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I am just a curious physics student. This question is about the nature of light.

In a single-photons double slits (or multiple slits) experiment, the interference pattern or the distribution of the landing positions of the photons shows the wave nature of light. However, each photon is only detected at a single location, showing the particle nature of light.

The photons are detected using a photodetector, e.g. photomultiplier tube. I think this detection method is only suitable for showing the particle nature of light, because the photodetector operates using the particle nature of light.

The question is: If we change our detection method, i.e. instead of using photodetectors, we use very sensitive antennas (because antennas operate using the wave nature of light), will we be able to detect signals at multiple locations at the same time? (Reminder: this experiment is carried out with single photons, i.e. one photon at a time.)

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  • $\begingroup$ On a related note: The wave nature of light was already known/established before quantum mechanics; here the experiment demonstrates a particle aspect of light. But consider also the flip side of the puzzle: what was surprising about the double slit experiment is that the interference pattern arises over time even when you shoot electrons, one at a time. So, this was very strange, and the puzzle was, if an electron is a classical particle, how each of the subsequent electrons "knows" where to hit so that it doesn't mess up the pattern. $\endgroup$ – Filip Milovanović Oct 18 '20 at 20:32
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Let me first comment a sentence from SuperCiocia’s answer.

The photodetector clicks ... are caused by the photoelectric effect, that is bound electrons in the photodetector are in quantised orbits and are only capable of discrete energy jumps. (1)

In addition to this statement, please recapitulate that any observation of the wave behaviour of light during the transit of an edge with its surface electrons is always an indirect measurement by interpretation of the stripes. Any direct measurement destroys the patterns. (2)

From (1) and (2) I feel free to conclude on another scenario of what is happening. The photon(s) with its (their) oscillating electric and oscillating magnetic field components interact with the fields of the electrons and this happens in discrete portions.(3) In analogy to the Stern-Gerlach experiment, the trajectory of the photon (as well as of an electron) is deflected with discrete values and from this the fringes with its intensity distribution of the photons result. 83 If we know how radio waves are generated, we can conclude how effective the proposed method is. A prerequisite is the recapitulation of the fact that photons are emitted by the relaxation of subatomic particles to lower energy levels or by acceleration processes. The high number of accelerated electrons in an antenna rod emits a high number of photons. These electrons are accelerated forward and backward in the rod (by the antenna generator), and this carrier frequency produces a stream of photons with a sinusoidal intensity.

If we change our detection method, i.e. instead of using photodetectors, we use very sensitive antennas (because antennas operate using the wave nature of light), will we be able to detect signals at multiple locations at the same time?

If we know how radio waves are generated, we can conclude how effective the proposed method is. A prerequisite is the recapitulation of the fact that photons are emitted by the relaxation of subatomic particles to lower energy levels or by acceleration processes. The high number of accelerated electrons in an antenna rod emits a high number of photons. These electrons are accelerated forward and backward in the rod (by the antenna generator), and this carrier frequency produces a stream of photons with a sinusoidal intensity.

The receiver uses the inverse process. The electric or the magnetic field of the (polarized!) photons induce in the metall rod tiny displacements of the surface electrons. If a single photon has enough energy to induce measurable phonons in the material (preferred in an ultra-cold rod to prevent the thermal noise). I think, photon detectors are the better way.

Another experiment could shed light on the phenomenon of intensity distribution behind edges. Electrons also have a magnetic and an electric field component and the interaction of the flying electrons with the surface electrons should induce the above mentioned phonons (oscillations) in the material. This will be a major experiment, which (3) will confirm or disprove.

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    $\begingroup$ Hi HolgerFiedler, thanks for the answer. What do you mean by 'transit of an edge'? $\endgroup$ – Steven Lee WW Oct 19 '20 at 9:41
  • $\begingroup$ He Steven. Means flying nearby. Sorry, not my mothers language. $\endgroup$ – HolgerFiedler Oct 19 '20 at 10:13
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The simple answer is no. For a photon to be observed, all its energy must be collected. You cannot observe half a photon, either you observe it or you do not. The observation or detection can only happen in one place. This is often referred to as "the collapse of the wave function".

As an electromagnetics engineer I sometimes monitored very faint signals, perhaps using an antenna. At the very bottom end of detectability, such a signal descends into what is called "shot noise". As each photon arrives there is a little measurement blip and the signal then goes quiet until the next photon arrives. It is like scattered particles, shots from a gun is where the name comes from, not at all like a wave. The wave only becomes apparent if you have a setup, such as Young's slits, which records the statistical scatter of the shots in space.

Ultimately, a photon is a photon, it is neither particle nor wave nor in all honesty a "wavicle" or "wave packet"; it references no such classical or pseudo-classical notions. It is a nonlocal, massless quantum of somewhat uncertain energy (aka a disturbance of the zero-point electromagnetic field) which propagates at a speed governed by the permeability and permittivity of the medium it is passing through. Its wave equation describes only the chance of it hitting your detector and its particulate energy describes only what threshold you need to get down to in order to notice it (for example if you are using a Geiger counter then your energy threshold will be in the ultraviolet or X-ray region and you will not detect visible light). And no, we don't know why.

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    $\begingroup$ Hi Guy Inchbald, thanks for the answer. Actually I expected to get such answer (i.e. we cannot observe fraction of a photon). To give some background of this question, the motivation of this question is actually to investigate the physical interpretation of the electric field. In QED, the vector potential of the EM field is quantized in a sense that the 'absolute square of its amplitude' is actually the number operator in disguise. So, in some sense, I can identify the magnitude of the electric field as the probability distribution of photons. (continue at the next comment) $\endgroup$ – Steven Lee WW Oct 19 '20 at 8:24
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    $\begingroup$ FYI, I use a chapter (in particular chapter 4) from the book 'The quantum theory of light' by Rodney Loudon as reference for the quantization of EM radiation field. On the other hand, we know that a particle with electric charge experience a force in a static electric field, despite there is no (real) photon being emitted in a static E field. This is also true for the charges in antenna. They move because they experience a force from the electric field, even when there is no photon being emitted. So, what I am trying to say is maybe this is also true for non-static field. (see next comment) $\endgroup$ – Steven Lee WW Oct 19 '20 at 8:54
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    $\begingroup$ In other words, maybe a photon does not really need to be absorbed for us to get a signal from an antenna. To be honest, I highly doubt this is possible. But I have not found any experiment that fully rule out this possibility. Therefore I asked this question. And therefore the problem of observing a fraction of a photon is avoided. $\endgroup$ – Steven Lee WW Oct 19 '20 at 9:06
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Light does not behave like a wave some times and like a particle some other times. Light behaves as light. Trying to categorise some behaviour as "wave-like" or "particle-like" is just an attempt to build an intuitive understanding for quantum phenomena by relating them to simpler everyday things like water waves or marbles hitting a wall.

The photodector clicks are not proof of the existence of photons. They are caused by the photoelectric effect, that is bound electrons in the photodetector are in quantised orbits and are only capable of discrete energy jumps. So if anything it's more to do with the quantum nature of matter than with the quantum nature of light. The photoelectric effect even works with classical (continuum, not quantised) a constant stream of light, not just single photons.

So using the photodetector measurement as a proof of the existence of photons is a bit of an abuse of the photon picture. And it's taking the "photon" picture to the classical extreme of "a billiard ball".

Really, the EM field is a quantum field obeying the wave equation and whose quantum is a photon. Even a single photon obeys the wave equation though, so the better question is:
how to reconcile the (seemingly) localised click of the photodetector with a delocalised photon wavepacket?

A photon is a wavepacket with some spatial extent and a wavefront. It also has a "direction", which we could define as the expectation value of the position operator over time. Hence, there's a spatially varying probability (and hence energy) density. When this gets close to the photodetector, the EM field and the quantum matter interact and cause the photon wavefunction to 'collapse' and to get position-localised (like when you measure the position of an electron in an atom). The position where the photon "localises" is random but follows the probability distribution of the incident photon field, which is a $\propto \sin^2$ and hence different photons cause clicks in different positions on the screen.

To put this in more "usual" quantum terms, then: as long as no measurement is performed, the photon is described by a delocalised wavefunction. When a measurement is performed, the photon localises. Measurements are destructive.

A measurement is performed both by the photodetector and by the antenna. In the antenna case, you'd see a transient discrete signal in one of the antennas, corresponding to where the electron was accelerated by the absorption of the photon.

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    $\begingroup$ "The photoelectric effect even works with classical (continuum, not quantised) light!" is not correct. Einstein won his Nobel prize not for Relativity but for explaining the photoelectric effect as a quantum phenomenon. Classical models had proved unable to explain it and as far as I am aware still can't. $\endgroup$ – Guy Inchbald Oct 18 '20 at 19:45
  • $\begingroup$ "A photon is a wavepacket with some spatial extent..." For a photon crossing intergalactic space, that can be quite some "some"! For example a wavefront emanating from a sufficiently distant quasar might diffract around a black hole and produce a ring image in both the Hubble telescope and another one on Earth. But only one point in one of those images will be detected. Other photons will contribute more points to one or other image, yet others will fly right on past to destinations billions more years in the future. $\endgroup$ – Guy Inchbald Oct 18 '20 at 19:53
  • $\begingroup$ @GuyInchbald, from the historical point of view you are certainly right. However, try to have a look at this: physicsoverflow.org/27774/…. PhysicsSE question: physics.stackexchange.com/q/68147/226902 $\endgroup$ – Quillo Oct 18 '20 at 19:54
  • $\begingroup$ @Quillo If you read the answers to those posts, there are some devastating criticisms. Many phenomena can be explained in many different ways; I can explain gravity perfectly well by appealing to the Flying Spaghetti Monster and the noodly appendage It keeps specially for me. Classical electromagnetism has had to bow to quantum electrodynamics for a myriad of reasons, and that is the end of it. $\endgroup$ – Guy Inchbald Oct 18 '20 at 20:07
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    $\begingroup$ Hi SuperCiocia, thanks for the answer. I agree that the photoelectric effect is not enough to prove the particle nature of light. However, the localization of the 'click' is a very 'particle' thing (Maybe I should also include my definition for 'particle'). The reason for using single photons is to show this localization. But thanks again for your answer. $\endgroup$ – Steven Lee WW Oct 19 '20 at 9:30
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You can also consider a photo detector as much more sensitive than a very sensitive antenna. Single photons are not typically detected with antennas, antennas use very large numbers of photons to generate voltage.

Historically (1801s) the DSE was said to show a pattern just like water thus the term "interference" and it must be due to waves. Modern statistical QM tells you that 2 photons can not interfere (violation of conservation of energy) and that every photon that is emitted is eventually absorbed. In the DSE darks areas are where no "photons" fall, bright areas get all the photons. The wave action that is really happening is better explained by Feynman (1960s), the photon must travel an integer multiple of its wavelength, like a note on a guitar string, thus paths are not probable or very probable. How would a photon know which path, likely as mentioned in another answer here, the field pattern is virtual before actual transit of the energy (or photon).

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