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All other neutral particles (e.g. neutrons, pions, atoms, etc.) have charged sub-particles in their composition, and, thus interact with photons.

We have to mention that, at elementary particle level, all the interactions are scattering. Namely, an elementary particle cannot absorb a photon, and remain the same. Composite particles that have excited bound states, may absorb a photon, and transit to an excited state. As they de-excite, they re-emit an incoherent photon.

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So what exactly happens when a photon is collided with a neutral elementary particle?^{0}vector boson and the gluons from the bosons. None interact with the photon directly (at tree level)

All other neutral particles (e.g. neutrons, pions, atoms, etc.) have charged sub-particles in their composition, and, thus interact with photons.

We have to mention that, at elementary particle level, all the interactions are scattering. Namely, an elementary particle cannot absorb a photon, and remain the same. Composite particles that have excited bound states, may absorb a photon, and transit to an excited state. As they de-excite, they re-emit an incoherent photon.

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Matterwave

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Do you mean by "incoherent photon" that there is no correlation between the photon's momentum vector (or spin) and the parent particle's momentum vector (or spin)?Composite particles that have excited bound states, may absorb a photon, and transit to an excited state. As they de-excite, they re-emit an incoherent photon.

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DrDu

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Not to forget the photon on that list.The only neutral elementary particles are the neutrinos from the fermions, and the Z^{0}vector boson and the gluons from the bosons.

In fact maybe the best studied interaction of a photon with a neutral particle is the interaction with another photon to produce e.g. electron positron pairs.

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Right. This is a tree level process, but a second order in the coupling [itex]e[/itex], whose transition amplitude is proportional to the fine-structure constant [itex]\alpha = \frac{e^2}{4 \pi}[/itex] (in natural units, and Lorentz-Heaviside units of electrodynamics), and the cross section is proportional to [itex]\alpha^2[/itex].Not to forget the photon on that list.

In fact maybe the best studied interaction of a photon with a neutral particle is the interaction with another photon to produce e.g. electron positron pairs.

In fact, Srednicki's QFT section 59 has the transition amplitude for the reverse process:

[tex]

e^{+} \, e^{-} \rightarrow \gamma \, \gamma

[/tex]

One needs to multiply by the Lorentz Invariant phase-space (LIPS) volume of the final states (See section 11 of the said book) to obtain a differential cross section.

Notice that this is quite different from a single photon pair production near a nucleus. Namely, this latter process is already accounted for in what I called scattering of a photon from a composite object (nucleus).

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DrDu

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Bound states of electrons and positrons are called positroniums and behave somewhat similar to hydrogen atoms although they decay within some ms again into two gamma quants. So it is difficult to store them.

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Note that there is a threshold energy of this reaction:

[tex]

\sqrt{s} \ge 2 m_e c^2 = 1.02 \, \mathrm{MeV}

[/tex]

Thus, the two photons should be

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Bound electron-positron pairs are called positronium, and have been extensively studied.So if two photons with enough energy collide, then they can be converted into the electron-positron pair. Can the electron-positron pair be used as a substitute for a hydrogen atom,

http://en.wikipedia.org/wiki/Positronium

I've never heard of positronium being produced by pair production from two photons. It seems to me that this would be an extremely delicate operation, because if the photons have too much energy, the e

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