\rightarrow\gamma$

is also allowed. This is because there is nothing in the fundamental QED Lagrangian, $L$, to tell us that this is not possible. Special relativity as represented by the above Lagrangian allows $E_\gamma=2m_ec^2$ as well as $2mc^2=E_\gamma.$ Quantum mechanics also allows, via the uncertainty principle, for the $e^++e^-$ pair to exist in virtual states, unless some sufficiently strong Coulomb field, or even uniform electric field, or strong gravitational field like just outside a balck hole's event horizon (Hawking radiation) separates them from each other.

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