From where does the photon reflect? (split from Finality of Bells Theorem)

[jlivingstonsg]

 

I would recommend Feyman's lectures on QED (videos or book). He explains a lot of this very clearly. One simple example is that the probability of a photon being reflected from the front surface of a sheet of glass depends on the thickness of the glass. This tells us that the photon somehow "knows" how thick the glass is.

 

http://www.vega.org.uk/video/subseries/8

 

 

The trouble is that there are good reasons for thinking that transferring information faster than light is impossible (and, indeed, entanglement cannot be used for communication). And in the photon reflection example above, what is doing the communication. That doesn't really work as an explanation. The "how" is simply non-locality.

 

Have anyone measured the time a photon take from emitting to detection do determine what surface it is reflected from?

Both in one surface and with two surface reflection.

Where is the photon reflected?

 

Can the time of the photon be measured that accurately?

 

Regards Magnus

Edited December 7, 2015 by jlivingstonsg

[Strange]

I don't think we could measure that. But also, I don't think it would tell you anything: we can tell whether the photon was reflected from the front surface (from the geometry of the reflected rays). What we can't explain without non-locality is how the photon "knows" whether the thickness of the glass will result in constructive or destructive interference.

[jlivingstonsg]

I don't think we could measure that. But also, I don't think it would tell you anything: we can tell whether the photon was reflected from the front surface (from the geometry of the reflected rays). What we can't explain without non-locality is how the photon "knows" whether the thickness of the glass will result in constructive or destructive interference.

 

They have done this two reflection experiment with a thickness of up to one meter.

See youtube Feynman video.

So it should be possible to determine if the photon comes from the last surface.

 

 

https://youtu.be/eLQ2atfqk2c?t=2457

[swansont]

 

They have done this two reflection experiment with a thickness of up to one meter.

See youtube Feynman video.

So it should be possible to determine if the photon comes from the last surface.

 

 

https://youtu.be/eLQ2atfqk2c?t=2457

 

Feynman doesn't actually say it's been done, just that the effect is there (at the 54-minute mark). And that experiment does not require that you time the photons.

[Strange]

The problem is not which surface if comes from. The problem is about the distance between the two surfaces and whether this results in interference or not.

[swansont]

The problem is not which surface if comes from. The problem is about the distance between the two surfaces and whether this results in interference or not.

 

The question in the OP is which surface the reflection comes from. Constructive interference is a requirement.

[jlivingstonsg]

I think it is interesting to ask if it is possible to detect with timing the photon, what surface it is reflected from.

 

Has it or has it not been done?

 

Link?

 

MagI

 

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[Strange]

I assume you could tell by the path it took (assuming it was incident at something other than 90˚).

[swansont]

I assume you could tell by the path it took (assuming it was incident at something other than 90˚).

 

I don't see how. You only get constructive interference along certain paths.

[Strange]

You are right. I hadn't thought it through.

If one could tell which surface each photon was reflected from, I assume would it stop interference occurring.

[robinpike]

I think it is interesting to ask if it is possible to detect with timing the photon, what surface it is reflected from.

 

Has it or has it not been done?

 

Link?

 

MagI

 

-

 

 

I assume you could tell by the path it took (assuming it was incident at something other than 90˚).

 

As Strange said, if the beam of light is aimed at an angle to the glass, some will be reflected from the front surface, and some will be reflected from the back surface. These two beams will be on separate paths, as the angle of the beam to the glass will cause the reflection from the back surface to not be below the point of reflection from the top surface.

 

If a low intensity beam of individually spaced photons are used, wouldn't the photons from the back surface take a longer time to get to the detector(s) than those from the front surface? thus showing that they really went to the back surface.

[imatfaal]

 

 

 

As Strange said, if the beam of light is aimed at an angle to the glass, some will be reflected from the front surface, and some will be reflected from the back surface. These two beams will be on separate paths, as the angle of the beam to the glass will cause the reflection from the back surface to not be below the point of reflection from the top surface.

 

If a low intensity beam of individually spaced photons are used, wouldn't the photons from the back surface take a longer time to get to the detector(s) than those from the front surface? thus showing that they really went to the back surface.

 

 

I wonder if the HUP would mean that if you measured the timing sufficiently accurately then you would lose enough precision in the position to stop interference. Trouble is with this idea is that I cannot quite work out whether these would be a non-commuting pair of observables (I am guessing the timing would give you direction which would give you momentum; the position and the momentum cannot be known with arbitrary accuracy at the same time) - and secondly not sure if it is enough to make any difference to interference pattern.

[swansont]

If a low intensity beam of individually spaced photons are used, wouldn't the photons from the back surface take a longer time to get to the detector(s) than those from the front surface? thus showing that they really went to the back surface.

 

That's the question — what happens with one photon?

 

 

I wonder if the HUP would mean that if you measured the timing sufficiently accurately then you would lose enough precision in the position to stop interference. Trouble is with this idea is that I cannot quite work out whether these would be a non-commuting pair of observables (I am guessing the timing would give you direction which would give you momentum; the position and the momentum cannot be known with arbitrary accuracy at the same time) - and secondly not sure if it is enough to make any difference to interference pattern.

 

I'm not coming up with a way off the top of my head of knowing when a photon was emitted. Perhaps a two-photon process with a beam splitter would work.

 

 

—————

 

OK, after consulting with my colleagues, here's a possibility: This is an interference effect, much like the double-slit experiment. In essence, it reflects off of both surfaces. Perhaps you lose the interference if you know what surface you reflect off of.

[imatfaal]

 

I'm not coming up with a way off the top of my head of knowing when a photon was emitted. Perhaps a two-photon process with a beam splitter would work.

 

Are all the methods you guys use to create photon sources in essence probabilistic? So when you get down to one photon you don't actually know when it will pop up.

 

Re the beam splitter - iirc delayed quantum eraser uses the matched timing of signal and idler photons to match pair and thus give which way information and thus destroy interference; so I presume that spontaneous parametric down conversion will have a fixed or predictable time match between the two photons produced

[swansont]

 

Are all the methods you guys use to create photon sources in essence probabilistic? So when you get down to one photon you don't actually know when it will pop up.

 

Re the beam splitter - iirc delayed quantum eraser uses the matched timing of signal and idler photons to match pair and thus give which way information and thus destroy interference; so I presume that spontaneous parametric down conversion will have a fixed or predictable time match between the two photons produced

 

Right - that's what I was thinking of for the two-photon source. Many other single-photon sources are just low-intensity via filters. The timing of the production of the photon doesn't come into play.