Wave and particle nature of light during detection in a single-photons double slit experiment

Question

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.)

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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).