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In a dark room there are two people and a very faint candle. Then the candle emits one photon. Is it true that only one person can see the photon? Why? And are there any experiments?

Edit 2019/4/23:

Thanks a lot. I was originally asking about quantum mechanical things. Because I believe such an experiment will turn the two people into Schrodinger's Cat. That's weird because it is not likely for a macroscope object get correlated so easily. Now thanks to the answers I realized that the efficiency and noise is as bad as in cases of other quantum mechanical processes.

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  • $\begingroup$ You should specify your question: are you asking from a purely physical point of view e.g. are you interested to know whether a single photon (forget about a candle and only a single photon!) can be detected by two different "sensors", which eyes in principle are, or rather on the full world realistic question including the energy needed to trigger something in our brain? $\endgroup$ – Mayou36 Apr 20 at 19:04
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    $\begingroup$ Photons are interchangeable. There is no such thing as "the same photon". There are just photons. $\endgroup$ – Stian Yttervik Apr 21 at 10:34
  • $\begingroup$ Related post here. Although a rod in the retina can respond to a single photon, such a photon must strike it in the right way, with the right frequency, after having made it past the air molecules, dust particles, eyelashes, tears, cornea, aqueous humour, lens, vitreous humour, retina membrane, blood vessels, nerve cells, other cells in the retina, and part of the rod cell itself as well. $\endgroup$ – user21820 Apr 21 at 14:26
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    $\begingroup$ Given the wavelength of any photon that can be seen, I figure they would have to stand very closely together to see the same photon. $\endgroup$ – EvilSnack Apr 21 at 17:08
  • $\begingroup$ google.com/search?q=eye+reflection&tbm=isch $\endgroup$ – Cœur Apr 22 at 10:47
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Candles do not give off single photons. Preparing light sources that can emit single photons is tricky.

The photon contains "one photon" (some small quantity of electronvolts) of energy. The energy in a photon is directly propotional to its frequency, so two photons of the same "color" have the same energy. The process of absorbing a photon transduces "one photon" of energy from the electromagnetic field to the detector. Consequently, if either human detects the photon, there is no energy left to be detected by the other human.

In "Direct detection of a single photon by humans", J.N. Tinsley et al. directly measure the event of conscious detection of single photons. Subjects in that experiment

  • did (barely) better than chance (51.6% (p=0.0545)) correctly identifying photon present and photon absent events) when observer confidence in event was excluded and
  • did better than chance (60.0% (p=0.001)) when confidence was included.

Interestingly, "the probability of correctly reporting a single photon is highly enhanced by the presence of an earlier photon within ∼5 s time interval. Averaging across all trials that had a preceding detection within a 10-s time window, the probability of correct response was found to be 0.56±0.03 (P=0.02)."

Of course, not every photon that strikes the retina is transduced. "Based on the efficiency of the signal arm and the visual system, we estimate that in ∼6% of all post-selected events an actual light-induced signal was generated ..." So we expect to see improvements over random chance in the neighborhood of 6%, and all numbers reported above are in that neighborhood.

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  • $\begingroup$ Surely candles give off lots of single photons, and at a sufficient distance they are spread out in space, and time, that they become singletons (unless you go with the 'always in pairs' view). The retina can be very sensitive - it's just a chemical version of the photo-electric effect (doen't mean the signal gets all the way to mk 1 brain..). $\endgroup$ – Philip Oakley Apr 22 at 12:03
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    $\begingroup$ @PhilipOakley : Why does a luminous flame emit localized (in space) photons? And I'm confused by your second point -- I have cited that single photon detection is observed in humans. $\endgroup$ – Eric Towers Apr 22 at 13:47
  • $\begingroup$ It is a matter of scale, both in time and space. The initial suggestion implied they come in massive bunches (seeing the beach rather than the gains of sand). I was simply pointing out that if we spread them all out (either space or time, or both) to a separation we can see them they are single photons. The single photon generation problem is one of perspective because of the 'random' emission nature. One cannot generate a single photon on demand (exact space/time), and cause real mental conflict for test engineers and metrologists who expect that one can ways create a reference! $\endgroup$ – Philip Oakley Apr 22 at 13:58
  • $\begingroup$ @Philip What is the 'always in pairs' view? $\endgroup$ – PM 2Ring Apr 22 at 13:59
  • $\begingroup$ @Eric We can produce IR emission at T=300K+/-0.01C with its mega-millions/cm^2/s (+/- a few) but to generate exactly 1.0 photon +/- 0.01 photon is a non-sense. You don't get fractional photons, rather sometimes you don't get a photon, and sometimes you get two (or maybe even three). we need to switch the probability measure from quantity to occurrence. $\endgroup$ – Philip Oakley Apr 22 at 14:03
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Seeing = detecting photons that happen to interact with your retina.

You can't see photons when they are just travelling nearby. Take lasers for example. When someone is using laser pointer, the only reason you see the beam is that photons collide with dust and air particles and therefore their direction is changed. For example into your eye. Otherwise you wouldn't see anything.

It isn't possible for two people to see the same photon.

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  • $\begingroup$ I don't think this tells the whole story. There's nothing to stop a photon (ie an excitation in the EM field) from spreading out in all directions before it's detected. Or, imagine a photon in a beam passing through a 50/50 beam splitter. It could have equal probability to collapse in two different locations, and only then would the path be determined. $\endgroup$ – Jay Apr 21 at 8:16
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    $\begingroup$ @Jay but you still can't distinguish "exists elsewhere" from "doesn't exist" by measurement. And if you measure it at your location, it automatically isn't anywhere else. $\endgroup$ – John Dvorak Apr 21 at 8:58
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    $\begingroup$ True. I agree with the conclusion but find the analogy to be a bit misleading. $\endgroup$ – Jay Apr 21 at 9:12
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In theory, in the most perversely contrieved case, and if you are willing to cheat a bit, it would be possible. In any half-reasonable, realistic setting, the answer is a clear, definite "No". Indeed, people cannot even see single photons at all (contrary to urban myths).

How does seeing a photon work? The photon has to hit your eye, specifically one of the billion rhodopsin molecules in one of the several-million retinal cells, then something-something, and then a nerve impulse maybe, if some conditions hold goes through the roughly-one-million ganglion network in the retina, and maybe makes it to the brain. Maybe. And maybe the visual cortex makes something of it.
The "maybe" part and the fact that a single cell has billions of G-proteins going active and inactive every second, and that there's a continuous flow of cGMP up and down is the reason why you cannot really see a single photon. That just isn't reasonably possible, if anything it's placebo effect or mere suggestion.

So what's that something-something mentioned previously? The photon flips the cis-bond at position 11 in retinal to trans. Which, well, takes energy, and absorbs the photon. This triggers a typical G-protein cascade, with alpha subunit going off and blah blah, resulting in production of cGMP at the end. If the cGMP concentration goes above some threshold, and if the cell isn't currently refractive, then the cell fires an AP. That's a big "maybe". Then comes something-something ganglion cells, which is the other big "maybe" part above.

The photon is "gone" after that. No second person could possibly see it.

Now of course, no absorption is perfect, there's an absorption maximum for each type of rhodopsin, and even at that it isn't 100%. Outside the maximum, the absorption is far from 100%. Which means that the photon is emitted again, and it could, in theory, in the most improbable case, hit another person's eye, why not. But of course we have to cheat a bit here because it strictly isn't the same photon.
Unless we are willing to cheat, the answer must therefore be "not possible".

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    $\begingroup$ Re "the photon is emitted again", is the emitted photon in any sense the SAME photon? I don't think so: the original photon is gone, and a new one is created. $\endgroup$ – jamesqf Apr 20 at 16:55
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    $\begingroup$ The photon is absorbed or not, so these absorption maxima are irrelevant here - even at 100% absorption you still have re-emission. @jamesqf Assuming same energy, I would consider it to be the same. Is the original photon also gone after its polarization changes? Or after it gets reflected/refracted? Or even after it travels in free space? $\endgroup$ – Zizy Archer Apr 20 at 21:59
  • $\begingroup$ I doubt that you can get a photon emission (other than reflection of an undetected photon) from a human retina. Do you have a source for that claim? $\endgroup$ – John Dvorak Apr 21 at 9:01
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    $\begingroup$ "Indeed, people cannot even see single photons at all (contrary to urban myths)." - According to this 2016 Nature-published paper, we can, sortof. Although the conclusion is criticized, I think casually dismissing it as an "urban myth" is oversimplifying. $\endgroup$ – marcelm Apr 21 at 9:40
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Somehow the exchange of energy between all objects must take place. It was found that this process takes place through the emission and absorption of photons (initially called energy quanta).

Photons are indivisible particles, they do not loose or gain inner energy during their life. The detection of a photon is possible only through the absorption of this photon.

Theoretically, it is possible to obtain information about an absorbed photon by observing secondary emitted photons with lower energy (and longer wavelength).

If you think of a laser beam that you have seen from the side, dust particles in the air are responsible. They reflect the laser light and you can see the beam. Of course, the photons reflected from the dust into the eyes do not arrive at the laser target.

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Two people cannot see the same photon. Only one person can see a specific photon.

To see a photon, it must be absorbed by a molecule in the retina [1]. The photon then no longer exists, so it is not available to be seen by another person.

[1] Mammalia retinas can respond to single photons

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Candles emit huge numbers of photons per second, and humans can't reliably detect single photons, so let's simplify your experiment to the bare essentials.

In the middle, we have an atom that we can excite (by firing a photon at it). Shortly after we excite this atom, it emits a single photon with a spherically symmetric radiation pattern, that is, there's an equal probability of detecting the photon in any direction. This is a standard example of an atom scattering a photon.

Now we place several identical photon detectors around our emitter atom, in various directions. After the photon is emitted, one of our detectors may detect it. Or the photon may miss all of our detectors and collide with something else.

We can model this as a spherical bubble centred on the emitter atom, expanding at the speed of light. When the bubble reaches a detector atom, that atom may detect the photon. When that happens, the bubble disappears, like a pin bursting a soap bubble. No other detector can detect the same photon (not even another detector at the exact same distance), all of the photon's energy was absorbed by the detector that was activated.

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    $\begingroup$ "When the bubble reaches a detector ... the bubble disappears, like a pin bursting a soap bubble" - a nice visualisation. $\endgroup$ – Philip Oakley Apr 22 at 13:48
  • $\begingroup$ Thanks, @Philip It's not perfect (what happens to the soap), but I'm fond of it. :) $\endgroup$ – PM 2Ring Apr 22 at 13:51

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