Light Interaction with Electrons in Glass

[thenewmans]

If I look through glass, the photons I see have encountered electrons. Is the photon that went into an electron the same one that comes out?

 

[Seele]

You can never truly state that any particle retains its individual identity, for that is a question demoted of any fundamental basis. Upon the absorption of a photon by an electron, a subsequent collision releases an identical photon traveling in a slightly altered path. Though however identical these photons may be to one another, that does not conclude that they are in fact the same photon.

 

[thenewmans]

So let’s say photons A and B are entangled and B goes through glass or bounces off a mirror and photon C pops out. (B and C might be the same.) Is C just as entangled to A as B was?

 

[humanino]

There is a problem anyway : photon only exist in vacuum. You can insist to describe the propagation of light with photons in the vacuum in the glass, but that is not the proper way to describe it : in the glass, you have another effective wave propagating.

There is no way to bounce photons. You can only absorb them and re-emit another one.

 

[RandallB]

So let’s say photons A and B are entangled and B goes through glass or bounces off a mirror and photon C pops out. (B and C might be the same.) Is C just as entangled to A as B was?

Entanglement experiments often route photons through wave filters, off mirrors, and beam splitters, etc. before sending them to polarization detection. And since correlations are maintained you have to say YES.

However I do not accept the absorption-reemission explanation. IMO it is the same photon; perturbed by, but not replaced when transiting those things it went through.

 

[humanino]

However I do not accept the absorption-reemission explanation. IMO it is the same photon; perturbed by, but not replaced when transiting those things it went through.

You may like it your may not, but wave coherence does not mean it's the same photon.

Otherwise, you have Afshar which-way experiment to explain.

 

[RandallB]

You may like it your may not, but wave coherence does not mean it's the same photon.

Otherwise, you have Afshar which-way experiment to explain.

I do not see in Afshart any requirement for absorption-reemission of photons by atoms and or the electrons in them. So I don’t see your Point.

I case it is involved some how; I do not accept the idea that the http://en.wikipedia.org/wiki/Principle_of_complementarity" , even though Afsher himself did not like it. And I will “in due course”, but in my priority and time, submit that to FOP as he requested.

 

[humanino]

I do not see in Afshart any requirement for absorption-reemission of photons by atoms and or the electrons in them. So I don’t see your Point.

I case it is involved some how; I do not accept the idea that the http://en.wikipedia.org/wiki/Principle_of_complementarity" , even though Afsher himself did not like it. And I will “in due course”, but in my priority and time, submit that to FOP as he requested.

Interesting. The way I see it, the photon is in a coherent sum of position eigenstates after the pinholes. Assuming the wires infinitely thin, and interferences there, then you have the lens, which is a momentum measuring device. It breaks the position coherence. Now you cannot assume this if you say this is the same photon before and after the lens. But I do not see any QED diagram of photon-non-interaction, so I don't understand how a single photon could be deflected by a lens if there was no atomic interaction (for the uses of lenses in most EPR experiments).

Note that this simple explanation is not contradictory to most of the other ones. I don't know of any experiment challenging elementary QM which cannot be simply explained once it is understood. Note also that the lens does not measure angular momentum.